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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 |
Quote:
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 |
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 dahl team 497 ] |
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 |
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.
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