dragonZ28

iTrader: (32)

Okay. I want to make a set of custom emblems for my hood, but I am wanting to have an accurate # on there.

I have a stock T-56 with a Spec 3 clutch, aluminum driveshaft, stock 10-bolt with 3.73 gears, and 17" Z06 motorsport wheels.

I made 522RWHP with a 78MM intake, and no tuning. I have added a 90MM intake setup, and will tune next time. I am hoping for 550RWHP. I want to put flywheel HP numbers on the hood, so I need to calculate loss. Does anyone know what percentage HP drain that these drivetrain components have?

Thanks guys.

Lithium

iTrader: (47)

TTT for you...nasty setup.

Arizona Mike

I know a lot of Ford people use a 19% drivetrain loss number with a T-5 and a Ford 8.8". The T-56 takes a little more. I have no idea about Chevy rears. Why not call it 20% loss (divide RWHP by 0.8 or multiply by 1.25 to get FWHP). It's conservative and probably accurate. 653 for your dynoed setup, 688 for what you are hoping to get.

Mike

99FormulaM6

iTrader: (21)

dont most people use 15% for m6's?

AintQik

iTrader: (3)

I normally use 17% for M6 but that pretty much a guess. Autos have a whole lot of variables in converters so its even more of a WAG. I think if you search around you will find something very close to 17%

JakeFusion

iTrader: (7)

Between 50-60HP.

It's not a percentage, it's an actual number. Dyno it on a engine dyno to see.

The reason? Why would it suddenly take MORE HP to drive the same components? If you replace the stock LS1 with a 408 making, oh, 522rwhp, the percentage difference at 15% is 51hp vs 92hp. Why did it suddenly eat 41hp more to drive the same rearend and transmission? It didn't. A 522rwhp engine is around 575 at the flywheel. 550rwhp is 600HP.

dragonZ28

iTrader: (32)

That's a pretty good point. I have never thought of it that way. However, I would could argue that it doesn't take a set amount of HP, but rather that it robs efficiency of engine power, which is the same as a parasitic percentage loss.

I like your point, though. That is the best input I have heard yet.

GuitsBoy

iTrader: (4)

Because the 408ci is going to accelerate the rotational mass inside the drivetrain much more quickly than the stock motor will, which requires more power to do so. This is the basic idea a dyno works off of. Accelerating the drivetrain more quickly puts more pressure on the gears, which creates more friction, which also takes power to overcome. This is why it is indeed a percentage and not a constant.

Still dont believe me? I can easily turn the drivetrain over by hand, and I can assure you that my hand is not producing 50 - 60 HP.

Jpr5690

iTrader: (13)

im sure its a deminishing percentage it its probaly only 10% on a 1000 hp motor but say 70% on 100 hp motor

dynocar

If it is a "fixed" amount, then expain this, on everything from a stock low HP VW to a stock high HP Viper, when we use our standard percentage multipliers for different drivetrains, such as manual Vs auto, we have always come very close to the advertised factory net SAE flywheel rating. This is based on 15 yrs of full time chassis dyno experiance on two different brands of measurable load dynos (Clayton and Mustang).

TeeKay

Jake, I can really understand why a person would intuitively think that was so. But, I'm sorry, it just ain't!

There are a number of real, physical reasons why it ain't, but let's just say that a drivetrain's efficiency is DEFINED as Power Out divided by Power In times 100% and the efficiency percentage for a given setup doesn't change all the way up to the physical limits of the drivetrain! Not an opinion, just the facts.

On topic: a stock converter A4 is about 17-18% loss (crank hp times .83 for whp) and the M6 is a little better (no converter loss) at about 14-15%. An aftermarket, higher stall converter goes about x80%x (oops, should have said 20% - chp x .80 for whp) for the A4. HTH

TOSTO RACING

iTrader: (37)

I've done both with the same motor rwhp flwhp and 15% was dead on actually on this application it was 518 rwhp and 613 flywheel. So figure 15% and you'll be fine .

dynocar

Just curious, if you happen to know, was the engine dyno using the SAE J1349 Rev 90 atmospheric correction factor that corrects to 77 deg F etc or the other one that is most commonly used by them that corrects to 60 deg f etc? Also did the engine dyno use the RPM step method or sweep? I'm assuming that the chassis dyno was a Dynojet with the normal 77 deg F etc correction, right?

