Light weight flywheel - good idea?

[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!

[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