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Why do you lose a % and not a given hp#????

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  • Why do you lose a % and not a given hp#????

    Why is it that when you have an engine that makes 100 flywheel hp it will only put down roughly 15% less at the rear wheels, 85whp? I would think that a drivetrain would require absorb a given amount of hp not a percentage.

    So if I have a 1000hp flywheel then I am only putting down 850whp. This is somthing that never made sence to me that maybe somone could exlplain a little more. I understand that friction could go up, but I wouldnt think it would be a percentage still. :?:
    85 Z31 6.0 LSX turbo 766whp/792wtq
    04 GTO, LS6, big cam, porting, N20... underway for summertime daily driver.

  • #2
    because you can grab the input shaft and turn it, and apply .001 horspower, low torque and low speed. if it took your straight 15 hp number, you couldnt possibly do it and would break your arm trying.

    or you can slap an engine on it, increase the torque by a bazillion times and rotate it a bunch faster as well.

    losses are due to frictions, caused by gear and lube frictional/shear forces (mating gears have to slide against each other, higher load = more friction), drag on the bearings (again based both on speed and loading due to torque causing tangential loading, as well as axial loading because of helical cut gears used in road car gear boxes) and then a little bit from the oil being flung about and squeezed out of gears etc.

    obviously some random straight percentage number is bogus, but its the easiest number for people to use, and 15% is just the standard for people who for some reason want to extrapolate engine horsepower backwards from a dyno test to enhance the size of their tesicles.

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    • #3
      nutsaq wrote: because you can grab the input shaft and turn it, and apply .001 horspower, low torque and low speed. if it took your straight 15 hp number, you couldnt possibly do it and would break your arm trying.

      or you can slap an engine on it, increase the torque by a bazillion times and rotate it a bunch faster as well.

      losses are due to frictions, caused by gear and lube frictional/shear forces (mating gears have to slide against each other, higher load = more friction), drag on the bearings (again based both on speed and loading due to torque causing tangential loading, as well as axial loading because of helical cut gears used in road car gear boxes) and then a little bit from the oil being flung about and squeezed out of gears etc.



      obviously some random straight percentage number is bogus, but its the easiest number for people to use, and 15% is just the standard for people who for some reason want to extrapolate engine horsepower backwards from a dyno test to enhance the size of their tesicles
      So then this being the case, If you were to run a dog box tranny you would have SLIGHT gains due to the fact that there are not helical cut gears? Is that correct or do dog boxes have helical gears too?
      85 Z31 6.0 LSX turbo 766whp/792wtq
      04 GTO, LS6, big cam, porting, N20... underway for summertime daily driver.

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      • #4
        SATAN wrote: So then this being the case, If you were to run a dog box tranny you would have SLIGHT gains due to the fact that there are not helical cut gears? Is that correct or do dog boxes have helical gears too?
        nope you're correct dog teeth = squared to the shaft

        also my understanding of physics says that drivetrain losses are not a fixed percentage but do vary over engine speed and whether speed is increasing or decreasing, there will be a fixed rotating mass behind the flywheel that will add or subtract mass because of inertia (momentum)

        also in the real world, road conditions, tires, wind direction etc will all play a role, but in purely test-lab conditions you have two things mass and friction to worry about, the friction coefficient shouldn't change, but een that does due to the fact that it is fluid based and fluid moves and (de)laminates at various speeds, and as I just explained the rotating mass "loss" will curve up or down based on the motion given to the objects in play engine vs. drivetrain

        and all that will plot on a graph and the power output of the engine will vary to the rpm as will the loses... in other words if you really calculated every detail as above, in theory you'd have a squiggly line... however add in the stabilizing factor of the car's net weight and that line will be oversampled and could be reduced to an adequate representation percentage.... is it 15? 20? no each car will vary there too...

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        • #5
          "Why do you lose a % and not a given hp#?"

          I asked my self that question once, and came up with a simple answer.
          The more power a motor makes, then the faster it will accelerate.
          It takes more power to accelerate something faster.
          So the faster you accelerate the drive train, the more is power is lost in it.
          imagination is a virtue

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          • #6
            craZed wrote: "Why do you lose a % and not a given hp#?"

            I asked my self that question once, and came up with a simple answer.
            The more power a motor makes, then the faster it will accelerate.
            It takes more power to accelerate something faster.
            So the faster you accelerate the drive train, the more is power is lost in it.
            That seems like a pretty good simple answer
            85 Z31 6.0 LSX turbo 766whp/792wtq
            04 GTO, LS6, big cam, porting, N20... underway for summertime daily driver.

