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Old 03-08-2011, 09:11 AM
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Default Brushless motors and gearing general questions

This is more of a general question to better understand these motors and why they get geared the way they do.

Let's say I have a 13.5 and a 17.5. Does the 17.5 have more torque and the 13.5 have a higher speed (total rpm)?

When the gearing is changed from the 17.5 to the 13.5 is it correct to go to a smaller pinion or larger spur in order to benefit from the lower torque but higher rpm?

Am I on the correct line of thinking here?

Thanks.
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Old 03-08-2011, 09:39 AM
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yes
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Old 03-08-2011, 09:44 AM
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Might be a good idea to look up FDR (final drive ratio) and start getting info on that.

Slower motor (17.5), better low end torque. You can gear higher (lower FDR)

Faster motor (13.5), less low end torque, but higher rpm. You have to gear lower to compensate (raise the FDR)


An easy site to play with gear ratios and get a general idea:

www.comeseethis.com (i usually play with the novak here)
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Old 03-08-2011, 09:49 AM
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Is the FDR the mechanical advantage obtained through the gearing and transmission or does it represent something else?
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Old 03-08-2011, 10:05 AM
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Originally Posted by GizmoJunkie
Might be a good idea to look up FDR (final drive ratio) and start getting info on that.

Slower motor (17.5), better low end torque. You can gear higher (lower FDR)

Faster motor (13.5), less low end torque, but higher rpm. You have to gear lower to compensate (raise the FDR)


An easy site to play with gear ratios and get a general idea:

www.comeseethis.com (i usually play with the novak here)

Kool site...
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Old 03-08-2011, 10:15 AM
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Originally Posted by JMYBFFT
Is the FDR the mechanical advantage obtained through the gearing and transmission or does it represent something else?
Your fdr is basically spur/pinion x trans ratio.
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