Originally Posted by dorkpunch
i dont know, i may be wrong in my thinking here, but i think that setup would have LESS flywheel efect than a standard set up. the clutch in a normal bike is efectively a flywheel. its not bolted directly on the crank, but it has momentum that will try and keep the crank turning just like a flywheel. big clutch= more flywheel=more torquey feel, little clutch, no flywheel, = snappy mx feel. anyone else, thoughts?
Not arguing with anyone, just want to clarify what I think you all are trying to say:
The fact that a flywheel has mass isn't what gives it torque. Torque is derived from the ability to rotate an object; so the further you move the applied force from its pivot point, the more energy (HP) is converted to torque.
Put another way, torque is the resulting rotational force when energy is applied at a distance from an objects rotational point. The further you move away from the rotational center, the more torque you create.
The flywheel effect is stored energy. What it delivers for torque is still dependent on where it is applying its force relative to its rotational point.
Essentially the simple answer is the bigger the crank diameter (where it engages the clutch), the greater the amount of energy (HP) is converted to torque.
The "Flywheel" effect is just mass in motion or stored energy. When a large mass is "connected" to the drive system, it takes torque to turn it and get it spinning, and once spinning, it wants to stay in motion until acted upon. So the flywheel effect is really storing whatever torque is delivered to it, and "smoothing" out the delivery of the torque.
Engines with lots of torque typically have big crank diameters (longer stroke). Engines with big flywheels, are smoother. Not all engines have both. Both of these characteristics add weight and cost more.
Heres an example taken from a university website
"Imagine pushing a door to open it. The force of your push (F
) causes the door to rotate about its hinges (the pivot point, O). How hard you need to push depends on the distance you are from the hinges (r
) (and several other things, but let's ignore them now). The closer you are to the hinges (i.e. the smaller r
is), the harder it is to push. This is what happens when you try to push open a door on the wrong side. The torque you created on the door is smaller than it would have been had you pushed the correct side (away from its hinges)."
Hope this helps......