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Machining Flywheels


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Flywheels utilise angular momentum to keep an engine rotating. So if Flywheel 1 has a larger angular momentum than Flywheel 2, the engine that Flywheel 1 is attached to will accelerate and decelerate slower than the engine Flywheel 2 is mounted on. This larger momentum is useful especially in road cars for going up and down hills. But if you want an engine that gains revs quicker, lightening the flywheel (lowering the momentum) will help. But torque and the radius of the flywheel is also related to the angular momentum of the flywheel.

 

Angular Momentum = Moment of Inertia x Angular Velocity

L = Iω

I (moment of Inertia) for a disc = 1/2mr2 where m is mass in kg and r is radius in meters.

ω (Angular Velocity) = θ/t where θ is the angle in radians and t is time in seconds.

(note: there are 6.28 radians in 360 degrees)

 

So if we have 2 Flywheels (discs) with different mass (eg 10kg and 9kg) but the same radius (eg 0.2m) the momentum of the flywheels will be different. If they both have a speed of 1000 rpm then

 

ω = ((1000/60) revs/second x 6.28 rads) / 1second = 104.67 radians per second

 

L1 (Angular Momentum of flywheel 1) = ½ x 10kg x 0.22 x 104.67 = 20.93 kgm2s-1

 

L2 (Angular Momentum of flywheel 2) =1/2 x 9kg x 0.22 x 104.67 = 18.84 kgm2s-1

 

That’s a fairly big difference in momentum between the two flywheels. But then there is also the torque of these flywheels;

 

Torque (τ) = I x Angular Accleration

 

In this case Angular Acceleration is zero, because they are rotating at a constant speed.

 

τ1 = ½ x 10 x 0.22 x 0 = 0.2 Nm

 

τ2 = ½ x 9 x 0.22 x 0 = 0.18 Nm

 

The torque is not the same! For the torque of each flywheel to be the same the radius must change. Using the Torque value of the first flywheel (where its radius was 0.2m);

 

0.2 Nm = ½ x 10 x r12 = ½ x 9 x r22

 

0.2 = ½ x 9 x r22

 

r2 = √(0.2 / (1/2 x 9)) = 0.211m

 

That is a difference of 11mm for the same torque value, which is quite large in engine terms.

So this shows that for a lighter flywheel the radius (or where the radius at which a large portion of the mass is) needs to be greater to retain the same amount of torque.

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the only stuff i really understand of lightening a flywheel is that

A> don't do it around the centre cause the bolts need the meat to hold it to the crank etc.

B> don't do too much around/behind where the clutch surface is cause its needed for heat removal, less chance of warpage, less prone to cracking etc

 

and

 

C> make sure its balanced

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Firstly, tell me what's wrong with this equation:

 

τ1 = ½ x 10 x 0.22 x 0 = 0.2 Nm

 

You're arguing a moment of inertia problem; two different flywheels with two different moments of inertia. Sure, they both require different input torques to overcome that MOI and accelerate them. In the reverse, when they are already spinning and you draw power from that rotation, each can then apply different torques.

 

You're correct in the sense that if you lighten a flywheel, you reduce the energy in a flywheel that is spinning at high revs. What you have forgotten is, where did you get all the energy to speed up the flywheel in the first place?

 

From your engine?

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I understand the logic of it, as i did physics in school.

 

To sum up what KENut was saying (I think).

 

If you want to lighten the flywheel and still retain your original torque, by however much you lighten it you have to figure out how much bigger your flywheel has to be in terms of radius.

 

So...

 

With my TODA racing flywheel of 3.9kg @ 200mm diatmeter, versus stock flywheel of 7kg @ 200mm diameter (this is on a 4age).

 

I will lose a shitload of torque, but if i got a lightweight flywheel of the 212mm (this size can fit in a T-50) i would approximately (because i CBF working it out) get the torque of say a 5ish kg flywheel if it was 3.9kg @ 212mm versus 7kg @ 200mm.

 

But since it is lighter the engine can spin easier and put more power out and possibly more torque?

 

I dunno, ill put it on the dyno and see when its done :rolla:

 

Travis

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In reponse to Philbey's question, yes the equation (and the other equations) τ1 = ½ x 10 x 0.22 x 0 = 0.2 Nm is wrong. The underlined 2 is meant to be the square of 0.2. I was typing this in Word and just copied it into the reply mewhatsit and didn't bother reading after I posted it, so sorry about that. (The other equations have squares in them too). I can fix it but meh, I'm sure no-one really cares that much. And basically I was referring to the flywheel as a single uniform disc, completely seperate from the engine, not taking into account the energy taken from the engine to rotate it. I was just saying that for the flywheel to continue to do its job (impart momentum on the engine to promote smooth rotation) properly, a lighter flywheel must have a larger radius. So what Trav_555 said, but I dunno bout the engine spinning easier/more torque output bit i.e. I'm sick and too lazy to even think about it.

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Nice

in a bad way.

 

Which more or less goes with what i said in my post #19

 

Which is also why drag racers have to have a shield over the bellhousing - you wouldn't even realize that your foot is missing till you look at it and wonder why all the blood

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