Re: [WheelCommander] RE: Encoder pulley placement

[Pete Skeggs <plskeggs@xxxxxxxxxxxxxxxx>]

> > 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?

> The encoder pulley is situated between the Wheel and the motor. The 
> belt is a double sided timing belt. The pulley is .7" in diameter. The 
> encoder pulley is in the drive belt loop between the wheel pulley and 
> the motor pulley
>  

OK, so,
motor rotations / sec * (0.9 / 0.7) = encoder rotations / sec
encoder rotations / sec * (0.7 / 0.9) = motor rotations / sec

so the encoder rotates between:

2.33 * (0.9 / 0.7) = 3 rotations / sec
and:
7.0 * (0.9 / 0.7) = 9 rotations / sec

> > 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.

At 256 encoder pulses per rotation, this is then between:
3 * 256 = 768 Hz.
and
9 * 256 = 2304 Hz.

Our firmware processes an interrupt for each edge of each of the two 
encoder channels, so this ends up being an interrupt frequency 4 times 
higher, or between 3072 Hz and 9216 Hz.

This may work, but will require testing and evaluation here.

Sorry for the long delay in replying.

Sincerely,
Pete Skeggs
Nubotics Tech Support