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Calculate the acceleration of a vehicle

  1. Mar 23, 2012 #1
    Hi all,

    I am trying to create a vehicle simulation and get the vehicle to accelerate realistically by calculating the wheel speed in rad/sec. I know the torques and masses of a real world vehicle and I have video showing how fast it accelerates, so i want to recreate that. I have independently calculated air drag and gravitational forces, and I am calculating the force applied by:

    force = wheelTorque/wheelRadius
    force = force - dragForce + gravitationalForce (this is signed appropriately)
    acceleration = force/mass of the vehicle * deltaTime
    angularAcceleration = acceleration / wheelRadius;

    The problem is that I am accelerating much too quickly (moving through the gears twice as fast as i should), and I think something big is missing. I decided that I wanted to try incorporating the moment of inertia for the wheels, but I'm not quite sure how to do this. I can calculate the moment of inertia just fine, but since I am converting everything to linear forces, I don't really know what to do with it.

    Does this sound like the missing piece? The vehicle seems like its accelerating much too quickly in the low gears, but in the higher gears, it seems closer to what it should be. Any thoughts??
  2. jcsd
  3. Mar 23, 2012 #2
    The gravitational force is vertical, and doesn't enter into the equation for a horizontal accelerating force.

    The total energy of the car at velocity v (ignoring rotating inertia of wheels) is
    [tex] E=\frac{1}{2}mv^{2} [/tex]
    The time derivative is accelerating power (not including drag force losses):
    [tex] P=\frac{\mathrm{d}E}{\mathrm{d} t} = mv\frac{\mathrm{d}v}{\mathrm{d}t } = mva [/tex]
    The input power is
    [tex] P_{wheels} = torque \space \cdot \space wheel \space angular \space speed)= T\frac{\mathrm{d}\theta}{\mathrm{d}t } [/tex]
    where T is Newton-meters and dθ/dt is radians per second
  4. Mar 23, 2012 #3
    Hi Bob, thanks for the reply. I am calculating the force of gravity based slopes as i must simulate hills, etc. That all seems to be working well.

    I understand the first equation, may I assume energy is in newtons?

    I am completely confused about what the second equation is

    I am also confused about the third equation. Am I to take the current rad/sec value times the torque to get Pwheels. What is Pwheels? I assume power at the wheels, but what is it exactly? Is it the Force in Newtons to apply to the vehicle mass?
  5. Mar 23, 2012 #4
    The first equation, representing the kinetic energy (not power) of the vehicle at velocity v, is in
    Joules = kilogram-meter2/second2 = Newton-meters.

    The second is the power required to change the energy an amount dE in a time interval dt, i.e., to accelerate the car. It has units
    kilograms·meters/sec·meters/sec2 = Newton-meters/sec = joules/sec = watts.

    Recall that 746 watts = 1 HP

    The third equation is the power applied to the road by the wheels, i. e., by torque at the wheels.
    Power is torque x angular speed, with units meter-newtons (also Newton meters) and radians per sec.

    2π x RPM/60 = radians per second.

    For example, 10 HP = 7460 watts. If the wheel RPM is 800 RPM, the wheel angular velocity is 83.8 radians per second, and the torque is 7460/83.8 = 89 N-m.
  6. Mar 26, 2012 #5
    Hi Bob, thanks for the reply again, but I am still missing something. I have looked over this time and time again and I just cannot seem to figure out how to make it do what i need. Soooo, let me try rephrasing my new question with an example...

    We know the following information:
    WheelRadius = 0.75 meters
    WheelTorque = 400 Nm
    AxleSpeed = 2 Radians/Second
    VehicleMass = 2500 kg

    What will the axle speed be in 0.033 seconds using this data? My approach to this was:
    AppliedForce = WheelTorque/WheelRadius (533.3 = 400/0.75)
    Acceleration = AppliedForce / VehicleMass * deltaTime ( 0.007 = 533.3 / 2500 * 0.033)
    AxleSpeed = AxleSpeed + (Acceleration / WheelRadius) (2.009 = 2 + (0.007 / 0.75)
  7. Mar 26, 2012 #6


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    Science Advisor

    You are assuming the wheels roll without slipping. Is this the case for the real vehicle? If you floor the accelerator, do the wheels lose traction?

