<p>The standard equation for aerodynamic forces is a half, times the density of air, times the speed squared, times the area, times an aerodynamic coefficient.</p>
<p><span class="math-container">$$L=\frac{1}{2}\cdot\rho\cdot v^2\cdot A\cdot\ C_L$$</span></p>
<p>You usually see it like this, for calculating Lift, but you can replace the 'L' with Drag or Moment if you replace the coefficient of drag or moment.</p>
<p>The half seems to just be tradition. Possibly the equation was originally derived by integrating for speed. It's quite handy though, because it means that the coefficient rarely exceeds 1.</p>
<p>You have to be careful with your units. Speed and density will use meters, so calculate your area in meters, not centimetres. 0.09 * 0.06 = 0.0054m^2</p>
<p>ρ (the Greek letter rho) is the density of air at sea level and is about 1.</p>
<p>250m/s is extremely fast for a model, but squared gives you 62,500</p>
<p>The coefficient varies depending on the aerofoil and angle of attack. There's more information <a href="https://aerotoolbox.com/lift-drag-moment-coefficient/" rel="nofollow noreferrer">here</a>. You could use 1 as a worst case or use the sample aerofoils in the chart at the end. I'll use 0.04. If you really think you'll reach transonic speeds, it could be higher.</p>
<p>That gives us 0.5 * 0.0054 * 1 * 62,500 * 0.04 = 6.75 Newton Meters.</p>
<p>A typical servo (Futaba S3115) has 3Kg-cm of torque (often mis-labelled as Kg/cm), which is 0.3 Nm. You can get <a href="https://www.modelsport.co.uk/product/372839" rel="nofollow noreferrer">high-torque servos</a> - this one has 16.2kg-cm, or 1.6NM</p>
<p>OTOH, if you are 'only' flying at 250mph (111m/s) the torque will be 1.3Nm</p>
<p>And you won't need 120 degrees of movement, so you could gain some mechanical advantage in the pushrod geometry.</p>

CC BY-SA 4.0
2022-11-17T12:51:13.850Z
Relevant question here: https://aviation.stackexchange.com/questions/51402/why-is-the-velocity-squared-in-the-lift-equation
Robin Bennett
CC BY-SA 4.0
2022-11-17T09:32:01.510Z
2022-11-17T12:50:36.947Z
2022-11-17T12:50:36.947Z
Robin Bennett
<p>I don't know that my question is belong to here or physics community.</p>
<p>I desire to make a supersonic micro-jet model plane also powered by model rocket engine to reaches less than Mach 3 and it has a pair of canard for pitch angle</p>
<p>THERE IS TWO CANARDS AND EACH CANARD HAS IT'S OWN SERVOMOTOR SO THERE IS TWO SERVOMOTOR FOR TWO CANARD (a servomotor for each canard).</p>
<p>The maximum angle of attack of canard is 15 degree and somehow, my the model plane can reaches 561 m/s (about mach 2)</p>
<p>The problem is:</p>
<p>How much torque do I need for rotating each canards?</p>
<p>Is there any formula to calculate it?</p>
<p>There is a picture of the canard and it's dimensions:</p>
<p><a href="https://i.sstatic.net/ZuGSV.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZuGSV.jpg" alt="enter image description here" /></a></p>
<pre><code>Root chord = 10 cm
Tip chord = 8 cm
Height = 6 cm
Sweep angle = 18.4 degrees
</code></pre>
<p>(Obviously leading edge is the leg of the shape which hasn't dimension)</p>
<p>The distance between the connection point of the servomotor shaft and the root chord from the leading edge of the canard is equal to the distance of the center of pressure from its leading edge.</p>
<p>How much torque do I need for rotating each canards?</p>
<p>Is there any formula to calculate it?</p>
<p>How much torque needs for 250 m/s (for it's normal speed)?</p>

CC BY-SA 4.0
2022-11-08T06:16:04.777Z
If the pivot point was at the canard's aerodynamic center, and the canard's aerodynamic center did not shift significantly when its AOA changed, then the I think the force required could potentially be quite small.
ThatCoolCoder
CC BY-SA 4.0
2022-11-08T15:03:36.550Z
And how can i calculate it for supersonic aerodynamic?
Curious guy
CC BY-SA 4.0
2022-11-10T16:09:58.173Z
Supersonic aerodynamics are not really something generall discussed on this SE, but as @ThatCoolCoder says, the minimal amount of torque required will be when the canard is aerodynamically balanced throughout its entire range. The rest of the torque can be calculated by measuring the amount of lift you expect the canard to generate, and the distance between that center of lift, and the center of rotation of the servo
Brydon Gibson
<aerodynamics><servo>
How much torque do I need for the servomotor which turns my canard?
Curious guy
# How much torque do I need for the servomotor which turns my canard?

