# Rodrigues, Olinde: “Des lois géométriques qui régissent les déplacements
d’un système solide..”, translation and commentary

Richard Friedberg

###### Abstract

I provide for the first time in English a line-by-line translation of the
entire text of the monumental 1840 memoir of Olinde Rodrigues, “On the
geometrical laws governing the motions of a solid system…” published in French
in the Journal de Mathématiques Pures et Appliquées. I accompany the
translation with copious notes in italics, in which I explicate some passages
whose meaning is obscure in the direct translation, supply detailed proofs,
where lacking, of assertions in the original, and clarify the overall
organization of the memoir and the relation of its sections to one another.
(In my notes, Rodrigues himself is consistently called “the author”.) I often
supply a rendering in modern vector notation, for equations and formulas in
the original text in which vectors are expressed laboriously in terms of their
projections on fixed axes.

Benjamin Olinde Rodrigues (1795-1851) was a successful banker and ardent
social critic descended from a Sephardic family long resident in France. He
associated himself for many years with the utopian philosophy of Henri de
Saint-Simon, supporting the latter’s movement both financially and
administratively. Physics students beyond the introductory level are likely to
know of his formula for the Legendre polynomial of any order, but unlikely to
know anything about its author. This formula was contained in his 1815 thesis
as Docteur ès-Sciences at the University of Paris, a thesis containing much
material of far greater depth. Among his subsequent writings for over twenty
years are articles favoring the equality of women, articles on Saint-Simonism,
articles on the reform of the banking laws, and other social issues, but none
on science or mathematics. Then in the last eight years of his life he
published a number of mathematical papers, including the magnum opus which is
the subject of this translation.

The paper is a double tour-de-force in which the same material is expounded
first geometrically in Sections 1-14 and then algebraically in Sections 15-22.
In reading the second part, one will necessarily experience some déja-vu, but
one should keep in mind that although the same assertions or formulas
reappear, the logic binding them is not necessarily the same, as the starting
assumptions may be different.

Then Sections 23-33 contain a kind of coda in which new issues are presented;
these sections are well worth study for themselves alone, especially Section
33 which gives a meaning to Lagrange parameters surely at odds with that
intended by Lagrange himself.

The paper now follows in English translation.

##  1 General idea of translation and rotation of a solid system.

I understand by a solid system any assemblage of points, either continuous or
discontinuous, that are mutually bound in a fixed way, such that if any three
of these points are located at positions not in a straight line, and all their
distances from other points of the system are given, the placement of the
system will be completely determined for any placement of the triangle formed
by these three points.

Such a system can actually exist, since on a given triangular base, with given
lengths of its sides, one can construct only a single pyramid identical to -
that is, superposable on - another given pyramid. A pyramid obtained by
reflecting the first pyramid through its base would not be considered
superposable. Accordingly, three noncollinear points of a solid being fixed,
no displacement of the solid is possible.

But if only two points of the system are required to stay unmoved, by fixing
the distances of all other points from these two, one assures, to begin with,
that all points in a line with these two are unmovable. This line becomes a
fixed axis, and any other point of the system can only move on the
circumference of a circle concentric with and normal to the axis. Since all
the points of the system are bound unchangeably to any one of them and to the
fixed axis, the rotation of one implies the rotation of all, and the amplitude
of this rotation is the same for all the points of the system.

Any displacement of a solid about two fixed points reduces, therefore, to a
rotation, of equal amplitude and in the same sense for all the points of the
system, about the axis formed by the two fixed points.

Here it must be remarked that any given rotation can be replaced by a rotation
of the opposite sense, of an amplitude complementary to that of the first
rotation with respect to $360^{\circ}$.

Different rotations about the same axis result in a rotation equal to their
sum; here one must take care to assign contrary signs to the amplitudes of
rotation that are effected in opposite directions, but otherwise the order of
succession of the rotations remains arbitrary.

If the amplitude of the rotations is infinitely small, the arcs described by
the points of the solid located at a finite distance from the axis become
indistinguishable from their chords, while the latter are variously inclined
according to the angles of the rays drawn from the axis to the points of the
system.

But suppose that the axis, while still firmly bound to the system under
consideration, is infinitely far removed from it, and that the system
undergoes an infinitely small rotation about the axis, of an infinitesimal
order reciprocal to that of the distance of the axis of the system. The effect
will be to make all those points describe equal and parallel straight lines,
so that the system will have simply undergone a translation, that is to say a
displacement resulting in an equal transport of all its points in a certain
direction.

Thus any translation of a system can be considered rigorously as a rotation of
infinitesimal amplitude about an axis infinitely far off and normal to the
direction of that translation.

It is no surprise, then, to find moreover that all the properties of
translations are implied by those of rotations, just as those of a straight
line are implied by those of a large circle to which the line is tangent. We
need not linger over this.

We shall complete this general exposition of the displacement about a fixed
axis by the following theorem, which is evident from the figure and whose
consequences will be of use in what is to follow.

###  1.1 Movement of the axis of rotation without changing its direction.

The rotation of a solid system about a fixed axis can be replaced by an equal
rotation about another parallel axis, followed by a translation of the system
equal and parallel to the chord of the arc described by a point of the second
axis about the first. Or, what comes to the same thing, to the chord of the
arc that would be described by a point of the first axis about the second,
except that the direction of the translation must be reversed.

The author apparently considers this theorem as having been proved by the
preceding discussion, along with the figure referred to. In fact, the theorem
is an immediate corollary to the theorem on couples to be discussed in 10.1.

##  2 Displacements by translation.

If two situations of the same solid are such that all the lines joining a
point of the solid in one situation to the corresponding point in the other
are equal and parallel, the solid may clearly arrive from the first to the
second situation by sliding parallel to itself along one of these lines. The
length and direction of the line will measure those of the translation of the
system.

###  2.1 General law of composition of successive translations.

If the system undergoes several consecutive translations, differing both in
direction and in extent, it is evident that all these translations sum up to a
single unique translation, equal and parallel to the line that joins a point
of the first situation to its corresponding point in the second. This line
would close the polygon traced by the successive translations of the point in
question, and its length and direction, as is well known, depend only on those
of the various other sides of the polygon and are independent of their order
of succession.

By means of this law, one can reciprocally decompose any given translation
into a succession of diverse translations, provided only that the sum of their
projections on three perpendicular axes, or more generally on any arbitrary
axis, is equal to the projection of the given translation on the same axis, or
to the sum of its projections on the three axes. The projections are here
being “added” in what we would call a “vectorial” sense.

One may call this law of composition the law of the polygon of translations.

##  3 On the displacement of a system about a fixed point.

What follows is a remarkably concise proof of a celebrated theorem due to
Euler.[1]

Let two situations of the same system be given, sharing a point $O$ which
remains fixed in passing from one situation to the other. It should be
understood that the displacement can be expressed in some way as a succession
of rotations, with no reflection. Consider two arbitrary points $A,B$ in the
first situation, different from $O$ and not collinear with it. In the second
situation the corresponding points are $A^{\prime},B^{\prime}$. Here and
elsewhere I have used letters to designate points, lines, etc. as an aid to
the reader’s comprehension. In the French original no letters are used, all
geometrical entities being described purely by words.

The isosceles triangles $AOA^{\prime}$, lying in plane $P_{a}$, and
$BOB^{\prime}$, lying in plane $P_{b}$, share a common vertex $O$. Through $O$
pass a plane $N_{a}$ which is normal to $P_{a}$ and bisects the vertex angle
$\theta_{a}$. See Fig. 1Ai and 1Aii. In Fig. 1Aii the points $A,A^{\prime}$
are superimposed, as the line $AA^{\prime}$ is normal to the plane of the
diagram. The line marked $x_{A}$ is actually the common perpendicular to
$AA^{\prime}$ and $L$, a line defined below.

Likewise a plane $N_{b}$ normal to $P_{b}$ and bisecting $\theta_{b}$. See
Fig. 1Bi and 1Bii. Again, $x_{B}$ denotes the common perpendicular to
$BB^{\prime}$ and $L$.

The intersection of $N_{a}$ and $N_{b}$ will be a line $L$ (See Fig. 1C)
passing through $O$ and normal to both $P_{a}$ and $P_{b}$. In Fig. 1C, it
must be understood, if we regard the line $L$ as ‘vertical’, that the triangle
$OAA^{\prime}$ is not “horizontal”, although its base $AA^{\prime}$ is
horizontal. Likewise $OBB^{\prime}$ is not horizontal, although its base
$BB^{\prime}$ is. Also, although $AA^{\prime}$ and $BB^{\prime}$ are both
horizontal, they lie in general in different horizontal planes and
consequently do not intersect.

Any point $S$ on $L$ (being equidistant from $A$ and $A^{\prime}$, as well as
from $B$ and $B^{\prime}$) can be considered as the common summit of two
identical or superposable pyramids $OSAB$ and $OSA^{\prime}B^{\prime}$ see the
second paragraph of Section 1, having as bases the triangles $OAB$ and
$OA^{\prime}B^{\prime}$; so that the line $L$, being invariably bound to the
displaced system, remains unchanged by the displacement. Hence this
displacement reduces to a rotation around the fixed axis $L$. ($L$ is
prevented from sliding along its length by the immobility of $O$. This short
passage is the whole proof, different from that given by Euler.)

Two pyramids are congruent if they agree in the lengths of all six edges. The
edges $OA,OB,AB$ are equal respectively to
$OA^{\prime},OB^{\prime},A^{\prime}B^{\prime}$ because the displacement is
rigid. The equality $SA=SA^{\prime}$ follows from the construction of the
plane $N_{a}$, and $SB=SB^{\prime}$ from that of $N_{b}$. The edge $OS$ is the
same in both pyramids because $S$ is defined as a point on $L$ a certain
distance from $O$. The line $L$, however, has been located by a construction
drawing on both the initial and the final configurations. But now the pyramid
$OSAB$ makes it possible to determine the position of $L$ in terms of the
initial configuration alone; and alternatively $OSA^{\prime}B^{\prime}$ by the
final configuration alone. The congruence of the two pyramids then ensures
that the location of $S$ is the same (both in the body and in space) in both
configurations, so that $L$ is unmoved by the displacement. (The author’s
primary emphasis on pyramids was already evident in the second sentence of 1.)

In the singular case (it could be avoided if we wished) in which the planes
$N_{a}$ and $N_{b}$ coincide in a single plane $N$, we see that the axis is
simply the line of intersection of the planes $Q,Q^{\prime}$ containing the
original triangles $OAB$ and $OA^{\prime}B^{\prime}$. Any other line, should
it lie in the plane $N$, forms with the sides of these triangles two trihedral
angles symmetrically related but not superposable.

To sum up, any displacement of a system (achievable by rotations) about a
fixed point reduces to a rotation about a fixed axis passing through this
point. Or, more generally, it reduces to an equal rotation about a different
fixed axis parallel to the first one, but otherwise located where one will,
provided that the rotation be followed by a translation of the same extent and
direction as the chord of the arc described by the original fixed point $O$
under the rotation about the new axis; but the sense of the translation must
be opposite to that of the chord. (This generalization follows from the
theorem in Section 1 relating to the parallel transport of an axis of
rotation.)

##  4 On the most general displacement of a solid system in space.

Now let us consider any two situations whatsoever of the same solid, and seek
the simplest mode of displacement that can bring the solid from one situation
to the other. Select any point $O_{1}$ of the solid in the first situation,
and imagine that for each point $A_{1}$, a straight line is drawn from $O_{1}$
that is equal and parallel to the line $O_{2}A_{2}$ in the second situation.
Denote by $A_{12}$ the termination of the line thus drawn. Thus $O_{1}A_{12}$
and $O_{2}A_{2}$ are opposite sides of a parallelogram, whence the same is
true of $A_{12}A_{2}$ and $O_{1}O_{2}$. We have thus constructed an
intermediate assemblage of points, $O_{12},A_{12},...$, forming a solid
entirely identical to the one under consideration, but lying in a situation
intermediate between the two given ones. Since $O_{12}=O$ (by considering the
case $A_{1}=O_{1}$), the intermediate situation can be derived from the first
situation (in view of the theorem of the preceding section) by means of a
certain rotation through an angle $\theta$ about a fixed axis $L$ passing
through $O_{1}$ (this is indeed the fixed point theorem of Euler, proved by
the author in 3), while the passage from the intermediate to the second
situation requires only a translation whose extent and direction are the same
as those of the line $O_{1}O_{2}$ (on account of the parallelogram
$O_{1}O_{2}A_{2}A_{1}$).

Moreover, we observe that in view of the theorem on parallel transport in
Section 1, nothing prevents us from supposing that this intermediate situation
of the solid is reached from the first situation by a rotation through the
same angle $\theta$ about an arbitrary line $L^{\prime}$ parallel to $L$,
followed by a translation equal to the chord of the arc that would have been
described by a point on $L^{\prime}$ in making the rotation about $L$ that
would take the first to the intermediate situation. The whole displacement
taking each $A_{1}$ to $A_{2}$ has already been decomposed as $R+T$ where $R$
is the rotation through $\theta$ about $L$ and $T$ is the translation taking
$O_{1}$ to $O_{2}$. By the theorem on parallel transport, $R$ can be
decomposed as $R^{\prime}+T_{LL^{\prime}}$ where $R^{\prime}$ is the rotation
through $\theta$ about $L^{\prime}$ and $T_{LL^{\prime}}$ is a translation
normal to $L$ and $L^{\prime}$.

But this translation and the following one taking the intermediate to the
second situation combine to make a single translation equal and parallel to
the line joining any point on $L^{\prime}$ to its corresponding point in the
second situation. Thus the whole displacement is
$R^{\prime}+T_{LL^{\prime}}+T=R^{\prime}+T^{\prime}$ where
$T^{\prime}=T_{LL^{\prime}}+T$ is the translation that would take
$O_{1}^{\prime}$ to $O_{2}^{\prime}$. Now, for any origin chosen instead of
$O$, there is only one axis of rotation possible; therefore we have completely
demonstrated the following theorem, indisputably one of the most beautiful in
geometry, which deserves to be considered the fundamental basis of the
geometric laws of the movement [of a rigid body in 3 dimensions].

###  4.1 Fundamental theorem.

However a solid has been transported from one place to another, the
displacement can always be considered as resulting from two consecutive
displacements, a rotation and a translation. The rotation takes place about a
fixed axis passing through an arbitrarily chosen point in the initial
situation and parallel to a certain direction. This direction, as well as the
amplitude and sense of the rotation, is invariably determined by the initial
and final situations. The translation takes place parallel to the line joining
a point on the said axis to its corresponding point in the second situation,
and its length is the length of that line.

The order of these displacements can be reversed: the translation can precede
the rotation, but the latter then takes place about an axis passing through
the point in the final situation that corresponds to the point that was taken
to be the origin in the initial situation. In addition, the direction of the
axis of rotation and its amplitude and sense are the same for all the points
of the system, whether before or after the translation.

In this succession of displacements, let us observe that the line joining any
point of the initial situation to its corresponding point in the final
situation, that is the resultant line really traced by this point, forms the
third side of a triangle, of which the first side (representing the effect of
the rotation) is variable for different points of the system but always normal
to the axis of rotation, and the second side, constant for all the points of
the system, measures the translation of the system.

The projection of this resultant line on the axis of rotation is therefore
constant for all the points of the solid. This constancy can be achieved only
relative to the direction of the axis of rotation, for the projection on an
arbitrary direction is the sum of two projections, that of the chord of the
arc of the rotation and that of the line traversed in translation. The first
of these is variable, the second is constant. Their sum cannot, therefore by
constant for any direction other than that which causes the first of these
projections to vanish, that, is, the direction of the axis of rotation.

All the points of a solid system displaced in an arbitrary manner are
therefore equally transported relatively to the (invariably determined)
direction of the axis of rotation.

If this constant projection vanishes, the displacement reduces to a rotation
about a certain (preferred) axis without any translation. The transverse
location of this axis is easily found by considering a plane normal to the
common direction of all the possible axes of rotation. The straight line from
any initial point in this plane to its corresponding final point will be the
base of an isosceles triangle lying in the plane, of which the angle at the
vertex will equal the amplitude of the rotation, and the vertex itself will
lie on the (preferred) axis to be found.

