diff --git "a/Astronomy/a_complete_system_of_astronomy-vince_1814.md" "b/Astronomy/a_complete_system_of_astronomy-vince_1814.md" new file mode 100644--- /dev/null +++ "b/Astronomy/a_complete_system_of_astronomy-vince_1814.md" @@ -0,0 +1,35525 @@ +. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + +A +COMPLETE SYSTEM +OF +ASTRONOMY; +BY THE +REV. S. VINCE, A.M. F.R.S. +PLUMIAN PROFESSOR OF ASTRONOMY AND EXPERIMENTAL PHILOSOPHY, +IN THE +UNIVERSITY OF CAMBRIDGE. + +ASTRONOMY LIBRARY +VOL. I. +NEW-YORK + +SECOND EDITION, +WITH ADDITIONS AND CORRECTIONS +LONDON: +PRINTED BY G. WOODFALL, ANGEL COURT, SKINNER STREET, +AND SOLD BY W. H. LUNN, SOHO SQUARE; J. DEIGHTON, AND +J. NICHOLSON, CAMBRIDGE. +1814. +35/2 + +A blank page with a faint grid pattern. + +TO + +THE REVEREND + +NEVIL MASKELYNE, D.D. F.R.S. + +ASTRONOMER ROYAL, + +THIS SYSTEM OF ASTRONOMY IS DEDICATED + +BY + +THE AUTHOR: + +AS WELL FROM MOTIVES OF PRIVATE FRIENDSHIP, AS FROM AN +EARNEST DESIRE TO PAY A PUBLIC TRIBUTE +OF SINCERE RESPECT, +JUSTLY DUE TO HIM +FOR THE IMPROVEMENTS WHICH HE HAS MADE IN THIS, +AND EVERY OTHER SUBSIDIARY BRANCH OF +PHILOSOPHY; +AND +FOR SO ABLY DISCHARGING THE ARDUOUS DUTIES OF HIS +HIGH AND IMPORTANT STATION, +BY WHICH HE HAS DONE SO MUCH HONOUR TO HIMSELF +AND +TO HIS COUNTRY. + +A scanned document with text on it. +783200 41H OT + +THE CONTENTS OF THE FIRST VOLUME. + +CHAPTER I. +DEFINITIONS +Page 1 + +CHAP. II. +On the Doctrine of the Sphere +9 + +CHAP. III. +On the Right Ascension, Declination, Latitude and Longitude of the Heavenly Bodies +30 + +CHAP. IV. +On the Equation of Time +43 + +CHAP. V. +On the Length of the Year, Precession of the Equinoxes, and Obliquity of the Ecliptic +50 + +CHAP. VI. +On Parallax +59 + +CONTENTS OF THE FIRST VOLUME. + +CHAP. VII. +On Refraction +Page 76 + +CHAP. VIII. +On the System of the World +93 + +CHAP. IX. +On Kepler's Discoveries +98 + +CHAP. X. +On the Motion of a Body in an Ellipse about the Focus +108 + +CHAP. XI. +On the Oppositions and Conjunctions of the Planets +113 + +CHAP. XII. +On the Mean Motions of the Planets +116 + +CHAP. XIII. +On the greatest Equation, Eccentricity, and Place of the Aphelia of the Orbit of the Planets +126 + +CHAP. XIV. +On the Motion of the Aphelia of the Orbits of the Planets +146 + +CHAP. XV. +On the Nodes and Inclinations of the Orbits of the Planets to the Ecliptic +151 + +CHAP. XVI. +On the Georgian Planet +161 + +CONTENTS OF THE FIRST VOLUME. +vii + +CHAP. XVII. +On the Apparent Motions and Phases of the Planets +Page 172 + +CHAP. XVIII. +On the Moon's Motion from Observation, and its Phenomena +Page 181 + +CHAP. XIX. +On the Rotation of the Sun, Moon, and Planets +Page 213 + +CHAP. XX. +On the Satellites +Page 231 + +CHAP. XXI. +On the Ring of Saturn +Page 297 + +CHAP. XXII. +On the Aberration of Light +Page 306 + +CHAP. XXIII. +On the Projection for the Construction of Solar Eclipses +Page 355 + +CHAP. XXIV. +On Eclipses of the Sun and Moon, and Occultations of the Fixed Stars by the Moon +Page 341 + +CHAP. XXV. +On the Transits of Mercury and Venus over the Sun's Disc +Page 394 + +CONTENTS OF THE FIRST VOLUME. + +CHAP. XXVI. +On Comets - - - - - - - - - - - - - - - - - - - - - - 425 + +CHAP. XXVII. +On the Fixed Stars - - - - - - - - - - - - - - 487 + +CHAP. XXVIII. +On the Longitude - - - - - - - - - - - 509 + +CHAP. XXIX. +On the Use of the Globes - - - 559 + +CHAP. XXX. +On the Division of Time - 562 + +A + +COMPLETE SYSTEM + +OF + +ASTRONOMY. + +CHAPTER I. + +DEFINITIONS. + +Art. 1. ASTRONOMY is that branch of natural philosophy which treats of the heavenly bodies. The determination of their magnitudes, distances and the orbits which they describe, is called plane or pure Astronomy; and the investigation of the causes of their motions is called physical Astronomy. The former is determined from observations on their apparent magnitudes and motions; and the latter from analogy, by applying those principles and laws of motion by which bodies on and near the earth are governed, to the other bodies in the system. The principles of plane Astronomy come first in order to be treated of; and in this, we shall begin with the explanation of such terms as are the fundamental principles of the science. + +2. A great circle of a sphere is that whose plane passes through its center; and a small circle is that whose plane does not pass through its center. + +3. A diameter of a sphere perpendicular to any great circle, is called the axis of that circle; and the extremities of the diameter are called its Poles. + +4. Hence, the pole of a great circle is $90^\circ$ from every point of it upon the surface of the sphere; but as the axis is perpendicular to the circle when it is perpendicular to any two radii, a point on the surface of a sphere $90^\circ$ distant from any two points of a great circle will be the pole. + +5. All angular distances on the surface of a sphere, to an eye at the center, are measured by the arcs of great circles; for they being arcs to equal radii, will be as the angles. + +VOL. I. + +9 + +DEFINITIONS. + +6. Hence, all triangles formed upon the surface of a sphere, for the solution of spherical problems, must be formed by the arcs of great circles. +7. All great circles must bisect each other; for passing through the center of the sphere their common section must be a diameter, which bisects all circles. +8. *Secondaries* to a great circle are great circles which pass through its poles. +9. Hence, secondaries must be perpendicular to their great circle; for if one line be perpendicular to a plane, any plane passing through that line will also be perpendicular to it; therefore as the axis of the great circle is perpendicular to it, and is the common diameter to all the secondaries, they must all be perpendicular to the great circle. Hence also, every secondary, bisecting its great circle (*7*), must bisect every small circle parallel to it; for the plane of the secondary passes through, not only the center of the great circle, but also of the small circles parallel to it. +10. Hence, a great circle passing through the poles of two great circles, must be perpendicular to each; and, vice versa, a great circle perpendicular to two other great circles must pass through their poles. +11. If an eye be in the plane of a circle it appears a straight line; hence in the representation of the surface of a sphere upon a plane, those circles whose planes pass through the eye are represented by straight lines. +12. The angle formed by the circumferences of two great circles on the surface of a sphere, is equal to the angle formed by the planes of those circles; and is measured by the arc of a great circle intercepted between them, described about the intersection of the circles as a pole. +A diagram showing two great circles intersecting at a point P. One circle is labeled PQE and the other PRE. The angles formed by these circles are shown as 90 degrees. +For let C be the center of the sphere, PQE, PRE two great circles; then as the circumferences of these circles at P are perpendicular to the common intersection PCE, the angle at P between them is equal to the angle between the planes, by Euc. B. XI. Def. 6. Now draw CQ, CR perpendicular to PCE; then the angle QCR is also the angle between the planes, and therefore equal to the angle at P formed by the two circles; and the angle QCR is measured by the arc QR of a great circle, which arc has (*2*) for its pole the point P, because PQ, PR are each 90°. +13. If at the intersection of two great circles as a pole, a great circle be described, and also a small circle parallel to it, the arcs of the great and small circles intercepted between the two great circles contain the same number of degrees. +For draw AB, AD perpendicular to PCE, then as AB, AD are parallel to CQ, CR, the plane BAD is parallel to the plane QCR, and therefore the small circle BD of which A is the center, is parallel to the great circle QR, and as + +* Figures included thus in a Parenthesis refer to the articles. + +DEFINITIONS. +3 + +each angle $BAD$, $QCR$, measures the inclination of the planes, they must be equal, and consequently the arcs $BD$, $QR$ contain the same number of degrees. Hence, the arc of such a small circle measures the angle at the pole between the two great circles. Also $QR : BD :: QC : BA :: \cos BQ$. Hence $QR$ is the greatest distance between the two circles. + +14. The *Axis* of the earth is that diameter about which it performs its diurnal motion; and the extremities of this diameter are called its *poles*. + +15. The *terrestrial Equator* is a great circle of the earth perpendicular to its axis. Hence, the axis and poles of the earth are the axis and poles of its equator. That half of the earth which lies on the side of the equator which we inhabit is called the northern Hemisphere, and the other the southern; and the poles are respectively called the north and south poles. + +16. The *Latitude* of a place on the earth's surface is its angular distance from the equator, measured upon a secondary to it. These secondaries to the equator are called *Meridians*. + +17. The *Longitude* of a place on the earth's surface is an arc of the equator intercepted between the meridian passing through the place, and another, called the first meridian, passing through that place from which you begin to measure. + +18. If the plane of the terrestrial equator be produced to the sphere of the fixed stars, it marks out a circle called the celestial equator; and if the axis of the earth be produced in like manner, the points in the heavens to which it is produced are called *poles*, being the poles of the celestial equator. The star nearest to each pole is called the pole star. + +19. Secondaries to the celestial equator are called *circles of Declination*: of these, 24 which divide the equator into equal parts, each containing 15°, are called *Hour circles*. + +20. Small circles parallel to the celestial equator, are called *parallels of Declination*. + +21. The *sensible horizon* is that circle in the heavens whose plane touches the earth at the spectator. The *rational horizon* is a great circle in the heavens, passing through the earth's center, parallel to the sensible horizon. + +22. *Almucantar* is a small circle parallel to the horizon. + +23. If the radius of the earth to the place where the spectator stands, be produced both ways to the heavens, that point vertical to him is called the Zenith, and the opposite point the Nadir. Hence, the zenith and nadir are (s) the poles of the rational horizon; for the radius produced being perpendicular to the sensible, must also be perpendicular to the rational horizon. + +24. Secondaries to the horizon are called *vertical circles*, because they are (9) perpendicular to the horizon; on these circles therefore the altitude of an heavenly body is measured. + +4 + +DEFINITIONS. + +25. A Secondary common to the celestial equator and the horizon of any place, and which therefore (10) passes through the poles of each, is the **celestial meridian** of that place. Hence, the plane of the celestial meridian of any place coincides with the plane of the terrestrial meridian of the same place. + +26. That direction which passes through the north pole is called **north**, and the opposite direction is called **south**. Hence, the meridian must cut the horizon in the **north** and **south** points. + +27. Hence, the meridian of any place divides the heavens into two hemispheres lying to the east and west; that lying to the east is called the **eastern hemisphere**, and the other lying to the west is called the **western hemisphere**. + +28. The vertical circle which cuts the meridian of any place at right angles, is called the **prime vertical**; and the points where it cuts the horizon are called the **east** and **west** points. Hence, the east and west points are 90° distant from the north and south. These four are called the **cardinal points**. + +29. The **Azimuth** of an heavenly body is its distance on the horizon, when referred to it by a secondary, from the north or south points. The **Amplitude** is its distance from the east or west points. + +30. The **Ecliptic** is that great circle in the heavens which the sun appears to describe in the course of a year. + +31. The ecliptic and equator being great circles must (7) bisect each other, and their angle of inclination is called the **obliquity of the ecliptic**; also the points where they intersect are called the **equinoctial points**. The times when the sun comes to these points are called the **Equinoxes**. + +32. The ecliptic is divided into 12 equal parts, called **Signs**: Aries $\varphi$, Taurus $\varphi$, Gemini $\varphi$, Cancer $\alpha$, Leo $\alpha$, Virgo $\pi$, Libra $\alpha$, Scorpio $\mu$, Sagittarius $\gamma$, Capricornus $\pi$, Aquarius $\varphi$, Pisces $\varphi$. The order of these is according to the motion of the sun. The first point of aries coincides with one of the equinoctial points, and the first point of libra with the other. The first six signs are called northern, lying on the north side of the equator; and the last six are called southern, lying on the south side. The signs $\varphi$, $\pi$, $\alpha$, $\gamma$, $\mu$, $\pi$ are called ascending, the sun approaching our (or the north) pole whilst it passes through them; and $\alpha$, $\beta$, $\pi$, $\alpha$, $\pi$ are called descending, the sun receding from our pole as it moves through them. + +33. The motion of the heavenly bodies which is according to the order of the signs, is called direct, or *in sequentia*; and the motion in the contrary direction is called retrograde, or *in antecedentia*. The real motion of all the planets is according to the order of the signs, but their apparent motion is sometimes in an opposite direction. + +34. The Zodiac is a space extending on each side of the ecliptic, within which the motion of all the planets is contained. + +35. The **right ascension** of a body is an arc of the equator intercepted be- + +DEFINITIONS. +5 + +tween the first point of aries and a declination circle passing through the body, +measured according to the order of the signs. + +36. The oblique ascension is an arc of the equator intercepted between the +first point of aries and that point of the equator which rises with any body, +measured according to the order of the signs. + +37. The ascensional difference is the difference between the right and oblique +ascension. + +38. The Declination of a body is its angular distance from the equator, mea- +sured upon a secondary to it drawn through the body. + +39. The Longitude of a star is an arc of the ecliptic intercepted between the +first point of aries and a secondary to the ecliptic passing through the body, +measured according to the order of the signs. If the body be in our system +and seen from the sun, it is called the heliocentric longitude; but if seen from +the earth, it is called the geocentric longitude; the body in each case being referred +perpendicularly to the ecliptic in a plane passing through the eye. + +40. The Latitude of a star is its angular distance from the ecliptic, measured +upon a secondary to it drawn through the body. If the body be in our system, +its angular distance from the ecliptic seen from the earth is called the geocentric +latitude; but if seen from the sun it is called the heliocentric latitude. + +41. Hence, if $\tau Q$ be the equator, $\tau C$ the ecliptic, $\tau P$ the first point of aries, +$\sigma$ a star, and the great circles $sr$, $sm$ be drawn perpendicular to $\tau C$ and $\tau Q$; +then $\tau P$ is its right ascension, $sm$ its declination, or its latitude and $\tau r$ its lon- +gitude. The circle $sr$ is called a circle of latitude. + +42. The Tropics are two parallels of declination touching the ecliptic. One, +touching it at the beginning of cancer, is called the tropic of cancer; and the +other touching it at the beginning of capricorn, is called the tropic of capricorn. +The two points where the tropics touch the ecliptic are called the solstitial +points. + +43. Colures are two secondaries to the celestial equator, one passing through +the equinoctial points, called the equinoctial colure; and the other passing +through the solstitial points, called the solstitial colure. The times when the +sun comes to the solstitial points are called the Solstices. + +44. The Arctic and Antarctic circles are two parallels of declination, the for- +mer about the north and the latter about the south pole, the distance of which +from the two poles is equal to the distance of the tropics from the equator. +These are also called polar circles. + +45. The two tropics and two polar circles, when referred to the earth, divide +it into five parts, called Zones ; the two parts within the polar circles are called +the frigid zones; the two parts between the polar circles and tropics are called +the temperate zones; and the part between the tropics is called the torrid zone. +Small circles in the heavens are referred to the earth, or the contrary, by lines + +Fig. 2. +2 + +6 + +DEFINITIONS + +FIG. +3. + +drawn to the earth's center. Thus, the small circle $A\cdot B$, in the heavens, is referred to $L\cdot M$ on the earth. Hence, if $A\cdot B$ be the tropic or the polar circle in the heavens, $L\cdot M$ will be the tropic or polar circle on the earth. These circles therefore retain the same relative situations, that is, the former is as far from the pole in the heavens, as the latter is from the pole of the earth. The planes of these corresponding small circles do not coincide; but when they become great circles, then the planes become coincident. + +46. A body is in **Conjunction** with the sun when it has the same longitude; in **Opposition**, when the difference of their longitudes is 180°; and in **Quadra- tures**, when the difference of their longitudes is 90°. The conjunction is marked thus $\varphi$, the opposition thus $\varphi$, and quadra-tures thus $\varphi$. + +47. Syzygy is either conjunction or opposition. + +48. The **Elongation** of a body is its angular distance from the sun when seen from the earth. + +49. The **diurnal parallax** is the difference between the apparent places of the bodies in our system when referred to the fixed stars, if seen from the center and surface of the earth. The **annual parallax** is the difference between the apparent places of a body in the heavens, when seen from the opposite points of the earth's orbit. + +50. The **Argument** is a term used to denote any quantity by which another required quantity may be found. For example, the argument of that part of the equation of time which arises from the unequal angular motion of the earth in its orbit about the sun, is the sun's anomaly, because that part of the equation depends entirely upon the anomaly; and the latter being given, the former is found from it. "The argument of a star's latitude is its distance from its node, because upon this latitude depends." + +51. The **Nodes** are the points where the orbits of the primary planets cut the ecliptic, and where the orbits of the secondaries cut the orbits of their primaries. That node is called **ascending** where the planet passes from the south to the north side of the ecliptic, and the other is called the **descending** node. The ascending node is marked thus $\alpha$, and the descending node thus $\beta$. The line which joins the nodes is called the **line of nodes**. + +52. If a perpendicular be drawn from a planet to the ecliptic, the angle at the sun between two lines, one drawn from it to that point where the perpendicular falls, and another to the earth, is called the angle of **Commutation**. + +53. The angle of **Position** is the angle at an heavenly body formed by two great circles, one passing through the pole of equator and the other through the pole of the ecliptic. + +54. **Apparent noon** is the time when the sun comes to the meridian. + +55. True or mean noon is 12 o'clock, by a clock adjusted to go 24 hours in a mean solar day. + +DEFINITIONS. +7 + +56. The Equation of Time is the interval between true and apparent time. + +57. A star is said to rise or set **comically**, when it rises or sets at sun rising; and when it rises or sets at sun setting, it is said to rise or set **achronically**. + +58. A star rises **heleically**, when, after having been so near to the sun as not to be visible, it emerges out of the sun's rays and just appears in the morning; and it sets **heleically**, when the sun approaches so near to it, that it is about to immerge into the sun's rays and become invisible in the evening. + +59. Cartate distance of a planet from the sun or earth, is the distance of the sun or earth from that point of the ecliptic where a perpendicular to it passes through the planet. + +60. Aphelion is that point in the orbit of a planet which is furthest from the sun. + +61. Perihelion is that point in the orbit of a planet which is nearest the sun. + +62. Apogee is that point of the earth's orbit which is furthest from the sun, or that point of the moon's orbit which is furthest from the earth. + +63. Perigee is that point of the earth's orbit which is nearest the sun, or that point of the moon's orbit which is nearest the earth. + +The terms aphelion and perihelion are also applied to the earth's orbit. + +64. Apsis of an orbit, is either its aphelion or perihelion, apogee or perigee; and the line which joins the apsides is called **the line of the apsides**. + +65. Anomaly (true) of a planet is its angular distance at any time from its aphelion, or apogee—(mean) is its angular distance from the same point at the same time if it had moved uniformly with its mean angular velocity. + +66. Equation of the center is the difference between the true and mean anomaly; this is sometimes called the precession. + +67. Nonageesimal degree of the ecliptic is that point which is highest above the horizon. + +68. The mean place of a body is the place where a body, not moving with an uniformly angular velocity about the central body, would have been, if the angular velocity had been uniform. The true place of a body is the place where the body actually is at any time. + +69. Equations are corrections which are applied to the mean place of a body in order to get its true place. + +70. A Digit is a twelfth part of the diameter of the sun or moon. + +71. Those bodies which revolve about the sun in orbits very nearly circular, are called Planets, or primary planets for the sake of distinction; and those bodies which revolve about the primary planets are called secondary planets, or satellites. + +72. Those bodies which revolve about the sun in very elliptic orbits are called Comets. The sun, planets and comets, comprehend all the bodies in what is called the Solar system. + +8 + +DEFINITIONS. + +73. All the other heavenly bodies are called fixed stars, or simply Stars. +74. Constellation is a parcel of stars contained within some assumed figure, as a ram, a dragon, an Hercules, &c. the whole heaven is thus divided into constellations. A division of this kind is necessary, in order to direct a person to any part of the heavens which we want to point out. + +Characters used for the Sun, Moon and Planets. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
A circle with a dot in the center. The Sun.Pallas.
