The following rules and examples show how the table is used:
Rule 1.--Given ---- number of teeth and pitch; to find ---- pitch diameter.
Select from table in columns opposite the given pitch--
First, the value corresponding to the number of units in the number of teeth.
Second, the value corresponding to the number of tens, and multiply this by 10.
Third, the value corresponding to the number of hundreds, and multiply this by 100. Add these together, and their sum is the pitch diameter required.
Example.--What is the pitch diameter of a wheel with 128 teeth, 1-1/2 inches pitch?
We find in line corresponding to 1-1/2 inch pitch--
Pitch diameter for 8 teeth 3.8197 " " 20 " 9.549 " " 100 " 47.75 --- ------- " " 128 " 61.1187
Rule 2.--Given ---- pitch diameter and number of teeth; to find ---- pitch.
First, ascertain by Rule 1 the pitch diameter for a wheel of 1-_inch pitch_, and the given _number of teeth_.
Second, divide _given pitch diameter_ by the _pitch diameter_ for 1-_inch pitch_.
Example.--What is the pitch of a wheel with 148 teeth, the pitch diameter being 72"?
First, pitch diameter for 148 teeth, 1-inch pitch, is--
8 teeth 2.5465 40 " 12.732 100 " 31.83 --- ------- 148 " 47.1085
Second, 72/47.1 = 1.53 inch equal to the pitch.
This is nearly 1-1/2-inch pitch, and if possible the diameter would be reduced or the number of teeth increased so as to make the wheel exactly 1-1/2-inch pitch.
Rule 3.--Given ---- pitch and pitch diameter; to find ---- number of teeth.
First, ascertain from table the _pitch diameter_ for 1 _tooth_ of the given _pitch_.
Second, divide the _given pitch diameter_ by the _value_ found in table.
Example.--What is the number of teeth in a wheel whose pitch diameter is 42 inches, and pitch is 2-1/2 inches?
First, the pitch diameter, 1 tooth, 2-1/2-inch pitch, is 0.7958 inches.
This gives a fractional number of teeth, which is impossible; so the pitch diameter will have to be increased to correspond to 53 teeth, or the pitch changed so as to have the number of teeth come an even number.
Whenever two parallel shafts are connected together by gearing, the distance between centres being a fixed quantity, and the speeds of the shafts being of a fixed ratio, then the pitch is generally the best proportion to be changed, and necessarily may not be of standard size. Suppose there are two shafts situated in this manner, so that the distance between their centres is 84 inches, and the speed of one is 2-1/2 times that of the other, what size wheels shall be used? In this case the pitch diameter and number of teeth of the wheel on the slow-running shaft have to be 2-1/2 times those of the wheel on the fast-running shaft; so that 84 inches must be divided into two parts, one of which is 2-1/2 times the other, and these quantities will be the pitch radii of the wheels; that is, 84 inches are to be divided into 3-1/2 equal parts, 1 of which is the radius of one wheel, and 2-1/2 of which the radius of the other, thus 84"/3-1/2 = 24 inches. So that 24 inches is the pitch radius of pinion, pitch diameter = 48 inches; and 2-1/2 × 24 inches = 60 inches is the pitch radius of the wheel, pitch diameter = 120 inches. The pitch used depends upon the power to be transmitted; suppose that 2-5/8 inches had been decided as about the pitch to be used, it is found by Rule 3 that the number of teeth are respectively 143.6, and 57.4 for wheel and pinion. As this is impossible, some whole number of teeth, nearest these in value, have to be taken, one of which is 2-1/2 times the other; thus 145 and 58 are the nearest, and the pitch for these values is found by Rule 2 to be 2.6 inches, being the best that can be done under the circumstances.