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 7940 Now , or = 59 feet, 141 141 four inches nearly, the distance of the centre of gravity 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 +IE,or= 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 distance between them ; and its centre of gravity will 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 TEX (4 TH + 5 KM) 1E (7 KM + 8NP), uppermost section by three times the number of sections El X (DF+TH + 12 KM +18NP + 24QS+14 third, three times the fourth, &c. the sum will be a first term. To balf the sum of upper and lower sec- tions add the intermediate ones, the sum will be a second keel. With regard to the centre of gravity of a ship, whe- 3N-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 vity of a ship of 74 Guns. I. Determination of the Centre of Gravity of the Up- per Horizontal Sectior. a a . . I 3897 • 25 X 10.03=70.5. 554 · 25 70.5 13.5 8g, is 84.0 Distance of the centre of gravity from the aft side of post, 8.42 0.58 9.0 Distance of the centre of gravity of this plane from the aft-side of the stern-post, 5.44 153.78 159.22 Distance of the centre of gravity of this trapezium from the aft side of the post, Pp 2 0.29 169.76 The Centre of 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 503037-7321 = 84.2, the distance of the centre of gravity of the whole section from the aft side of 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 70.79 523.95 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 8.38 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 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 0.29 269.76 The Centre of Centre of 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 = 0.77 the area of the section of the sternpost, and its momentum 0.77%0.29 = 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 = 71.44 469.87 13.5 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 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 0.29 169.76 Centre of Centre of The areas of these several planes will be found to be as follow : 93.84 the area of double the trapezium AR 3 e 88, and its momentum 93.84 x 8.615 0.77 the area of the section of the post, and its momentum 0.77 x 0.29= 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 2883.916 X 10.03= 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 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 4.83 153.78 158.61 Distance of the centre of gravity of the trapezium from the aft side of the post 0.29 169.76 The |