A crescent moon. The Moon.Juno.
A circle with a line through it. Mercury.= Vesta.
A circle with a line through it. Venus.Jupiter.
A circle with a line through it. The Earth.Saturn.
A circle with a line through it. Mars.Georgian.
A circle with a line through it. Ceres.
+ +Characters used for the Days of the Week. + + + + + + + + + + + + + + + + + + +
A circle with a dot in the center. Sunday.Thursday.
A crescent moon. Monday.Friday.
A circle with a line through it. Tuesday.Saturday.
A circle with a line through it. Wednesday.
+ +CHAP. II. + +ON THE DOCTRINE OF THE SPHERE. + +ART. 75. A SPECTATOR upon the earth's surface conceives himself to be placed in the center of a concave sphere in which all the heavenly bodies are situated; and by constantly observing them, he perceives that by far the greater number never change their relative situations, each rising and setting at the same interval of time and at the same points of the horizon, and are therefore called fixed stars; but that a few others, called planets, together with the sun and moon, are constantly changing their situations, each continually rising and setting at different points of the horizon and at different intervals of time. Now the determination of the times of the rising and setting of all the heavenly bodies; the finding of their position at any given time in respect to the horizon or meridian, or the time from their position; the causes of the different lengths of days and nights, and the changes of seasons; the principles of dialling, and the like, constitute the doctrine of the sphere. And as the apparent diurnal motion of all the bodies have no reference to any particular system or disposition of the planets, but may be solved, either by supposing them actually to perform those motions every day, or by supposing the earth to revolve about an axis, we will suppose this latter to be the case, the truth of which will afterwards appear. + +76. Let $p_{e}^{i}$ represent the earth, $O$ its center, $b$ the place of a spectator, $H_{ZRN}$ the sphere of the fixed stars; and although the fixed stars do not lie in the concave surface of a sphere of which the center of the earth is the center, yet, on account of the immense distance even of the nearest of them, their relative situations from the motion of the earth, and consequently the place of a body in our system referred to them, will not be affected by this supposition. The plane $a e c$ touching the earth in the place of the spectator is called (91) the sensible horizon; as it divides the visible from the invisible part of the heavens; and a plane $HOR$ parallel to $a e c$, passing through the center of the earth, is called the rational horizon; but in respect to the sphere of the fixed stars, these may be considered as coinciding, the angle which they make subtends at the earth becoming then insensible from the immense distance of the fixed stars. Now if we suppose the earth to revolve daily about an axis, all the heavenly bodies must successively rise and set in that time, and appear to describe circles whose planes are perpendicular to the earth's axis, and consequently parallel to each other; thus all the stars would appear to revolve daily about the earth's axis, as if they were placed in the concave surface of a sphere having the earth in + +Fig. 4. + +VOL. I + +10 +**ON THE DOCTRINE OF THE SPHERE.** + +the center. Let therefore $pp$ be that diameter of the earth about which it must revolve in order to give the apparent diurnal motion to the heavenly bodies, then $p$, $p'$, are called its poles ; and if $pp$ be produced both ways to $P$, $P'$ in the heavens, these points are called (18) the poles in the heaven, and the star nearest to each of these is called the pole star. Now, although the earth, from its motion in its orbit, continually changes its place, yet as the axis always continues parallel to itself, the points $P$, $P'$ will not, from the immense distance of the fixed stars, be sensibly altered ; we may therefore suppose these to be fixed points*. Produce $Ob$ both ways to $Z$ and $N$, and $Z$ is the zenith and $N$ the nadir (25). Draw the great circle $PZHNR$, and it will be the celestial meridian (23), the plane of which coincides with the terrestrial meridian $pob$ passing through the place of the spectator. Let $eq$ represent a great circle of the earth perpendicular to its axis $pp$, and it will be the equator (15), and if the plane of this circle be extended to the heavens it marks out a great circle $EQ$ called the celestial equator (18). Hence, for the same reason that we may consider the points $P$, $P'$ as fixed, we may consider the circle $EQ$ as fixed. Now as the latitude of a place on the earth's surface is measured by the degrees of the arc $be$ (16), it may be measured by the arc $Ze$ ; hence as the equator, zenith, and poles in the heaven, correspond to the equator, place of the spectator, and poles of the earth, we may leave out the consideration of the earth in our further enquiries upon this subject, and only consider the equator, zenith and poles in the heavens, and $HR$ the horizon to the spectator. + +77. Let therefore figure the fifth represent the position of the heavens to $Z$ - 5- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- 5- + +the zenith of a spectator in north latitude, $EQ$ the equator, $P$, $P'$ its poles, $HOR$ the rational horizon, $PZHPR$ the meridian, and draw the great circle $ZON$ perpendicular to $ZPRH$, and it is the prime vertical (29); $R$ will be the north point of the horizon and $H$ the south (26), and $O$ will be either east or west points (28) according as this figure represents the eastern or western hemisphere. Draw also a great circle $POP$ perpendicular to the meridian, its pole must be in each ($8$, $9$), therefore their common intersection $O$ is the pole of the meridian. Draw also the small circles $\omega H_1$, $\omega E_1$, $\omega Z_1$, $\omega P_1$ parallel to the equator; and as the great circle $POP'$ bisects $EQ$ in $O$, it must also bisect the small circles $\omega H_1$, $\omega E_1$, $\omega Z_1$, $\omega P_1$ in each ($8$, $9$), for as $\omega E_0 = \omega H_0$, therefore (18) $\omega E_0 = \omega H_0$. As all of heavenly bodies, in their diurnal motion, describe either the equator, or small circles parallel to the equator, according as the body is in or out of the equator, if we conceive this figure to represent the eastern hemisphere, $\omega E_0$, $\omega E_1$, $\omega M_1$, may represent their apparent paths from under explained in proper place. + +* This is not accurately true, as earth's axis varying a little from its parallactic, as will be ex- +plained in proper place. + +ON THE DOCTRINE OF THE SPHERE. + +the horizon to the meridian above, and the points $b$, $O$, $s$ are the points of the horizon where they rise. And as $ae$, $QE$, $mt$, are bisected in $e$, $O$, $r$, $eb$ must be greater than $ba$, $QO = OE$, and $a$ less than $m$. Hence, a body on the same side of the equator with the spectator will be longer above the horizon than below, because $eb$ is greater than $ba$; a body in the equator will be as long above as below, because $QO = OE$, and a body on the contrary side will be longer below than above, because $me$ is greater than $mt$. The bodies describing $ae$, $mt$, rise at $b$ and $s$; and as $O$ is the east point of the horizon, and $A$ and $H$ are the north and south points, a body on the same side of the equator with the spectator rises between the east and the north; and a body in the equator rises in the east at $O$. When the bodies come to $d$ and $n$, they are in the prime vertical, or in the east; hence, a body on the same side of the equator with the spectator comes to the east after it is risen, and a body on the contrary side, before it rises. The body which describes the circle $aev$, or any circle nearer to $P$, never sets; and such circles are called circles of *perpetual apparition*; and the stars which describe them are called *circumpolar* stars. The body which describes the circle $WH$, just becomes visible at $H$, and then it instantly descends below the horizon; but the bodies which describe the circles nearer to $P$ are never visible. Such is the apparent diurnal motion of the heavenly bodies when the spectator is situated any where between the equator and poles; and this is called an *oblique sphere*, because all the bodies rise and set obliquely to the horizon. As this figure may also represent the western hemisphere, the same circles $ae$, $mt$ will represent the motion of the heavenly bodies as they descend from the meridian above the horizon to the meridian under. Hence, a body is at the greatest altitude above the horizon, when on the meridian, and at equal altitudes when equidistant on each side from it, if the body have not changed its declination. This is the foundation of finding the time of passing the meridian, from equal altitudes of a body on each side. + +79. If the spectator be at the equator, then $E$ coincides with $Z$, and consequently $EQ$ with $ZN$, and therefore $PP$ with $HR$. Hence, as the equator $EQ$ is perpendicular to the horizon, the circles $ae$, $mt$, parallel to $EQ$, must also be perpendicular to it; and as these circles are always bisected by $PP$, they must now be bisected by $HR$. Hence, all the heavenly bodies are as long above the horizon as below, and rise and set at right angles to it, on which account this is called a *right sphere*. + +80. If the spectator be at the pole, then $P$ coincides with $Z$, and consequently $PP'$ with $ZN$, and therefore $EQ$ with $HR$. Hence, the circles $mt$, ae', parallel to the equator, are also parallel to the horizon; therefore as a body in its diurnal motion describes a circle parallel to the horizon, those fixed bodies in the heavens which are above the horizon must always continue above, and + +Fig. 6. +Fig. 7. + +11 + +12 + +**ON THE DOCTRINE OF THE SPHERE.** + +those which are below must always continue below. Hence, none of the bodies by their *diurnal* motion can either rise or set. This is called a *parallel* sphere, because the diurnal motion of all the bodies is parallel to the horizon. These apparent diurnal motions of the fixed stars remain constant, that is, each always describes the same parallel of declination. + +81. The *ecliptic*, or that circle in the heavens which the sun appears to describe in the course of a year, does not coincide with the equator, for during that time it is found to be only twice in the equator; let therefore *COL* represent the ecliptic, which being a great circle must cut the equator into two equal parts (?). Hence, as this apparent motion of the sun is nearly uniform, the sun is nearly as long on one side of the equator as on the other. When therefore the sun is at $q$ on the *same* side of the equator with the spectator, describing the parallel of declination $ae$ by its diurnal motion, the days are longer than the nights, and it rises at $b$ from the east towards the north; but when it is on the *contrary* side, at $p$, describing $mt$, the days are shorter than the nights, and it rises at $s$ from the east towards the south, the spectator being on the north side of the equator; but when the sun is in the equator, at $O$, describing $QE$, the days and nights are equal, and it rises in the east at $O^{\circ}$. If $ae$, $mt$ be equidistant from $EQ$, then will $be = ms$ and $ab = ut$; hence, when the sun is in these opposite parallels, the length of the day in one is equal to the length of the night in the other; and the mean length of a day at every place is 12 hours. Hence, at every place, the sun, in the course of a year, is half a year above the horizon and half a year below. When the spectator is at the equator, $tm$, $ea$ being bisected by the horizon, the sun will be always as long above as below the horizon, and consequently the days and nights will be always 12 hours long. + +There will however be some variety of seasons, as the sun will recede $23\frac{1}{2}$° on each side from the spectator. When the spectator is at the equator, the sun will be vertical to him at noon when it is in the equator. And when the spectator is any where between the tropics, the sun will be vertical to him at noon when its declination is equal to the latitude of the place, and of the same kind, that is, both north, or both south. When the spectator is at the pole, the sun at $p$ or $q$ is carried by its *diurnal* motion parallel to the horizon; hence it never sets when it is in that part of the ecliptic which is above the horizon, nor rises + +A diagram showing a celestial sphere with lines representing different positions of Earth's axis relative to Sun's path (ecliptic) and Earth's equator. +*The different degrees of heat in summer and winter do not altogether arise from the different lengths of time when the sun is above or below; but from different situations of his rays above or below; he higher he sun is above or below he greater number of rays which fall on any given space, and he greater also is force of those rays. From all these circumstances arise different degrees of heat in summer and winter. The increase of heat also as you approach equator arises from two latter circumstances.* + +† This is not accurately true, because though sun's motion in ecliptic is not quite uniform, so much as to make it appear as if it rose on one side of equator as on another. If major axis of earth's orbit coincided with line joining equatorial points, times would be equal. + +ON THE DOCTRINE OF THE SPHERE. + +13 + +when in that part which is below; consequently there is half a year day and half a year night. Hence, the variety of seasons arises from the axis of the earth not being perpendicular to the plane of the ecliptic, for if it were, the ecliptic and equator would coincide, and the sun would then be always in the equator, and consequently it would never change its position in respect to the surface of the earth. If $QR = EH = 25°$. 25 the sun's greatest declination, then on the longest day the sun would describe the parallel $Re$, which just touching the horizon at $R$, shows that the sun does not descend on that day below the horizon, and therefore that day is 24 hours long. But when the sun comes to its greatest declination on the other side of $EQ$, it describes $WH$ and consequently does not ascend above the horizon for 24 hours, and therefore that night is 24 hours long. This therefore happens when $EH$, the complement of $EZ$ the latitude ($16$) is $25°$. 28, or in latitude $66°$. 52. If $EH$, the complement of latitude, be less than $35°$. 28, the sun will be above the horizon in summer, and below in winter, for more than 24 hours, and the longer above or below as you approach the pole, where, as before observed, it will be 6 months above and as long below the horizon. The orbits of all the planets, and of the moon, are also inclined to the equator, and consequently their motions amongst the fixed stars must be in circles inclined to the equator; therefore the same appearances will take place in each, in the time they make one revolution in their orbits. All these different appearances is the motion of the moon must therefore happen every month. It is evident also, that these variations must be greater or less as the orbits are more or less inclined to the equator; hence they must be greater in the moon than in the sun*. This apparent motion of the sun, and real motion of the moon and planets amongst the fixed stars, is from west to east; and therefore contrary to their apparent diurnal motion. + +82. Hitherto we have considered the motion of the heavenly bodies in the eastern hemisphere; but if this figure represent the western hemisphere, all the reasoning will equally apply; hence, the bodies will be just as long in descending from the meridian to the horizon as in ascending from the horizon to the meridian, the paths described will be similar, and they will set in the same situation in respect to the west point of the horizon as they rise in respect to the east; that is, if a body rise from the east towards the north or south, it will set at the same distance from the west towards the north or south. + +83. Having thus explained all the apparent diurnal motions of the heavenly bodies, with cause of the variety of seasons, we shall proceed in the next place to shew *the method of determining* **the positions** of **the different cir-** + +*On account of the continual change of declination of the sun, moon, and planets, their appa- rent diurnal motions will not be accurately parallel to the equator; in such therefore whose declina- tion alters sensibly in the course of a day, and in cases where great accuracy is required, we must, in our computations, take into consideration the change of declination.* + +14 + +**ON THE DOCTRINE OF THE SPHERE.** + +cles, and the situation of the bodies in respect to the horizon, meridian or any other circles, at any given time, and to find the time from their position. + +84. The altitude $PR$ of the pole above the horizon is equal to the latitude of the place. For the arc $ZE$ is (16) the measure of the latitude, but $PE = ZR$, each being = 90°, take away $ZP$ which is common to both, and $EZ = PR^*$. + +85. The latitude of a place may be found by observing the greatest and least altitude of a circumpolar star, and then applying the correction for refraction, and half the sum so corrected will be the altitude of the pole. For if $y$ be the circle described by the star, then, as $Py = Py$, $PR = \frac{1}{2} \times Ry + Rx$. The latitude may also be found thus. Let $O$ be the ecliptic, then when the sun comes to $e$ its declination is the greatest, and $cH$ is the greatest meridian altitude; when the sun comes to the ecliptic at $t$, let $ti$ be the parallel described on that day, and then $zH$ is the least meridian altitude; and as $Ee$, $\frac{1}{2} \times He + Hr = HE$ the complement of the latitude. + +86. Half the difference of the sun's greatest and least meridian altitudes is equal to the inclination of the ecliptic to the equator. For half $He - Hs$, or half $se$, is equal to $Ee$ which (12) measures the angle $EO$, i.e. the inclination of + +* From hence arises the method of measuring the circumference of the earth; for if a man travel upon a meridian till he height of the pole has altered one degree, he must then have travelled one degree; hence by measuring that distance and multiplying it by 360, we get the circumference of the earth. This was undertaken in England by our countryman Mr. Norman, who measured the distance between London and York, and observed the different altitudes of the pole at those places. Afterwards the French mathematicians measured a degree. Cassini measured one in France. After that, Clairaut, Maupertuis and several others made similar observations; but they did not find out that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into account that since they had not taken into accounted for this measurement. Eratosthenes makes the distance of Syene from Alexandria (which he says be under the same meridian) to be fifth part of the circumference of the earth; and he also makes it 5000 stades; hence, the circumference is 250000 stades. But as we do know nothing about length of stade, we cannot say how accurate this is. Clementus Memnon, lib. ii. ch. 18. + + + + + + +
8.
