Does a lighter Flywheel = more HP?

[moersfelng08]

Was having this discussion yesterday and someone was convinced that a new lighter flywheel does not actually provide a HP gain. This seemed counter intuitive to me, since you're reducing mass and therefore freeing up more power. For instance on my SR20 I have a JGY 8.8lb flywheel, considerably lighter than the stock one and while I haven't been able to afford dyno time to check it feels like more power at the wheels.

So this is a two part question. Does it actually change power output, and by how much (i know this is dependent on the actual weights so use hypothetical).

[raremotive]

Was having this discussion yesterday and someone was convinced that a new lighter flywheel does not actually provide a HP gain. This seemed counter intuitive to me, since you're reducing mass and therefore freeing up more power. For instance on my SR20 I have a JGY 8.8lb flywheel, considerably lighter than the stock one and while I haven't been able to afford dyno time to check it feels like more power at the wheels.

So this is a two part question. Does it actually change power output, and by how much (i know this is dependent on the actual weights so use hypothetical).

Accelerating, yes results you with more recovered power. At constant speed, no.

We are dealing with a mass. To be more accurate putting a mass in motion which results in the inevitable moment of inertia.

Lets start with a concept of a bowling ball. You can choose a 8 lb bowling ball, it accelerates quickly, it's easier to control, but however at the end of the lane it loses a lot of it's momentum. You also don't feel tired tossing it around. Next is a 14 lb bowling ball, it's hard to push, it resists change it's direction making it hard to control, but however when it reaches at the end of the lane it still has it's velocity/momentum to give you your desired strike.

Now to bring things in picture as in car terms. With a lighter flywheel, you will experience that it has a lost of torque starting from rest requiring you to gas it more. Same with downshifting to engine brake, the light flywheel will not resist change resulting the engine reving up immediately...you won't feel the car slow down much. However with a heavier flywheel, moving from rest is easier because of the flywheel maintains it's momentum. Same for downshifting, the heavier flywheel will resist change in speed which slows the engine revs up to speed of the wheels resulting a more effective engine brake.

Then at constant velocity with a light flywheel, since it loses it's momentum quickly it requires more energy to maintain that speed. However, with the heavier, it's resistance in change maintains your speed. This is why your fuel economy suffers when you switch to a lighter flywheel.

Now..let's say I wanted to accelerate from rest to 60, the lighter flywheel will allow the engine rev up quicker, requiring less energy as I am constantly changing it's speed. This is where power recovery comes in play, from lighter rotational mass. Where the heavier flywheel will be constantly resisting you. If you are city driving with constant changing, your fuel economy would prove to be better with a lighter flywheel.

You will also notice with a lighter fly wheel, you will be able to control the engine with your throttle a bit better; more responsive in that matter. The lighter flywheel allows shifts to be better/quicker since it lets the rpm match the transmission quicker. Lighter flywheel lets your engine start up quicker.

In perspective. Your friend is right, there is no power gain because it's all a wash. However if you look a single aspect of what the car will be doing, you can be losing power or recovering lost power with a lighter flywheel.

And from what you have read. There isn't one way that the best way. What I mean is, you have to decide what you want to use your car for and then choose the parts... heavier fly wheel or lighter flywheel with in mind the pros and cons of each.
Modified by Rare_f8 at 2:40 PM 1/24/2009

[breadbox]

^quite the answer.

Lighter flywheels rev faster and drop revs faster.

Heavier Flywheels do not drop revs as fast therefore easier to keep it the torque band.

[c72c]

agreed... quite the answer. now i gotta ask. if a lighter flywheel in terms 'reduces' torque, whats the best way to go? which would be best to use it for; drift, drag, touge, ghymkana? not as the preference of what you wanna do but, when would it be more effective/usefull?

