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.

π

11. 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.

12. A locomotive weighing 30 tons draws a train weighing 65 tons at 50 miles per hour, resistance 33 lbs. per ton. Find the II. 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.

Fig. 56.

To find the effect of this we resolve R, by the laws of Statics, into two components, R cos 0 and R sin 0, then the the same as the effect of two separate forces, as in Fig. 57.

effect of R is

In considering these forces separately, we see first that the magnitude of R sin can, in the

R cos

Fig. 57.

absence of friction, have no effect on the Rsin 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.

Neglecting friction then

Effort R cos 0,

and for a movement s,

Energy exerted = R cos 0 × s,

= Resolved resistance x distance moved.

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.

The constant resistance is W lbs.

vertically.

Draw now BC vertical and AC horizontal.

Then, whatever be the shape of AB, Fig. 59. the total motion is composed simply of AC and CB, and from the preceding we say

Energy exerted or work done = W x BC.

We have not strictly followed the preceding because ab there was a straight line; but we can do that by dividing AB into a large number of small parts as ab, each of which is straight if we take them small enough. Then for ab

Energy or work=W × bc,

.. Total energy or work=W × sum of all the small heights, =Wx BC.

We are now able to examine in another way the case of turning, which we considered in the last chapter, viz. when the resistance or effort is not at right angles to the radius.

The figure (Fig. 49) on page 65 is here reproduced (Fig. 60).

Now resolve P along, and at right angles to, the radius

A

Fig. 60.

B

OA. Then the part along OA is always at right angles to the motion, this being at the instant along the tangent to the circle in which A moves. Therefore, as we have seen, it can exert no energy, and therefore the Effort is simply P sin OAM at right angles to OA.

.. Turning moment = P sin OAM × OA,

which is of course identical with P x OM, the result obtained on page 65.

The part P cos OAM will, of course, practically affect the motion, because it will pull the shaft against its bearing, and thus create friction. So actually it becomes a source of, not effort, but resistance. Whether there be friction or no then, the Effort will be only P cos OAM.

Similarly we can treat the resistance R, and obtain

as the resistance against which work is done, the force R cos OBN.

Variable Forces.—In all practical cases, the forces are not constant, but they vary. We require then to calculate the Work done against a varying resistance, or Energy exerted by a varying effort.

Let now AB represent the path of one piece of a sliding pair against a direct resistance.

A a b c d e f g h i j k l B

Fig. 61.

Divide AB into a number of parts as shown, Aa, ab, . . ., /B, and let the resistance given be constant during each small movement; its values being R1 during Aa, R, during ab, etc., to R11 during /B. Then

Work done=R1 × Aa + R2 × ab+ ... + R11 × /B.

Now we will apply the graphic method of the preceding chapter.

To do this set up ordinates on Aa, ab, etc., to represent R1, R2 etc.

We thus obtain a number of rectangles Ca, etc., to ID, each representing the work done during the movement represented by its base. Therefore

Total work done = sum of areas of rectangles

(1).

The tops of the rectangles constitute a stepped curve CD, and this curve has the property that—at any point of the movement its ordinate represents the value of the resistance. This we know by the manner in which we constructed it. Hence then we call CD the Curve of

Resistance.

In the above the variations of R occur at definite