Just wanted to know the HP/TQ numbers are worth the purchase of this lighter flywheel.
Negligible to you may not be negligible to someone else...But, each tire / wheel will weigh, say 35 lbs, plus wheight of brakes, say 15 lbs per corner, Â
plus axles, about 10 lbs a peice, plus gearbox, say 25lbs of rotating mass, plus clutch / presure plate, say 10 lbs, plus the crank / rods / cams / etc, about 20 lbs. The total comes to 275 lbs. Â
going from a 10lb flywheel to a 5lb flywheel come out to a 1% change in rotating mass. No granted, none of the earliner items have the same radius or RPM, therefore, none of them will count equally twords the total rotational inertial of the driveline, so loosing 5 lbs of the flyweel may acount possibly for 5%, but even this is still (in my opinion) negligble.
So lightweight tires and wheels will only prevail against heavier wheels and tires if their radius and revs/mi is different?Originally posted by badgerman
none of the earliner items have the same radius or RPM, therefore, none of them will count equally twords the total rotational inertial of the driveline,
I = aMR^2, thats all you need to calculate rotational inertia, where the 'a' depends on the shape of the object (for a disc, I believe a = 0.5). Reduce the mass of a similar sized object, you reduce the rotational interia, its that simple.Originally posted by badgerman
To calculate rotational inertia, you need to know a lot of things, such as radius, RPM, and most importantly weight distubution. (is the weight distubuted at the center or the outside) I do not know the values of all of these for every single rotating item, I was just trying to come up with a rough guess, and error on the light side (hence my estimate of 5% instead of 1%), and see what proprotion to the total rotating mass a flywheel contributes.
Haha we just covered this in physics class, you are correct it is I = (.5)MR^2Originally posted by Mazda_Gremlyn
I = aMR^2, thats all you need to calculate rotational inertia, where the 'a' depends on the shape of the object (for a disc, I believe a = 0.5). ...
Not sure what you wrote here, but 1 RPM (rev/min) converts to 60 rev/s, and to convert rev/s to rad/s you multiply by, I believe, 2pi, giving you 120pi rad/s for each RPM, that may change the numbers a bit, if you want to rework, but you get the point, less hp is used up turning the flywheel, so you get those few extra ponies to your wheels.Originally posted by badgerman
and to convert RPM to rads / second: RPM = pi /30 rads / second
In a perfect world this would be true, but you're negelecting friction, gravity, etc. Plus you're only analyzing the flywheel itself, and forgetting it has to turn a whole bunch of other stuff that will also cause it to slow down if a force is not applied to it.
To keep the flywheel, or any other mass rotating at a constant velocity, neglecting air resitance requires no power, power is only required to accelerate or decelerate the rotating mass. Â
No problem, this should make more sense:Not sure what you wrote here, but 1 RPM (rev/min) converts to 60 rev/s, and to convert rev/s to rad/s you multiply by, I believe, 2pi, giving you 120pi rad/s for each RPM,
Why?Wouldn't you lose revs quickly too? I would think that a lightened flywheel would be annoying on the highway,m