WHEEL TRAINS 393 
| Problems on epicyclic gears may however be solved by aid of a 
formula constructed as follows :— 
Let a=number of revolutions of the arm EF in a given time. 
m=number of revolutions of the wheel A in the same time. 
n=number of revolutions of the wheel L in the same time. 
The speed of A in relation to the arm is m—a, and the speed of L in 
relation to the arm is m—a, hence the value of the train or its velocity 
ratio is ane =e. In using this formula it is most important that the 
proper signs be given to the values of a, m,n, and e. For example, if 
the arm makes 20 revolutions in a direction taken as positive, a= + 20, 
and if A makes 30 revolutions in the opposite or negative direction, 
then m= — 30 and m—a= —30-20=-—-50. Again, the value of e is 
positive or negative according as A and L rotate in the same or in opposite 
directions respectively. 
n-—a 
m-a 
shown in Fig. 611, and let the wheels A and L be equal. Here e= — 1. 
Let L be Pera from rotating about its axis, then n=0, and by 
—a 
m—a 
as the arm in the same direction. This is the well-known sun and planet 
motion used by Watt as a substitute for the ordinary crank in the steam- 
engine. A was fixed to the fly-wheel shaft, and L was bolted to the 
connecting-rod. 
As an example on the use of the formula ¢= 
take the gear 
formula —1= 
or m= 2a, that is, the wheel A rotates twice as fast 
In using the formula e=”~—* to solve the problem on Humpage’s 
gear, already worked out, two applications have to be made. First 
consider the train made up of A, B, C, and L (Fig. 616). Here 
e= px i=? m the speed of A=0, and n is the speed of L. Hence 
.-" =< and Amz Next consider the train made up of A, B, and D. 
“Here e= -S- —4, m the speed of A=0, and n is the speed of D. 
Hence — 4=0—, and n=5a. Therefore the speed of L is to the speed 
of D as G54, or as 1; 20, as before. 
Exercises XXIV. 
1. The axes of two spur wheels in gear are 37 inches apart. One wheel 
rotates four times as fast as the other. Find the diameters of the pitch circles 
of the wheels. 
2. It is required to connect two shafts, whose axes are tg be as nearly as 
possible 40 inches apart, by spur wheels so that the velocity ratio may be 
exactly 9:2. Find the number of teeth in each of the two wheels and the dis- 
tance between the axes of the shafts, to the nearest hundredth of an inch, if the 
pitch of the teeth is 2} inches. 
3. The crank of a direct double-acting steam-engine is 15 inches long. A 
spur wheel 9 feet in diameter on the crank shaft drives a pinion 2 feet in 
