Barneyfife911

Have a 2001 vert and just had a dyno run with new SC/cam/headers. RWHP went from 305 to 545. Just curious what horsepower that would translate to at the flywheel. Is there a formula? No real reason to know, other than just curious and would like to say a number even bigger!

nickolbag

assuming 12% drivetrain loss its 610.4 flywheel horsepower.

itchynackers

If its a manual, I've found that 12.5% is very accurate.

Barneyfife911

Thanks, but wouldn't it be right at 620 if you lose 12%? 610.4 minus 12% translates to 537, unless my math is just wrong or I'm misunderstanding. No big deal though. Still a big number. Thanks again for the feedback.

nickolbag

It was late at night

irun4cops

adding 12 percent to wrhp is different than subtracting 12 percent from a flywheel hp. Whoever knows the number needs to clarify which way they did the math on the car that they knew both hp numbers, make sense?

irun4cops

my guess is its flywheel minus 12 percent... so the easiest way is to do guess and check math bc im too tired to think of the equation that gets you up to the flywheel hp number. Wrhp +12 percent is not same as flywheel hp -12 percent. Adding 12 percent of the rwhp number will be a smaller number. Subractin 12 percent from flywheel is a larger number. If you just add 12 percent of rwhp, you will get a short number for your flywheel power. Sorry if im confusing

mtm87tx

divide your rwhp by .875 to get your crank hp assuming you want a 12.5% loss

i come up with almost 623 rwhp

tboneinsc

OK I understand that a stock C5 loses 12% of its HP between the crank and the rear wheels, but shouldn't that HP loss be a constant number no matter how much HP you make at the crank?

irun4cops

no, friction increases with more power I think. Ill put it this way, i know an 850 wrhp car was said to of had over 1000 bhp by road and track. Thats a 12 percent rule.

vrybad

Unless Road and Track actually dynoed the engine out of the car and then redynoed it in the car, there is no way to know if they are correct.
Why are so many people hung up on flywheel hp numbers?!!
It's what hits the ground that matters.
Maybe LPE has an answer.
They are most likely the only shop that may have done just what I mentioned above.

AtlBlkZ06

Err you mean 15+% rule?

I think 15-20% is right for most everyday boring cars. They have the same parasitic loss (similar RPMS/MPH) but with much lower power numbers, so the % number is inflated.
So ~12 sounds right to me.

Pardon my unscientific and vague talk...
You're right, friction isn't constant, but the quadratic "rule" in fluid dynamics doesnt apply either so the resistance is just "slightly" higher at higher velocities. (Where is a physics prof when you need one)
So you increase the power by 2x and increase the friction losses by less than that and end up with 12%. High HP cars have heavy duty components to work with so its only fair that you expect a little bump to 15% or so.

glass slipper

It depends on the type of dyno used. A Mustang dyno can put a load on a car and control the rate of engine acceleration (250 RPM/sec, 500 RPM/sec or whatever) or it can hold the RPM constant. That way, HP loss due to the inertia of engine components, drivetrain components and wheels/tires/rotors are held constant. On a Dynojet dyno, there is only the inertia of the drum to resist the torque at the wheels so when we add more HP or dyno in a lower gear, the time for the dyno run is less which means more HP loss to inertia.

Drivetrain HP loss on a chassis dyno is composed of the following:
1) second order or exponential losses like friction
2) first order or linear losses like inertia and gears
3) constants like windage of gears through the oil in the transmission/differential, shafts rotating through seals, and sidewall flex all at a particular RPM. In other words, those losses are independent of HP transmitted.
That leaves us with the equation HP (loss)= ax^2+bx+c where a, b, and c are constants and x is FWHP. "c" is the predominate term meaning most of the HP loss is a constant.

Other variables to consider is when a dyno run is done in a lower gear, the tire, transmission shafts/gears, and differential shafts/gears RPM is less reducing the HP loss to sidewall flex/windage and giving more RWHP. On a Dynojet, this is countered by a shorter dyno run increasing loss to inertia, not a factor on a Mustang dyno though since it keeps the time of the dyno run constant. Conversely, a dyno run in 5th or 6th on a Dynojet will have less loss to inertia (longer dyno run time) but more loss to windage/tires...but there's also a speed limitation on chassis dynos which prevents a run in 6th (unless you have some 4.56s in the rear). On the road, quicker acceleration in the lower gears results in more loss (as a percentage) to inertia, the same effect as a Dynojet.

As far as the 1:1 ratio of 4th gear being the best to dyno in, there is fact as a basis to that statement. Quality involute gears have a HP loss of ~1%, so when we dyno in 1st, 2nd, 3rd, 5th, or 6th, part of the drivetrain loss is ~2% of the HP transmitted through any of those gears (it's 2% because power flows through one set of gears from the input shaft to the countershaft then through another set of gears from the countershaft to the output shaft). You may notice the distinct absence of 4th gear in the list above...that's because there are no "gears" for 4th gear, something a lot of people don't know. When you make the shift to 4th gear, the synchro mechanism mechanically couples/locks the input shaft to the output shaft and power flows straight through the transmission making 4th gear the most efficient "gear" to dyno a car in as far as the transmission is concerned. It doesn't have anything to do with the fact the "ratio" is 1:1 since we could also put the power through gears at a 1:1 ratio and achieve the same output RPM.

Hopefully this shines a little light on the subject, it's neither a percentage or a constant. But if you had to pick one, you'll be closer to FWHP if you use a constant. I.E. if you dyno a stock C6Z06 and come up with 450 RWHP for a 55 HP/10.9% loss, then mod it to 600 RWHP on the same dyno, you'll be much closer to the true FWHP if you add the 55 HP loss from stock rather than back calculate using the percentage number. And the guys mod their cars to 600+RWHP that just pull a percentage number out of their *** (like 15%) to back calculate their FWHP are just looking to stroke their ego with a big number and impress people. The only way to know the percentage is to remove the engine and put it on an engine dyno to get the FWHP. But then the first time you mod the engine, that number is not valid anymore. The bottom line: there is no formula to calculate FWHP from RWHP...too many variables and too many unknowns. It won't stop people from making up something in the interest of stroking their egos.