Tuesday Feb 16, 2010 (updated 2/21/10)

**A quick introduction** to motor selection from an industrial
site is here (Aveox). Somehow
this page seems to finesse the Matlab calculation below. I don't understand
how. I'd
like to have the 10 or 100 page version of that page, but don't know
what book has that information. Let me know if you find it.

**Matlab code of selection
of voltage and gears for a motor**. This
is basically the solution to the HW due Thursday Feb 11. Please
excuse my pathetic computer algebra skills. If you know how to do it better,
I'll post that! (code corrected on 2/21/10).

Here is the problem:

Given a motor with given k, R and c.

Given a desired output angular velocity omega.

Given a desired output power Pout.Find the best voltage and gear box so as to minimize the electrical power to the motor.

In class today I eliminated I and T and solved with V and omega. In the code posted hereI redid it writing all in terms of T and omega. Seems more sensible. Doesn't change the answer from class though. In class I only found the peak efficiency. In the code here I carried the calculation all the way to the end.

Here are some observations that guide the calculations.

1) The motor equations are linear in V, I, omega and T. So if you have one solution of the motor equations you have a family of solutions by multiplying all 4 variables by, say, 17.6 or 89.8 or whatever.

2) The power in and power out are both products of two of the terms from V I, omega and T. So the efficiency won't change if they are all multiplied by 17.6, 89.8 or whatever. So the efficiency can only depend on the ratio of two of the terms. And the peak efficiency can be had at any power.

3) The big result is that the key motor parameter is D = k/sqrt(RC) or k^2/RC. The
peak efficiency is a complicated (lots of square roots) function of D that
gives a peak efficiency of zero when D=0 and peak efficiency of 1 when D=infinity.

**The answer**. We use motor (Faulhaber
2657 - 012 CR) we are
now using in our Ranger robot is with omega desired at 5 rad/sec and the desired
power at 30 watts.

V = 17.3 volts, Gear reduction = 183:1 and we will then get 84.5% efficiency with motor at 917 radians/s.

**Caveats**

* This uses the motor specs for R which we know are wrong by a factor of 2 (from bench top measurements). Putting in the right R = 1.3 ohms changes the battery voltage to 20.7 volts, the gear ratio to 213:1 and the motor angular velocity to 1065 rad/s. The efficiency drops to 80%.

* We didn't check for overheating.

* We didn't check if the motor exceeded its speed spec. And it does. The motor spec sheet gives a max allowed speed of 6000 rpm which iis about 625 radians/s. Its a "12 Volt" motor, so running it at 20V without a heavy load is, well, out of spec. 20 V by itself is not a problem, but the fast spinning, due to low loading, is. So the matlab gear selection solution doesn't actually work, it makes the motor spin too fast.

* There are other friction terms that are not in standard linear motor models
that might be important. For exampl, in bench tests we have measured

about 10% loss of power in our gear boxes.

**More motor information**

Eric Sopp, the webmaster of the following site, claims it is a good resource
for information about motors:

http://www.ohioelectricmotors.com/resources