Horsepower rating question...
 

Is there a ratio of what RWHP converts to HP at the motor like factory ratings? What I am getting at, how much rear wheel horsepower do I need to hang with a stock C6 Z? If a stock LS7 is rated at 505hp what does that end up being at the wheels?

I am trying to decide whether to modify what I have or simply buy a C6Z...

Go over to the C6Z section and see what they are putting down (~440) at the wheels and then you wil know where you need to be. If you are converting from BHP to RWHP, every car is different, but I would say to use about a 15% loss from BHP to be overly conservation. That means BHP*0.85 = RWHP. Hope this helps.

My car is putting down 440 to the wheels. If you figure 15% driveline loss that would be 506 at the flywheel.

The "generic" formula I've always used for RWHP from BHP is BHP*.83=RWHP Using 15% loss is fairly generous in most cases. 17% loss may be an over estimate on the Z06, but all of these formulas are just estimates anyway. Without actually dynoing the engine at the crank on a specific car, it's just a guesstimate, as loss is going to vary from car to car based on lots of different factors.

For most of us, that's really all you need, a guess or estimate. Rather than trying to pin down a fixed number, it might even make more sense to just say that the car makes "somewhere around XXX hp at the crank". But folks get so hung up on how big their numbers are.

:cheers:

Thanks guys. My question has been answered. I wasn't looking for an exact answer but more of a range.

:iagree: Based on all the info I've seen over the years of dyno tests with motor out of the car, then installed and tested at the rear wheels.....15% loss is pretty close. And I think the number goes up a little if you have really big HP (more frictional losses).

Close, but you have to keep in mind that you are not adding 15% to the RWHP (440 * 1.15 = 506) you need to subtract 15% from the BHP. The formula for RWHP to BHP is RWHP/.85 which in your case would be 440/0.85 = 517....Good numbers. :thumbs:

I calculate it like this and iv been told by many to do it this way. For example you have 505 hp X .85= 429 rwhp. Or you can do it the harder way 505X, 15%, 75.75,-from 505= 429rwhp. I like the first way better, lol.

Either way I have a bunch of modding/spending ahead of me to run with the C6Z...

shoot for 460+ rwhp

Sounds about right. Do you finance?

I'm in the minority here, but I don't believe in using a percentage when determining flywheel HP, versus RWHP. Not getting into chassis dyno variabilities, which is a whole other discussion, I think whatever losses your particular platform has to begin with, are the same after mods, as before. In other words.... most stock 405 rated Zs have about 345-360 RWHP, translated to approx/50 HP driveline loss. I'm of the opinion that a modded Z, with 475 RWHP has approx 525 flywheel HP. For another example, a highly modded 745 RWHP Z doesn't really have 900 crank HP,(@15%) losing an extra 105 HP to driveline loss? I really doubt it.

Gear efficiencies are rated in percentage for a reason. The amount of hp lost through the drivetrain does not stay the same when you get to higher hp levels. It may not be a perfectly linear relationship but it's close.

How can you state such a thing as factual? Is there a non-scientific explanation that would support your OPINION? I'm just a simple working man, and logic seems easier for me to grasp.

Heat, vibration, etc all increase as HP increases. There is more power wasted. A Top Fuel Dragster's supercharger takes like 400 HP or something just to spin while it makes 7000 HP. That is one example of higher HP using more HP in operation. It is an increase in wasted energy.

Is it exactly 15% at all times for every application? No, of course not, but, as mentioned above, it is the closest approximation available over the widest range of power levels without taking an engine out and dynoing it separately from the car.

It is certainly not a constant by any stretch of the IMAGINATION....

It's similar to the reason the relationship between HP and MPH is exponential and not linear. It might take 100 HP to get from 13 seconds to 12 seconds but it will take 350 HP to get from 9 to 8 (for example).

The law of diminishing returns.

lb for lb, all things being equal, the C6Z will have more driveline losses than a C5Z, due to the heavier clutch/FW assembly, heavier wheels/tires, and heavier rear rotors.

Wish I had 440 at the wheels.......plus your car sounds bad a$$ too with that exhaust :yesnod:

You also have to figure in torque due to the fact that the C6Z is a 427 and you are working with a 346.

This is true, Higgs, but for our applications (i.e. modified 346-NA) I don't believe they're relevant. Add power-robbing changes, such as heavier driveline components, belt driven superchargers, and other such things, and yes, the overall loss would increase. However, if you take a stock LS6, put it on a hypothetical "calibrated" chassis dyno, then take same engine and run it on another hypothetically "calibrated" engine dyno; then take that difference, and modify that same engine with the usual bolt-ons-perform the same two dyno runs, under the same conditions, I believe the DIFFERENCE in the two dynos would be near enough to identical, to be statistically irrelevant. Of course, I don't know this to be a fact, but I'd say these results would be probable.

usual bolt ons don't produce major hp differences (maybe 50-70) so why would you even waste time making this argument?

you are missing a concept because you are caught up in essentially immeasurable changes. think about it.

when the hp change becomes large enough to NOTICE a difference between a constant argument and a variable (%loss) argument the variable based measurement will be closer to reality.

you're getting hung up on very small differences in calculation at these power levels and it isn't even worth arguing about.

did you not say earlier that you did not think a 900 HP car is 750 at the wheels? that hardly seems like a bolt on 346 argument to me....