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20% HP lost through drive train how does it work

Discussion in 'The Bench' started by bgs455, Oct 8, 2003.

  1. bgs455

    bgs455 OIF OEF HOA ONE

    Hi all, can some body school me on this? How does the standard 20% loss stay constant. In other words a 500HP engine would be 400HP at the wheels a 100HP loss but, a 1000HP engine would be 800HP at the wheels a 200HP loss. How does this work? Shouldn't there be a set HP loss for a given trany, drive shaft, rear end, wheels, and tires? If so, what would be the loss for a 400 trany, 3 in. drive shaft, 12 rear, 15X8 steel wheel and 10 in. slicks? I just don't understand why the more power you make the more power it takes too run the drive train.:Do No:
     
  2. ricknmel67

    ricknmel67 Well-Known Member

    I'm no expert, but....

    I would venture a guess that it's not a linear loss in HP in relation to starting HP. I'm sure there's some type of "curve"

    Similar to the old "100 lbs equals a tenth off your ET" equation.

    100 pounds might make a 15.20 car run 15.10

    But 100 lbs taken off a 9.50 car might make it run 9.35?

    (Or maybe it curves the other way?)

    Anyways... my point is that I'm sure there's a curve in the equation one way or another.
     
  3. bobc455

    bobc455 Well-Known Member

    20% is just an approximation, and depends entirely on the running conditions and 100 other things.

    For example if you have an automatic transmission, when you are sitting at the stoplight the engine might be producing 20 HP, but 100% of that is being absorbed by the drivetrain.

    On the other hand, driving down the highway at steady speed, you might have only about 5% loss in the drivetrain.

    There are several factors to consider, and they all kinda interrelate. The torque convertor slips, which is a "loss", different amounts depending on how much torque the engine is producing, the stall speed of the convertor, the rearend ratio, etc. Also, u-joints are flexing and the rearend is spinning. If you have ever had a rearend out of the car and spun it by hand, you'll notice that it doesn't keep spinning forever and ever- there is some friction in there that wants to make it stop.

    Some people shy away from the TH400 because it "absorbs" more HP- this is a bit misleading. The TH400 has heavier rotating components, and therefore mpre HP is needed to accelerate those components from stationary to full speed that lighter components in other tranmissions. Since the components in the TH400 are heavier, they take more force/torque/HP to accelerate up to speed. It would be the same if you had a 400# driveshaft- a shaft that heavy would take more force just to get up to a higher RPM, although the driveshaft doesn't really "absorb" any torque.

    I'm sure that could have been worded better, but it's only 5: 30 in the morning. :)

    -Bob Cunningham
     
  4. gsxnut

    gsxnut Well-Known Member

    Most of the losses in your drivertrain are due to friction in some way shape or form. Friction is not a constant force but a multiplier. i.e. When you have a friction force, the bigger the initial force used the bigger the friction force. So if you are making 100HP your friction force is 20HP. meaning 20% of the energy you are expending is being used on friction and it turned into some other type of energy form such as heat. The more horsepower you create the more force is used.

    The 20% is a rough estimate used.

    Mark
     
  5. Nailheaded48

    Nailheaded48 Well-Known Member

    I was thinking about this same thing the other day and the only thing I could come up with is that since it takes a certain amount of hp to accellerate the drive train parts at a given rate, it must take more to accellerate them even faster. So as your engine hp goes up and you have the potential to more quickly accellerate the car, it would also take more to get the drivetrain moving at a quicker rate. So I imagine the 20% is a number that only applies to wot situations. That make sense to anyone but me?
     

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