+ +ON THE DOCTRINE OF THE SPHERE. + +15 + +the ecliptic to the equator. To determine these altitudes without a quadrant, a gnomon $AB$ has been erected perpendicular to the horizon $AC$, and the length of the shadow $AD$ has been observed on the longest and shortest day, and knowing $AB$, the angle $BDA$ is known, which is the sun's altitude. Sometimes instead of erecting a gnomon, a small hole is made in the wall or roof of a building for the sun to shine through, and a plumb line let fall from the hole to the floor, the length of which is measured, and which answers to a gnomon $AB$; and the length of the shadow $AD$ from $A$ being also measured, ($A$ being the point where the plumb line meets the floor) the angle $BAC$ becomes known as before. Dr. Lovo made use of this method to find the latitude of Pembroke Hall in this University, from the known declination of the sun : He made the latitude 59°, 10' 55"; see his Astronomy, page 513. + +87. The angle which the equator makes with the horizon, or the altitude of that point of the equator which is on the meridian, is equal to the complement of the latitude. For $ZH$ is 90°, and therefore $EH$ is the complement of $EZ$; and as $OE=OH=90°$, $EH$ measures ($\sin$) the angle $EOH$. + +88. Let abdebe be parallel of declination described by any heavenly body in the eastern hemisphere, and draw the circles of declination $P_4$, $P_4$, $P_4$, and the circles of altitude $Z_4$, $Z_4$, $Z_4$. Now, as has been already explained, when the body comes to $b$ it rises, at $c$ it is at the middle point between $a$ and $e$, and at $d$ it is due east; and let $x$ be its place at any other time. Let us first suppose this body to be the sun, and not to change its declination* in its passage from $a$ to $e$, and let us suppose a clock to be adjusted to go 24 hours in one apparent diurnal revolution of the sun, or from the time it leaves any meridian till it returns to it again, then the sun will always approach the meridian which it has left, at the rate of 15° an hour; so that each circle of declination ; also, the angle which the sun describes about the pole will be at the same rate, because ($15^\circ \times 60 = 900^\circ$), are $\pi x$, which the sun at $x$ has to describe before it comes to the meridian, measures the angle $\pi P_4$, called the hour angle. If therefore we suppose the clock to show 12 when the sun is on the meridian at $a$ or $e$, it will be 6 o'clock when it is at $b$. And as the sun describes angles about the pole $P_4$ at the rate of 15° in an hour, the angle between any circle $P_4$ of declination passing through the sun at $x$ and the meridian $P_4x$, converted into time at the rate of 15° for an hour, will give the time from apper- rend noon, or when the sun comes to this meridian. + +89. Given the sun's declination and latitude of the place, to find the time of + +* Knowing the longitude and time nearly at any place, the sun's declination may be found at that time by first taking the declination for noon on the given day from the nautical Almanac, and then cor- recting it for the difference of the meridians of the place and Greenwich, and for the hour of the day, by th e 6th of Requisite Tables. These tables were computed to facilitate computations from +the nautical Almanac, and were published by th e Board of Longitude. + +Fig. 9. +Fig. 8. + +18 + +ON THE DOCTRINE OF THE SPHERE. + +time when twilight begins. Twilight is here supposed to begin when the sun is 18° below the horizon ; hence draw the circle $hk$ parallel to the horizon and 18° below it, and twilight will begin when the sun comes to $y$, and $Zy=108°$; hence, sin. $yP = \sin ZP = \cos PZ + Py + 108° \times \sin \frac{1}{2} \times PZ + Py - 108° \times \cos \frac{1}{2} \times PZ$, hence $yPZ$ is known, which converted into time gives the time from apparent noon. + +95. To find the time when the apparent diurnal motion of a fixed star is perpendicular to the horizon. Let $ye$ be the parallel described by the star; draw the vertical circle $Zh$ touching it at $a$, and when the star comes to $o$ its motion is perpendicular to the horizon ; and as the angle $ZoP$ is a right one, we have, (Trig. Art. 212.) rad. : tan. $oP = cot. PZ = cos. ZPo$, that is, rad. : cot. dec. :: tan. lat. :: cot. $ZPo$, which converted into time (Tab. 1) gives the time from the star's being on the meridian. Hence, the time of the star's coming to the meridian being found by Art. 103, the time required will be known. + +96. To find the time of the shortest twilight. Let $ab$ be the parallel of the sun's declination at the time required, drawn ed indefinitely near and parallel to it, and $TwA$ parallel to the horizon 18° below it ; then $cPe$, $dPz$ measure the twilight on each parallel of declination, and when the twilight is shortest, the increment of the hour angle being zero, these must be equal; hence, $vPe=vPz$, therefore $ve=ze$, and as $rv=ts$, and the angles $r$ and $z$ are right ones, $rv=zt$; but $Pv-90°=2ze$, take $Ze$ from both, and $Pv=-ze$; for the same reason $Pu-Z=zt$, hence, $PuZ=PwZ$. Take ve=$zE-90°$, then as $Pu=Pw$, and the angle $Pv=PwZ$, therefore $Pe=PZ$; let fall the perpendicular $Py$ and it will bisect the base $eZ$. Then (Trig. Art. 212.) cos. $Py=\cos vy=\cos Pv$; also, cos. $Py=\cos ey=\cos ez$; hence, cos. $Py=\cos PZ=\cos ze$; or sin. ey $\cdot$ sin. ez $\cdot$ sin. ey $\cdot$ sin. ez $\cdot$ cos. PZ $\cdot$ tan. eZ, e.g., hence, rad. :: cos. PZ, or sin. lat.: tan. ey=9° : sin. ke the sun's declination at the time of shortest twilight. Because $PZ$ is always less than 90°, and $Zy=9°$, therefore $y'Y$ is always less than 90°, and therefore its cosine is positive ; also, vy is always greater than 90°, therefore its cosine is negative, hence, cos. Fe ($\equiv$cos Py $\times$ cos ey) is negative, consequently Fe is greater than 90°, therefore the sun's declination is south. This is M. Cagnoli's Investigation. + +97. To find the length of the shortest twilight. As $ePZ=ePe$, therefore $ZePe=Pwz$, measuring the shortest time Now sin. PZ, or cos. lat.: rad.: $\sin Zy=9° : \sin ZFy$, which doubled gives ZePe or Fwz, which converted into time gives the length of the shortest twilight. + +A diagram showing geometric relationships between angles and distances. + +ON THE DOCTRINE OF THE SPHERE. + +19 + +Ex. To find the time of the year at Cambridge, when the twilight is short- +est ; and the length of that twilight. + + + + + + + + + + + + + + + + + + + + + + + + + + +
Rad.- - - - - - - - -- - - - - - - - -10,000000
Sin. $22^{\circ}$, $12^{\circ}$, $35^{\circ}$- - - - - - - - -- - - - - - - - -9,877695
Tan. $9^{\circ}$- - - - - - - - -- - - - - - - - -9,1997125
Sin. $7^{\circ}$, $11^{\circ}$, $25^{\circ}$ dec.- - - - - - - --9,0974820
+ +This declination of the sun gives the time about March 2, and October 11. +Cos. $52^{\circ}$, $12^{\circ}$, $35^{\circ}$ A.C. +Sin. $9^{\circ}$ - +Rad. - +Sin. $14^{\circ}$, $47^{\circ}$, $27^{\circ}$ +- 9,4070328 + +The double of this gives $29^{\circ}$, $54$, $54$, which converted into time gives 14. +$58$, $30$ for the duration of the shortest twilight, it being supposed to end when +the sun is $18^\circ$ below the horizon. + +98. To find the sun's declination when it is just twilight all night. Here the +sun at a must be $18^\circ$ below the horizon ; hence, $18^\circ + \text{dec.} = RQ = EII = \text{comp. lat. of place},$ and the sun's $\text{dec.} = \text{comp. lat.} = 18^\circ$. Look therefore into the Nautical Almanac, and see on what days the sun has this declination, +and you have the time required. The sun's greatest declination being $25^\circ$, $28^\circ$, +it follows, that if the complement of latitude be greater than $41^\circ$, $28^\circ$, or if the +latitude be less than $46^\circ$, $32^\circ$, there can never be twilight all night'. If the sun +be on the other side of the equator, then its $\text{dec.} = 18^\circ$, -- comp. lat. + +99. If the spectator be between E and I., and the sun's declination Ee be +greater than EZ, then the sun comes to the meridian at e to the north of its +zenith ; and if we draw the secondary Zym touching the parallel ae of declina- +tion described by the sun, then Rm is the greatest azimuth from the north +which the sun has that day, the azimuth increasing till the sun comes to g, +and then decreasing again, and the sun has the same azimuth twice in the morn- +ing. If therefore we draw the straight line Ze perpendicular to the horizon, +the shadow of this line, being always opposite to the sun, would first recede +from the south point H and then approach it again in the morning, and there- +fore would go backwards upon the horizon. But if we consider P'F' as a +straight line, or the earth's axis produced, the shadow of that line would not +go backwards upon that plane, because the sun always continues to revolve +about that line, and therefore its shadow must always go forwards; whereas +the sun does not revolve about the perpendicular Ze. Hence it appears, that + +FIG. 5. + +FIG. 11. + +20 + +**ON THE DOCTRINE OF THE SPHERE.** + +the shadow of the sun upon a dial can never go backwards, because the gnomon of a dial is parallel to $PP$, and therefore the sun must always revolve about the gnomon. + +The time when the azimuth is greatest is found from the right angled triangle $PQZ$, by saying, rad. : tan. $gP$ : cot. $PZ$ : cos. $ZPq$, or rad. : cot. dec. :: tan. lat. : cos. $PZq$ the hour angle from apparent noon. + +100. It has hitherto been supposed, that it is 12 o'clock when the sun comes to the meridian, and that the clock goes just 24 hours in the interval of the sun's passage from any meridian till it returns to it again. But if a clock be thus adjusted for one day, it will not continue to show 12 o'clock every day when the sun comes to the meridian, because the intervals of time from the sun's leaving any meridian till it returns to it again, are not always equal; this difference between the sun and the clock is called the **Equation of Time**, as will be explained in Chap. IV. Hence, when the clock does not agree with the sun, any arc $\alpha r$ is not the measure of the time from 12 o'clock, but from the time when the sun comes to the meridian, or from **apparent noon**. + +101. In the same manner as we find the hour angle for the sun, we may also find it for any fixed star or planet, its altitude and declination being given ; but when the hour angle is thus found, it is necessary to know the time when the body is upon the meridian in order to find the time from thence, the hour angle being the distance from the meridian ; also the method of reducing the hour angle into time will be different. + +For let $E$ be the earth, $rmax$ the equator, $ar$ a meridian passing through a fixed star $S$ reduced to the equator ; then as the meridian returns to the star in $236$. $56$. $4^{\circ}$ after leaving it (187), we have $360^{\circ}$ : hour angle :: $236$. $56$. $4^{\circ}$ : time from the meridian. Now let $P$ be a planet, and the meridian $ar$ pass through it ; then the meridian will return to that position again in $236$. $56$. $4^{\circ}$; now let $Pv$ or $Pw$ be the planet's motion in right ascension in one day, according as its motion is direct or retrograde, and reduce this into time ($t$) at the rate of $1^\circ$ for an hour, which will be sufficiently exact for so small an arc, then the meridian returns to the planet again after an interval of $236$. $56$. $4^{\circ} \pm t$; hence, the meridian, after leaving the planet, approaches it at the rate of that time for $360^{\circ}$ because when the meridian leaves the planet it is then approaching a point $360^{\circ}$ from it; hence, $360^{\circ}$ : hour angle :: $236$. $56$. $4^{\circ} \pm t$ : time from the meridian. + +Fig. 10. +Fig. 12. + +* The conversion of the hour angle into time for the sun at the rate of 15' for an hour, by a clock adjusted to mean solar time, is not accurate, because the solar days are not all accurately equal to 24 hours, but to 244$\frac{1}{2}$-the variation $(e)$ of the equation of time for that day, according as the equation is increasing or decreasing; hence, to reduce the hour angle to give accurately the time from apparent noon, say, $300^{\circ}$ : hour angle ($a$) :: 244.$\pm e$: time from apparent noon; say, 300$\frac{1}{2}$ : hour angle ($a$) :: 244.$\pm e$: time from apparent noon; say, 300$\frac{1}{2}$ : hour angle ($a$) :: 244.$\pm e$: time from apparent noon; say, 300$\frac{1}{2}$ : hour angle ($a$) :: 244.$\pm e$: for, in this case, the meridian instead of returning to the sun in 244. returns to it in 244.$\mp e$. This quantity $e$ is sometimes 20', and therefore if $a = 300^{\circ}$, the correction would be 5'. A clock is adjusted to mean solar time, when it is adjusted to go 24 hours in a mean solar day. See Art. 127. + +Art. 127 + +ON THE DOCTRINE OF THE SPHERE. + +21 + +02. The hour angle which we have hitherto found for the time at which a body rises, has been upon supposition that the body is upon the rational horizon when it first appears; but all bodies in the horizon are elevated by refraction 35° above their true places; this therefore would make them appear when they are 35° below the rational horizon, or 90° + 35° from the zenith; also, all the bodies in our system are depressed below their true places by parallax, as will be afterwards explained, therefore from this cause they would not appear till they were elevated above the rational horizon by a quantity equal to their horizontal parallax, or when distant from the zenith 90° - hor. par. Hence, from both causes together, a body becomes visible when its distance $ZV$ from the zenith = 90° + 35° - hor. par. $V$ being the place of the body when it becomes visible; $Z$ the zenith and $P$ the pole; hence, knowing $ZV$, also $ZP$ the complement of latitude and $PV$ the complement of declination, we can find the hour angle $ZPV$. A fixed star has no parallax, therefore $ZV = 90° \cdot 35°$. + +103. If the body sensibly alter its declination in a few hours, as the moon does, the time of its rising may be thus found. Let $w$ be the place of the moon on the meridian, $v$ when in the horizon, and $d$ the point when it becomes visible; draw $adc$ parallel to $EQ$, and $\Delta w$ is the change of declination in the time from rising to the meridian. Now from knowing the time (105) of passing the meridian, and the declination at noon, with the change of declination in the interval of the passages of the moon over the meridian by the Nautical Almanac, compute the change of declination in the interval between noon and the time of the moon's transit, and you will get the moon's declination at the time of its transit. To that declination compute the hour angle upon supposition that the declination continued the same as on the meridian, which will be nearly the angle $\pi Pd$. From the Nautical Almanac find the change ($\varphi$) of declination in the interval ($t$) of time from the moon's passage over the meridian till it returns to it again; then say, $860^\circ$: hour angle:: $\varphi$: the change of declination in describing that angle, which added to or subtracted from the declination at the time of passing the meridian gives very nearly the declination at rising; to which compute the hour angle and convert it into time as before and subtract it from the time of passing the meridian, and it gives very nearly the time of rising; and if greater accuracy should be required, the operation may be repeated by taking this hour angle. + +Ex. To find at what time the moon rose at Greenwich on July 1, 1767. The latitude of Greenwich is 51° 28' 40", and (105) the moon passed the meridian at 4h. 2'. 9"; now $t=24h$. 40', and $v=5^\circ$. 28'; hence, $24h$. 40': 4h. 2'. 9": $5^\circ$. 28' : $5^\circ$. 38" : The change of declination in 4h. 2'. 9", which, as the declination is decreasing, subtracted from $5^\circ$. 28', the moon's north declination at noon, leaves $4^\circ$. 38'. 22" for the moon's declination when it was on the meridian; hence we take $Pd=85^\circ$. 31'. 38", also $PZ=38^\circ$. 31'. 20"; and as + +22 + +**ON THE DOCTRINE OF THE SPHERE.** + +moon's hor. parallax=54° 21', and refraction 33', we have $Zd = 89°$, $38°$, $89°$, +hence the angle $ZPd = 95°$. $S$. $80°$. Hence, $360°$: $95°$. $S$. $80°$: $5°$, $28'$. $1''$. +$26'. 37'$ the change of declination in the time of describing $95°$. $S$. $80°$, which +added to $4°$. $28'$. $22'$ gives $5°$. $54'. 59'$ for the declination at the time of rising, +very nearly; hence, $Pd = 86°$. $5'. 1''$, therefore the angle $ZPd = 96°$. $54'. 2''$; +hence, $360°$: $96°$. $54'. 2''$: $244'. 40'$. $64'. 38'. 22'$ the time of describing the +angle $ZFd$, which subtracted from $44'. 2''. 9''$, the time when the moon was on +the meridian, gives the time of rising $21h$. $25'. 47''$, answering to $9h$. $47''$ +in the morning apparent time. + +104. In determining the time when any body rises, or when it is at any known +altitude or position, it has been supposed that we know the time at which it +comes to the meridian ; the determination of this circumstance must therefore +be next explained. + +105. Let a clock be adjusted to mean solar time, which we may therefore +consider as the time from the sun's leaving the meridian till it returns to it +again, where great accuracy is not required, the difference being only the vari- +ation of the equation of time in 24 hours. Let $S$ and $P$ be the places of the +sun and a planet reduced to the equator; then the meridian $\pi$ approaches the +sun at the rate of $1^\circ$ in an hour; for when it leaves the sun at $S$ it may be +considered as approaching a point at that time $360^\circ$ from it, and which it comes +up to in 24 hours; hence if any other point were moving forwards with the ve- +locity of the sun, the meridian would approach it at the same rate. Therefore +if the planet at $P$ move forwards with a different velocity from that of the sun, +the interval of their passages over any meridian will be the same as if we sup- +posed the sun to be at rest and the planet to move with its own proper motion +minus that of the sun, the planet's motion in right ascension being greater +than that of the sun. + +Let $x$ be the difference of their motions in right ascen- +sion in 24 hours reduced into time, and $t = SP$ reduced also into time in like +manner, the planet being at $P$ at the time the meridian passes through the sun at $S$; and let $\nu$ be the place of the planet when the meridian overtakes it, +and $\epsilon$ be the arc $\nu P$ in time; then the motions of the meridian will be $\frac{2\pi}{t}$ and +$t + \epsilon$, and of the planet in the same times $\nu$ and $\epsilon$; hence, as we may consider +each motion as uniform, $\frac{2\pi}{t} : t + \epsilon : \nu : \epsilon = \frac{2\pi}{t} : t + \epsilon : x : t : e = \frac{tx}{24 - x}$. This is +the case if the planet's motion be greater than the sun's, but if the sun's be +greater, then $\nu$ itself becomes negative, and therefore $\nu$ will be positive; +hence $\epsilon = \frac{-tx}{24 + x}$; therefore $t + \epsilon = t \pm \frac{tx}{24 + x} = \frac{24t}{24 + x}$, the time from apparent +noon when the planet passes the meridian, where the upper or lower sign +prevails according as the planet's or sun's motion is greatest. If the motion +of the planet in right ascension be retrograde, it is manifest that $\nu$ is the + +Fig. 12. + +ON THE DOCTRINE OF THE SPHERE. + +23 + +sum of the motions of the planet and sun in 24 hours, for the bodies moving in opposite directions they approach each other with the sum of their motions; let therefore $v$ be the place of the planet when it comes to the meridian, then the motion of the meridian from its passage through the sun to the planet will be $t - e$; hence $\frac{tx}{24+x} = \frac{tx}{24+x} = \frac{24x}{24+x}$ But as the division by $24+x$ is not so convenient as it would be by 24, therefore resolve $\frac{24t}{24+x} = \frac{tx + tx^2}{24+x}$. where the two first terms will be sufficient for all cases except the moon, where it will be necessary to take the third. For a fixed star, $x$ will represent the increase of the sun's right ascension in 24 hours, and the time required $\frac{24t}{24+x} = t - \frac{tx}{24}$. By this method we find, very nearly, the time at which any body comes to the meridian, and hence, by the last articles, we may find the time of its rising, or the time at any given altitude. + +Ex. To find the time of the moon's passage over the meridian at Greenwich on July 1, 1767. The sun's AR* when on the meridian that day was 6h. 40' 25", and its daily increase 4' 48"; also, the moon's AR was 10h. 36' 8", and its daily increase 4' 28". Hence, $t = 10h. 36'. 8' - 6h. 40'. 25" = 3h. 55'. 43 = 3,928 (Tab. 3.), also, $x = 42'. 28' - 4' 48" = 37'. 40" = 0,6277; hence, $\frac{tx}{24} = 6'. 10'$; $\frac{tx^2}{24} = 10'$; therefore $t + \frac{tx}{24} + \frac{tx^2}{24} = 4h. 2'. 5"$ the apparent time of passing the meridian. + +Where great accuracy of time is required from an observed altitude, the body made use of must be the sun or a fixed star. The method of finding the time by the sun has been already explained (92); and the time by a star may be found by the following method. + +106. Find the star's true altitude, and take its declination from the 7th of the Requisite Tables, or from any other tables if it be not there; then in the triangle ZPz ($z$ representing the place of the star) we have $ZP$ the complement of latitude, $Pz$ the complement of declination and $Zz$ the complement of the star's altitude, to find the angle ZPz, the star's distance from the meridian, which convert into time. Now the point of the equator which is upon the meridian at any time, is called the mid-heaven; therefore the angle ZPz measures the star's distance from the mid-heaven. Hence, if the star be to the east of the meridian, subtract its distance from the meridian from its AR. (adding, if necessary, 24 hours to its AR.) and the difference is the AR. of the mid-heaven: But if the star be to the west, add them together (sub- + +A diagram showing a triangle with points labeled ZPz and Pz. +- S. + +*AR means right ascension.* + +24 + +**ON THE DOCTRINE OF THE SPHERE.** + +tracting 24 hours from the sum, if greater,) and the sum gives the $AR.$ of the mid-heaven*. Then find the sun's $AR$ at the preceding noon at Greenwich from the Nautical Almanac, and from thence at noon at the given place by the 23d of the Requisite Tables, and subtract it from the $AR.$ of the mid-heaven (adding 24 hours to the latter, if necessary), and the difference would be the apparent time from the preceding noon, or the estimate time, if the sun had no motion in that time; but as it has moved, find that motion by the 23d of the Requisite Tables, and subtract it, and it gives the apparent time required.—Hence, if we apply the equation of time it gives the true time, which compared with the watch, shows how much it is too fast or too slow; and by repeating the observations, the rate of going of the watch may be determined; but this will be further explained in Chap. IV. + +Ex. On April 14, 1780, lat. 48°. 56' N. lon. 66° W. the true altitude of Aldebaran west of the meridian was 22°. 17'. 50'; to find the apparent time. + +Sun's $AR.$ for noon at Greenwich by the Nautical Almanac 1h. 31. 1' +Corrected for the Long. by the 23d of the Requisite Tables† +41 + +Sun's $AR.$ at noon at the given place - - - - - - - - 1.91.42 + +Also by Requisite Table 7. the star's dec. is 16°. 3' N. Hence $ZP = 41^\circ$. 4', $Zx = 67^\circ$. 42', $xP = 78^\circ$. 57', hence by sph. trig. +$$Pz = 78^\circ. 57'. \text{ o' arith. comp. of sine } 0.017804$$ +$$ZF = 41^\circ . \text{ o arith. comp. of sine } 0.182476$$ +$$Zx = 67^\circ . 42'. \text{ 10}$$ +Sum = 182 . 43 . 10 + +Fig. 15. +* That this is true for every position of the point aries and place of the star, may be thus shown. +Let $ES$ represent the equator, $E$ the point on the meridian, $\varphi$, $\psi$, $\vartheta$, different positions of the point aries, in respect to the place $A$, $A'$ of the star referred to the equator, $A$ on the western side of the meridian, and $A'$ on the eastern; $B$ the point to which the sun is referred; $\varphi EB\beta$ the direction in which the right ascension is measured. +Now suppose the star at $A$, to the east of the meridian; then, I. $\varphi A - AR = \varphi E$. II. $\varphi A - AR = \varphi B - \varphi E$. III. $\varphi A + AR = \varphi E + \varphi B$. Now suppose the star at $A'$ on the west side; then I. $\varphi A + AR = \varphi E$. II. $\varphi A + AR = \varphi E + \varphi B$. III. $\varphi A - AR = \varphi E + \varphi B$, because $\varphi A + AR = \varphi E$. +† For I, $\varphi EB\beta - \varphi B = EB\beta$. II. $\varphi E + \varphi B$. III. $\vartheta B = EB\beta - \varphi B$, because $\vartheta E + EB\beta = \vartheta B$. + +The daily variation of the sun's $AR,$ with which you enter the Requisite Tables, is taken from +the Nautical Almanac. + +§ + +ON THE DOCTRINE OF THE SPHERE. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
