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Thread: Is drivetrain loss a linear function?

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    Forum User verified Feedback Score 1 (100%) thor'svr4's Avatar
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    Is drivetrain loss a linear function?

    For example:
    in general a stock 1st gen with 300 crank hp lays down about 230 whp.

    So does that mean for every 300 crank hp we loose ~70 hp in the drivetrain? So would a 600hp motor will put down 460 awhp with ~23% DTL, or will a 600 hp motor put down 530 awhp with ~11.7% DTL?

    Obviously we prob cant tell for sure without a ton of engine dyno testing and chassis dyno testing, but I’m just curious. I’ve always heard of ‘percent drivetrain loss’, but does this ‘% DTL’ vary with hp? Or is it a fairly constant ~23% for most awd 3S’s?
    Last edited by thor'svr4; 03-15-2011 at 10:56 AM.
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    it seems to be a % but its hard to say without using an engine dyno. the big power cars lose a lot of power it seems

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    Forum User supporter Feedback Score 1 (100%) Rocket's Avatar
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    I can't say for sure, but have always heard that any AWD drive will be ~25% lossy to the wheels. RWD cars somewhere around 13-17% and FWD around 10%. But I have no numbers to back these up, just what I remember reading over the years.

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    Forum User Feedback Score 1 (100%) lawdogg's Avatar
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    I had a similar thought a while back ... did some probably wrong calculations.

    3000GT/Stealth International Message Center

    Quote Originally Posted by lawdogg
    --------------------------------------------------------------------------------

    Well I did some back-of-the-envelope thinking about drivetrain loss while watching a Scrubs re-run this weekend.

    Unfortunately I can't scan it in so you'll have to bear with me and try to follow along.

    Let's set up the system as a mass-spring-damper 2 node system with an input and output torque.

    To node 1, we have the input torque from the engine (Te) and the output torque through the damper (b) and inertia (J) to ground as well as the output toqrue throug the spring (Tk) to the second node. From the second node, we have the output torque (Tf) which is essentially the power-to-the-ground.

    Te = Tb + Tj + Tk

    Tf = Tk

    Now if we subititute Te-Tb-Tj for Tk, we get Tf = Te - Tb - Tj.

    This makes sense, right? The torque to the wheels is the torque from the engine minus the torque lost though the bearings and inertia. The spring factor disappears because it would only be a factor of the time-response of the system.

    Now if we substitute in for Tb and Tj their equivalent torque loss ratings as a function of the rotaional velocity, we get:

    Tf = Te - b(Omega1g) - J (dOmega1g/dt)

    As you can see, the torque to the wheels equals the torque from the engine minus the friction loss times the rotational velocity of the drivetrain minus the mass of the drivetrain times the acceleration of the drivetrain.

    I'll stop short of the laplace domain to say:

    And it is proven - drivetrain loss is not a constant number. It is a function of the total coefficient friction throught the drivetrain, the total mass of the rotating drivetrain, and the acceleration/velocity of the drivetrain.

    I apologize for my lack of graphics and hope you were able to follow along.



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    One day....it will be on 4 wheels. supporter Feedback Score 2 (100%)
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    woulden't it be as easy as finding two identical cars with two separate engines that have been on engine dyno's and seeing what the final HP figures are? before and after for each car? would be conclusive in my eye's.
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    LW fears my posts Not Verified Feedback Score 1 (100%)
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    It is definitely NOT a fixed percentage. As was stated above, there are a lot of factors that contribute to the total drivetrain loss. Road and Track did an article on this long ago, the tires are considered "static" friction (since you obviously can't change tires while driving down the road, and their width and air pressure is basically constant while driving, so they are given a fixed value, a certain amount of horsepower is needed just to move the car against the friction of the road bringing it to a halt.

    The transmission itself however, will vary in the amount of drivetrain loss based on what gear you are in, what rpm you are at, and heat build up. This number could equal 30-60hp depending on conditions. Frictional losses in gear are higher with higher speeds, and lower with lower speeds.

    Other things are also a fixed measurement. Switching to a CFDS and lighter wheels will enable you to spin the tires faster, showing an increase on the dyno. That is a fixed amount of horsepower you will gain, not a variable one. Your total drivetrain losses are variable, within a fixed range. You'll never get your losses down below a certain horsepower number, and you'll never suddenly increase your drivetrain loss simply because you bolted on a new set of turbos and are now making more power. (However, a higher horsepower car may see some increase in total drivetrain loss due to the increased friction in the tranny/xfer case resulting from the elevated acceleration rate.)

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    Member verified Feedback Score 1 (100%) B-Man's Avatar
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    I use that site all the time when this comes up.
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    One day....it will be on 4 wheels. supporter Feedback Score 2 (100%)
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    fwiw I'm not saying it's fixed...just saying engine in and engine out will tell you exactly how much your loosing.

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