300bhp/ton

umm not too sure on this, I mean you can turn the drive train over at a very low speed with no load on it.

But could you spin it over at 500rpm or 5000rpm? And that's really what HP is all about. As you don't measure HP you derive it from torque at a particular speed.

Sure there must be more loss at 5000rpm than there is at 500rpm because an engine at idle can still move a car along a level gradient even if it isn't producing much power. But 5000rpm with 300bhp or 500bhp I can't see it making any odds as the rpm is the same as is the rotating mass.

I am a strong beleiver that HP is not a fixed percentage loss, but I also don't think it can be 100% static loss either. A fixed amount plus a deminishing percentage is the only answer I have been able to come up with.

I think it is for this reason too many people always think their engine produces more power than they think. But there's no way a 520rwhp car is making ~650bhp at the engine.

Example's being (using a Mustang Dyno as it is load bearing and correcting to SAE):

A 260bhp Mustang will tend to dyno ~220rwhp HP so we can see a loss of about 40bhp

A 345bhp Corvette or Fbody will tend to dyno ~295bhp so a loss of 45-50bhp

C5 Z06 405bhp and ~350rwhp again ~50bhp loss

C6 Z06 512bhp with ~450-460rwhp giving 50-60bhp loss

So if a car genuinly dyno's 550rwhp (Mustang Dyno) then it would be logical to assume 60-70bhp loss at the most this would give you a real world figure of 620-650bhp SAE Net The 15% rule is not far off and the 12% + 10bhp rule also coinsides.

But it's better to be more conservative than over the top so if you claimed 620bhp SAE Net then you could be pretty certain no matter the weather or altitude it is probably going to be making that number as a minimum.
Of course none of this has taken into account varaiton in dyno type, correction factors, standards and graph smoothing.

GuitsBoy

iTrader: (4)

300bhp/ton, ill have to take your word on that. I havnt had my coffee yet and my brain is still mushy. It looks like you atleast agree that its not a fixed HP loss. Good 'nuff for me.

Patrick G

iTrader: (14)

It is closer to a fixed number than it is a percentage. Like JakeFusion said, with your particular setup, it's probably close to 50-60 rwhp. You think your motor is going to be at 550 rwhp (possibly optimistic). Keeping that in mind, a badge that says 600 hp would be perfectly appropriate. But a 15% loss on a 550 rwhp engine is 97 hp. I think that's a little on the high side. 60 hp loss at your power level is probably more like it. If you think 600 hp looks too common, then give it an oddball number like 605 or 610.

Just don't put your ***** size on your car too, the hp emblem is probably enough bragging, LOL.

Jpr5690

iTrader: (13)

so 400rwhp would = what if the car was a automatic and stalled

also does a locked up converter car have about the same efficnecy as a manual?

i was looking at my number this way 500bph x .85 (15% drivetrain loss in a m6) = 425-20 (unlocked ssf 3500) =405rwhp is this correct?

the car has 17x 11 wheels and a steel driveshaft bulit 10bolt 3.73 rear gears

JakeFusion

iTrader: (7)

Maybe my example was not the greatest. So let's assume a stock LS1 vs a typical head/cam LS1. Same rotating assembly, same weight of pistons/rods/crank. Lighter clutch assembly on the head/cam engine.

300rwhp vs 450rwhp at similar RPM (which is possible). 15% rule means, 352BHP and 529BHP--differences of 52 and 79. The engine didn't get heavier at the same RPM, nor did it suddenly start making more friction. If anything, the new clutch assembly is helping it make more power on the dyno, meaning it is now eating less engine brake power. So, is it 500BHP or 530BHP? I don't really care, but I'll say it's closer to 500.

Forced370GTO

look at it this way, the equation would look more like this: y = 1.01x + 40 rather then y = 1.25x. this yeilds the idea of diminishing return per horsepower. y= engine horsepower, x= rwhp. with a constant horsepower (exaple:40) derived from the weight and centripital forces associated with the rotational masses.

Patrick G

iTrader: (14)

That formula is probably closer to reality than anthing else I've seen. Yes, frictional losses do increase SLIGHTLY as hp increases, but just very gradually. More like this example above...not a fixed percentage (regardless of hp level).