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Unread 23-06-2002, 22:38
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Join Date: May 2001
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While Yer (Going) Up

Posted by Dodd Stacy at 1/10/2001 8:46 AM EST


Engineer on team #95, Lebanon Robotics Team, from Lebanon High School and CRREL/CREARE.


In Reply to: Re: Mu Master 2000 - follow-up question
Posted by Raul on 1/10/2001 8:01 AM EST:



All the right questions, now how about the assumptions? Since our train is 3' long (maybe 2 1/2') and each 75 lb car (goal) is 3' long (still waiting for drawing dimension clarification) and our 14 degree incline is only 8' long ....... So, who's hanging off the end? Which end? Are we pushing or pulling? Cars on one end, or one on each end? And by the way, where is the train when the system cg causes the bridge to tip the other way?

I think it's going to be a lot of fun watching the various ways that multiple robots, and other "vehicles," make it over the bridge together - or don't. It's probably a good idea to make sure the bots don't get hung up when two wheels drop off the step at the end of a still-raised bridge. Hmmm, maybe some of us should make "wrecker"/towbots to clear the bridge.

Dodd

: Here is a slightly tougher but relevent question. Where does the CG have to be to ensure that with that much friction, you will actually pull the cars and not do a wheelie. The better question is: what is the minimum ratio of the distance from the floor to the hitch vs. from the center of the back wheels to the CG? And of course the height of the CG also matters on a 14 deg incline.

: Raul

: : Along with this question, I have another.

: : If a train weighing 130 pounds is headed up a 14 degree incline pulling two 75 pound cars, what is the mininum coefficient of friction (mu) that the train can have between its wheels and the tracks.

:
: : My answer: .53 assuming all wheel drive -- it is worse if only one set of wheels is driving.

: : Teams that plan on just driving up that sea-saw without much thought are in for a surprise!

: : Joe J.


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