<p>The yaw effect created by the same effect that would cause a helicopter to spin if it didn't have a tail rotor.</p>

<p>On a multirotor, half the propellers spin clockwise (CW) and half counter-clockwise (CCW). This 50/50 split evens out the rotational forces for straight and level flight. To yaw, these forces need to be unbalanced.  To spin CW for example, the CW motors spin faster and/or the CCW motors spin slower.</p>

<p>To minimise other movements, the CW and CCW motors alternate around the aircraft's frame. If all CW motors were on one side, a yaw motion would also cause the aircraft to tilt and move sideways.</p>

CC BY-SA 4.0
2020-04-17T11:32:34.173Z
2020-04-17T11:32:34.173Z
Kralc
<p>It’s all about inertia.</p>

<p>By changing the speed of the rotors spinning in one direction, due to the conservation of momentum, the quad moves in the other direction.</p>

<p>That’s quite a confusing way to put it, so imagine this:</p>

<p>You’re facing a friend, both of you are in an office chair with wheels. You put out your right hand and push off your friend’s left hand. Even though you’re pushing your friend to make them rotate, (like the motor spinning the propeller), you also end up rotating. You’re the drone, so by you pushing the propeller at different speeds, you yourself end up rotating.</p>

CC BY-SA 4.0
2020-07-19T01:36:47.307Z
But according to conservation of momentum, wouldn't the props spinning back up to normal speed (finishing the yaw movement) cause the quad to yaw the opposite way? Sorry I might not be understanding your answer.
Galaxy
CC BY-SA 4.0
2021-12-02T18:21:16.140Z
It would cause the _momentum_ or the _kinetic energy_ to be conserved, however (assuming no drag), the quadcopter can start the yaw, wait, and then stop the yaw (initial and final rotational velocities are the same), but in between these moments, the quadcopter has changed orientation. Also important to note rotational momentum is less of a factor here than the torque created by the propellers
Brydon Gibson
CC BY-SA 4.0
2020-04-17T11:16:25.227Z
2020-04-17T11:16:25.227Z
Drones and Whatnot
<p>As it transitions back from yawing one set of propellers are losing momentum while the other set of propellers are gaining until the momentum are both equal.  The one losing are the propellers that generated the yaw condition in the first place and the propellers gaining are the ones that lost momentum during the yaw.  When the yaw stops is when the pairs are now equal in momentum.</p>

