a

stein or stein.

63

Centre of vertical section passing through the keel, and dividing ber of ordinates minus four ; then the second ordinate, Centre of Gravity. the ship into two equal and similar parts, at a certain di- twice the third, three times the fourth, &c. the sum Gravity.

stance from the stern, and altitude above the heel. will be a first term. Then to half the sum of the ex

In order to determine the centre of gravity of the treme ordinates add all the intermediate ones, and the immersed part of a ship's bottom, we must begin with sum will be a second term. Now the first term divided

determining the centre of gravity of a section of the ship by the second, and the quotient multiplied by the inFig. 56.

parallel to the keel, as ANDFPB (fig. 56.), bounded terval between two adjacent perpendiculars, will be the

by the parallel lines AB, DF, and by the equal and distance sought. 66 similar curves AND, BPF.

Thus, let there be seven perpendiculars, whose vaDistance

If the equation of this curve were known, its centre lues are 18, 23, 28, 30, 30, 21, o, feet respectively, and of the centre of

of gravity would be easily found : but as this is not the the common interval between the perpendiculars 20 gravily case, let therefore the line CE be drawn through the feet. Now the sixth of the first term 18 is 3 ; and as from the middle C, E, of the lines AB, DF, and let this line the last term is o, therefore to 3 add 23, twice 28 or

CE be divided into so great a number of equal parts 56, thrice 30 or 90, four times 30 or 120, five times
by the perpendiculars TH, KM, &c. that the arches of 21 or 105; and the sum is 397. Then to the half of
the curves contained between the extremities of any two 18+o, or 9, add the intermediate ordinates, and the
adjacent perpendiculars may be considered as straight
lines, The momentums of the trapeziums DTHF,

7940
sum will be 141.

Now
397 x 20

, or = 59 feet,
TKMH, &c. relative to the point E, are then to be

141 141
found, and the sum of these momentums is to be divided

four inches nearly, the distance of the centre of gravity
by the sum of the trapeziums, that is, by the surface from the first ordinate.
ANDFPB.

Now, when the centre of gravity of any section is The distance of the centre of gravity of the trape- determined, it is easy from thence to find the centre of * Bezord's zium THFD from the point E is={IE X (DF+2TH) gravity of the solid, and consequently that of the bottom

DF+TA* of a ship. nique, art. For the same reason, and because of the equality of the The next step is to find the height of the centre of g

Height of 379 lines IE, IL, the distance of the centre of gravity of gravity of the bottom above the keel. For this pur-the centre

the trapezium TKMH from the same point E will be pose the bottom must be imagined to be divided into of gravity
TEX(TH+2KM)

+IE,or=
HIEX (4TH+SKM) sections by planes parallel to the keel or water-line, above the

keel. TH+KM

TH + KM.

(figs. 57, 58.). Then the solidity of each portion conIn like manner, the distance of the centre of gravity of

tained between two parallel lines will be equal to half
the trapezium NKMP from the point E will be the sum of the two opposed surfaces multiplied by the
FIE X (KM + 2NP) TEX(7KM+8NP)

distance between them ; and its centre of gravity will
+21E, or
KM+NP

KM+NP

be at the same altitude as that of the trapezium abcd, &c.

(fig. 58.), which is in the vertical section passing Now, if each distance be nultiplied by the surface of through the keel. It is hence obvious, that the same the corresponding trapezium, that is, by the product of

rule as before is to be applied to find the altitude of the half the sum of the two opposite sides of the trapezium

centre of gravity, with this difference only, that the into the common aluitnde IE, we shall have the momen

word perpendicular or ordinate is to be changed into

section. Hence the rule is, to tlie sixth part of the
tums of these trapeziums, namely, :|Epx (DE+2TH), lowest section add the product of the sixth part of the

TEX (4 TH + 5 KM) 1E (7 KM + 8NP), uppermost section by three times the number of sections
&c. Hence the sum of these momentums will be 5 minus four; the second section in ascending twice the

El X (DF+TH + 12 KM +18NP + 24QS+14 third, three times the fourth, &c. the sum will be a
AB). Whence it may be remarked, that if the line

first term.

To balf the sum of upper and lower sec-
CE be divided into a great number of equal parts, the

tions add the intermediate ones, the sum will be a second
factor or coefficient of the last term, which is here 14, term. Divide the first term by the second, and the
will be = 2+3 (1–2) or 3 n--4, n being the number quotient multiplied by the distance between the sections
of perpendiculars. Thus the general expression of the will give the altitude of the centre of gravity above the
sum of the momentums is reduced to IE X (GDF+

keel.

