04-04-2006, 01:02 PM
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CC Member
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Join Date: Aug 2000
Location: Chicago,
IL
Cobra Make, Engine: ERA 454 S.O.
Posts: 1,684
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Quote:
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Originally Posted by Clois Harlan
If your 428 flywheel came from McLeod there was a small rectangular weight (28 oz) that bolted to the top of the flywheel on the clutch side. This weight can be removed to achieve zero balance. Also, your harmonic balancer will have a 28 oz weight inside for balance. I know several people that are running an aluminium flywheel in their Mustang.
Clois
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Not to be picky, but it is not a 28 oz. weight. What 28 oz. refers to is 28 inch ounces. At a 1 inch radius, the weight would truly weigh 28 ozs. Seeing that this weight is bolted on the end of the flywheel (about a 6" radius), the weight actually weighs 28/6 = 4.67 ozs. If you don't believe me, pick up the weight, it is nowhere near 1.5 lbs.
It is entirely possible to remove weight from the opposite side of the wheel to simulate the extra weight needed on the counter balance side (inline with the crank lugs). 4.67 ozs. = 1 cu. in. of steel or 2.85 cu. in. of aluminum. To achieve this with drilling, there would have to be about 10 .5 in. holes x .5 in. deep in steel or 28 holes of the same dims. in aluminum. These calculations are close, they do not take into account the tip of the drill point if they are not through holes.
If you need help with the calcs. reply with your number, dia., and depth of the holes and I can help you.
__________________
Jeff
“If you can make black marks on a straight from the time you turn out of a corner until the braking point of the next turn, then you have enough horsepower.”
Mark Donahue
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