[Z06] LS6 underrated???
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Safety Car
Joined: Jun 2006
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From: Ashland, OH
LS6 underrated???
according to Wikipedia GM underrated the engine by 20hp giving it a true hp rating of 425 (02-04 versions)......hmmmmm 
i wonder if this was done for the the then new C6 rolling out?
i wonder if this was done for the the then new C6 rolling out?
Burning Brakes
Joined: Nov 2004
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Back when GM High Tech Performance magazine tested the new 2004 Z06, they estimated their test car at 424 FWHP, using their observed RWHP# and doing the math. They got the car into the 11.9s after only a few 1/4 mile runs, bone stock--tires and all, so they had no reason not to believe such figures.
As for the new LS7 cars, we only have to see what the tuners are getting out of them by only re-tuning on the dyno. If you look at the LS7 head flow numbers, the only thing holding back the 427 is the relatively 'mild' cam timing/phasing, and GM's attempt at a "politically correct" tune aimed at the obvious emissions and fuel economy constraints.
As for the new LS7 cars, we only have to see what the tuners are getting out of them by only re-tuning on the dyno. If you look at the LS7 head flow numbers, the only thing holding back the 427 is the relatively 'mild' cam timing/phasing, and GM's attempt at a "politically correct" tune aimed at the obvious emissions and fuel economy constraints.
Le Mans Master
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From: Winchester TN
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St. Jude Donor '03-'04-'05-'06-'07-'11
Hi Cjunkie -
Wikipedia can be updated by anyone. I do not believe that any of the "facts" on Wiki are verified.
An excerpt from the Wiki "about us" page.....
Most 02-04 Z06's were, in experience, putting out pretty near 405 hp factory, and putting 350-360 hp to the rear wheels.
The article that the Wiki "quotes" shows that the people who tested the GM provided "press" car achieved 363rwhp on a dyno - using 15% drivetrain loss, that comes to 417 fwhp - not 425 as they somehow calculated (and they used 13% drivetrain loss factor - so I don't know where they came up with 425....).
Moral of the story - don't believe everything you read on the internet - or Wiki.....
best rergards -
mqqn
Wikipedia can be updated by anyone. I do not believe that any of the "facts" on Wiki are verified.
An excerpt from the Wiki "about us" page.....
Originally Posted by Wikipedia
....its articles can be edited by anyone with access to the Internet, simply by clicking the edit this page link.
The article that the Wiki "quotes" shows that the people who tested the GM provided "press" car achieved 363rwhp on a dyno - using 15% drivetrain loss, that comes to 417 fwhp - not 425 as they somehow calculated (and they used 13% drivetrain loss factor - so I don't know where they came up with 425....).
Moral of the story - don't believe everything you read on the internet - or Wiki.....
best rergards -
mqqn
Last edited by mqqn; Sep 9, 2007 at 12:44 PM.
Le Mans Master
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St. Jude Donor '03-'04-'05-'06-'07-'11
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Burning Brakes
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From: Douglasville Ga
the calculation should be:
360/0.85 = 423.529-----this is the same as (423.529 * 0.85 = ? RWHP)
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St. Jude Donor '03-'04-'05-'06-'07-'11
Thanks - that makes sense then, and my car is making (according to this) 458.8 fwhp.
And I still cannot get it in the 11's.
best regards -
mqqn
Melting Slicks
Joined: Oct 2006
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From: Gainesville Georgia
My 02' did 356 when It was stock. I think they generate a little more than the factory rated 405 figure. Every same kind of car is different though obviously.
Moderator
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I think the 405 number represents an average from GM; I remember reading that they tested a string of LS6's and the lowest powered average they got became the rating. I also believe that they are all pretty close to 405 410 hp.
Last edited by Millenium Z06; Sep 9, 2007 at 06:23 PM.
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St. Jude Donor '03-'04-'05-'06-'07-'11
Hi Guys -
After thinking about this for a while today, I had the following thought.
Drivetrain loss for a given car should be static - the cost of driving a specific drivetrain would not have a linear correlation to the output of the engine driving it.
In other words - if it takes x hp to turn the drivetrain, why would it take x+ to turn the same drivetrain if the force driving it was increased?
To look at this yet another way, say my car makes 390 rwhp, and doing the rwhp / (100 - drivetrain loss percent) it raises the number to 458. That seems to indicate that it takes roughly 68 hp to turn the drivetrain and accessories.
So why, if I increased (or decreased) the output of the engine, would the drivetrain take more or less power to drive it? If my car were making 490 rwhp, using the "generally accepted" equation, my exact same drivetrain now takes ~86 hp to turn it. I just picked up 18 free brag-powers.