(Asked by: username_0 on 2022-11-04)

I don't know that my question is belong to here or physics community.
I desire to make a supersonic micro-jet model plane also powered by model rocket engine to reaches less than Mach 3 and it has a pair of canard for pitch angle
THERE IS TWO CANARDS AND EACH CANARD HAS IT'S OWN SERVOMOTOR SO THERE IS TWO SERVOMOTOR FOR TWO CANARD (a servomotor for each canard).
The maximum angle of attack of canard is 15 degree and somehow, my the model plane can reaches 561 m/s (about mach 2)
The problem is:
How much torque do I need for rotating each canards?
Is there any formula to calculate it?
There is a picture of the canard and it's dimensions:

Root chord = 10 cm
Tip chord = 8 cm
Height = 6 cm
Sweep angle = 18.4 degrees

(Obviously leading edge is the leg of the shape which hasn't dimension)
The distance between the connection point of the servomotor shaft and the root chord from the leading edge of the canard is equal to the distance of the center of pressure from its leading edge.
How much torque do I need for rotating each canards?
Is there any formula to calculate it?
How much torque needs for 250 m/s (for it's normal speed)?

**Tags:** aerodynamics, servo
> **Comments:**
> **username_1**: If the pivot point was at the canard's aerodynamic center, and the canard's aerodynamic center did not shift significantly when its AOA changed, then the I think the force required could potentially be quite small.  
> **username_0**: And how can i calculate it for supersonic aerodynamic?  
> **username_2**: Supersonic aerodynamics are not really something generall discussed on this SE, but as @ThatCoolCoder says, the minimal amount of torque required will be when the canard is aerodynamically balanced throughout its entire range. The rest of the torque can be calculated by measuring the amount of lift you expect the canard to generate, and the distance between that center of lift, and the center of rotation of the servo  

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## 1 answers

### Answer by username_3 on 2022-11-17. Score: 1 | Accepted answer

The standard equation for aerodynamic forces is a half, times the density of air, times the speed squared, times the area, times an aerodynamic coefficient.

$$L=\frac{1}{2}\cdot\rho\cdot v^2\cdot A\cdot\ C\_L$$

You usually see it like this, for calculating Lift, but you can replace the 'L' with Drag or Moment if you replace the coefficient of drag or moment.

The half seems to just be tradition. Possibly the equation was originally derived by integrating for speed. It's quite handy though, because it means that the coefficient rarely exceeds 1.

You have to be careful with your units. Speed and density will use meters, so calculate your area in meters, not centimetres. 0.09 \* 0.06 = 0.0054m^2

ρ (the Greek letter rho) is the density of air at sea level and is about 1.

250m/s is extremely fast for a model, but squared gives you 62,500

The coefficient varies depending on the aerofoil and angle of attack. There's more information [here](https://aerotoolbox.com/lift-drag-moment-coefficient/). You could use 1 as a worst case or use the sample aerofoils in the chart at the end. I'll use 0.04. If you really think you'll reach transonic speeds, it could be higher.

That gives us 0.5 \* 0.0054 \* 1 \* 62,500 \* 0.04 = 6.75 Newton Meters.

A typical servo (Futaba S3115) has 3Kg-cm of torque (often mis-labelled as Kg/cm), which is 0.3 Nm. You can get [high-torque servos](https://www.modelsport.co.uk/product/372839) - this one has 16.2kg-cm, or 1.6NM

OTOH, if you are 'only' flying at 250mph (111m/s) the torque will be 1.3Nm

And you won't need 120 degrees of movement, so you could gain some mechanical advantage in the pushrod geometry.

> **Comments:**
> **username_3**: Relevant question here: https://aviation.stackexchange.com/questions/51402/why-is-the-velocity-squared-in-the-lift-equation  

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