$\frac{1}{2}$ Sum = 91.21.35 sine - - - - 9.99987425
Zr = 67.42.10
Dif. = 23.39.25 sine - - - - 9.603425
2)19.803079
9.901539 the cosine of
37°. 8°. 29°; hence the angle $xPZ$ (or in fig. 15. the arc $AE$) = 74°. 16°. 58°,
or in time = 4h. 57°. 8°; hence,
Star west of merid. - 4h. 57°. 8° Estimate Time - - - 7h. 48°. 46°
Star's AR by Req. Tab.7.4 . 23. 20 Correc. from Req. Tab.23. - 1. 12
AR of mid-heaven - 9 . 20 . 28 Apparent Time required 7 . 47 . 34
+ +fig. 16. + +107. The time of the passage of a star over the meridian may be found (78) from taking the times at which it had equal altitudes on each side of the meridian, and bisecting the interval. If equal altitudes be taken at $8$ and $11$ o'clock, the star was upon the meridian at half past $9$ o'clock. But for the sun this will want a correction, owing to its change of declination, on which account it is not at equal altitudes when equidistant from the meridian. If $be$ be the diurnal arc described by the sun in its ascent to the meridian, and $ed$ in its descent from it, and $mn$ be drawn parallel to $HOR$, then the sun is at equal altitudes at $m$ and $n$, and the angle $mPn$, or the arc $gr$, measures the difference of the times at $m$ and $n$ from the meridian; when we therefore bisect the interval of the times at which the sun was at $m$ and $n$, we must correct it by half $mPn$, or half $gr$, in order to get the time at which it comes to the meridian. This correction is called the equation of equal altitudes. Now (Trig. Art. 264.) if $d=$ the variation of the sun's dec., in the interval of the observations, $t=$tan lat., $v=$tan.decl.at noon, $s=$sin e, $r=$tan.of the hour angle from noon at the time of observation, taking the half interval of times for the measure of that angle; then $\frac{1}{2}t^{\prime}=\frac{d}{s}\times\frac{v}{r}$ radius being unity; or as the value of $d'$ in time is $\frac{d}{15'}$ seconds, estimated at the rate of $15'$ for $1$ hour, or $15'$ for $1$ second of time, therefore the correction is $\frac{d}{30}\times\frac{T_{\text{sec}}}{s}\times\frac{v}{r}$ seconds of time, where the sign — is to be used when the lat. and decl. are both north or both south, and + when one is north and the other south. Now in north latitude, when the sun approaches the north pole, or is in the 9th. 10th. 11th. etc., signs, it is manifest + +vol. I. +E + +26 + +**ON THE DOCTRINE OF THE SPHERE** + +from the figure, that the sun, after passing the meridian, will not come to the same altitude as at the observation before, until it be at a greater distance from the meridian; therefore the middle point of time between the observations must be, when the sun has passed the meridian, and the correction must be subtracted. When the sun is in the other signs, receding from the north pole, it comes to the same altitude at a less distance from the meridian; therefore the middle point of time must be, before the sun comes to the meridian, and consequently the correction must be added. To facilitate this computation, Mr. WALKER constructed and computed a set of tables which were published in the Nautical Almanac for 1778; these tables are called *Equation to corresponding altitudes*. + +To find the Time the Sun is passing the Meridian, or the horizontal or perpendicular Wire of a Telescope. + +FIG. +8- + +108. Let $m_x$ be the diameter $d'$ of the sun, estimated in seconds of a great circle; then, (as the minutes in $m_x$, considered as a small circle, must be greater in proportion as the radius is large, because, when the arc is given, the angle is inversely as the radius), sin. $P_x$, or cos. dec. $x : rad.:$ second $d'$ in me of a great circle: the seconds in $m_x$ of the small circle $e_a$, which is equal to $(19)$ the seconds in $gr$: the angle $rFg = d'$ divided by cos. dec. (rad. being unity) = $d' \times sec. dec.$, which measures the time the sun is passing over its diameter, and consequently the time the diameter would be in passing over the meridian; hence (as in Art. 107), the time of passing the meridian = $\frac{d'}{\cos x \times \sec x}$. + +Hence $gr$, the sun's diameter in right ascension, is equal to $d' \times \sec x$. If therefore the sun's diameter = $38^\circ = 1920'$ and its dec. $20'$, its diameter in right ascension = $1920' \times 1,064 = 34', 2', 96'$. The same will do for the moon, if $d'$ is its diameter. + +109. By Art. 93. $gr = nx \times \frac{rad.}{\cos lat. \times sin. azi} = (\text{if } nx = d' \text{ the sun's diam.)}$ + +$d' \times \frac{rad.}{\cos lat. \times sin. azi} ;$ hence, as before, the time of describing $gr$, or the time in which the sun ascends perpendicularly through a space equal to its diameter, or the time of passing an horizontal wire, is equal to $\frac{d'}{\cos lat. \times sin. azi}$. The same expression must also give the time which the sun is in rising. + +If $m_x = 38^\circ$ be the horizontal refraction, then $d'$ divided by $15^\circ$ = $192'$; hence, + +refraction accelerates the rising of the sun by $192' - \cos lat. \times sin. azi$. + +A diagram showing a celestial sphere with lines representing different angles and measurements. + +ON THE DOCTRINE OF THE SPHERE. + +The $\sin \alpha \cdot \cos \beta = \sin \alpha \cdot \cos \beta$ : $m n = m n \times \sin \alpha \cdot \cos \beta$ ; hence (38), $gr =$ +$\frac{mn}{\sin \alpha} \times rad.$ +and if $mn = d$, we +time, in which the horizontal motion of the sun is equal to its diameter, +$x \times cos 2F \times cos dec.$ +which is therefore the time in which the sun +pass the vertical wire of a telescope. + +MASKELYNE'S Rules to find the Time of the Passage of a Star or Planet from one Wire to another of a transit Instrument. + +For a fixed Star. Multiply the equatorial interval of time by the secant +star's declination, and you have the time required. For an arc of the +tor, measured on a small circle parallel to it, subtends a greater angle +the earth's axis, in the proportion of rad. : cor. dec. or sec. dec. : + +the Sun. Increase the equatorial time of a star by the 365th part (owing +sun's motion in that time) and you have the equatorial time by the sun, +proceeds as for a star. + +or a Planet, except the moon. Take the difference ($d$) of 25h. 56', and +interval of two successive transits of the planet over the meridian, as given +Nautical Almanac; then say, 24h. $d$: the time of the passage of a star +the same declination : a fourth number, which added to or subtracted: +the time of the passage of a star, according as the interval of the two +cessive transits is more or less than 25'. 56", gives the time of the planet's +age. + +For the Moon. Put $n =$ the equatorial interval by a star, $r =$ daily retard- +on of the moon's passage over the meridian in minutes; then allowing for +moon's motion, 28h. 56' : $1440 + r$ : $n \times 1440 + r$ : $23h$. 36' +time in the equator from +wire, seen from the earth's center. Now the time of the image from wire- +are, is $\pi$eris paribus, as the angle subtended by the interval of the wires at +ject glass, or as its vertical angle, or the angle described by the moon about +opposed place of observation; but the velocity of the moon and the angle +being given, the arc, and therefore the time, is as the distance; +the time seen from the center of the earth $(n \times 1440 + r)$ : time at the +$e'$s dist. from center : $e'$s dist. from spectator : $\sin$ ap. Zen. dist. +true zen. dist. therefore the interval of time ($t$) at the spectator= + +#8 + +**.ON THE DOCTRINE OF THE SPHERE.** + +$$x \times 1440 + r' \times s.\mathrm{tr.\,zen.\,dist.} \times \sec. e' \text{'s dec.};\; \text{hence, Log. } t = 6.84273 + l.\; n + 23h.56 \times s.\mathrm{ap.\,sen.\,dist.}$$ + +$$L.(1440+r) + L.\; \text{Req. Tab. IX.-l sec. 'e' dec. -30.}$$ + +**On the Principles of Dialling.** + +112. As the apparent motion of the sun about the axis of the earth is at the rate of $1^\circ$ in an hour, very nearly, let us suppose the axis of the earth to project its shadow into the meridian opposite to that of the sun, and then this meridian will move at the rate of $1^\circ$ in an hour. Hence, let $zPRpH$ represent a meridian on the earth's surface, $PO$ its axis, $z$ the place of the spectator, $HKRV$ a great circle of which $z$ is the pole; draw the meridians $P1$, $P2$, $\delta c.$ making angles with $PRp$ of $1^\circ$, $50^\circ$, $\delta c.$ respectively; then supposing $PR$ to be the meridian into which the shadow of $PO$ is projected at 12 o'clock, $P1$, $P2$, $\delta c.$ are the meridians into which it is projected at 1, 2, $\delta c.$ o'clock, and the shadow will be projected on the plane $HKRV$ in the lines $OR$, $O1$, $O2$, $\delta c.$, and the arcs $R1$, $R2$, $\delta c.$ will measure the angles $RO1$, $RO2$, $\delta c.$ between the 12 o'clock line and the 1, 2, $\delta c.$ o'clock lines. Now in the right angled triangle $PRI$, we have $PR(34)$ the latitude of the place, and the angle $RP1 = 1^\circ$; hence, rad. : tan. $1^\circ$: sin. $PR$: tan. $R1$; in the same manner we may calculate the arcs $R2$, $R3$, $\delta c.$ In this case we make the earth's axis the gnomon, and the shadow is projected upon the plane $HKRV$. But if we take a plane abcd at a parallel to $HKRV$, and consequently parallel to the horizon at $z$, and draw ze parallel to Po9, then on account of the great distance of the sun we may conceive it to revolve about $z$ in the same manner as about Pp, and consequently the shadow will be projected upon the plane abcd in the same manner as the shadow of PO is projected upon the plane HKRV, and therefore the hour angles are calculated by the same proportion. This is an horizontal dial. + +113. Now let NLzK be a great circle perpendicular to PRpHzn, and consequently perpendicular to the horizon at z, and the side next to H is full south; Then, for the same reason as before, if the angles Np1, Np2, $\delta c.$ be $1^\circ$, $50^\circ$, $\delta c.$ the shadow of pO will be projected into the lines O1, O2, $\delta c.$ at 1, 2, $\delta c.$ o'clock, and the angles NO1, NO2 will be measured by the arcs N1, N2 $\delta c.$ Hence, in the right angled triangle pNt, pN = complement of latitude; and the angle Np1 = 1°; therefore rad. : tan. 1°: sin. pN : tan. Nt; in the same manner we find Np2, Np3, $\delta c.$ Hence, for the same reason as for the horizontal dial, if abcd be a plane coinciding with NLzK, and st be parallel to Op, st will project its shadow in the same manner on the plane abcd as Op does on the plane NLzK, and therefore the hour angles from the 12 o'clock line are computed by the same proportion. This is a vertical south dial. In + +A diagram showing a spherical model with various lines and angles labeled. +fig. +17. + +fig. +18. + +ON THE DOCTRINE OF THE SPHERE. +29 + +same manner the shadow may be projected upon a plane in any position, and the hour angles be calculated. + +114. In order to fix an horizontal dial, we must be able to tell the exact time of the sun's coming to the meridian; for which purpose, find the time (92) by the sun's altitude when it is at the solstices, because then the declination does not vary, and set a well regulated watch to that time; then when the watch shews 12 o'clock, the sun is on the meridian ; at that instant therefore set the dial to 12 o'clock, and it stands right. + +115. Hence we may easily draw a meridian line upon any horizontal plane. Suspend a plumb line so that the shadow of it may fall upon the plane, and when the watch shows 12, the shadow of the plumb line is the true meridian. The common way is to describe several concentric circles upon an horizontal plane, and in the center to erect a gnomon perpendicular to it with a small round well defined head, like the head of a pin; make a point upon any one of the circles where the shadow of the head, by the sun, falls upon it on the morning, and again where it falls upon the same circle in the afternoon; draw two radii from these two points, and bisect the angle which they form, and it will be a meridian line. This should be done when the sun is at the tropic, when it does not sensibly change its declination in the interval of the observation; for if it do, the sun will not (107) be equidistant from the meridian at equal altitudes. This method is otherwise not capable of very great accuracy, as, from the shadow not being very accurately defined, it is not easy to say at what instant of time the shadow of the head of the gnomon is bisected by the circle. If, however, several circles be made use of, and the mean of the whole taken, the meridian may be gotten with sufficient accuracy for all common purposes. + +116. To find whether a wall be full south for a vertical south dial, erect a gnomon perpendicular to it and hang a plumb line from it; then when the watch shows 12, if the shadow of the gnomon coincide with the plumb line, the wall is full south. + +85 + +$$x 1440 + r \times s. b.z. \\ +93h. 56', \times s. a.p. \\ +L (1440 + r) + 7. R.e. - 2$$ + +112. As the apparent rate of $15^\circ$ in an object its shadow meridian will move on the meridian on the HKRF great circle making angles with the meridians $P_1$, $P_2$, &c., and the shadows $O_2$, &c., and the times between the 12 o'clock angled triangle $RP_1 = 15^\circ$; hence we may calculate the gnomon, and the plane abcd at z, and draw the sun we must Pp, and in the same manner therefore the horizontal dial. + +Fig. 113. Now by sequentially perpendicularly for the shadow, &c., the o'clock, and the Hence, in the angle A and the same manner we will project Op, st does on the plane line are computed by their error. + +A page from a book, possibly a scientific treatise on astronomy or navigation, with text discussing the movement of objects and shadows relative to a meridian and a gnomon. +Mr. Flamsteed says that this star with that of explanation thereof, + +| No. | Description | +|---|---| +| 17. | The meridian will move on the meridian on the HKRF great circle making angles with the meridians $P_1$, $P_2$, &c., and the shadows $O_2$, &c., and the times between the 12 o'clock angled triangle $RP_1 = 15^\circ$; hence we may calculate the gnomon, and the plane abcd at z, and draw the sun we must Pp, and in the same manner therefore the horizontal dial. | + +| No. | Description | +|---|---| +| 18. | By sequentially perpendicularly for the shadow, &c., the o'clock, and the Hence, in the angle A and the same manner we will project Op, st does on the plane line are computed by their error. | + +TO DETERMINE THE RIGHT ASCENSION, &C., OF THE HEAVENLY BODIES. + +together with an example. Let $AGCKE$ be the equator, $ABCEW$ the ecliptic, $S$ the place of the star, and $Sm$ a secondary to the equator, and let the sun be at $P$, very near to $A$, when it is on the meridian, and take $CT = PA$, and draw $JL$, $TQ$ perpendicular to $ACG$ and $QL$ parallel to $AC$; then the sun's declination is the same as $T$ at $P$. Observe the meridian altitude of the sun when at $P$, and also the time of the passage of its center over the meridian; observe also what time the star passes over the meridian, and then (318) find the apparent difference $Lm$ of their right ascensions. When the sun approaches near to $T$, observe its meridian altitude for several days, so that one of them, at $i$, may be greater and on the next day, $i'$, it may be less than the meridian altitude at $P$, so that in the intermediate time it may have passed through $T$; and drawing $\alpha\beta$, perpendicular to $ACGE$, observe on those two days, the differences $\delta m$, $\delta m'$ of the sun's right ascension and that of the star; draw also $\alpha\gamma$ parallel to $QQ'$. Hence, to find $QQ'$ we may consider the variation both of the right ascension and declination to be uniform at some time, and consequently to be proportional to each other; hence, $\alpha\beta$ (the change of meridian altitudes in one day); $\delta\beta$ (the difference of the meridian altitudes at $i$ and $i'$, or the difference of declinations); $\delta\alpha$ (the difference of $\alpha\beta$, as found by observation); $\alpha Q_0$, which added to $\delta\beta$, or subtracted from it, according to the situation of $\alpha\beta$, gives $Qm$, to which add $\delta m$, or take their difference, according to circumstances; and we get $QT$, which subtracted from $ACC$ or 180°, half the remainder will be $\Delta L$ the sun's right ascension at the first observation, to which add $\delta m$ and we get the star's right ascension at the same time. Instead of finding $QQ'$ we might have found $QQ_1$ by taking $TQ - \epsilon s$ for the second term, and from thence we should have gotten $Qm$. Thus we should get the right ascension of a star, upon supposition that the position of the equator had remained the same, and the apparent place of the star had not varied, in the interval of the observations. But the intersection of the equator with the ecliptic has a retrograde motion, called the Precession of the Equinoxes; also, the inclination of the equator to the ecliptic is subject to a variation, called the Nutation; and from the Aberration of the star, its apparent place is continually changing. The effects of all these circumstances in changing the right ascension of the star will be explained and investigated in their proper places. + +Now Tables VII. and VIII. (see Vol. II.) contain these corrections for so principal cases; that is, if the mean right ascension of any star be taken for the beginning of the year, and these corrections be applied to it, according to their signs, for any day, the result gives the apparent right ascension of the star for that day. + +120. Let therefore $ABCE$ be the ecliptic, $AGCE$ the position of the equator at the first observation when the sun was at $P$, and aged the position of this equator at the time of the observation at other equinoxes, and take $TC = 27^{\circ}4'$, + +Fig. 19. + +Fig. 20. + +32 + +TO DETERMINE THE RIGHT ASCENSION, DECLINATION, + +and draw $TQ$ perpendicular to $AGCE$, as before, and draw $Qq$ parallel to $ABC$, and $tqr$ perpendicular to $AGCE$; let $Ae$ be also perpendicular to $agcd$. Now as the position of the equator and the apparent place of the star are altered in the time between the two observations, let $m$ be the point where a secondary from the apparent place of the star to the equator at the first observation would cut it, and $v$ the place at the second observation, and draw $vw$ perpendicular to $AGCE$; then $Am$ is the apparent right ascension of the star at the first observation, and $av$ at the second. Also, the sun must be at $t$ when it has the same declination $tg$ at the second observation as it had at the first, and consequently $qv$ is the apparent difference of right ascensions of the sun at $t$ and star, which difference is found by observation in the same manner as the difference at $T$ was before found, when the equator was fixed. Also, as $Qg=Cc=Aa$, and the angle $qdr=Cd=Ae$, we have $Qr=ae\cos Ae$. Now if we put $M$ for the mean right ascension of the star at the beginning of the year, and $S$ for the sum of all the corrections due at the time of the first observation, and $s$ for the sum due at the second; then, from what we have already explained in the last article, $M+S=Am$, $M+S-s=mv$, hence, if we take the former from the latter, supposing $s$ to be greater than $S$, we have $s-S=av-Am=ae+ev-Aw-wm$ ($m$ lying beyond $w$), but $\omega v-Aw$; hence, $s-S=ae-wm$, consequently $\omega w=ae-t-S$. Now $qv$, or $\omega w$, is known, hence we know $\omega m$, and as $Qr$ is known, $Qm$ will be known; and as we also know $\omega Lm$, we get the value of $\Omega L^{\prime}$, with which we proceed, as before, to get the star's right ascension. The great advantage of this method, is, that it does not depend upon any determination of the latitude of the place, declination of the sun or accuracy in the divisions of the instrument. If the latitude be known, we may find the declination from the meridian altitude, it being, from Art. 87., equal to the difference between the meridian altitude and the complement of latitude, and then one observation at the second equinox will be sufficient, because the daily variation of the declination and right ascension may be taken from the Nautical Almanac. Having thus determined the right ascension of one star, the right ascension of all the heavenly bodies may from thence be found (118). + +If the right ascension of a star, which is not in these tables, should be required, the corrections must be computed by the Rules which we shall give in their proper places. If the right ascension of the star be first computed without considering these corrections, it will be sufficiently accurate to compute. The corrections from, and then they may be applied. + +* In all these cases, if you draw the figure and put the star in its proper place, and put $\omega w$ and $\omega m$, in their proper situations, which may be done by observing whether $\omega w$ or $\omega m$ be the greater, you will immediately see what quantities are to be added together, and what subtracted. This figure is drawn for the Example. + +8 + +LATITUDE AND LONGITUDE OF THE HEAVENLY BODIES. + +Ex. Let it be required to find the right ascension of Pollux on March 24, in the year 1768, from Dr. Maskelyne's observations. +On March 24, Pollux passed the meridian at 7h. 31'. 38"; and on the 25th, at 7h. 31'. 37'. 66", on the same day the sun passed at Oh. 16. 35". 5 ; hence, the apparent difference of the $A R$'s of the sun and Pollux on the 24th, allowing for the error of the clock (192), was 7h. 15'. 2'. 46 = 108°. 45. 36'. 0 = Lm. Now on March 24, +Appar. zen. dist. $\odot$ L. L. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 49°. 58'. 58".7 +Semidiam. -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- +Appar. zen. dist. $\odot$ cen. --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- +Parallax ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- +Refr. cor. for Bar. and Ther. + 1. 10,4 +True zen. dist. $\odot$ cen. ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- +True meridian altitude ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ -------- + +To find when the sun had the same meridian altitude, or declination, just before it came to the next equinox, let us take Sept. 18, on which we find, +Appar. zen. dist. $\odot$ L. L. ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- -------- + +Semidiam. ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- +Appar. zen. dist. $\odot$ cen. --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- --------- +Parallax -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- -------------- +Refr. cor. for Bar. and Ther. + 1. 5,8 +True zen. dist. $\odot$ cen. ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- +True meridian altitude ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- ------------- + +As this altitude is less than that on March 24, the instant of time when the sun had the same declination as on the 24th must be before the 18th; therefore as the sun on the 18th had gotten beyond that point where its declination was the same as at $P$, we must, from the difference of the right ascensions of the sun and star observed on that day, subtract the increase of the sun's right ascension between the 18th and that point of time when the declination was the same as at $P$, in order to get the difference of the apparent right ascensions at the time when the sun's declination was the same as at $P$. We may also observe, that the difference of any two true meridian altitudes is the same as the difference of the declinations at the same times. Now as the sun's altitude was not observed on the 17th, we will take the change of declination for that day from the Nautical Almanac, which is $25^{\circ}.20'$; also, the increase of the sun's $AR$ for that day was $9^{\circ}.96'$ in time, or $54'$ in space. The difference of the + +Vol. I. + +A page from a book with text discussing celestial navigation. + +54 + +TO DETERMINE THE RIGHT ASCENSION, DECLINATION, + +true meridian altitudes, or the difference of declinations on March 24, and Sept. 18, was $9'.39',5$; hence, $25'.20':9'.39',5::54':22'.21',4$, the increase of the sun's right ascension from the time before the 18th at which the declination was the same as on March 24, to the 18th. On Sept. 18, Pollux passed the meridian at $76'.50'.39',9$, and on the 19th at $76'.50'.40'$. On the 18th the sun passed at $114'.44'.53',35$; therefore the apparent difference of the $AR's$ of the sun and Pollux on that day, allowing for the error of the clock (122), was $44'.14'.19':5=63'.35'.22',5$, from which subtract $22'.21',4$, and we have $63'.11'.1',1=eqn$. Now to get the correction in Table VIII. we must have the place of the moon's ascending node, which, from the Lunar Tables, is found to be $9'.17'.45'.28'$ on March 24, and $9'.8'.19'.54'$ on Sept. 18. Hence, + +March 24, Correction from Table VII. $\{\}$ red. to space + $19',3$ + +VIII.$\{\}$ + $19',8$ + +Sept. 18, Correction from Table VII. $\{\}$ red. to space + $38',7=S$ + +VIII.$\{\}$ + $31',8$ + ++ $20$ + ++ $51',8=S$ + +Hence, $S- S=12',6$. + +Prec. of Equin. from March 24, $\{\}$ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 24',9 + +to Sept. 18, Table XV. + +Variation of the equat. of equinoxes, Table XVI. + $0$,7 + +True Precession in the interval --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- 25,$6=Ae$ + +Cos. $28^o$, $28^o$ ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ ------ -------- -------- -------- -------- -------- 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+ +LATITUDE AND LONGITUDE OF THE HEAVENLY BODIES. +35 + +121. But the method made use of by Dr. Maskelyne in settling the right ascensions of the stars, though founded upon the same principle as this of Mr. Flamstead, is different in its process, and procured him the advantage of a greater number of observations, both of the sun and stars, in the same time, and consequently enabled him to fix the right ascension of the stars with greater accuracy in a shorter time. He took a Aquile for his fundamental star, and assumed its right ascension as settled by Dr. Bradley, reducing it to the time of his observations by the mean precession, and afterwards making the following correction. By comparing a great many observed transits of such stars as he thought proper to select, with that of Aquile, in various parts of the year, and applying the proper equations, he obtained their mean right ascensions relative to that of a Aquile assumed, or affected with the same error; and comparing the transits of the sun near the equinoxes with those of the above mentioned stars observed on the same day, he obtained the sun's right ascension relative to that of a Aquile assumed. From the observed zenith distances of the sun on the same days, corrected for refraction, parallax and the error of the line of collimation, with the apparent obliquity of the ecliptic at the time, he deduced the sun's right ascensions; and then by comparing the sun's right ascensions deduced from the observed transits with those deduced from his observed zenith distances at equal or nearly equal declinations of the same kind near both equinoxes, he deduced the error of the assumed right ascension of a Aquile, which came out 3',8 additive. He observed further, that in the interval of 12 years, which passed between the settling of Dr. Bradley's Catalogue about 1755 and his own about 1767, the precession in right ascension was diminished by 2',16 by the action of the planets. Therefore if this had been allowed in assuming the right ascension of a Aquile from Dr. Bradley's determination, the correction of the right ascension of a Aquile would have come out 5',96 additive, or at the rate of $\frac{1}{2}$' a year, which agrees very well with the annual proper motion of a Aquile deduced from other observations. Dr. Maskelyne has also given the following method. + +Assume the mean $AR$ of the star at the beginning of the year, and thence, by applying the equations, compute its apparent $AR$ on two days of the year when the sun has nearly equal declinations on the same side of the equator, from two declinations observed; and then by the observed difference of the transits of the sun and star, compute the two apparent $AR'$s of the sun and star; call this by *the star*. Correct the observed zenith distances of the sun by the correction of the line of collimation (if necessary), refraction and parallax, and you will obtain its apparent zenith distances, affected only by an error in latitude of place, making an error in its declination. To the mean obliquity of the ecliptic at the beginning of the year, apply the proportional part of the annual diminution, the correction for the day of year, + +A page from a book discussing astronomical methods. + +36 + +TO DETERMINE THE RIGHT ASCENSION, DECLINATION, + +and the equation depending on the place of the moon's node, and you will have the apparent obliquity, with which and the two declinations of the sun before found, compute the two $AR$'s by the sun; call this by the declination. Subtract the sun's $AR$ by the star from his $AR$ by the declination near the ver- nal equinox, and call the difference $a$ put down with its proper sign. Do the same for the autumnal equinox, and call the difference $b$. Then $\frac{1}{2}(a+b)$ is the correction of the mean $AR$ of the star at the beginning of the year. This correction being applied to the two $AR$'s of the sun by the star, will give the apparent $AR$'s of the sun at those times. For let $A = opp. AR$ of $\odot$ at $P$ by the star, $A$ that at $T$, $B = \odot s AR$ at $P$ by the declination, $B$ that at $T$; $y=$ correction to be applied to correct the computed declination of the sun, and let $1:n:\odot s$ error $(y)$ in decl.; corresponding error in $AR = ny$. Now an increase of declination, increases the $AR$ in the first quadrant, and decreases it in the second; hence, an increase $(ny)$ of $AR$ in the first quadrant, makes it $B+ny$, and in the second, $B-ny$; these we may consider as the true $AR'$s of the $\odot$ from the declination; also, the true $AR'$s from the star (putting $x=$the correction of the mean $AR$ of the star at the beginning of the year) are $A+x$ and $A+x$, hence, $A+x-B+ny$, $A+x=B-ny$, and $x=\frac{1}{2}(B-A-B+A)$; but $a=B-A$, $x=B-A$; therefore $x=\frac{1}{2}(B-A-B+A)$ Further, $y=\frac{1}{2n}(B-B+A-A)$ the error in declination. But 1 : n : $\frac{1}{2}PL:\frac{1}{2}AP$: now sin. AP = tan. PL × cot. A, therefore sin. AP = $\overline{AP}$ × cos. AP = $\overline{AP}$ × cos. PL × cot. A, and 1 : n : cos. AP : sec. PL × cot. A; hence, y = $\frac{1}{2}(B-B+A-A)$ × cos. AR × cot. dec. × tan. obl. ecl. + +By making a great number of observations of this kind, and taking th mean, the $AR$ of a star may be very accurately determined. Dr. MARKLEY observed, that this method is more simple than that of Dr. BRADLEY, or LACATTE, though on the same principle, first introduced by FLAMSTEAD. + +122. The practical method of finding the right ascension of a body from a fixed star, by a clock adjusted to sidereal time, is thus. Let it begin its motion from 0h. G. O' at the instant first point of Aries is meridian; then, when any star comes to the meridian, the clock would show the apparent right ascension of the star, the right ascension being estimated at the rate of 15° an hour, provided the clock was subject to no cause it would then show at any time how far the first point of Aries is meridian. But as the clock is necessarily liable to err, we must take any time to ascertain what its error is, that is, what is that difference between its apparent right ascension shown by th clock and its right ascension shown by its apparent right ascension with the right ascension shown by its equator which is at that time on the meridian. To do this, a star whose apparent right ascension is known, passes through its apparent right ascension with the right ascension shown by its equator which is at that time on + +LATITUDE AND LONGITUDE OF THE HEAVENLY BODIES. +37 + +the difference will show the error of the clock. For instance, let the apparent right ascension of *Aldebaran* be 4h. 25° 50' at the time when its transit over the meridian is observed by the clock, and suppose the time shown by the clock to be 4h. 25° 52', then there is an error of 2' in the clock, it giving the right ascension of the star 2' more than it ought. If the clock be compared with several stars* and the mean error taken, we shall have, more accurately, the error at the mean time of all the observations. These observations being repeated every day, we shall get the rate of the clock's going, that is, how fast it gains or loses. The error of the clock, and the rate of its going, being thus ascertained, if the time of the transit of any body be observed, and the error of the clock at the time be applied, we shall have the right ascension of the body. This is the method by which the right ascension of the sun, moon and planets are regularly found in Observatories. + +Ex. On April 27, 1774, the following observations were made at Greenwich: + +*Serpentis* passed the meridian at 15h. 31° 28' 76", the moon's second limb passed at 15h. 39° 7' 76', and *Antares* at 16h. 18° 55' 02 sidereal time ; to find the moon's right ascension. + +First, to find the error of the clock by the transit of the stars. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Mean AR. of a serpentis at begin. of 1790 by Tab. VI.15° 33' 55" 84-46° 94
Mean AR. at begin. of 177415. 33. 8, 90
Cor. for aber. and prec. to April 27, by Tab. VII.+ 2, 12
Cor. for nutation by Tab. VIII.- - - -- - - -- - - -
App. AR. by the tables15. 33. 10, 79
App. AR. by the clock15. 31. 28, 76
Error of the clock by a serpentis too slow1. 42, 03
Mean AR. of Antares at begin. of 1790 by Tab. VI.16. 16. 33, 24
Precession in 16 years by Tab. VI.- - - -- - - -- - - -
Mean AR. at begin. of 177416. 15. 34, 79
Cor. for aber. and prec. to April 27, by Tab. VII.+ +, 38
Cor. for nutation by Tab. VIII.- - - -- - - -- - - -
*The stars used for this purpose at the Observatory at Greenwich are those in Tab. VI whose AR's Dr. Maskelyne settled to a very great degree of accuracy.* As many of these as conveniently can, are observed every day, in order to ascertain the going of the clock, and for no other purpose. +
+ +*The stars used for this purpose at the Observatory at Greenwich are those in Tab. VI whose AR's Dr. Maskelyne settled to a very great degree of accuracy.* As many of these as conveniently can, are observed every day, in order to ascertain the going of the clock, and for no other purpose. + + + + + + + + + + + + + + + + + + + + + + + + +
38TO DETERMINE THE RIGHT ASCENSION, DECLINATION,
App. $AR.$ by the tables- - - -- - - -16. 15. 97, 08
App. $AR.$ by the clock- - - -- - - -16. 13. 55, 02
Error of the clock by antares too slow1. 42, 06
+ +The mean of these two errors gives $1'42''$,045 for the error at the middle between the times of the transits of the two stars, or at $15h$. $52'.41'',89$. Now from knowing the error of the clock at this time, and the rate of its going, we must find the error at the time the moon passed, which may, in this case, be considered the same, the times being nearly equal. Hence, + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Moon passed the meridian by the clock- - - -- - - -19'. 59''.775
Error of the clock, too slow- - - -- - - -1' 42,045
True $AR.$ of the moon's 2d limb16. O. 49,795
Do. in degrees- - - -- - - -8'. O'. 12'. 26'.9
Moon's semid. in $AR.$ (109)- 17. 19,5
True $AR.$ of the moon's center7. 29. 55. 13,4
+ +The error of the clock is generally determined by a greater number of stars, when they can be observed; and the mean error from day to day gives the rate of its going, from which we may find the error at any other time. For instance, on August 8, 1769, I found, from taking the mean of the errors of four stars, that the mean error of the clock was $2'32$, too fast, at $16h$. $21'.18'$, being the mean of all the times when the stars were observed; and on the 9th the error was $3'09$, too fast, at $15h$. $52'.58'$, the mean of all the times. Also Jupiter passed the meridian on the 9th at $14h$. $49'.10'.4$. Now the interval between the $8d$. $16h$. $91'.18'$ and $9d$. $18h$. $52'.58'$ is $21h$. $31'.40'$ in which time the clock lost $0'23$; also the interval between $13h$. $52'.58'$ and $14h$. $49'.10'.4$ is $56'.12'.4$; hence, $21h$. $31'.40'$ : $56'.12'.4$:::$0'23$ : $0'009$, which is what the clock lost in the second interval; therefore when Jupiter passed the meridian, the clock was $2'09-0'009=2'08$ too fast, which subtracted from $14h$. $49'.10'.4$ gives $14h$. $49'.32$, the apparent right ascension of Jupiter. To the apparent AR apply the aberration in A.R. and you get the true AR. + +$183$. The right ascension of the heavenly bodies being thus ascertained, the next thing to be explained is, the method of finding their declinations. Take the apparent altitude of the body, when it passes the meridian, by an astronomical quadrant, as explained in my Treatise on Practical Astronomy; correct it for parallax and refraction, and for the error of the line of collima- + +LATITUDE AND LONGITUDE OF THE HEAVENLY BODIES. +39 + +tion of the instrument, if necessary, and you get the true meridian altitude, +the difference between which and the altitude of the equator (87) (which is +equal to the complement of the latitude, previously determined) is the decli- +nation required. + +Ex. On April 27, 1774, the zenith distance of the moon's lower limb when +it passed the meridian at Greenwich was $68^{\circ}.19',97',3$; its parallax in al- +itude was $56'.19'',2$, allowing for the spheroidal figure of the earth; the +barometer stood at $29,58$, and the thermometer at $49$; to find the decli- +nation. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Observed zenith distance of L. L.$68'.19'.97',3$
Refr. cor. for bar. and ther. Tab. XL XII.$+ 2.23$
Parallax$68'.22.00,3$
$- 56'.19',2$
True zenith distance of L. L.$67'.25'.41,1$
Semidiameter$- 16'.35$
True zenith distance of the center$67'.9'.6,1$
Latitude$51'.28'.40$
Declination south$15'.40'.26,1$
+ +The horizontal parallax and semidiameter may be taken from the Nautical +Almanac; and the parallax in altitude may be found, as will be explained when +we come to treat of the Parallax, and then the correction is to be applied to +the semidiameter, from Table XIII. + +124. To find the latitude and longitude from the right ascension and +declination, or the converse, we have the following admirable Rules, given by +Dr. Maskelyne. + +Given the Right Ascension and Declination of an Heavenly Body, and the Obliviquity +of the Ecliptic, to find its Latitude and Longitude. + +1. The *sine of $AR$. cotang. decl.* - $10$, = cotang. of arc $A$, which call +north or south, according as the declination is north or south. 2. Call the obli- +quity of the ecliptic south in the 6 first signs of $AR$, and north in the 6 last. +Let the sum of arc $A$ and obl. eclip. according to their titles, = arc $B$ with its +proper title*. $3$. The arith. comp. of cos. arc $A + \cos$. arc $B + \tan$. $AR$— +* By sine, tang., etc. is meant log. sine, log. tang., etc. +† If one be north and the other south, the proper title is that belonging to the greater of the two, +and in this case, arc $B$ is their difference, one being considered as negative to the ether. + +A table showing observed zenith distance of L.L., refr. cor., parallax, true zenith distance of L.L., semidiameter, true zenith distance of center, latitude, declination south. + +40 + +TO DETERMINE THE RIGHT ASCENSION, DECLINATION, + +$10.=\tan.$ of the longitude, of the same kind as $AR$, unless arc B be more than 90°, in which case, the quantity found of the same kind as $AR$ must be subtracted from 12 signs or 360°. 4. The sine of longitude + tan. arc B = 10, =tan. of the required latitude, of the same title as arc B. N. B. If the longitude come out near 0°, or near 180°, for the sine of long. in the last operation, substitute $\tan$. long. + cos. long. - 10, *; or the last operation will be, $\tan$. long. + cos. long. + tan. arc B - 20, =tan. lat. The $\tan$. long. is already given. + +Given the Latitude and Longitude of an Heavenly Body, and the obliquity of the Ecliptic, to find its Right Ascension and Declination. + +1. Sine long. + cot. lat. - 10, =cot. arc A, which call north or south, according as the lat. is north or south. +2. Call the obliquity of the ecliptic north in the first semicircle of longitude, and south in the second. Let the sum of arc A and obl. eclipt. according to their titles, =arc B with its proper title. +3. The arith. comp. of cos. arc A + cos. arc B + tan., long. - 10, =tan. of right ascension, of the same kind as the longitude, unless arc B be more than 90°, in which case, the last quantity found of the same kind as the longitude, must be subtracted from 12 signs or 360°. 4. The sine of $AR$ + tan., arc B - 10, =tan. of the required declination, of the same title as arc B. +N. B. If $AR$ come out near 0°, or near 180°, for the sine $AR$, in the last operation, substitute $\tan$. $AR$ + cos. $AR$ - 10; or the last operation will be $\tan$. $AR$ + cos. $AR$ + tan. arc B - 20, =tan. declination. +The $\tan$. $AR$, is already given. + +DEMONSTRATION. Let s be the body, $\tau C$ the ecliptic, $\tau Q$ the equator, +$\sigma r$, sin perpendicular to $\tau C$, $\tau Q$. Then rad.: sin $\tau n$: cot m : cot $\sigma r n$, hence, log.sin $\tau n$: log.cos $\sigma r n$ - 10, =log.cos $\sigma r n$ arc A. +Hence, $\sigma r n$ = $\tau Q \cdot \tau C = \sigma r n$ arc B. +Also + +Fig. +cos $\sigma r n$: rad.: tan $\tau n$: tan $\sigma r n$ Trig Art. 219. + +21. +rad.: cos $\sigma r n$: tan $\sigma r n$: tan $\tau n$ + +$\therefore$ cos $\sigma r n$: cos $\sigma r n$: tan $\tau n$ = cos $\sigma r n$: cos $\sigma r n$; hence, ar.co. + +log.cos $\sigma r n$: log.cos $\sigma r n$: tan $\tau n$ - 10, =log.tan.$\tau n$ the longitude. +And (Trig Art. 310), rad.: sin $\tau n$: tan.$\tau n$: tan.$\sigma r n$; hence, log.sin.$\tau n$ + log.tan.$\tau n$ - 10, =log.tan.$\sigma r n$ arc latitude. +And in whatever position we take $s$, these conclusions will give the rule as stated above. +If we consider $\tau C$ as the equator and $\tau Q$ the ecliptic, the demonstration will do for the second rule. + +* For the reason of this correction in extreme cases, see Dr. Maskelyne's excellent Introduction to Taylor's Logarithms. + +2 + +LATITUDE AND LONGITUDE OF THE HEAVENLY BODIES. + +41 + +Ex. Given the true $A.R.$ of the moon's center 75. 29°, 55.13',4, and its declination 15°, 40. 26',1 south, as determined in the two last Examples; to find its latitude and longitude*. + +By Dr. MASELYNE's observations, the mean obliquity of the ecliptic at the beginning of the year 1784, was 23°, 28'. 0',2, and as its gradual diminution is at the rate of $\frac{1}{2}$ second in a year, the mean obliquity at the beginning of 1774 was 29°, 28'. 5',2, which corrected by Tab. IX. X. gives 23°, 27'. 55',8 for the obliquity at the time of observation. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Sine of right asc.-7°. 29°. 55. 13',4-9.9971817
Cotan. of decl.-15 . 40. 26,1-10.5519183
Cotan. arc A south-17 . 57. 57,8-10.4891000
Obliq. ecl. north-23 . 27. 53,8-
Arc B north-5 . 29. 38,0 cos.9.9979064,
Arith. comp. of log. cos. arc A---0.0217102
Tang. of right asc.---10.2871744
Tang. of longitude-8°, 1°, 2°, 7',4-10.2568810
Sine of longitude---9.9419678
Tang. of arc B---8.9835328
Tang. of latitude north4°, 48'. 54',1--8.9255006