Seems to be a flaw in the logic here - or is it just me....again
I assert that the higher the rwhp number plugged into the equation, the more invalid upward skew is built in.
best regards -
mqqn
After thinking about this for a while today, I had the following thought.
Drivetrain loss for a given car should be static - the cost of driving a specific drivetrain would not have a linear correlation to the output of the engine driving it.
In other words - if it takes x hp to turn the drivetrain, why would it take x+ to turn the same drivetrain if the force driving it was increased?
To look at this yet another way, say my car makes 390 rwhp, and doing the rwhp / (100 - drivetrain loss percent) it raises the number to 458. That seems to indicate that it takes roughly 68 hp to turn the drivetrain and accessories.
So why, if I increased (or decreased) the output of the engine, would the drivetrain take more or less power to drive it? If my car were making 490 rwhp, using the "generally accepted" equation, my exact same drivetrain now takes ~86 hp to turn it. I just picked up 18 free brag-powers.
Seems to be a flaw in the logic here - or is it just me....again
I assert that the higher the rwhp number plugged into the equation, the more invalid upward skew is built in.
best regards -
mqqn
Melting Slicks
Joined: Nov 1999
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From: Madison Alabama
Here's the run sheet on my Z16 - bone stock with about 11K miles on the clock. Previous owner dynoed the car about 3 months before I bought it...as he picked up a C6 Z.
The numbers from the run file: RunFile_005.drf - 11/18/05 6:46:28 PM Run Type RO Run Conditions 60.65 degrees f, 29.66 in-Hg, Humidity 9%, SAE 0.97
This was a SAE corrected pull. The Raw numbers can be obtained by dividing the HP & TQ numbers by the SAE factor (0.97) i.e., 355.42/.97=366.41 HP. Either way that's about a 10 to 12 percent loss through the drive train.
The numbers from the run file: RunFile_005.drf - 11/18/05 6:46:28 PM Run Type RO Run Conditions 60.65 degrees f, 29.66 in-Hg, Humidity 9%, SAE 0.97
This was a SAE corrected pull. The Raw numbers can be obtained by dividing the HP & TQ numbers by the SAE factor (0.97) i.e., 355.42/.97=366.41 HP. Either way that's about a 10 to 12 percent loss through the drive train.
Melting Slicks
Joined: Dec 2006
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From: Woodinville WA
Hi Guys -
After thinking about this for a while today, I had the following thought.
Drivetrain loss for a given car should be static - the cost of driving a specific drivetrain would not have a linear correlation to the output of the engine driving it.
In other words - if it takes x hp to turn the drivetrain, why would it take x+ to turn the same drivetrain if the force driving it was increased?
To look at this yet another way, say my car makes 390 rwhp, and doing the rwhp / (100 - drivetrain loss percent) it raises the number to 458. That seems to indicate that it takes roughly 68 hp to turn the drivetrain and accessories.
So why, if I increased (or decreased) the output of the engine, would the drivetrain take more or less power to drive it? If my car were making 490 rwhp, using the "generally accepted" equation, my exact same drivetrain now takes ~86 hp to turn it. I just picked up 18 free brag-powers.
Seems to be a flaw in the logic here - or is it just me....again
I assert that the higher the rwhp number plugged into the equation, the more invalid upward skew is built in.
best regards -
mqqn
After thinking about this for a while today, I had the following thought.
Drivetrain loss for a given car should be static - the cost of driving a specific drivetrain would not have a linear correlation to the output of the engine driving it.
In other words - if it takes x hp to turn the drivetrain, why would it take x+ to turn the same drivetrain if the force driving it was increased?
To look at this yet another way, say my car makes 390 rwhp, and doing the rwhp / (100 - drivetrain loss percent) it raises the number to 458. That seems to indicate that it takes roughly 68 hp to turn the drivetrain and accessories.
So why, if I increased (or decreased) the output of the engine, would the drivetrain take more or less power to drive it? If my car were making 490 rwhp, using the "generally accepted" equation, my exact same drivetrain now takes ~86 hp to turn it. I just picked up 18 free brag-powers.
Seems to be a flaw in the logic here - or is it just me....again
I assert that the higher the rwhp number plugged into the equation, the more invalid upward skew is built in.
best regards -
mqqn
There may be some truth to that as far as frictional losses, but a lot of the HP loss is in accelerating the mass of the drivetrain. The more HP the faster you accelerate the parts, but the faster they accelerate the more the inertia of the parts resists the acceleration, so it is pretty constant.