Spacing of wheels?

[archiver]

Posted by Eric Reed at 1/14/2001 8:51 PM EST


Coach on team #481, NASA Ames / De Anza High School, from De Anza High School and It could be you!.



Assuming the CG is centered perfectly over the drive train, does the spacing of wheels have any effect on traction?

Thanks,

Eric Reed
Team 481

[archiver]

Posted by Joe Johnson at 1/14/2001 9:21 PM EST


Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.


In Reply to: Spacing of wheels?
Posted by Eric Reed on 1/14/2001 8:51 PM EST:



To answer your question, I need more information, but
the short answer is yes.

In a Freshman Statics class, you learn more than you
will ever want to know about drawing free body diagrams
which help you noodle out the equations that let you
determine the answer to your question.

To really answer your question a number of things have
to be addressed,

How high is the CG? Are you going uphill or downhill?
Are you pulling something? If so, what type of joint
is the hitch? Knowing that, what is the relationship
between the CG, the hitch point, and the wheels?

There are a lot of complicated issues surrounding this
year's game.

Joe J.

[archiver]

Posted by bill whitley at 1/14/2001 9:48 PM EST


Student on team #70, Auto City Bandits, from Powers Catholic High School and Kettering University.


In Reply to: This is what Freshman Statics Class is for...
Posted by Joe Johnson on 1/14/2001 9:21 PM EST:



Some teams don't have the advantage of someone who took statics, much less understands/remembers it. I am a senior in high school, and I understand the torque, force, gear ratios, etc better than anyone else on my team, including the teachers, coaches, mentors, (whatever you want to call them). We have 0 engineers. Just a prospective of some other FIRST teams

Bill


: To answer your question, I need more information, but
: the short answer is yes.

: In a Freshman Statics class, you learn more than you
: will ever want to know about drawing free body diagrams
: which help you noodle out the equations that let you
: determine the answer to your question.

: To really answer your question a number of things have
: to be addressed,

: How high is the CG? Are you going uphill or downhill?
: Are you pulling something? If so, what type of joint
: is the hitch? Knowing that, what is the relationship
: between the CG, the hitch point, and the wheels?

: There are a lot of complicated issues surrounding this
: year's game.

: Joe J.

[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.

[archiver]

Posted by Eric Reed at 1/15/2001 10:48 AM EST


Coach on team #481, NASA Ames / De Anza High School, from De Anza High School and It could be you!.


In Reply to: Original Question too broad
Posted by Lloyd Burns on 1/15/2001 2:23 AM EST:



Thanks, Lloyd. Like Bill's team, we have nobody who is beyond high school physics (including myself). Your explanation helps to identify the factors we need to look at.

Good Luck!

Eric.