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Old 02-17-2015, 10:42 PM   #161
Sypher
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Quote:
Originally Posted by stugray View Post
I simplified it as best I could:

Energy required to accelerate a 2800 lb car to 100 MPH
1270 kg @ 44.7 m/s

Total kinetic energy = ½ MV^2 = 0.5*1270 * 44.7^2 = 1268787.15 joules

Rotational kinetic energy of a spinning cylinder
E-rotational = ½ I w^2 (w is angular velocity in radians/sec I is moment of inertia )

30 lbs driveshaft = 13.6 kg 3" dia (1.5” radius = 3.81 cm = .0381 m) =
I = .0381 * .0381 * 13.6 = .01974 kg.m^2


Stock wheel = 1964.44 mm circumference

@ 100 MPH our wheels are spinning at
@ 44.7 m/s => 44.7/1.964 = 22.75 RPS @ wheels
Final drive ratio = 4.1:1

Driveshaft RPS = 22.75 * 4.1 = 93.275 RPS = 5596 RPM = 35160.7 radians/min = 586 radians/sec

Rotational energy of the spinning driveshaft = ½ I w^2 = 0.5 * .01974 * 586^2 = 3389.32 joules

Energy stored in moving car: 1268787.15 joules
Energy stored in spinning driveshaft: 3389.32 joules

Total energy (NOT COUNTING the energy stored in the REST of the rotating mass) = 1272176.47 joules

For a whopping ratio of 375/1 = .0027 or .27% of the energy in the system.

SO while the engine has stored up a total of 1272176.47 joules in the system, only 3389.32 joules are in the spinning up of the driveshaft.


BRZ gets ~ 16.2 seconds 0-100 mph
That’s 1272176.47 joules / 16.2 seconds or 78529.4 joules/second

OR ~105.36 hp Which for back of napkin calcs is in the ballpark enough (holy shit!).


If you had no energy stored in the spinning of the driveshaft at all: total system energy would be 1268787.15 joules at 100 MPH
If the car delivered the same 78529.4 joules/second it would achieve 1268787.15 joules in 16.156 seconds

So I guess you were all right!
If you reduce the 30 lbs driveshaft to 0 pounds (removing the energy stored in it entirely!) and do otherwise equal runs 0-100 MPH you WILL notice a 44 millisecond difference in a 16.2 second run! Congratulations you are superhuman!

And I was being generous by leaving things out such as the REST of the rotating mass so the actual difference of removing the shaft would be even less.

I never said that it wouldnt make a difference I said that a driver wouldnt be able to TELL the difference.


Shall I repeat the calcs for a 4 pound reduction of the drive pulley?
Which coincidentally would be MORE than the driveshaft since IT spins to ~6000 RPM

And @Shutter beat me to it:
i love the math here but i do have a question about the moment of inertia for the drive shaft. did you figure it as a hollow cylinder? when i figure it as a hallow cylinder with an internal and external radii i get a very different moment of inertia. 0.07574. of course this is with the assumption of a 3" external radius and a 2.875 internal radius and weighing 30lbs.
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