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Vehicle dynamics

  1. Oct 14, 2008 #1
    A vehicle's engine produces 100 hp at 4773 rpm. If the wheel radius is 12 inches, the final drive ratio is 3 and the transmission ratio is 3.00, calculate the tractive force at the wheel, in steady state.
  2. jcsd
  3. Oct 14, 2008 #2
    1. Convert engine power and rpm to engine torque:
    hp * [33000 (lb*ft/min)/hp] * [1 min / engine revolutions] * [1 rev / 2 (pi) rad] = engine torque

    2. Multiply engine torque by transmission ratio and final drive ratio to get axle torque
    engine torque * trans ratio * final drive ratio = axle torque

    3. Divide axle torque by wheel radius to get tractive force
    axle torque / wheel radius = tractive force

    I hope I wrote the formulae clearly enough. Pay close attention to unit conversions. If you're not sure where I got the conversion factors, please let me know.
  4. Oct 15, 2008 #3

    thanks, but could u please explain just the 1st step again?
  5. Oct 15, 2008 #4
    A vehicle's engine rotates at 5640 rpm and produces 62 hp. If the final drive ratio is 4.00 and the transmission gear ratio is 3.00, calculate the vehicle speed, in mph. The tire radius is 10.1 inches. Vehicle is at constant speed.
  6. Oct 15, 2008 #5
    The first step is basically just a conversion from horsepower to torque. RPM provides a time unit for the conversion.

    (1 horsepower) = 33000 lb*ft/min (this is just unit conversion)

    To convert rotating power to torque, you need to multiply by time. Since we don't have a direct time unit, we can divide by RPM (rev/min). This gives us a unit of lb*ft/revolution.

    We then need to adjust from revolutions to radians, so we divide by 2 * pi radians/revolution.

    If you've ever seen the formula "torque = hp * 5252/engine rpm", that is just a simplification of this step (33000 / 2*pi = 5252).

    Hope that clears things up.
  7. Oct 15, 2008 #6
    This one is actually a bit simpler, just filter out the unnecessary information.

    We assume the engine's rotation is directly coupled to the wheel rotation (if one turns, the other turns - no slipping). Since you have the engine speed, you do not need the power - you just need to know how many times the wheel will turn per engine revolution.

    Engine RPM * Transmission ratio * final drive ratio = wheel rpm

    After that, it's just a matter of multiplying the wheel rpm by the wheel circumference to get the distance traveled per minute. Convert the units to mph and you will have the answer.
  8. Oct 15, 2008 #7
    when i multiply wheel rpm by wheel circumference i will get a value in terms of rpm*inches.
    How do i convert that into mph?
  9. Oct 15, 2008 #8
    revolutions/minute * (pi*2*radius) inches / revolution = inches/minute

    inches/minute * 1 mile / 63360 inches * 60 minutes / 1 hour = miles/hour
  10. Oct 15, 2008 #9
    A vehicle with a weight of 3000 lbf is accelerating at .7 gs in first gear. The vehicle shifts from first to second at 1721 rpm (speed before shift). If the final drive ratio is 3.56, first gear transmission ratio is 3, and second gear transmission ratio is 2.4, calculate the engine speed in rpm after the shift.
  11. Oct 15, 2008 #10
    A lot of unnecessary information in this one. If you assume the same vehicle speed immediately before and immediately after the shift, you will have the same rotational speed on the output side of the transmission before and after the shift.
  12. Oct 16, 2008 #11
    A 6 speed standard transmission has a 6th gear ratio of 1:1. if the first gear ratio is 4.86:1, calculate the 4th gear ratio, based on a geometric progression. Answer must be expresses as a single number, two decimal places.
  13. Oct 16, 2008 #12
    This one is a math problem. If you're familiar with geometric progression, it's fairly straightforward. Basically, there is a common factor that can be multiplied by each progressive gear ratio that, when starting with 4.86, will result in 1.00 in the 5th step (6th gear).

    x*4.86 = 2nd gear
    x*2nd gear = 3rd gear
    x*5th gear = 6th gear = 1.00

    Solve for x, the 3rd iteration (4th gear) will be the answer.

    I'm not sure the best mathematical approach for this, but there are few enough steps that trial-and-error is a reasonably practical way to get the answer.
  14. Oct 21, 2008 #13
    A vehicle has a vertical cg height of 20.5 inches, a wheelbase of 108.5 inches, a wheel radius of 12.11 inches, and a total weight of 4074.0 lbf. 57.7 percent of the weight is on the front wheels and both the front and rear brakes have the same gain of 20.0 in-lbf/psi. Assuming the front and rear brakes both receive the same applied pressure, and that there are two brakes per axle, calculate the pressure in psi required to achieve a rear braking coefficient of 0.6.
  15. Oct 21, 2008 #14
    Ques: A disk brake consists of a 5.3 inch radius rotor and a 2 piston caliper, with each piston having a diameter of 1.1 inches. If the brake pad has a surface area of .5 square inches and a friction coefficient of .37, calculate the brake gain, in inch-lbf/psi. Note: the definition of a two piston caliper is two pistons per each side. Create a free body diagram of the caliper if necessary.
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