The Real Science About Flywheel Lightening

[Holyman]

Hi Dave -

Things have been going well at Princeton

 

Your flywheel question is pretty interesting - I've listed some of the basic theory behind flywheels/rotating masses on the first page, the second page should show you how to apply it to an ATV engine.

 

1) The performance of a flywheel is not determined by mass alone, but rather the distribution and location of the mass with respect to the axis of rotation. The most important characteristic of a flywheel is its Moment of Inertia, which determines its resistance to changes in RPM.

 

The equations governing the behavior of rotating objects can involve some pretty complex calculus and differential equations. Below is the equation used to determine moment of inertia for a mass of any shape, but this simplifies greatly for simple geometric shapes.

 

 

If we assume the flywheel is perfectly balanced and uniform the equations for a round object can easily be solved.

 

 

 

 

For a round, uniform disk flywheel, the moment of inertia is given as J = 0.5*M*R^2 (one half mass times (disk radius)squared). The most efficient flywheel is one that has 100% of its mass located at the rim of the wheel, since this gives the highest moment of inertia per unit mass of 1*MR^2.

 

"J" = Moment of inertia in kg*m^2

M=Mass in kg

R=Radius in Meters

 

So to get the best performance, you need to adjust and tune the moment of inertia, not the mass.

 

As you know, piston engines only produce power during the short power stroke following combustion, the rest of the engine cycle requires the engine to be externally driven during intake and compression. The purpose of the flywheel is to store some of the energy from the power stroke to be used to turn the engine during the rest of the cycle. If the flywheel's moment of inertia is too low, the engine will run extremely rough since it will be on the verge of stall during compression. Vibrations will be very high since the engine will accelerate rapidly during the power stroke, and greatly slow during the compression stroke. Under these conditions, the engine will produce virtually no power, and the slightest load will stall it. An engine with an oversized flywheel will be difficult to start since it will not want to accelerate, but eventual operation will be very smooth and powerful, and it will resist any sudden change in RPM. This is why power plant generator engines usually have massive flywheels weighing several tons so they output a constant smooth RPM even if there are sudden changes in loading. However, ATV's require a balance between the rapid acceleration of a small "J" flywheel and the power and smoothness of a high "J" flywheel. A perfectly optimized flywheel will store enough energy to power the compression stroke of the engine, but will have a moment of inertial low enough to allow rapid acceleration. At low revs, the compression stroke takes longer so a larger "J" is required than at high revs where the compression stroke is only a few milliseconds and only a small "J" is needed.

 

 

Flywheel performance summary:

Higher moment of inertia: Increased low rpm power, less vibration, smoother operation, but: reduced acceleration, increased engine weight, harder to start

Lower moment of inertia: Faster acceleration, reduced engine weight, easier to start, but: slightly reduced power at low rpm, increased vibrations, can stall easily if clutch engaged too quickly

 

Also note that by increasing the number of cylinders in an engine, the time periods where power is not being produced by any cylinder are greatly reduced, so less flywheel is needed. If an engine contains 9+ cylinders, it can operate decently well without any flywheel since the time periods where no power is produced are reduced to zero, but small flywheels are still used to improve operating smoothness. This is one of the reasons V-10 and V-12 engines are used in high performance sports car engines instead of 4 big cylinders.

 

Mathematically calculating the ideal flywheel "J" involves numerous differential equations and measurement of over 50 engine parameters, so I would recommend selecting a flywheel through experimentation.

 

So getting back to question #2, removing mass from the edge is much different than removing mass from the halfway point between axle and rim. Using the 0.5MR^2 equation, taking 5ounces from the rim will reduce your "J" by 4 times more than taking 5ounces from the halfway point. Theoretically, if it was possible to remove 5 ounces from the exact center of the wheel it would have no effect on "J", since the radius would be 0. Just make sure the wheel remains balanced or the performance will suffer since it will no longer be rotating around its center off mass, invalidating the equation. Likewise, removing some mass from 1/3 the distance from axle to rim would have 1/9 the effect as removing the same mass from the rim.

 

The ideal flywheel is a hoop of mass, but this would be difficult to mount on a shaft, so the realistic ideal flywheel is one that uses a thin disk with a thickened outer rim. This gives the best "J" per mass. Mass located near the axis of rotation has little effect on "J" and just adds to engine weight.

 

APPLYING TO ATV's:

 

You need to optimize "J" for desired acceleration, power ranges, smoothness, and RPM.

 

The flywheel included on stock motors is usually pretty well matched to the engine, but reducing "J" often can increase performance to some extent. When the clutch is engaged and the vehicle is moving, the drivetrain and wheels function as a virtual flywheel and can prevent the engine from stalling if the engine flywheel was a bit small, but problems can arise when accelerating from a a stop or shifting gears - when the drivetrain is not engaged as an additional flywheel, the engine is much more likely to stall under a sudden increase in load.

 

Conclusion:

An engine with a low "J" flywheel will rev faster, improve acceleration, improve cornering, but will be a bit more prone to stalling and will require a bit more skill to drive, especially when engaging the clutch from a standstill. As long as the engine revs are kept high and the wheels are in gear, the ATV should perform better.

