dsm-onster
DSM Wiseman
- 8,592
- 124
- Jul 11, 2004
-
Bloxom,
Virginia
This can be seen as a bit off topic. So I'll just link to some math I've already done. 15% drivetrain loss: 500 X 15% - 40hp rotational energy loss in the tranny if it were frictionless. Do you know how MUCH heat energy equals the additional 35whp loss? You really should measure the heat difference in the transmission from say 200whp to 500whp, then tell everyone how much horsepower was actually lost to heat.
Just have to add . . .
KE of rotation = 1/2Iω^2
KE of rotation = 1/2(τ/α)ω^2
I (moment of inertia) is a constant for each mass or colection of different masses because I = mr^2. m is mass and r is radius. Also, I = τ/α. As α (angular acceleration) goes up so does the τ (net external torque) to keep I constant. KE stays the same. So apparently it does take the same amount of energy to freewheel a drivetrain to 7500rpms no matter what horsepower you're engine has.
Just have to add . . .
KE of rotation = 1/2Iω^2
KE of rotation = 1/2(τ/α)ω^2
I (moment of inertia) is a constant for each mass or colection of different masses because I = mr^2. m is mass and r is radius. Also, I = τ/α. As α (angular acceleration) goes up so does the τ (net external torque) to keep I constant. KE stays the same. So apparently it does take the same amount of energy to freewheel a drivetrain to 7500rpms no matter what horsepower you're engine has.