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Challenging engineering problem

Discussion in 'After Hours Lounge (Off Topic)' started by Michael Varacin, Sep 30, 2003.

  1. Michael Varacin

    Michael Varacin Stunt Coordinator

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    All,

    I'm trying to estimate (for my own personal benefit) gains that can be had by shaving two pounds off the crank shaft in the engine used in my race car.

    To simplify the problem, I am estimating the weight savings to be on the outer edge of the crankshaft. I also know from on board data that it takes approximatly 4.5 seconds to go from 5500 to 6000 RPM in top gear.

    So I am trying to calculate the horespower required to accelerate a ring with a WEIGHT of 2lbf, 3" OD and 2.5" ID, from 5500 to 6000 RPM in 4.5 seconds.

    I keep coming up with an answer of .011 Horsepower, but that just seems too small.

    Anyone else want to try?

    My method:

    I calculated the rotational energy of the ring at both 5500 and 6000 rpm. I took that difference in enegry over the 4.5 seconds to get the work needed. Convert to HP's.
     
  2. Greg_R

    Greg_R Screenwriter

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    It takes you 4.5 seconds (measured) to go from 5500 to 6000 with a load on the crankshaft (drivetrain + car, etc.). It looks like you have used just the load/weight of the crankshaft in your calculation. Also, you have neglected friction (not sure how significant it is in this calculation).

    In order to get the correct number, you will need to calculate the energy expended by the vehicle as it accelerates from 5500 to 6000 (and then work backward to determine the energy difference at the crank). Although I'm not an ME, it sounds like an integral (Calculus) problem (area under the curve). Enjoy...
     
  3. Garrett Lundy

    Garrett Lundy Producer

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    Don;t forget to multiply by the number of combustion cylinders[​IMG]
     
  4. BrianW

    BrianW Cinematographer

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  5. Michael Varacin

    Michael Varacin Stunt Coordinator

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    Brian,

    You're right...I am trying to just find the difference in the engine performance. The fact that the car took 4.5 seconds is in some ways trivial. It seems like a simple problem...but somewhere I must be screwing up my calculations...or the improvment is actually small. The .011 HP answer I keep getting just doesn't seem right. I'm going to try solving it differently. I'll post a different answer if I come up with one.

    To add to the confusion, using the same method, I calculated the HP required to accelerate the 1025 lb car 10 mph in 4.5 seconds and got about 13 HP. Which seems right, neglecting friction and air drag. So if anything, I must be off in the inertia calculations.
     
  6. Artur Meinild

    Artur Meinild Screenwriter

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    I think I have the solution to this - somebody please correct me, I was never *that* good at mechanics! [​IMG]

    First you have to find the angular velocity (w) - and you already have that since you have the rpm before and after.

    Then you have to use the shafts Moment of Inertia (I) which is M*r^2 for a circular ring.

    My suggestion would then be to calculate the kinetic energy of the shaft at 5500 rpm and at 6000 rpm by using the formula: Ekin = 1/2*I*w^2

    Then you will end up with the energy difference in Joules. If you divide this amount of Joules with 4.5 seconds you will get the desired effect in Watts (J/s).

    I'm 100% certain that you have to use the Moment of Inertia, you always do when you have rotation, but I'm not quite sure about the rest.
     
  7. Artur Meinild

    Artur Meinild Screenwriter

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    ok, and here are the calculations: (why can't I edit my post???)

    5500 rpm = 575,9 rad/s
    6000 rpm = 628,3 rad/s

    I = (0,9 kg)*(0,0381 m)^2 = 0,00131 kg*m^2

    1 kW = 1,34 HP

    Ekin(before) = 0,5*(0,00131 kg*m^2)*(575,9 rad/s)^2 = 217,23 J

    Ekin(after) = 0,5*(0,00131 kg*m^2)*(628,3 rad/s)^2 = 258,57 J

    Ekin(delta) = 41,34 J ~

    P = (41,34 J)/(4,5 s) = 9,19 W = 0,00919 kW = 0,012 HP

    The same result as originally posted and using the same method. So at least I think we can conclude that the calculations are correct.
     
  8. Michael Varacin

    Michael Varacin Stunt Coordinator

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    Artur,

    You and I did arrive at the same answer.

    Thanks!


    I am still going to think about this a little bit. Only a .01 HP change doesn't seem right. But perhaps it is. If that is correct, then it's safe to assume the making the crankshaft lighter will have no real advantage. (Other then ignored affects like reduced drag in the oil...ect.)

    Perhaps I will try it and test on our dyno to verify results.


    Mike.
     
  9. Edwin-S

    Edwin-S Producer
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    I'm curious is that 4.5 second figure with the engine mounted in a test stand and no drive train attached? Wouldn't inertia in the transmission and drive train have an effect on the time taken to go from 5500 to 6000 rpm?
     
  10. Artur Meinild

    Artur Meinild Screenwriter

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    In a way, I think it makes perfect sense. If you got a much higher answer (like 13 HP) then that would be the amount of energy required to rotate the shaft alone, and that absolutely doesn't make any sense at all! [​IMG]
    It's reasonably to assume that only a very small fraction of the total effect goes to moving and rotating the motor parts, while the biggest fraction goes to moving the car itself, and overcome what friction there might be underway...
     

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