In the more general case in which the displacement is not a pure rotation,
this constant projection is the measure of the absolute translation of the
system, by which is meant the minimum translation among all those associated
with the possible axes of rotation, variously located but all having a common
direction; it is none other than the translation of those points of the system
that are displaced parallel to [this common direction]. If any of these points
is chosen as the origin $O$, the axis of rotation, which we shall distinguish
by calling it the central axis of the displacement, is also the axis of
translation.

Thus the displacement reduces, with respect to its central axis, to turning
about this axis while sliding parallel to its direction: a kind of movement
that has been compared to that of a screw turning in its nut. This is the
simplest expression of the fundamental theorem, in which the two displacements
of rotation and translation take place [simultaneously and] orthogonally. Here
the author attributes the theorem to M. Chasles[2].

This theorem may be viewed as the natural generalization of Euler’s fixed
point theorem - a generalization in which the displacement is assumed to
result from an arbitrary series of rotations (excluding reflections) about
axes that are not assumed to have a common point of intersection.

##  5 Locating the central axis.

But we have now to find this central axis, that is to find those points of the
system carried by the displacement along a line parallel to the common
direction of the possible axes of rotation. Now, one may arrive at this by the
following construction.

Let $A$ be an arbitrary point in the first situation, and $A^{\prime}$ its
corresponding point in the second. Let $L$ be the line through $A$ in the
common direction which is the (known) direction of the central axis. Let us
think of this direction as “vertical”. Let $P$ be the (vertical) plane
containing $L$ and $AA^{\prime}$. (See Fig. 2A.) Within $P$, erect from $A$ a
(horizontal) perpendicular to $L$ which terminates at the point $Z$ chosen so
that $AZA^{\prime}$ is a right angle. We may introduce Cartesian coordinates
$x,y,z$, with origin at $A$, and let $L$ be the $z$-axis and $AZ$ the
$x$-axis. Then $P$ is the $(x,z)$ plane.

Now let $R$ be the plane containing $AZ$ and normal to $L$. $R$ is the $(x,y)$
plane. See Fig. 2B. Note that $Z$ and $A^{\prime}$ are both in the $(x,z)$
plane P and that the line $ZA^{\prime}$ is perpendicular to the $x$-axis $AZ$;
therefore $ZA^{\prime}$ is vertical, parallel to the $z$-axis $L$. Hence
$ZA^{\prime}$ is normal to the horizontal plane $R$. In other words, $Z$ is
the foot of the perpendicular dropped from $A^{\prime}$ to $R$.

Within $R$, construct an isosceles triangle having $AZ$ as base and vertex
angle equal to the (known) amplitude $\theta$ of the rotation. The vertex $V$
of this triangle will lie on the central axis, provided only that the
isosceles triangle is placed, relatively to $AZ$, in the sense of the
rotation. On a given base in a given plane, there are two ways to erect an
isosceles triangle with a given vertex angle, related by reflection in the
base. Only one is right.

For it is evident that this vertex $V$, turning through $\theta$ about $L$,
will reach a point $\bar{V}$ in the plane $R$, such that the chord from $V$ to
$\bar{V}$ is equal and parallel to the line from $Z$ to $A$ because the
triangle $VA\bar{V}$ is congruent to the triangle $ZVA$, and moreover that the
translation from $\bar{V}$ to the image $V^{\prime}$ of $V$ in the second
situation is equal and parallel to the line from $A$ to $A^{\prime}$. The
whole motion from $V$ to $V^{\prime}$ is made by the rotation carrying $V$ to
$\bar{V}$, followed by the translation from $\bar{V}$ to $V^{\prime}$; whereas
the motion of $A$ to $A^{\prime}$ consists entirely of the translation. Since
a translation affects all points equally, it follows that $\bar{V}V^{\prime}$
is equal and parallel to $AA^{\prime}$. See Fig. 2C. Hence the resultant
motion from $V$ to $V^{\prime}$ is equal and parallel to the resultant motion
from $Z$ to $A^{\prime}$ that is, $V\bar{V}+\bar{V}V^{\prime}=ZA+AA^{\prime}$
which in turn is parallel to $L$ as noted in previous paragraph but one. But
this is just the condition satisfied by points on the central axis.

And reciprocally, taking $VV^{\prime}$ as the axis of rotation, the point $A$
on rotation through $\theta$ about $V$ will travel to $Z$, and then by the
translation parallel to $VV^{\prime}$ will travel from $Z$ to $A^{\prime}$,
reaching its given destination.

And this construction shows that when the axis of relative translation
$AA^{\prime}$ is normal to the axis of rotation $L$, the whole displacement
reduces to a simple rotation about the central axis, since then $Z$ coincides
with $A^{\prime}$ and so $V$ and $V^{\prime}$ are the same.

Also, if a displacement of the solid is such that all the points of the solid
remain in mutually parallel planes, the displacement reduces to a rotation
about some fixed axis normal to these planes.

Although the foregoing construction correctly locates the central axis, the
information required to carry it out as well as the reasoning to justify it
are drawn from the whole of the preceding paragraphs of the essay. But if one
studies the construction in this light, one is forced to keep in mind much
redundant material. If, on the other hand, we disregard the preceding
paragraphs, we are in danger of drawing false inferences. For example, it
looks as if this construction requires only one pair $A,A^{\prime}$ of
corresponding points. But that is not so: the fixed axis theorem of Euler,
proved in 3, requires two such pairs, and the result is a necessary part of
the proof of the author’s construction to locate the central axis. Therefore
he gives another construction, entirely self-contained and requiring only the
location of two pairs $A,A^{\prime}$ and $B,B^{\prime}$ as well as the
direction of the central axis. He attributes this construction to “mon ami M.
Lévy”.

Drop from $A$ a line perpendicular to the central axis at a point $C$, and
likewise from $A^{\prime}$ a perpendicular at $C^{\prime}$, and consider the
quadrilateral $ACC^{\prime}A^{\prime}$. Let $A^{m},C^{m}$ be the midpoints of
$AA^{\prime}$, $CC^{\prime}$; then the line $A^{m}C^{m}$ will be perpendicular
to both $AA^{\prime}$ and $CC^{\prime}$. This follows from the symmetry of the
figure with respect to a $180^{\circ}$ rotation about $A^{m}C^{m}$. This
property gives the following construction of the central axis, being given
only the points $A,A^{\prime}$, another corresponding pair $B,B^{\prime}$, and
the direction of the central axis:

Through each point $A_{m}$ (resp. $B_{m}$), pass a line $L_{a}$ (resp.
$L_{b}$) parallel to the central axis, as well as a line $N_{a}$ (resp.
$N_{b}$) normal to both $AA^{\prime}$ and $L_{a}$ (resp. $BB^{\prime}$ and
$L_{b}$). Let $P_{a}$ (resp. $P_{b}$) be the plane formed by $L_{a}$ and
$N_{a}$ (resp. $L_{b}$ and $N_{b}$). Then the planes $P_{a}$ and $P_{b}$ will
intersect precisely on the central axis.

To justify the construction, we must refine our notation to take into account
the use of two pairs: the quadrilateral based on $A$ must now be called
$AC_{a}C^{\prime}_{a}A^{\prime}$, and that based on $B$ must be called
$BC_{b}C^{\prime}_{b}B^{\prime}$. The points $C_{a},C_{b}$ are not necessarily
the same, but both lie on the central axis, as well as
$C^{\prime}_{a},C^{\prime}_{b}$.

Since $L_{a}$ is parallel to $C_{a}C^{\prime}_{a}$, $N_{a}$ is normal to
$C_{a}C^{\prime}_{a}$ as well as to $AA^{\prime}$; thus $N_{a}$ is the very
line of which $A^{m}C_{a}^{m}$ is a segment. Therefore $C_{a}^{m}$ lies on
$N_{a}$. But then the plane $P_{a}$ can equally well be described as formed by
$AA^{\prime}$ and $C_{a}C^{\prime}_{a}$ instead of by $AA^{\prime}$ and
$L_{a}$, so that it contains the whole line $C_{a}C^{\prime}_{a}$. that is the
whole central axis. Likewise the plane $P_{b}$ is formed by $BB^{\prime}$ and
$C_{b}C^{\prime}_{b}$ and also contains the central axis. Hence $P_{a}$ and
$P_{b}$ intersect on the central axis, q.e.d.

There are degenerate cases, which will be left to the reader’s study.

##  6 Consequences of the Fundamental Theorem.

It now behooves us to set forth the principal corollaries that follow from the
fundamental theorem, relating to the particular displacements of the points,
the lines, and the planes of a solid system.

(a) The distances separating each point of the solid in the first situation
from its corresponding point in the second all have equal projections on the
direction of the central axis; this common projection is the measure of the
absolute translation of the system.

(b) As any line belonging to the displaced system does no more, with respect
to its direction, than turn about the axis of rotation, there results a very
simple relation between the angle formed by this line with the axis of
rotation and that formed between the initial and final directions of this
line, to wit:

“The sine of the half-angle of displacement of any line belonging to a
displaced system is equal to the sine of the half-rotation of the system,
multiplied by the sine of the angle between this line and the axis of
rotation.”

The author here is concerned only with the directions of the line in the
initial and final configurations. Now, the set of all possible directions can
be mapped onto the points on the surface of a sphere (say of radius $r$) by
mapping each direction $D$ to the unique point $P$ for which the line from
$O$, the center of the sphere, to $P$ has the direction $D$. The initial
direction of the line in question thus maps to a point $A$ on the surface of
the sphere, and its final direction to a point $B$. The direction of the
central axis (the “axis of rotation”) is mapped to a point $C$, which we shall
call the North Pole so as to make use of the ideas of latitude and longitude.
The central axis is then the diameter from North to South Pole, and the
rotation about this axis clearly does not change the angle made with it by the
line in question. Thus $A$ and $B$ lie on the same circle of latitude, which
we shall call $L$, and the arc from any point $P$ on $L$ to the North Pole is
of a fixed length, making a constant angle $POC$ which we shall call $\chi$.
The “angle of rotation of the system” is the difference in longitude between
$A$ and $B$, which we may call $\phi$, and the “angle of displacement” of the
line in question is the angle $\theta=AOB$, which measures the geodesic
distance from $A$ to $B$.

With these definitions, the proposition rendered above in words may be
expressed in symbols as

$\sin(\theta/2)=\sin(\phi/2)\sin\chi.$ (1)

We can understand the rôle of $\chi$ by noting that when $\chi$ becomes small,
the geodesic distance from $A$ to $B$ becomes small because of the pinching of
the base of the isosceles triangle $ABC$, even though the difference $\phi$ in
longitude is kept constant.

The equation (1) can be derived as a restriction of the spherical Law of
Cosines to isosceles triangles; but the author gives no indication of having
such reasoning in mind. Instead, he says briefly that the proposition “is made
evident by observing the figure”. Unfortunately, the figures originally
appearing in the Journal de Mathématiques have been lost, but I believe I have
closely reconstructed the one referred to in this passage, with the aid of my
associate Dr. Familton.

The easy demonstration of (1) depends on the construction of planes and
straight lines in the interior of the sphere (see Fig. 3A), particularly the
straight line $\bar{AB}$. On the one hand, this line is a chord of the great
circle $G$, of radius $r$ and center $O$ (see Fig. 3B) upon which both $A$ and
$B$ are located; since this chord subtends an angle $\theta$ at $O$, we have

$\bar{AB}=2r\sin(\theta/2).$ (2)

On the other hand, this same line is a chord of the small circle of latitude,
$L$ (see Fig. 3C), whose center $K$ lies on the axis $OC$ and whose radius we
shall call $\rho$. Since the angle subtended at $K$ is $\phi$, we have

$\bar{AB}=2\rho\sin(\phi/2).$ (3)

Comparing (3) with (2), we find

$\frac{\sin(\theta/2)}{\sin(\phi/2)}=\rho/r.$ (4)

Finally we determine $\rho/r$ by considering the right triangle $OKP$ (see
Fig. 3D) for an arbitrary point $P$ on $L$. The hypotenuse $OP$ has length
$r$, the angle $POK$ is $\chi$, and the side $KP$ opposite this angle has
length $\rho$. Therefore

$\rho/r=\sin\chi.$ (5)

Substituting (5) into (4), we obtain (1).

(c) Any line parallel to the axis of rotation is transported parallel to
itself, while any line normal to that axis suffers an angular displacement
equal to the amplitude of the rotation. (Special cases of (b)]]

(d) Any plane invariably bound to the displaced system, and normal to the axis
of rotation, is therefore transported into a plane parallel to the initial
one, at a distance equal to the absolute translation of the system.

(e) The midpoint of the line that joins any point of the system to its
correspondant i. e., to its final position after displacement is the point of
that line that approaches most closely to the central axis of the
displacement.

(f) the midpoints of all the lines that join the various points of a plane
figure to their correspondants after an arbitrary displacement lie in a single
plane, which also contains the midpoints of the lines joining any point
outside the plane figure to its symmetric correspondant.

This plane makes equal angles with the planes of the two plane figures, as
well as with the corresponding lines bound to the two figures, but if not
within their respective planes, then symmetrically inclined.

The meaning of “symmetrically” in (f) is “by reflection in the special plane
under consideration.” I leave the study of (f) to the most ambitious of
readers.

##  7 The decomposition of any displacement into two pure rotations.

Having presented the fundamental geometric law of the passage of a solid from
one given situation to another, also given in an arbitrary way, we have now to
study the law of composition of successive displacements; by means of this law
one can construct or calculate the elements of the composite displacement,
that is the position of its central axis, the amplitude of its rotation, and
the extent of its translation.

We have already presented the law of composition of translations; we shall
next give that of rotations about different fixed axes, and finally that of
arbitrary displacements, each resulting from a combined translation and
rotation.

From this law of composition of rotations about different axes, we shall
deduce an important transformation of the fundamental theorem (4.1), to wit:

“Any displacement of a solid system can be represented, in an infinite number
of ways, as the composition of two successive rotations of the system about
two nonintersecting fixed axes. The product of the sines of half these
rotations, multiplied by the sine of the angle between the two axes and by the
minimum distance between them, is equal - for each of these conjugate pairs of
axes - to the product of the sine of half the angle of rotation of the system
about the central axis of the (total) displacement with half the length of its
absolute translation.”

Let the arbitrary (total) displacement be characterized by a central axis
$\cal C$, a rotation angle $\theta_{C}$, and a translation distance $T$. It is
asserted that there are infinitely many “conjugate pairs”, each of which
consists of a rotation $\theta_{A}$ about an axis $\cal A$ followed by a
rotation $\theta_{B}$ about an axis $\cal B$, such that the composition of
these two rotations is equivalent to the total displacement under
consideration. Each of these conjugate pairs is related to the total
displacement by the equation

$D\sin\nu\sin(\theta_{A}/2)\sin(\theta_{B}/2)=(1/2)T\sin(\theta_{C}/2),$ (6)

where $D$ is the minimal distance between the two axes and $\nu$ is the angle
between their directions.

The author only states this remarkable theorem here, deferring its proof to
11.

To put it another way, the volume of the tetrahedron of which two opposite
edges lie anywhere along the respective conjugate axes, provided that the
length of each of these edges is proportional to the sine of the corresponding
half-angle of rotation, is the same for all conjugate pairs of rotations whose
composition is equivalent to a given displacement.

The equivalence of this second statement of the theorem to the first is based
on a theorem of geometry, that the volume of any tetrahedron is one-sixth the
product of the lengths of any two opposite edges, times the minimal distance
between them, times the sine of the angle between them. This formula can be
established by a variety of methods; further study is left to the reader.

Thus, any displacement of a solid system reduces to a rotation about one or
two fixed axes.

In the case where one of these axes is parallel to the central axis (of the
total displacement), it follows from the law of composition of rotations that
its conjugate is situated at infinity and that the rotation corresponding to
it becomes infinitely small and therefore amounts to a simple translation.
This leads to the original version (4.1) of the fundamental theorem, so that
that version is no more than a particular case of the theorem we have just
given.

##  8 The composition of two given rotations.

We have now to present the law of composition of successive rotations of a
solid about different axes. The author now begins the succession of
composition theorems that will lead ultimately to the decomposition theorem
stated in the previous section.

Let us begin by considering only two intersecting axes, and let us seek to
determine the resultant axis of these two rotations - the one about which the
given solid will be finally found to have turned, in order to arrive from the
initial to the final situation. (A problem closely related to this was studied
in [3].)

