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HP loss @ the rw

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Old Oct 3, 2002 | 01:27 PM
  #1  
Black91Z28's Avatar
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HP loss @ the rw

how much HP is lost between the flywheel and the rear wheels?
Old Oct 3, 2002 | 03:01 PM
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It all depends. Meaning the strength of your drivetrain. A built tranny with a nice tourque converter and will save more hp/and tourqe at the rear wheels then a stock tranny would. Its can be anywhere from 30-50. Motors can be rated at a certain horsepower too, but then can choke around 25-30 more becuase of the stock intake, these TPI's need all the help they can get to breathe.
Old Oct 3, 2002 | 05:45 PM
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It can almost always be expressed as a percentage of SAE net hp.

An engine on a regular dyno will be read to produce, say, 300 hp. To find its SAE or net hp, it is necessary to deduct the number of hp used in running the engine's accessories. These are the alternator, the water pump, the power steering pump, the AC, etc. Usually the large manufacturers like GM will deduct 30 to 40 hp for this reason.

Thus the engine's SAE hp would be perhaps 270 hp.

From this number it has been found to be a very consistent loss of about 15% of the 270 hp for a manual transmission equipped car, and about 20% for one with an automatic.

Therfore the manual would have 270 - 15% (40 hp) = 230 rw/hp

And the auto equipped would have 270 - 20% (54 hp) = 216 rw/hp
Old Oct 4, 2002 | 09:07 PM
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From this number it has been found to be a very consistent loss of about 15% of the 270 hp for a manual transmission equipped car, and about 20% for one with an automatic






That is what i always heard
Old Oct 4, 2002 | 09:57 PM
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Moving to the bigwigs at the tech forum.

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Old Oct 5, 2002 | 05:25 PM
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As the hp goes up, the % loss goes down. If we are talking about measuring rwhp on a chassis dyno, many of the losses are not proportional to hp at all and are actually inertial losses. IOW they are not proprotional to hp but depend on the rotational mass of the drivetrain components. So the 15-20% range that is often quoted may be pretty accurate for a stock or nearly stock car, but as hp goes up the percentage loss will go down.

Rich Krause
Old Oct 5, 2002 | 06:17 PM
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That's because the increase in HP means an increase in torque. Torque is the work that's being done. HP is how quickly the work can be done. If the torque is increased, less HP is required to do the same amount of work at the same time.

If you can reduce the amount of friction losses that torque has to work against then again less HP is required.

Friction losses in a driveline depend on many factors. Transmission choice, torque converter stall speed, tranny oil type, differential type, posi choice and diff oil will all effect how much driveline losses a car can have.

Having an engine dynoed by itself is only good for torque converter selection. A chassis dyno that measures RWHP is what you need to use since that's the final output that pushes the car down the road. RWHP can be changed by simply changing driveline components. A 7.5" 10 bolt takes less energy to turn than a 9". A powerglide tranny takes very little HP to turn compared to a TH400. Even a TH400 can be modified to take less HP to turn by using lightweight internal parts.
Old Oct 6, 2002 | 03:54 AM
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For a stock or near stock car the usual 15% and 20% are pretty safe bets, but once you up the power those numbers will change, thus proving the fixed percentage idea incorrect.

If you care to read (long):
http://www.ls1.com/forums/showthread...hreadid=151557
Old Oct 7, 2002 | 12:32 PM
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arnie's Avatar
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Originally posted by Stephen 87 IROC
That's because the increase in HP means an increase in torque. Torque is the work that's being done. HP is how quickly the work can be done. If the torque is increased, less HP is required to do the same amount of work at the same time.
Say what? I suggest you study some engine dyno results/graphs and refresh yourself on the formula used to calculate hp based on measured torque readings.
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