CC BY-SA 4.0
2021-12-02T10:08:15.547Z
2021-12-02T10:08:15.547Z
Andrew Wise
<p>Any vehicle yaws (i.e. turns) by having a net torque applied. What's interesting about a quadcopter is not just how it yaws, but how it yaws and <em>doesn't roll, pitch, or climb at the same time</em>.</p>
<p>To understand how this works, we need to briefly look at the math. We'll use a plus configuration, but really any <a href="https://drones.stackexchange.com/a/323/46">mutirotor configuration works</a>.</p>
<p><a href="https://i.sstatic.net/bGidS.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bGidS.png" alt="enter image description here" /></a></p>
<p>The thing to keep in mind is that thrusts and torques are related to propeller speeds. If you speed a propeller up, it's intuitively obvious that it will create more thrust. And likewise, if you spin it faster, you need more torque. So changing motor speeds changes the net forces and torques on the airframe.</p>
<p>(Pedantically, it goes with the square of speed. So if you double the speed you quadruple the thrust and torque. But that's not important to this analysis.)</p>
<p>Here's the high-level driving equation. If you've ever messed around with mixers, you'll notice that the 4x4 matrix in the middle looks really familiar:</p>
<p><a href="https://i.sstatic.net/NDLu8.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NDLu8.gif" alt="Reaction speed relation" /></a></p>
<p>What this does is it maps rotor speeds (squared) to torques about the roll, pitch, and yaw axes, as well as the net vertical thrust.</p>
<p>For hover, let's assume all motors are spinning at the same speed, <code>W</code>. So <code>W = w1 = w2 = w3 = w4</code></p>
<h3>Yaw</h3>
<p>What happens if we <em>speed up the first and third and slow down the second and fourth</em> by the same (squared) amount <code>dW</code>?</p>
<pre><code>torque_x = 0*(W^2 + dW) + 1*(W^2 - dW) + 0*(W^2 + dW) - 1*(W^2 + dW) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 - dW) - 1*(W^2 + dW) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + dW) - 1*(W^2 - dW) + 1*(W^2 + dW) - 1*(W^2 + dW) = 4*dW
F_z      = 1*(W^2 + dW) + 1*(W^2 - dW) + 1*(W^2 + dW) + 1*(W^2 + dW) = 4*W^2
</code></pre>
<p>So the net force doesn't change (all the <code>dW</code> cancel out), and neither do the net rolls and pitches, but voila we have <code>4*dW</code> worth of torque!</p>
<hr />
<p>For completeness, here's what happens when you want to change the other axes as well.</p>
<h2>Pitch</h2>
<p>Let's change the front and back motors by the same (squared) speed, but we'll leave the other two motors alone:</p>
<pre><code>torque_x = 0*(W^2 + 0) + 1*(W^2 - dW) + 0*(W^2 + 0) - 1*(W^2 + dW) = 2*dW
torque_y = 1*(W^2 + 0) + 0*(W^2 - dW) - 1*(W^2 + 0) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + 0) - 1*(W^2 - dW) + 1*(W^2 + 0) - 1*(W^2 + dW) = 0
F_z      = 1*(W^2 + 0) + 1*(W^2 - dW) + 1*(W^2 + 0) + 1*(W^2 + dW) = 4*W^2
</code></pre>
<p>Notice that, again, z-thrust stays constant, but this time only a pitching torque appears.</p>
<h2>Roll</h2>
<p>Let's change the left and right motors by the same (squared) speed, but we'll leave the other two motors alone:</p>
<pre><code>torque_x = 0*(W^2 + dW) + 1*(W^2 + 0) + 0*(W^2 + dW) - 1*(W^2 + 0) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 + 0) - 1*(W^2 + dW) + 0*(W^2 + 0) = 2*dW
torque_z = 1*(W^2 + dW) - 1*(W^2 + 0) + 1*(W^2 + dW) - 1*(W^2 + 0) = 0
F_z      = 1*(W^2 + dW) + 1*(W^2 + 0) + 1*(W^2 + dW) + 1*(W^2 + 0) = 4*W^2
</code></pre>
<p>As always again, z-thrust stays constant, but this time only a rolling torque appears.</p>
<h2>Thrust</h2>
<p>Finally, what happens if we speed up all four motors by the same (squared) speed?</p>
<pre><code>torque_x = 0*(W^2 + dW) + 1*(W^2 + dW) + 0*(W^2 + dW) - 1*(W^2 + dW) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 + dW) - 1*(W^2 + dW) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + dW) - 1*(W^2 + dW) + 1*(W^2 + dW) - 1*(W^2 + dW) = 0
F_z      = 1*(W^2 + dW) + 1*(W^2 - dW) + 1*(W^2 + dW) + 1*(W^2 + dW) = 4*W^2 + 4*dW
</code></pre>
<p>So only in this case do we see an increase in vertical thrust (by <code>4*dW</code>). Notice how the net torques about each axis cancel out.</p>

CC BY-SA 4.0
2020-04-17T18:56:12.583Z
I don't think I understand this. When entering a yaw turn, two diagonal opposite props (say, CW) spin up, creating lift, while the CCW props spin down relative to the throttle set point (creating less lift). When ending the yaw turn, the CW props spin down and the CCW spin up to end the yaw rotation. Both entering and exiting the yaw turn create net lift, because the down-spinning props cannot go arbitrarily low to counteract the lift from the up-spinning props. Don't they? Maybe this is true for 3D-enabled quads?
mcenno
CC BY-SA 4.0
2020-04-17T20:35:09.957Z
@mcenno, you're correct that the props cannot go arbitrarily slowly. However, they're already going quite quickly so there's typically enough room to slow down, while still remaining with positive speed, to give yaw control authority. However, your point is very valid and is exactly the reason why multirotors can't yaw very quickly compared to their roll and pitch rates.
Kenn Sebesta
CC BY-SA 4.0
2020-04-17T12:42:54.367Z
2021-03-30T06:18:55.417Z
2021-03-30T06:18:55.417Z
Kenn Sebesta
<p>I've always wondered how a quadcopter actually yaws, when all propellers are horizontal. I know that two of the motors spins faster, but I don't understand how that generates thrust in the horizontal direction (I assume it has to?) to make the quadcopter turn.</p>

<quadcopters><propellers>
How does a quadcopter yaw?
Hannes Hultergård
# How does a quadcopter yaw?