With regard to the centre of gravity of a ship, whe-
TH + 2 KM + 3 NP + 4 QS +, &c. - +

3N-4
4

tber it is considered as loaded or light, the operation 6

becomes more difficult. The momentum of every difXAB.

ferent part of the ship and cargo must be found sepaThe area of the figure ANDFPB is equal to JE

rately with respect to a horizontal and also a vertical X(DF + TH + KM + NP +, &c..... +1 AB); plane. Now the sums of these two momentums being hence the distance EG of the centre of gravity G

divided by the weight of the ship, will give the altitude from one of the extreme ordinates DF is equal to 167

of the centre of gravity, and its distance from the ver

39-4 Rule for IEX (DF+TH+2KM+3NP+,&c.

X AB)

tical plane; and as this centre is in a vertical plane pasthe dis

6

sing through the axis of the keel, its place is there fore ADF+11+KM +NP+, &c. +? AB determined. In the calculation of the momentums, it the centre Whence the following rule to find the distance of the of gravily

must be observed to multiply the weight, and not the from one

cen're of gravity G from one of the extreme ordinates magnitude of each piece, by the distance of its centre of the ex. DF. To the sixth of the first ordinate add tlie sixth

of gravity. of the last ordinate multiplied by three times the num- A more easy method of finding the centre of gravity dinates.

of

2

а

a

tance of

trenie or

a

A mecha

thod for

gravity of

skip

Centre of of a ship is by a mechanical operation, as follows: Con- other convenient point in the middle line ; and another Centre of Gravity. struct a block of as light wood as possible, exactly similar line is to be drawn on the block parallel to the line sus- Gravity.

to the parts of the proposed draught or ship, by a scale pending it, as before. Then the point of intersection 60

of about one-fourth of an inch to-a foot. The block is of this line with the former will give the position of the nical me

then to be suspended by a silk-tbread or very fine line, centre of gravity on the block, which may now be laid

placed in different situations until it is found to be in a down in the draught. escertaine state of equilibrium, and the centre of gravity will be jag the

pointed out. The block may be proved by fastening CHAP. V. Application of the preceding Rules to the centre of

the line which suspends it to any point in the line join- Determination of the Centre of Gravity and the
ing the middles of the stem and post, and weights are Height of the Metacenter above the Centre of Gra-
to be suspended from the extremities of this middle line

vity of a ship of 74 Guns.
at the stem and post. If, then, the block be properly
constructed, a plane passing through the line of suspen- In fig. 59. are laid down the several sections in a Fig. 39
sion, and the other two lines, will also pass through the horizontal direction, by planes parallel to the keel, and
keel, stem, and post. Now, the block being suspended at equal distances from each other, each distance be-
in this manner from any point in the middle line, a line ing 10 feet o inches 4 parts.
is to be drawn on the block parallel to the line of sus-
pension, so that the plane passing through these two

I. Determination of the Centre of Gravity of the Up-
lines may be perpendicular to the vertical plane of the

per Horizontal Sectior.
ship in the direction of the keel. The line by which To find the distance of the centre of gravity of the
the block is suspended is then to be removed to some plane 8 g o G from the first ordinate 8 g.

a

a

.

.

I
3897 3
Now

3897 • 25
XIO
4=

X 10.03=70.5.
554 4 3

554 · 25
Hence the distance of the centre of gravity of double the plane 8 go G from the first ordinate, Feet.

70.5
Distance of this ordinate from the aft side of stern-post,

13.5

8g, is

84.0

Distance of the centre of gravity from the aft side of post,
Distance of the centre of gravity of double the trapezium AR g 8 from its ordinate AR,
Distance of this ordinate from the aft side of the stern-post,

8.42 0.58

9.0

Distance of the centre of gravity of this plane from the aft-side of the stern-post,
Distance of the centre of gravity of double the trapezium G o go from its ordinate G 0,
Distance of this ordinate from the aft-side of the post,

5.44 153.78

159.22

Distance of the centre of gravity of this trapezium from the aft side of the post,
Distance of the centre of gravity of the section of the stern-post from the aft part of the post,
Distance of the centre of gravity of the section of the stern from the ast-side of the post,

Pp 2

0.29 169.76

The

Centre of
Gravity.

Centre of The areas of these several planes, calculated by the common method, will be as follow :

Gravity. 5558.90 for that of the plane, and its momentum 5558.9 x 84

466947.6000 199.13 for that of double the trapezium AR g 8, and its momentum 199.13 X9 = 1792. 1 700 214.59 for that of double the trapezium Gogy, and its momentum 214.59 X 159.22= 34167.0236 0.77 for that of the section of the stern-post, and its momentum 0.77 x 0.29 =

0.2233 0.77 for that of the section of the stem, and its momentum 0.77X169.76 =

130.7152
5974.16 Sum

503037-7321
Now
503037.7321

= 84.2, the distance of the centre of gravity of the whole section from the aft side of
5974.16
the stern-post.

