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Effect of GCW on vehicle speed

  1. Nov 13, 2015 #1
    Hi all.

    I am currently working on calculating vehicle performance. I have found a formula to calculate max theoratical velocity that is:

    V= engine RPM x Wheel circumference / transmission Ratio x Rear Axle Gear Ratio

    but I could not be able to calculate my actual speed and its effect on increasing my GCW.

    What I am know are

    Max Torque at given RPM
    Power at given RPM
    Gear Ratios
    Rear Axle Ratio
    Wheel Diameter
    Load on Rear Axle

    Is there any solution to my situation?
  2. jcsd
  3. Nov 13, 2015 #2

    Ranger Mike

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    this should help

    Attached Files:

  4. Nov 13, 2015 #3

    jack action

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    Vehicle weight usually have very little effect on maximum speed. But it does have one on acceleration (which dictates how long it takes to reach maximum speed).

    There are two things that can limit maximum speed: Traction and power.


    The vehicle's tires must transfer enough traction force to sustain aerodynamic drag and rolling resistance. In a simplified version, the maximum speed related to traction is:
    [tex]v_{max} = \sqrt{\frac{(\mu - f_r)mg}{0.5\rho C_d A}}[/tex]

    [itex]\mu[/itex] is the tire-ground friction coefficient;
    [itex]f_r[/itex] is the rolling resistance coefficient;
    [itex]mg[/itex] is the weight of the vehicle;
    [itex]\rho[/itex] is the air density;
    [itex]C_d[/itex] is the aerodynamic drag coefficient of the vehicle;
    [itex]A[/itex] is the frontal area of the vehicle.

    This is for an All-Wheel-Drive. For a RWD or FWD, it will be less. More info here.


    It is one thing to have the tires being able to transfer the necessary force to the ground, but it is the engine that has to produce it. With the use of a gear ratio, any force can be created at the wheel. But the power at the wheel must be the same as the power required to fight aerodynamic drag and rolling resistance at the vehicle speed. In a simplified version, the maximum speed related to wheel power is:
    [tex]v_{max} = \sqrt[3]{\frac{P_w}{0.5\rho C_d A}}{}[/tex]
    Where [itex]P_w[/itex] is the available power at the wheels. Usually, it is about 85% of the engine power, due to losses in transmission components.

    Most of the time, the engine power is the limiting factor for maximum speed.

    More detailed info here.
  5. Nov 15, 2015 #4

    What I have got from your reply is that

    the vehicle speed will be either limited by traction or engine power.

    In my case it seems that vehicle speed (calculated by traction formula) is a limiting value as it is lower than vehicle speed calculated by power.

    Is it true..??

    and if it is true than always traction will be a limiting factor???
  6. Nov 16, 2015 #5

    jack action

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    That is rarely the case, unless you have slippery conditions like snow for example. The equations must be used with SI units. It might not be clear, but the power equation is a cubic root, not a square root.

    A typical car (say m = 1500 kg, Cd = 0.30, A = 1.5 m², ρ = 1.23 kg/m³) on asphalt (µ = 0.9, fr = 0.013) will give:
    [tex]v_{max} = \sqrt{\frac{(0.9 - 0.013)(1500)(9.81)}{0.5 (1.23) (0.3) (1.5)}}[/tex]
    [tex]v_{max} = 217 m/s = 782 km/h[/tex]

    From the second equation, to reach that speed you need an engine producing 827 918 W (= 217³ * 0.5 * 1.23 * 0.3 * 1.5) which is 1100 hp !

    Even at those speeds (> 400 km/h), these equations might be too simple to get appropriate values.

    If you put a more common engine power like 200 hp (149 200 W), the maximum speed of a typical car would be:
    [tex]v_{max} = \sqrt[3]{\frac{149 200}{0.5 (1.23) (0.30) (1.5)}}[/tex]
    [tex]v_{max} = 81 m/s = 293 km/h[/tex]
  7. Nov 16, 2015 #6
    from this equation it can be seen that greater the mass greater will be the maximum velocity..

    Is it practically true..??

    It is clear to me that for greater mass acceleration will be lower.

    also in the mentioned equation, Should I take mass as whole vehicle mass or mass on driver axle??
  8. Nov 17, 2015 #7

    jack action

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    Yes, a greater mass will give a larger friction force, which can fight a greater aerodynamic force, resulting in a greater speed. But you still need the power to reach that speed.

    In the real world - when designing for high speed and acceleration - we introduce aerodynamic downforce to push down on the car such that we don't rely as much on the weight to create the friction force. The good thing about aerodynamic downforce is that it increases with the speed squared just like the aerodynamic resistance. So, in theory, it is possible to design a vehicle that has no speed limit.

    In the link I gave you earlier (see below), you have a more precise equation that take into account the layout of the vehicle and even the aerodynamic downforce. When you are considering RWD or FWD, you cannot just take the static weight on the driven axle. The aerodynamic resistance will create a moment that will transfer the weight from the front axle to the rear axle.

  9. Nov 25, 2015 #8
    I have seen a truck specification, they mentioned a 265 PS engine (261 hp almost)
    and max speed 93 km/hr.

    Calculating with above given formula and keeping all other factors same as you have given, gives 163 km/hr almost twice as of specification.

    Is there anything wrong with it..??
    or should Ihave to consider some other factors that affect the speed??
  10. Nov 25, 2015 #9

    jack action

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    The equation you gave in your OP is still valid:
    If you replace engine RPM with the maximum RPM of the engine and transmission Ratio with the highest gear value (i.e. lowest number), you will get a theoretical maximum speed. Whether you can reach that speed or not will depend if you have enough power at that engine RPM (and if you have also enough traction, of course).

    Often, especially with working trucks, reaching the potential top speed is not a concern for the vehicle manufacturer and it designs the drivetrain to reach its desired top speed only. The vehicle might still need high power to produce the desired accelerations at lower speeds. Working trucks are usually very heavy and need a lot of power to produce acceptable acceleration.
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