of the teeth in A and B, the planetary pinions merely acting as intermediate gears without any influence on the speed ratio. When the highest speed is required it is usual to employ a clutch and cause the whole of the gear to revolve at the same speed as the engine shaft. That is to say, the gear is idle, no teeth being exchanged, and the only reduction in speed between the motor and the road wheels is that due to the ratio of the number of teeth in the sprocket wheels used.

Epicyclic gearing has been employed in place of a main clutch in some cars, the driving pinion A being secured to the motor shaft and the plate D to the speedgear shaft. By checking the ring B slowly the load is taken by the engine very gradually, and the car can be started without any shock. A further advantage is that there is no end thrust on either the motor or the gear shafts as with a conical clutch. The gear should be so designed as to be capable of being filled with lubricant, when the wear will be but slight. By employing gear wheels with the teeth cut at a slight angle, i.e. helical gears, epicyclic gearing can be made practically noiseless in action.

BRAKES.

PROBABLY no part of an automobile requires more careful design than the brakes, in view of the probability that efficient stopping power may on occasion prevent loss of life. A high factor of safety is desirable, and in calculating strengths the weakest places should receive chief consideration.

The duty of the brakes is to dissipate the energy stored in the moving vehicle, and in the shortest possible time. Hence, before we can calculate the strength required for any part of the brake gear, the energy contained in the moving mass must be ascertained.

Denoting the energy in foot-pounds by E, the weight of the vehicle in pounds by W, and the velocity in feet per second by s, we have

g representing the acceleration due to gravity = 32.2 feet per second. It will be more convenient to modify the expression to agree with miles per hour, as the speed of motor vehicles is generally expressed in these terms. Therefore, as 1 mile is = 5280 feet, and 1 hour is = 3600 1.466 feet per

seconds, 1 mile per hour is =

5280
3600

second. Hence for 1 mile per hour we have, substituting in 54—

and putting Smiles per hour, the expression 54 becomes

As an example, we will assume a vehicle weighing, including its load, 16 cwt., running at a speed of 15 miles per hour. Ignoring road resistance, and substituting known values in 55, we obtain—

E 1792 x 225 x 0.0334 13466 foot-pounds

A factor of importance to be considered when calculating the stopping power of the brakes is the coefficient of friction between the tyre of the wheel and the road surface. For want of precise information on this head we may assume this coefficient to be 04 for iron tyres and 0·7 for rubber tyres. These values will vary according to the condition and material of the road surface. For wood and asphalte roads, when the surface is greasy, the coefficients should be taken as not more than one-half the above values. On the majority of automobiles the brakes only act upon two of the wheels, usually the driving wheels. Therefore we shall require to know the proportion of the total weight of the vehicle carried by the wheels to which the brakes are applied. For the example selected above we will assume this to be one-half of the total weight 896 lbs. Then for the minimum distance in which the car can be stopped we have the expression

=

in which L = the minimum distance in which the car can be stopped, k = the coefficient of friction between the tyre and the road, and w = the weight in pounds carried by the wheels to which the brakes are applied; the other factors being as before. Substituting known values in 56, we have

By dividing the energy E stored in the moving car by the distance L, we obtain the mean resistance required on the periphery of the tyres. Thus—

tyres, or P =

13466
21.4

say 630 lbs. for rubber tyres; onehalf of these amounts being required on each of the two wheels. From this it will be seen that the coefficient of friction for rubber tyres, being greater than for iron, allows of a stronger breaking effort being applied without skidding the wheels, and hence the car is stopped in a much shorter distance.

When band brakes are used, their diameter is of necessity less than that of the road wheels, and the pull on the brake rods will be increased in inverse proportion to the diameter of the road wheel and brake drum. Thus

D

(58)

p = kw /

I

in which D the diameter of the road wheel, d = the diameter of the brake drum, p = the pull on the brake rod; and the other factors as before. Assuming the road wheels of the car in the example to be 30 inches diameter, and the brake drums 10 inches diameter, and rubber tyres

30 10

on the wheels, we have, from 58, p = 630 × = 1890 lbs.

for the pair of brake bands, or 945 lbs. mean pull on each. In the case of sudden applications of the brakes, such as cause the wheels to skid, this pull of 945 lbs. will be much exceeded.

For shoe brakes, acting directly on the tyres, the pressure can be found by dividing the mean resistance P by the coefficient of friction, or which in our example

P

k'

= 450 lbs. on each tyre.

In all the above calculations we have assumed the car to be travelling on a level road. Gradients will tend to lengthen or shorten the distance in which the car can be stopped, according to whether it is an up or down grade. Down grades will reduce the resistance P, and may be expressed as a fraction = f. The distance L in which the car can be stopped going downhill may then be calculated from

Substituting known values from our example, and assuming a gradient of 1 in 20, we have