 

A larger "J" will be much easier to drive, and more forgiving when engaging the clutch, but the acceleration and performance at high RPM's will be reduced.

 

Mass is only a minor factor in determining moment of inertia - The radius of the mass from the axis is much more significant. For example: A 10kg flywheel can actually have a smaller moment of inertia than a 2kg flywheel. Assume Flywheel 1 is a plain disk of 10kg, with a 10cm(0.1m) radius and Flywheel 2 is a plain disk of 2kg with a 30cm(0.3m) radius:

 

Flywheel 1 "J" = 0.5*10kg*0.1m^2 = 0.05kgm^2

Flywheel 2 "J" = 0.5*2kg*0.3m^3= 0.09kgm^2

 

In this situation, the 2 kg flywheel would have nearly twice the moment of inertia of the 10kg flywheel, even though it weighs 5 times less. The engine would not know the difference, it only responds to moment of inertia, not mass. It would seem that cars would use lightweight flywheels of enormous radius to take advantage of this fact, but there are also realistic limits - Larger radius flywheels have higher rim velocities, and can develop stress/rupture problems at high rpm - the centripetal forces also quadruple with a doubling of radius. Also, a large flywheel at high rpm has more air resistance. A giant flywheel could also start to introduce gyroscopic effects and could fight rapid turing of the vehicle, but this is insignificant for any realistically sized flywheel that would fit in an ATV.

 

If you dyno test an ATV with a standard flywheel vs a lightened flywheel at a constant speed, there should be no difference at all, but if you dyno test starting from zero to full speed, the lightened flywheel engine will produce significantly more average power than a standard engine in the first 10 seconds.

 

I would recommend machining a selection of flywheels starting at the same "J" as the factory, then 0.95* " factory J", 0.90*" factory J", etc... and testing each one.

Just remember the equations are non-linear, so

 

Solid, plain disk "J" = 0.5*M*R^2; M=PI*R^2*T*DENSITY; "J" = 0.5*PI*T*DENSITY*R^4; where PI~ 3.14, T= flywheel thickness in Meters, Density = kg/M^3, R=radius in meters

For aluminum, DENSITY=2700kg/m^3

For Steel, DENSITY=7870kg/M^3

 

But if you make disks of constant thickness and density, everything cancels out and the "J" ratio will be simply related to R^4.

So if you want to make a flywheel disk that has 90% the "J" of another, just take the fourth root of the percentage/100, which will give you the radius required. For example for 90%:

90% /100 = 0.90 taking the fourth root is the same as raising to the 1/4 power so (.90)^0.25 = 0.974. So to create a disk with 90% the J of another, you only need to reduce the radius to 0.974 times that of the original - removing mass at the very edge has a very significant effect! An 80% disk would require a radius of 0.946 times that of the original, and reducing the radius to 0.841 times the original cuts the "J" in half, so be careful when lightening a flywheel - The optimal performance point probably only involves the removal of a very small amount of mass. Most real flywheels are more complex shapes than a simple disk, but the principle is essentially the same - a flywheel with a thickened rim will be even more sensitive to mass removal from the edge, so only take tiny amounts off at first. Be careful not to weaken the rim too much - this is the highest stress area of the wheel.

 

I hope this info helps you understand how a flywheel really works and why it is "J", not mass that you need to adjust - there are also many more equations which explain how much power is stored in a flywheel, and how much torque it takes to spin it up - let me know if any of this would be helpful.

[raxen]

Radius of Gyration

Moment of Inertia

Center of Mass of a distibution

 

Damn it has been a few years since that statics and dynamics class but it is pretty familiar and good stuff.

[Justintoxicated]

oh man.... you reminded me of that nightmare I had in college called Physics 195.

 

LOL just kidding, makes sense. But what about the simple reduction of spinning mass in general? Is it negated since the weight of the flywheel itself is more critical?

[Holyman]

The way I read it, the reduction of mass is better taken from as close to the center of the mass as possible. His example of a 2oz flywheel that is large is better than a 10oz flywheel that is small seems to show that. I never went to physics 195 though so you'll have to draw your own conclusions

[Holyman]

bump

[FORMULASPEED]

Then create a fly wheel set up like a sprocket hub setup.

The hub mounts on the crank the outer rim mounts to the hub.

Now If I had a machine shop and Dyno. Dam

[FORMULASPEED]

Did anyone else see the Chopper build off, on the learning channel.

That one builder created the centerless back wheel. Its was trick as all hell.

Just brought that up cause it kinda relates.

[blk02banshee]

what I got out of that, is the weight of the outer part of the flywheel and the diameter has more effect than the weight of the flywheel as a whole.

Basically, if you could make the center of the flywheel weightless, it would still function the same, because it's the outer diameter mass that does the work.

for example, if you could shave the same weight by elongating the holes on a banshee flywheel, as shaving the outer part like boonman does, it would have less effect on rotating mass, because the holes are closer to the center of the flywheel.