Newsgroups: comp.robotics
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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 21:23:45 GMT
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In article <29fles$7kn@skates.gsfc.nasa.gov> nstn@quercus.gsfc.nasa.gov 
(Nathan Stratton) asks:
>
>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) replies:
>
        [... trig lesson deleted ...]
>
>If A=10 degrees then sin(10)=.1736 so you need 52.08 lbs of force to balance 
>gravity. You will need a little more to move the robot because of friction.
>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.

Remember that 'friction' encompasses two things: internal and external 
motion resistance.  Many robotic drive components have significant internal
motion resistance, so you would need to rate one for _net_ torque output of
52.08 lb-ft.  External motion resistance depends on running gear and terrain, 
and is normally rated in terms of vehicle weight - about 5% for hard rubber
wheels on pavement, and 30% for balloon tires in sand or snow.  Also don't
forget that your running gear has to be able to use that driving force - if any
slip occurs, the net driving torque drops.

In article <29h3u8$lus@skates.gsfc.nasa.gov> nstn@quercus.gsfc.nasa.gov 
(Nathan Stratton) further asks:
>
>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 the product of force and the perpendicular distance at which it is 
applied:
               T = F x R

F is a function of vehicle weight W and grade angle A, and motion resistance 
MR is a percentage of W.  So if your wheel has an 8 inch radius, and the 
terrain is firm soil:

               T = W [ (sin A * R) + MR ]
               T = 300 lb * (16 oz/lb) * [ (sin(10) * (8 in)) + (0.10) ]     
               T = 7148 oz-in

Therefore, find a drive with 7148 oz-in net output torque.

--------------------------------------------------------------------------
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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