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Inertia Dynomometer equations

  1. Mar 3, 2013 #1
    I'm trying to figure where I've gone wrong in my calcs. the physics of the inertia dyno are simple enough however I can't seem to get the proper results.

    [tex]\alpha=\frac{d\omega }{dt}[/tex]
    [tex]\tau =I*\alpha[/tex]

    these are some of the equations I'm using, the torque should be in Nm when [tex]MOI=kg/m^2[/tex]

    the information I'm using comes from data sets off the ECU. I'll explain a bit, a test development engine has been run on the dyno using a chassis, the numbers are good. what I was asked is if it's possible to back out and use the ECU information to output similar numbers to the dyno, first blush I figure sure it shouldn't be that difficult, famous last words, right?

    I have the rpm vs time data in milliseconds, specifically the rpm is counted every 20 milliseconds, I have the dyno numbers for each recorded run so I can compare results, i.e. the data set for one pull has a recorded ECU run for all sensors, rpm,time, temp etc...

    I have the gear ratios for all gears, wheel dia etc..I have put together a spread sheet with the input rpm to the gear box, output for each gear in rpm and mph to the drum roller dynomometer, drum dyno rpm and mph, and delta between them. the rpm on the dyno drum is converted to radians per second.
    step 1:
    engine rpm to drum rpm (calculated via gear ratios)
    step 2:
    drum rpm to radians [tex]\omega /sec[/tex]
    step 3:
    [tex]\alpha =(\omega _{1}-\omega _{2})/(t_{1}-t{2})[/tex]
    step 4:
    [tex]\tau Nm=I*\alpha[/tex]

    the numbers are wildly high, the curve looks about right, peaks near the dyno's output. however there are massive spikes due to the time count where the rpm is recorded in whole numbers to the millisecond count so the time moves forward but the rpm holds a bit, on avg it's 60~80 milliseconds per revolution count then it jumps a fair amount in the next step. resolution could be better but it's all I've got to work with.

    I'm assuming at this point I've calc'd wrong somewhere but can't seem to figure where. I have the drum mass and radius data for a correct MOI. using the gear reduction I have the drum rpm to engine rpm and assume that should be accurate enough neglecting slip and friction loss.

    the reason I'm going to this trouble is the inertia drum mass exceeds the chassis weight by a factor of 2.2, the loads are very high for the final working loads the engine will be under. If I can re-calc the dyno run then I should be able to factor the delta in the dyno's inertia to the chassis inertia and approx the acceleration of the chassis in the real world by using this equation:
    [tex]\I =[\frac{Dm-Cm}{Cm}*Dm]+Dm[/tex]
    Dm=drum mass
    Cm=Chassis mass

    Hopefully someone can point out my error, Thanks..
  2. jcsd
  3. Mar 3, 2013 #2


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    Moment of inertia has units of kg*m^2
  4. Mar 3, 2013 #3
    As mentioned above, you have the wrong units for your angular moi.

    You can derive the drum's moi from your dyno data since you know the output power and how fast the rpms (angular velocity) is changing. But you need to make sure that there are not any other braking forces being applied to the dyno drum during the run which would make your calculations essentially useless.

    One other thing to think about: If you know what your power curve looks like from the dyno data, it's fairly easy to apply the data in a simulated run. (I'm assuming your doing this for as part of a racing program.) Remember that power is how fast energy can be delivered to a system and that E = 1/2 m V^2. Just total up the energy delivered and solve for V. (You'll need to add a factor to account for aerodynamic drag.)
  5. Mar 3, 2013 #4


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    You also have another possible problem with the formula for mass moment of inertia of the dynamometer drum.

    Assuming the drum is a hollow cylinder, the MOI is given below:

    MOI = (m/2)*(Router^2 + Rinner^2)

    It's counterintuitive, but accurate

    Attached Files:

  6. Mar 3, 2013 #5
    syntax error on my post, the slash shouldn't be there. the drum weighs in at 2950kg, the MOI I've calculated is 1008.71 kg m^2, it's not a solid cylinder or it would be 548.128kg m^2

    The runs where done on an inertia drum dyno, no eddy or brake drag at any point, other then frictional drag of the bearings and the rotating engine assy, gearbox etc...

    the runs are on avg 7~7.5 secs long, a bit shorter then the 10sec avg. if I avg the angular acceleration over 7.3 sec for example, then I get 7,799 Nm.

    let me walk thru a step,
    engine rpm to drum rpm:
    drum rpm to radians per second:
    the .02 being the 20 milliseconds between each point

    I'm assuming I've goofed a step somewhere...
  7. Mar 3, 2013 #6
    that's the same MOI I posted isn't ?, the drum is a hollow cylinder. the first equation in my post I=1/2*m*(r_{1}^{2}-r_{2}^{2}
  8. Mar 3, 2013 #7


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    Look carefully at your formula for MOI versus what is in my attachment:

    Your formula has the difference of the radii squared. For a hollow cylinder, the MOI is proportional to the sum of the radii squared.
  9. Mar 3, 2013 #8
    thanks, I need to double check my posts, fortunately the equation in my spread sheet is the sum and not the difference.
  10. Mar 3, 2013 #9
    the only way I've been able to get similar results to dynomometers output is bytaking the Nm of torque dividing it by 10, seems to be high by that factor then summing time point t1+t2 for t2, then t2+t3 for t3 etc.. it's a bit jittery more then likely due to the resolution on the rpm counts.

    I did a calculation est on the overall run time to mass of the drums as the chassis mass, using the delta in time points to velocity I could plot the Tq points. that matched the dyno numbers.

    the only thing off with the first calcs seems to be with the Nm tq force. possibly that the MOI is kgf which is not equal to Nm 1:1 ? though MOI of 1kg m^2 = 1 Nm
  11. Mar 4, 2013 #10


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    What are the details of the dyno roller? (mass, dimensions, etc.)
  12. Mar 4, 2013 #11
    A Nm can be a measure of work or it can be a torque depending on what you're doing. It's -not- a measure of moi.

    One kgf will be equal 9.8N.
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