This resultant axis must be placed in such a way that in being subjected to
the two rotations indicated about the supposed intersecting axes, it comes
back to its initial position. If, therefore, through each of the given axes
one passes a plane that makes an angle with the plane of the two axes that is
equal to half the rotation about that same axis, the intersection of these two
planes will be the resultant we seek, as it arrives by virtue of the first
rotation at the position symmetric by reflection in the plane of the two axes,
and returns by the second rotation to its original position.

Call the two intersecting axes ${\cal A}$ and ${\cal B}$; they determine a
plane that we shall call $\cal AB$. The resultant axis we call ${\cal C}$, and
let $\cal AC,BC$ denote the planes formed respectively by ${\cal A}$ and
${\cal C}$, and by ${\cal B}$ and ${\cal C}$. The angles of rotation about
${\cal A}$ and ${\cal B}$ are $\theta_{A}$ and $\theta_{B}$.

It is assumed that the axes ${\cal A}$ and ${\cal B}$, intersecting at the
origin $0$, as well as the amplitudes of rotation $\theta_{A}$ and
$\theta_{B}$, are given, and the problem posed is to determine the axis ${\cal
C}$ and rotation amplitude $\theta_{C}$ such that the rotation $\theta_{A}$
about ${\cal A}$ followed by the rotation $\theta_{B}$ about ${\cal B}$ will
produce as resultant the rotation $\theta_{C}$ about ${\cal C}$. By Euler’s
theorem (3) such a ${\cal C}$ exists and passes through the origin $0$.

In understanding the solution proposed by the author, it is necessary to
distinguish the angle between two axes (e. g. between ${\cal C}$ and ${\cal
B}$) from the angle between two planes (e. g. between $\cal AC$ and $\cal
AB$).

The solution proposed is that $\cal AC$ should make an angle $\theta_{A}/2$
with $\cal AB$, and $\cal BC$ should make an angle $\theta_{B}/2$ with $\cal
AB$. In this way the planes $\cal AC$ and $\cal BC$ are determined, and ${\cal
C}$ is their intersection.

The argument is that if ${\cal C}^{*}$ is defined as the reflection of ${\cal
C}$ in $\cal AB$, then the angle between $\cal AC$ and $\cal AC^{*}$ will be
twice that between $\cal AC$ and $\cal AB$, that is twice $\theta_{A}/2$, so
that the first rotation (of $\theta_{A}$ about ${\cal A}$) will bring $\cal
AC$ to $\cal AC^{*}$ and hence ${\cal C}$ to ${\cal C}^{*}$ (since the angle
between ${\cal C}$ and ${\cal A}$ is unchanged by the rotation). Then by a
similar argument, the second rotation (of $\theta_{B}$ about ${\cal B}$) will
take ${\cal C}^{*}$ back to ${\cal C}$. So the combined effect of the two
partial rotations will be to leave ${\cal C}$ unaltered. But this is what is
required in order that ${\cal C}$ be the resultant axis.

(It may well be objected that it has not been shown that the axis ${\cal C}$
is restored to its original position in the same sense in which it began. But
if not, the situation may be remedied by taking the appropriate sign of
$\theta_{C}$.)

At the same time one sees that the angle between the two planes ($\cal AC$ and
$\cal BC$) will be half the angle of the resultant rotation ($\theta_{C}$).
For the first axis, which does not move under the first rotation, is displaced
only by the second, and it describes about the resultant axis, determined as
shown above, an angle twice that between the two planes.

The phrase “determined as shown above” refers to the determination of $\cal C$
as the intersection of two planes, of which in particular the plane $(\cal
BC)$ forms an angle $\theta_{B}/2$ with the plane $(\cal AB)$. It follows that
a rotation about $\cal B$ through an angle $\theta_{B}$ brings the plane
$(\cal AB)$ to its reflection in $(\cal BC)$. and in particular it brings
${\cal A}$ to its reflection ${\cal A}^{*}$ in $(\cal BC)$.

But since ${\cal A}$ was unchanged by the first rotation about itself, this
second rotation about $\cal B$ must have the same effect on $\cal A$ as does
the composite rotation through the angle $\theta_{C}$ (to be found) about
$\cal C.$ That is, this composite rotation must take $\cal A$ to ${\cal
A}^{*}$. But this requires that $\theta_{C}$ be twice the angle between $(\cal
AC)$ and $(\cal BC)$, as asserted by the author.

Here let us note that the half-rotation (of a plane) about each axis can be
measured equally well by the interior or the exterior angle of the two planes
passing through that axis, only the sense of the rotation depending on which
measure one adopts, since any rotation (of a point) about an axis in one sense
is equivalent to a rotation in the opposite sense with an amplitude
complementary to the first by 360∘.

The conscientious reader may wish to ascertain that if a consistent sign
convention be followed, whereby the angle of rotation about an axis is
measured either always clockwise or always counterclockwise looking along the
direction of the axis, this construction will yield the correct sign of
$\theta_{C}$ in relation to those of $\theta_{A}$ and $\theta_{B}$.

Furthermore, as to the order of succession of these rotations, it comes about
that if the two rotations are supposed to take place in a certain order,
leading to a particular resultant axis ${\cal C}$, then by reflecting this
axis in the plane of the of the two given axes one obtains the resultant
${\cal C}^{*}$ of the same two rotations in the inverse order. From this we
see that the amplitude of the resultant rotation is independent of the order
of the two given ones, but that the position of the resultant axis depends on
this order, and that in the composition of more than two rotations about
arbitrary intersecting axes, the order cannot be modified without altering
both the position of the resultant axis and the amplitude of the resultant
rotation.

The last statement is a bit too strong: the resultant amplitude will be
unaltered if the order is completely inverted, as from 1234 to 4321, no matter
how long the succession is. Likewise the amplitude is preserved under a cyclic
permutation as from 12345 to 34512. A corollary of these two facts is that the
amplitude cannot be altered by any permutation of the composing displacements
unless they number at least 4.

Such is the characteristic difference between the composition of rotations and
that of successive translations. In fact, these two kinds of composition are
analogous in a way similar to the properties of a plane triangle and those of
a spherical triangle. For if one compares the translations parallel to the
three sides of a planar triangle to the sines of the half-angles of rotations
effected around the three sides of a trihedral angle, the values of these
translations and those of these sines are in equal manner proportional to the
sines of the angles opposite to the respective sides of the planar triangle
and of the trihedral angle.

The author is essentially comparing the Law of Sines for a planar triangle to
that for a spherical triangle. But he adds the complication of associating the
sides of the triangle to the corresponding translations or rotations of a
solid system.

##  9 Composition of infinitesimal rotations.

But these two resultant axes ${\cal C}$ and ${\cal C}^{*}$, corresponding to
the same two rotations in two different orders of succession, will coincide in
the plane of the two axes if the rotations become infinitely small, and from
this there follow two important consequences:

First, the order of succession of infinitesimal rotations about two
intersecting axes (and, as it follows, about as many such axes as one wishes)
is immaterial. And second, the axis and amplitude of the infinitesimal
rotation resulting from the succession of two infinitesimal rotations ${\cal
A}$ and ${\cal B}$ about two intersecting axes are determined in the same way
as the axis and translation length that would result from two successive
translations proportional to the given rotations and parallel to their axes.

The author is referring here to the theorem enunciated in Section 1:

“Thus any translation of a system can be considered rigorously as a rotation
of infinitesimal amplitude about an axis infinitely far off and normal to the
direction of that translation.”

Note that this theorem concerns a single axis, and in the present context it
applies separately to ${\cal A}$ and to ${\cal B}$. Note also that the angle
between ${\cal A}$ and ${\cal B}$ is finite, unlike that between ${\cal C}$
and ${\cal C}^{*}$.

Since, by removing the axes of rotation far away, one may transform the
infinitesimal rotations into finite translations perpendicular to these axes
and inclined one to the other just as the axes are to each other, one achieves
all the generality of the law of composition of finite rotations, which by
mediation of the infinitesimal rotations includes also the law of composition
of translations.

The author evidently does not mean that the two axes are removed far from each
other, since they continue to intersect. Rather, he is observing the action of
the infinitesimal rotations at points far from both of the two axes. But these
points are regarded, for the present purpose, as “here”, while the two axes
with their intersection are “there”, that is removed to infinity.

The author’s point is that the law of composition of finite rotations is so
powerful that by suitable applications of it one can derive that of
composition of finite translations as well.

##  10 Of the composition of rotations about two parallel axes.

All the points of the system displaced by two consecutive rotations about two
parallel axes remain within parallel planes normal to these axes. Therefore
the displacement reduces to a simple rotation about a certain axis parallel to
the first two. This being admitted, the mode of determination and of
construction of the resultant axis of two intersecting axes applies equally
well to this case, and yields a resultant axis parallel to the first two and a
composite rotation equal to the sum or the difference of the given rotations,
according as they act in the same or opposite senses.

###  10.1 Couples of parallel rotations.

But here we encounter a remarkable case, that in which the two rotations are
equal and of opposite sense. The composite rotation is then null and the
composite axis is placed at infinity, which causes the displacement to amount
to a simple translation. To be precise, each point of the displaced solid has
traversed a line of length and direction constant for all points of the
system; the direction of this line is normal to the two axes but makes an
angle with the normal to the plane of the two axes, equal to half the given
angle of rotation about each axis; and its length is the product of the
distance between the axes by twice the sine of half the rotation angle.

Using rectilinear coordinates $x,y,z$, let the two axes point in the
$z$-direction and their separation $d$ in the $x$-direction, so that they lie
in the $x$-$z$ plane; then the normal to their plane is in the $y$-direction.
Since the coordinate $z$ is unchanged in each rotation, we may regard the
whole operation as confined to the $x$-$y$ plane. Take any initial point $P$
in this plane. Under the rotation about the first axis ${\cal A}$ it describes
an arc of amplitude $\theta$ ending at a point $P^{\circ}$. Then the rotation
about the second axis ${\cal B}$ carries the point along an arc also of
amplitude $\theta$ but in the opposite sense, from $P^{\circ}$ to its final
position $P^{\prime}$. The straight line from $P$ to $P^{\prime}$ is asserted
by the author to have the same length and direction for all points $P$ \- a
fact by no means obvious. The following proof can best be followed by
consulting Figs. 4a and 4b.

Let the axes ${\cal A}$ and ${\cal B}$ cut the $x$-$y$ plane at the points $A$
and $B$, so that the distance $d$ between the axes is the line segment $AB$.
And let the distances from $P^{\circ}$ to the axes $A$ and $B$ be respectively
$r_{A}$ and $r_{B}$. Then the isosceles triangles $PAP^{\circ}$ and
$P^{\circ}BP^{\prime}$ are similar, having the same vertex angle $\theta$; and
their scale ratio is $r_{A}/r_{B}$. Therefore
$PP^{\circ}/P^{\circ}P^{\prime}=r_{A}/r_{B}$.

The base angles of the isosceles triangles pertaining to various initial
points $P$ are all equal to $\phi=(180^{\circ}-\theta)2$. Therefore the angles
$AP^{\circ}B$ and $PP^{\circ}P^{\prime}$ are equal, being both equal to
$\phi+\chi$ where $\chi$ is the (undetermined) angle $AP^{\circ}P^{\prime}$.
(This assertion merits close examination of Fig. 4a.)

Hence the triangles $AP^{\circ}B$ and $PP^{\circ}P^{\prime}$ are similar,
having two sides in the same ratio
$AP^{\circ}/BP^{\circ}=PP^{\circ}/P^{\prime}P^{\circ}=r_{A}/r_{B}$ and
agreeing in the included angle.

It follows that $PP^{\prime}/AB=PP^{\circ}/AP^{\circ}$ or
$PP^{\prime}=(AB)(PP^{\circ}/AP^{\circ})=AB(2\sin(\theta/2))$, as asserted by
the author. Note that this equation does not involve the angle $\chi$.

Furthermore, the angle between the directions $PP^{\prime}$ and $AB$ is the
angle through which the triangle $PP^{\circ}P^{\prime}$ must be turned about
$P^{\circ}$ so as to make the angles $AP^{\circ}B$ and $PP^{\circ}P^{\prime}$
coincide. Clearly this angle is $\phi$, so that $PP^{\prime}$ makes an angle
$\phi=(180^{\circ}-\theta)/2=90^{\circ}-\theta/2$ with the $x$-axis. Therefore
it makes an angle $\theta/2$ with the $y$-axis, as claimed.

Figs. 4a and 4b show two realizations of this construction. Using the line
segment AB as the reference for length and direction, one sees that the point
$P$ is placed differently in the two diagrams, and that the ratio
$r_{A}/r_{B}$ is also quite different, as well as the size of the angle
$\chi$. Nevertheless, the dotted line from $P$ to $P^{\prime}$ has the same
length as well as the same direction in both diagrams, and in each diagram the
triangle $PP^{\circ}P^{\prime}$ is similar to the triangle $AP^{\circ}B$ as
found in the above proof.

It may help the understanding to distinguish between determinate ratios, which
for a given $\theta$ are independent of the choice of $P$, and indeterminate
ratios, which are affected by that choice. The two isosceles triangles in the
diagram have determinate shape, which yields the ratio $2\sin(\theta/2)$ (base
to sides), but their relative size is indeterminate, depending on the ratio
$r_{A}/r_{B}$. For the two symmetric triangles, however, the situation is
reversed: the shape of these triangles involves $r_{A}/r_{B}$ and is hence
indeterminate, while the ratio of their size is determinate as the diagram
exhibits it as tied to the shape of the isosceles triangles.

Once it is established that the direction and length of $PP^{\prime}$ is the
same for all $P$, the author’s statements about this direction and length can
be derived very easily by setting $P=A$. For then one has obviously
$P^{\circ}=P$, and $PP^{\prime}=P^{\circ}P^{\prime}$ which is the base of an
isosceles triangle with vertex angle $\theta$ at $B$ and side $AB$. The
author’s statements follow.

The order of succession of the two rotations makes a difference; if the order
is reversed, the composite line of translation is reflected about the normal
to the two axes.

All this follows easily from a comparison of similar triangles (as shown
above); and then if the rotations are infinitesimal, the order of succession
becomes immaterial, and the translation acts along the normal to the plane of
the two axes.

Thus any couple of parallel rotations (not necessarily infinitesimal) is
equivalent to a simple translation, and reciprocally, any translation can be
replaced in an infinite number of ways by a couple of this kind. The word
“couple” is meant to imply that the two rotations are equal and opposite.

These couples of parallel rotations compose and decompose, in accordance with
the law of translations, in an arbitrary order of succession, acting in all
the positions that correspond in length and direction to a particular
translation; compositions and decompositions which can be found by
substituting for the couples the translations that they represent. The order
of different couples is arbitrary; the order of the two rotations forming a
single couple is not, if the rotations are finite.

Thus we have generalized to couples of finite rotations the law of composition
which M. Poinsot, I believe, was the first to state for couples of
infinitesimal rotations.

##  11 Proof of the general decomposition theorem

As any displacement of a solid system can be reduced to a rotation followed by
a translation (see 4), and this translation can always be replaced by a couple
of rotations (see 10.1), one of whose axes intersects the given rotation axis
of the system, and the rotations about these two intersecting axes can be
composed (see 8) into a single rotation, there results immediately the proof
of the transformation stated above (see 7) of the fundamental theorem (see
4.1), to wit: that any displacement of a solid system can be accomplished in
an infinite number of ways by the succession of two rotations about two fixed
nonintersecting axes.

The axes will be nonintersecting unless the translation is null, in which case
one rotation suffices. This one-sentence proof can benefit, as usual, by some
expansion of the reasoning and naming of the geometrical entities.

Let the axis of the given rotation be called $\cal C$ and the accompanying
translation be called $T$. As shown in 10.1, $T$ can be replaced by a couple
of equal but opposite rotations about two parallel axes which we may call
${\cal B}^{\prime}$ and $\cal B$, of which ${\cal B}^{\prime}$ may be located
so as to intersect $\cal C$. (The axis ${\cal B}^{\prime}$ is the one that was
called $\cal A$ in 10.1; here we shall define an entirely different axis as
$\cal A$.)

