Does anyone know why drivetrain loss is calculated as a dynamic number as a percentage of RWHP instead of being a fixed value? Assuming that the transmission, clutch, axles, diff, driveshaft, etc were all a constant, why would the frictional and intertial losses be dynamic? So a car making 400rwhp operating on the same car can have a drmatically different actual drivetrain loss number than a car only making 100rwhp or one making 1200rwhp. Why? Who knows?
well HP is a function of RPM and torque and is only modified by a correction factor. Aerodynamic drag is not an issue on a chassis dyno...unless for some reason the force of friction increases or decreases with increased and decreased HP respectively...? How can that be the case?
The secret is to understand that your drivetrain is effectively a load bearing surface. Friction is related to velocity (faster movement yields more friction per unit time), and input torque (since it's load bearing, input torque increases the normal force, thus increasing friction).
Imagine dragging a board across a floor. Now stack bricks on it, and it's harder to drag. Dragging the board is akin to the work your engine performs. Adding bricks corresponds to increased friction on the drivetrain. It takes more work to offset the additional friction you're producing by loading up the contact surfaces.
I posted a thread on this very topic a few weeks ago on the org site - posted up the best and most scientific data i could find and there was some good comments made by others.
Bottom line - there are some fixed loses and variable losses - as you would expect - therefore a flat percentage is not always accurate....however, as a rule of thumb for rear wheel drive this formula works better than anything else we've found so far...
1 - Take corrected rwhp from an industry standard chassis dyno
2 - add 10hp
3 - Divide result by 0.88
= Engine Hp aprox.
This will give a lower figure than many popular conversions but is more realistic to actual numbers run.
To reverse:
1 - Take engine dyno corrected hp
2 - subtract 10hp
3 - multiply by 0.88
= RWHP.
Here's an example - the 649rwhp John just dyno'd + 10 = 659/0.88 = 749hp at crank.
Yours Smokin - 777rwhp + 10/.88 = 895hp!
Paolo's - 888rwhp + 10/.88 = 1020hp!
engine makes say 650hp - 10 x 0.88 = 563rwhp
This formula appears to be quite close to what a Viper will produce.
Torquemonster, it looks like that just works out to a more or less flat 14% -- mainly because the numbers are very close together. In fact, you're really just using a flat 12% (100% x .88 = 12%) combined with a static 10HP fudge factor. Cool if it works, I guess, but it seems a lot easier to just go with 14% and get practically the same number -- closer to the same than a realistic margin for dyno error, anyway.
I think S. Roe did a dyno test of a stock engine in and out of the car. He came up with a 12% loss. At the sametime each car is going to be VERY different.
Some may have 6th gear removed, dry sump and reduction pullies etc. These would certainly reduce engine drag and increace WHP without increasing BHP.
Bottom line, each component of the car will have its own loss factor and I think 12% loss is probably pretty close for a stock Viper engine/transmission.
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