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acceleration... Posted by JimBlake [Email] (#141) [Profile/Gallery] (more from JimBlake) on Thu, 19 Sep 2002 11:53:31 In Reply to: Another Followup..., Mr. Science, Wed, 18 Sep 2002 21:06:29 Members do not see ads below this line. - Help Keep This Site Online - Signup |
Forget drivetrain losses for a moment, look at it from the other side of the problem. Pretend the tires are on the verge of spinning...
F(normal) * mu(guess)
7500N * 0.9 = 6750 N tractive force.
acceleration = F/M = 6750N / 1500kg = 4.5 meters/sec^2
tires' rotational acceleration = 14.2 radians/sec^2
crank's rotational acceleration = 195 radians/sec^2
Here's where I have to guess...
Flywheel is maybe 40 lb, maybe 14" diameter?
So its moment of inertia is maybe 0.29 N-m-sec^2
So it takes 56 N-m of torque just to accelerate the flywheel at 195 radians/sec^2
Maybe somebody knows what the flywheel assembly actually weighs, & you also gotta add the moment of inertia of all the spinning parts in the transmission, too.
posted by 131.184.14...
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