Go to Post In most cases on FIRST robots, we have more material than in we need in some places and less than we need in others. That single phenomenon is the leading cause of robot failures. - Paul Copioli [more]
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Unread 12-01-2002, 08:32
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Turning

As stated a few times, but is worth stating again, turning at a decent speed while holding the goal (and while having decent traction against being pushed sidways) is the real design challenge.

The wonder wheels (wheel with side rollers offering side motion as well as the wheels rolling) worked pretty well last year for some, but this leaves you very vunerable to being pushed sideways.

It will be interesting to see what people come up with.

Scott358
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Unread 12-01-2002, 10:00
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Quote:
As stated a few times, but is worth stating again, turning at a decent speed while holding the goal (and while having decent traction against being pushed sidways) is the real design challenge.


You need to think outside the box.


Turning if done right is very very easy. we can turn now with all three goal.

the fun part will be pulling other robots and still pull all three goals.

Watch the KSC in March 7-9
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Unread 17-01-2002, 14:39
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coeff of friction

by using the above numbers-
the coeff of friction (for hard rubber vs carpet) for the goals on the carpet to be used in the comp
it is
Force of friction=mu * normal force

181= mu * 35 (assuming the max needed to move goal)

mu=35/181

mu (or the coeff of friction)=.19337

This sounds about right for the material in question



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Unread 17-01-2002, 19:19
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That equation doen not seem accurate

You put the weight down..not the force pushing down

Te acutal force on the casters is as follows

181 pounds = 82.1 Kilograms

so 82.1*9.8m/s^2 = 804.58 Newtons of force total

Now... to find the coefficient of friction...

Friction = Mu*Mass*Gravity
or in this case ----> Mu(804.58)

so in order to move this goal...

You plug in the following

Force of Robots Push or Pull - MU(804.58) = 82.1* Acceleration

Now...to fin dthe minmum force needed to move the goal...

You set acceleration to zero...

So you get... Force of Robots Push or Pull - Mu(804.68) = 0

Force of Ropbots Push or Pull = Mu(804.68)

Force of Robots Push or Pull / 804.68 = MU

There you go...just plug in the force from your robot and you should find mu.

I need to check some notes..but there might be a better way...

And remember Its in Metric

G.Goldman
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Unread 17-01-2002, 20:09
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we usda s spring scale designed for bows and got somewhere around 28-30 lbs of force to get it moving when all 8 casters were pointing sideways.

Last edited by foursixnine : 17-01-2002 at 20:14.
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