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Finding velocity/time graph for a car?

  1. May 11, 2005 #1
    I have been recently playing around with some figures in an attempt to find a velocity versus time graph under ideal conditions including air resistance and rolling resistance for a car that I am in the process of fixing. What better incentive to get it running than knowing (at least a rough estimate) of how fast it will be? However, I have run into some problems.

    I have Approximated a Power versus velocity graph based on a graph that I got from finding a cubic regression from a dynamometer chart of the engine in the car. I then altered tha graph so that instead of having power versus rpm of the flywheel to power versus velocity given the gear ratios and radius of the tire.
    v=(2pi*x/60)/(Ratio of gear * Ratio of Differential)*Radius of Tire
    where v is velocity and x is the RPM of the engine.

    Once this has been found, I determined that the derivative of P(v) would equal F(v) due to the engine. I also looked up the formula for F(v) of air resistance to be (1/2)*Coefficient of drag*Frontal area*Air density*velocity^2. Rolling resistance's would be Coefficient of rolling friction*Normal.

    From this I get a differential equation that m(dv/dt)=Fengine(v)-Fair(v)-Ffriction(v).

    However, upon solving this, I get a result that does not make sense at all. Could someone verify that I am doing this correctly, and if not, give me some instructions on how to do this the correct way? Any and all help is appreciated.
  2. jcsd
  3. May 11, 2005 #2


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    The force given by engine is not the derivative dP/dv. It is P/v.
  4. May 12, 2005 #3
    But I thought that the integral of Force versus Velocity gives you Power?

    Can you explain why the force of an engine would be power/velocity?
  5. May 12, 2005 #4
    [tex] P = \frac{dW}{dt} [/tex]

    [tex]F = \frac{dW}{dx} [/tex] so [tex] dW = F dx [/tex]


    [tex] P = \frac{F dx}{dt} [/tex]

    If F is assumed to be constant it simplifies to the very nice

    [tex] P = F \frac{dx}{dt} = Fv [/tex]
  6. May 12, 2005 #5
    The thing about this is that the force of my engine is not constant. So I need a way to allow for this in my equations
  7. May 12, 2005 #6
    If the force is varying with time, just differentiate the second last expression, if it is varying with some other variable, then use the chain rule to find the net rate of change WRT to time.
  8. May 12, 2005 #7


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    That's all true except the last sentence. F needn't be constant.
    [tex] P = \frac{F dx}{dt} = F \frac{dx}{dt} = Fv [/tex]
  9. May 12, 2005 #8
    Thank you guys, I will try and play around with that formula. I was playing around in Derive (very nice once you get the hang of it) and I got a velocity curve that made no sense at all. After thinking about that I remember reviewing that Power = Force * Average Velocity in my Physics C class.
  10. May 12, 2005 #9


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  11. May 12, 2005 #10
    I am getting a force of 19k Newtons. This seems unreasonably high to be pulling about 1.4 G's in a 3000 pound car with only 160 hp to the wheels. What would cause this?
  12. May 12, 2005 #11


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    160hp can get you this force, but at only 14 mph.
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