Newsgroups: comp.robotics
Path: brunix!sgiblab!sdd.hp.com!usc!cs.utexas.edu!uunet!butch!netcomsv!netcom.com!jfox
From: jfox@netcom.com (Jeff Fox)
Subject: Re: 300 lb robot on 10 degree grade?
Message-ID: <jfoxCEtLnM.9Ir@netcom.com>
Summary: simple jr high algegbra and trig
Sender: jfox@netcom.com (Jeff Fox)
Organization: Netcom Online Communications Services (408-241-9760 login: guest)
References: <1993Oct5.191036.26711@CSD-NewsHost.Stanford.EDU>
            <29fles$7kn@skates.gsfc.nasa.gov>
Distribution: usa
Date: Wed, 13 Oct 1993 05:23:45 GMT
Lines: 35

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

               /!
 Hypotenuse  /  !
           /    !
         /      ! Opposite
       /        ! 
     /  A       ! 
   <__)_________!

        Adjacent

For your robot on a slope think of a right triangle with your robot on the
hypotenuse (slope).  The ratio of the opposite side to the hypotenuse on the
triangle is the sine of the angle A.  If A=90 degrees then sin(90)=1, so you
would need a force of 300 lbs to make it go straight up.  If A=45 degrees
then sin(45)=.707 so you need a force of 210 lbs to get it up a 45 degree 
slope.  If A=30 then sin(30)=.5 so you need a 150 lb force.  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.

I hope this helps.
