FLYWHEEL MATH

This is a simplified explanation of one way to think about rotational mass in a flywheel, and what effect can be expected from reducing the mass.  Being a sort of conceptual approach, it leaves out a lot of real world factors, but should provide an idea of the principles involved.

Consider a simplified automotive system where a source of torque (the engine) spins a massive flywheel that is also connected through a reduction gear train to a drive wheel in contact with the ground.  The drive wheel converts the torque it receives from the gear train to a linear force on the ground, which can accelerate the car.

If the engine tries to accelerate the car, it must overcome two kinds of resistance to changes in speed:  The flywheel resists angular acceleration through its moment of inertia, and the car itself resists linear acceleration through its mass.  If either one of these kinds of resistance is reduced, the car will accelerate faster for a given torque from the engine.

It may be instructive to find a relationship between these two kinds of resistance.  That is, what reduction of the car's mass will lead to the same acceleration increase as a given reduction of moment of inertia in the flywheel?

To get an idea of this relationship, we can look at two simplified cases:  One with only the effect of the flywheel in play, and the other with only the car's mass considered.

First, let's first look at the system without the flywheel.  We'll have to imagine that the flywheel's smoothing action on crankshaft impulses is accomplished some other way.  For simplicity, we will also ignore the mass of all other rotating parts, too.

The engine's torque undergoes a multiplication in the reduction gear train, so the torque at the drive wheel is

,

where

 is the torque at the drive wheel,

 is the torque from the engine, and

 is the total reduction in the gear train (transmission and differential).

The drive wheel converts the torque to a force on the ground depending on its radius (radius of wheel & tire):

where

 is the linear force due the torque at the wheel, and

 is the drive wheel radius.

Finally, the acceleration due to this force will be

where

 is the acceleration in the no flywheel case, and

 is the mass of the car.

Next, we look at the situation where the flywheel resists acceleration, but we will assume the car is massless, and so doesn't contribute any opposition to change in speed.

A torque on a flywheel will accelerate it:

where

 is the angular acceleration of the flywheel (that's an alpha, and is different from the "" above), and

 is the moment of inertia for the flywheel.

Now an angular acceleration is a change in angular velocity (rotation), and we know that in a reduction gear train, the rotation speed gets slower, so the angular acceleration at the drive wheel is

where

 is the angular acceleration at the drive wheel.

This angular acceleration of the drive wheel will produce a linear acceleration of the massless car

where

 is the acceleration in the flywheel only case.

Since we want to know how these two cases can produce the same acceleration, we'll set them equal to each other, and then see how the mass of the car and the moment of inertia of the flywheel would be related for the same acceleration.

Rearranging, we get

And since the equation is linear in m and I, we can say

This gives us a way to evaluate a change in the moment of inertia of a flywheel to a lightening of the car itself in terms of acceleration.  The  factor represents a multiplier that relates the effect of MI reduction in terms of car mass. 

To think deeper into this, it's easiest to use metric units so we can avoid the whole pound force vs pound mass confusion.

The MI of a rotating mass has units of kg-m².  It is conceptually calculated by adding up the MI of very many very small parts of the entire mass.  The MI of each little part is its mass times the square of its distance from the axis of rotation.  This implies that heavier parts contribute more to total MI, but that parts that are more distant from the axis contribute disproportionally more MI.  This explains why we want to remove most of the mass of a flywheel near the perimeter--the material at the perimeter accounts for much more MI than material near the center of rotation.   The amount of MI reduction then gets multiplied by the  factor to get what mass would have to be removed from the car to get the same acceleration benefit. 

Of course, one sticking point here is that it isn't usually easy to know how much MI reduction we get from relieving a flywheel of a certain number of pounds (or kilograms).  Without getting into pretty detailed calculations involving the shape and size of the removed volume, a rough guess is probably the best we can do.

For example, if we remove three kilograms from a flywheel and that mass had an effective distance from the center of four inches (about 0.1 meter), the MI removed would be about 0.03 kg-m².  In a car with a first gear ratio of 4.0 and a rear end ratio of 4.0 (picking easy numbers), and a tire radius of 0.3 m, the multiplying factor would be (16/0.3)² = 2844.  The units of the factor are m^-2 since the gear ratio is dimensionless.  Taking this times the MI reduction gives about 85 kilograms (about 188 pounds), which is the calculated equivalent  car mass that would have to be removed to get the same benefit.  (It's actually slightly better than this since by lightening the flywheel, you've also lightened the car by three kg already, so the total benefit would be the same as lightening the car by 88 kg, or 194 pounds).  This basically like negating the weight of a substantial passenger.