In like manner, the right ascensions and declinations of the fixed stars being found from observation, their latitudes and longitudes may be computed, and thus a catalogue of all the fixed stars may be made for any time. But as both the equator and ecliptic are subject to a change in their positions, the right ascension, declination, latitude and longitude of all the fixed stars will vary Hence, if their annual variations be computed, as will be afterwards explained, their right ascensions, &c., may be found at any other time. + +125. If the body be the sun at $t$, whose right ascension and declination are given, to find its longitude; then sin $w'n$ : rad.: sin $s'n$ : sin $w'x$, that is sin $obl$. ecl.: rad.: sin $decl.$ : sin longitude. Or, cos $w'n$ : rad.: tan $w'n$ : tan $w'x$, that is, cos $obl$. ecl.: rad.: tan right asc.: tan longitude. The sun, being always in the ecliptic, has no latitude. + +To find the angle of Position. + +126. Let $p$ be the pole of the ecliptic $w'P$, $P$ the pole of the equator $w'C_1$ + +A diagram showing a line segment from P (pole of the equator) to C (pole of the ecliptic), with angles labeled for calculation. + +* In making trigonometrical calculations, it will save time when the same arcs occur, to take out all their logarithms at once, to avoid the trouble of turning them again. The Computer therefore before he begins his operation, should put it down in its proper order, leaving it to be filled up by the logarithms; he will then see what arcs are repeated, and he may, at one opening of the table, take out all their logarithms and put them down in their proper places. +FIG. +22. + +**Note** + +A small note or symbol indicating "Note" or "Note." This is typically used to indicate additional information or clarification about a previous point in a text. + +42 + +**TO DETERMINE THE RIGHT ASCENSION, DECLINATION, &C.** + +*In a star, draw the great circles $pPLC$, $pSD$, $PSBA$, and (53) $PSp$ is the angle of position. Now the angle $PpS$, or (12) $DL$, is the complement of longitude $\varphi D$; the angle $pPS$ is the supplement of $APC$, or of $AC$ (12), which is the complement of the right ascension $\varphi A$ of the star; $pP$ is the obliquity of the ecliptic; $PS$ is the complement of declination, and $pS$ the complement of the latitude of the star. Hence, if the longitude and declination of a star be given, we have, $\sin PS : \sin PpS = \sin PP : \sin PSp$, that is, $\cos \text{star's dec.} : \cos \text{its long.} :: \sin \text{obl. ecl.} : \sin \text{angle of Position}$. If the latitude and declination of the star be given, we know $pS$ and $PS$ their complements, and $PP$; hence, $\sin pS \times \sin PS : rad. :: \sin \frac{1}{2} \times SP + Sp + Pp \times sin. \frac{1}{2} \times SP + Sp - Pp : cos. \frac{1}{2} PSp^{\circ}$. Or of the right ascension, declination, latitude and longitude of the star, any two being known, we shall know three parts of the triangle $PpS$, and consequently the angle $PSp$ may be found. If $S$ be the sun, $pS = 90^{\circ}$, and the triangle may be solved by the circular parts.* + +CHAP. IV. + +ON THE EQUATION OF TIME. + +Art. 197. HAVING explained, in the last Chapter, the practical methods of determining the place of any body in the heavens, we come next to the consideration of another circumstance not less important, that is, the irregularity of time as measured by the sun. The best measure of time which we have, is a clock regulated by the vibration of a pendulum. But however accurately a clock may be made, it must be subject to go irregularly, partly from the imperfection of the workmanship, and partly from the expansion and contraction of the materials by heat and cold, by which the length of the pendulum, and consequently the time of vibration, will vary. As no clock therefore can be depended upon for keeping time accurately, it is necessary that we should be able to ascertain at any time, how much it is too fast or too slow, and at what rate it gains or loses. For this purpose it must be compared with some motion which is uniform, or of which, if it be not uniform, you can ascertain the variation. The motions of the heavenly bodies have therefore been considered as most proper for this purpose. Now the earth revolving uniformly about its axis, the apparent diurnal motion of the fixed stars about the axis must be uniform. If a clock therefore be adjusted to go 24 hours from the passage of any fixed star over the meridian till it returns to it again, its rate of going may be at any time determined by comparing it with any fixed star, and observing whether the interval continues to be 24 hours; if not, the difference shows how much it gains or loses in that time. A clock adjusted to go 24 hours in this interval is said to be adjusted to sidereal time. But if we compare a clock with the sun, and adjust it to go 24 hours from the time the sun leaves the meridian on any day, till it returns to it the next day, which is a true solar day, the clock will not, even if it go uniformly, continue to agree with the sun, that is, it will not show 12 when the sun comes to the meridian. + +128. For let $P$ be the pole of the earth, $v$ its equator, and let the earth revolve about its axis in the order of the letters $veg$, $vDLE$ the celestial equator, and $vCL$ the ecliptic, in which the sun moves according to that direction. Let $a$, $m$, be the sun when on the meridian of any place on two successive days, and draw $Pve$, $Prmh$, secondaries to the equator, and let the spectator be at $a$ on the meridian $Pv$, with the sun at $a$ on his meridian. Then when the earth has made one revolution about its axis $Pav$ is come again into the same position; but the sun having moved forward to $m$, the earth has still + +Fig. 23. + +44 + +ON THE EQUATION OF TIME. + +to describe the angle $vPr$ in order to bring the meridian $Pov$ into the position $Pr$, so that the sun may be again in the spectator's meridian. Now the angle $vPr$ is measured by the arc $ch$, which is the increase of the sun's right ascension in a true solar day; hence, the length of a true solar day is equal to the time of the earth's rotation about its axis - the time of its describing an angle equal to the increase of the sun's right ascension in a true solar day. Now if the sun, moved uniformly in the equator $DLE$, this increase $ch$ would be always the same in the same time, and therefore these solar days would be always equal; but the sun moves in the ecliptic $CJL$, and therefore if its motion were uniform, equal arcs upon the ecliptic would not give equal arcs $ch$ upon the equator*. But the motion of the sun is not uniform, and therefore am, described in any given time, is subject to a variation, and which also must necessarily make $ch$ variable. Hence, the increase $ch$ of the sun's right ascension in a day varies from two causes, that is, from the obliquity of the ecliptic to the equator, and from the unequal motion of the sun in the ecliptic. The length therefore of a true solar day, is subject to a continual variation; consequently a clock adjusted to go 24 hours for any one true solar day, would not continue to show 12 when the sun comes to the meridian; because the intervals by the clock would continue equal (the clock being supposed neither to gain nor lose), whiles the intervals of the sun's passage over the meridian would vary. + +139. As the sun moves through $360^\circ$ of right ascension in 365$\frac{1}{4}$ days very nearly, therefore 365$\frac{1}{4}$ days : 1 day :: $360^\circ$: 39'. 8".2 the increase of right ascension in one day, if the increase were uniform, or it would be the increase in a mean solar day, that is, if the solar days were all equal. If therefore a clock be adjusted to go 24 hours in a mean solar day, it cannot continue to coincide with the sun, that is, to show 12 when the sun is on the meridian; but the sun will pass the meridian, sometimes before 12 and sometimes after. This difference is called the Equation of Time. A clock thus adjusted is said to be adjusted to mean solar time. The time shown by the clock is called true or mean time, and that shown by the sun is called apparent time. What we call apparent time the French call true. + +* For draw $\mu$ parallel to $ch$, and suppose $\mu$ to be indefinitely small; then by plain trigonometry, +$$\tan \mu = \sin \alpha \cdot \cos \beta$$ +or +$$\tan \alpha = \cos \beta$$ + +As $\mu$ approaches zero, $\tan \mu$ approaches zero; hence, $\tan \alpha$ approaches $\cos \beta$. Therefore +$$\tan \alpha = \cos \beta$$ + +† As the earth describes an angle of $360^\circ$. 39'. 8".2 about its axis in a mean solar day of 24 hours, and an angle of $360^\circ$ in a sidereal day, therefore 39'. 8".2 : 360° = 944 : 834. 567. 4".0989 +the length of a sidereal day in mean solar time, or the time from passage of a fixed star over the meridian till it returns to it again. + +ON THE EQUATION OF TIME + +130. A clock adjusted to go 24 hours in a mean solar day, would coincide with an imaginary star moving uniformly in the equator with the sun's mean motion $59.8^{\circ}2$ in right ascension, if the star were to set off from any given meridian when the clock is 12; that is, the clock would always show 12 when the star came to the meridian, because the interval of the passages of this star over the meridian would be a mean solar day. This star therefore, if we reckon its motion from the meridian in time at the rate of 1 hour for $15^{\circ}$, would always coincide with the clock; that is, when the clock shows 1 hour, the star's motion would be 1 hour; when the clock shows 2 hours, the star's motion would be 2 hours; and so on. Hence, this star may be substituted instead of the clock; therefore when the sun passes the given meridian, the difference between its right ascension and that of the star, converted into time, is the difference between the time when the sun is on the meridian and 12 o'clock, or the equation of time; because the given meridian passes through the star at 12 o'clock, and its motion in respect to that star is at the rate of $15^{\circ}$ in an hour (139). + +131. Now to compute this equation of time, let $APLS$ be the ecliptic, $ALv$ fig. 24. the equator, $A$ the first point of aries, $P$ the sun's apogee, $S$ any place of the sun, draw $Su$ perpendicular to the equator, and take $Am = AP$. When the sun sets out at $P$, let the imaginary star set out at $n$ with the sun's mean motion in right ascension, or longitude, or at the rate of $59.8^{\circ}2$ in a day, and when $n$ passes the meridian let the clock be adjusted to 12, as described in the last Article: These are the corresponding positions of the clock and sun, as assumed by Astronomers. Take $mn = Ps$, and when the star comes to $m$, the place of the sun, if it moved uniformly with its mean motion, would be at $s$, but at that time let $S$ be the place of the sun. Now let the sun $S$, and consequently $v$, be on the meridian; then as $m$ is the place of the imaginary star at that instant, $mv$ is the equation of time. The sun's mean place is at $s$, and as $Am = AP$, and $mn = Ps$.:. $Am = ApS$, consequently $mv = Av - Am = Av - ApS$. Let $a$ be the mean equinox, and draw $ax$ perpendicular to $AL$, then $Am = Ax + zm = Ax \cos ALx + zm = \frac{1}{2}Ax + zm$; hence, $mv = Av - am = \frac{1}{2}Av$. But $Av$ is the sun's true right ascension, $zm$ is the mean right ascension, or mean longitude, and $\frac{1}{2}Ax (As)$ is the equation of the equinoxes in right ascension; hence, the equation of time is equal to the difference of the sun's true right ascension, and its mean longitude corrected by the equation of the equinoxes in right ascension. When Am is less than Av, mean time precedes apparent; and when greater, apparent time precedes mean; for as the earth turns about its axis in the direction Avs, or in the order of right ascension, that body whose right ascension is least must come to the meridian first. That is, when the sun's true right ascension is greater than its mean longitude corrected as above, we must add the equation of time to apparent; to get the mean time; and when it is less, we must subtract. To convert mean. + +A diagram showing an imaginary star moving uniformly in an equatorial plane with respect to a given meridian. + +36 + +ON THE EQUATION OF TIME. + +time into apparent, we must subtract in the former case and add in the latter. +This Rule for computing the equation of time was first given by Dr. MARKE- +LYNE in the Phil Trans. 1784. + +132. As a meridian of the earth, when it leaves $m$, returns to it again in 24 hours, it may be considered, when it leaves that point, as approaching a point at that time 360° from it, and at which it arrives in 24 hours. Hence, the relative velocity with which a meridian accedes to or recedes from $m$ is at the rate of 15° in an hour. Therefore when the meridian passes through $v$, the arc $vm$ reduced into time at the rate of 15° in an hour, gives the equation of time at that instant. Hence, the equation of time is computed for the instant of apparent noon. Now the time of apparent noon in mean solar time, for which we compute, can only be known by knowing the equation of time. To compute therefore the equation on any day, you must assume the equation the same as on that day four years before, from which it will differ but very little, and it will give the time of apparent noon, sufficiently accurate for the purpose of computing the equation. If you do not know the equation four years before, compute the equation for noon mean time, and that will give apparent noon accurately enough. + +Ex. To find the equation of time on July 1, 1792, for the meridian of Greenwich, by Mayer's Tables. + +The equation on July 1, 1788, was, by the Nautical Almanac, s. 28", to be added to apparent noon, to give the corresponding mean time; hence, for July 1, 1792, at oh. s. 28" compute the true longitude*. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Mean Long.Long. o's Apog.N°.1.N°.2.N°.3.N°.4.
Epoch for 17929°. 16' .50'0° .79'9° .23'. 46'241227123478
Mean Mot. July 1,33. 29. 28. 16', 23316845631227
57;
25'1;1;
Mean Longitude3. 10. 13. 25', 43.9. 24. 19.404689485505
Equat. of Center-1. 37', 13.10. 13. 25', 4
Equat. I.+;4;
e II.-;4;7;49.; 6;4;Mean Anomaly.
e III.+;3;63;
e IV.-;0;6;
True LongitudeS. 70. 11. 51', 15'
+ +* The reason of this operation will appear, when we come to the construction and use of the Polar Tables. + +ON THE EQUATION OF TIME. +49 + +what we call, mean time nearly; corresponding therefore to this time, take out the correction from Table XVIII, which is 45.5, and add it to the given mean solar time, and we get 44. 25. 38',14 correctly for what we call mean time nearly; add this to 64. 54. 35',36, the sun's mean longitude at noon, and it gives 11h. 20'. 16" the sidereal time required. + +136. Whenever the time is computed from the sun's attitude, that time must be apparent time, because we compute it from the time when the sun comes to the meridian, which is noon, or 12 o'clock, apparent time. Hence also, the time shown by a dial is apparent time, and will differ from the time shown by a well regulated watch or clock, by the equation of time. A clock or watch may therefore be regulated by a good dial, by applying the equation, as before directed, to the apparent time shown by the dial, and it will give the mean time, or that which the clock or watch ought to show. + +137. Mr. Wollaston has proposed to regulate a watch or clock by a dial constructed to show mean noon, or 12 o'clock by a watch or clock. A ray of light through a small hole being let into a dark chamber upon the floor, draw a meridian upon the floor corresponding to the hole, on which therefore the sun's rays will always fall when the sun comes to the meridian. On each side of this line, for every day of the year, make a point where the image of the sun is at 12 o'clock mean time, by a clock or watch regulated for that purpose; through all these points draw a curve, and then you may regulate your clock or watch by setting it to 12 when the image of the sun falls on that curve. To prevent any mistake, put the months against the different parts of the curve on which the ray falls in them. Or the same may be done on any horizontal plane, by erecting a piece of brass, and making a small hole for the sun to shine through. The curve may also be laid down by calculation, as Mr. Wolaston has shown; and if it be drawn with great care, it will be sufficiently accurate for regulating all common clocks; and it has this advantage over that of correcting them by a common sun dial, that as the months are put to the curve, you cannot easily make a mistake; whereas, in applying the equation of time to a dial, a person, ignorant of these matters, is very apt to apply it wrong. + +138. The Equation of Time was known to, and made use of by PROLEMY. Tycho employed only one part, that which arises from the unequal motion of the sun in the ecliptic; but KEPLER made use of both parts. He further suspected, that there was a third cause of the inequality of solar days, arising from the unequal motion of the earth about its axis. But the Equation of Time, as now computed, was not generally adopted till 1672, when FLAMSTEAD published a Dissertation upon it, at the end of the works of HORROX. + +VOL. I. +H + +CHAP. V. + +ON THE LENGTH OF THE YEAR, THE PRECESSION OF THE EQUINOXES FROM OBSERVATION, AND THE OBLIQUITY OF THE ECLIPTIC. + +Art. 139. FROM comparing the sun's right ascension every day with the fixed stars lying to the east and west, the sun is found constantly to recede from those on the west, and approach to those on the east; and the interval of time from its leaving any fixed star till it returns to it again is called a sidereal year, being the time in which the sun completes its revolution amongst the fixed stars, or in the ecliptic. But the sun, after it leaves either of the equinoctial points, returns to it again in a less time than it returns to the same fixed star, and this interval is called a solar or tropical year, because the time from its leaving one equinox till it returns to it, is the same as from one tropic till it comes to the same again. This is the year on which the return of the seasons depends. + +On the Sidereal Year. + +140. To find the length of a sidereal year. On any day take the difference between the sun's right ascension when it passes the meridian and that of a fixed star; and when the sun returns to the same part of the heavens the next year, compare its right ascension with the same star for two days, one when their difference of right ascensions is less and the other when greater than the difference before observed; and let $D$ be the increase of the sun's right ascension in this interval of one day; then take the difference ($d$) between the differences of the sun's and star's right ascensions on the first of these two days and on the day when the observation was made the year before; and let $t$ be equal to the exact time between the transit of the sun over the meridian on the two days; then $D : d : t$: the time from the passage of the sun over the meridian on the first day to the instant when it had the same difference of right ascension compared with the star which it had the year before; and the interval between these two times gives the length of a sidereal year. The best time for these observations is about March 25, June 20, September 17, December 20, the sun's motion in right ascension being then uniform. Instead of observing the difference of the right ascensions, you may observe that of their longitudes. If instead of repeating the second observations the year after, there be an interval of several years, and you divide the observed interval of time when the difference of their right ascensions was found to be equal, by the number of years, you will have the length of a sidereal year more exact. Or the length may be found thus. + +ON THE LENGTH OF THE YEAR, &C. +51 + +141. Take the time ($t$) of a star's transit over the meridian by a clock adjusted to mean solar time; then the year after, take the time again on two days, one ($m$) when it passes the meridian before, and the other ($n$) after the time $t$; then $m - n : m - t : 258$. 6. 4°: the time from $m$ till the difference between the star's and sun's right ascension was the same as at the first observation; and the interval of these two times is the length of a sidereal year. Cassini's *Elem. d'Astron.* pag. 202. + +Ex. On April 1, 1669, at 0h. 3'. 47" mean solar time, M. Picard observed the difference between the sun's longitude and that of Procyon to be 3°. 6". 59". 36", which is the most ancient observation of this kind whose accuracy can be depended upon; see *Hist. Celeste, par M. le Monnier*, pag. 37. And on April 2, 1745, M. de la Caille found, by taking their difference of longitudes on the 2d and 3d, that at 11h. 10". 45" mean solar time, the difference of their longitudes was the same as at the first observation. Now as the sun's revolution was known to be nearly 365 days, it is manifest that it had made in this interval 76 complete revolutions in respect to the same fixed star in the space of 76 years to 11h. 6". 59". But in these 76 years, there were 58 of 365 days, and 18 bisextiles of 366 days; that interval therefore contains 27793d. 11h. 6". 58", which being divided by 76, the quotient is 365d. 6h. 8'. 47" the length of a sidereal year. + +Ex. M. Cassini observed the transit of Sirius over the meridian on May 21, 1717, to be at 2h. 38'. 58"; on May 21, 1718, it passed at 24. 40', and on the 2nd at 2h. 36'; to find the length of the sidereal year. + +In this case $t = \text{2h}. \text{38'.} \text{58'}$, $m = \text{2h}. \text{40'}$, $n = \text{2h}. \text{36'}$, hence, $4' : \text{1}' :: \text{23h}. \text{56'.} \text{4'} : \text{6h}. \text{10'.} \text{39'}$, which added to $\text{2h}. \text{40'}$ the time it passed on May 21, 1718, gives $\text{sh}. \text{30'.} \text{39'}$ for the time on that day when the difference between the sun's and star's right ascensions was the same as on May 21, 1717. Hence this interval is $\text{365d}. \text{6h}. \text{10'.} \text{39'}$ for the length of a sidereal year. The mean of these two, gives the length $\text{365d}. \text{6h}. \text{9'.} \text{53'}$. But the length of a sidereal year has generally been determined from the length of a tropical year, found as we shall now proceed to explain. + +On the Tropical Year. + +142. Observe the meridian altitude ($a$) of the sun on the day nearest to the equinox; then the next year take its meridian altitude on two following days, one, when its altitude ($m$) is less than $a$, and the next when its altitude ($n$) is greater than $a$, and $n - m$ is the increase of the sun's declination in $24$ hours; hence, $n - m : a - m : \text{24 hours}$: the interval from the first of the two days till the sun has the same declination as at the observation the year before, because + +A diagram showing a star's transit over a meridian. + +54 + +**ON THE LENGTH OF THE YEAR, THE PRECESSION OF THE EQUINOXES** + +to get the length of a tropical year between the mean equinoxes in order to get the length of a mean tropical year. But in taking a long interval of time, the difference, whether we take the true or mean equinox, will be insensible. Another correction might also be added, when we compare the modern observations with the ancient ones, on account of the precession of the equinoxes being greater now than it was then. From the modern observations the length of a mean solar year appears to be $2^{\circ}.6$ less than that which is deduced from comparing the same observations with those of Hipparchus. + +144. As the sun's declination at the equinoxes changes about $2^{\circ}$ in 24 hours, an error of $10'$ in the altitude of the sun will cause an error of 10 minutes in the determination of the time of the equinox, and consequently the same error in the length of the year, if it were determined by 2 observations at the interval only of 1 year; but if the interval were 60 years, the error would be only 10 seconds. As the accuracy therefore is very much increased by taking a long interval, let us compare the most ancient observations with the modern ones. + +HIPPARCHUS, in the year 145 before J. C. found the time of the equinox to be on March 24, at 11h. $55'$ in the morning at Alexandria. In the year 1735, at the Royal Observatory at Paris the time of the equinox was found to be on March 20, at 14h. $20'$. Now the difference of the meridians between Paris and Alexandria is, in time, $1h. 51'.46'$, which, as Alexandria lies to the east of Paris, being added to $14h. 20'$. $40'$ gives $16h. 12'.26'$ the time at Alexandria. Reduce this time to the Julian year, by subtracting 11 days by which the Gregorian is before the Julian, and we have the time of the equinox by this style, on March 10, at $4h. 12'.26'$ in the morning. Between these two observations there was an interval of 1880 Julian years, except $14d.7h.$ $42'.34'$. In these years there were 470 bissexiles and the rest common Julian years of 365 days. Therefore if we divide $14d.7h.$ $42'.34'$ by 1880 it gives $10'.58'.10''$, showing how much the apparent solar year is less than 365 days 6 hours; hence, the length of the apparent solar year is $365d.5h.49'.1''.50''$, to which add $6'.90''$, being what the apparent is less than the mean solar year, found as before, and we get $365d.5h.49'.8''.30''$ the length of the mean solar year from these observations. The mean of 10 results from different observations made by HIPPARCHUS, compared with the modern ones, gives the length of the mean solar year $365d.5h.48'.49''$. + +FIG. +23. + +145. The length of the year may also be found by finding the time when the sun comes to the tropic. For let $ADL$ be the equator; $ASL$ be the ecliptic; A arises; find the time (119) when the sun has the same declination $m_n$ on each side of the tropic $S$, and at the same times find also the differences of its eight ascension and that of a fixed star $\alpha$, s sum or difference of which $\omega_r$, $\tau_z$, according to the position of $z$, measures the motion $\omega_w$ of the sun in right ascension; th half of which is $\omega D$ ($SD$, $\epsilon z$ being perpendicular to $AL$); + +FROM OBSERVATION, AND THE OBLIQUITY OF THE ECLIPTIC. +58 + +the spring 1716 is greater by 1° than in 1672, and this answers to 5°. 16' in time ; in this interval of time therefore (44 years), there have been 44 mean revolutions + 5°. 16', and consequently 44 apparent solar years are greater by 5°. 16' than 44 mean ; divide this by 44, the number of years in the interval, and it gives 7°. 11' for the length of the apparent above the mean solar year. +Now the length of the apparent solar year was determined to be 365d. 5h. 49'. +0'. 53'; hence, from these observations, the length of the mean solar year is +365d. 5h. 48'. 58'. 42". + +149. The length of a tropical year may also be found by observing the exact time of the equinoxes. To do this we must previously know the latitude of the place, from which we shall know the altitude of the point of the equator on the meridian, it being equal (87) to the complement of latitude. Take the meridian altitude of the sun's center on two days, one when it is less than the complement of latitude and the other when greater; then the sun must have passed the equator in the intermediate time. Take the difference ($D$) between these altitudes and it gives the increase of the sun's declination in 24 hours; take also the difference ($d$) between the altitude on the first day and the complement of latitude, and then say, $D : d = 24$ hours : to the time from noon on the first day till the sun came to the equator. Repeat this when the sun returns to the same equinox, and the interval of the times gives the length of a tropical year. +If an interval of several years be taken, and you divide by the number, it will give the time more accurately. If we take a difference of two days, the third term must be $48h$. The same may be done by one observation, if we know the rate at which the sun changes its declination in 24 hours, which at the equinox in spring time is found, by the mean of a great number of observations, to be $25^{\circ}$. $40'$, and in the autumn to be $25^{\circ}$. $28'$. +Cassini's *Elem. d'Astr.* pag. 207. + +Ex. On March 20, 1673, the sun's meridian altitude at the Royal Observatory at Paris was observed to be $41^{\circ}$. $25^{\circ}$. $36'$, from which subtract $41^{\circ}$. $9$. $60'$ -the meridian altitude of the equator, and there remains $16^{\circ}$ for the sun's declination; hence, $25^{\circ}$. $40'$ : $16^{\circ}$ :: $24$ hours : $16^{\circ}$. $19'$, the sun's distance in time from the equinox, which, as the sun was past the equinox, subtracted from the 90th gives the 19th day $7h$. $41'$ for the time of the equinox. And in 1731 the time of the equinox was found, in the same manner, to be on Mar. 20, at $14h$. $45'$. In this interval of 39 years there were 13 bisestiles, and consequently the whole number of days in the 39 years was $91548$, and therefore the whole interval between the two equinoxes was $91549$. $7h$. $4'$, which divided by 39 gives the length of the apparent solar year $365d$. $5h$. $48'$. $53'$; from this subtract $7'$, the variation of the equation of orbit in this interval of observations, and we have the mean length of the solar year $365d$. $5h$. $48'$. $46'$. +The interval has here been taken between true equinoxes, whereas we want + +58 + +ON THE LENGTH OF THE YEAR, THE PRECESSION OF THE EQUINOXES, &C. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
