hakcenter
15+ Year Contributor
- 43
- 28
- Jun 9, 2004
-
Riverside,
California
Ok I think I see where you are getting confused. The flywheel does not multiply engine torque and is not multiplied by engine torque, rather the RPM at which the motor is spinning the rotating assembly is what multiplies the torque. It is the speed of the rotating assembly that multiplies the torque produced by the rotating assembly. The torque it produces has nothing to do with the torque the engine is producing. The only thing the engine can do to increase the torque the rotating assembly puts out is spin faster. The faster it spins the more inertia the more torque is produced completely independent of torque produced from the engine itself.
Okay cool, now we're on the same page completely.
[...]
The flywheel torque is enough to get a factory dsm a 1.7 60 foot with a 5500 rpm launch. So a car launching with about 100 instant lbft plus flywheel weight can overcome the 3200 lbs of potential energy and drivetrain loss because of this instantaneous torque spike.
I want to reply to your entire post but save space on the page, haha. We're all at least understanding each other well now, .
The potential energy in the flywheel isn't going to change regardless of how it is applied, either by the edge, or near the edge or even at the center. However in a clutch system, the flywheel is going to transfer its potential into the clutch, where the clutch is spinning around the input shaft at the exact center of the inertia forces.
When I take a wheel and spin it, stopping it from the outside is of course harder than stopping it in the middle. To me, something running on the inside (gear type) is a reduction type.
You must be logged in to view this image or video.
But the clutch disc really isn't a reduction.. so help me here??