This investigation was undertaken to determine the effect upon the compass of a pronounced change in trim due to the flooding of a large compartment or of a number of compartments at one end of the ship, resulting in an abnormal trim, in which condition it might become necessary to work into port. The derivation of the change of trim equation follows similar lines to the derivation of that for heel, as given on page 131, et seq, Admiralty Manual, edition of 1901. In the case of change of trim, however, the problem is more complicated and the resultant equation less simple than for heeling, for reasons which will appear as the work progresses.
Let X, Y and Z represent, as they do when the deck is horizontal, the two first the components of the earth's magnetic force alone to head and to starboard, in the plane of the horizon, and the third the component vertically downward.
Let Xtand Zt represent the components of the earth's magnetic force alone, the first to head in a plane parallel to the deck when inclined, and the second to keel when inclined, and let t represent the angle at which the deck is inclined to the horizontal due to the abnormal trim.
Let X't and Z’t represent the components of the combined magnetic force of earth and of inclined ship, the first to head, in a plane parallel to the deck, and the second to keel.
Let X', Y' and Z' represent, with the ship down t degrees by the head, the components of the combined magnetic force of earth and ship, the first two to head and to starboard in the horizontal plane, and the third vertically downward.
Then we desire to obtain expressions for X', Y' and Z', in terms of X, Y and Z, when the ship is down by the head, which expressions will represent for the inclined condition what Poisson's equations do for the ship in the horizontal trim.
Poisson’s equations are:
- X’ = X + aX + bY + cZ + P.
- Y’ = Y + dX + eY + fZ + Q.
- Z’ = Z + gX + hY + kZ + R.
If now we rotate the ship about the horizontal, thwartships line through the compass, by dropping the bow, we get, by the ordinary rules for the transformation of equations under such conditions:
- Xt = X cos t + Z sin t.
- Zt = Z cos t — X sin t.
- X’t = Xt + axt + bY + cZt+ P.
- Y’ = Y + dXt + eY + ƒZt + Q.
- Z’t = Zt + gXt + hY + kZt + R.
- X’ = X’t cos t — Z’t sin t.
- Z’ = Z’tcos t + X’t sin t.
From (4), (5) and (6) we get:
- X’t + X cos t + Z sin t + aX cos t + aZ sin t + bY + cZ cos t — cX sin t + P.
From (4), (5) and (7) we get:
- Y’ = Y + dX cos t + dZ sin t + eY + ƒZ cos t — ƒX sin t + Q.
From (4), (5) and (8) we get:
- Z’t = Z cos t — X sin t + gX cos t + gZ sin t + hY + kZ cos t — kX sin t + R.
From (9), (11) and (13) we get:
- X’ = X cos2 t + Z sin t cos t + aX cos2 t + aZ sin t cos t + bY cos t + cZ cos2t — cX sin t cos t + P cos t — Z sin t cos t + X sin2t — gX sin t cos t — gZ sin2t — hY sin t — kZ sin t cos t + kX sin2t — R sin t.
From (10), (11) and (13) we get:
- Z’ = Z cos2t — X sin t cos t + gX cos2t + gZ sin t cos t + hY cos t + kZ cos2t — kX sin t cos t + R cos t + X sin t cos t + Z sin2t + aX sin t cos t + aZ sin2 t + bY sin t + cZ sin t cos t — cX sin2t + P sin t.
Now in (14) and (15) substitute for cos t its value from
cos2t = 1 — sin2t
and we get—
From (14):
- X’ = X — X sin2 t + Z sin t cos t + aX — aX sin2 t + aZ sin t cos t + bY cos t + cZ — cZ sin2t — cX sin t cos t + P cos t — Z sin t cos t + X sin2t — gX sin t cos t — gZ sin2t — hY sin t — kZ sin t cos t + kX sin2t — R sin t.
From (15):
- Z’ = Z — Z sin2t — X sin t cos t + gX — gX sin2t + gZ sin t cos t + hY cos t + kZ — kZ sin2t — kX sin t cos t + R cos t + X sin t cos t + Z sin2t + aX sin t cos t + aZ sin2t + bY sin t + cZ sin t cos t — cX sin2t + P sin t.
Simplifying the above we get—
From (16):
- X’ = X + (b cos t — h sin t) Y + {a—(c + g) sin t cos t — (a — k) sin2t} X + {c + (a — k) sin t cos t — (c + g) sin2t} Z + P cos t — R sin t.
From (17):
- Z’ = Z + (h cos t + b sin t) Y + {g + (a — k) sin t cos t — (c + g) sin2t}X + {k + (c + g) sin t cos t + (a — k) sin2t}Z + P sin t + R cos t.
And from (18), (12) and (19) we get:
- X’ = X + {a — (c + g) sin t cos t —(a — k) sin2 t}X + (b cos t — h sin t) Y + {c + (a — k) sin t cos t — (c + g) sin2t}Z + P cos t — R sin t.
- Y’ = Y + (d cos t — ƒ sin t) X + eY + (d sin t + ƒ cos t) Z + Q.
- Z’ = Z + {g + (a — k) sin t cos t — (c + g) sin2t}X + (b sin t + h cos t) Y + {k + (c + g) sin t cos t + (a — k) sin2t}Z + P sin t + R cos t.
These are the form of:
- X’ = X + atX + btY + ctZ + Pt.
- Y’ = Y + dtX + etY + ƒtZ + Qt.
