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Ideal 1/4 mile time from Average Power

  1. Apr 1, 2014 #1
    I'm wanting to get the time it would take a vehicle to travel a 1/4 mile coming out of a corner in the ideal world. By ideal I mean that I'm assuming there is no energy loss to friction, wheel slippage, drag, or heat. I want to avoid using gear ratios, and since it is an ideal world I'm assuming that top speed is not an issue, and wheel power is the same as engine power. This is the information I know.

    Peak Power: 140 HP
    Power Peak RPM: 6400 RPM
    Peak Torque: 130 pound-foot
    Torque Peak RPM: 3800 RPM
    Average Engine Power: 137.8 HP
    Average Engine Torque: 115.8 pound-foot
    Average Engine RPM's: 6250 RPM
    Weight: 2161 lbs.
    Drive: RWD
    Weight percentage carried on rear axel: 46%
    Full wheel diameter: 22.7 inches
    Exit Speed (wheels straight): 44 MPH

    Is it possible to find the maximum ideal time this quarter mile would be run in without calculating out the power curves on each gear?
     
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  3. Apr 1, 2014 #2

    Simon Bridge

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    Welcome to PF;
    It takes no power to move at a constant speed in the ideal world you describe.
     
  4. Apr 1, 2014 #3

    SteamKing

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    All of this information is superfluous except for the last line: exit speed: 44 mph, which I am assuming means exit from the turn.

    In order to travel 1/4 mile, 44 mi/hr * 5280 ft/mi / 3600 s/hr = 64.53 ft/s

    1/4 mile = 1320 ft, so time t = 1320 / 64.53 = 20.45 s

    Of course, it's not clear what you mean by 'maximum ideal time' to run a quarter mile. Most people try to find the minimum time it takes to cover a set distance.
     
  5. Apr 1, 2014 #4
    equation of motion for a body under uniform acceleration:
    ut + 1/2 a tt - s = 0

    the solution that you're interested in:
    t = u/a - sqrt(uu + 2as)/a

    where:
    a = T/mr

    u = initial speed
    s = distance
    T = average torque
    r = wheel radius
    m = mass

    Note that this isn't quite what you asked for in the title, since there isn't a constant power output to achieve a uniform acceleration. The equation for motion for constant power is a little more involved.
     
    Last edited: Apr 1, 2014
  6. Apr 1, 2014 #5

    pervect

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    I assume that the driver "floors it", transfering 140 horsepower to the wheels, as soon as the car's wheels straighten out? And we assume all the energy goes into linear motion of the car (ignoring the small amount that would get stored in the wheels by virtue of their moment of inertia)?
     
  7. Apr 1, 2014 #6
    None of this is quite what I'm looking for, so I think I'll start with the direction I'm trying to go in mathematically. I'm making this as a perfect world problem, because I want to start with the basic equation before I start adding in power loss. I know that my starting point is:
    A(t) = Power(t) / ( Mass * V(t))

    I'm hoping it is legitimate to sub in the average power rather than trying to come up with the exact power curves on each gear. If that is the case, I would assume my equations is.

    A(t) = (Avg. Power) / (Mass * V(t))

    Now I assume my first step is to take the integral of A(t), but my calculus is rusty, so I get confused about what to do with (Avg. Power) / (Mass) LN(V(T)) + C from 0 to an unknown time. I begin to think that maybe the answer is to figure acceleration in respect to distance, since I know we are going from 0 to 1320 ft, but then I get lost on how to figure out V(d). Am I going in a completely wrong direction? Is it completely illegitimate to substitute avg. power for P(t)?
     
  8. Apr 1, 2014 #7
    If you can't remember the maths then just do it numerically, in fact I'd recommend that way if you plan to add in later things like torque curves, transmission losses, gear change delays and air drag.

    Work in small time steps of 0.1 seconds or less, start at t=0. At t=0 you know the velocity (44 mph) and the distance (zero), so you can calculate the acceleration using the equation you quoted. Using that acceleration you can then work out what the speed will be 0.1 seconds later, and how much distance you covered. At the next time step you can then calculate the new acceleration based on your updated velocity, repeat until your distance counter gets to 1/4 mile. It's pretty easy to set this up in Excel.
     
  9. Apr 1, 2014 #8
    If you're happy to use the constant power assumption then that subsitution is exactly what you're looking for. All you need to do is solve the differential equation and get time as a function of distance, in a similar way to which I showed for the constant acceleration case, and you're done.

    The reality is that engines exhibit power curves under different loads and rotation rates, but I think you want to ignore that for now.
     
  10. Apr 1, 2014 #9
    [tex]a = \frac{c}{v},[/tex] where c is a constant. [tex]\frac{d a}{d t} = - \frac{c a}{v^2} = - \frac {a^3}{c}[/tex] [tex]\frac{d a}{a^3} = - \frac {dt}{c}.[/tex] Integrating both sides [tex]\frac{- 1/2}{a^2} = - \frac {t}{c} + C_1.[/tex] from the original equation we get
    [tex]v^2 = \frac {c^2}{a^2} = 2 c^2 [\frac {t}{c} - C_1] .[/tex] Using the initial condition [itex]v = 0[/itex] at [itex]t = 0[/itex], we get [itex]C_1 = 0[/itex], and
    [tex]v^2 = \frac {c^2}{a^2} = 2 c^2 \frac {t}{c} = 2ct.[/tex]

    Can you take it from here?