Then, by the method of 8, the rotations about $\cal C$ and ${\cal B}^{\prime}$
can be composed to make one about an axis we shall call ${\cal A}$; the two
rotations about the nonintersecting axes $\cal A$ and $\cal B$, performed
successively, generate the same displacement as $\cal C$ and $T$.

This is quite clear as far as it goes, but it gives no clue as how to derive
eq. (6) of 7. This will be done in 13; it is unnecessary to carry out the
demonstration also in the present context, as it would involve the same steps
sometimes done in reverse order.

##  12 Rotations about an arbitrary number of fixed nonintersecting axes.

Finally, there is the composition of rotations about an arbitrary number of
fixed nonintersecting axes. Let us take a point in the space, upon which we
shall study the effect of all these rotations in their order. We have seen
that any rotation about a fixed axis can be replaced by another equal
rotation, accomplished about another axis parallel to the first, followed by a
translation equal to the chord of the arc described by a point of the new axis
about the first in consequence of the rotation given at the outset. We have
also seen that a translation followed by a rotation about an axis passing
through the endpoint of the axis of translation can instead be preceded by it,
if the axis of rotation passes through the origin of the axis of translation.

This last statement deserves careful examination. Let a translation $T$ be
followed by a rotation $R$ whose axis is $L$. Decompose $T$ into $T_{l}$
parallel to $L$ and $T_{t}$ transverse to $L$. Since $T_{l}$ commutes with
$R$, it suffices to consider $T_{t}$ and to project the whole situation onto a
transverse plane.

We may represent points on this plane by complex numbers, and for simplicity
let us represent the displacement $T_{t}$ by the number $1$. Take $R$ to be a
rotation of the plane through an angle $\theta$ about the point $1$, and
identify the author’s phrase “axis of translation” as the line from $0$ to
$1$. Thus the endpoint of the axis of translation is at $1$, and its origin is
at $0$. The claim is that the effect of $T_{t}$ followed by $R$ is the same as
that of $R^{\prime}$ followed by $T-t$, where $R^{\prime}$ is the rotation
through $\theta$ about $0$.

But this is easily proven. Start with a point $z$ and first perform
$R^{\prime}$; this takes $z$ to $ze^{i\theta}$. Then perform $T_{t}$; the
result is $ze^{i\theta}+1$. On the other hand, $T_{t}$ acting first on $z$
produces $z+1$, and the radius vector from $1$ to $z+1$ is $z$ so that the
second transformation $R$ replaces the term $z$ by $ze^{i\theta}$, yielding
the final result $ze^{i\theta}+1$ as before.

This being given, if, through the point of origin we have chosen for study, we
pass axes parallel to each of the given nonintersecting axes, the displacement
of the system operates successively about these axes, by means of the
transport of the rotations to the intersecting axes respectively parallel to
the original ones. By virtue of the successive replacement of rotations about
axes passing through the endpoints of the translation lines by rotations about
axes passing through their origins, the displacement of the system will be
partitioned into a series of rotations respectively equal to those originally
given, taking place successively about intersecting axes parallel to the first
series, followed by a series of translations resulting from the chords
successively traversed by the chosen point about the original nonintersecting
axes in the order ascribed to the rotations.

One may ask whether the final composite translation would be the same if a
different point of origin “$p$” had been chosen for study. The answer is yes,
because a change in $p$ can be simulated by keeping $p$ unchanged and rigidly
changing the positions of all the rotation axes.

(The composition of rotations about intersecting axes and that of translations
will take place in the manner described above; in this case the composite
displacement will reduce to a rotation and a translation whose axes both pass
through the point of intersection.)

We see from (the general) construction that the elements of the final
composite rotation depend only on the amplitude and direction of the
individual rotations, and are not changed by any parallel movement of their
axes; while the length and direction of the composite translation depend, as
well, on the positions of the individual rotations, seeing that the chords
successively described by the chosen point vary in length and direction
according to the successive positions that the displaced point takes
relatively to the various given axes.

If, in the system of these axes, there are found consecutive pairs that form
couples of parallel rotations, it is evident that these couples do not
contribute anything to the determination of the direction and amplitude of the
resultant rotation, and that they influence only the length and direction of
the resultant translation, as the point whose successive rotations determine
this translation will be found, upon completion of each couple, to have
described the translation equivalent to the couple.

##  13 The case of only two nonintersecting fixed axes.

The author intends in this section to give a second proof of the “two-axis”
theorem of 7, which has already been proved in 8-11. But in that first proof,
he started with a displacement described in screw form with the central axis
${\cal C}$ and rotation angle $\theta_{C}$ given, and constructed the two
(usually) nonintersecting axes ${\cal A},{\cal B}$ of rotations
$\theta_{A},\theta_{B}$ whose composition gives this displacement. In this
second proof he assumes that ${\cal A},{\cal B}$ are given, along with
$\theta_{A},\theta_{B}$, and constructs $\cal C$ and $\theta_{C}$. Thus the
decomposition theorem of 7 is replaced in the present section by a composition
theorem, a generalization of the theorem of 8 in that the axes ${\cal A}$ and
${\cal B}$ are no longer required to intersect. And this time the author
carries through the derivation of eq. (6) in 7, which was omitted in 11.

Consider two nonintersecting axes and their shortest distance $D$, and take as
the origin of the displacement the end $A$ of that shortest distance lying on
the first axis of rotation ${\cal A}$ \- that is, the rotation to be first
executed. Then, on passing through that origin an axis ${\cal B}^{\prime}$
parallel to the second one ${\cal B}$ given, the two intersecting axes will be
composed into a third, which will be the axis of rotation of the displacement
relative to that origin $A$.

Let the symbol $D$ refer equally to the shortest distance between ${\cal A}$
and ${\cal B}$ and to the line segment of that length, pointing from its
intersection $A$ with ${\cal A}$ to its intersection $B$ with ${\cal B}$. The
author has introduced a substitute axis ${\cal B}^{\prime}$, parallel to
${\cal B}$ but passing through $A$. Regarding $A$ as Euler’s (3) fixed center,
the fixed-point theorem says that the rotations
$\theta_{A},\theta_{B^{\prime}}$ about the intersecting axes ${\cal A}$,
${\cal B}^{\prime}$ can be composed to make a rotation $\theta_{C^{\prime}}$
about a third axis ${\cal C}^{\prime}$, also passing through $A$. The author
calls ${\cal C}^{\prime}$ the “axis of rotation of the displacement relative
to A.”

At the same time (see 1.1 and 4), the rotation about $\cal B$ can be
accomplished by performing the rotation about ${\cal B}^{\prime}$ followed by
a translation, which we may call $T^{\prime}$. Thus the entire displacement
$({\cal A},\theta_{A})$ followed by $({\cal B},\theta_{B})$ is equivalent to
the three actions $({\cal A},\theta_{A})$, $({\cal
B}^{\prime},\theta_{B^{\prime}})$, $T^{\prime}$ taken in sequence, which in
turn is the same as $({\cal C}^{\prime},\theta_{C^{\prime}})$ followed by
$T^{\prime}$. More compactly, we can write this result as

$({\cal A},\theta_{A})({\cal B},\theta_{B})=({\cal
C}^{\prime},\theta_{C^{\prime}})T^{\prime}.$ (7)

The accompanying axis of translation $T^{\prime}$ will be given (1.1) as the
chord of the arc described by this same origin $A$ in consequence of the
rotation about the second (${\cal B}$) of the two axes given.

It is desired, however, to express the whole displacement in terms of the
central axis $\cal C$ and its associated quantities; thus we must have

$({\cal A},\theta_{A})({\cal B},\theta_{B})=({\cal C},\theta_{C})T,$ (8)

where $T$, called the absolute translation of the displacement, is directed
along the axis $\cal C$.

Comparing (8) to (7), we find that the rotation $({\cal C},\theta_{C})$
differs from the rotation $({\cal C}^{\prime},\theta_{C^{\prime}})$ only by a
translation, which may be written $T^{\prime}-T$ since translations compose by
addition. It follows immediately that $\cal C$ and ${\cal C}^{\prime}$ are
parallel and that $\theta_{C}=\theta_{C^{\prime}}$. Furthermore, both these
rotations move points only within planes perpendicular to $\cal C$ and to
${\cal C}^{\prime}$. Therefore the difference $T^{\prime}-T$ is perpendicular
to $\cal C$, hence perpendicular to $T$ which lies along $\cal C$. It follows
that $T$ is the projection of $T^{\prime}$ onto $T$, as stated in the author’s
next remark:

The projection of this chord onto the composed axis of rotation, determined as
above, measures the absolute translation of the (composite) displacement. It
is equal to the sum of the projections of the two sides of the isosceles
triangle of which it is the base. The two sides are equal (in length) to the
shortest distance between the two given nonintersecting axes.

Let the whole displacement under study carry the point $A$ to its final
position $A_{f}$. Since the first rotation about $\cal A$ left $A$ unmoved,
the movement from $A$ to $A_{f}$ is accomplished entirely by the second
rotation, about $\cal B$. But the point $B$, lying on $\cal B$, is unmoved by
this rotation; since the displacement is rigid, the distances $AB$ and
$A_{f}B$ are equal. That is, the triangle $ABA_{f}$ is isosceles, with vertex
at $B$ and sides $AB$, $A_{f}B$ both equal to $D$ as defined above.

The base of this triangle is the chord $AA_{f}$, which has previously been
identified (author’s comment after eq.(7)) as giving the translation
$T^{\prime}$. Its projection on $\cal C$ is equal to the sum of the
projections of $AB$ and $BA_{f}$. (This is readily understood by thinking of
$AB,BA_{f},AA_{f}$ as vectors.)

It is easily demonstrated in addition that the two sides are equally inclined
to the composed axis.

This is an important claim. We see at once that the projection of $AB$ on
$\cal C$ is $D\cos S$ where $S$ is the angle between $\cal C$ and $D$. The
author wishes to establish that $BA_{f}$ makes the same angle with $\cal C$
and therefore has the same projection. This will establish the important
equation $T=2D\cos S$.

In fact, this shortest distance ($D$) is normal to the plane ${\cal
AB}^{\prime}$ of the two intersecting axes. Now, in considering the angle
formed by this normal $D$ with the reflection ${\cal C}^{*}$ of the resultant
axis $\cal C$ in this same plane ${\cal AB}^{\prime}$, one sees that this
angle does not change when one supposes it mobile and displaced rigidly by the
second rotation (about $\cal B$), which brings the reflected line $\cal
C^{*}$of which we speak into coincidence with the resultant axis $\cal C$.

But, in this rotation, the normal $D$ is rotated through a plane perpendicular
to the second axis $\cal B$ (sweeping out a cone with vertex at $B$) so as to
become parallel to the second side $A_{f}B$ of the isosceles triangle we are
considering, and as it is evident that the angle of the normal $D$ with the
resultant axis $\cal C$ is supplementary to that which it forms with the
reflection $\cal C^{*}$ of that axis, one sees that the resultant axis $\cal
C$ is, as we have just said, equally inclined with respect to the two sides
$AB$, $BA_{f}$ of this isosceles triangle. The projection of $\cal C^{*}$ onto
the initial position of $D=AB$ is the negative of the projection of $\cal C$;
that is, it equals the projection of $\cal C$ onto $BA$. Therefore the $\cal
B$ rotation sweeps the angle under discussion to the angle between $\cal C$
and $BA_{f}$ as asserted.

The preceding argument is best understood by comparing it with the
construction in 8, where $\cal A$ and $\cal B$ intersect; $\cal B$ and ${\cal
B}^{\prime}$ coincide, $D$ vanishes but its direction is still defined as the
normal to the plane $AB$; $A_{f}$ is identical to $A$ since both rotations
leave $A$ fixed; and there is no translation $T^{\prime}$ or $T$. The key
specification is that $\theta_{A}$ is twice the angle between the planes $\cal
AC$ and $\cal AB$, and $\theta_{B}$ is twice the angle between the planes
$\cal BC$ and $\cal AB$, so that $\cal C$ is reflected in $\cal AB$ by the
$A$-rotation and reverse-reflected by the $B$-rotation. The author’s
determination of $\theta_{C}$ depends on computing the final position of a
line whose initial position was $\cal A$, deduced on the one hand from the
effect of the $B$-rotation and on the other hand from that of the
$C$-rotation.

In the present case the plane of reflection is taken as ${\cal AB}^{\prime}$,
but it could be any plane parallel to both $\cal A$ and $\cal B$, without
changing the angles. Instead of $\cal A$, the initial position of the moving
line is taken as $D$, and again a comparison is made between two ways of
finding its final position. The required agreement between the two ways, as in
8, yields the result, which in this case is the equality of the angles made by
$AB$ and $BA_{f}$ with the resultant axis $\cal C$.

From which one finally concludes that the absolute translation of a solid
system arising from the succession of two rotations about two fixed
nonintersecting axes is equal to double the distance between the two axes,
projected on the direction of the composite, or resultant, axis. That is, if
the side $AB$ of the isosceles triangle $ABA_{f}$ is extended to twice its
length, the projection of this doubled side onto the direction of $\cal C$
will fall on $A_{f}$. The resulting equation is

$T=2D\cos S$ (9)

as anticipated above.

But it is evident that the cosine of the angle $S$ of this distance $D$ with
the composite axis $\cal C$ is equal to the sine of the angle made by $\cal C$
with the plane of the two composing axes $\cal A,B^{\prime}$ (since $D$ itself
is normal to this plane). which is found to be equal to the product of the
sines of the given half-rotations by the sine of the angle between the two
axes, divided by that of the composite half-rotation. That is,

$\cos S=\sin
h=\sin(\theta_{A}/2)\sin(\theta_{B}/2)\sin\nu/\sin(\theta_{C}/2),$ (10)

where $h=\angle(\cal C,\cal AB)$ and $\nu=\angle(\cal A,\cal B)$.

Equation (10) follows from the law of proportion of the sines of the half-
rotations to those of the angles included between the opposing axes.

Here again the author has compressed many steps into one. We can understand
(10) more readily by noting that $D$ and $T$ no longer appear, so that the
equation involves only directions. This enables us to associate the directions
$\cal A,\cal B,\cal C$ with points $A,B,C$ on a sphere, forming the vertices
of a spherical triangle. Since we are dealing only with directions, we need
not distinguish between $C$ and $C^{\prime}$ or between $B$ and $B^{\prime}$.
The angle $\nu=\angle(\cal A,\cal B)$ is just the arc $c$, the side of the
triangle opposite to $C$. Thus (10) becomes

$\sin h=\sin(\theta_{A}/2)\sin(\theta_{B}/2)\sin c/\sin(\theta_{C}/2).$ (11)

where $h$ is what we might call the altitude of the triangle, that is the arc
running from $C$ to $c$ and making a right angle with the latter.

The author asserts that (11) is a consequence of the “law of proportion…”
which can be stated as

$\frac{\sin(\theta_{A}/2)}{\sin a}=\frac{\sin(\theta_{B}/2)}{\sin
b}=\frac{\sin(\theta_{c}/2)}{\sin c}.$ (12)

We must ask how this law is arrived at, and also how it leads to (11).

Let us adopt the custom, with spherical triangles, of allowing the letters
$A,B,C,$ to stand also for the spherical angles at the respective vertices,
thus $B=\angle(\cal AB,\cal BC)$, etc. Now referring to the law of composition
in 8, we see that $\angle(\cal AB,\cal BC)$ is set equal to half the rotation
angle $\theta_{B}$, and likewise for the other axes. Therefore (12) is
equivalent to

$\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c},$ (13)

which is just the Law of Sines for a spherical triangle. (I am unable to see
how the author could have arrived at (12) without relying on this law from
spherical trigonometry.)

Now we must show how (12) leads to (11). If $h$ were the altitude of a planar
triangle we would obviously have $h=a\sin B$. The spherical analogue is $\sin
h=\sin a\sin B$, also an application of the Law of Sines to the special case
of a right triangle. But since $B=\theta_{B}/2$ from 8, we have

$\sin h=\sin
a\sin(\theta_{B}/2)=\sin(\theta_{A}/2)\sin(\theta_{B}/2)\frac{\sin
a}{\sin(\theta_{A}/2)}.$ (14)

Now, applying (12), we may replace $\frac{\sin a}{\sin(\theta_{A}/2)}$ by
$\frac{\sin c}{\sin(\theta_{C}/2)}$, obtaining (11) as asserted by the author.

