|Calculating wheel revolutions over distance|
by Christopher Bischoff
|Date Added: 3/12/2003|
|!div style="display:none"!fjrigjwwe9r2content:tip!/div!!div style="display:none"!edf40wrjww2content:tip!/div!A question was posed to the Stratalist: " (If) I had a) speed of vehicle and b) diameter of tires, what is the formula I would use to calculate revolutions per second?|
Francis Drouillard was kind enough to provide the answer and I thought it important enough to preserve here:
The perimeter of a circle is PI times its diameter, or P = pi x diameter.
Say a wheel tracks a distance X in time t, which is an average linear rate of X per t. The number of wheel revolutions, N, is X divided by P, or N = X / P. Thus, the average angular velocity, w, of the wheel is N per t, or w = ( X / P ) / t.
For example, if the distance tracked is X = 48 feet in time t = 6 seconds, the linear velocity is 8 feet per second. If the wheel diameter is 2 feet then the perimeter is 6.28 some odd feet. The number of wheel revolutions needed to travel that distance is N = 48 / 2pi = 7.634 revolutions. Then angular rate is then w = N / t = 1.273 revolutions per second.
He then added:
A better way to keep track of those formulas is to keep the units in your equations. Going back to the earlier wheel problem (distance X = 48 feet in time t = 6 seconds, wheel diameter D = 2):
perimeter P = pi x D = 6.283 feet, thus
distance traveled in 1 revolution, c = 6.238 feet per revolution
rate of travel, r = 48 feet / 6 seconds = 8 feet per second.
Now, to convert the linear rate r to the angular rate w, divide r by c (or multiply by 1/c):
8 feet 1 rev
w = -------- x -------------- = 1.273 revs per second
sec 6.283 feet
Notice how the units of feet cancel each other out to leave the desired units of revs per sec.
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