[WheelCommander] RE: WheelCommander Digest, Vol 2, Issue 1

[Pete Skeggs <peters@xxxxxxxxxxxx>]

Hi, Jeff,

For some reason your messages are being discarded by the email list 
system, though I get dicarded copy as an administrator of the list.  I 
just made some changes to the content filtering rules to hopefully clear 
up the problem.

Anyway, to answer your question...

> I'm not really sure how to figure speed based on mechanics but here is 
> what I'm working with maybe it would give you an idea...or maybe you 
> could send me the formula for computing the speed variables...
>
> The average speed of the robot would be about .5 - 1.5 mph.
>
> The encoder is in-line between pulleys.
>
> 1.  Wheel base = 15" in diameter
>
> 2.  Wheel size = 4" in diameter
>
> 3.  Wheel timing pulley = 3" in diameter
>
> 4.  Motor timing pulley = .9" in diameter

Well, first, convert mph to inches per second:
X (mi/h)  *  (1h/3600s) * (5280 ft/mi) * (12 in / ft) = X * 17.6 in/sec

With a wheel size of 4 inches diameter, the circumference is 4 * pi = 
12.6 inches

So, the wheels rotate between:
0.5mph * 17.6 * (1 rotation / 12.6 inches) = 0.5 mph * 1.4 = 0.7 
rotations / sec
to 1.5 mph * 1.4 = 2.1 rotations / sec

The relationship between wheel and motor rotations is related to the 
size of the pulleys.

motor rotations / sec * (0.9 / 3) = wheel rotations / sec,
or:
wheel rotations / sec * (3 / 0.9) = motor rotations / sec

So, the motor rotates between:
0.7 * (3 / 0.9) = 0.7 *  3.33 = 2.33 rotations / sec
and:
2.1 * 3.33 = 7.0 rotations / sec

Now, the question is, how is the encoder driven by the belt "in-line 
between pulleys"?  Is it on its own pulley?   If so, what is the 
diameter of that?  If not, how is it driven by the pulleys?

With that, we can calculate the estimated rotation rate of the encoder, 
and so knowing the encoder resolution is 256, we can figure out the 
signal frequencies that the WheelCommander would need to handle.

-Pete