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of the end ones, add four times the odd and twice the even ordinates. The total sum multiplied by one third the common interval between these ordinates, will produce the area. It should, however, be stated that the number of equal parts need not necessarily be even, and as it is sometimes desirable to calculate the area to an odd ordinate by taking the sum of the first ordinate and adding to it four times the odd ones, and twice the last as well as the even ordinates into one third the common interval, the area

culated accurately. In the foregoing definition it should be noted that the first ordinate is numbered “0," and that the number of intervals multiplied by 3 should equal the sum of the multipliers.

may be

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Area of ABCD = (Yo + 441 + 2 y2 + 443 + y4).

3 And if half ordinates be inserted between yo and y1 and between Y3 and Y4 we should then have :

Area =(19+2 y + 1} 41+4 y,+1} Y3+2 434 + } ya). Should, however, we desire to calculate the area embraced within the limits of y: only, omitting the half ordinate yf, then :

Area = (40+ 4 y2 + 2 y2+2 ya). So that it is immaterial what subdivision of parts we may use as long as the multiplier is given the relative value to the space it represents as exemplified in the subjoined table. It will be obvi. ous that we may also give multiplier only half its value, as

1 Yo + 2 yı+1 42+2 y: + 44,

and multiply the sum by of x, which will be found the more convenient way to use the rule, involving as it does figuring with smaller values.

Multipliers for Subdivided Intervals.

1

4

Ordinates, 0 1 2 3 4 5 51
Multipliers, 11 4 2 4 11 1
Ordinates, 0 221 3 3731 33
Multipliers, 1 4 112
Ordinates, 0 1 112 213 4
Multipliers, | 1 12" 12114 11 | 2
Ordinates, 0 1 2 2 23 3 31 34 5 6° 637
Multipliers, 1 4 13 ij 1} 1j|14 11 2 1

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As proof of the rule let us deal with an example :

2 Area ABCDE

3

(yo + 4yı + y2). Assume curve DFC is part of a common parabola ; area DKCFD is area of parallelogram. Join DC, and draw parallel

H

F

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A

B

FIG. 2. to GH touching curve. If DFC be part of parabola area, DFC is of parallelogram DCHG.

yo+y2 EK = (Yo + yı).

FK = Yu

2 Parallelograms on same base and between same parallels are equal. Draw through G and H two lines parallel to base as GM and DL, then area

DCHG = area DLMG

= 2 X X DG
= 2 X X FK

Yo + y2

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4 X

+

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Yo + y2
Area DFC = { of above = Yi

3

2

Yo Y2
Area ABCKD = 2 X +

2 2
Yo , Y2 4 x Yo + y2
Whole area 2 X

.
3

2 -3

(40 + 4 + y2). Simpson's second rule for determining areas bounded by a parabola of the third order and the “five eight” rule applicable to the calculation of one of the subdivided areas are given in most text-books, but are omitted here as superfluous, Simpson's first rule being adaptable to either of these cases, so that for all ship calculations where areas, volumes, or moments are required, the first rule, or as hereafter explained Tchiby scheff's rule, are recommended.

We have seen, then, how the area or surface may be calculated by this rule, and as the volume is the area by the thickness, it will be evident that if the areas be calculated at various levels or water lines, as shown in the figure, and these areas in turn treated as a curve and integrated by means of the rule, that the result will be the volume of the body.

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Let the Figs. 3 and 4 represent the immersed half longitudinal body of a vessel 100 feet long by 12 feet broad submerged to 5 feet draught as represented by L. W.L. It is required to calculate the volume of water displaced by Simpson's first rule. The base line length between perpendiculars should be divided into an equal number of intervals, and as advocated in the chapter on Design, it will be well to have a definite number and retain same for all designs, as by so doing it will facilitate comparisons and working from one design to another. Ten such intervals with half-end ordinates is a very convenient division, and in this case will give a common interval of 10 feet. The draught of 5 feet must likewise be subdivided into a certain number of equal intervals, which in this case we will fix at 4, so that

5 ft. draught

: 1.25 ft.

4 interval between water lines. These divisions of water lines must be drawn across the body plan of ten sections, and the half breadths read off with a scale and tabulated as in table on following page.

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It should be stated in connection with the subdivision of the base line that the length taken for displacement is measured by some designers from the after side of body post i.e., ignoring the propeller aperture ; and by others from the fore side of body post to the after side of stem omitting the moulded size of these forgings. Both of these methods are inaccurate besides leading to confusion, as, in the first case, the displacement of the propeller with its boss will equal the displacement cut out for aperture not to mention the volume of the rudder, which is rarely, if ever, taken into account. And in the second case the tiny amount of displacement added at the knuckle formed by the bearding line of plating when the length is taken to forward and after sides of stem and stern post respectively, is compensated for by the gudgeons on stern post. Therefore the most correct and also the most convenient length is from after side of rudder post to forward side of stem at load water line.

Where vessels have a very flat floor line a half water line should be taken between base line

1st W.L. and first water plane, and the keel or bottom half-breadth

Keel given a value proportioned to the rise of floor line as in Fig. 5.

FIG. 5.

Rise Line

Required the half-breadth x at the keel for the displacement sheet, where 10 feet is the actual scaled length L, 6” the rise of floor,' 7" the distance from the rise line to first water line at moulded half-breadth of ship and, of course, 13 inches the water line interval, then :

13": 7":: 10 feet : x.
..x = 5.38 feet = bottom breadth.

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(} W.L. interval) (ordinate interval) x 2 (both sides)

=coeff. 35 (cub. ft. of S.W. in a ton) (1.25 x ş) X (10 x {) 2

2.315.

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