Example of Analyzation according to the method of Mr. Scott Russell. Diameter of paddle-wheel outside floats, as taken from the draw ing, in feet 36.5 Diameter of wheel to journals in feet or effective diameter. 32 Indicated horse-power 2207.98 Consumption of coals, in lbs. per hour... Coefficient of fineness of midship section or 653÷Bxd= performance VX D÷I. H. P. W in this formula means consumption of coals in cwt. per hour. To find the velocity of the paddle-wheel, the effective diameter has been multiplied by 3.1416; this product has been multiplied by the number of revolutions per minute, this product again by 60, and the last product has been divided by 6082-66 to find as quotient the velocity in knots per hour. Slip in knots per hour equal to 14.29-11.22 Dip of paddle-board, measured from lower edge, in feet Area of paddle-race 12 × 9 × 2, in square feet Area of midship section: Area of paddle-race :: 1: = = 3:07 5.19 0.27 9 12 216 .33 7936 0-869 body or 3979 × 35÷Bxdx L= ends or D in cubic feet÷xL= 0.5512 0.6339 160-62 5005 419-66 14.42 14.29 Resistance due to area of paddle-race equal to speed of slip, in feet per second, multiplied by the area of paddle-race, or 5.192 x 216 = 5818 lbs. Resistance due to length of paddle-race equal to resistance due to area of paddle-race multiplied by the speed of the ship÷speed of slip, or 21,263 lbs. Coefficient of diminished resistance.-This coefficient belongs to a puremathematical wave-line bow. The question offers itself, what is the length of this bow? The length of the bow of the ship can be denoted 1. By the speed, 2. By the place of half the beam in the light water-line, 3. By the coefficient of fineness of ends, and 4. By the actual length as found by the lines. For the afterbody the length would be of the above quantities, with the exception of 4, where it is given by the lines. 1, 2, 3, 4 worked out for the 'Atrato' would give a length of bow as computed by The quantities which differ immensely. But whatever the length of forebody, afterbody, and middlebody may be as computed above, the three together must form the given displacement with the given draught of water. wave-line method supplies formula by means of which the exact length of bow answering both conditions may be found; and in order to prove which of the four quantities is correct we proceed as follows: Let P denote the coefficient of fineness of midship section, q the coefficient of fineness of body, the coefficient of fineness of ends equal to q÷p; then Dividing this by the parallelopiped, BLd, will give the coefficient of fineness of body, or 7, ·51 +·51′ +·19635B-l"=?L=rL; but 1+l'=L—l"; hence 5L+5l′′ +·19635B=rL, Р from which equation l', or the length of the middlebody, may be determined; and having the length of the middlebody deducted from the total length of the ship, six-tenths of the remainder will give the length for the forebody, and four-tenths of the remainder will give the length of the afterbody. Working the above formula out for the Atrato' we shall get •5 x 336.5+57"+·19635 × 40·92 = ·6339 × 336·5, hence length of forebody equal to 6(336.5-74.04)=157.476, •4(336·5—74·04)=104.984. It will be seen from these quantities that already twice as long a bow has been obtained as is necessary for a speed of 11.22 knots. Let us now test these quantities for the displacement. A curve is appended at the end of this Report, by which for any given speed in statute miles the length of bow and stern might be measured. 1469+979-3+149.9+1381=3979.2 tons, or just two-tenths of a ton more than the actual displacement. Now there is no length of forebody, afterbody, and middlebody possible that will fulfil the conditions required; and it would therefore be wrong to compute the length of forebody in any other way than through means of the coefficient of fineness of ends. The 'Atrato' has no actual middlebody; but we see that she really could have had a parallel middlebody of 74 feet without in the least injuring her qualities. The length of forebody, afterbody, and middlebody, through means of the mentioned formulæ, have been calculated for several ships, and the result has been appended in a Table which follows the Table of Analysis according to Mr. Scott Russell's method. It will easily be seen that this length of middlebody varies with the draft of water; the lighter the vessel is, the shorter the middlebody, and the deeper the vessel the longer the middlebody, the different coefficients of fineness necessarily becoming smaller when light, and larger when laden. The coefficient of diminished resistance is therefore (40.92÷157-476)* =0.0675. Resistance due to ship's way, equal to area of midship section multiplied by the square of the speed of the ship in feet per second; and this product multiplied by the coefficient of diminished resistance gives 15850 pounds' resistance due to ship's way. Girth at midship section in feet. 62.80 This item, when the lines are in hand, is not immediately necessary, although, when such is the case, that girth must be measured, in order to lay down the surface of the skin; but in the absence of the lines of the vessel the girth at the midship section becomes a great function of the surface of the skin. The 66 feet, as above, has been actually measured from the body-plan; but where a certain proportion exists between the beam and the draft of water, the girth may be found to a close approximation by multiplying the beam plus twice the draft with a certain coefficient, which coefficient may be found from a Table at the end of this Report, in which the girths at the midship section have all been found from the lines of the ship. In the case of the 'Atrato,' where the proportion between breadth and draft is as 1 : 448, the coefficient would be like H.M.S.Warrior' and 'Achilles,' or the mean between the two would give ·8; and this multiplied by 40.92 +2×18·35=62-09, or 71 feet shorter than is actually the case. The cause of this is that the Warrior' and Achilles' both have more rise of floor than the Atrato,' and this rise of floor may be judged from the coefficient of fineness of midship section; therefore, by using a little caution in the use of this Table, the girth at the midship section may be found to a very close approximation. Surface of skin in square feet 17233 This surface has been exactly measured from the lines of the ship; and where this has not been possible, the coefficient of 77 may be judiciously used from a comparison with other ships, of which the surface of skin has been laid down and calculated, and given at the end of this Report. Resistance due to skin equal to the surface of the skin multiplied by the square of the speed of the ship, in knots, and the product divided by 100, supposing that the skin of the ship is clean and smooth, or =27300 pounds skin resistance. Total resistance equal to 15850+27300=43150 pounds. Horse-power required for ship's way equal to the resistance due to ship's way multiplied by the speed of the ship, in feet per minute, and the product divided by 33000, or =546 horse-power for ship's way. Horse-power required for skin-resistance equal to the resistance due to skin multiplied by the speed of the ship, in feet per minute, and the product divided by 33000, or = 940 horse-power for skin-resistance. Horse-power required for slip equal to the total resistance multiplied by the slip, in feet per minute, and the product divided by 3300, or =406 horse-power for slip. Total horse-power required 546+940+406= 1892 2207-1892=315 •24 Percentage of total horse-power employed in driving the skin •42 Percentage of total horse-power expended on slip.. .13 and propeller •14 Consumption of coals per nominal horse-power, per hour... 9.9 Consumption of coals per indicated horse-power, per hour 3.6 |