David426

I have a 99 A4 (3.15) that on the dyno ran 299 rwhp and 312 rwhp torque. How do I convert this to crank HP. I think I have to multiply by 0.15 or 0.18 its different on Manual transmissions due to less drivetrain power loss. The motor made 345 crank HP and 350 crank hp stock (factory rating). Thanks for the help. And yes there was a thread on this moons ago but I can't find it using the search.. Thanks

allanlaw

Multiply by 1.18 and you'd be there.

SweetSavannah

If parasitic losses on an A4 are assumed to be 18%, then divide (not multiply!) 299 & 312 by 1.00-.18 or .82

'06 Quicksilver Z06

90 droptop

T_Vert

hes trying to find crank HP from RWHP. If you divide 299 by .82 you come out to 245 HP. You multiply by 1.18 299 RWHP = 353 crank HP.

Dave

90 droptop

You may want to check your math,

kewlbrz

Thats incorrect (assuming 18%) is the loss from the crank to the rear wheels.

You cant simply add back the percentage lost to the end result. You have to take the inverse proportion.

correct:
299 rwhp divided by .82 = 364.63 crank HP

incorrect
299 X 1.18 = 352.82

T_Vert

oh blah nevermind, been too long sinze i did math.

Mitch C

illmac77

so lets say you have 370rwhp on a manual 15% loss you would divide .85 by 370 and get 435 crank hp right

AFVETTE

So does .82 = an 18% loss drive train loss. Then is .85 correct for a manual?

Tom

AU N EGL

Manuals about 15% ?? So 400 RWHP be about 460 FLHP ?? That seems a bit high.

6Speeder

Most people use 15% loss for M6's, and 20% loss for A4's. This is just for bragging rights on HP so what does it matter but... some people believe 12% loss is more correct for M6's, and 18% loss for A4's. That would give a multiplication factor of .88 or .82 respectivly to get from rwhp to fwhp.

90 droptop

WHy do you say that.

If a stock m6 dynos on avg. 300 rwhp and you use 12% loss, that = 340 fwhp. Just about what the ls1 is rated.

nwdanner

When going from RWHP to FWHP you *divide* rather than multiply. For example:

If a given engine puts out 405hp at the flywheel, and the drivetrain "eats" 15% of that output, then the formula for calculating RWHP would be:

FWHP * .85 = RWHP (because RWHP is %85 of total flywheel output).

To get FWHP from RWHP, you just move the ".85" to the right-hand side of the equation by using Algebra and dividing both sides by ".85". This means that the left-hand side is left with FWHP and the right-hand side is RWHP / .85, such that:

FWHP = RWHP / .85

So using this formula for a car that dyno'd 350rwhp, and considering a 12% drivetrain loss for a manual transmission, equates to approximately 398fwhp (350 / .88).

03LingenfelterC5

Wow, this is just like when I talk to people about mark-up vs. margin...

Gordy M

There is no formula, just a range. There are too many variables involved but you can get reasonably close. I had several friends who worked at the proving grounds when the 01 and 02 Z06 was developed. They reported less than +/- 2 hp on the engine dyno. from the reported 385 and 405. Different dyno's give different results i.e mustang dyno, etc.

David426

Thanks for explaining this. It now makes sense to me. The guy who did my dyno test told me to multiply 299 X 1.18, which would not be correct because the 299 is the number showing the 18% powertrain loss. So instead of 299 x 1.18 = 352.82.. The actuall equation is 299(rwhp) / (divided) by .82 =364.6(fwhp). Thanks guys

nwdanner

That's what we're here for...to help each other.