2.74 22.34 22.36 11.88 23.42 16.00 57. 48 5.50 Jeta.' square of the distance between the two axes will equal the moment eat the about ordinate 5. I equalplane, and consequently at the B.M. without the labor of the middle = foregoing calculation by multiplying the Length by the Breadth by à coefficient, which coefficient will be determined by a and Apps selected from the table given on page 48. By referring to this e are & table, we find for a (value .694) that the coefficient “ į" (inertia coefficient) is equal to .0414, whence we get I=LX B3 Xi = 100 7154 2583.7 7191 =.0416. 100 X 123 angles to the previous case, viz., at right angles to the centre line 8.00 : through the centre of gravity of water plane, which will be where the original and new water planes cross one another in a longitu- 11 of Water Plane about its C.G.. Volume of Displacement The moment of inertia about the midship ordinate we shall call 1.00 5.11 1 6.6 7.07 1.38 3.70 Je to ded br 1 ne, we water head specti pivota ship's produ there height be cal Area of water plane = 62.41 x (x 10) X 2. =832.14 square feet. Distance of centre of flotation abaft ordinate 5 (67.57 — 52.41) 10 = 2.42 feet. 62.41 Moment of inertia of water plane about ordinate 5 = 324.13 X (3 10) X 102 2 = 432,172 =1. Moment of inertia of water plane about axis through its centre of flotation. = 432,172 — (832.14 x 2.422) = 427,304 = 11. Longitudinal metacentre above C.B. I 427,304 = 165 feet = Longitudinal B.M. RODUCTS FOR IOMENTS OF VERTIA An excellent approximate formula for the longitudinal B.M. is given by J. A. Normand in the 1882 transactions of the I.N.A. Taking the symbols we have been using : A2 X I L.B.M. = .0735 B XV Applying this formula to the vessel with which we are dealing, we find : 832.142 x 100 L.B.M. = .0735 = 164.12 feet. 12 X 2583.7 which is a very close appróxrimation to the calculated result of 165 feet. We may also use the aphr-eximate formula which we applied in the case of the transverse B.M. altered to suit the new axis with a modified coefficient, as : L.B.M. = 13 x B xii. Moment to Change Trim (M1). D= Weight of ship including weight P = 73.82 tons. P = 5 Tons. 1 = 50 feet (distance moved). L = 100 feet (length of vessel). entre In the figure we have the centre of gravity G to Gı, and the centre of buoyancy from B to B1, due to the shifting of the weight P forward for a distance represented by I, giving a moment PX1 D WW1 The new water line is at WiLi and BiG1 are in the same verti cal and at right angles to it, and the point of intersection of the original and new water line at “0” equal to the centre of gravity (flotation) of water plane, therefore the triangles GMG1, WOW, and LOL1, are of equal angle, so that GG1 LLI WW1 + LLI. change of trim WW1 + LL WO +LO but we have seen that GM X change of trim ΡΧί D PXL XL feet. D X GM = 2.116 feet = 24] inches. Calling this change of trim 24 inches, and assuming that the point of intersection “0” is at the centre of the length, we should have the stem immersed 12 inches and the stern raised 12 inches from the original water line, the sum of these figures equalling the total change. Moment to Alter Trim One Inch (M”). From the foregoing it will be seen that the total change of trim being known for a given moment, inversely we may get the amount necessary to alter the trim for one inch only, this being a convenient unit with which to calculate changes of trim when a complexity of varying conditions are being dealt with. As we have seen Px1=Mthe moment to change trim, and In designing preliminary arrangements of vessels, it is necessary that we should know fairly accurately the moment which it will take to alter the trim one inch (M") to enable us to arrange the principal weights in the ship, and the varying effects on the trim consequent on their alteration in position or removal. For this purpose a close approximation to this moment (M'') is desirable and may be calculated from Normand's formula as follows: M" .0001726, or on2 x 30.9 B B Where A2=the square of the water plane area, and B=the greatest breadth of water plane. Applying this approximate formula to the foregoing example, we have: 832.142 x .0001725 = 9.95 foot-tons, 12 as against 9.84 foot-tons found by actual calculations, a difference too insignificant to affect noticeably the change in trim. This inoment is useful to have for various draughts, and consequently should be calculated for light and load conditions, and for one or two intermediate spots and a curve of M“ run on the usual sheet of “ Curves of Elements." M" = |