    (PS: note that it is possible to get to the same result by use of the work-energy theorem W=ΔK)
     
  11. Apr 1, 2014 #10
    That last bit from Dauto is exactly what I was looking for. I think I'll play around with the work-energy theory on my own to see what I get there. Power is work over time, so I think I see where you are going with that. Thanks.
     
  12. Apr 1, 2014 #11
    Ok, I did several things wrong. I'll make a new post with the corrected math.

    This is perfect. The application of this is that I have three cars with similar engine compartments, so they have similar possibilities when it comes to power. They don't have similar losses in acceleration due to the many things that affect acceleration. I can observe how fast they are exiting the corner in question, and then I can easily get a ratio between observed time and ideal time with this equation without having to give the cars similar power. That way I can have a rough measuring stick to choose the car I want to put time and money into without first bringing them to similar power capabilities or raw launch capabilities in a drag test.
     
    Last edited: Apr 1, 2014
  13. Apr 1, 2014 #12
    The starting equation should be v2 = 2ct + v02 where c is power/mass, not power/weight.
     
  14. Apr 1, 2014 #13
    Something is still wrong. with that last change, I end up with negative time. I'll have to check C1 again when I get another chance in a couple hours.
     
  15. Apr 1, 2014 #14
    That should do it. C1 needed to be -v02/2c2. When you plug it back into the equation the 2c2 cancels out. Now I just have to be careful to remember pound force in power assumes 32.174 ft/s2 over it's 550 pound feet per second. That's the biggest problem with pounds is remembering if you are looking at force or mass. Am I correct in thinking a perfect system could convert that average power to linear motion, or should I still be converting this power to torque on the wheel of the diameter I gave?
     
  16. Apr 1, 2014 #15
    All those different torque and power measurements are given because the engine isn't ideal. They're related to the power curves.

    If you did try to convert your power to a torque, you'd need an angular speed, which you'd derive from the velocity at the wheel, which would change nothing.

    It's worth trying to find out what the average engine power specification actually means. I would imagine that it would be the power output averaged over time, under maximum acceleration with optimal gear changes, upto a certain speed. That would be the case where it would be most relevant to your calculation, but since the figure is so close to the peak power measurement, then it seems unlikely. Car engines deliver much less power at lower revs, which is inevitable after shifting up a gear. It also seems very odd that the average engine RPM is so close to the peak power RPM.
     
    Last edited: Apr 1, 2014
  17. Apr 1, 2014 #16
    I can always work on low gear ratios later to improve low rev power delivery. At the moment, my key concern is power loss at the speed and revs the car is normally going to be moving through. This is all a rough rationalization fro a starting point. I'm just trying to kick out the types of things I can easily change later. I think this equation is what I want, I'll just have to go over my work a couple times, I have a feeling something is still off based on the units.
     
  18. Apr 1, 2014 #17
    Looking at a typical car engine it seems like the power would only drop down by 10-15% after shifting up a gear, so the *average* power whilst accelerating through the gears up to the red-line is going to be roughly 5-8% down on the peak power. Assuming that this constant average power accelerates the car directly is OK for a first approximation, there is no need to convert to torque at the wheels as you'll just get the same result (the gear ratios and tyre radius will multiply up the engine torque and divide down the speed, the power is the same). Especially as you're starting from 44mph I don't think you need to worry about low-RPM power, the driver will presumably be in the lowest gear possible at that point anyway.
     
  19. Apr 1, 2014 #18
    Ok, let me try this again. I want to be able to enter the appropriate V0 every run, so I need to work that into the equation.

    v2 = 2c2 [t/c −C1]

    With t = 0, I should have:

    v2 = -2c2C1

    Meaning:

    C1 = v2/-2c2

    Plugging that back into the equation gives us:

    v2 = 2c2 [t/c − v2/-2c2] = 2tc + v2

    Now I have:

    (dx/dt)2 = 2ct + v02

    dx/dt = (2ct + v02)1/2

    dx = (2ct + v02)1/2dt

    Integrate both sides:

    x = 1/3c (2ct + v02)3/2+C2

    When x = 0 and t = 0 we have:

    0 = 1/3c(v02)3/2+C2

    C2 = -v03/3c

    Which plugged back in gives us:

    x = 1/3c (2ct + v02)3/2-v03/3c

    Solving for when x = 1320 feet gives us:

    1320 = 1/3c (2ct + v02)3/2-v03/3c

    [3c(1320 + v03/3c)]2/3 = (2ct + v02)

    ([3960*(Power / Mass) + v03]2/3 - v02)/(2 * Power / Mass) = t

    A quick sanity check of the units seems to suggest I might finally have this:

    [3960*(Power / Mass) evaluates out to ft3/s3

    v03 evaluates out to ft3/s3

    The sum of those two should evaluate out to ft2/s2 after taking the cube root then squaring.

    We can then subtract v02, which is also in ft2/s2.

    (2 * Power / Mass) should evaluate out to ft2/s3. If you divide that with what we got earlier you get just plain seconds, which is what we are looking for.
     
    Last edited: Apr 1, 2014
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