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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? |
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.
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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. |
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".
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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 |
Golly, David....
Permit me to return us to the very beginning of this thread; and please stick with me, ok?:
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But then, that's just my perspective. |
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...
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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. |
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. |
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. |
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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! |
Secret to turning fast lap times." Size 14 shoe, size 2 hat." Buddy Baker former Daytona 500 winner.
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So the street answer is ???
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. |
Quoting myself from the 7th post on this thread...
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Scott |
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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. |
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:JEKYLHYDE ... I will try to look it over again during the daylight hours ;) Otherwise, let me know if you find problems. Thanks |
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. |
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