[archiver]

Posted by Lloyd Burns at 1/15/2001 2:23 AM EST


Other on team #188, Woburn Robotics, from Woburn Collegiate and Canada 3000, ScotiaBank, Royal Bank Financial.


In Reply to: Re: This is what Freshman Statics Class is for...
Posted by bill whitley on 1/14/2001 9:48 PM EST:



Dr J is correct, but partly because the original question was about "any effect".

View this in a monospaced font (Courier, perhaps).

Perfectly level, with no forces, the free body diagram shows the forces acting on the body, including the reaction to the weight, and the force of gravity acting down (the direction of gravity being the definition of down). for example:

............100#...(CG).................
..............|........................^
.....^........|........^...............C
.....|........v........|.50# each......v
.....O..A.........B....0..(wheels)......
.....|||..distances from "down from CG"

[The '.'s are there to force the CD system to allow me to space the diagram - but the carats are to wide]

If you take wheel A and compute the sum of the moments (force times [horizontal] distance) of all the forces, taking CW as positive, you get
-50 x 0 + +100 x A + -50 x (A+B) = +100A -50(2A) if A=B. The sum is 0.

If you apply a 20# pull sideways on the CG which is C distance (let's say C = A) above the line joining the wheels, you now get a +20C moment. To keep the robot from tipping, (still with reference to the front wheel), the back wheel must provide the -20C moment, so it must provide 10# more force (the effective weight on the back wheels increases by 10#) giving more traction.

If you look at it from the back wheel, however, the "weight" on the front wheel decreases by the same 10# (since the sum of the weight thru the CG and the reaction forces up thru the wheels must be zero). You dont wantr to be behind the robot when the pull makes the weight on the front wheels go below zero !

Going up a 45 degree incline, if the CG is C above the wheels when level, then drawing the FBD in ASCII gets hard. The weight thru the CG still goes "down", but the line for "down" now goes thru the back wheel. Compute the moments using the *horizontal* distance from the front wheel and lo-and-behold, the "weight" on the back wheel and the weight thru the CG must be equal for the moment sum to be equal (no flipping backward ... yet).

Calculate the "weight" on the front wheel now, however; when the distance of the CG force from the rear wheel is zero, the "weight" (the reation, going upwards, actually) must be zero. More pull, and the robot enters the Olympics diving competition, back flip division.

To reduce "weight" on the front, pull from above the wheels, or send the vehicle up a ramp. When this "weight" is less than zero, turning will result.

This is why Dr J didn't give you a definitive answer; its a loooooong story.