DNK

06-17-2008, 12:32 PM

Thought everyone would like this about their lightened flywheel. It was posted on Woody's site. It's long, but good info

A while back I ran the numbers for switching from an iron to aluminum

flywheel for my Triumph TR8. 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 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

car 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. If that 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 2.7Cool. 2.78 * 180 is 500 lbs equivalent weight

reduction. A non-trivial effect.

I don't recall whether the TR8 flywheel has the big hub like the

GM 215 flywheels so I ran the numbers both ways, 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

If the stock flywheel has the big ring, then lightening it is similar

to removing 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 10

pounds from the perimeter is equivalent to 493 lbs weight

reduction in first gear while across the face was equivalent

to 177 lbs. Reducing from the hub is even less effective since

the effect is a function of the diameter squared.

Dan Jones

St. Louis, MO

1980 TR8 Convertible

1977 TR7 Coupe

A while back I ran the numbers for switching from an iron to aluminum

flywheel for my Triumph TR8. 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 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

car 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. If that 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 2.7Cool. 2.78 * 180 is 500 lbs equivalent weight

reduction. A non-trivial effect.

I don't recall whether the TR8 flywheel has the big hub like the

GM 215 flywheels so I ran the numbers both ways, 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

If the stock flywheel has the big ring, then lightening it is similar

to removing 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 10

pounds from the perimeter is equivalent to 493 lbs weight

reduction in first gear while across the face was equivalent

to 177 lbs. Reducing from the hub is even less effective since

the effect is a function of the diameter squared.

Dan Jones

St. Louis, MO

1980 TR8 Convertible

1977 TR7 Coupe