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well fitted to withstand blows by their shape, are supported inside the double thickness of plating by closely spaced vertical girders. Behind armour, where men are likely to be employed in action, the inside of the framing is covered in with 10-lb. plating (see

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Fig. 146.

Fig. 145.—Armour bolt.

main deck in Fig. 13). When armour is struck, rivets are likely to break and the heads to fly off, so that this lining forms some protection to men inside. This plating itself should be secured by screw rivets to the framing (see K, Fig. 9).

Armour Bolts. — With wroughtiron armour, the bolts for securing the armour to the ship's structure were carried right through with a large conical head flush with the surface of the armour. With hardfaced armour, however, the surface must not be pierced for bolts because the surface would then be liable to crack badly from hole to hole when struck. Armour bolts are now screwed into the back of the plate (Fig. 145), and about one bolt to every 7 square ft. is allowed. Fig. 146 shows the holes in a specimen plate; it is important to have good security in order to keep fragments together, even if the plate is badly cracked. Experiments have shown that the pieces are still very efficient, provided they are held up to the backing.

In order to diminish the liability of bolts breaking under the impact of projectiles, the shank of the bolt is made slightly less in diameter than at the bottom of the thread. The bolt will then stretch at this weakest part rather than break under the thread. A sleeve is fitted to provide sufficient length for this weakest portion; also, to provide some elasticity to take the shock, the nut securing the bolt to the ship is fitted with an elastic washer of india-rubber. This is placed inside a "cup washer" to keep the

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rubber washer in place. In some places, as barbettes, it is not possible to get sufficient room for an ordinary armour bolt; in such places, the bolt, as in Fig. 147, has to be used.




1. Area of rectangle, as ABCD (Fig. 148).—The area is given by AB x BC. The length and breadth must be of the same denomination, i.e. if the length is in feet the breadth must be in feet, the area then being square feet.

2. Area of triangle, as ABC (Fig. 149).—We draw CD perpendicular to the base AB, meeting it, or it produced, in D. Then the area = £ X AB X CD, or one-half the base into the height.

3. Area of trapezoid.—This is a four-sided figure, in which

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two sides only are parallel, as ABCD (Fig. 150). Calling the parallel sides a and b respectively, and h the distance between them, the area = £ (a + b) h, or one-half the sum of the parallel sides multiplied by the distance between them.

4. Circle.(a) Length of circumference is n times the diameter, where n = 31416, or -272- nearly.

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5. Curvilinear figure, as ABCD (Fig. 151).—The area of figures of this character are continually required in ship calculations.

(a) Trapezoidal ride.—This rule has found considerable favour, especially in France and the United States, on account of its great

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two, then all the remaining ordinates are added and the total sum is multiplied by the common interval.

The following example will illustrate the use of the rule and how near the result obtained is to the real area required.

The curve, whose ordinates 2 ft. apart are 0, 2'2, 4-0, 5-4, 6-4, 7*0, and 7"2 ft. respectively, is a portion of a common parabola, and the exact area enclosed is 57'6 square ft. Find the area by using the trapezoidal rule.


= 57-2 square ft.


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There is thus an error of nearly 1 per cent. The error involved in using this rule is lessened by spacing the ordinates closely, but it is not used in Admiralty calculations on account of the approximate nature of the results obtained. The rule that is employed is—

(b) Simpson's first rule.—Take a figure as ABCD (Fig. 152),

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