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From: amccor@dres.dnd.ca (Alan McCormac)
Subject: Re: 300 lb robot on 10 degree grade?
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References: <1993Oct5.191036.26711@CSD-NewsHost.Stanford.EDU> <29fles$7kn@skates.gsfc.nasa.gov> <jfoxCEtLnM.9Ir@netcom.com> <29h3u8$lus@skates.gsfc.nasa.gov>
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Date: Wed, 13 Oct 1993 21:00:27 GMT
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>>In article <29fles$7kn@skates.gsfc.nasa.gov> nstn@quercus.gsfc.nasa.gov (Nathan Stratton) writes:
>>>
>>>This is a little, ok a lot off the topic, but can anyone tell me how to fiend
>>>the torque is necessary on a drive axle to move a 300 pound robot up a 10 ^
>>>incline??? I would like to now the formula.

>In article <jfoxCEtLnM.9Ir@netcom.com> jfox@netcom.com (Jeff Fox) writes:
>>
         [... trigonometry lesson deleted ...]
>>
>>Now if you have a wheel with a 1 foot radius then 52.08 foot pounds of torgue
>>will balance gravity on a 10 degree slope for your 300 lb robot.  You will
>>need a 52 lb force regardless of the wheel size.  The wheel size and gearing
>>determine the torgue required by the motor.

The 52.08 lb-ft assumes perfect traction and zero friction.  You will also
need a tractive surface on the wheel if you want to go anywhere at all.
Most robot platforms also have significant internal motion resistance, and
if the terrain is soft, external motion resistance as well - about 5% for hard 
rubber wheels on pavement and about 30% for balloon tires in sand.

In article <29h3u8$lus@skates.gsfc.nasa.gov> nstn@quercus.gsfc.nasa.gov 
(Nathan Stratton) writes:

>Yse is dose help. So is 52 lb of force 52 Foot/Pounds ? And can you change
>in/oz to ft/lb by / by 12 and 16????

Torque is a product of force and the perpendicular distance at which it is 
applied:


>Thank You for your help.

>Nathan Stratton
>nstn@quercus.gsfc.nasa.gov


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Alan McCormac                                        |  amccor@dres.dnd.ca
Vehicle Concepts Group - DRES                        |  Tel: (403)544-4832
Box 4000, Medicine Hat, Alberta, Canada    T1A 8K6   |  FAX: (403)544-3761
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   " The cost of research is a full garbage can "    -     A wise guy