[ravera]

That's going to depend heavily on a lot of things, mostly torque (HP really doesn't matter because we're talking about inertia here), application, and the weight of the car. I'm willing to bet if more S1x owners left heavy flywheels on their car they'd see 60ft times reduced because it can store more energy while you stage it so when you let off the clutch you'll get a harder launch to get that thing moving with a little turbo 4. Drifting and road racing both depend on being able to change direction and RPM quickly so for that I'd tend to say a lightened flywheel will assist. For street you'd probably want something a little lighter than stock for response but at the same time you really want to keep a little weight in it so you can get off the line without having to rev the motor and be able to make smoother starts and shifts.

[seang]

For reference, the flywheel in my truck (ka24e) is like 22 or 25 lbs.

[RBbugBITme]

Then at constant velocity with a light flywheel, since it loses it's momentum quickly it requires more energy to maintain that speed. However, with the heavier, it's resistance in change maintains your speed. This is why your fuel economy suffers when you switch to a lighter flywheel.

I'm here to pick a fight

I don't agree with this statement. Once your engine is up to cruising speed there is no energy transfer to or from a flywheel. It is an energy storage device and any change in the energy of that device requires an acceleration or deceleration event which obviously isn't occuring if you are traveling at a constant speed.

[raremotive]

I'm here to pick a fight

I don't agree with this statement. Once your engine is up to cruising speed there is no energy transfer to or from a flywheel. It is an energy storage device and any change in the energy of that device requires an acceleration or deceleration event which obviously isn't occuring if you are traveling at a constant speed.

"it's resistance in change maintains your speed" is the same as your "It is an energy storage device and any change in the energy of that device."

The principle is from Newton's law of once an object is in motion, it stays in motion.

Notice I used momentum for lighter flywheel? Momentum is based on mass and velocity. Ever spin a heavy bowling ball and a light one? Let's say we put the same amount of energy into spinning both balls. Which one stops first? The light one, right? Well, in order to keep the light one spinning to match the heavier ball duration of spin we have to give the lighter ball more energy. Because it lacked momentum.

Now back to flywheels, at constant velocity, you are going to have to put in more energy for the lighter flywheel to make up for the energy it did not store.

Also, remember you are moving a constant mass (the car), the flywheel is not a free body.
Modified by Rare_f8 at 8:53 AM 10/30/2009

[RBbugBITme]

I think I see what you're saying, if you talking about cruising then letting off the gas and getting back on the gas to the original cruising speed.

To clarify, I was talking about how energy from combustion to the wheels "passes through" the flywheel while cruising as if it wasn't there.

Your comment raises another question in my head but I'll wait till you respond to see if its worth asking.

[raremotive]

I edit some more.

[RBbugBITme]

Quote »Now back to flywheels, at constant velocity, you are going to have to put in more energy for the lighter flywheel to make up for the energy it did not store.[/quote]This doesn't make sense. You obviously know that a car can accelerate faster with a lighter flywheel but you're saying it requires more energy input as compared to the heavier flywheel. Where is this extra energy coming from that increases the acceleration rate of the car? Less energy is going into a lighter flywheel, all dimensions kept the same, and more is passing straight through to the wheels.

An 12" flywheel at 11.5lbs and 3000 rpms stores about 2/3 the energy of a 15lb flywheel of the same dimensions and rpm. You need to clarify where this extra energy you're talking about is going and in what driving situation. Transient or steady state?

[raremotive]

Acceleration needs less energy for the light fly wheel because it it has a lower inertia force resisting the change in motion.

However, in steady state it will require more energy to maintain a constant velocity compare to a heavy flywheel.

Let's just say. You're not losing anything from the light flywheel because it weights next to nothing. But you are not benefiting from the rotational momentum of the heavy flywheel maintains, which in turn allows the engine to work less. Now we are talking how we are loading the engine.

[RBbugBITme]

But that goes against what is in motion tends to stay in motion. You're saying that it requires constant energy input to maintain its rotational velocity, just less so for the one with more momentum. This violates Newton's law as I see it.

[raremotive]

hahaha, welcome to the real world.

Why does the spinning bowling ball stop?

[RBbugBITme]

However, in steady state it will require more energy from the engine to maintain a constant vehicle velocity compare to a heavy flywheel.

