The diameter of the addendum or outer circle of a wheel has no influence upon the velocity of the wheel. Suppose, for example, that we have a pair of wheels of 3 inch arc or circular pitch, and containing 20 teeth, the driver of the two making one revolution per minute. Suppose the driven wheel to have fast upon its shaft a pulley whose diameter is one foot, and that a weight is suspended from a line or cord wound around this pulley, then (not taking the thickness of the line into account) each rotation of the driven wheel would raise the weight 3.1416 feet (that being the circumference of the pulley). Now suppose that the addendum circle of either of the wheels were cut off down to the pitch circle, and that they were again set in motion, then each rotation of the driven wheel would still raise the weight 3.1416 feet as before.
It is obvious, however, that the addendum circle must be sufficiently larger than the pitch circle to enable at least one pair of teeth to be in continuous contact; that is to say, it is obvious that contact between any two teeth must not cease before contact between the next two has taken place, for otherwise the motion would not be conveyed continuously. The diameter of the pitch circle cannot be obtained from that of the addendum circle unless the pitch of the teeth and the proportion of the pitch allowed for the addendum be known. But if these be known the diameter of the pitch circle may be obtained by subtracting from that of the addendum circle twice the amount allowed for the addendum of the tooth.
Example.—A wheel has 19 teeth of 3 inch arc pitch; the addendum of the tooth or teeth equals 3⁄10 of the pitch, and its addendum circle measures 19.943 inches; what is the diameter of the pitch circle? Here the addendum on each side of the wheel equals (3⁄10 of 3 inches) = .9 inches, hence the .9 must be multiplied by 2 for the two sides of the wheel, thus, .9 × 2 = 1.8. Then, diameter of addendum circle 19.943 inches less 1.8 inches = 18.143 inches, which is the diameter of the pitch circle.
Proof.—Number of teeth = 19, arc pitch 3, hence 19 × 3 = 57 inches, which, divided by 3.1416 (the proportion of the circumference to the diameter) = 18.143 inches.
If the distance between the centres of a pair of wheels that are in gear be divided into two parts whose lengths are in the same proportion one to the other as are the numbers of teeth in the wheels, then these two parts will represent the radius of the pitch circles of the respective wheels. Thus, suppose one wheel to contain 100 and the other 50 teeth, and that the distance between their centres is 18 inches, then the pitch radius or pitch diameter of one will be twice that of the other, because one contains twice as many teeth as the other. In this case the radius of pitch circle for the large wheel will be 12 inches, and that for the small one 6 inches, because 12 added to 6 makes 18, which is the distance between the wheel centres, and 12 is in the same proportion to 6 that 100 is to 50.
A simple rule whereby to find the radius of the pitch circles of a pair of wheels is as follows:—
Rule.—Divide number of teeth in the large wheel by the number in the small one, and to the sum so obtained add 1. Take this amount and divide it into the distance between the centres of the wheels, and the result will be the radius of the smallest wheel. To obtain the radius of the largest wheel subtract the radius of the smallest wheel from the distance between the wheel centres.
Example.—Of a pair of wheels, one has 100 and the other 50 teeth, the distance between their centres is 18 inches; what is the pitch radius of each wheel?
Here 100 ÷ 50 = 2, and 2 + 1 = 3. Then 18 ÷ 3 = 6, hence the pitch radius of the small wheel is 6 inches. Then 18 - 6 = 12 = pitch radius of large wheel.
Example 2.—Of a pair of wheels one has 40 and the other 90 teeth. The distance between the wheel centres is 321⁄2 inches; what are the radii of the respective pitch circles? 90 ÷ 40 = 2.25 and 2.25 + 1 = 3.25. Then 32.5 ÷ 3.25 = 10 = pitch radius of small wheel, and 32.5 - 10 = 22.5, which is the pitch radius of the large wheel.