DNK
Great Pumpkin
Offline
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