From which we arrive at the modified Fundamental Theorem in the form (7) in
which we have already pronounced it, namely:

“any displacement of a solid system can always arise, in an infinite number of
ways, from the succession of two rotations about two nonintersecting fixed
axes, provided that the product of the sines of the successive half-rotations
by the distance between the two conjugate axes and by the sine of the angle of
these axes is equal to the product of the absolute half-translation of the
displaced system by the sine of the resultant half-rotation.”

That is,
$\sin(\theta_{A}/2)\sin(\theta_{B}/2)D\sin\nu=(1/2)T\sin(\theta_{C}/2)$, where
$D$ is the distance and $\nu$ the angle between the two axes. (Having
established (11), we can discard the spherical triangle representation,
writing once more $\nu$ in place of $c$, and substitute (10) into (9). This
gives indeed

$2D\sin(\theta_{A}/2)\sin(\theta_{B}/2)\sin\nu=T\sin(\theta_{C}/2)$ (15)

in agreement with eq. (6) of 7.)

##  14 Composition of general displacements.

We are now in a position to resolve completely the following general problem,
in which one considers the succession of (an arbitrary number of)
displacements of the same solid.

Being given the axes of rotation and translation as well as the amplitude of
the rotations and extent of the translations for each successive displacement
of a system, it is required that we construct the axes [[and amplitudes]] of
rotation and translation of this system relative to a given origin.

The solution of this problem is evidently the same as that of the previous
one, where it was only a matter of rotations about fixed axes, since the
translations can be replaced by couples of rotations about fixed axes. We have
briefly indicated the solution in the last paragraph of 12. We therefore need
not linger over it further.

###  14.1 The particular case of infinitely small displacements.

The solution is considerably simplified when one considers only infinitely
small displacements. First of all, the order of the rotations is indifferent,
and their composition by whatever number around intersecting axes operates
like that of translations proportional to these rotations and parallel to
these axes. Second, the order of the rotations and translations successively
accomplished by the origin of the displacement is equally indifferent, and
each of these rotations and translations can be established directly and
separately, as though the point to be displaced were displaced only
alternately and not successively, which follows from the fact that the space
traversed by each of these displacements is infinitely small. The composition
of these partial translations resulting from withdrawing from the given axes,
or from the translations themselves that are joined to the rotations, acts in
accordance with the same law as that of the rotations.

## 15

We have now to apply calculation to the geometric laws that we have just
presented concerning the general displacements of a solid system. We shall
start by deriving the formulas for change of coordinates of points in the
solid system, which hold such a large place in analytical mechanics.

Let $x,y,z$ and $x+\Delta x,y+\Delta y,z+\Delta z$ be the coordinates of two
points of which the first is moved to the second by the displacement of the
system, and let $\xi,\eta,\zeta$ be the coordinates of the midpoint of the
line joining the two, so that

$\xi=x+(1/2)\Delta x,\;\;\;\;\;\eta=y+(1/2)\Delta
y,\;\;\;\;\;\zeta=z+(1/2)\Delta z.$ (16)

Furthermore, let $g,h,l$ be the angles formed by the direction of the axis of
rotation with the three coordinate axes, $\theta$ the amplitude of the
rotation, $t$ the absolute size of the translation, and $X,Y,Z$ the
coordinates of an arbitrary point on the central axis of the displacement. In
much of what follows, the origin of coordinates may be assumed to lie on the
central axis; that is, we can take $X=Y=Z=0$, or in the vector notation to be
introduced, $\vec{W}=0$.

Consider the right triangle whose hypotenuse is formed by the line joining the
initial and final point and whose sides are given, one by the arc of the chord
described by the initial point under the rotation $\theta$, and the other by
the line traversed in a translation by this same point after undergoing the
rotation. Clearly, the changes $\Delta x,\Delta y,\Delta z$ are respectively
equal to the projections of this hypotenuse, that is to the sum of the
projections of the other two sides of this triangle on the respective
coordinate axes.

Now, the side equal and parallel to the absolute translation $t$ gives the
three projections $\cos g,\cos h,\cos l$; the other side is equal to
$2u\tan(\theta/2)$, $u$ denoting the distance from the central axis to this
same side (the one formed by the chord). Let us call $G,H,L$ the angles
between this side and the coordinate axes. Then we have immediately

$\displaystyle\Delta x=t\cos g+2u\tan(\theta/2)\cos G$ $\displaystyle\Delta
y=t\cos h+2u\tan(\theta/2)\cos H$ $\displaystyle\Delta z=t\cos
l+2u\tan(\theta/2)\cos L.$ (17)

This and the following equations can be better understood if translated into
modern vector notation. Let us define a right-handed orthonormal system
$\hat{t},\hat{u},\hat{v}$ where $\hat{t}$ points along the central axis;
$\hat{v}$ along the chord; and $\hat{u}$, perpendicular to both, points to the
midpoint of the chord from the base of the perpendicular dropped from that
midpoint to the central axis. The whole displacement $(\Delta x,\Delta
y,\Delta z)$ may be designated as $\vec{\Delta}$; then the above equations say
that

$\vec{\Delta}=\vec{t}+2\hat{v}u\tan(\theta/2),$ (18)

where $\vec{t}=\hat{t}t$ is the translation vector.

Since one has necessarily

$\cos g\cos G+\cos h\cos H+\cos l\cos L=0,$ (19)

(this says that $\hat{t}\cdot\hat{v}=0$) one deduces

$\Delta x\cos g+\Delta y\cos h+\Delta z\cos l=t$ (20)

$(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}=t^{2}+4u^{2}\tan^{2}(\theta/2).$
(21)

That is, $\vec{\Delta}\cdot\hat{t}=t$ and
$\vec{\Delta}\cdot\vec{\Delta}=t^{2}+(2u\tan(\theta/2))^{2}$.

The first terms (of (15)) $t\cos g,t\cos h,t\cos l$ represent the part of the
changes that arise from the absolute translation displacement; the second (set
of three) terms, the part due to the rotation performed by the displacement.
In comparing these first terms to the second, one finds that the first, which
measure the effect or moment of the translation of the system, have for value
the projections of this translation on each axis of the coordinates, while the
second, which represent for each point the effect or moment of the rotation of
the system, have for value the projection upon each coordinate axis of (the
area of an isosceles) triangle whose vertex is the midpoint of the line
finally traversed by the point considered, and whose base is a line [segment]
directed along the central axis and of length $4\tan(\theta/2)$.

In the case of an infinitely small displacement, this midpoint (the vertex of
the triangle) coincides with the initial point, and consequently the moment of
the rotation, relative to any given direction, is equal to double the
projection on that direction of (the area of) a triangle whose vertex is the
point (under consideration) and whose base taken on the central axis is equal
to the rotation of the system.

This explains how the theory of projections applies to the laws of translation
through linear projections, and to those of rotation through the projection of
areas. A possible influence of Grassmann here? or independent? Let us
continue.

(Wordy explanation here omitted: terms in $t$ represent the effect of
translation, those in $u$ the effect of rotation.)

The chord $2u\hat{v}\tan(\theta/2)$ being normal both to the central axis and
to the perpendicular dropped from the point $\xi,\eta,\zeta$ to this axis,
which has length $u$, we have

$\displaystyle u\cos G=(\eta-Y)\cos l-(\zeta-Z)\cos h$ $\displaystyle u\cos
H=(\zeta-Z)\cos g-(\xi-X)\cos l$ $\displaystyle u\cos L=(\xi-X)\cos
h-(\eta-Y)\cos g.$ (22)

Here we need to define more vectors. Let $\vec{W}=(X,Y,Z)$ represent the point
on the central axis that has been identified as locating that axis with
respect to the immovable space within which the solid exists. Let $\vec{w}$
stand for the base of the perpendicular dropped from the midpoint of the chord
of rotation to the central axis. ($\vec{W}$, although chosen arbitrarily along
the axis, is fixed for a particular displacement of the solid, whereas
$\vec{w}$ slides along the axis as we consider the trajectories of different
points of the solid.) And let $\vec{\omega}=(\xi,\eta,\zeta)$ represent the
position of the midpoint of the chord. Then (15) tells us that

$\hat{v}u=(\vec{\omega}-\vec{W})\times\hat{t}.$ (23)

I note here that all the author’s equations starting with (15) are consistent
with $(\hat{t},\hat{u},\hat{v})$ being a left-handed system, whereas the
motion described in the Chasles theorem at the end of 4 (“mouvement … de la
vis dans son écrou”, movement of a screw in its nut) makes it definitely
right-handed. I choose to write vector equations in the original right-handed
notation. When equation arrays are written out in the original in terms of the
components, I shall reproduce them without change; but in writing these arrays
in vector notation I shall reverse the order of all cross-products. Thus, I
shall interpret (15) as

$\hat{v}u=\hat{t}\times(\vec{\omega}-\vec{W})$ (24)

instead of as (23). I shall do this consistently without further comment.

Moreover, $\hat{t}\times(\vec{w}-\vec{W})=0$ since $\vec{w}-\vec{W}$ lies
along the central axis. Therefore $\vec{\omega}-\vec{W}$ may be replaced by
$\vec{\omega}-\vec{w}$ which is just $\vec{u}=\hat{u}u$. So (15) becomes

$\hat{v}u=\hat{t}\times\vec{u}$ (25)

which need not surprise us.

And in consequence,

$\displaystyle\Delta x=A+p\eta-n\zeta,$ $\displaystyle\Delta y=B+m\zeta-p\xi,$
$\displaystyle\Delta z=C+n\xi-m\eta,$ (26)

$A,B,C,m,n,p$ being six constants that depend on the position of the central
axis, the length of the translation, and the amplitude of the rotation, as
follows:

$\displaystyle A=t\cos g+2\tan(\theta/2)(Z\cos h-Y\cos l),$ $\displaystyle
B=t\cos h+2\tan(\theta/2)(X\cos l-Z\cos g),$ $\displaystyle C=t\cos
l+2\tan(\theta/2)(Y\cos g-X\cos h),$ (27)

and

$\displaystyle m=2\tan(\theta/2)\cos g$ $\displaystyle n=2\tan(\theta/2)\cos
h$ $\displaystyle p=2\tan(\theta/2)\cos l.$ (28)

By including “the position of the central axis” as a variable, the author
signals that the following calculations do not assume that this axis passes
through the origin. Indeed, much of what follows in this Section becomes
trivial if that assumption (the “null central axis” assumption or NCA) is
made. For example, (15) becomes $A=t\cos g$, etc.

Let us put $\vec{\Gamma}=(A,B,C)$, $\vec{q}=(m,n,p)$; then these definitions
become

$\vec{\Gamma}=\vec{t}+2\vec{W}\tan(\theta/2)\times\hat{t}$ (29)

and

$\vec{q}=2\hat{t}\tan(\theta/2);$ (30)

under NCA, (29) becomes $\vec{\Gamma}=\vec{t}$.

Going back to (15), in vector notation it reduces to

$\vec{\Delta}=\vec{\Gamma}+\vec{q}\times\vec{\omega}$ (31)

($\vec{\Delta}=\vec{\Gamma}$ under NCA).

Applying (29) and (30), this becomes

$\displaystyle\vec{\Delta}$ $\displaystyle=$
$\displaystyle\vec{t}-2\hat{t}\times\vec{W}\tan(\theta/2)+2\hat{t}\tan(\theta/2)\times\vec{\omega}$
(32) $\displaystyle=$
$\displaystyle\vec{t}+2\hat{t}\times(\vec{\omega}-\vec{W})\tan(\theta/2),$

agreeing with (18) in view of (24).

If we denote by $\alpha,\beta,\gamma$ “les variations des coordonnées de
l’origine des axes coordonnés” (the coordinates of the point to which the
origin of coordinates is carried by the displacement), we have the following
relation:

$\displaystyle\alpha=A+(1/2)(p\beta-n\gamma)$
$\displaystyle\beta=B+(1/2)(m\gamma-p\alpha)$
$\displaystyle\gamma=C+(1/2)(n\alpha-m\beta).$ (33)

Let us introduce the vector $\vec{\delta}=(\alpha,\beta,\gamma)$.Then (15)
becomes

$\vec{\delta}=\vec{\Gamma}+(1/2)\vec{q}\times\vec{\delta}.$ (34)

To arrive at this equation, consider (31) and recall that
$\omega=\vec{r}+(1/2)\vec{\Delta}$, where the displacement carries $\vec{r}$
into $\vec{r}+\vec{\Delta}$. Now take the special case $\vec{r}=0$ (the origin
of coordinates). In this case $\vec{\Delta}$ takes the value of
$\vec{\delta}$, by definition of the latter. Thus (31) reduces to
$\vec{\delta}=\vec{\Gamma}+(1/2)\vec{q}\times\vec{\delta}$ which is exactly
(34). (Under NCA, one has simply $\vec{\delta}=\vec{\Gamma}$.)

Without NCA, equation (34) looks indeterminate since $\vec{\delta}$ is defined
in terms of itself. But actually the system (15) provides three linear
equations in the three unknowns $(\alpha,\beta,\gamma)$ which are perfectly
determinate.

The solution is obtained transparently by vector algebra. Let $\vec{U}$ be the
position vector of the point on the central axis closest to the origin, and
$\hat{V}=\hat{t}\times\hat{U}$, then $\hat{t},\hat{U},\hat{V}$ are
orthonormal. Moreover, (29) reduces to

$\vec{\Gamma}=\vec{t}+2\vec{U}\times\hat{t}\tan(\theta/2)$ (35)

since $\vec{W}$ and $\vec{U}$ are both on the central axis. This can be
written

$\vec{\Gamma}=\vec{t}-2U\hat{V}\tan(\theta/2)=\vec{t}-Uq\hat{V}$ (36)

in view of (30). From this we have

$\vec{\Gamma}\cdot\vec{U}=0$ (37)

and also

$\vec{\Gamma}\cdot\hat{t}=t.$ (38)

Then from (34) we find

$\vec{\delta}\cdot\hat{t}=t,$ (39)

and also, in view of (37),

$\vec{\delta}\cdot\vec{U}=-(1/2)\vec{\delta}\times\vec{q}\cdot\vec{U}=-(1/2)\vec{\delta}\cdot\vec{q}\times\vec{U}=-(1/2)\vec{\delta}\cdot
Uq\hat{V}.$ (40)

which can be rewritten as

$\vec{\delta}\cdot\hat{U}=-\vec{\delta}\cdot\hat{V}\tan(\theta/2).$ (41)

On the other hand, (36) also gives

$\vec{\Gamma}\cdot\hat{V}=-Uq$ (42)

whence

$\vec{\delta}\cdot\hat{V}=-Uq+\vec{\delta}\cdot\hat{U}\tan(\theta/2).$ (43)

Substituting (41) into (43) gives

$\vec{\delta}\cdot\hat{V}=-Uq-\vec{\delta}\cdot\hat{V}\tan^{2}(\theta/2),$
(44)

or

$\vec{\delta}\cdot\hat{V}=-Uq\cos^{2}(\theta/2)=-2U\cos^{2}(\theta/2)\tan(\theta/2)=-U\sin\theta$
(45)

whence by (41)

$\vec{\delta}\cdot\hat{U}=+2U\sin(\theta/2)\cos(\theta/2)\tan(\theta/2)=U(1-\cos\theta).$
(46)

Combining (39),(45), and (46),we obtain the formula

$\vec{\delta}=\vec{t}+\vec{U}-U[\hat{U}\cos\theta+\hat{V}\sin\theta],$ (47)

which tells us that the endpoint of $\vec{\delta}$ can be located by passing
through the origin $0$ a circle in the $\hat{U},\hat{V}$ plane with center at
$\vec{U}$ on the central axis, moving counterclockwise on this circle from $0$
through an arc subtending an angle $\theta$ at the center, and erecting on the
endpoint of this arc the translation vector $\vec{t}$. Indeed, this is just
the operation (rotation $\theta$ about the central axis followed by
translation $\vec{t}$) that takes the origin of coordinates into its image
under the displacement considered.

