Is a lightened flywheel overrated? (I think so)

[MikeK]

I guess (and I do mean guess!) it depends on where you want the crank to break. The traditional place is for the rear to snap off. Any piece of metal can only take X number of stresses before fracture. That number is a function of the degree of flex and the number of cycles. An 80 year old crank is already way towards that point. Adding throw counters and reducing the flywheel weight changes the dynamics of the shock waves, or flexes. You now have less at the rear flange and higher stress in other places. (Here a front dynamic element would be nice.)

As a result, you will get more remaining run time on an already 80 year old crank than would remain with the heavier wheel in place. Simply, you are gambling that your chances of crank failure will be reduced. Now, if you can find a NOS crank (keep dreaming) with zero flexes, you'll gain almost no confidence against failure unless you live another 80 years. It would be someone else's problem.

OK, now let's look at a NEW crank. Not NOS without counterweights, a new Scat, Crane, Burlington, etc. with counterweights. None (Well, maybe a $3K custom) is 100% countered for each throw. That crank will STILL always be doing a harmonics dance back and forth along it's length while running. ANY inertial dynamic introduced at either end will make the stresses unequal along the length. Now remember X number of stresses before fracture, that number a function of the degree of flex and the number of cycles. For maximum life of an I-4 flat crank, on each external end of the crank you need a dynamic(inertial) mass equivalent of sq.root of 2 (1.414) times the inertia of the adjacent crank element (throw). Nobody runs a front flywheel (solid damper) that big, and nobody has a rear flywheel that small. The best compromise is a rear flywheel as light as you can get. Of course, even with a heavy stock fly-anchor at the back, the average street A will never approach that cumulative X stress point within that owner's use time if you have a NEW crank to start. The only thing that owner will get is a marginally quicker acceleration.

Now, another consideration- If you reduce the inertial damping (actually, you are reducing the energy conservation of the shock wave) by lightening the flywheel, that shock wave continues down the drivetrain! No flywheel on the back would be like taking a #3 pneumatic rivet gun to the tranny teeth!

Everything is a compromise. Henry made his, based on NOS parts and then- conditions. There probably is no definitive right or wrong, it is only what you want to happen, or how you wish to prioritize and distribute probabilities of failure at various points in the entire system.

[Pete]

Originally Posted by MikeK

I guess (and I do mean guess!) it depends on where you want the crank to break. The traditional place is for the rear to snap off. Any piece of metal can only take X number of stresses before fracture. That number is a function of the degree of flex and the number of cycles. An 80 year old crank is already way towards that point. Adding throw counters and reducing the flywheel weight changes the dynamics of the shock waves, or flexes. You now have less at the rear flange and higher stress in other places. (Here a front dynamic element would be nice.)

As a result, you will get more remaining run time on an already 80 year old crank than would remain with the heavier wheel in place. Simply, you are gambling that your chances of crank failure will be reduced. Now, if you can find a NOS crank (keep dreaming) with zero flexes, you'll gain almost no confidence against failure unless you live another 80 years. It would be someone else's problem.

OK, now let's look at a NEW crank. Not NOS without counterweights, a new Scat, Crane, Burlington, etc. with counterweights. None (Well, maybe a $3K custom) is 100% countered for each throw. That crank will STILL always be doing a harmonics dance back and forth along it's length while running. ANY inertial dynamic introduced at either end will make the stresses unequal along the length. Now remember X number of stresses before fracture, that number a function of the degree of flex and the number of cycles. For maximum life of an I-4 flat crank, on each external end of the crank you need a dynamic(inertial) mass equivalent of sq.root of 2 (1.414) times the inertia of the adjacent crank element (throw). Nobody runs a front flywheel (solid damper) that big, and nobody has a rear flywheel that small. The best compromise is a rear flywheel as light as you can get. Of course, even with a heavy stock fly-anchor at the back, the average street A will never approach that cumulative X stress point within that owner's use time if you have a NEW crank to start. The only thing that owner will get is a marginally quicker acceleration.

Now, another consideration- If you reduce the inertial damping (actually, you are reducing the energy conservation of the shock wave) by lightening the flywheel, that shock wave continues down the drivetrain! No flywheel on the back would be like taking a #3 pneumatic rivet gun to the tranny teeth!

Everything is a compromise. Henry made his, based on NOS parts and then- conditions. There probably is no definitive right or wrong, it is only what you want to happen, or how you wish to prioritize and distribute probabilities of failure at various points in the entire system.

Right on on all that...
I have had very good success using a 11 lb aluminum flywheel with an aluminum clutch cover and a 12 lb big block Chev. front damper.
This is about as close to equal weight on each end of the crank as is practical.. It torsional vibration you are reducing.
We spin these engines 6000 and have never broken a crank.

[Marco Tahtaras]

Originally Posted by MikeK

I guess (and I do mean guess!) it depends on where you want the crank to break. The traditional place is for the rear to snap off.

I'm sure there are examples to the contrary, but the ONLY broken cranks I've seen were improperly reground. I still have a Burlington crank on the shelf for a future project however.

[SteveB31]

It has always been my understanding that we remove the weight off the flywheel BECAUSE we add the weights on the crankshaft to counter balance the crankshaft (like Ford did on the later B motors). Therefore, we still have the SAME amount of total weight (crankshaft + weights + flywheel) like Ford did (they reduced the flywheel weight when they added weight to the crankshaft).

Your thoughts?

[Jim/GA]

Originally Posted by SteveB31

It has always been my understanding that we remove the weight off the flywheel BECAUSE we add the weights on the crankshaft to counter balance the crankshaft (like Ford did on the later B motors). Therefore, we still have the SAME amount of total weight (crankshaft + weights + flywheel) like Ford did (they reduced the flywheel weight when they added weight to the crankshaft).

Your thoughts?

When it comes to rotational inertia of a rotating system (also known as moment of inertia), not all mass added or removed from the system is the same. It also depends on the SQUARE of the distance that mass is from the center of rotation.

Here is a simple example:

1 lb. of mass that is removed 10" from the center of rotation (1*10*10=100 lb-sq.in.) is equivalent to 4 lbs. of mass added to a point 5" from the center of rotation (4*5*5=100 lb-sq.in.) to maintain the same total rotational inertia of the system.

So, to remove as many pounds from the flywheel as you add to the crank in counterweights is probably over correcting by quite a bit, because the counterweights added are probably closer to the crankshaft main bearing centers than the weight you removed from the flywheel. You need to keep track of where you are removing and adding the weight.

This is classical mechanics.

[Marco Tahtaras]

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Originally Posted by Jim/TX

When it comes to rotational inertia of a rotating system (also known as moment of inertia), not all mass added or removed from the system is the same. It also depends on the SQUARE of the distance that mass is from the center of rotation.

Here is a simple example:

1 lb. of mass that is removed 10" from the center of rotation (1*10*10=100 lb-sq.in.) is equivalent to 4 lbs. of mass added to a point 5" from the center of rotation (4*5*5=100 lb-sq.in.) to maintain the same total rotational inertia of the system.

So, to remove as many pounds from the flywheel as you add to the crank in counterweights is probably over correcting by quite a bit, because the counterweights added are probably closer to the crankshaft main bearing centers than the weight you removed from the flywheel. You need to keep track of where you are removing and adding the weight.

This is classical mechanics.

I was going to comment similarly but you beat me to it! Several folks have mentioned "total weight" without regard to the distance of that weight from the axis.