Help with gear ratio

[Wayne C.]

This is not wayne c. this is his son. It wont let me sign him out.

Ok im trying to think up a gear ratio. So im taking the direct drive and compairing it to what I want it to drive at. What I did is I took the rpms and i multiplied it by the circumfrence of our tires and then i divided by 60 to convert it to inch/minute/sec. but its coming out way to large....

What do the CIM motors normally run at(RPMs)!?

4"/12"(convert to feet)=.333... * 3.14(Pi)(circumference)=1.246666ft...

1.04666ft*5000rpm=5233.3333ft/min.../60sec = 87.222... ft/min./sec?????

This is wrong why?

[Swan217]

RPM/60=Rev/sec
Rev/sec*(D(in)*Pi)/360=in/s

I think you have to divide by 360 degrees

[Wayne C.]

5000/60=83.333.

83.333"(4"*3.14)/360= 2.90inchs/s = 14.5 ft/sec

Is that rite?

[Kevin Sevcik]

well you have your unit conversion a little technically wrong. it should go like this:
1.04666 ft * 5000 rpm = 5233 ft/min

5233 ft/min * (1 min/60 sec) = 87.2222 ft/sec

Otherwise your math is correct. a 4" diameter wheel on a CIM WOULD run that fast. 5000 rpm is ridiculously fast, after all. The real point is that running something straight of the CIM means it puts out very little torque. with a 4" wheel, you could stall the motor with your pinky finger. This is why all CIM drivetrains have gear systems.

[Wayne C.]

What rpms do they run at im looking at spec sheet and i understand what its saying but i dont no what to use for rpms

[Mike Betts]

What rpms do they run at im looking at spec sheet and i understand what its saying but i dont no what to use for rpms

The free speed is valid only at no load (including no frictional losses).

Look in the white paper section for a few excellent papers on how to calculate speed.

In general, design to 1/2 of the free speed and 1/2 of the stall torque.

[Greg Needel]

i made this spreadsheet last year to help someone with their drive maybe this can help you....this is for a double reduction drive train