By means of (15) one may replace, if one wishes, the constants $A,B,C$ by
their values in terms of $\alpha,\beta,\gamma$. This leads to

$\displaystyle\Delta x=\alpha+2\tan(\theta/2)(\eta-\beta/2)\cos
l-(\zeta-\gamma/2)\cos h)$ $\displaystyle\Delta
y=\beta+2\tan(\theta/2)(\zeta-\gamma/2)\cos g-(\xi-\alpha/2)\cos l)$
$\displaystyle\Delta z=\gamma+2\tan(\theta/2)(\xi-\alpha/2)\cos
h-(\eta-\beta/2)\cos g)$ (48)

where the first terms $\alpha,\beta,\gamma$ express the translation relative
to the origin of coordinates (the length being
$\sqrt{\alpha^{2}+\beta^{2}+\gamma^{2}}$) and those pertaining to the rotation
express the rotation about an axis passing through the origin.

To understand the last remark, let us call the direction of the central axis
“vertical” and recall the statement of the “Fundamental Theorem” in 4. There
it is pointed out that any vertical axis can be chosen as the rotation axis,
and that the accompanying translation, while constant for all points
considered, is not vertical unless the central axis is the one chosen. In
general, it is only the projection of this translation on the axis of rotation
that is vertical. There is an additional horizontal translation that
compensates for the change of rotation axis.

The equation (15) can be written in vector notation as

$\vec{\Delta}=\vec{\delta}+\vec{q}\times(\vec{\omega}-(1/2)\vec{\delta}),$
(49)

where $\vec{\delta}$ is given by (47). In (47) the term $\vec{t}$ can be taken
as a vertical translation and the remaining term as the trajectory of the
initial point $\vec{r}=0$ under rotation about the central axis. But now the
author desires to consider the vertical axis through the origin (call it the
0-axis) as the axis of rotation. Then the initial point $\vec{r}=0$ does not
move under the rotation, and hence the entire expression (47) must be viewed
as translation, containing both a vertical and a horizontal part.

For a general point $\vec{r}$, still taking the 0-axis as the axis of
rotation, the whole of $\vec{\delta}$ is still translation and therefore the
second term of (49) gives the rotation about the 0-axis.

Of course, under NCA there is no change of (47) and one still has
$\vec{\Delta}=\vec{\delta}=\vec{t}$.

We could have established these formulas directly, as well as those which
precede, and which we have constructed on the central axis.

##  16 Equations of the central axis.

The equations of the central axis follow in their turn from the above
formulas, with the greatest simplicity; for the rotation has no effect on any
point lying on that axis, and one then has $\Delta x=t\cos g,\;\;\;\Delta
y=t\cos h,\;\;\;\Delta z=t\cos l$. (That is, $\vec{\Delta}=\vec{t}$.) The
coordinates $x,y,z$ describe some point on that axis (I shall denote this
point by $\vec{r}$) and taking this into account one has the equations sought,
expressed by means of $\alpha,\beta,\gamma$:

$\displaystyle py-nz+A=t\cos
g=\frac{m(Am+Bn+Cp)}{m^{2}+n^{2}+p^{2}}=\frac{m(\alpha m+\beta n+\gamma
p)}{m^{2}+n^{2}+p^{2}},$ $\displaystyle mz-px+B=t\cos
h=\frac{n(Am+Bn+Cp)}{m^{2}+n^{2}+p^{2}}=\frac{n(\alpha m+\beta n+\gamma
p)}{m^{2}+n^{2}+p^{2}},$ $\displaystyle nx-my+C=t\cos
l=\frac{p(Am+Bn+Cp)}{m^{2}+n^{2}+p^{2}}=\frac{p(\alpha m+\beta n+\gamma
p)}{m^{2}+n^{2}+p^{2}}.$ (50)

In vector notation this array becomes

$\vec{\Gamma}+\vec{q}\times\vec{r}=\vec{t}=\vec{q}\,\vec{\Gamma}\cdot\vec{q}/q^{2}=\vec{q}\,\vec{\delta}\cdot\vec{q}/q^{2},$
(51)

where the first equality is simply a rearrangement of (29) taking account that
both $\vec{r}$ and $\vec{W}$ are arbitrary points on the central axis, and
also using (30); the second equality follows from (30) which gives
$\vec{q}\,\vec{\Gamma}\cdot\vec{q}=q^{2}\hat{t}\,\vec{\Gamma}\cdot\hat{t}$;
and the third from (39) with (38).

This can be simplified by eliminating the constants $A,B,C$; we find (using
(15))

$\frac{x-(1/2)\alpha-\frac{p\beta-n\gamma}{m^{2}+n^{2}+p^{2}}}{m}=\frac{y-(1/2)\beta-\frac{m\gamma-p\alpha}{m^{2}+n^{2}+p^{2}}}{n}=\frac{z-(1/2)\gamma-\frac{n\alpha-m\beta}{m^{2}+n^{2}+p^{2}}}{p}.$
(52)

To derive this, we combine (51) with (34), and eliminating $\vec{\Gamma}$,we
have

$\vec{\delta}-(1/2)\vec{q}\times\vec{\delta}+\vec{q}\times\vec{r}=\vec{q}\,\vec{\delta}\cdot\vec{(}q)/q^{2}$
(53)

which can be rearranged, using the identity
$(\vec{\delta}\times\vec{q})\times\vec{q}=\vec{q}\vec{\delta}\cdot\vec{q}-q^{2}\vec{\delta}$,
to yield

$(\vec{r}-\vec{r_{0}})\times\vec{q}=0$ (54)

where

$\vec{r}_{0}=(1/2)\vec{\delta}-\vec{\delta}\times\vec{q}/q^{2};$ (55)

these two equations are exactly the content of (52), which says that
$\vec{r}-\vec{r}_{0}$ is parallel to $\vec{q}$ when $\vec{r}_{0}$ is given by
(55).

In these equations the coordinates subtracted from $x,y,z$ (that is, the
components of $\vec{r}_{0}$) are precisely those of the vertex of an isosceles
triangle normal to the plane of the two relative axes of translation and of
rotation passing through the origin of the coordinate axes, raised on a base
perpendicular to the relative axis of rotation, cutting at its midpoint the
relative axis of translation and equal in length to the translation itself,
the vertex angle being equal to the amplitude of rotation $\theta$; all this
in agreement with the construction given previously.

I find this description difficult to visualize. Again I resort to the
equations to find the geometric meaning of $\vec{r}_{0}$.

From (39), (45), and (46), let us extract the half-angle formula for
$\vec{\delta}$:

$\vec{\delta}=\vec{t}+2\vec{U}\sin^{2}(\theta/2)-2U\hat{V}\sin(\theta/2)\cos(\theta/2).$
(56)

Then, since $\vec{q}/q^{2}=\hat{t}/(2\tan(\theta/2))$,

$\displaystyle[\vec{\delta}\times\vec{q}/q^{2}$ $\displaystyle=$
$\displaystyle\vec{\delta}\times\hat{t}/(2\tan(\theta/2))$ (57)
$\displaystyle=$
$\displaystyle[2\vec{U}\sin^{2}(\theta/2)-2U\hat{V}\sin(\theta/2)\cos(\theta/2)]\times\hat{t}/(2\tan(\theta/2))$
$\displaystyle=$
$\displaystyle[\vec{U}\sin(\theta/2)\cos(\theta/2)-U\hat{V}\cos^{2}(\theta/2)]\times\hat{t}$
$\displaystyle=$
$\displaystyle-U\hat{V}\sin(\theta/2)\cos(\theta/2)-\vec{U}\cos^{2}(\theta/2)$

Now (57) and (55) combine harmoniously to yield

$\displaystyle\vec{r}_{0}$ $\displaystyle=$
$\displaystyle(1/2)\vec{\delta}-[-U\hat{V}\sin(\theta/2)\cos(\theta/2)-\vec{U}\cos^{2}(\theta/2)]$
(58) $\displaystyle=$
$\displaystyle(1/2)[\vec{t}+2\vec{U}\sin^{2}(\theta/2)-2U\hat{V}\sin(\theta/2)\cos(\theta/2)]+[U\hat{V}\sin(\theta/2)\cos(\theta/2)+\vec{U}\cos^{2}(\theta/2)]$
$\displaystyle=$
$\displaystyle(1/2)\vec{t}+\vec{U}[\sin^{2}(\theta/2)+\cos^{2}(\theta/2)]+U\hat{V}(\sin(\theta/2)\cos(\theta/2)-\sin(\theta/2)\cos(\theta/2)]$
$\displaystyle=$ $\displaystyle(1/2)\vec{t}+\vec{U}$

saying that $\vec{r}_{0}$ is the midpoint of the line joining the point
$\vec{U}$, the point on the central axis closest to the origin, to the point
$\vec{U}+\vec{\Delta}=\vec{U}+\vec{t}$ to which the point $\vec{U}$ is carried
by the displacement. Under NCA, $\vec{U}=0$ and $r_{0}=\vec{t}/2$.

These equations (referring to (52)) can also be written as follows, on
introducing the rotation angle and the direction of the axis of rotation:

$\displaystyle\frac{x-(1/2)\alpha-(1/2)\cot(\theta/2)(\beta\cos l-\gamma\cos
h)}{\cos g}$ $\displaystyle=\frac{y-(1/2)\beta-(1/2)\cot(\theta/2)(\gamma\cos
g-\alpha\cos l)}{\cos h}$
$\displaystyle=\frac{z-(1/2)\gamma-(1/2)\cot(\theta/2)(\alpha\cos h-\beta\cos
g)}{\cos l}$ (59)

These equations follow immediately from (52) by using (15). In modern notation
they become just another way of saying that $\vec{r}_{0}$ is given by (55).

###  16.1 General equation of the central axis

.

But one may represent these three equations for the projections of the central
axis (it is not clear which three equations are meant, since (16) consists of
only two equations) by a single equation with undetermined coefficients,
namely

$\cos a\Delta x+\cos b\Delta y+\cos c\Delta z=t\cos(t,a,b,c)$ (60)

where

$\cos(t,a,b,c)=\cos a\cos g+\cos b\cos h+\cos a\cos l,$ (61)

$a,b,c$ being the three angles formed by any direction whatever with the
coordinate axes. That is, if $\hat{d}=\cos a,\cos b,\cos c$ is a unit vector
in any direction whatsoever, then
$\vec{\Delta}\cdot\hat{d}=t\hat{d}\cdot\hat{t}=\vec{t}\cdot\hat{d}$. But this
is tantamount to saying that $\vec{\Delta}=\vec{t}$, which is just the
property of initial points $\vec{r}$ that lie on the central axis.

##  17 The case of an infinitesimal displacement.

If one considers only an infinitesimal displacements, that is where $\Delta
x,\Delta y,\Delta z$ are infinitely small and can be replaced by $dx,dy,dz$
and correspondingly the constants $\alpha,\beta,\gamma,\theta$ are small of
the same order, then, neglecting terms of the second or higher order, the
equations (15) and (15) reduce to

$dx=\alpha+py-nz,\;\;\;\;\;dy=\beta+mz-px,\;\;\;\;\;dz=\gamma+nx-my;$ (62)

$m=\theta\cos g,\;\;\;\;\;n=\theta\cos h,\;\;\;\;\;p=\theta\cos l;$ (63)

and for the equations of the central axis,

$\frac{\theta x+\gamma\cos h-\beta\cos l}{\cos g}=\frac{\theta y+\alpha\cos
l-\gamma\cos g}{\cos h}=\frac{\theta z+\beta\cos g-\alpha\cos h}{\cos l}.$
(64)

## 18

Let us return to the general formulas (15). From them one derives

$(\vec{\Delta}-\vec{\delta})^{2}=4\tan^{2}(\theta/2)\;[(\vec{\omega}-(1/2)\vec{\delta})^{2}-((\vec{\omega}-(1/2)\vec{\delta})\cdot\hat{t})^{2}].$
(65)

I shall henceforth omit some of the more complex equations given in the
original text, supplying only my transcription into modern notation. The
derivation of (65) may proceed as follows: (15) can be transcribed to

$\vec{\Delta}=\vec{\delta}+2\tan(\theta/2)\hat{t}\times(\vec{\omega}-(1/2)\vec{\delta})$
(66)

where $\vec{\delta}$ is given by (47). To see how (66) leads to (65), let us
first define $\vec{\,\omega}^{\prime}=\vec{\omega}-(1/2)\vec{\delta}$. Then
(66) can be written

$\vec{\Delta}=\vec{\delta}+2\tan(\theta/2)\hat{t}\times\vec{\,\omega}^{\prime}$
(67)

which leads immediately to

$(\vec{\Delta}-\vec{\delta})^{2}=4\tan^{2}(\theta/2)\;(\hat{t}\times\vec{\,\omega}^{\prime})^{2},$
(68)

whereas (65) becomes

$(\vec{\Delta}-\vec{\delta})^{2}=4\tan^{2}(\theta/2)\;[(\vec{\,\omega}^{\prime
2}-(\vec{\,\omega}^{\prime}\cdot\hat{t})^{2}].$ (69)

But (68) and (69) are identical, from the familiar identity
$(\vec{a}\times\vec{b})^{2}=a^{2}b^{2}-(\vec{a}\cdot\vec{b})^{2}$. Thus (65)
is a consequence of previously derived formulas.

To the equation (69) (which gives the magnitude of
$\vec{\Delta}-\vec{\delta}$) one may adjoin the two equations

$(\vec{\Delta}-\vec{\delta})\cdot\hat{t}=0$ (70)

$(\vec{\Delta}-\vec{\delta})\cdot\vec{\,\omega}^{\prime}=0$ (71)

(which give its direction, normal to both $\hat{t}$ and
$\vec{\,\omega}^{\prime}$; these equations follow directly from (67).)

But now let us replace $\vec{\omega}$ by its value
$\vec{r}+(1/2)\vec{\Delta}$. We thus obtain from (67)

$\vec{\Delta}-\vec{\delta}=2\tan(\theta/2)\hat{t}\times\vec{r}+\tan(\theta/2)\hat{t}\times(\vec{\Delta}-\vec{\delta})$
(72)

(I have dissected my own composite $\vec{\,\omega}^{\prime}$ as well as making
the author’s substitution) and from this in turn

$\vec{\Delta}=\vec{\delta}+(\sin\theta)\hat{t}\times\vec{r}+2\sin^{2}(\theta/2)[\hat{t}\vec{r}\cdot\hat{t}-\vec{r}],$
(73)

of which only the first two terms will survive on passing from finite to
infinitesimal displacements.

In order to derive (73), one must substitute (72) into itself as follows:
Evaluate

$\displaystyle\hat{t}\times(\vec{\Delta}-\vec{\delta})$ $\displaystyle=$
$\displaystyle\hat{t}\times[2\tan(\theta/2)\hat{t}\times\vec{r}]+\hat{t}\times[\tan(\theta/2)\hat{t}\times(\vec{\Delta}-\vec{\delta})]$
(74) $\displaystyle=$ $\displaystyle
2\tan(\theta/2)[\hat{t}\hat{t}\cdot\vec{r}-\vec{r}]+\tan(\theta/2)[\hat{t}\hat{t}\cdot(\vec{\Delta}-\vec{\delta})-(\vec{\Delta}-\vec{\delta})]$
$\displaystyle=$ $\displaystyle
2\tan(\theta/2)[\hat{t}\hat{t}\cdot\vec{r}-\vec{r}]+\tan(\theta/2)[0-(\vec{\Delta}-\vec{\delta})].$

Now substitute this expression into the last term of (72):

$\vec{\Delta}-\vec{\delta}=2\tan(\theta/2)\hat{t}\times\vec{r}+2\tan^{2}(\theta/2)(\hat{t}\hat{t}\cdot\vec{r}-\vec{r})-\tan^{2}(\theta/2)(\vec{\Delta}-\vec{\delta}).$
(75)

Transposing the last term to the left side of the equation, and noting that
$1+\tan^{2}(\theta/2)=1/\cos^{2}(\theta/2)$, we find

$(\vec{\Delta}-\vec{\delta})/\cos^{2}(\theta/2)=2\tan(\theta/2)\hat{t}\times\vec{r}+2\tan^{2}(\theta/2)(\hat{t}\hat{t}\cdot\vec{r}-\vec{r})$
(76)

Finally multiplying the equation by $\cos^{2}(\theta/2)$, we obtain(73).

