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If gears multiply torque then why does a Dyno read less?

  1. Apr 4, 2015 #1
    If the selected gear ratio and final drive ratio multiply the torque and reduce the speed then why does the dyno read less torque?

    e.g a car with 200 Nm (@5000 RPM) of torque with a 2nd gear selected (Ratio = 2) and final drive ratio 3.5

    200*2*3.5 = 1400 Nm of torque
    5000/2/3.5 = 714.3 RPM

    Why would the dyno read close to 200 Nm then? is it the corrected torque?
  2. jcsd
  3. Apr 4, 2015 #2

    jack action

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  4. Apr 4, 2015 #3

    Randy Beikmann

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    knight92, you didn't mention it, but it sounds like you are measuring your output on a chassis dynamometer driven by your tires. In that case, the dyno only "knows" two things about your car: the traction force exerted on the dyno rolls (in newtons, say), and the speed at the tire patch (in meters/sec, say). But if you type your gear ratios into the controller, and the tire diameter, the software in the controller can back-calculate the torque at the engine (assuming some level of friction in the driveline). That may be what it's doing.
  5. Apr 5, 2015 #4
  6. Apr 5, 2015 #5
    Thats what I was wondering whether the torque shown is back calculated but then not every car's gear ratios are known so how can someone put a random car on the dyno and know what torque the engine is delivering.
  7. Apr 5, 2015 #6

    Randy Beikmann

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    knight92, if you don't know the gear ratios, you can use the first law of thermodynamics (or at least the time-derivative of it) by saying "power at the tire patch equals power from the engine." (Of course the engine power must be more, because of the mechanical losses in-between.) So you can write force X velocity = engine torque X engine angular velocity, that is, Ptire = Pengine. The dyno should either read out force (surface force at the tire patch) and velocity, or power at the tire patch (Ptire). You should also be recording engine speed, so at that point it's just engine torque = Ptire/omega, where omega is the engine's angular velocity.
    To be more proper, you would multiply the engine power by eta, the combined mechanical efficiency of the driveline and tires, which is somewhere around 0.85. Using that, the engine torque would be Ptire/(omega*0.85), increasing the back-calculated engine torque by about 18%.
    Note that without accounting for the mechanical efficiency eta, the engine torque predicted from the power at the tire patch would be less than the engine's actual torque, just as you originally said!
    And you're right, the more general way to do this is from a power-balance standpoint. Then you don't have to worry about gear ratios, or even having a torque converter in the system.
  8. Apr 5, 2015 #7


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    In the unlikely event you can't find ratio data online, Measure the engine rpm (at the dizzy or via ecu) : engine rpm over dyno rpm gives drive train (gear box and diff) ratio.
    You can also calc drive train ratio on the open road from the rpm, speed and wheel radius. Error in your tacho and speedo mean it won't be very accurate through, I'd guess ~5% for an average car.
  9. Apr 5, 2015 #8


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    or simpler still, simply attach a sensing wire / clamp to one of the spark plug wires (this is done at some dyno shops).
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