I don't have a problem with bogging out of the hole. The tires spin fine, but when they hook, I bog for a split second, if that makes any sense. I don't know if that's technique, revs, flywheel mass, or too little/too much tire for my application.

 

It all depends on what you want. If you plan on doing any road racing (track events on road courses etc.) then I'd consider the lighter flywheel for its ability to rev up and down quicker, especiall when rev-matching downshifts. It will feel very precise at higher engine speeds. But if you plan to drag race your car I'd go against it. Because lighter flyhweels definitely make it harder to consistently launch at the track.

 

Yeah, that's what I meant. The car comes off the line well, but when the momemtum of the flywheel catches up to that of the tires that grab, the car lays over for a split second then takes off. That slight bog could be costing about .2 seconds in the first 100'.

 

Very informative thread. I have the JWT combo at home and still waiting to install it.

 

Plz dont hate on me but can somebody explain this to me a little better.
i guess i dont understand exactly what the flywheel does.

thanks!!!

 

A flywheel is bascily an energy storage device. It's mass helps the engine rev smoother and helps smooth out the shifting. ie.. keeps the engine revs up during an upshift. When you start off in 1st, it's what makes it easier to engage the clutch smoothly vs bogging it and shutting the engine off.

Now the heavier the flywheel, the smoother the shifts. But that also makes it harder to turn. Once it's in motion, it's harder to slow down but that comes later.

If you lighten the FW, it's easier to turn and you will note the car reving faster. But shifting will be a bit more difficult and your 1st gear launch will be more difficult. Less mass = less kinetic energy stored = harder to get going.

Now for everyday driving, it's nice as you can rev quicker. The launch issue is easily adapted to.

But if you drag race alot or if 1/4 times are important to you, then a light FW makes getting a good launch more difficult. In fact, drag racers use heavier FWs as they can store more energy at the line. They can rev the engine to what they want before the light goes and the engine is much less prone to bogging and the revs won't drop as fast inbetween shifts.

Imagine spinning something light such as a bicycle wheel and then let it hit the ground. Note how fast it stops. Now spin something like a car wheel/tire to the same rpm and let it hit the ground. Should take off! Harder to get to the same rpm yes. But once in motion it tends to stay in motion.

 

i think im going to keep my stock flywheel and just do the jwt clutch. until i boost of course.

 

thanks for the lesson... i think i understand it alot more now. but i dont really drag race but if i wanted to, i guess i will have to get use to the new launch...
im still debating on weather to do it or not.

what to do what to do???

i am buying the clutch today so i am running out of time to think...

 

The only thing that I don't like is the chatter at idea. I got the Fidanaz Liteweight Flywheel because the steel plate can be replaced for less than $100, which beats buying a new flywheel when you replace the clutch.

 

another positive of a lightweight flywheel is that it puts less stress on the crankshaft due to the less weight.

 

I don't know about that. If you have a big heavy weight turning in one direction, it will want to continue to turn in that direction. It also will resist a force that wants to prevent that. ie... engaging clutch. I would "guess" that as the weight of the FW gets lighter, more force is being transmitted into the crank, not the other way around. Either way, I would say the effect is minimal. Any significant transfer of negative forces would probably require a FW weight so light, it wouldn't be practical anymore.

 

Quote:

Originally Posted by in2win

lol show me the math at showing the 50 lb. lighter per pound. It makes the car engine rev faster not drop weight. I'm just tryin to weight the bad with the good. High revs to get moving and harder to rev match on the street vs. faster rpm's.

HAHA you asked for it so here it is:
A flywheel is a rotating disk used as a storage device for kinetic energy. Flywheels resist changes in their rotational speed, which helps steady the rotation of the shaft when a fluctuating torque is exerted on it by its power source such as a piston-based (reciprocating) engine, or when the load placed on it is intermittent (such as a piston pump).


Rotational energy

The rotational energy or angular kinetic energy is the kinetic energy due to the rotation of an object and is part of its total kinetic energy. Looking at rotational energy separately in an object's centre of massframe, one gets the following dependence on the object's moment of inertia:




<DL><DD> </DD></DL>where




<DL><DD>ω is the angular velocity<DD> is the moment of inertia. </DD></DL>The energy required for / released during rotation is the torque times the rotation angle. The instantaneous power of an angularly accelerating body is the torque times the angular frequency.
Note the close relationship between the results for linear (or translational) and rotational motion; the formula for the




<DL><DD> </DD></DL>In the rotating system, the moment of inertia, I, takes the role of the mass, m, and the angular velocity, ω, takes the role of the linear velocity, v. The rotational energy of a rollingcylinder varies from one half of the translational energy (if it is massive) to the same as the translational energy (if it is hollow).
As an example, let us calculate the rotational kinetic energy of the Earth. As the Earth has a period of about 23.93 hours, it has an angular velocity of 7.29×10<SUP>-5</SUP> rad·s<SUP>-1</SUP>. Assuming that the Earth is perfectly spherical and uniform in mass density, it has a moment of inertia, I = 9.72×10<SUP>37</SUP> kg·m<SUP>2</SUP>. Therefore, it has a rotational kinetic energy of 2.58×10<SUP>29</SUP> J.
Part of it can be tapped using tidal power. This creates additional friction of the two global tidal waves, infinitesimally slowing down Earth's angular velocity ω. Due to conservation of angular momentum this process transfers angular momentum to the Moon's orbital motion, increasing its distance from Earth and its orbital period

 

Equations involving the moment of inertia

The rotational kinetic energy of a system can be expressed in terms of its moment of inertia. For a system with N point masses m<SUB>i</SUB> moving with speeds v<SUB>i</SUB>, the rotational kinetic energy T equals

<DL><DD> </DD></DL>where ω is the common angular velocity (in radians per second). The final formula also holds for a continuous distribution of mass with a generalisation of the above derivation from a discrete summation to an integration.
In the special case where the angular momentumvector is parallel to the angular velocity vector, one can relate them by the equation

<DL><DD> </DD></DL>where L is the angular momentum and ω is the angular velocity. However, this equation does not hold in many cases of interest, such as the torque-free precession of a rotating object, although its more general tensor form is always correct.
When the moment of inertia is constant, one can also relate the torque on an object and its angular acceleration in a similar equation:

<DL><DD> </DD></DL>where N is the torque and α is the angular acceleration.