(Asked by: username_0 on 2020-04-17)

I've always wondered how a quadcopter actually yaws, when all propellers are horizontal. I know that two of the motors spins faster, but I don't understand how that generates thrust in the horizontal direction (I assume it has to?) to make the quadcopter turn.

**Tags:** quadcopters, propellers


---

## 4 answers

### Answer by username_6 on 2020-04-17. Score: 17 | Accepted answer

Any vehicle yaws (i.e. turns) by having a net torque applied. What's interesting about a quadcopter is not just how it yaws, but how it yaws and *doesn't roll, pitch, or climb at the same time*.

To understand how this works, we need to briefly look at the math. We'll use a plus configuration, but really any [mutirotor configuration works](https://username_2.stackexchange.com/a/323/46).

[![enter image description here](https://i.sstatic.net/bGidS.png)](https://i.sstatic.net/bGidS.png)

The thing to keep in mind is that thrusts and torques are related to propeller speeds. If you speed a propeller up, it's intuitively obvious that it will create more thrust. And likewise, if you spin it faster, you need more torque. So changing motor speeds changes the net forces and torques on the airframe.

(Pedantically, it goes with the square of speed. So if you double the speed you quadruple the thrust and torque. But that's not important to this analysis.)

Here's the high-level driving equation. If you've ever messed around with mixers, you'll notice that the 4x4 matrix in the middle looks really familiar:

[![Reaction speed relation](https://i.sstatic.net/NDLu8.gif)](https://i.sstatic.net/NDLu8.gif)

What this does is it maps rotor speeds (squared) to torques about the roll, pitch, and yaw axes, as well as the net vertical thrust.

For hover, let's assume all motors are spinning at the same speed, `W`. So `W = w1 = w2 = w3 = w4`

### Yaw

What happens if we *speed up the first and third and slow down the second and fourth* by the same (squared) amount `dW`?

```
torque_x = 0*(W^2 + dW) + 1*(W^2 - dW) + 0*(W^2 + dW) - 1*(W^2 + dW) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 - dW) - 1*(W^2 + dW) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + dW) - 1*(W^2 - dW) + 1*(W^2 + dW) - 1*(W^2 + dW) = 4*dW
F_z      = 1*(W^2 + dW) + 1*(W^2 - dW) + 1*(W^2 + dW) + 1*(W^2 + dW) = 4*W^2

```

So the net force doesn't change (all the `dW` cancel out), and neither do the net rolls and pitches, but voila we have `4*dW` worth of torque!

---

For completeness, here's what happens when you want to change the other axes as well.

## Pitch

Let's change the front and back motors by the same (squared) speed, but we'll leave the other two motors alone:

```
torque_x = 0*(W^2 + 0) + 1*(W^2 - dW) + 0*(W^2 + 0) - 1*(W^2 + dW) = 2*dW
torque_y = 1*(W^2 + 0) + 0*(W^2 - dW) - 1*(W^2 + 0) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + 0) - 1*(W^2 - dW) + 1*(W^2 + 0) - 1*(W^2 + dW) = 0
F_z      = 1*(W^2 + 0) + 1*(W^2 - dW) + 1*(W^2 + 0) + 1*(W^2 + dW) = 4*W^2

```

Notice that, again, z-thrust stays constant, but this time only a pitching torque appears.

