driveline loss formula..

[Eddie N]

even though no one agrees on one driveline loss percentage, for the sake of my question, lets assume its %15...

what is the correct formula for measuring a flywheel horsepower number from the rwhp figure from the dyno?

ive seen:

rwhp*(1.15)

and

rwhp/(.85)

both result in totally different numbers...

thanks,

- eddie -

 

[onerareviper]

I remember reading another post on this topic, and the conclusion was:

rwhp/(.85)

Some legnthy explaination was included, but my memory is sh*t..

Later

 

[HP]

Another way of looking at it, is to think of the equation as
driven only one way, from motor to rear wheels. The motor, or
flywheel, loses 15% before it reaches the rear wheels.
So 100fwhp-15hp=85rwhp - the 15hp comes from 15% of 100fwhp and the equation is correct fwhp=rwhp/.85
If you start from the rear wheels 85rwhp and add 15%, or 12.75hp
you would get 97.75fwhp. which would be incorrect. This would be
represented by fwhp=rwhp x 1.15 and is wrong. HP(Hugh Perkins) rwhp?

 

[Hostile]

Correct formula

200FWHP less15%=170RWHP to return calculate 170/.85=200
works all the time,divide by the reciprocal

Skeeter

 

[HP]

Ok, let me take a stab at it. I've never seen it explained, but
looking at it, the explaination seems to be in a simple mistake
of setting up the problem. This is best explained by setting the flywheel hp(fwhp) at 100. So if you lose 15% in the drive train, you have 100hp-15hp=85hp Here you can see that rwhp=fwhp x 85% or fwhp=rwhp/85%
But when the mistake of thinking you can add 15hp to 100rwhp
to get 115fwhp is made, the equation changes to fwhp=rwhp x 1.15 Because, there, you are adding 15hp not 15% and you no longer have the right percentage. 100 is not 85% of 115. Its kind of like an optical illusion, it plays on faulty wiring.