CHAPTER XVII. On the Longitudinal Metacentre of a Ship. THE effect of shifting a large weight in a fore-and-aft direction, of putting a large weight on board, or of taking a large weight out of a ship, must be dealt with by the naval architect, and he must take into account the form of the transverse sections of the ship between the original and new water-lines, and the calculations must be made by a tentative process. With this I shall not now occupy myself; my object will be to show that the effect of a moderate weight on the trim of a ship may be readily calculated, when it is either shifted in a fore-and-aft direction, or put on board, or taken out of that ship. To proceed first with the case where a weight in the ship is shifted in a fore-and-aft horizontal direction. Let ABWL, Fig. 16, represent a ship floating at the water-line, WL; B her centre of buoyancy; G her centre of gravity; BGM the vertical line through these points; and M (a point in BGM) the longitudinal metacentre-all corresponding to the above draught of water. Let now some of her weights be shifted in a horizontal direction -say further aft. The effect of this removal is that the centre of gravity of the ship (G) moves horizontally aft to G1, and the centre of buoyancy from B to some point B1; and when the ship has reached the new position of equilibrium (WL, being the new water-line) and a vertical line BG be drawn through the new centres of buoyancy and gravity, this line will cut the original vertical BGM in some point M, and will make with it the angle BMB equal to the angle LSL-between the two waterlines. Now, when the weight moved is very small, and also the distance through which it has been moved is very small, the angle between the water-lines WL and WIL1, and therefore between the verticals BM and BM, is very small also, and the point M in which the two verticals intersect is called the longitudinal “metacentre," corresponding to the water-line, WL. Through B draw BQ perpendicular to BM, cutting BM in the point Q. BQ=BM tan I, if I represent the inclination of the two verticals. Then Since the displacement of the ship to the water line WL, is the same as that to the water-line W1L1, take away the common part AWSLB; and the wedge WSW1, remaining in one case, is equal to the wedge LSL1, remaining in the other case. Also by a well-known property of the centre of gravity of bodies, since the wedge LSL, may be conceived to have been shifted to the position WSW1, therefore BQ (the distance through which the centre of buoyancy moves in the direction WL) is equal to either of the wedges WSW1 or LSL1 multiplied by the horizontal distance between their centres of gravity, divided by the whole volume of the displacement, or BQ, which is equal to BM tan I = Moment of wedge LSL1 about S+Moment of WSW1 about S Volume of the displacement 1 1 : To find the moments of the wedges WSW1 and LSL1, about S. Let PQMN, Fig. 17, represent the section of the wedge LSOL, by a plane perpendicular to the load-water-section WL, and also to the longitudinal plane which divides the ship into two equal and symmetrical parts. Let MN be the section of the ship's side by this plane. Let the distance SP of this plane from the intersection of the load-water-sections be represented by x, and PM by y. Let also another plane, parallel to this one, be drawn at a very small distance PP1, from it, and let PP, be represented by dx; let M1N1 be the section of the ship's side by this plane. Then since PQ and PP1 are both very small, the volume of the small prism PM, is nearly equal to PQ.PP1y. Now PQ = x tan I where I is the very small inclination of the load-water-sections. Therefore the volume of the prism PM1 is equal to tan I y x dx. But y x dx is the moment of the small area PP,MM1 about the axis SO; therefore the volume of the small prism PM, is equal to the moment of the small area forming its base on the load-watersection about the axis SO multiplied by the tangent of the small inclination of the water-lines. Again, if the whole wedge LSL1 be imagined to be cut up into a very large number of small prisms similar to PM1, their sum, or the volume of the whole wedge, would be equal to the sum of the moments of all the very small areas into which the load-water-section would be divided about the axis SO multiplied by tan I; that is, the volume of the wedge LSL1, is equal to the moment of the fore-part SOL of the loadwater-section about SO. 1 In the same manner it can be shown that the volume of the wedge WSW, is equal to the moment of the after-part WOS of the load-water-section about SO. But the wedge LOSL, has been shewn to be equal to the wedge WOSW1, therefore the moment of the part SOL of the load-water-section before SO is equal to the moment of the part WOS abaft SO; that is, the two watersections intersect in a line passing through the centre of gravity of either. Again, the moment of the small prism PM1 about SO is equal to its volume multiplied by its distance from SO 1 tan I y x dx x = tan I y x2 dx = moment of inertia of the small portion PN1 of the load-water-section forming its base about the axis SO multiplied by tan I; and the moment of the whole wedge LSOL1 is equal to the sum of the moments of inertia of all such small portions into which the water section is divided; that is, it is equal to the whole moment of inertia of the part LOS of the load-water-section before SO about the transverse axis passing through its centre of gravity. This may be represented by the symbol tan I y2 x d x. In the same manner it can be shown that the moment of the wedge WOS about OS is