## Roll

Let's change the left and right motors by the same (squared) speed, but we'll leave the other two motors alone:

```
torque_x = 0*(W^2 + dW) + 1*(W^2 + 0) + 0*(W^2 + dW) - 1*(W^2 + 0) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 + 0) - 1*(W^2 + dW) + 0*(W^2 + 0) = 2*dW
torque_z = 1*(W^2 + dW) - 1*(W^2 + 0) + 1*(W^2 + dW) - 1*(W^2 + 0) = 0
F_z      = 1*(W^2 + dW) + 1*(W^2 + 0) + 1*(W^2 + dW) + 1*(W^2 + 0) = 4*W^2

```

As always again, z-thrust stays constant, but this time only a rolling torque appears.

## Thrust

Finally, what happens if we speed up all four motors by the same (squared) speed?

```
torque_x = 0*(W^2 + dW) + 1*(W^2 + dW) + 0*(W^2 + dW) - 1*(W^2 + dW) = 0
torque_y = 1*(W^2 + dW) + 0*(W^2 + dW) - 1*(W^2 + dW) + 0*(W^2 + dW) = 0
torque_z = 1*(W^2 + dW) - 1*(W^2 + dW) + 1*(W^2 + dW) - 1*(W^2 + dW) = 0
F_z      = 1*(W^2 + dW) + 1*(W^2 - dW) + 1*(W^2 + dW) + 1*(W^2 + dW) = 4*W^2 + 4*dW

```

So only in this case do we see an increase in vertical thrust (by `4*dW`). Notice how the net torques about each axis cancel out.

> **Comments:**
> **username_7**: I don't think I understand this. When entering a yaw turn, two diagonal opposite props (say, CW) spin up, creating lift, while the CCW props spin down relative to the throttle set point (creating less lift). When ending the yaw turn, the CW props spin down and the CCW spin up to end the yaw rotation. Both entering and exiting the yaw turn create net lift, because the down-spinning props cannot go arbitrarily low to counteract the lift from the up-spinning props. Don't they? Maybe this is true for 3D-enabled quads?  
> **username_6**: @username_7, you're correct that the props cannot go arbitrarily slowly. However, they're already going quite quickly so there's typically enough room to slow down, while still remaining with positive speed, to give yaw control authority. However, your point is very valid and is exactly the reason why multirotors can't yaw very quickly compared to their roll and pitch rates.  

---

### Answer by username_1 on 2020-04-17. Score: 12

The yaw effect created by the same effect that would cause a helicopter to spin if it didn't have a tail rotor.

On a multirotor, half the propellers spin clockwise (CW) and half counter-clockwise (CCW). This 50/50 split evens out the rotational forces for straight and level flight. To yaw, these forces need to be unbalanced. To spin CW for example, the CW motors spin faster and/or the CCW motors spin slower.

To minimise other movements, the CW and CCW motors alternate around the aircraft's frame. If all CW motors were on one side, a yaw motion would also cause the aircraft to tilt and move sideways.


---

### Answer by username_2 on 2020-04-17. Score: 3

It’s all about inertia.

By changing the speed of the rotors spinning in one direction, due to the conservation of momentum, the quad moves in the other direction.

That’s quite a confusing way to put it, so imagine this:

You’re facing a friend, both of you are in an office chair with wheels. You put out your right hand and push off your friend’s left hand. Even though you’re pushing your friend to make them rotate, (like the motor spinning the propeller), you also end up rotating. You’re the drone, so by you pushing the propeller at different speeds, you yourself end up rotating.

> **Comments:**
> **username_3**: But according to conservation of momentum, wouldn't the props spinning back up to normal speed (finishing the yaw movement) cause the quad to yaw the opposite way? Sorry I might not be understanding your answer.  
> **username_4**: It would cause the _momentum_ or the _kinetic energy_ to be conserved, however (assuming no drag), the quadcopter can start the yaw, wait, and then stop the yaw (initial and final rotational velocities are the same), but in between these moments, the quadcopter has changed orientation. Also important to note rotational momentum is less of a factor here than the torque created by the propellers  

---

### Answer by username_5 on 2021-12-02. Score: 0

As it transitions back from yawing one set of propellers are losing momentum while the other set of propellers are gaining until the momentum are both equal. The one losing are the propellers that generated the yaw condition in the first place and the propellers gaining are the ones that lost momentum during the yaw. When the yaw stops is when the pairs are now equal in momentum.


---