CASSINI (the Father) in 1656-23°. 29. 2'
CASSINI (the Son) in 1672-23°. 28. 54
FLAMSTEAD in 1690-23°. 28. 48
De la CAILLE in 1750-23°. 28. 19
Dr. BRADLEY in 1750-23°. 28. 18
MAYER in 1750-23°. 28. 18
Dr. MASKELYNE in 1769-23°. 28. 8,5
M. de la LANDE in 1786-23°. 28. 0
+ +The observations of ALBATEGIUS, an Arabian, are here corrected for re-fraction. Those of WALTHEURUS, M. de la CAILLE computed. The obliquity by Tycho is here put down as correctly computed from his observations. +Also the obliquity, as determined by FLAMSTEAD, is corrected for the nutation of the earth's axis. These corrections M. de la LANDE applied. + +152. It is manifest from the above observations, that the obliquity of the ecliptic keeps diminishing; and the irregularity which here appears in the diminution we may ascribe to the inaccuracy of the ancient observations, as we know that they are subject to greater errors than the irregularity of this variation. +If we compare the first and last observations, they give a diminution of $70^\circ$ in $100$ years. If we compare the last with that of Tycho, it gives $45^\circ$. The last compared with that of FLAMSTEAD gives $50^\circ$. If we compare that of Dr. MASKELYNE with Dr. BRADLEY's and Mayer's it gives $50^\circ$. The comparison of Dr. Maskelyne's determination, with that of M. de la LANDE, which he took as the mean of several results, gives $50^\circ$. We may therefore state the secular diminution of the obliquity of the ecliptic, at this time, to be $50^\circ$, as determined from the most accurate observations. This result agrees very well with that deduced from theory, as will be shown when we come to treat of the physical cause of this diminution. +It must however be observed, that some eminent Astronomers use $50^\circ$, $25^\circ$. + +**CHAP. VI.** + +ON PARALLAX. + +Art. 153. **THE** center of the earth describes that circle in the Heavens which is called the ecliptic; but as the same object would appear in different positions in respect to this circle, when seen from the center and surface, Astronomers always reduce their observations to what they would have been, if they had been made at the center of the earth, in consequence of which, the places of the heavenly bodies are computed as seen from the ecliptic, and it becomes a fixed point for that purpose, on whatever part of the earth's surface the observations are made. + +154. Let $C$ be the center of the earth, $A$ the place of the spectator on its surface, $S$ any object, $ZH$ the sphere of the fixed stars, to which the places of all the bodies in our system are referred; $Z$ the zenith, $H$ the horizon; draw $CSm$, $ASn$, and $m$ is the place seen from the center, and $n$ from the surface. Now the plane $SAC$ passing through the center of the earth must be perpendicular to its surface, and consequently it will pass through the zenith $Z$, and the points $m,n$ lying in the same plane, the arc of parallax $mn$ must lie in a circle perpendicular to the horizon, and hence the azimuth is not affected, if the earth be a sphere. Now the parallax $mn$ is measured by the angle $mSn$ or $ASC$, and by trig. $CS : CA :: \sin SAc = \sin ASC$. The parallax = $\frac{CA \times \sin SAZ}{CS}$. As $CA$ is constant, supposing the earth to be a sphere, the sine of the parallax varies as the sine of the apparent zenith distance directly, and the distance of the body from the center of the earth inversely. Hence, a body in the zenith has no parallax, and at $s$ in the horizon it is greatest. If the object be at an indefinitely great distance, it has no parallax; hence the apparent places of the fixed stars are not altered by it. As $n$ is the apparent place, and $m$ is called the true place, the parallax depresses an object in a vertical circle. For the same body at different altitudes, the parallax varies as the sine ($s$) of the apparent zenith distance; therefore if $p=$the horizontal parallax, and radius be unity, the sine of the parallax = $\sin p$. To ascertain therefore the parallax at all altitudes, we must first find it at some given altitude. + +155. **First** method, for the sun. Aristarchus proposed to find the sun's parallax, by observing its elongation from the moon at the instant it is dichotomized, at which time the angle at the moon is a right angle; therefore we should know the angle which the distance of the moon subtends at the sun; + +Fig. 26. + +3 + +60 + +ON PARALLAX. + +which diminished in the ratio of the moon's distance from the earth's center to the radius of the earth, would give the sun's horizontal parallax. But a very small error in the time when the moon is dichotomized, (and it is impossible to be very accurate in this) will make so very great an error in the sun's parallax, that nothing can be depended upon from it. VENDELINUS determined the angle of elongation when the moon was dichotomized to be $89^{\circ}. 45'$, from which the sun's parallax was found to be $15'$. But P. Riccioli found it to be $28'$ or $30'$ from like observations. + +![Fig. 27.] + +156. Second method. HIPPARCHUS proposed to find the sun's parallax from a lunar eclipse, by the following method. Let $S$ be the sun, $E$ the earth, $Ev$ the length of its shadow, $mE$ the path of the moon in a central eclipse. Observe the length of this eclipse, and then, from knowing the periodic time of the moon, the angle $mE\tau$, and consequently $nE\tau$, will be known. Now the horizontal parallax $ErB$ of the moon being known, we have the angle $Evr = ErB - nE\tau$; hence we know $EA.B = AES - Evr = AES - ErB + nE\tau$; that is, the sun's horizontal parallax = the apparent semidiameter of the sun - the horizontal parallax of the moon = the semidiameter of the earth's shadow where the moon passes through. The objection to this method is, the great difficulty of determining the angle $nE\tau$ with sufficient accuracy; for any error in that angle will make the same error in the sun's parallax, the other quantities remaining the same. By this method PROXIMUS made the sun's horizontal parallax $9'. 50'$. Tycho made it $5'$. + +157. Third method, for the moon. Take the meridian altitudes of the moon, when it is at its greatest north and south latitudes, and correct them for refraction; then the difference of the altitudes, thus corrected, would be equal to the sum of the two latitudes of the moon, if there were no parallax; consequently the difference between the sum of two latitudes and the difference of the altitudes will be the difference between the parallaxes at the two altitudes. Now to find from thence the parallax itself; let $S$, $s$ be the sines of the greatest and least apparent zenith distances, $P$, $p$ the sines of the corresponding parallaxes; then as, when the distance is given, the parallax varies ($154$) as the sine of the zenith distance, $S : s : P : p$, hence $S - s : S - s : P - p = \frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{P - p}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{2\pi}{S - s} : \frac{2\pi}{S - s}$ = $\frac{2\pi}{S} : \frac{\cancelto{\infty}}{\cancelto{\infty}} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +\\ +}\\ + +This supposes that the moon is at the same distance in both cases; but as this will not necessarily happen, we must correct one of the observations in order to reduce it to what it would have been, had the distance been the same. If the observations be made in those places where the moon passes through the zenith in one of the observations, the difference between the sum of two latitudes and the zenith distance at that altitude will be reduced to zero. + +ON PARALLAX. +61 + +158. Fourth method. Let a body $P$ be observed from two places $A$, $B$ in the same meridian, then the whole angle $APB$ is the effect of parallax between the two places. The parallax $(154) APC = hor.\cdot par.\cdot \sin{PAL}$, taking $APC$ for sin. $APC$, and the parallax $BPC = hor.\cdot par.\cdot \sin{PBM}$; hence $hor.\cdot par.\cdot \sin{PAL} + sin.\ PBM = APB$, i.e. $hor.\cdot par.\cdot = APB$ divided by the sum of these two sines. If the two places be not in the same meridian it does not signify, provided we know how much the altitude varies from the change of declination of the body in the interval of the passages over the meridians. + +Ex. On Oct. 5, 1751, M. de la Caille, at the Cape of Good Hope, observed Mars to be 1° 25' 8" below the parallel of $\lambda$ in aquarius, and at 23° distance from the zenith. On the same day at Stockholm, Mars was observed to be 1° 57' 7" below the parallel of $\lambda$ and at 66° 14' zenith distance. Hence the angle $APB = 31° 1'$, and the sines of the zenith distances being 0.4886 and 0.9887, the horizontal parallax was 25' 8". Hence, if the ratio of the distance of the earth from Mars to its distance from the sun be found, we shall have the sun's horizontal parallax. Now from comparing the altitudes on the northern limit of Mars with stars nearly in the same parallel observed on the same days at the Cape and at Greenwich, Bologna, Paris, Stockholm, Upsal, Hernosand, the mean of the whole gave 10' 2" for the horizontal parallax of the sun ; and rejecting those results which differed the most from the rest, the mean was 9° 84'. From the mean of another set of observations, the result was 9° 57'. From the mean of several observations on Venus made in like manner, the parallax came out 10' 3". The mean of all these last gave 9° for for the horizontal parallax of the sun. Excepted, from an observation on Mars, concluded that such a parallax could not be more than 10°. Maraldi found it same. From the observations of Pound, and Dr. Bradley, Dr. Halley found it never greater than 12° nor less than 9°. Cassini, from his observations on Mars, found it to be between 11° and 15°. But the most accurate method of determining the sun's parallax is from the transit of Venus over its disc, as will be explained when we treat on that subject. + +159. If the earth be a spheroidal ball $E$ be the equator; draw $GAH$, $HBc$ perpendicular to this surface, and compute the angles $CAv$ or $LAG$, and $CBv$ or $MBH$ by the Rule which we shall give, when we treat of the figure of the earth ; subtract these from the observed zenith distances $PAG$, $PBH$, and we have the angles $PAL$, $PBM$. Now $CP : CA :: \sin{CAP} : \sin{PAL} = \frac{CA \times \sin{PAL}}{CP}$; also, $CP : CB :: \sin{CBP} : \sin{PBC} = \frac{CB \times \sin{PBM}}{CP}$; and as the parallax is very small, the sum of the two sines will be very nearly the sine of the sum, therefore the sine of $APB =$ + +A diagram showing a spherical Earth with lines representing latitude and longitude. +2 + +62 + +ON PARALLAX. + +$$CA \times \sin{PAL} + CB \times \sin{PBM}$$ +; hence, $$CP = \frac{CA \times \sin{PAL} + CB \times \sin{PBM}}{\sin{APB}}$$ + +A diagram showing the parallax calculation between two stars. + +**ra.** +160. Fifth method. Let $EQ$ be the equator, $P$ its pole, $Z$ the zenith, $v$ the true place of the body and $r$ the apparent place as depressed by parallax in the vertical circle $ZK$, and draw the secondaries $Pm$, $Prb$; then $ab$ is the parallax in right ascension, and $rs$ in declination. Now $vr : ev : 1 (rad.) : sin. vs or ZsP, and vs : ab : cos. va : 1 (18); hence, vr : ab : cos. va : sin. ZsP, : + +$$ab = \frac{vr \times sin. ZsP}{cos. va};$$ but vr = hor. par. x sin. vZ (164), and (Trig. Art. 291.) sin. vZ : sin.ZP :: sin. ZPe : sin. ZsP = sin. ZP x sin. ZPe, therefore by substitution, + +$$ab = \frac{hor. par. \times sin. ZP \times sin. ZPe}{cos. va}.$$ Hence for the same star, where the hor. par. is given, the parallax in right ascension varies as the sine of the hour angle. + +Also the hor. par. = $$\frac{ab \times cos. va}{sin.ZP \times sin.ZPe}.$$ For the eastern hemisphere, the apparent place $b$ lies on the equator to the east of $a$ the true place, and therefore the right ascension is diminished by parallax; but in the western hemisphere, $b$ lies to the west of $a$, and therefore the right ascension is increased. + +Hence, if the right ascension be taken before and after the meridian, the whole change of parallax in right ascension between the two observations is the sum $(s)$ of the two parts before and after the meridian; and the hor.par. = $$\frac{s \times cos.va}{sin.ZP \times S},$$ where $S =$ sum of sines of the two hour angles. + +161. To apply this Rule, observe the right ascension of the planet when it passes the meridian, compared with that of a fixed star, at which time there is no parallax in right ascension; about 6 hours after, take the difference of their right ascensions again, and observe how much the difference $(d)$ between the apparent right ascensions of the planet and fixed star has changed in that time. Next observe the right ascension of the planet for 3 or 4 days when it passes the meridian, in order to get its true motion in right ascension; then if its motion in right ascension in the above interval of time between taking of the right ascensions of the fixed star and planet on and off the meridian be equal to $d$, the planet has no parallax in right ascension; but if it be not equal to $d$, the difference is the parallax in right ascension; and hence, by the last Article, the horizontal parallax will be known. Or one observation may be made as long before the planet comes to the meridian, by which a greater difference will be obtained. + +Ex. On August 15, 1719, Mars was very near a star of the 5th magnitude in the eastern shoulder of aquarius, and at 9h. 18' in the evening, Mars fol- + +ON. PARALLAX. + +68 + +lowed the star in 10°, 17', and on the 16th at 4h. 21' in the morning it followed it in 10°, 1", therefore in that interval, the apparent right ascension of Mars had increased 16" in time. But according to observations made in the meridian for several days after, it appeared, that Mars approached the star only 14" in that time, from its proper motion, therefore 2' in time, or 30" in motion, is the effect of parallax in the interval of the observations. Now the declination of Mars was 15°, the co-latitude 41°, 10', and the two hour angles 49°, 15 and 56°, 39'; therefore the $hor.\ por. = \frac{30^\circ \times \cos(15^\circ)}{\sin(41^\circ) \cdot 10^\circ \times \sin(49^\circ) \cdot 15 + \sin(56^\circ) \cdot 39'} = 27\frac{1}{2}''$. +But at that time, the distance of the earth from Mars was to its distance from the sun as 37 to 100, and therefore the sun's horizontal parallax comes out 10°, 17'. + +168. When Dr. Maskelyne was at St. Helena and Barbadoes, he made several observations of this kind on the moon, in order to determine her horizontal parallax; and he further observes, "that if the like observations were repeated in different parts of the earth, it would probably afford the best means, yet proposed, for ascertaining the true figure of the earth, as they would determine the ratio of the diameters of the parallels of latitude to each other, the horary parallaxes being in proportion thereto: For though the earth affords but a small base at the moon, yet, by repeating these trials, and comparing the results, we may hope to attain that degree of exactness, which we could never expect from fewer observations." + +169. But besides the effect of parallax in right ascension and declination, it is manifest that the latitude and longitude of the moon and planets must also be affected by it; and as the determination of this, in respect to the moon, is in many cases, particularly in solar eclipses, of great importance, we shall proceed to show how to compute it, supposing that we have given the latitude of the place, the time, and consequently the sun's right ascension, the moon's true latitude and longitude, with her horizontal parallax. + +164. Let HZR be the meridian, $\tau EQ$ the equator, $p$ its pole; $\tau E$ the ecliptic, $P$ its pole; $\tau$ first point of aries, $HQR$ the horizon, $Z$ the zenith, $ZL$ a secondary to the horizon passing through the true place $r$ and apparent place $t$ of the moon; draw $Pt$, $Pr$, which produce to $c$, drawing the small circle $\alpha$ parallel to $\omega$; then let $rn$ be perpendicular to $Pt$, and draw the small circle $ra$ parallel to $\omega$; then $rs$, or $ta$, is the parallax in latitude, and $\omega c$ the parallax in longitude. Draw the great circles $Fv$, $PZA$, $Pde$, and $ZH$ perpendicular to $pc$; then as $\tau P=90^\circ$, and $\wp p=90^\circ$, $\tau v$ must ($\alpha$) be the pole of $Pde$, and therefore $\delta r=90^\circ$; consequently $d$ is one of the solstitial points $\omega$ or $\sigma$; also draw $Zz$ perpendicular to $Pr$, and join $Zv$, $\wp v$. Now $\tau E$, or the angle $\wp E$, or $Zp\wp$, is the right ascension of the mid-heaven, which is known (108); + +A page from a book with text discussing celestial mechanics. +68 +SO. + +64 +ON PARALLAX. + +$PZ = AB$ (because $AZ$ is the complement of both) the altitude of the highest point $A$ of the ecliptic above the horizon, called the nonagesimal degree, and $\varphi A$, or the angle $\varphi PA$ is its longitude. Now in the right angled triangle $ZpW$, we have $Zp$ the co-latitude of the place, and the angle $ZpW$, the difference between the right ascension of the mid-heaven $\varphi E$ and $w d$; hence, (Trig. Art. 212.) cot. $p Z$: rad.: cos. $p$: tan. $p W$; therefore $PW = p W^{\prime} p P$, where the upper sign is to be taken when the right ascension of the mid-heaven is less than $180^{\circ}$, and the under, when greater. Also, in the triangles $WZp$, $IWZp$, (Trig. Art. 231.) sin $H_{P}$ : sin $WP$ : cot. $WPZ$ : cot. $WPZ$, or tan. $AP_{P}$; and as we know $\gamma_0$, or $\varphi P_0$, the true longitude of the moon, we know $AP_0$, or $ZP_0$. Also (Trig. Art. 219.) cos. $WPZ$, or sin. $APZ$: rad.: tan. $HP$: tan. $ZP$. Hence, in the triangle $ZrP$, we know $ZP$, $Pr$ and the angle $P$, from which the angle $ZrP$ or $\tau r s$, and $Zr$ may be found ; for in the right angled triangle $ZPr$, we know $ZP$ and the angle $P$, to find $Pr$; therefore we know $\tau r s$; and hence (Trig. Art. 231.) we may find the angle $Zr z$, with which, and $\tau r s$, we may find $Zr$ the true zenith distance; to which, as if it were the apparent zenith distance, find the parallax (154) and add to it, and you will get very nearly the apparent zenith distance, corresponding to which, find the parallax $\tau r t$; then in the right angled triangle $\tau r s t$, which may be considered as plane, we know $\tau r t$ and the angle $\tau r s$ to find $\tau r s$ the parallax in latitude ; find also $\alpha s$, which multiplied (108) by the secant of $\nu v$, the apparent latitude, gives the arc $\omega v$, the parallax in longitude. + +Ex. On January 1, 1771, at 9h. apparent time, in lat. 58°N. the moon's true longitude was 9s. 18° 27' 35", and latitude 4° 5'. 50'S., and its horizontal parallax 61'. 9'; to find its parallax in latitude and longitude. + +The sun's right ascension was 282° 22' 2" by the Tables, and its distance from the meridian 153"; also (106) the right ascension $\varphi E$ of the mid-heaven was 57° 22' 2"; hence, the whole operation for the solution of the triangles may stand thus. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
$\begin{array}{l}{ZpW = 92° 23' 57}\\ {Zp = 97° 0' 0'}\end{array}$- - -cos.9.9258864
Tri.$\begin{array}{l}{pW = 92° 23' 57}\\ {Pp = 23° 28' 0}\end{array}$- - -tan.9.9871144
$\begin{array}{l}{PW = 55° 51' 57}\\ {PW = 55° 51' 57}\\ {ZpW = 92° 23' 57}\\ {APW = 67° 29' 8}\end{array}$- - - -A.C.sin.0.3709855
Tri.$\begin{array}{l}{PW = 92° 23' 57}\\ {PW = 55° 51' 57}\\ {ZpW = 92° 23' 57}\\ {APW = 67° 29' 8}\end{array}$- - - -sin.9.9178865
Tri.$\begin{array}{l}{PW = 92° 23' 57}\\ {PW = 55° 51' 57}\\ {ZpW = 92° 23' 57}\\ {APW = 67° 29' 8}\end{array}$- - - -cot.10.1935941
Tri.$\begin{array}{l}{PW = 92° 23' 57}\\ {PW = 55° 51' 57}\\ {ZpW = 92° 23' 57}\\ {APW = 67° 29' 8}\end{array}$- - - -tan.10.3824661
+ +A table showing calculations for finding a moon's parallax in latitude and longitude based on its true longitude, latitude, and right ascension. + +ON PARALLAX. +65 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
oPA = 108°. 27. 35"
oPA = 40. 58. 27 + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ + TPZ +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +


































































































<
A.C.sin.
