997 2005-2012 911 C2, C2S, C4, C4S, GTS, Targa and Cabriolet Model Discussion.
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Old May 26, 2006 | 08:54 AM
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It is a low 4.5 lbs of boost..
 
Old May 26, 2006 | 09:14 AM
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So, Are you still gonna sell it then? I would'nt
 
Old May 26, 2006 | 10:53 AM
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...not sure, I'm in love all over again
 
Old May 26, 2006 | 12:33 PM
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Great power!!! Have fun with it!
 
Old May 26, 2006 | 09:29 PM
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Originally Posted by MexicoBlue
Actually Billy, that would indicate 543 chp (you multiply by 1.22, not 1.18 to get back up to your crank hp if you lost 18% to drivetrain - common mistake). Even better! Sweet . . . I'm going to claim your dyno sheet as my own!!! (j/k sort of)
how did you do the math on that one? what would 1.18 translate to as far as driveline loss?
 
Old May 26, 2006 | 11:30 PM
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Originally Posted by deputydog95
how did you do the math on that one? what would 1.18 translate to as far as driveline loss?
OK, fairly abstract math concept, but hopefully easy to explain. Let's pick round numbers. Let's say you have a motor that makes 500 hp at the crank, and 400 hp at the wheels. Therefore the 100hp loss due to drivetrain is 20% (100hp is 20% of your 500chp, so you have a 20% loss due to drivetrain). However, 100 hp is 25% of 400hp (again, same drivetrain loss, but when you are starting with whp, the same 100hp is 25% of the smaller number, whereas it is 20% of the larger crank #). So, when you know your % loss due to drivetrain, and you have your wheel hp number, you are actullay multiplying by a larger number to gross back up to crank hp, because it is the percentage of the smaller number now that you are multiplying by. So in our example, if our car measures 400whp, and you multiply by 1.2 because of the 20% drivetrain loss, you would wrongly think the true chp is 480, but in fact you have to multiply by 1.25 to get back to the true chp of 500hp, again because the actual loss due to drivetrain is a much larger percentage of your wheel hp measurement, than of your crank, because wheel is a smaller number. And it's particularly relevant because most people can approximate loss to drivetrain by comparing actual wheel hp to published crank hp, so they know the actual hp loss from crank to wheel, but you have to remember that is a higher percentage of your wheel hp, so that after mods when you are trying to gross back up to crank hp, you have to use the larger conversion % of your actual hp loss as a % of your smaller wheel hp.

No idea if that made any sense. I'm operating on a lot of wine here. But the underlying principle is that actual hp loss is a larger % of your wheel hp than your crank hp, yet drvetrain loss is usually quoted as a reduction FROM crank hp, but the conversion back TO crank hp requires a higher multiple as a result.

Sorry for the thread jack, but it's a frequent mistake, that almost defies common sense, so its understandable why the conversion back to crank hp is so often done with the wrong multiple.

- Russ
 

Last edited by MexicoBlue; May 27, 2006 at 12:17 AM.
Old May 27, 2006 | 03:05 AM
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1 - From Russ's example
2 - HP at the crank = 500 = X
3 - Hp at the wheels = 400
4 - loss due to drivetrain = 500 – 400 = 100 = Y
................................................( X - 400 = Y )
5 - 100 / 500 = 0.20 (20% loss due to drivetrain)
......Y / X = 0.20
6 - 100 / 400 = 0.25 (percentage of loss to the wheels HP ) = Z
7 - so 400 x 1.25 = 500

8 - let’s apply this on BPANOS’s car
9 - we don’t know the HP at the crank lets call it = X......................#2
10 - HP at the wheels = 444.88
11 - We don’t know the loss due drivetrain let’s call it = Y which is = (X – 444.88 = Y )...............#4
12 - Y / X = 0.18 (18% loss due to drivetrain ) .............#5
13 - Y / X = Z we don’t know the percentage due to the wheels HP #6