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            • #7
              [quote]SATAN wrote:
              Originally posted by craZed
              "Why do you lose a % and not a given hp#?"

              I asked my self that question once, and came up with a simple answer.
              The more power a motor makes, then the faster it will accelerate.
              It takes more power to accelerate something faster.
              So the faster you accelerate the drive train, the more is power is lost in it.
              That seems like a pretty good simple answer
              exactly, accelerating an object is like increasing it's mass, it gets harder to push the more you push, that's why I said it's a curve

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              • #8
                [quote]G-E wrote: [quote=SATAN]
                Originally posted by craZed
                "Why do you lose a % and not a given hp#?"

                I asked my self that question once, and came up with a simple answer.
                The more power a motor makes, then the faster it will accelerate.
                It takes more power to accelerate something faster.
                So the faster you accelerate the drive train, the more is power is lost in it.
                That seems like a pretty good simple answer
                exactly, accelerating an object is like increasing it's mass, it gets harder to push the more you push, that's why I said it's a curve
                Gotcha. Thanks guys
                85 Z31 6.0 LSX turbo 766whp/792wtq
                04 GTO, LS6, big cam, porting, N20... underway for summertime daily driver.

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                • #9
                  come on now guys, I cent let this be dropped with such a silly last post and goofy conclusion. If you buy into the mass acceleration theory, then what happens when you are going a contant speed? Angular acceleration of drivetrain components goes to zero and they suddenly start to operate at 100% efficiency with no losses? You guys are smarter than that

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                  • #10
                    nutsaq wrote: come on now guys, I cent let this be dropped with such a silly last post and goofy conclusion. If you buy into the mass acceleration theory, then what happens when you are going a contant speed? Angular acceleration of drivetrain components goes to zero and they suddenly start to operate at 100% efficiency with no losses? You guys are smarter than that
                    Right,

                    You guys are right on the money in most respects.

                    The one thing I will add is that the percentage does change a lot with different RPM and torque applied. It's not a flat CoF like you find with a friction lining, it's dynamic. While a stock turbo Z may lose around 15% torque to the wheels at redline, a modified car may lose 13% or even 17% of total engine torque; the ballpark is the same but the figure is not. The only way to know for sure would be to put you engine on an engine dyno and then drop it in the car and run it on a chassis dyno.

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                    • #11
                      I'll toss in another rule of thumb I've heard over the years:

                      AWD cars: 20% drivetrain loss



                      I'm curious on FWD cars. Seeing that there is a bunch less weight to rotate between the engine and the wheels....maybe 10% loss would be a solid estimate?

                      How about mid-engine rwd cars?

                      This could be a job for the mythbusters 8)
                      2008 BMW E92 335i 6MT

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                      • #12
                        Signal 12 wrote: This could be a job for the mythbusters 8)
                        That would be a great episode!

                        (I reject your reality, and substitute my own.)

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                        • #13
                          Signal 12 wrote: I'll toss in another rule of thumb I've heard over the years:

                          AWD cars: 20% drivetrain loss



                          I'm curious on FWD cars. Seeing that there is a bunch less weight to rotate between the engine and the wheels....maybe 10% loss would be a solid estimate?

                          How about mid-engine rwd cars?

                          This could be a job for the mythbusters 8)
                          again, stop putting weight into this equation, there are no losses due to weight. FWD will benefit from slightly fewer bearings and gears, thats it. At the same time it can be troubled by the need to be compact.

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                          • #14
                            I don't see why this thread continued past the first reply with the "coefficient" theory. To simplify this even more, lets say you have a box that you are pushing and there is friction coefficient of k. If you push it with x newtons, then you will lose kx of the force you applied due to friction. If you push it again with x+25 newtons, you will lose k(x+25) of the pushing force. There is a coefficient similar to this friction coefficient in your drivetrain. You could think of x as your drivetrain loss in the case at hand. I know this disregards Jason's post on different RPMs, but i was just trying to simplify this as much as possible.

                            440whp on pump fuel and the STOCK MAF

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                            • #15
                              enjoyrth wrote: I don't see why this thread continued past the first reply with the "coefficient" theory. To simplify this even more, lets say you have a box that you are pushing and there is friction coefficient of k. If you push it with x newtons, then you will lose kx of the force you applied due to friction. If you push it again with x+25 newtons, you will lose k(x+25) of the pushing force. There is a coefficient similar to this friction coefficient in your drivetrain. You could think of x as your drivetrain loss in the case at hand. I know this disregards Jason's post on different RPMs, but i was just trying to simplify this as much as possible.
                              the conversion from energy into useful work is non-linearly efficient, just like drag doesn't linearly increase either

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