Page:Practical Treatise on Milling and Milling Machines.djvu/66

60 Principle same as for Change Gears of a Lathe. In principle, these calculations are the same as for change gears of a screw cutting lathe. The compound ratio of the driven to the driving gears equals in all cases, the ratio of the lead of the required spiral to the lead of the machine. This can be readily demonstrated by changing the diameters of the gears.

Gears of the same diameter produce, as explained above, a spiral with a lead of 10 inches, which is the same lead as the lead of the machine. Three gears of equal diameter and a driven gear double this diameter, produce a spiral with a lead of 20 inches, or twice the lead of the machine; and with both driven gears, twice the diameters of the drivers, the ratio being compound, a spiral is produced with a lead of 40 inches, or four times the machine's lead. Conversely, driving gears twice the diameter of the driven produce a spiral with a lead equal to $1/4$ the lead of the machine, or $2 1/2$ inches.

Expressing the ratios as fractions, the $${ \text{Driven Gears} \over \text{Driving Gears} } = { \text{Lead of Required Spiral} \over \text{Lead of Machine} }$$ or, as the product of each class of gears determines the ratio, the head being compound geared, and as the lead of the machine is ten inches, the $${ \text{Product of Driven Gears} \over \text{Product of Driving Gears} } = { \text{Lead of Required Spiral} \over 10 }$$     That is, the compound ratio of the driven to the driving gears may always be represented by a fraction whose numerator is the lead to be cut and whose denominator is 10. In other words, the ratio is as the required lead is to 10; for example, if the required lead is 20, the ratio is 20:10. To express this in units instead of tens, the ratio is always the same as one-tenth of the required lead is to 1. And frequently this is a very convenient way to think of the ratio; for example, if the lead is 40, the ratio of the gears is 4:1. If the lead is 25, the gears are 2.5:1, etc.

To illustrate the usual calculations, assume that a spiral of 12 inch lead is to be cut. The compound ratio of the driven to the driving gears equals the desired lead divided by 10, or it may be represented by the fraction $12⁄10$. Resolving this into two factors to represent the two pairs of change gears, $12⁄10$ = $3⁄2$ X $4⁄5$. Both terms of the first factor are multiplied by such a number (24 in this instance) that the resulting numerator and denominator will correspond with the number of teeth of two of the change gears furnished with the machine (such multiplications not affecting the value of a fraction) $3⁄2$ X $24⁄24$ = $72⁄45$. The second factor is similarly treated: $4⁄5$ X $8⁄8$ = $32⁄40$, and the gears with