0.1600743
sin.
9.8863144
tan.
9.9887076
cot.
9.9812846
cot.
9.8754949
log.
3.5645477
log.
3.4674839
log.
3.9075042
log.
3.5645477
+
oPAtan. oPAsin.9.9655709
oPA = 67. 29. 8- - -tan. +1020.1688210
oPA = 55. 51. 57- - -tan.10.2032510
ZP = 57. 56. 36- - -tan.10.2032555
ZP = 40. 58. 27- - -cos.9.8779500
Pz = 50. 19. 33- - -tan.10.0812055
Pz = 94. 5. 30- - -
+ + + +
The value of $tv$ is $ro-rr+rs$, according as the moon has N. or S. latitude.The Figure is drawn for north latitude, but the Example is for south latitude.This is the direct method of solving the problem from the triangles ; but the vol.
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat.ev par., in long.=206°=34°, 27'.
+ + + +
sec.log.=3.8144416
+ + + +
true lat. + +66 + +**ON PARALLAX.** + +operation may be rendered easier by the following Rule (the most convenient of any yet given) discovered by Dr. Maskelyne, but communicated without the demonstration. The investigation here given, is by the Rev. Dr. Bainklev, Professor of Astronomy at Dublin. + +Let the height $H$ of the nonage-simal degree, or $PZ$, and the angle $ZPr$ ($a$), the moon's true distance from the nonagesimal, be computed as before. Put $P=$the parallax $\omega v$ in longitude, $Q=$the parallax $\alpha t$ in latitude, depressing the moon southwards, $L=$the true latitude, $l=$the apparent latitude, $h=$the horizontal parallax. Now + +$$P : \tan : \tan . \sin Pr$$ + +$$\therefore P : h : \sin . ntr \times \sin Zt : \sin Pr,$$ + +$$rt : h : \sin Zt : rad.$$ + +radius being unity; + +hence, $P = \frac{h \times sin . ntr \times sin Zt}{\sin Pr} = (\text{as } \sin . ntr \times \sin Zt = \sin ZPt \times sin PZ)$ + +$$h \times sin . PZ \times sin ZPt = h \times sin . H \times sin . l + nP$$ + +$$\therefore \cos L = \frac{\sin Pr}{\sin ZPt},$$ + +the parallax in Longitude. + +Also, $ta : br : cos . rta : rad. :: sin . rta : tan . rtan$ + +$$\therefore ta : h : \sin . rta : sin . Zt : tan . rta \times rad. :: sin . PZ \times sin . ZPt :$$ + +$$\sin . Pt \times cot.ZP - cos.Pt \times cos.ZPP$$ + +substituting their values; hence, $m = h \times sin . PZ = sin . Pt \times cot.ZP - h \times sin . PZ$ $\times$ cos. $Pt \times cos.ZPt = h \times cos.H \times cos.l - h \times sin.H \times sin.l \times cos.n + P.$ + +Now as the angle $cPr$ is very small, we have $an = 2\tan Pr = (from the first proportion above)$ + +$$P' \times sin.Pr = 2\tan Pr = 2P' \times sin.Pr \times cos.Pr = 2P' \times P' \times sin.Pr \times$$ + +$$cos.Pr = (as., from above, P' \times sin.Pr = h \times sin.PZ'z.sin.ZPt') \\ h'P'xh'x.sin.$$ $$H\times sin.n+P\times sin.L_1$ or $sin.L_1$ nearly; hence, $q=ta=im-an=h\times cos.H\times$$ $$cos.l-h\times sin.H\times cos.l\times cos.n+P+h\times sin.H\times 1+P\times sin.n+P\times sin.L.$$ But as $P'$ is very small, we may call $\frac{1}{2}P'$ the sine of $\frac{1}{2}P$, and its cosine we may put $=tan.\frac{1}{2}$; hence, for cos.$n+P$ we may substitute cos.$n+\frac{1}{2}P\times cos.\frac{1}{2}P$, and for $\frac{1}{2}P\times sin.n+P$ we may put $\sin.n+\frac{1}{2}P\times\frac{1}{2}P$; hence, $q=h\times cos.$$H\times cos.l-h\times sin.H\times cos.n+\frac{1}{2}P\times cos.\frac{1}{2}P +\sin.n+P\times\sin.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}P=(be-$$ put by plane Trig. Art. 105. cos.$n+P=\frac{1}{2}\cos.(P+\sin.n+P)\cos.\frac{1}{2}F(=be-)$ + +ON PARALLAX. + +we must first suppose $P = \frac{h \times \sin H \times \sin n}{\cos L}$, which will give a near value of $P$; then put that value into the numerator, and you will get a very accurate value of $P$. Also, in the expression for $Q$, we have the apparent latitude, which cannot be known without knowing $Q$; hence we must first get a near value of $Q$ and apply it to the true latitude to get the apparent nearly; to do this, we may omit the second part as being small, on account of $\sin I$ being small for the moon, and suppose $Q = h \times \cos H \times \cos I - h \times \cos L$ nearly; or when the latitude is very small, as is the case of the moon in solar eclipses, we may suppose $Q = h \times \cos H$, from which we shall get the apparent latitude with sufficient accuracy. + +In the application of this Rule, regard must be had to the signs of the quantities; if $n + P$ be greater than 90° its cosine becomes negative, in which case $Q$ will be the sum of the quantities, unless the apparent latitude $l$ is south, in which case, its cosine will be negative, which makes the first term negative. In general, $Q$ will be the sum of the two parts, when $n + P$ and the moon's apparent distance from $P$ are, one greater and the other less than 90°; otherwise $Q$ will be the difference. The parallax in longitude increases the longitude, if the body be to the east of the nonagesimal degree, and decreases it, if it be to the west. This Rule is more correct than the other, because in that we took the small circle $t$, instead of a great circle from $t$, as the perpendicular from $t$ upon $Pr$ produced. This error, for the moon, may sometimes amount to about 2'. It may be corrected by applying an found above. + +To apply this Rule to the last case, we have $H = 57°. 58°. 96°$, $n = 40°$. 28°. 27°. $L = 4°$. 5°. 30° south, $h = 61°. 9' = 3669'$; hence, + + + + + + + + + + +
Log. $h$----3.5645477
Sin. $H$----9.9281518
Cos. $L$----A.C. 0.0011084
Sin. $n$----3.4938079
Log. 204° = 34°. 4' = P nearly----9.8167176
3.3105255
Therefore $n + P = 41°$. 32'. 31'; hence,----3.4938079
9.8216397
3.3154316
3.5645477
9.7248563
3.2894440
Sin. $\overline{n+P}$----3.5645477
9.8216397
3.3154316
3.5645477
9.7248563
3.2894440
Log. 2067° = 34'. 27' par. in Longitude----3.5645477
9.7248563
3.2894440
Cos. $H$----3.2894440
3.5645477
9.7248563
3.2894440
+ +Logarithms of Parallaxes. +67 + + + + + + + + + + + + + + +
4°. 5°. 30°
4°. 41°. 3 app. lat. nearly.
Log. $h$ - - - - - - - - 3.5645477
Cos. $H$ - - - - - - - - 9.7249868
Cos. $l = 4°$. 41°. 3' nearly - - - 9.9985470 first part of Q.
Log. $1941^{\circ} = 32^{\circ}. 21'$ - - - 3.2879910
Log. $h$ - - - - - - - 3.5645477
Sin. $H$ - - - - - - - 9.9281518
Sin. $l$ - - - - - - - 8.9120258 second part of Q.
Log. $n + \frac{1}{2}P$ - - - 9.8759999
Log. $191^{\circ} = 3^{\circ}. 11'$ - - - 2.2806552
+ +32. 21 + +35. 32 par. in Latitude. + +The sum of the two parts is here taken, because $P\ell$ is greater than $90^\circ$, and $n + \frac{1}{2}P$ less than $90^\circ$. + +165. Hitherto we have considered the effect of parallax, upon supposition that the earth is a sphere; but as the earth is a spheroid, having the polar diameter shorter than the equatorial, it will be necessary to show how the computations are to be made for this case. The following method is given by CLAIRAUT. + +166. Let $EPQp$ be the earth, $EQ$ the equatorial and $Pp$ the polar diameters, $O$ the place of the spectator, $HCR$ the rational horizon, to which draw ZONK perpendicular to $LK$. Now to compare the apparent places seen from $O$ and $C$, let us compare the places seen from $O$ and $K$, and from $K$ and $C$. Put $\alpha_1$-the horizontal parallax to the radius OC, or ON which is very nearly equal to it, on account of the smallness of the angle CON. Let CO = 1, and CN (the sine of CON to that radius) = $\alpha_1$, i.e., the angle KON = h; hence, as h=the angle under which ON (which we may consider as equal to unity) appears when seen directly at the moon, we have h x $\alpha_1$=the angle under which NK would appear; therefore h x (1 + $\alpha_1$)=the horizontal parallax of OK; considering therefore K as the center of a sphere and KO the radius, compute the parallax as before. Now as the planes of all the circles of declination pass through Pp, in estimating the parallax either from K or O, the parallax in right ascension must be the same, because K and O lie in the plane of the same circle of declination; the only + +ON PARALLAX. + +difference therefore between the effect of parallax at $K$ and $O$ must be in declination. Now at $K$, the angular distance of the moon from the pole $P$ is $LKP$, and the angular distance from $C$ is $LCP$; the difference of these two angles therefore, or $CLK$, is the difference between the parallax in declination at $K$ and at $C$, and this angle $CLK$ is always to be added to the polar distance seen from $K$ to get the polar distance from $C$. Now $CLK = h \times CF$; but the angle $FCK (= LCE)$ is the moon's declination, therefore $CF = CK \times \cos \text{dec.}$ also, $CK = CN = \frac{a}{\cos \text{lat.}}$ hence, $CLK = h \times a \times \cos \text{dec.}$ This therefore is the equation of declination for the spheroid, to be applied to find the parallax in declination seen from $C$, after having calculated the effect of parallax in declination for a sphere whose center is $K$ and radius $KO$. There is no equation for the parallax in right ascension. To find how this equation in declination will affect the latitude, let $P$ be the pole of the equator, $p$ the pole of the ecliptic, $L$ the place of the moon seen from $K$, and $\theta$ seen from $C$; then $\delta L$ is the equation in declination; draw $La$ perpendicular to $pb$, and $\alpha ba$ is the equation in latitude, and the angle $\alpha pL$, the equation in longitude. Now considering $\delta L$ and $\alpha ba$ as the variations of the two sides $Pb$, $\delta pb$, whilst $PP$ and the angle $P$ remain constant, we have $\delta L : \alpha ba : (Trig. Art. 202.) \tan : \cos b_1$ or cos $L = (Trig. Art. 243.) \cos Pp - \cos Pb \times \cos pb_1 = h \times a \times \cos Pp - \cos Pb \times \cos pb_1 = h \times a \times \sin Pp \times sin pb_1$ + +$$\cos Pp - \cos Pb \times cos pb_1 = h \times a \times cos lat. sin pb_1$$ + +$$\cos Pb - cos Pb \times cotan pb_1 = h \times a \times cos lat. cos 23^{\circ} 28' - sin dec. x tan. moon's lat.$$ moon's lat. But if $CP$ be to CE as 1 : 1 + m, and $x, y_1$, are the sine and cosine of the latitude of the place, then $a = 2m x y_1$, as shown in Chapter on the Figure of the Earth; hence, $\delta ba = 2hmx \times cos 23^{\circ} 28' - sin dec. x tan. moon's lat.$ The sign becomes + if the declination and latitude of the moon be of different affections, that is, one south and the other north. The latitude here used, is that seen from the center of the earth. This correction increases the moon's distance from the pole $p$ of the ecliptic. + +167. To find the correction of the longitude, or the angle $Lpa$, we have (13) $La = Lpa \times sin. PL$, hence, $Lpa = La : sin PL ;$ but $\delta L = bL \times sin b_1$, and by spher. trig. sin. $Pb : sin Pp :: sin Pp : sin b_1 = sin p : sin Pp ;$ also, $\delta b = 2hmx ;$ hence, $Lpa = 2hmx \times sin p : sin Pp = 2hmx \times cos lon. c : sin 23^{\circ} 28'$ cos dec. c : cos lat. c + +70 + +ON PARALLAX. + += (as the cos. of the moon's latitude may be considered equal to unity) $2\sin x \times$ $\sin 25° 28' \times \cos$ lon. $x$. In north latitude, we must add this correction to the longitude seen from $K$, when the moon is in the descending signs 3, 4, 5, 6, 7, 8, but subtract it, when in the ascending signs 0, 1, 2, 9, 10, 11, to have the longitude seen from $C$; and the contrary when the latitude of the place is south. + +168. According to the Tables of Mayer, the greatest parallax of the moon, (or when she is in her perigee and in opposition) is $61° 22'$; the least parallax (or when in her apogee and conjunction) is $55° 35'$, in the latitude of Paris; the arithmetical mean of these is $57° 42'$; but this is not the parallax at the mean distance, because the parallax varies inversely as the distance, and therefore the parallax at the mean distance is $57° 24'$, an harmonic mean between the two. M. de Lambe recalculated the parallax from the same observations from which Mayer calculated it, and found it did not exactly agree with Mayer's. He made the equatorial parallax $57° 11'4''$. M. de la Lande makes it $57° 5'$ at the equator, $56° 53'8$ at the pole, and $57° 1'$ for the mean radius of the earth, supposing the difference of the equatorial and polar diameters to be $\frac{1}{300}$ of the whole. From the formula of Mayer, the equatorial parallax is $57° 11'4''$ with the following equations, according to M. de la Lande. + +$$ +\begin{array}{r} +57° 11'4'' - 3° 7'7" \cos$ ano. $x \\ ++ 10° \cos$ ano. $x \\ +- 0° \cos$ ano. $x \\ +- 37° \cos$ arg. evetion \\ ++ 0° \cos$ arg. evetion \\ ++ 26° \cos$ dist. $x \oplus \\ +- 1° \cos$ dist. $x \oplus \\ ++ 0° \cos$ dist. $x \oplus \\ ++ 2° \cos$ (apo. $x - \ominus) \\ ++ 0° \cos$ (apo. $x - \ominus) \\ ++ 1° \cos$ (arg. evetion + ano. $\oplus) \\ ++ 0° \cos$ (8 arg. lat. $x \oplus -$ ano. $\oplus) \\ +- 0° \cos$ (8 dist. $x \oplus +$ ano. $\oplus) \\ +- 0° \cos$ (arg. evetion - mean ano. $\ominus) \\ ++ 0° \cos$ (apo. $x - \ominus) \\ ++ 0° \cos$ (apo. $x - \ominus) \\ ++ 0° \cos$ (mean ano. $\ominus -$ mean ano. $\oplus) \\ ++ 0° \cos$ (dist. $x \oplus +$ mean ano. $\ominus) +\end{array} +$$ + +ON PARALLAX. +71 + +169. Let $r = \frac{1}{2}$ the semiaxis major, $p = \frac{1}{2}$ the semiaxis minor, $n = \sin$ the sine, $m$ the cosine of the angle $OCE$; then, from conics, the sine of the horizontal polar parallax : sine of the hor. parallax at $O$: $\sqrt{r^2 + p^2} : r_p$; hence the sine of the hor. par. at $O = \frac{r_p}{\sqrt{r^2 + p^2}} \times$ the sine of the hor. polar parallax. If $r : p :: 230 : 229$, we have the following Table for the horizontal parallax for every degree of latitude, that at the pole being unity. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
LatHor. Par.Lat.Hor. Par.Lat.Hor. Par.
0°10048851°10032161°100103
11004883910031462100097
21004573310030763100091
31004363410030064100085
41004353510029365100079
51004343610028666100073
61004323710027967100067
71004283810027368100062
81004283910026769100057
91004264010025770100059
101004244110025071100047
111004214210024372100042
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+ + + +```html + + + + Polar Parallax Table Example - HTML Version > + > + > + +``` +```html + + + + Polar Parallax Table Example - HTML Version > + > + > + +``` +```html + + + + Polar Parallax Table Example - HTML Version > + > + > + +``` +```html + + + + Polar Parallax Table Example - HTML Version > + > + > + +``` +```html + + + + Polar Parallax Table Example - HTML Version > + > +