 # Servo Question about Torque (Physics/Mechanics Problem)

Hi,

I am a total beginner. Just bought my arduino a month ago. My first projekt will be an automated Nerv-Gun turret. So far I have got the basic setup and coding covered. Next step is getting the parts and building it: So the big question right now is: How much torque should the 2 servos have?

So far my specs: [u]Size of gun is about 0.8m; weight is 1.4kg[/u] I want to control rotation (180°) and tilt (90°) with one servo each.

After some google/wikipedia/etc. i came up with this: Torque = I * "alpha" The moment of inertia I = 1/12 m * L² (i figuered to assume the gun as a long stick with equal mass distribution would serve my needs of accuracy. Also I plan on mounting it at its center of gravity)

So if i want the gun to turn fast and to be stopped quickly, without the wobbling i have observed in many similar projects, how strong a servo must i get?

Torque = I * "alpha"

"alpha"?

The greek Alpha, i can not write it here (?).
The formula is as follows:
Torque = Inertia x Angular acceleration(alpha)

With some more Google I am now this far:
Assuming that the gun turns 90° in a second and manages to start/stop in 0.10 seconds that should get me an angular acceleration:
Velocity would be: V = pi/4 * 1/t
Differentiating this leads to acceleration: a = -pi/4 * 1/t²
And torque: t = 0.11666 * pi/4 * 1/0.1² = 0.916…Nm
I have no idea if this is right or if these are realistic values…?

The most realistic thing to do is see what sized servos have been used by others making the same type device. If the mount is well designed to reduce friction and no weight holding by the servos, then smaller servos could be used.

There's no simple answer to your question. Without reading a physics book, it's probably too hard to try to explain how to design this properly. Your best bet is therefore trial and error. Here are some basic statements though:

tau = I alpha isn't the appropriate equation to be thinking about. Except for starting and stopping, almost all of your operation will actually be at alpha ~ tau ~ 0 (steady angular speed). The point, though, is that tau is the net torque. The servo exerts some torque, and this has to counteract torques exerted by gravity and friction. It's really these that you need to worry about.

The torque exerted by gravity on a massless (!) rod of length L with a weight w on the end is Lw when the rod is horizontal. Less torque is needed if the rod is already raised. Less torque is also needed if the weight is distributed along the rod and not all hanging off the end. This is, however, the torque exerted about the tip of the rod (torques are always in reference to particular points). It would be what you care about if the rod were directly mounted to the output shaft of the servo. Different specifications would be required if you use pushrods or something like that, which I recommend. Think about the principle of a lever: An appropriate mechanical setup can always be used to trade speed and travel distance for torque if needed (or vice versa).

One more tip is that holding something in place with a servo against a sustained torque (gravity) isn't a particularly good idea. It puts a lot of strain on the servo, possibly causing it to overheat. This wastes a lot of power, and can easily end up "wobbly" as you've observed. It would be much better to design a system that requires power only to move, but is otherwise mechanically locked. The usual way to do this is with worm gears. The input end of a worm drive system can be driven with a motor, but friction usually prevents the output end from being turned by external torques. So an elevated rod would remain in place and stable even with no power sent to the motor.

Assuming that the gun turns 90° in a second and manages to start/stop in 0.10 seconds that should get me an angular acceleration: Velocity would be: V = pi/4 * 1/t Differentiating this leads to acceleration: a = -pi/4 * 1/t²

No, thats wrong. You are proposing that the gun accelerates, during 0.10 seconds, from zero to some speed V, and then turns at some constant speed V for 0.8 seconds ( during which the acceleration is zero ), and then slows down during 0.10 seconds from V to zero speed.

If you assume that the acceleration rate during the first and last part of that is constant, and the same for acceleration and deceleration, and has some value "a", then

V = 0.1a ( the speed reached after 0.1 seconds ), and the total distance travelled during the three phases will be

0.5Vx0.1 + Vx0.8 + 0.5Vx0.1 = 0.005a + 0.08a + 0.005a = 0.09a = 90 degrees.

Therefore a = 1000 degrees/sec/sec acceleration and V = 100 degrees/second rate of rotation.