###  18.1 Expressions for finite displacements as rational functions of
$\vec{\delta}$ and $\vec{q}$

If one retains in these formulas the primitive constants $m,n,p$, and directly
extracts the values of $\Delta x,\Delta y,\Delta z$…, (the author now gives
formulas which I shall abbreviate by recalling that these three $\Delta$’s
form a single vector $\vec{\Delta}$, that $\alpha,\beta,\gamma$ form the
vector $\vec{\delta}$, and that $m,n,p$ are the components of
$\vec{q}=q\hat{t}$, and which I shall derive as follows: Let us write all the
trigonometric functions in (76) in terms of $\tan(\theta/2)$, thus:

$\vec{\Delta}-\vec{\delta}=[2\tan(\theta/2)\hat{t}\times\vec{r}+2\tan^{2}(\theta/2)(\hat{t}\hat{t}\cdot\vec{r}-\vec{r})]/(1+\tan^{2}(\theta/2)).$
(77)

Then recalling the definition (30), we have

$\vec{\Delta}=\vec{\delta}+[\vec{q}\times\vec{r}+(1/2)(\vec{q}\vec{q}\cdot\vec{r}-q^{2}\vec{r})](1+(1/4)q^{2})$
(78)

in which the displacement $\vec{\Delta}$ is given as a _rational_ function of
$\vec{\delta}$ and $\vec{q}$. That is, in the words of the author),

one has the following expressions (for $\Delta x,\Delta y,\Delta z$) in
rational functions of the six constants $\alpha,\beta,\gamma,m,n,p$: (the
expressions he gives are essentially those of (78) written out in components.)

###  18.2 Important consequence relating to the formulas for transforming
rectangular coordinates.

Comparing these expressions with those that one would obtain by considering
the transformation of rectangular coordinates, as this will be indicated in
26, one obtains a way of reducing the nine coefficients that enter into the
formulas for this transformation to three independent variables $m,n,p$,
entirely free of irrationals, which I believe has not been given before.

Until now,the author has been occupied with calculating $\vec{\Delta}$, the
straight-line displacement from an initial point $\vec{r}$ to its destination
$\vec{\,r}^{\prime}$. He now turns his attention to the _transformation_ of
$\vec{r}$ to $\vec{\,r}^{\prime}$, in the case of a pure rotation. Nowadays we
think of this transformation as given by a $3\times 3$ matrix transforming
$(x,y,z)$ to $(x^{\prime},y^{\prime},z^{\prime})$. The author frames this in a
complementary way as making a transformation of the _coordinate system_
$\hat{x},\hat{y},\hat{z}$, such that if $\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$
then
$\vec{\,r}^{\prime}=x\hat{\,x}^{\prime}+y\hat{\,y}^{\prime}+z\hat{,z}^{\prime}$
\- the coordinates, not the components, being altered. The “nine coefficients”
he speaks of are the nine elements of the matrix

$(M)=\left(\begin{array}[]{ccc}\hat{x}\cdot\hat{x}^{\prime}&\hat{y}\cdot\hat{x}^{\prime}&\hat{z}\cdot\hat{x}^{\prime}\\\
\hat{x}\cdot\hat{y}^{\prime}&\hat{y}\cdot\hat{y}^{\prime}&\hat{z}\cdot\hat{y}^{\prime}\\\
\hat{x}\cdot\hat{z}^{\prime}&\hat{y}\cdot\hat{z}^{\prime}&\hat{z}\cdot\hat{z}^{\prime}\end{array}\right).$

The author gives formulas for these nine elements which have three notable
features: (1) they involve no input other than the three parameters $m,n,p$;
(2) they contain no irrational expressions; and (3) the three parameters are
completely independent. He simply lists these nine formulas; I shall display
them in matrix form, but I shall write each matrix element exactly as it
appears in the original text except that in order to save space I shall write
$mn/2$ rather than $\frac{mn}{2}$, etc., and I shall place the common
denominator $1+(m^{2}+n^{2}+p^{2})/4$ outside the matrix as a prefix. Here is
the result:

$(M)=[1+(m^{2}+n^{2}+p^{2})/4]^{-1}\left(\begin{array}[]{ccc}\
1+(m^{2}-n^{2}-p^{2})/4&mn/2-p&pm/2+n\\\
mn/2+p&1+(n^{2}-p^{2}-m^{2})/4&np/2-m\\\
pm/2-n&np/2+m&1+(p^{2}-m^{2}-n^{2})/4\end{array}\right).$

On eliminating $m,n,p$ from the (off-diagonal) coefficients, one obtains the
formulas of Monge, in irrational functions, for the three (diagonal)
coefficients.

It is of some interest to decompose $(M)=(M)_{1}+(M)_{2}+(M)_{3}$ where

$(M)_{1}=[1-(m^{2}+n^{2}+p^{2})/4]/[1+(m^{2}+n^{2}+p^{2})/4]\left(\begin{array}[]{ccc}\
1&0&0\\\ 0&1&0\\\ 0&0&1\end{array}\right),$

$(M)_{2}=[1+(m^{2}+n^{2}+p^{2})/4]^{-1}\left(\begin{array}[]{ccc}\
m^{2}/2&mn/2&pm/2\\\ mn/2&n^{2}/2&np/2\\\
pm/2&np/2&p^{2}/2\end{array}\right),$

$(M)_{3}=[1+(m^{2}+n^{2}+p^{2})/4]^{-1}\left(\begin{array}[]{ccc}\ 0&-p&n\\\
p&0&-m\\\ -n&m&0\end{array}\right).$

Passing over to modern vector notation, this gives us

$\vec{\,r}^{\prime}=(M)\cdot\vec{r}=(1+q^{2}/4)^{-1}[(1-q^{2}/4)\vec{r}+(1/2)\vec{q}\vec{q}\cdot\vec{r}+\vec{q}\times\vec{r}].$
(79)

Now if we readmit the trigonometric functions via (30), we have

$[1-q^{2}/4]/(1+q^{2}/4)=(1-\tan^{2}(\theta/2))/(1+\tan^{2}(\theta/2))=\cos\theta,$
(80)

$(1/2)\vec{q}\vec{q}/(1+q^{2}/4)=2\hat{t}\hat{t}\tan^{2}(\theta/2)/\sec^{2}(\theta/2)=2\hat{t}\hat{t}\sin^{2}(\theta/2)=\hat{t}\hat{t}(1-\cos\theta),$
(81)

$\vec{q}/(1+q^{2}/4)=2\hat{t}\tan(\theta/2)\cos^{2}(\theta/2)=\hat{t}\sin\theta;$
(82)

and so we arrive at the celebrated Rodrigues rotation formula

$\vec{\,r}^{\prime}=\vec{r}\cos\theta+\hat{t}\hat{t}\cdot\vec{r}(1-\cos\theta)+\hat{t}\times\vec{r}\sin\theta.$
(83)

##  19 On the composition of two displacements.

The following three paragraphs summarize the entire essay, looking backward to
the beginning as well as forward to the end.

From a geometric consideration of the displacement of a solid system, we began
(this essay) by deducing the characteristic properties or general laws of the
displacement, which always reduces to a rotation followed by a translation, or
equivalently to a single couple of rotations about two fixed axes. (The two
axes may be either parallel or not parallel; in the first case the
displacement reduces to a simple translation, provided that the two rotations
are equal and in opposite directions.)

From these properties we have now derived the analytic expression for the
transformation, either finite or infinitely small, of the coordinates of a
solid system undergoing an arbitrary displacement.

The author seems to be referring here to the formula for the matrix $(M)$,
which turned out in modern notation to be the rotation formula (83). But in
the reasoning to follow, he works not from (83) but from the earlier formula
(49), which gives the simple displacement $\Delta$ in terms of $\vec{\delta}$
and $\vec{q}$. One readily sees that if one sets $\vec{q}=0$ one obtains
$\Delta=\delta$, which is a pure translation since the components
$\alpha,\beta,\gamma$ of $\delta$ do not depend on the location of $\vec{r}$;
and that if $\delta=0$ then $\Delta=\vec{\omega}\times\vec{r}$, which
describes a pure rotation since it vanishes at the origin $\vec{r}=0$.

We shall now deduce, from this expression, the laws of composition of
rotations and of translations that we previously exhibited synthetically. And
finally we shall establish these same formulas directly, by three distinct
analytic procedures, making use exclusively of the invariance of the distances
between points of this system.

Let us designate by $\vec{\Delta^{\prime}},\vec{\Delta^{\prime\prime}}$ the
two successive changes of positions of a point in the displaced system, and by
$\vec{\Delta}$ the resultant change. Likewise by
$\vec{\,\omega}^{\prime},\vec{\,\omega}^{\prime\prime}$ the positions of the
midpoints of the two straight-line displacements of the point, and by
$\vec{\omega}$ the midpoint of the resultant displacement. Thus one has

$\vec{\,\omega}^{\prime}=\vec{r}+(1/2)\vec{\Delta^{\prime}},$ (84)

$\vec{\,\omega}^{\prime\prime}=\vec{r}+\vec{\Delta^{\prime}}+(1/2)\vec{\Delta^{\prime\prime}},$
(85)

and

$\vec{\omega}=\vec{r}+(1/2)\Delta\vec{r}.$ (86)

Moreover,

$\vec{\Delta}=\vec{\Delta^{\prime}}+\vec{\Delta^{\prime\prime}}$ (87)

I am routinely putting all equations in modern form.

Furthermore, let us designate by $\hat{t}^{\prime},\hat{t}^{\prime\prime}$ the
absolute translations, and by $\theta^{\prime},\theta^{\prime\prime}$ and
$\hat{q}^{\prime},\hat{q}^{\prime\prime}$ the rotations and the directions of
the rotation axes of each of the two consecutive displacements under
consideration; and by $\hat{T},\Theta.\hat{Q}$ the analogous elements of the
composite displacement.

First we shall examine separately the composition of simple translations and
that of rotations without translation. If we suppose that the displacements
consist purely of translations, we have

$\Delta^{\prime}=\vec{t}^{\prime},\Delta^{\prime\prime}=\vec{\,t^{\prime\prime}},\Delta=\vec{T}=\vec{t^{\prime}}+\vec{\,t^{\prime\prime}},$
(88)

from which we see that the composite translation is nothing other than the
third side of the triangle formed by the successive passages of a point of the
system by reason of the two given translations. To compose more than two given
translations, one may easily generalize from the triangle to the polygon of
translations.

Now consider the composition of two rotations, without translation, about two
axes that intersect at the origin of coordinates. We shall have

$\displaystyle\vec{\Delta^{\prime}}$ $\displaystyle=$
$\displaystyle\vec{\,q}^{\prime}\times\vec{\,\omega}^{\prime},$
$\displaystyle\vec{\Delta^{\prime\prime}}$ $\displaystyle=$
$\displaystyle\vec{\,q}^{\prime\prime}\times\vec{\,\omega}^{\prime\prime},$
$\displaystyle\vec{\Delta}$ $\displaystyle=$
$\displaystyle\vec{\,Q}\times\vec{\,\omega},$ (89)

where

$\displaystyle\vec{q^{\prime}}$ $\displaystyle=$
$\displaystyle\hat{q}^{\prime}2\tan(\theta^{\prime}/2),$
$\displaystyle\vec{q^{\prime\prime}}$ $\displaystyle=$
$\displaystyle\hat{q}^{\prime\prime}2\tan(\theta^{\prime\prime}/2),$
$\displaystyle\vec{Q}$ $\displaystyle=$
$\displaystyle\hat{Q}^{\prime}2\tan(\Theta/2),$ . (90)

To understand these equations, consider the third line of (19). The part of
$\omega$ parallel to $\vec{Q}$ does not contribute to
$\vec{Q}\times\vec{\,\omega}$. Therefore the expression
$\;\hat{Q}\times\vec{\,\omega}$ simply rotates $\vec{\,\omega}$ through a
right angle in the plane perpendicular to $\vec{Q}$. This gives a vector with
the right direction (along $\vec{\Delta}$) but the wrong magnitude. The
magnitude of $\vec{\Delta}$ is obtained by inserting a factor
$2\tan(\Theta/2)$, that is by replacing $\hat{Q}$ with $\vec{Q}$ as indicated
in the third line of (19). Likewise for the first and second lines of these
equations.

In all these discussions $\hat{q}$ is the direction of the rotation axis, and
$\hat{t}$ of the translation. But sometimes, as in 18, it is assumed that one
is dealing with the central axis, so that $\hat{q}=\hat{t}$, Now, however, we
have two partial rotation axes intersecting at $0$, which does not necessarily
lie on the central axis of either partial displacement; therefore $\hat{q}$
must be distinguished from $\hat{t}$, for each partial displacement and for
the resultant. Indeed, for the remainder of this Section the translations
$\vec{t}^{\prime},\vec{t}^{\prime\prime},\vec{T}$ are taken as zero, so that
$\hat{t}^{\prime},\hat{t}^{\prime\prime},\hat{T}$ are indeterminate.

We need to determine the composite parameters $\Theta,\vec{Q}$ as functions of
the partial ones $\theta^{\prime},\vec{\,q}^{\prime}$ and
$\theta^{\prime\prime},\vec{\,q}^{\prime\prime}$. It will be useful to begin
by eliminating the variables
$\vec{\,\omega}^{\prime},\vec{\,\omega}^{\prime\prime}$ from (19), using the
relations (84) through (87). This elimination is the key to the author’s
solution of the problem posed. He writes down the elimination formulas without
showing a derivation; here I shall present a geometrical derivation.

Let $A,B,C$ denote respectively the points whose position vectors are
$\vec{r},\vec{r}+\vec{\Delta^{\prime}},\vec{r}+\vec{\Delta}$. Then
$\vec{\,\omega}^{\prime},\vec{\,\omega}^{\prime\prime},\vec{\omega}$ are the
position vectors of the midpoints of $AB,BC,AC$. To eliminate
$\vec{\,\omega}^{\prime}$, consider the triangle formed by
$A,\vec{\,\omega}^{\prime},\vec{\omega}$. It is similar to $ABC$ but half as
large in linear dimension. Therefore the vector
$\vec{\,\omega}^{\prime}-\vec{\omega}$ is just half the vector from $C$ to $B$
which is $-\vec{\Delta^{\prime\prime}}$. So we have

$\vec{\,\omega}^{\prime}=\vec{\omega}-(1/2)\vec{\Delta^{\prime\prime}}.$ (91)

To eliminate $\vec{\,\omega}^{\prime\prime}$, consider the triangle formed by
$\vec{\omega},\vec{\,\omega}^{\prime\prime},C$. It is also similar to $ABC$
but half as large in linear dimension. Therefore the vector
$\vec{\,\omega}^{\prime\prime}-\vec{\omega}$ is just half the vector from $A$
to $B$ which is $+\vec{\Delta^{\prime}}$. So we have

$\vec{\,\omega}^{\prime\prime}=\vec{\omega}+(1/2)\vec{\Delta^{\prime}}$ (92)

Note the opposite placement of ′ and ′′, as well as the change of sign of the
displacement term between (91) amd (92). These features will have important
consequences later on.

Substituting (91) and (92) into (19), we obtain

$\vec{\Delta^{\prime}}=\vec{\,q}^{\prime}\times\vec{\omega}-(1/2)\;\vec{\,q}^{\prime}\times\vec{\Delta^{\prime\prime}}$
(93)

$\vec{\Delta^{\prime\prime}}=\vec{\,q}^{\prime\prime}\times\vec{\omega}+(1/2)\;\vec{\,q}^{\prime\prime}\times\vec{\Delta^{\prime}},$
(94)

from which we deduce the following values for the partial displacements: The
author writes down the “following values” without showing the derivation,
which is far from trivial. It will be observed that (93) and (94) constitute a
system of two linear equations in the two unknowns $\vec{\Delta^{\prime}}$ and
$\vec{\Delta^{\prime\prime}}$. They can be solved by substituting each one
into the other.