    Also, are you sure the torque numbers are accurate at the speeds you are working with? As far as I know, the engine torque depends on rotation velocity.
  8. Mar 26, 2012 #7
    We equate the second and third equations in my post #2:
    [tex] P= mva=T\frac{\mathrm{d}\theta}{\mathrm{d}t } [/tex]
    yielding an acceleration a :
    [tex] a=\frac{T} {mv} \frac{\mathrm{d}\theta}{\mathrm{d}t } \space \space \text{ meters per sec}^{2} [/tex]
    In this example, the power is 400 N x 2 radians/sec = 800 watts ≈ 1 HP
    mass = m = 2500 kg
    velocity = v= 2 radians/sec x 0.75 meters per radian = 1.5 meters per sec (this is very slow)
    Torque = T = 400 Nm
    angular velocity = dθ/dt = 2 radians per second

    So acceleration = a = (400/(2500·1.5)·(2) = 0.213 meters per sec2
    vehicle speed after 0.033 sec is v = 1.5 + 0.033 x 0.213 = 1.507 meters per sec
    ( or 2.009 radians per second)
  9. Mar 26, 2012 #8
    Hi Bob, I really appreciate you helping me out. I am just trying to understand this and I appreciate your patience.

    And Khashishia, slip may exist but is negligible at this point (none visible on the real vehicles). I am not sure that the torque values are very accurate, they are guesses based on ONE data point, We are using the known max torque value of the engine across the entire RPM range which I know is wrong but it is all I have as of now. I wouldn't expect that there is so much error to cause it to look as far off as it does, but maybe there is.

    So Bob, when you say a = (400/(2500*1.5)*(2) = 0.213...
    I think a parenthesis is missing so I am just ignoring the first one. I hope that is right. Anyway, what are the 1.5 and 2 variables? I assume it is linear and angular velocity yes?

    If either of these are velocities, then I won't be able to ever accelerate from a stop, so something seems to be missing. For example:
    a = 400/(2500*0)*(2) (DIVIDE BY 0)
    a = 400/(2500*1.5)*(0) (0 ACCELERATION)

    I am curious though, it appears that we still came to the same result. Is my math sound even though it is achieved differently?
  10. Mar 26, 2012 #9
    What I mean was

    acceleration = a = (400/(2500·1.5))·(2) = 0.213 meters per sec2

    I attach a BSFC (brake specific fuel consumption) plot of a typical 2.7 liter automobile engine. Note that the peak torque output is about 200 Nm at 4000 RPM, which would be 800 Nm and 1000 RPM with a 4:1 differential.

    Power = watts = torque x radians/sec = 800 x 1000 x 2π/80 = 84,000 watts = 112 HP.

    A typical wheel might have a diameter of 0.7 meters, so 1000 RPM is 105 radians/sec so speed is 36.7 meters/sec = 82 MPH.

    Attached Files:

  11. Mar 26, 2012 #10
    Hi Bob,

    I am still not seeing how I would accelerate from a stop with that equation though. Can you help me understand that? Are the 1.5 and the 2 values velocities (linear and angular respectively)?
  12. Mar 26, 2012 #11
    If a wheel has an angular velocity of 2 radians per second, and a radius of 0.75 meters, the linear velocity is 2 x 0.75 = 2 meters per second (about 4.47 MPH).
  13. Mar 27, 2012 #12
    I understand what you are saying, but if I start with a velocity of 0, your math will never allow for acceleration to have a valid non zero value (if i am understanding what the variables represent). See below

  14. Mar 27, 2012 #13


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    Science Advisor

    That's what the clutch is for, and if your simulation needs to be that precise, you'll have to model it. Clutch is what allows non-zero engine RPM at zero vehicle velocity, and that's the only way you'll get a non-zero torque to your wheels.
  15. Mar 28, 2012 #14
    I'm sorry to be an annoyance, but this still doesn't make sense. From what I can judge, all a clutch does is transition the amount of torque being applied to the transmission over time, and allows the speed of the transmission to accelerate to the speed of the engine. But acceleration cannot occur from a stop with this equation. I don't think a clutch should change the acceleration equation entirely, but just the torque variable being applied to accelerate the transmission.

    BUT, lets eliminate the clutch from the equation all together. Lets say that I want to calculate the acceleration of a wheel when I am using a 1 meter bar to apply torque. No clutch, just a torque applied to an object with 0 velocity. The equation for this should be the same as for my wheels. If the equation is the one I am questioning:
    Acceleration = (Torque/(WheelMass·MetersPerSecond))*(RadiansPerSecond)

    Then no matter what my torque is on that 1 meter bar, no acceleration will occur. This makes no sense. With a mass of 0.01kg, and a torque of 1000000000000000000 Nm, no acceleration would occur from a dead stop, and that just doesn't make sense to me.
    0 = (100000000000000000/(0.01*0))*0
  16. Mar 28, 2012 #15
    The torque is equal on both sides of the clutch plate, even if the engine is 1500 RPM on one side, and the other side is 0 RPM. So torque can be transferred to the wheels. This means that power is being dissipated in the clutch - I burned a clutch once backing a trailer up a hill in reverse.
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