best understood in simple cases, and the gain will be seen when we come to more complicated ones.

Power-Horse Power. The definitions of Energy and of Work done contain no reference to the speed at which the operations considered have to be carried on; thus to lift one lb. through one ft. requires the same amount of energy to be exerted whether it be done in one second, or one minute, or one hour.

[The speed of a certain operation has an effect generally on the energy required, because it affects the values of the useless resistances, i.e. those which waste energy; but this does not affect what is stated above where we take the resistance as given.]

But to compare the value of different working agents, we must consider the energy they can respectively exert in a given time. This quality of an engine, or other energy exerting agent, we call its Power. The Power of an engine, for example, then means the energy it can exert in a given time.

To measure power we require some Unit Power, and this we take as the capacity for exerting 33,000 ft.-lbs. in one minute. The unit is called a Horse Power, and is due to Watt; being introduced by him, so that purchasers of engines should be able to compare their values with those of horses. The unit is much above the power of any horse in continuous work; but the extra margin was allowed to prevent any possibility of a mistake in the other direction.

The Horse Power then of any agent is equal to the number of ft.-lbs. of energy it can exert in a minute divided by 33,000.

If, for example, a steam engine work with constant steam pressure p lb. per sq. in., piston area A sq. ins., stroke s ft., revolutions per minute N. The calculation of the Horse Power proceeds thus

Effort=px A=pA lbs.

Distance moved in one revolution = 2s ft.

... Energy exerted in one revolution=þA × 2s ft.-lbs. minute =pA x 25 x N ft.-lbs. PA x 25 x N

... Horse power=

33,000

and similarly for any other agent.

EXAMPLES.

1. A man weighing 150 lbs. carries loads of 144 lbs. to a height of 30 ft. Find the work done in each journey. If the man exerted the same amount of energy in lifting the weights by means of a winch, which wasted one-third of the energy applied, how much more useful work could he do?

Ans. 8820 ft.-lbs.; .3 times. 2. Iron pigs six inches square lying originally on the ground, are built into a stack 6 ft. by 5 ft. by 4 ft. high. Find the work done. Ans. 90720 ft. -lbs. 3. A bicyclist and machine weigh 180 lbs. Find the H. P. he exerts when riding at 20 miles per hour, on a track the resistance of which is I per cent of the weight. Ans. .096. 4. A colliery engine raises loads weighing 21 cwt. from a depth of 1200 ft. in 45 seconds. Find the H. P. required.

Ans. 114.

5. Compute the nett H. P. required to pump out a basin with vertical sides in 48 hours, the area of the water surface being 50,000 sq. yds., and depth of water 20 ft., the water being delivered at a height of 26 ft. above the bottom of the basin. (35 c. ft. of sea water weigh I ton.) Ans. 97.

6. A boiler 12 ft. diameter, 10 ft. long, full of water, is to be pumped out, the water being delivered at the sea level. The ship's draught is 21 ft., and the bottom of the boiler is ft. above the keel. Find how long two men each capable of exerting 2500 ft.-lbs. per minute would take, using a hand pump in which of the energy is wasted. Ans. 3 hrs. 28 min.

7. A pumping engine 9 ins. diameter, 6 ins. stroke, making 80 revolutions per minute, is fitted in the above ship. Assuming a constant steam pressure (effective) of 40 lbs. per sq. in.; find how long the engine would take in pumping out the boiler, assuming that engine and pump together waste one half the energy. Ans. 8 minutes.

8. The resistance of a ship at 15 knots is 70,000 lbs. Of the energy exerted by the steam 15 per cent is wasted before reaching the crank shaft, and 50 per cent altogether. Find the H. P. of the engines. There are two propellers, and the revolutions are 80 per minute. Find the moment exerted on each shaft.

Ans. 6430, 960 tons-ins.

9. Each engine in the preceding has three cylinders, the work being equally divided between them. The stroke is 3 ft. 6 ins., and the diameters are 70 ins., 48 ins., and 32 ins. Find the effective steam pressure in each cylinder.

Ans. 7.73, 16.44, 36 lbs. per sq. in.

10. An anchor weighs 5 tons, and the cable 1 ton per 8 fathoms. The ship is anchored in 9 fathoms, and the deck is 10 ft. above the water line. The cable hangs in the quadrant of a circle, touching the ground where the anchor lies. Find the work done,-Ist, while hauling in the slack; 2d, while lifting the anchor. Neglect the buoyancy of the water. Distance of the C. G. of the arc of a quadrant from either bounding Ans. 33.58, 474.66 ft.-tons.

2

radius times the radius.

II. A water-power engine of 10 H. P. is supplied from a tank 12 ft. by 8 ft. by 6 ft., at a height of 120 ft. Supposing the tank be full but no supply, find how long the engine could run.

Ans. 13 minutes.

Find

12. A locomotive weighing 30 tons draws a train weighing 65 tons at 50 miles per hour, resistance 33 lbs. per ton. the H. P. required. If the train became detached, the engine still exerting the same power, what speed would it attain, assuming the resistance to vary as the speed squared.

Ans. 418, 73.5 miles per hour. 13. In question 8, assume the resistance to vary as the cube of the speed, and find the H. P. required to propel the ship at 18 knots. Ans. 13,300.

14. A crank shaft, diameter 12 ins., weighs 12 tons, and is also pressed against the bearings by 36 tons horizontal. Find the H. P. lost in friction at 90 revolutions. Coefficient .06.

Ans. 45.6.

CHAPTER IV

OBLIQUE AND VARIABLE FORCES INDICATOR

DIAGRAMS

IN the preceding chapter the efforts and resistances considered have been constant, and in the line of motion. We now proceed to consider more general cases. Oblique Action. In the motion of a sliding pair as A and B (Fig. 56), the resistance may not be directly against the motion, but inclined, as R at an angle 0.

B

Fig. 56.

effect of R is

| Rsin 0

R cos 0

To find the effect of this we resolve R, by the laws of Statics, into two components, R cos and R sin 0, then the the same as the effect of two separate forces, as in Fig. 57. In considering these forces separately, we see first that the magnitude of R sin can, in the absence of friction, have no effect on the effort required to move A along B. We can never of course practically get rid of the effect of R sin 0, but we can, by successive improvements in smoothness of surface and in lubrication, continually diminish it; and the only obstacle to complete elimination of its effect is imperfection of surface and of lubrication. We can thus conceive that, in itself, it is incapable of offering any resistance at all to the motion, but can only do so by the friction it can excite.

Fig. 57.

We can express this in a slightly different manner, which will be found useful.

For let ab be the movement (Fig. 58). From 6 let fall bc perpendicular to R's direction, then

Energy exerted or work done = R cos @ xs=Rx s cos 0,

We can now apply this to the case in which motion takes place in any curve, against a resistance constant in direction. Such a resistance would be, for example, the pull of the earth on a weight.

Let a weight, W lbs., be lifted along the path shown from A to B.

Fig. 59.

B

C

The constant resistance is W lbs.

vertically.

Draw now BC vertical and AC horizontal.

Then, whatever be the shape of AB, the total motion is composed simply of

AC and CB, and from the preceding we say

Energy exerted or work done = W x BC.