If that is what you meant then I think I buy that.

[RBbugBITme]

I've been neglecting friction and other nasty forces from the start.

[raremotive]

Also, remember you are moving a constant mass (the car), the flywheel is not a free body.

I been too, but I been thinking it as a resisting force, or a degrading vehicle velocity that requires input.

The example above is friction.

But the whole problem is road to tire, wind resistance, friction... all these that can try to stop the flywheel from having Newton's law. But all held at a constant.

Yes, I been talking about maintaining a vehicle 60 mph on the highway, with the applied resistance constant above. That is the final intended result, vehicle dynamics, no?
Modified by Rare_f8 at 9:37 AM 10/30/2009

[RBbugBITme]

I take it back, I don't buy it... Lets break it down Hammer style.

Ok, picture a shaft that is rotating at 50 rad/s. At one point energy was input into it and there are no frictional forces or driving forces anymore. It will continue to spin at that speed forever.

Now add your friction, wind resistance, etc. to one end and an engine on the other with an equivalent input force. It is still rotating at 50 rad/s correct?

Now take that same shaft with a motor and a frictional force, and throw a flywheel on right in the middle. What happened to the system? There is more total energy in the system because some was stored initially in the flywheel but now it is still rotating at 50 rad/s. There is no energy going into or out of the flywheel and the motor, at this steady state condition, does not have to output more energy than before to maintain that velocity.

If you turn the motor off now, the energy from the flywheel will resist the frictional force at the other end of the shaft because the flywheel doesn't want to slow down. It will do so until 0 rad/s. The rate that this happens at (50-0 rad/s) will be dependent on the amount of energy and momentum that is stored in the flywheel to combat it.

[raremotive]

Now add your friction, wind resistance, etc. to one end and an engine on the other with an equivalent input force. It is still rotating at 50 rad/s correct?

That's your flaw. Engine and Resistance Force are not the same, because the power generated is not independent from the resistance force, it's depended on it, plus whatever resistance the flywheel produces, a variable. Your power generated also depends on how the pistons are loaded. Which converts rotational to linear motion. Greater momentum reduces the on the pistons on the downward stroke, therefore less energy wasted. From there it's thermodynamics, fluid dynamics, and the moving boundary layer.
Modified by Rare_f8 at 5:50 PM 11/1/2009

[RBbugBITme]

plus whatever resistance the flywheel produces, a variable.

There is your flaw. Please show me why in physics terms how a flywheel is a variable in a steady state system. It is a damper on the overall system and you just brought in a whole bunch of irrelevant fun facts to a discussion about a very simple object.

If you're cruising along a flat stretch of road and start driving up a hill all of a sudden, you will decelerate more quickly with a lighter flywheel. That is a TRANSIENT situation. You will need to more quickly react with your foot to counter the increased load and more slowly with a heavier flywheel. This is really simple physics stuff, pull your book back open and think about my simple and accurate system I described above. Cut out the unnecessary crap and take a macro look at it (no piston loads, fluid dynamics, and other silly stuff). You have power delivery at the crank, resistance and the wheels (and drag if you feel so inclined to include that which is also a constant at constant velocity), and a damping mechanism in between.

[raremotive]

You are missing the point. Variable also means you can change it to what ever set value you want it to be.(You know, stuff you learn in math and computer science) I never meant it's was changing on it's own.

Your simple system doesn't work. Because the flywheel has a mass in motion, and a body in space being affected upon. The power produced does NOT SIMPLY EQUAL the resisting forces before the flywheel, even given in the most ideal situation. In energy analysis, Ein-Eout=Etot. Since steady state, Etot is equal zero. So Ein=Eout. Which you stated. Engine in.. and Resisting out.. the same. But, one crucial fact is missing. At some point the flywheel had energy put into it, it has mass and it's moving! This energy is called kinectic energy, 1/2mV^2... So when you plug it in the Ein+KE=Eout... it's now a inequality because you naturally assumed Engine and resisting energies are the same.

