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When submerged, a submarine may be considered to be suspended from a single point called the center of buoyancy. The condition of stable equilibrium requires that the center of gravity be located in a vertical line from the center of buoyancy downward. Following this natural law, a suspended body will always assume this position. In the submarine, however, this result does not meet the requirements necessary for stable equilibrium on an even keel.

The conditions necessary for equilibrium at even keel follow the laws of the simple balance. The products of the weights on each side of the support times their moment arms, or distance from the support, must be equal. Before a submarine leaves port, it is trimmed for surface operation by the

19B2. Examples. Some typical examples will demonstrate the method of compensation by moments.

Two torpedoes weigh 6,354 pounds. The distance of the center of gravity of the forward torpedo racks from the center of the auxiliary tank is 94.5 feet. The distance of the center of gravity of the forward trim tank from the center of auxiliary tank is 113.5 feet. The moment of two torpedoes about the center of gravity is 6,354 X 94.5=

Therefore 5,290 pounds are flooded into the forward trim tank from the sea, preserving the fore-and-aft trim, but the total weight of the ship would be 1,064 pounds light. This amount, 1,064 pounds, is therefore flooded into the auxiliary tank from sea and the trim is correct over-all and fore and aft.

The process is the same in more complicated problems as illustrated below A submarine in perfect trim takes aboard 3 torpedoes in the forward racks. Ten thousand gallons of fuel oil are taken into each of the normal fuel oil tanks No. 1 and No. 2. Compensation is desired using the forward trim and auxiliary tanks. Three torpedoes, at 3,177 pounds each, weigh 9,531 pounds; fuel oil weighs approximately 7.13 pounds per gallon; sea water weighs 8.56 pounds per gallon. Using moment arms, it is found that to compensate for the torpedoes, the weight of sea water to be pumped from forward trim tank to sea is:

Therefore, 5,670 pounds {14,300-8.630} are flooded into the auxiliary tank. The weight used in the compensation for the No. 2 fuel tank is found to be:

Therefore, 6,930 pounds are flooded into the forward trim and 7,370 pounds (14,300 - 6,930) into the auxiliary. The vessel now has the same trim fore and aft and over-all. In actual practice, ballast is pumped or flooded to the nearest point that can be read on the gages.

The ballast disposal may be summarized as shown in the following table:

Therefore 6,182 pounds out of the after trim tank will maintain the fore-and-aft trim. However the ship is now light over-all and ballast must be added to the auxiliary tank to compensate for the net loss of 19,069 pounds plus the amount pumped from after trim. Total ballast added to the auxiliary would be 19,069 + 6,182 = 25,251 pounds.

The center ordinate representing the auxiliary tank is laid out in pounds. The axis of abscissas is laid out in thousand-pound feet. The curves for the various tanks are drawn from the data available. A rapid and accurate way to construct the curves is by locating points as far from the center of origin as possible and drawing the lines to the point of origin.

The curves in Figure 19-1 are located as follows:

The moment arm of the forward torpedo tubes is 118 feet from the center of the auxiliary. The number of pounds required to produce a moment of 200,000 pounds is

200,000 / 118 = 1,695

This point is located on the proper ordinate and the line drawn.

200,000 / 113.5 = 1,762

19C2. Use of curve. In using the curve, the point is taken on the center ordinate that represents the weight to be compensated. It is projected to the line representing the location of the change. A line is dropped from this point to the line of the forward or after trim tank and from this intersection back to the center ordinate. The point thus found gives the number of pounds to be added or taken from the trim tank to preserve the fore-and-aft trim. If the line is dropped from the intersection on the tank line to the base line, the moment in thousand-pouted feet is read directly.

As the summary of the changes in ballast to effect compensation is the algebraic sum of the weight added to or taken from the vessel, some care must be used to avoid any confusion of signs. In the use of the curves, all tanks and moment arms to the right of the center ordinate are positive and those to the left of the ordinate are negative. Any weight added to or taken from the vessel, which tends to depress the bow or raise the stern, is positive. If the result of

As the charts are laid out with the bow to the right, any change which tends to rotate the boat clockwise about its center of gravity is positive.

The preceding example in compensation is worked out with the curve as follows:

Three torpedoes with a total weight of 9,531 pounds are placed in the forward racks. Ten thousand gallons of fuel oil are poured into the No. 1 fuel tank and 10,000 gallons additional into the No. 2 fuel tank.

As the 9,531-pound weight of the torpedoes extended from the center ordinate does not meet the torpedo rack curve, the weight may be divided until the intersection is on the chart. Dividing by 5 and extending the 1,906 point to the rack curve and from there down to the base line, the moment is read as +180 thousand-pounds feet. Multiplying by 5 gives 900,000 as the moment of the 3 torpedoes.

The fuel oil taken aboard displaces an equal amount of sea water with the result that 14,300 pounds are removed from each of the two fuel tanks. The 14,300 pounds extended runs off the chart, so 1/5, or 2,860, is extended to intersect with the tank curves. Doing this, and extending to the base line,

Summarizing these moments:

This shows that the ship is light forward and ballast must be flooded into the forward trim tank to produce an 865,000-pound moment.

As 865 is off the chart it is divided by 5. Projecting upward from 173 on the base line to the forward trim tank curve and then left to the center ordinate, the amount is read as 1,520. Multiplying by 5 gives 7,600 pounds as the amount to be flooded into the forward trim tank.

As 9,531 pounds were added to the vessel and 28,600 pounds were lost as a result of the displacement of sea water by the fuel oil, the net loss in over-all weight is 28,600 - 9,531 = 19,069 pounds. Since 7,600 pounds were added to the forward trim tank, 11,469 pounds must be flooded into the auxiliary tank to maintain the original over-all trim.

19D2. Use of curve. This curve is used by projecting the point, indicating the tank in which the change in weight is made, up to an intersection with the percentage curve. This intersection gives the percentage of the

The previous example may be worked out with this curve as follows:

Weight in forward torpedo racks is 9.531 pounds

Weight out (net) No. 1 fuel tank is 11,300 pounds

The forward torpedo rack intersection with the forward trim tank line shows that 83.5 percent of the 9,531 pounds should come out of the trim tank and the remainder, 16.5 percent, out of the auxiliary.

7,958 pounds out of trim tank.

The 14,300 pounds out of No. 1 fuel tank reads 60.5 percent or 8,650 in the trim tank, the remaining 39.5 percent of 5,650 in the auxiliary tank.

The 14,300 pounds out of No. 2 fuel tank reads 48.5 percent or 6,935 pounds in the trim tank, the remaining 51.5 percent or 7,365 pounds in the auxiliary.

If the location of the weight change is in the after part of the ship, compensation is effected by using the after trim tank and the auxiliary tank.