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Who's done the math?
I was just curious..
Our team is in the middle of doing the math, trying to figure out the maximum force and acceleration that can be applied to the wheels before loosing traction. Obviously it isn't going to be much, but we are trying to make a decision with our drive-train and gear box's. So given the coefficients of friction.. Quote:
Good Luck Teams! THANKS! |
Re: Who's done the math?
Maximum force due to wheel friction f = umg (u = coef of friction)
Max force for 1 wheel = umg/4 Max torque T = Fr = umgr/4 For u = 0.05 mg = (50kg)(9.8m/s^2)= 500N r = 0.05m T = 1.25 N-m Much less than the CIM can supply... |
Re: Who's done the math?
Thank You Very Much!!
I will share this with my team, hopefully it will help. Good Luck! |
Re: Who's done the math?
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It doesn't make much of a difference, though, in this example, and may have even been a typo. The coefficient used was .05, and the proper one per bhsrobotics1761 is .06. The number is closer to .37, by my calculations, which is somehow much smaller and even less than dmcguire's number. I found the torque, by his numbers, to be .31 N-m. In the end, your answer is "a very small number." :). |
Re: Who's done the math?
okay so in the end, the CIM is still powerful enough itself without a gearbox to accelerate, what is limiting you is the max velocity without the gearbox?
Thanks For Your Reply! |
Re: Who's done the math?
You guys need to check your assumptions. For example 50kg is only a 110 lb robot. I think a robot with battey and bumpers will be closer to 150 lbs or 68 kg. For a four wheel chassis with weight equally distributed each wheel would exert 2.25 lb (150x0.06/4) before breaking free. That equates to 0.56 ft-lbs or 108 oz-in of torque.
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Re: Who's done the math?
Very good point. We have been using 145lbs in most of our math.
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Re: Who's done the math?
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Re: Who's done the math?
Sorry, arithmetic is not my forte... The equations are correct. I agree with the updated numbers - in any case much smaller than a CIM's max torque.
I did use the kinetic rather than tha static coefficient of friction. One of the secrets in this game will be to NOT to accelerate faster than allowed by the coefficient of friction, in this case 0.06g or .6m/s^2. Thanks! |
Re: Who's done the math?
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Re: Who's done the math?
If you add more wheels to the robot that are powered, you would get more force right? So then here's a couple questions.
1) Why not use every motor in the kit powering as many wheels? Obviously not very feasible and a bunch of other reasons why not to do this, but what would the maximum force be that you could generate? 2) Or, just have your drive motors connected to more than 1 wheel, like a 4 wheel-drive system with 2 wheels (or more). What would the benefits be? Just curious, I understand about the higher complexity/weight and all that stuff. |
Re: Who's done the math?
You're only going to get about 9 lbs of tractive force regardless of how many wheels you use. 150 lbs x 0.06 = 9 lbs. More wheels just divide it up.
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Re: Who's done the math?
Not directly related to the Friction stats now.. but more towards the battery power.
Some team members have brought up battery drainage issues if we were to go straight CIM to wheel. Others have argued that the low resistance level will require less of an elec. current for the motors. Our other concern points to the new electrical system. Does anybody know if this system will require more or less power than the old system? Our main concern is the wireless router... Has anyone else brought up battery concerns? With our experience with the batteries, they seem to hold up pretty well. |
Re: Who's done the math?
Robot = 150lb (120lb + battery + bumpers)
Max force available to move robot: 0.06 * 150lb = 9lb Trailer ~ 40lb (from eye witness reports) Drag on robot from trailer: 0.06 * 40lb = 2.4lbs Total: 9 - 2.4 = 6.6lbs force Total mass to move forward: 190lbs I'll leave the acceleration calculations for you ... Now tell me the top speed from rest when you slam into the wall after 50ft? What's the maximum speed before you need to start braking so you don't crash? How long did the traversal of the whole field take? While braking, did you account for the center of mass moving forward over the front wheels? If so how fast could you brake? What happens if the trailer jack-knifes into the robot while braking? So many questions, so little time! :rolleyes: |
Re: Who's done the math?
tell me about it. thanks for your help. i will be bringing this up with my team tomorrow.
this will definitely help with our calculations, i hope we will be able to put it all together and make a decision. thanks again! |
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