Light weight flywheel - good idea?

[mylesdw]

Originally Posted by quickjack

So, instead of 7,000 rpm you need 7,100 to move from rest?

I doubt whether stalling would be an issue at either!

[DavidNJ]

Originally Posted by mylesdw

A 'quick' calculation shows that lightening the flywheel should have little effect on moving from rest. If you went from a 40lb to a 20lb flywheel you would need to raise your rpm by just 100rpm when moving from rest to have exactly the same energy stored in the rotating assembly. I doubt whether one would notice the difference.

My quick calc says the energy stored in a 45# flywheel clutch assembly at idle is about twice as much as a 2800# car (with driver) moving 5mph. Going from a 45# 11" assembly to a 18# 8.5" assembly will drop the stored energy by over 75%.

Yes, if you have 700gm pistons and 900gm rods with a 60# crank, you will still have quite a lump. However if you have 400gm pistons with 520gm rods and a 40# crank, it will be significant.

BTW, do you have a formula for calculating the inertia of the pistons and the rods?

[RACER X #99]

All these facts and figures are really impressive and sombody has spent a lot of time reading up on all this crap. My experience has been that on the racetrack the car that contains the driver with the biggest balls is usually turning the fastest lap times.

[turnpike boy]

Originally Posted by RACER X #99

All these facts and figures are really impressive and sombody has spent a lot of time reading up on all this crap. My experience has been that on the racetrack the car that contains the driver with the biggest balls is usually turning the fastest lap times.

Amen to that.

Not that it will matter one whit......my car weighs 2390 lbs, has a 9:1 compression 302, B cam and 650 Demon; it has 4.09:1 gears and the Tremec .68 hi gear 5 speed. It also has a 12 lb. (weighed) Fidenza aluminum wheel. I can let the clutch out (a stock 10.5", by the way) at the 800 rpm idle, without touching the gas, and drive away comfortably without slipping anything.

And it revs like a sprint car. Big bore Jap bikes are surprised regularly. Most cars don't stand a chance, altho I admit ZO6 Corvettes are awesomely fast.

The only thing given away is hi gear torque and that has absolutely NOTHING to do with the flywheel.

[DavidNJ]

If you have a stock clutch, the aluminum flywheel reduced the weight from maybe 45# to 35#. Which is why you can "I can let the clutch out ... at the 800 rpm idle, without touching the gas, and drive away comfortably without slipping anything".

[DavidNJ]

Originally Posted by RACER X #99

All these facts and figures are really impressive and sombody has spent a lot of time reading up on all this crap. My experience has been that on the racetrack the car that contains the driver with the biggest balls is usually turning the fastest lap times.

It is my experience the driver with the biggest balls end up in the walls. The fastest drivers are the ones who through natural talent or experience have a fine feel for their cars.

[RACER X #99]

Dave,
Thanks for your expert advise. You seem to be an authority on so many topics around here I was wondering about your automotive/racing background. Have you ever driven on a racetrack? Do you own a racecar? Do you know anybody who owns a racecar? Can you identify a racecar?
Thanks,
RD

[turnpike boy]

Permit me to return us to the very beginning of this thread; and please stick with me, ok?:

Originally Posted by Cobra 29

Are these light weight flywheels a good selection for spirited street driving? I would like to see quicker revs.

Somehow, I just can't get around the thought that a $1500 multi-disc clutch setup, versus a $150 single disc, FOR THE STREET just might be overkill. Particularly in the context of the original question.

But then, that's just my perspective.

[SACoupe]

I have a 347 Stroker with a lightweight aluminium flywheel and centerforce clutch, and i drive my car every day....it does rev very quickly but is only really close ot unmanagable in the rain. in the dry its is fine.........heheheh if you dont mind having to catch it every time you leave a Traffic light...

[DavidNJ]

Originally Posted by RACER X #99

Dave,
Thanks for your expert advise. You seem to be an authority on so many topics around here I was wondering about your automotive/racing background. Have you ever driven on a racetrack? Do you own a racecar? Do you know anybody who owns a racecar? Can you identify a racecar?
Thanks,
RD

What do they look like?

[mylesdw]

Originally Posted by DavidNJ

My quick calc says the energy stored in a 45# flywheel clutch assembly at idle is about twice as much as a 2800# car (with driver) moving 5mph. Going from a 45# 11" assembly to a 18# 8.5" assembly will drop the stored energy by over 75%.

Yes, if you have 700gm pistons and 900gm rods with a 60# crank, you will still have quite a lump. However if you have 400gm pistons with 520gm rods and a 40# crank, it will be significant.

BTW, do you have a formula for calculating the inertia of the pistons and the rods?

I may of course be wrong but I think that your quick calc is WAY out. Did you perhaps forget to convert the units? If you use rpm,inches and lbs to calculate the flywheel stored energy then the speed (5mph) needs converting to inches/minute.

The 'guesstimate' that I included in mine was to assume that a 40 lbs flywheel represents about half of the total rotating mass. I don't know how true that is.

I don't have a formula for piston inertia, I made the assumption that it is fairly neutral because they spend an equal amount of time accelerating and decelerating.

[Mike Simard]

Here's some food for thought on rotational inertia:
http://www.tiltonracing.com/content.php?page=faq&view=3
From that page:"Doubling the weight of an object without changing how far the weight (technically mass) is from the axis of rotation doubles the MOI, which follows common sense. Doubling how far the weight is from the axis of rotation quadruples the MOI,"
BTW, I just started using a 7.25 twin disc with Tilton cerametallic discs. It weighs over 20 lbs less than the previous 10.5 street twin setup and that weight is on a smaller radius. Whew!, what a difference. It's a treat to match rpm's during shifting and it revs quickly. The cerametalic discs are much thicker than the ones normally used on small racing clutches, they can therefore absorb more heat and slip well.

