Gear ratio calculation question

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SUMMARY

The discussion focuses on calculating the gear ratio necessary for a scooter with a 5-inch diameter wheel to achieve a speed of 25 mph using a motor with specific power and torque ratings. The user calculated the required RPM of the wheel to be 1680 RPM and determined that with a 17-volt supply, the motor would run at 21,420 RPM. This leads to a gear reduction ratio of 12.75, suggesting a required gear size of 0.39 inches, which is confirmed to be a diameter. Potential issues include high continuous RPM leading to friction losses that could damage the drive surfaces.

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david90
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I have a scooter with a 5 inch diameter wheel and I want to make it go 25mph with a motor that has this specs.
====
Power 0.35 hp @12v 1.43 hp @24v
Torque 0.47 lbf*ft @12v 0.97 lbf*ft @24v
Efficiency 70% @12v 76% @24v
Speed 1260 RPM / Volt
Voltage 12 volts or 24 volts*
Weight 11 oz.
Output Shaft 5mm dia x .75" long
Diameter 1.7"
Length 2.8" (from face)
Face Mount 2 M4 holes 1.14" apart
===
I use this equation to figure out the RPM of the wheel for 25mph.
RPM*(((D*3.14)/12)*60)/5280)=MPH
Calcuated RPM for 25mph is = 1680 RPM

i'm planning to use 17 volts, thus the motor rpm would be 1260*17=21420 RPM

Now I'm finding the gear reduction. 21420/1680= 12.75
5"/12.75 thus .39" is the required gear size that is attached to the motor.

Is the .39" gear size the diameter, radius or circumference?

i'm not using a sprocket because it is friction drive.

Are there any potential problems with this setup? Is my calculation correct?
 
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david90 said:
Now I'm finding the gear reduction. 21420/1680= 12.75
5"/12.75 thus .39" is the required gear size that is attached to the motor.

Is the .39" gear size the diameter, radius or circumference?
Since the 5" wheel is a diameter, the 0.39" must also be a diameter.
One problem I see is the high continuous RPM of the motor.
I would expect friction losses in the drive will damage the friction surfaces.
 

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