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How is this measured/calculaed? I have always seen it as a percentage of the given engines power at the crank, but why a PERCENTAGE? Shouldn't it be a fixed amount seeing as how all the drivetrain components' rotating wieghts are fixed values? Maybe it's just late and I'm not thinking clearly.
Ohsigmachi '96 GL MTX Zetec SS (SuperSleeper)
Suffice it to say my mod list has become so long that it is just ridiculous
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Most rolling road dynos should be able to 'motor' the trans in each gear to establish exact HP loss to drive...few do!
V6 MTX
'Don't p**s up my back then tell me it's raining!!!"
'Its only nuts & bolts!'
'If I build it,fix it,upgrade it or modify it...MAYBE they will come....!
Haines Motor Sports Inc,
Dealer for 'Quaife America' & 'Autotech Sport Tuning'
SOLE USA Dealer for the American Axle 'AUSSIE BAR'...
Get a Turbo for you Zetec from HMS Inc...by 'The Demon' ...www.DemonDynamics.co.uk
..don't talk about it DO IT !!!
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Terry, do you know the real % loss to drivetrain with the MTX75 or does it vary car to car?
1999 Amazon Green SVT Contour (#554/2760)
Stock SVT Duratec V6 with:
Intake- K&N filter/75mm MAF meter
Exhaust- MSDS Y-pipe/Bassani catback
Durability-Ford "dual mode" damper, Mobil 1/K&N oil filter
179.2 FWHP at 6900 RPM
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MTX-75 is about 18%, IIRC.
-Mark
You can do something for Love...
You can do something for Money...
But there is nothing quite so satisfying as doing something out of Spite.
GTExtreme1@aol.com
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Why a PERCENTAGE though ,thats my real question. OK I will try to explain myself more clearly w/ an example. For ease of Calculation I just say the MTX-75 has a 20% parasitic loss. My car came from the factory with 125hp so 100 hp@ the wheels right. A 25 HP Net LOSS. Now if I modify my zetec (and belive me it is) should I subtract: 25 HP OR 20%...Am I making sense at all???
Ohsigmachi '96 GL MTX Zetec SS (SuperSleeper)
Suffice it to say my mod list has become so long that it is just ridiculous
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Ok here is a simple formula of how to figure out your crank hp after you dyno:
FWHP divded by remainder (if 20% is the loss then 80% is the remainder, so you would divide by .8)
sticking with the 20% loss,
you dyno 100fwhp
100/.8 =125
Like I said above 18% is a little more accurate
so if you dynoed 100fwhp
100/.82 = 121.95
my last good dyno pull was 293.6fwhp
293.6/.82 = 358.05
See how that works?
now to go the other way you would
take your crank hp and multiply it by the % of loss then subtract the answer from the crank hp number to get the whp.
BHP-(BHP x %) = WHP
358-(358 x .18)=294
-Mark
You can do something for Love...
You can do something for Money...
But there is nothing quite so satisfying as doing something out of Spite.
GTExtreme1@aol.com
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Hey, thanks for the added info... I was under the impression that (using the previous example) 20% loss on a 125HP engine meant losing 25HP in the drivetrain... Now turbocharge that engine, you would still lose 25HP in the drivetrain because you didn't change anything else... Your equation says I'm wrong but I fail to understand why...
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Originally posted by Terry Haines:
Most rolling road dynos should be able to 'motor' the trans in each gear to establish exact HP loss to drive...few do! Got a quick explaination as to how this is done?
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yeah unfortunatly the loss does increase with power... the an excuse to make more power  -Mark
You can do something for Love...
You can do something for Money...
But there is nothing quite so satisfying as doing something out of Spite.
GTExtreme1@aol.com
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yeah unfortunatly the loss does increase with power That is my entire question... How does the number of horse power loss increase linearly with the increase in engine HP if the drivetrain stays the same 
Ohsigmachi '96 GL MTX Zetec SS (SuperSleeper)
Suffice it to say my mod list has become so long that it is just ridiculous
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