Well, I have to say that it is my feeling that for a given MANUAL trannied car you will not have an accurate estimation of rwhp for a 1500 hp car if you first determine the % drivetrain loss for the vehicle at say 300hp.
What is drivetrain loss? I can think of four things that contribute, though there are probably more:
Well, a torque applied at one end of a static object will apply an identical torque on the other end. However, we're not dealing with static objects, but accelerating objects. The acceleration in and of itself "consumes" some of that force meaning that the force applied at the other end of the object will now be less the torque required to accelerate that object's mass into rotational velocity. I think it's somewhat difficult to determine the force required because while we could estimate the rotational inertia of a given drivetrain, the rest of the equation would depend on the rate of acceleration of each component of the drivetrain, each of which would depend on the rate of acceleration of said vehicle. They would each be proportional and dependant on the gear ratios/tire radius involved. This would, therefore, not necessarily be a constant percentage of the power applied, but would depend on where you are in the powerband, which gear you're in, and . As a general rule, more force equals more acceleration, so this requirement as Mike mentioned should increase with the drivetrains rate of acceleration. Inertial dynos, therefore, would show an increase in measured torque loss that increases with the amount of torque produced. However, I'm uncertain whether or not this could accurately be represented with a constant percentage of the rate of acceleration. On a dyno that held a constant velocity, however, the loss of force due to this phenomenon would be 0.
Another loss in the drivetrain would be the loss of torque due to friction between the gears in the transmission, and the friction between the ring/pinion gears in the rear-end housing. This too, would increase proportionally with the amount of torque applied, and would be accurately represented by using a constant percentage of that torque. That's because the kinetic friction coefficient is that constant, and the only other information required to determine the friction involved is the normal force which would be proportional to the torque applied to the system, and which would account for the angles of the gears and the gear ratios used as gear ratios directly multiply the normal force.
There would be a some amount of loss in the hysterisis due to deflection in the tire, too. This loss of force is the reason that some tires allow a car to get better gas mileage than the previous tires did. With most street tires, the force loss should not be noticeably different when more force is applied. The biggest factors in tire deflection include the weight of the vehicle on the tire, and the speed of the tire. Neither of these need change in a higher HP application. However, non-standard tires like bias ply tires used for racing will deflect significantly due to the torque applied and as a result the force loss here should increase as the torque applied to the tires does. However, I would expect this loss to be relatively small by comparison to the other factors already mentioned because the tire only deflects once as the power is applied and then begins to recover its normal shape slowly as torque decreases through the powerband, and then more so when the driver gets out of the throttle. However, this is really only one long event for each molecule whereas for the other factors discussed (speed and weight) each rotation of the tire causes a deflection and recovery.
The last loss of force that occurs to me is the force lost in the friction of the bearings of the moving componenets in the drivetrain. Though it's a relatively insignificant force-loss overall, some of it will also increase with the force applied, but some of it will not. That's because the force applied also adds weight to some of these bearings like the ones in the transmission, u-joint, and rear-end. As load is applied to the gears, they in turn add load to the shaft that they are on, which adds weight to the rollers on their bearings. Other bearings do not take the added weight because the force is not applied directly to them, only a rotational force. When that happens, no extra weight/load is applied to the bearing, and only the twisting force (torque) is applied to the component itself.
I'm sure this is getting to be a bit of a novel now, but it's interesting to me because I believe one side of the argument, though I've never actually thought through the forces and losses involved. This is not an argument so much as an open discussion. Feel free to add any input or correct my observation.
yeah, what he said ^^^

I will say that is the best (if not, slightly overstated
) explanation of it though, and if you can read between teh lines, makes the most sense...to me anyways.Thanks.



.....having a SC whine would drive me crazy in the DD.