II. Determination of the Centre of Gravity of the Second Horizontal Section.

To find the distance of the centre of gravity of double the plane 8 fn G from its first ordinate 8 f.

273 2 3 546 4 6

3698 5 3

523 11

6
Hence the distance of the centre of gravity of double the plane 8 fn G from its first ordinate 8 n is
3698 5 3 3698.43
X 10.0.45
X 10.03=

70.79
6

523.95
Distance of this ordinate from the aft side of the stern-post

13.5

523 11

Distance of the centre of gravity of the above plane from the aft side of post

84.29

Distance of the centre of gravity of double the trapezium ARf8 from its ordinate AR
Distance of this ordinate from aft side of stern-post

8.38
0.57

Distance of the centre of gravity of the trapezium from the aft side of the

post

8.95

Distance of the centre of gravity of the trapezium before the ordinate G n from that ordinate
Distance of that ordinate from the ast side of the post

5.74 153.78

Distance of the centre of gravity of the trapezium from the aft side of the post

159.52

Distance of the centre of gravity of the section of the stern-post from the aft side of the post
Distance of the centre of gravity of the section of the stem from the aft side of the post

0.29 269.76

The

Centre of
Gravity.

Centre of
Gravity.

The areas of these several planes being calculated, will be as follow:

5255.22 for that of the plane 8 fn G, and its momentum 5255.22 x 84.29 =
153.11 for that of double the trapezium AR f8, and its momentum 153.11 X 8.95 =
182.40 the area of the trapezium before, and its momentum 182.40 x 159.52=

0.77 the area of the section of the sternpost, and its momentum 0.77%0.29 =
0.77 the area of the section of the stem, and its momentum 0.77X169.76 =

442962.4938

1370.3345 29096.4480

0.2233 130.7152

5592.27 Sum

473560.2148

Now 473560.2148

= 84.68, the distance of the centre of gravity of the whole section from the aft-side of the

59.52.27 stern-post

III. Determination of the Centre of Gravity of the Third Horizontal Section.

Distance of the centre of gravity of double the plane 8 e m G from its first ordinate 8 e.

3347.04 x 10.03=

6Ҳio

Hence the distance of the centre of gravity of double the plane 8 em G from its first ordinate 8e is =
4347 o 6
4=

71.44
469 10

469.87
Distance of this ordinate from the aft side of the post

13.5
Hence the distance of the centre of gravity of this plane from the aft side of the post is

84.94 Distance of the centre of gravity of double the trapezium AR e 8, from its ordinate AR

8.03 Distance of this ordinate from the aft side of the post

0.58

8.61

Distance of the centre of gravity of this trapezium from the aft side of the post
Distance of the centre of gravity of the foremost trapezium from its ordinate Gm
Distance of this ordinate from the aft side of the post

5.19 153.78

Distance of the centre of gravity of this trapezium from the aft side of the post

158.97

Distance of the centre of gravity of the section of the post from the aft side of the post
Distance of the centre of gravity of the section of the stem from the aft side of the post

Centre of
Gravity.

Centre of
Gravity.

The areas of these several planes will be found to be as follow :
4712.7961 for that of double the plane 8 e m G, and its momentum 4712.7961 X 84.94=

93.84 the area of double the trapezium AR 3 e 88, and its momentum 93.84 x 8.615
131.1 for the area of foremost trapezium, and its momentum 131.1 X158.97=

0.77 the area of the section of the post, and its momentum 0.77 x 0.29=
0.77 the area of the section of the stem, and its momentum 0.77% 169.76=

400304.9007

807.9624 20840.967

0.2233 130.7152

4939.2761 Sum

422084.7706

Now 422084.7706

4939.2716

85.45, the distance of the centre of gravity of the whole section from the aft side of

the post.

IV. Determination of the Centre of Gravity of the Fourth Horizontal Section.

Distance of the centre of gravity of double the plane 8 d/ G from its first ordinate 8 d.

Hence the distance of the centre of gravity of double the plane 8 d 1G from its first ordinate 8 d is
2883 11

2883.916
XIO
4=

X 10.03=
402 6 9

71.85 402.56

Distance of this ordinate from the aft side of the post

13.5

Distance of the centre of gravity of the place from the aft side of the post

85-35

Distance of the centre of gravity of double the trapezium AR d 8 from its ordinate AR
Distance of this ordinate from the ast side of the post

7.89 0.58

Distance of the centre of gravity of the trapezium from the aft side of the post

8.47

Distance of the centre of gravity of the foremost trapezium from its ordinate GI
Distance of this ordinate from aft side of the post

4.83 153.78

158.61

Distance of the centre of gravity of the trapezium from the aft side of the post
Distance of the centre of gravity of the section of the post from its aft side
Distance of the centre of gravity of the section of the stem from the aft side of the post

0.29 169.76

The