Substituting (94) into (93), we have

$\displaystyle\vec{\Delta^{\prime}}$ $\displaystyle=$
$\displaystyle\vec{\,q}^{\prime}\times\vec{\omega}-(1/2)\;\vec{\,q}^{\prime}\times[\vec{\,q}^{\prime\prime}\times\vec{\omega}]+(1/2)\;\vec{\,q}^{\prime\prime}\times\vec{\Delta^{\prime}}$
(95) $\displaystyle=$
$\displaystyle\vec{a}^{\prime}+\vec{b}^{\prime}-(1/4)\;\vec{\,q}^{\prime}\times[\vec{\,q}^{\prime\prime}\times\vec{\Delta^{\prime}}]$

where

$\vec{a}^{\prime}=\vec{\,q}^{\prime}\times\vec{\omega},\;\;\vec{b}^{\prime}=-(1/2)\;\vec{\,q}^{\prime}\times[\vec{\,q}^{\prime\prime}\times\vec{\omega}]$
(96)

and substituting in the reverse direction,

$\displaystyle\vec{\Delta^{\prime\prime}}$ $\displaystyle=$
$\displaystyle\vec{\,q}^{\prime\prime}\times\vec{\omega}+(1/2)\;\vec{\,q}^{\prime\prime}\times[\vec{\,q}^{\prime}\times\vec{\omega}]-(1/2)\;\vec{\,q}^{\prime}\times\vec{\Delta^{\prime\prime}}$
(97) $\displaystyle=$
$\displaystyle\vec{a}^{\prime\prime}+\vec{b}^{\prime\prime}-(1/4)\;\vec{\,q}^{\prime\prime}\times[\vec{\,q}^{\prime}\times\vec{\Delta^{\prime\prime}}]$

where

$\vec{a}^{\prime\prime}=\vec{\,q}^{\prime\prime}\times\vec{\omega},\;\;\vec{b}^{\prime\prime}=+(1/2)\;\vec{\,q}^{\prime\prime}\times[\vec{\,q}^{\prime}\times\vec{\omega}].$
(98)

The exchange of order of the two partial displacements carries $a^{\prime}$
into $a^{\prime\prime}$ but $b^{\prime}$ into $-b^{\prime\prime}$; this change
of sign is inherited from the one between (91) and (92). We can express the
change by rewriting
$b^{\prime}=+(1/2)\;\vec{\,q}^{\prime}\times[\vec{\omega}\times\vec{\,q}^{\prime\prime}]$.

In view of (87), we now have

$\displaystyle\vec{\Delta}$ $\displaystyle=$
$\displaystyle\vec{a}+\vec{b}-(1/4)[\vec{\,q}^{\prime}\times[\vec{\,q}^{\prime\prime}\times\vec{\Delta^{\prime}}]+\vec{\,q}^{\prime\prime}\times[\vec{\,q}^{\prime}\times\vec{\Delta^{\prime\prime}}]$
(99) $\displaystyle=$
$\displaystyle\vec{a}+\vec{b}+(1/4)\vec{\,q}^{\prime}\cdot\vec{\,q}^{\prime\prime}[\vec{\Delta^{\prime}}\
+\vec{\Delta^{\prime\prime}}]=\vec{a}+\vec{b}+(1/4)\vec{\,q}^{\prime}\cdot\vec{\,q}^{\prime\prime}\vec{\Delta}$

where we have used the fact that
$\vec{\,q}^{\prime}\cdot\Delta^{\prime}\vec{r}=\vec{\,q}^{\prime\prime}\cdot\Delta^{\prime\prime}\vec{r}=0$,
and where

$\vec{a}=\vec{a}^{\prime}+\vec{a}^{\prime\prime}=(\,\vec{\,q}^{\prime}+\,\vec{\,q}^{\prime\prime})\times\vec{\omega}$
(100)

and

$\vec{b}=\vec{b}^{\prime}+\vec{b}^{\prime\prime}=(1/2)\;[\vec{\,q}^{\prime}\times(\vec{\omega}\times\vec{\,q}^{\prime\prime})+\vec{\,q}^{\prime\prime}\times(\vec{\,q}^{\prime}\times\vec{\omega})]=-(1/2)\;\vec{\omega}\times(\vec{\,q}^{\prime\prime}\times\vec{\,q}^{\prime}).$
(101)

In the last step I have used the Jacobi identity for the cyclic sum of a
double cross product.

The last term in (99) can be transposed to the left side, giving
$\vec{\Delta}(1-(1/4)\;\vec{\,q}^{\prime}\cdot\vec{\,q}^{\prime\prime})=\vec{a}+\vec{b}$
or

$\vec{\Delta}=\frac{(\,\vec{\,q}^{\prime}+\,\vec{\,q}^{\prime\prime})\times\vec{\omega}+(\vec{\,q}^{\prime\prime}\times\vec{\,q}^{\prime})\times\vec{\omega}}{1-(1/4)\;\vec{\,q}^{\prime}\cdot\vec{\,q}^{\prime\prime}}.$
(102)

The appearance of the common factor $\vec{\omega}$ in (102) is the major fruit
of (91) and (92).

Now by comparison of this expression for $\Delta\vec{r}$ with the postulated
expression for the same quantity in terms of $\Theta,\hat{T}$, we arrive at
the following relation: The author wishes to compare (102) with the third
equation in (19). This comparison yields the equation

$\vec{Q}\times\vec{\omega}=\vec{s}\times\vec{\omega}$ (103)

where

$\vec{s}=\frac{\vec{\,q}^{\prime}+\vec{\,q}^{\prime\prime}+(1/2)\vec{\,q}^{\prime\prime}\times\vec{\,q}^{\prime}}{1-(1/4)\vec{\,q}^{\prime}\cdot\vec{\,q}^{\prime\prime}}.$
(104)

One must now remember that $\vec{Q}$, as well as the variables entering into
$\vec{s}$, are fixed for a particular displacement of the _whole solid_ ,
whereas $\vec{r}$, which enters into $\vec{\omega}$, can be _any point_ in the
solid. Therefore (103) holds for every possible $\vec{\omega}$. This implies
the author’s “following relation”, namely

$\vec{Q}=\vec{s},$ (105)

from which one deduces, for the value of the resultant rotation,

$\cos(\Theta/2)=\cos(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)-\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\cos\nu,$
(106)

where $\nu$ is the angle between the two axes of rotation, thus
$\cos\nu=\hat{q}^{\prime}\cdot\hat{q}^{\prime\prime}$.

The deduction may proceed as follows. First we replace the three $q$-vectors
by their definitions: generically, $\vec{q}=2\hat{q}\tan(\theta/2)$. Thus
$\tan^{2}(\Theta/2)=Q^{2}/4$, and

$\cos^{2}(\Theta/2)=(1+(Q^{2}/4))^{-1}.$ (107)

Meanwhile we may write $\vec{s}=2\vec{N}/D$, so that substituting (105) into
(107) we have

$\cos^{2}(\Theta/2)=(1+(s^{2}/4))^{-1}=\frac{D^{2}}{N^{2}+D^{2}}$ (108)

where

$\vec{N}=\vec{q}^{\prime}/2+\vec{\,q}^{\prime\prime}/2+\vec{q}^{\prime\prime}\times\vec{q}^{\prime}/4,$
(109)

$D=1-(1/4)\cos\nu.$ (110)

From (108) we see that if $N^{2}+D^{2}$ can be exhibited as a perfect square
we can obtain $\cos(\Theta/2)$ without radicals. The third term of (109) is
orthogonal to both of the first two; therefore

$\displaystyle N^{2}$ $\displaystyle=$
$\displaystyle[\vec{\,q}^{\prime}/2+\vec{\,q}^{\prime\prime}/2)]^{2}+(\vec{\,q}^{\prime\prime}\times\vec{\,q}^{\prime})^{2}]/16$
(111) $\displaystyle=$
$\displaystyle[\tan^{2}(\theta^{\prime}/2)\\!+\\!\tan^{2}(\theta^{\prime\prime}/2)+2\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2)\cos\nu]$
$\displaystyle+$
$\displaystyle\tan^{2}(\theta^{\prime}/2)\tan^{2}(\theta^{\prime\prime}/2)(1-\cos^{2}\nu);$

of course
$(\hat{q}^{\prime\prime}\times\hat{q}^{\prime})^{2}=\sin^{2}\nu=1-\cos^{2}\nu$.
At the same time (110) gives us

$D^{2}=1-2\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2)\cos\nu+\tan^{2}\theta^{\prime}/2)\tan^{2}(\theta^{\prime\prime}/2)\cos^{2}\nu.$
(112)

The terms in $\cos\nu$ and $\cos^{2}\nu$ obligingly cancel between (111) and
(112), leaving

$\displaystyle
N^{2}+D^{2}=1+\tan^{2}(\theta^{\prime}/2)+\tan^{2}(\theta^{\prime\prime}/2)+\tan^{2}(\theta^{\prime}/2)\tan^{2}(\theta^{\prime\prime}/2)$
$\displaystyle=(1+\tan^{2}(\theta^{\prime}/2))(1+\tan^{2}(\theta^{\prime\prime}/2))=1/[\cos^{2}(\theta^{\prime}/2)\cos^{2}(\theta^{\prime\prime}/2)].$
(113)

This enables us to take the square root of (108):

$\displaystyle\cos(\Theta/2)$ $\displaystyle=$
$\displaystyle\cos(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)(1-\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2)\cos\nu)$
(114) $\displaystyle=$
$\displaystyle\cos(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)-\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\cos\nu,$

in agreement with (106). This startling relation was deduced from (105) by the
author, who shows no intermediate steps in the text. One can only wonder at
his ability to navigate the maze of substitutions without the help of our
vector relations.

\- and for the inclination of the resultant axis,

$\hat{Q}\sin(\Theta/2)=\hat{q}^{\prime}\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)+\hat{q}^{\prime\prime}\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)+\hat{q}^{\prime\prime}\times\hat{q}^{\prime\prime}\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2).$
(115)

Recalling the definitions given after (19),

$\vec{Q}=2\hat{Q}\tan(\Theta/2),\;\;\vec{q\,}^{\prime}=2\hat{q}^{\prime\prime}\tan(\theta^{\prime}/2)\,\;\;\vec{q\,}^{\prime\prime}=2\hat{q}^{\prime\prime}\tan(\theta^{\prime\prime}/2),$
(116)

we can write (105) as

$\hat{Q}\tan(\Theta/2)=\frac{\hat{q}^{\prime}\tan(\theta^{\prime}/2)+\hat{q}^{\prime\prime}\tan(\theta^{\prime\prime}/2)+\hat{q}^{\prime\prime}\times\hat{q}^{\prime}\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2)}{1-\hat{q}^{\prime\prime}\cdot\hat{q}^{\prime}\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2))}$
(117)

and (106) as

$\cos(\Theta/2)=\cos(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)(1-\hat{q}^{\prime\prime}\cdot\hat{q}^{\prime}\tan(\theta^{\prime}/2)\tan(\theta^{\prime\prime}/2)).$
(118)

Multiplying these two equations, we obtain (115).

In these formulas one notices immediately that the order of succession of the
rotations $\theta^{\prime},\theta^{\prime\prime}$ has no effect on the
amplitude $\Theta$ of the resultant rotation (see (114)), but that it does
affect the direction of the axis of that rotation (see the last term of
(115)), unless the rotations $\theta^{\prime},\theta^{\prime\prime}$ are
infinitely small.

Now, the expression for $\cos(\Theta/2)$ holds if $\Theta/2$ is an angle of a
spherical triangle, of which the opposing side is $\nu$ and the two other
angles are $\theta^{\prime}/2,\theta^{\prime\prime}/2$. This situation is dual
to the one described by the usual spherical law of cosines, which gives a side
in terms of the opposite angle and the two other sides. That is why the
formula (114) has a minus sign before the second term instead of a plus sign.
To specify more precisely the position of the resultant axis in relation to
the two given axes, let us suppose, which is always possible, that

$\cos l^{\prime}=\cos l^{\prime\prime}=0,\cos h^{\prime}=0,\cos
g^{\prime}=1,\cos g^{\prime\prime}=\cos\nu,\;\cos h^{\prime\prime}=\sin\nu.$
(119)

Generally, the author has regarded the $x,y,z$ coordinate system as fixed and
independent of what displacement is being applied to the solid system.
Occasionally he has made slight departures: in the previous section he
discussed linear orthogonal transformations of the coordinates, and in the
present inquiry he has already specified that the origin of coordinates is at
the intersection of the two given (and hence also of the resultant) axes of
rotation. Now he chooses an orientation of the coordinate system completely
tailored to the problem at hand. He has the $x$-axis coinciding with the axis
of the rotation $\theta^{\prime}$, and the $z$-axis perpendicular to both
rotation axes $\theta,\theta^{\prime\prime}$. Thus the axis of rotation
$\theta^{\prime\prime}$ lies in the $x$-$y$ plane making an angle $\nu$ with
the $x$-axis. This will simplify his calculations, but at the cost of making
it impossible to apply modern vector notation to his formulas; they will have
to be written out in components as in the original text.

One now has

$\displaystyle\sin(\Theta/2)\cos G$ $\displaystyle=$
$\displaystyle\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)+\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)\cos\nu,$
$\displaystyle\sin(\Theta/2)\cos H$ $\displaystyle=$
$\displaystyle\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)\sin\nu,$
$\displaystyle\sin(\Theta/2)\cos L$ $\displaystyle=$
$\displaystyle\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\sin\nu.$
(120)

These equations can be obtained by resolving (115) into its $x$-,$y$-, and
$z$-components, and applying (119).

If the order of rotations is reversed, the only change in (19) is that the
sign of $\cos L$ becomes negative, from which it follows that the new
resultant axis is placed in a _symmetric_ position to the old one relative to
the plane of the two given axes.

This result can be obtained directly from (115) more easily than by struggling
through (19). The sum of the first two terms on the right side of (115) is
symmetric in $\hat{q}^{\prime}$ and $\hat{q}^{\prime\prime}$, and both lie in
the $\hat{q}^{\prime}$-$\hat{q}^{\prime\prime}$ plane; the last term is
antisymmetric and perpendicular to this plane. Therefore the interchange of
$\hat{q}^{\prime}$ with $\hat{q}^{\prime\prime}$ causes $\hat{Q}$ to be
reflected in the $\hat{q}^{\prime}$-$\;\hat{q}^{\prime\prime}$ plane.

If we denote by $H^{\prime}$ the angle formed by the resultant axis with the
axis of the rotation $\theta^{\prime\prime}$, we shall have

$\cos H^{\prime}=\cos G\cos g^{\prime\prime}+\cos H\cos
h^{\prime\prime}=\frac{\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)+\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)\cos\nu}{\sin(\Theta/2)}$
(121)

(the first equality comes from the decomposition
$\hat{t}^{\prime\prime}=\hat{x}\cos g^{\prime\prime}+\hat{y}\cos
h^{\prime\prime}$, the second by applying (119) to (19)); besides this we have

$\displaystyle\sin^{2}(\Theta/2)=\sin^{2}(\theta^{\prime}/2)\sin^{2}\nu+[\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)+\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)\cos\nu]^{2}$
(122) $\displaystyle=$
$\displaystyle\sin^{2}(\theta^{\prime\prime}/2)\sin^{2}\nu+[\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)+\sin(\theta^{\prime\prime}/2)\cos(\theta^{\prime}/2)\cos\nu]^{2}$

and consequently

$\sin^{2}G=\frac{\sin^{2}(\theta^{\prime\prime}/2)\sin^{2}\nu}{\sin^{2}(\Theta/2)},\;\;\sin^{2}H^{\prime}=\frac{\sin^{2}(\theta^{\prime}/2)\sin^{2}\nu}{\sin^{2}(\Theta/2)},$
(123)

equations that establish the proportionality of the sines of the half-
rotations to those of the angles formed by the resultant axis with the given
axes inversely corresponding (that is, $G$ to $\theta^{\prime\prime}$ and
$H^{\prime}$ to $\theta^{\prime}$), and which lead to the construction that we
have indicated from the outset for the composition of rotations.

The first line of (122) is obtained by summing the squares of the three
equations in (19). The second line follows from the first in view of (114)
which gives $\Theta$ as a symmetric function of $\theta^{\prime}$ and
$\theta^{\prime\prime}$. In regard to (123), note that $G$ bears the same
relation to $\theta^{\prime}$ as $H^{\prime}$ to $\theta^{\prime\prime}$, in
view of the condition $\cos g^{\prime}=1$ from (119). Of course the author
means to consider the square root of (123). The resulting formula is the one
already described in Section 6, corollary # 2.

If we were to follow an analogous procedure for the composition of rotations
about an arbitrary number of intersecting axes, the resulting formulas would
be rendered exceedingly complex by the terms of second order; hence we omit
that subject.

##  20 On the analytic composition of rotations about nonintersecting axes.