Reply to "Aluminum flywheel?"

It's not a Ford 351C in a Pantera but the physics still apply. A while
back I ran the numbers for switching from an iron to aluminum flywheel
for my Triumph TR8. The effect was surprisingly large. There are a
couple of approaches to doing the math. The more rigorous approach is
to calculate the polar moment of inertia for the two different flywheels,
adjust for the square of the overall gearing (transmission, final drive
and tires) and convert to an equivalent linear inertia. The second
method (the one I chose) is to start with a known linear to rotational
equivalent and ratio from there. The known relationship I used is a
solid disk rolling on its edge. It has an effective inertia exactly
1.5 times what it would be if it wasn't rotating. That means the
rotational component is 50% of the linear component. Adjust for the
square in gearing and you have the answer. I wrote a little Fortran
program to do the calculations. I assumed a 12" diameter flywheel
which is the Buick/Rover diameter, less the ring gear. The circumerence
of a circle is the diameter multiplied by pi. So if you roll the
flywheel along the ground it will move 37.7 linear inches per
revolution (= pi * 12). A 205/50/15 has a diameter of approximately
23.1 inches. My TR8's final drive ratio is 3.45:1 and first gear is
3.32:1 so one revolution of the flywheel results in the Triumph TR8
moving around 6.3 inches. Ratio the squares and take half
((37.7/6.3)**2)/2 = 17.9. So each pound removed from the flywheel
(equally across the face) is the same as about 18 pounds of weight
removed from the car when in first gear. So if you remove ten pounds
from the flywheel (equally across the face), the result is equivalent
to removing 180 pounds of vehicle weight in first gear. The effect
goes down for each higher gear, of course. Removing weight farther
from the rotational axis has a more pronounced effect. If the weight
is removed from the outside of the flywheel only, the effect is about
2.78 times as strong since a solid disk has a radius of gyration of
0.6 times the radius (1.0/0.6)**2 is 2.78). 2.78 * 180 is 500 lbs
equivalent weight reduction. A non-trivial effect, particularly
in a lightweight car. I ran the numbers a couple of ways to
illustrate. For my TR8, assuming a 3.45:1 final drive ratio,
205/50/15 tires and LT77 gear ratios of:

1st 3.32:1
2nd 2.09:1
3rd 1.40:1
4th 1.00:1
5th 0.83:1

along with flywheel weights of:

stock flywheel - 32 lbs
lightened steel - 22 lbs
aluminum - 11 lbs

The engine in the TR8 is essentially a Buick 215 aluminum V8 from the
early 1960's. The stock flywheels in those had a big ring around the
perimeter. Lightening the flywheel by milling off the ring is similar
to removing the mass from the perimeter (from 32 to 22 lbs). In the
numbers below, I didn't do it that way but a more accurate approach for
the aluminum flywheel would be to assume a reduction of 22 to 11 lbs
equally across the face and add that to the difference of the 32 to
22 lbs across the perimeter. In any event, a lighter flywheel looks
like a good thing to do for performance. Here are the numbers:

32 to 22 lbs (across face assumption):
1st 177.5 lbs
2nd 70.3 lbs
3rd 31.6 lbs
4th 16.1 lbs
5th 11.1 lbs

32 to 22 lbs (perimeter reduction assumption):
1st 493.4 lbs
2nd 195.5 lbs
3rd 87.7 lbs
4th 44.8 lbs
5th 30.8 lbs

32 to 11 lbs (across face assumption):
1st 372.7 lbs
2nd 147.7 lbs
3rd 66.3 lbs
4th 33.8 lbs
5th 23.3 lbs

32 to 11 lbs (perimeter reduction assumption):
1st 1036.1 lbs
2nd 410.6 lbs
3rd 184.2 lbs
4th 94.0 lbs
5th 64.8 lbs

Rotational inertia is mass multiplied by the distance from the
rotational axis (integrated over the surface). The effect is
stronger farther away from the hub. The best is from the
perimeter. Equally across the face is less effective and near
the hub is the least effective. In my example, dropping 21 lbs
from the perimeter is equivalent to over 1000 lbs reduction in
weight in first gear. Dropping the same mass the face is equivalent
to 372.7 lbs.

Reducing the flywheel inertia does reduce the stored energy for
start from a stop. Torque follows displacement. Little engine
in big car with tall gearing needs more stored inertia at start.
Big engine in little car with short gearing can get away with
much less stored inertia. On the street with a ligter flywheel,
you may need to use more RPM and clutch slip. On the strip, you
may bog if you don't have enough excess torque at the rear tires
(more traction than engine/gearing). Remember that HP is the
measure of how much potential torque you can have at the rear
tires via gearing. If you have enough power to overcome your
traction, then a heavy flywheel is a loser.

Dan Jones