14 – .........................................Y = X – 444.88 from .....................#11
15 – Y / X = 0.18 that means Y = 0.18X..............#12........................mult iply both side by X
16 – Y = X – 444.88 ...................= 0.18X .............................from #14 & #15
17 - X – 444.88 ...........................= 0.18X
18 - X – 444.88 – 0.18X .............= 0.18X – 0.18X ..................add - 0.18 for both side
19 - X – 444.88 – 0.18X .............= zero ...................................0.18 - 0.18 cancel eachother
20 - X- 444.88 +444.88 – 0.18X= zero + 444.88 ....................add 444.88 for both side
21 - X – 0.18X ............................= 444.88 ............................. - 444.88 + 444.88 cancel
22 - X(1 – 0.18) ..........................= 444.88 ...............................commen factor X
23 - X( 0.82 ) ..............................= 444.88 .................................divide by 0.82
24 - X( 0.82 ) / (0.82) .................= 444.88 / (0.82)
25 - X = 542. 54 which is 543 = X the HP at crank ......... .........#2
from #4 ..........X - 444.88 = Y which is 543 - 444.88 = 98.12 = Y
lose due drivtrain = 98.12
from #6 ..........98.12 / 444.88 = 0.22 (22% percentage of loss due to the wheels HP)
here is why we multiply by 1.22
444.88 x 1.22 = 542.7 which is 543HP ..........#7



I know russ you did all that to get the 1.22 I really happy that I learn very nice lesson from you thank you so much and sorry guyz it's a lot of math but this's how it works
 

Last edited by cayenne_ksa; May 27, 2006 at 03:16 AM.
Old May 27, 2006 | 06:48 AM
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anybody got an apirin? my head is killing me

i've got a pretty decent amount of dyno time under my belt now.
from what i've told, it's fairly accepted to use 15% for driveline losses on a C2.

the formula i have been working off is RWHP/.85=crank HP

that's what i was told to use anyway. there is another way to do it, but i don't remember right now.

is this way oversimplifying it? it seems to work.
 
Old May 27, 2006 | 09:01 AM
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Originally Posted by deputydog95
anybody got an apirin? my head is killing me

i've got a pretty decent amount of dyno time under my belt now.
from what i've told, it's fairly accepted to use 15% for driveline losses on a C2.

the formula i have been working off is RWHP/.85=crank HP

that's what i was told to use anyway. there is another way to do it, but i don't remember right now.

is this way oversimplifying it? it seems to work.
That formula is the same as what has been said above

RWHP/0.85=crank HP=RWHP*1.176 Both correct, just the method of how to get thereis different.
 
Old May 27, 2006 | 09:12 AM
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Whew. Lots of red wine last night. This all was so much clearer then! And thanks to Abdullah for explaining the mathematical steps much better than I ever could have, truly! But for most of our purposes, probably simpler just to use measured rwhp, which is why I have it in my sig line . . .
 

Last edited by MexicoBlue; May 27, 2006 at 09:15 AM.
Old May 27, 2006 | 09:38 AM
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Hey Russ, Did you ever get the dyno graph for your car? It would be interesting to compare since your car is running the oem cats.
 
Old May 27, 2006 | 10:41 AM
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bpanos: how did you determine how much boost you're running? do you have an aftermarket gauge?


here's my dyno chart after i had the suncoast headers installed. not to shabby for a 2002 3.6 liter it was measured on a dynojet as well.

 
Old May 27, 2006 | 11:03 AM
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great numbers deputydog,.. w/regards to boost I am just going by what VF told me.
 

Last edited by bpanos; May 27, 2006 at 11:05 AM.
Old May 27, 2006 | 11:51 AM
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funny thing is that VF builds in a +-1 psi variance. i would have it checked if i were you to make sure it's running the advertised boost. if you're a little on the low side you may be able to tap into 30 or so more HP.
 
Old May 27, 2006 | 10:46 PM
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Too much math in this thread, lol. Your car looks great, nice mods you have.
 


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