Even if we have a constant speed system, the resisting forces are producing a negative moment(Mr) about the axis, and since the heavy flywheel still wants to spin because of it's own momentum in a positive direction it applies a positive moment(Mf) about the axis..now in which naturally based on road resistance and wind drag and other components, you can assume Mr > Mf.. which in turns produces a smaller negative moment for the engine to match. I am sure you have taken statics at some point to recognize that you can have resisting moments in a steady state.
Modified by Rare_f8 at 4:50 PM 11/3/2009

[RBbugBITme]

In energy analysis, Ein-Eout=Etot. Since steady state, Etot is equal zero. So Ein=Eout. Which you stated. Engine in.. and Resisting out.. the same. But, one crucial fact is missing. At some point the flywheel had energy put into it, it has mass and it's moving! This energy is called kinectic energy, 1/2mV^2... So when you plug it in the Ein+KE=Eout... it's now a inequality because you naturally assumed Engine and resisting energies are the same.

Well we can throw out this whole long incorrect diatribe because if you read what I wrote... Quote » Now take that same shaft with a motor and a frictional force, and throw a flywheel on right in the middle. What happened to the system? There is more total energy in the system because some was stored initially in the flywheel but now it is still rotating at 50 rad/s.[/quote]You have your kinetic energy stored initially during acceleration. THE WHOLE POINT OF THIS IS THAT IT TOOK LESS FROM THE ENGINE TO DO IT and it doesn't leave the system until you're off the gas and decelerating. This is essentially the only point I was trying to make and you have just demonstrated that you have trouble grasping it. I have wasted enough time on here with you, if another engineer would like to chime in please do so.

And yes I understand a heavy flywheel has a smoothing effect you can actually feel at idle speeds. I'm not getting into the minute energy input differences you're talking about that only get smaller and smaller to the point of being negligible as engine speed rises, especially when you don't understand the system and you're talking about a different system with different and more dynamic resistant forces (solely from the engine) that aren't negligible at 0 vehicle velocity.

Quote »Even if we have a constant speed system, the resisting forces are producing a negative moment(Mr) about the axis, and since the heavy flywheel still wants to spin because of it's own momentum in a positive direction it applies a positive moment(Mf) about the axis..now in which naturally based on road resistance and wind drag and other components, you can assume Mr > Mf.. which in turns produces a smaller negative moment for the engine to match. I am sure you have taken statics at some point to recognize that you can have resisting moments in a steady state.[/quote]You got it right here! And you won't find anywhere in here where I said this situation would be wrong, in fact, from my clarification of your earlier statement I said the same thing in a single sentence... Quote »However, in steady state it will require more energy from the engine to maintain a constant vehicle velocity compare to a heavy flywheel.[/quote]. This is Conservation of Momentum as you have spoken about so much. Less from the flywheel means more from the engine.

Before I leave lets recap how we got here. You something that didn't make any sense in a constant velocity situation.Quote »Then at constant velocity with a light flywheel, since it loses it's momentum quickly it requires more energy to maintain that speed. However, with the heavier, it's resistance in change maintains your speed. This is why your fuel economy suffers when you switch to a lighter flywheel.[/quote]I hope at the very least you realize that the flywheel does not lose momentum at constant velocity.

Good luck in school.McG,Over and out
Modified by RBbugBITme at 9:51 AM 11/3/2009

[raremotive]

Good luck, as well.


Modified by Rare_f8 at 4:29 PM 11/3/2009

[flinterman2000]

Does it actually change power output, and by how much (i know this is dependent on the actual weights so use hypothetical).

I read the whole thread and Rare_f8 and RBbugBITme strayed from the answer, although both have their points. The answer is NO there is no change in power output. It only feels so because there is less weight through the drive train for the same power of the engine to convert from reciprocal movement to circular motion. In other words you will get faster engine acceleration feeling as more power but you will lose economy as the engine has to work harder to maintain speed just like both of them had said. The more work the engine does the more heat it generates, more heat means more fuel and so on.