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Wheel speed at tyre compared to at the centre

  1. Sep 19, 2014 #1
    does anyone know the formula for calculating the speed a wheel has to be travelling to cover a certain amount of rotations within a minute eg 30rpm,

    if 2m diameter implies a 3.142 perimeter (x 30)
    therefore do the wheels move at 94 meters p m (x 60 minutes , in time)
    = 5.64 kmph or 3.5 mph? essentially small wheels to require faster speed hence gearing etc
    ? if yes does anyone have a simpler calculation?

    Also the calculation to turn an item such as a water wheel different size for different torques on the crank shaft

    The leverage required at different size to achieve a desired rpm at different levels of torque?
  2. jcsd
  3. Sep 19, 2014 #2


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    For two wheels turning at the same rate, i.e. RPM is constant, the wheel with the larger diameter will roll farther in the same amount of time than the wheel with the smaller diameter.

    So, if you take the stock 13" wheels and tires off your ride and replace them with 22" rims, you will either run the engine at a lower RPM (easy) to go 60 mph, or you'll have to change the final drive ratio in the drive train (a little harder). Also, the odometer and the speedometer are driven by a gear mounted in the transmission and this gear is calibrated for a certain tire size. If you increase the tire size, the mileage reading on the odometer will be off, as will the reading on the speedometer, so be sure to tell the traffic cop this if you get pulled over for speeding.
  4. Sep 20, 2014 #3
    yeah i get what you mean ground it covers would be more due to larger area etc, so larger wheels a larger distance so if the size changes the speed of the wheel and the distance it covers is altered due to requiring more rpm to travel a certain distance, tyre sizes change yet one rotation is measured and calculated etc

    So irrespective of the size of the wheel etc the main i thing is rpm and therefore to get a certain desired speed thing coming from the wheel, essentilly the alrger the wheel allows for a greater gearing later, eg 2m increase the gearing to the lower therefor increased speed

    do you know hot to calculate gearing ratios? eg 2m diameter requires 200 rpm at output there the gearing has to change 30 rpm to 2000,

    just found the below calculation in regards to generating eg pairing to get the desired out put, yet to calculate it etc, but in regards to the radians do you know if, as its divided by time after diameter x 2 x pie is per 60 seconds. hour etc

    The shape of the gear teeth are the some on both the input and output gears, and thus the larger gear has more teeth on it. The Pitch distance, Pd, is the distance between gears. Thus the number of teeth on the gear, n, times the Pitch is equal to the circumference of the gear. Accordingly,

    Pd nin = 2 p rin

    Pd nout = 2 p rout

    nin/ nout = rin/ rout

    The gear pair is analyzed with the following assumptions:

    Quasi-static analysis (it is assumed that the gears are rotating at a constant speed, and thus acceleration torques can be neglected)

    Frictional losses are neglected (friction can be significant, and should be considered separately!)

    The gear teeth mesh with each other (no jumping of gears!)

    Since there are no frictional losses, the input and output power can be set equal to each other as:

    Pin = tin win

    Pout = tout wout

    tin win = tout wout

    We now need to consider the relative velocity of the two gears, which is determined by the meshing of the teeth. Since the teeth mesh, we know that the same number of teeth must mesh from both gears. For each revolution of the input gear the following number of teeth pass through the mesh area, where nrevin is the number of revolutions of the input gear:

    number of teeth that mesh = nrevin 2 p rin / Pd

    Applying the same equation to the output gear, and setting the number of meshed teeth equal to each other provides:

    nrevout 2 p rout / Pd = nrevin 2 p rin / Pd

    The above equation simplifies to:

    nrevout / nrevin = rin / rout

    If we multiple the number of revolutions by 2p, we get the angle of rotation of both gears in radians, which gives:

    rin dqin = rout dqout

    If we divide the angle of rotation by time, dt, then we get the ratios of angular velocities in radians per second

    wout / win = rin / rout

    An alternative interpretation is that the angular velocity at the mesh point is the same for both gears. Since velocity of a point on a rotating object is given by rw. The velocity equality at the mesh point is given by:

    rin win = rout wout

    And we see that the two previous equations are identical.

    Since the radius of a gear is proportional to the number of teeth, the velocity relationship can be given in terms on numbers of teeth on the input and output gears. Simply substitute into the above equation that nPd=2pr for both gears, to give:

    wout / win = nin / nout

    We can now combine the power equation with the velocity equation to get the ratio of input and output torques:

    tin win = tout wout (power equation)

    tout / tin = win / wout

    tout / tin = rout / rin (substituting in velocity relationship)

    Thus when the input gear is smaller than the output gear:

    The output torque is higher than the input torque

    The output velocity is lover than the input velocity (i.e. the smaller gear needs to make more revolutions than the larger gear)

    In Summary

    The fundamental equations for a gear pair are:

    tin win = tout wout (power equality)

    wout / win = rin / rout (velocity relationship in terms of radiuses)

    wout / win = nin / nout (velocity relationship in terms of number of teeth)

    tout / tin = rout / rin (torque relationship in terms of radiuses)

    tout / tin = nout / nin (torque relationship in terms of number of teeth)

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