Drivetrain Losses

[GMC_DUDE]

Question: When estimating drivetrain losses in order to calculate RWHP from Crank HP or vice versa (and same for torque), should we assume constant percent loss or constant HP (torque) loss? For example, suppose a truck has a 200 hp engine, 20% drivetrain loss (that would be 40 hp), and thus would have 160 RWHP. Now suppose the owner increases the engine to 300 CHP. Would the RWHP be 300 - 20%(60 hp) = 240 RWHP, or 300 - 40 = 260 RWHP? Assume that we are doing all measurements over the same RPM range (e.g. a dyno run). My guess would be that since the losses are due mostly to parts friction and rotational moments of inertia, and thus are speed related and independent of the amount of torque passing through them, then a constant hp model would apply. The big exception, of course, is the torque converter, where an increase in torque passing through the unit would create more slippage, and thus increase the losses, so a constant % model may be more accurate. Any Mechanical Engineers or other such experts out there who can shed some light on this?

 

Peace and Coolness,

 

GMC_DUDE

[snoman]

I have seen figures from 20 to 30% and no way to be 100% sure without pulling engine. I would tend to think that as you increase the power of engine that the loss may increase do to more heat and slipage in a automatic trannyand about 25% the norm. I would think that a stick shift would have the lowest loss and be fairly consistant too. My two cents anyway.