jsbwac

Does anyone know the percentage of HP lost on a c6? I know the motor is rated at 400 HP, but I heard the HP to the rear wheels is only in the 350 range.

Also, if a supercharger is installed, what is the percentage difference between the motor's HP and the RWHP?

Thanks!!!!

meshoooo

Its about %15

jsbwac

Thanks!

6Speeder

Try 12% for manual trannys, 15% for autos.

haljensen

Why worry about percentages?

LS2; Dynojet 248 SAE corrected = Manual 340-345 RWHP
" '' '' '' '' '' '' '' '' '' '' " '' '' '' '' '' '' '' = Auto 330-335RWHP

55-60 HP loss manual and
65-70 HP loss automatic.

Same should hold true for the LS3 since it's pretty much the same drivetrain. Worked for my LS3, it dyno'd 373RWHP for an A6 w/NPP
436 Advertized
373 Dyno
63 HP loss to the rear wheels from advertized.

HITMAN99

Also depends on the dyno used to measure the RWHP. Dynojets typically measure higher than Mustang dyno's, so they will show the lowest amount of drivetrain loss.

jrnorman

That is about right and maybe specifically be better numbers for a C5 but I have generally used 15% for manuals and 17% for automatics for all cars and it generally is about as close as you can get because of the uncertainty of the actual measurements.

HuskerBullet

Don't look at it as a percentage of engine HP loss. It's a straight forward loss. The drive train doesn't increase in HP loss if you increase the engine up to 800HP. For example: 400HP x 15% = 60HP loss, which is true, but 800HP x 15% = 120HP loss, which is false.....the drive train is still only a 60HP loss. RWHP; 400 - 60 = 340rwhp, and 800 - 120 = 680rwhp (wrong), 800 - 60 = 740rwhp (right). Make sense?

jsbwac

That makes a lot of sense, because like you said, why would the drivetrain take up more on a high HP engine (if anything you would think it might even be less). Thanks guys for the advice. I am getting a SC from ECS put on and it is expected to produce about 550 HP at the rear wheels, so I think it is safe to say the engine HP should be in the 600 range.

6Speeder

While it sounds straight forward it isn't right. HP lost through the drivetrain is a percentage of what's produced. Look at it this way, we can motor down the freeway using less than 20hp. IF the loss was a fixed amount (say 60 hp) that would be impossible.

mirage2991

Physics 101 will explain why it's a % loss and not a fix number

glass slipper

Really??? Please enlighten us with your explanation...and don't hesitate to use Physics 201 if needed, I'll try to keep up. Throw as many formulas/equations out there as needed too, I can envision over 50 right off the top of my head so don't think you're over doing it when you come up with over 100.

Hint: You'll need Physics at the Doctoral level for your answer. But then again, if you had that knowledge you would not have made your statement above. I'll give you the basic formula for HP loss through the drivetrain just to get you started:
FHP=Ax^2+Bx+C where FHP is friction HP, A is associated with exponential losses, B is associated with linear losses, and C is constant losses and x is the HP of the engine at the crank.

TrenAman

I am not an expert, nor claim to be one, but this just does not make sense at ALL!! YOua re telling me if fuel management was on the car, so only half the cylinders "like the new LS3 in the Camaro for example" is ti would be running 200hp, you are trying to tell me that it would still lose 60hp, or if I had a 1,000 rwhp engine I would still only lose 60hp? This does not make sense to me, I am not saying you are wrong. I have just never heard this and in my mind, it sounds wrong..

KB9GKC

Ditto!

Douglas in Green Bay

haljensen

20 RWHP PLUS 60HP lost to the drivetrain means 80HP at the flywheel is all thats needed.

Seems believable, my 80HP 48 Ford would almost do 100MPH.

FireFalcon

You didn't explain the forumla very well, so you must not have a Ph.D. in physics, lol. By assertion, that HP loss through the drivetrain is fairly constant (only a fool would say it is purely constant), all terms but C would have to be zero. If this is the case, I doubt we would need to resort to second order equations to solve the problem.

My way of reasoning about the problem is this. We know that the primary source of power loss through the drivetrain is kinetic friction. So, the question reduces to the replationship between the force of kinetic friction versus the force working against it. Is this force constant or not?


---Matthew Hicks

FireFalcon

... by the way, did the OP mean C6? The C5's engine wasn't rated near 400HP.


---Matthew Hicks

Tzzird

My LS2 dynoed at 346 rwhp stock, which is a 13.5% loss.

jrnorman

The basic physics is that the drag from the bearings and gears is roughly proportional to the power being transmitted. Take just gears - the teeth are usually at an angle and when power is being transmitted, it causes drag on the gear teeth and that drag is directly proportional to the amount of force or power. Lubrication helps reduce some of it but that drag is what makes the lube hot and why racing cars need additional coolers and street cars don't.

The more power being transmitted the more drag and the more heat generated. All gears slip a little on the tooth face (friction) and the angles cause thrust forces on the shafts that put additional load on the bearings (more friction). Absolutely all gears cause a parasitic loss - less power coming out than when in and it is roughly proportional to the power.

Speed which includes tooth face velocities and tooth meshing frequency is also a factor – the faster a given gear or bearing turns the more losses and that is generally by the square – not proportional. This is a primary reason why the LS2 and LS3 have such good gas mileage because everything is turning a little slower than most cars. My C6 does about 85 at less than 2000 rpms – you gotta love that!

jsbwac

Wow, this has become quite the debate. I appreciate all the help and advice. It seems that after all of the above, the best way to figure it out is based on percentages, with the overall consensus being in the 10 to 12% range. While I was a math major, for some reason physics always eluded me. If anyone can explain the physics portion in a simple way and give me a simple formula, that would be appreciated. At first the constant seemed to make some sense, but after reading the above, I can see both the constant and variable arguments. My C6 (sorry, I put C5 in the original posting) came with 400 HP but allegedly has 345 to the rear wheels. That is a about a 14% loss. I would think that as HP goes up it would have a direct relationship with the percentage, with the percentage slowly dropping in some type or ratio between the two. I guess without doing a study of different cars comparing engine and rear wheel HP, this would be a difficult ratio to come up with so the answer was more concrete.