[DavidNJ]

I used MathCad, which defaults to Joules. I change the units to lbf-ft, and it does the coversion. The quick part was assuming the flywheel clutch is a unitform disc, etc. I may have overcalculated the energy in the flywheel/clutch at idle...but not by an order of magnitude.

In Engine Analyzer Pro, the flywheel/clutch change could be worth 2-4 lb-ft² of inertia. That is worth maybe 5-7 lbf-ft of torque at 600 rpm/sec acceleration, or upto 9 hp at 6500. On no load situations, that is more dramatic, as Mike indicated.

[mylesdw]

Originally Posted by DavidNJ

I used MathCad, which defaults to Joules. I change the units to lbf-ft, and it does the coversion. The quick part was assuming the flywheel clutch is a unitform disc, etc. I may have overcalculated the energy in the flywheel/clutch at idle...but not by an order of magnitude.

Go on then, put me out of my misery! What have I done wrong:

We have a 40 lb flywheel 11" in diameter at 800 rpm,

Rotational mass (I)= 0.5 * m * r^2

so

I = 2420 = 0.5 * 40 * 11^2

Energy = 0.5 * I * w^2

so

774400000 = 0.5 * 5280 * 800^2

We also have a 2800 lb car travelling at 5 mph,

to make a comparison we need to convert mph to inches per minute

v = 5 / 60 * 63359 = 5280

Energy = 0.5 * m * v^2

39028528010 = 0.5 * 2800 * 5280^2

My energies are a factor of 50 apart!

[RACER X #99]

Secret to turning fast lap times." Size 14 shoe, size 2 hat." Buddy Baker former Daytona 500 winner.

[Cobra 29]

Given that I want to retain a 11 or 10.5 inch diameter single disk clutch for the STREET. Will a light weight flywheel, such as the 12 to 15 pounder, give me much quicker revs with little to no negative impact on accelerating from a complete stop?

I do appreciate a lot of the discussion but reduced diameter, multi-disks clutch setups for the STREET seems way too serioes and expensive to me.

[scottj]

Quoting myself from the 7th post on this thread...

Originally Posted by scottj

...the aluminum flywheel, though lighter, still retains the stock diameter so the effect will not be that noticable...I've run an aluminum flywheel on the street and found no drivability issues at all.

and no, I wouldn't say that it gave much quicker revs.

Scott

[DavidNJ]

Originally Posted by mylesdw

Go on then, put me out of my misery! What have I done wrong:

We have a 40 lb flywheel 11" in diameter at 800 rpm,

Rotational mass (I)= 0.5 * m * r^2

so

I = 2420 = 0.5 * 40 * 11^2

Energy = 0.5 * I * w^2

so

774400000 = 0.5 * 5280 * 800^2

We also have a 2800 lb car travelling at 5 mph,

to make a comparison we need to convert mph to inches per minute

v = 5 / 60 * 63359 = 5280

Energy = 0.5 * m * v^2

39028528010 = 0.5 * 2800 * 5280^2

My energies are a factor of 50 apart!

Units,units, units...

Your inertia is in lb*in². It is 16.8 lb*ft².

However, your rotational speed should be in radians/second not revolutions/min. That is a factor of 377.

So you energy becomes 1/2 * 16.8 lb*ft² * (800/(60*2*pi) rad/sec)². This equals 1800 lbf-ft.

At 5mph your car is traveling 7.33 ft/sec. So your equation is 1/2 * 2800 lb * (7.33 ft/sec)² which equals 2340 lbf*ft.

The 40# figure is a little light. 45-50# is probably closer.

[PDUB]

It's been a long time since physics class, but recall that the formulas in play here are based on the mass of an object (the m in the formula). The US measure of mass (unit) is the slug (short for "sluggishness??").

Pounds are a measure of force... the force due to gravity on the flywheel or car in this case (acceleration due to gravity is 32.2 ft/s^2 in US units).

Additionally, r is for radius, rather than diameter, d.

Correct me if I am wrong, but it looks like it should be:

I = kmr^2, where k is the inertial constant (1/2 for a uniform rotating disk).

= 0.5[(40lb)/(32.2ft/s^2)][5.5in/(12in/ft)]^2 = 0.5(1.24slug)(0.45833ft)^2 = 0.13045 slug-ft^2


KE = 0.5(I)(w)^2, where w = angular velocity = (800 rev/m)(2*pi rads/rev)/(60s/m) = 83.77 rad/s

= 0.5(0.13045 slug-ft^2)(83.77rad/s)^2 = 457.78ft-lb = 620 J

For the car:

KE = 0.5mv^2 = 0.5(2800/32.2)(7.33)^2 = 2338 ft-lb = 3169.8 J

It is pretty late, so this could be in error ... I will try to look it over again during the daylight hours

Otherwise, let me know if you find problems.

Thanks

[DavidNJ]

I actually used kilograms and coverted back...never could keep the English units straight. Matcad does the unit conversions for you though, In this case the constant, g, is the same on both sides so the ratio holds.

However, you are correct, I forgot to convert diameter to radius, so I was off by a factor of 4. This would make the energy equivalent to around 2mph. I knew the number should have been around 4-6 lb*ft²...so I must have been asleep at the wheel here. There is probably another 1-1.5 lb*ft² from the crank, rods, pistons, pins, rings, and bearings.