1
$\begingroup$

I’m building a pitching machine that has an 8” wheel (16” OD tire) that is directly mounted to a 1/2” motor shaft.

Most similar questions reference a gearbox or pulpy system, but my situation is directly mounted to a 12v motor shaft.

Due to motor/fabrication design, the closest I can get my magnetic rpm sensor is a radius of 5” from the shaft.

So known variables for the calculation will be RPM at a diameter of 10” and the OD of the tire at 16” all on the same wheel.

How do I calculate the speed (MPH) of the tire?

I would imagine the OD of the tire runs slower than the shaft.

I hope this makes sense.

$\endgroup$
1
$\begingroup$

Whatever RPM your sensor reads is the RPM of the 16" tire as well. You say the tire is directly mounted on the shaft.

E.g., if your sensor is reading 100RPM your tire is turning the same and the ball is ejected theoretically, depending on geometry of the shoot and ignoring spitting added acceleration,

$ v= 16/12 *\pi*100= 133*\pi (ft/m) $

$\endgroup$
3
  • $\begingroup$ Not sure if this is correct. The wheel makes 100 revolutions in 60 seconds (RPM!) and each revolution travels 16/12 x $\pi$ feet. $v = 16/12 [ft] \times \pi \times 100 / 60 [s] \approx 7 ft/s$ 1ft/s is 0.681818 mph $\endgroup$
    – D Duck
    Mar 10 '20 at 23:57
  • $\begingroup$ Ok, I just assumed the 16” diameter was spinning slower than the 10”, even though they are part of the same wheel. It is hard for me to wrap my head around it. $\endgroup$ Mar 10 '20 at 23:59
  • $\begingroup$ @DDuck the answer gives feet per minute, you calculated feet per second. Both numbers are correct. $\endgroup$
    – alephzero
    Mar 11 '20 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.