equal to the moment of inertia of the after-part WOS of the load-water-section about SO multiplied by tan I. This may be represented by tan IW y x2 dx. Therefore the moment of the two wedges LOSL1 and WOSW1 about OS is equal to the moment of inertia of the whole load-watersection about SO multiplied by tan I; observing that the axis OS passes through the centre of gravity of the load-water-section. Therefore the value of BQ above given Moment of inertia of the load-water-section about SO × tan I Volume of the displacement But BQ=BM tan I, therefore dividing each side of the equation by tan I, BM Moment of inertia of the load-water-section about SO Σ y x2 d x + ZW y x2 d x where D represents the displacement of the ship. It would be very inconvenient in practice, if it were necessary, first, to find the centre of gravity of the load-water-section, and then to arrange the ordinates, for making the calculations, with reference to this point; but if the moment of inertia be obtained with reference to any axis, the moment of inertia about an axis parallel to it through the centre of gravity can be readily found. Let y x2 dx + y2 de represent the moment of inertia of the load-water-section about a transverse axis, distant a feet from the centre of gravity; and let X represent the distance of the ordinate y from the centre of gravity, then for one part of the water-section x = x a, and for the other part of the watersection x=x+a. W + y x2 dx ...y x2 dx + = dx} = = y (x—a)2 dx 2y (X+a) da (taken between the same limits, =zy x2 dx-2a yx dx+a2 zy dx + 2y x2 dx 2 ay x dx + a2 zy d x2 = zy3 dx + Σy x2 dx+2a {y x2 dx-Σy x dx} +a2 { Σy dx + Σy dx}, all taken between the same limits respectively. Now the two first terms of the last line of the above equation represent the moment of inertia of the load-water-section about a transverse axis through the centre of gravity: the next two terms the difference of the moments of the fore and after portions of the load-water-section about an axis passing through the centre of gravity, which is zero: and the last two terms the whole area of the load-water-section multiplied by the square of the distance between the axes. The moment of inertia of the water-section, therefore, about a transverse axis through its centre of gravity is equal to its moment of inertia about any other transverse axis, diminished by the product of its area and the square of the distance between the axes. To find the moment of inertia of the load-water-section about any transverse axis: Like all other calculations in naval architecture, this is found by obtaining a curvilinear area, the ordinate to which at any point represents the product of the ordinate of the load-water-section at the same point, and the square of its distance from the axis of moments. Let WL, Fig. 18, be taken as the axis of co-ordinates, where WL is equal to the length of the ship, and let any ordinate SO be taken as the axis of moments, the parts before and abaft SO being divided into equal intervals, and their number such as Simpson's rules can be applied to them. Let SNL represent the curvilinear area required, then any ordinate PN represents the ordinate to the load-water-section at the point P multiplied by SN2. Let m represent the common interval between the ordinates, then the several ordinates of the load-water-section must be multiplied by 0, m3, (2m)3, (3m), (4m)3, etc., to obtain the ordinates of the area SNL. m2 But, since m is common to all, this may be suppressed until the final result is obtained; and the ordinates of SNL would, therefore, represent the ordinates of the load-water section multiplied by the square of the number of intervals between the the ordinate and the axis of moments SO. It will, however, be more convenient to proceed in the manner hereafter described, since in order to obtain the curvilinear area required, we can also obtain at the same time the area of the load-water-section, and the distance of its centre of gravity from the axis of moments. This will be best described by means of the following table, in which the necessary calculations are made for the armor-plated ship Warrior, at her load-draught of water. The ordinate No. 12, of the load-water-section is taken as the axis of moments; and the ordinates of the water-section before and abaft it are so arranged that one of Simpson's Rules can be applied to find the area. Taking the part before ordinate No. 12; in the first column are written down the numbers of the several ordinates. In the second column, the lengths of the corresponding ordinates of the water-section. The third column contains the multipliers according to Simpson's rule for finding the area. The fourth column contains the products of the ordinates and these multipliers: and it is evident that the sum of the quantities in this column, if will give one-half the area of the fore-part of M multiplied by "g, 39 m the load-water-section. The fifth column contains the number of the intervals between the several ordinates and the ordinate No. 12. The sixth column contains the continued product of the several quantities in No. 4 column, and the respective numbers in No. 5 column. It is clear, also, that the sum of the quantities in this column, if multiplied by m, and again by the result will be the moment of one-half of the load-water-section about No. 12 ordinate. The seventh column also contains the number of intervals between the several ordinates and the ordinate No. 12. The eighth column contains the continued products of the quantities in the sixth column, and the numbers opposite to them in the 3' |