- Z’ = Z + gtX + htY + ktZ + R.
Where
- at= a — (c + g) sin t cos t — (a — k) sin2t.
- bt = b cos t — h sin t.
- ct = c + (a — k) sin t cos t — (c + g) sin2t.
- dt= d cos t — ƒ sin t.
- et = e.
- ƒt = d sin t + ƒ cos t.
- gt = g + (a — k) sin t cos t — (c + g) sin2t.
- ht = b sin t + h cos t.
- kt= k + (c + g) sin t cos t + (a — k) sin2t.
- Pt= P cos t — R sin t.
- Qt = Q.
- Rt = P sin t + R cos t.
Now, if ?t, A’t, B’t, C’t, D’t, E’t represent the altered values of ?, A’, etc., we will have:
- ?t = 1 + at + et = ? — c + g sin t cos t — a —k sin2t.
2 2 2
- ?tA’t = dt — bt = ?A’ cos t — f — b sin t.
2 2
- ?tB’t = ct tan ? + Pt = ?B’ + {(a — k) sin t cos t.
H
— (c + g) sin2t — R sin t} tan ? — P versin t.
2 H
- ?tC’t = ƒt tan ? + Qt = ?C’ + (d sin t — ƒ versin t) tan ?.
H
- ?tD’t = at — et = ?D’ — ½ {(c + g) sin t cos t + (a — k) sin2t}.
2
- ?tE’t = dt + bt = ?E’ cos t — ƒ + h sin t.
2 2
Now if the soft iron be symmetrically arranged on each side of the amidships fore and aft line, we may assume to be the case, we have:
b = o; d = o; ƒ = o; h = o; A’ = o; E’ = o.
Also if, as we may also assume, t be so small that its square and higher powers may be neglected, the
sin t = t; cos t = 1; versin t = o; sin2 t = 0;
and we will have:
- ?t = ? — c + g t.
2
- ?tA’t = ?A’.
- ?tB’t = ?B’ + (a — k) t tan ? — R t tan ? = ?B’
z
+ {(a — k) — R } t tan ?.
- ?tC’t = ?C’.
- ?tD’t = ?D’ — ½ (c + g) t.
- ?tE’t = ?E’.
Taking now the equation
- ? = A + B sin z’ + C cos z’ + D sin 2z’ + E cos 2z’, as is done in the deduction of the equation for heeling error, and substituting in it from the above, we get:
From (44) and (53), if the values of the parameters be known, we can determine the deviation resulting from a given change of trim. To further simplify, and put in more workable shape, however, we have:
(58) ?t + 2?G? + 2?GMt sin z’ — G (c + g) t sin 2z’.
From a consideration of the above we see that we have caused the following changes in the deviation by changing the trim:’
First. We have changed the value of ?, and have therefore changed the power of the compass to resist deviation. This change is represented by the term which shows that the change is caused by the c and g rods, and that the change varies with the degree of inclination. It is apparent from the normal position of the c and g rods that to incline them in a fore and aft direction results in the introduction of a new a rod, and in a change in the old values of c and g. The introduction of the new a of course directly affects the value of ?, which is the directive force of the compass.
Second. We have introduced a new semicircular force involving the quantities G and M, which in turn depend, the first upon c and g, and the second upon a, k, R, Z and ?, and both upon the angle of inclination. Evidently the c rod, which originally caused semicircular deviation, will suffer change when inclined. Also the g rod, which caused no deviation when horizontal, will now act to produce such deviation, in other words will create a small resultant vertical (c) rod, which will cause semicircular deviation. Similarly a, when inclined, produces a resultant vertical rod, and likewise k, when inclined, produces a resultant vertical rod forward or abaft the compass. Also R, which formerly acted downward, and produced no deviation, now has its pole thrown forward or aft of the compass, causing a resultant pole in the horizontal plane, and produces semicircular deviation. It is evident at once that all these causes would properly produce a semicircular deviation, as indicated by their presence in the equation in the term involving sin z'.
Third. We have introduced a new quadrantal force, involving G = _____1_____ (c + g), and the angle of inclination. 2? — (c +g) t’
Plainly, the c and g rods, when inclined, not only produce resultant vertical rods with semicircular deviation, as described above, but also horizontal, fore and aft rods in the plane of the compass, or a rods, thus causing quadrantal deviation, as shown by their presence in the equation in the term involving sin 2z'.
It is therefore clear that, by inclining the ship, as she changes trim:
The c and g rods produce a resultant a rod, modify the value of the existing c rod, and affect the semicircular and quadrantal deviations, and the value of ?.
The a and k rods produce a resultant c rod, modify the value of the existing a rod, and affect both semicircular and quadrantal deviations, and the value of ?. The change in the a rod, however, is not sufficient to affect the quadrantal deviation, and it disappeared from the quadrantal term of the equation accordingly. See equation No. (42).
The R force changes its effect, its pole shifting its position relative to the compass, and becoming active in the horizontal plane to produce semicircular deviation, P and Q theoretically act in a similar way, but their change is so small as to be negligible, under our assumptions, and they accordingly disappear from the equations.
It may also be noted that these changes are all caused by the inclination of rods in the fore and aft vertical plane, in other words by changing the position of horizontal fore and aft rods and vertical rods relative to the inducing forces that act upon them, namely H and Z. As Z acts equally on every heading, its effect will not change with change of heading, while the effect of H will show as a maximum on north and south, and disappear on east and west. This does not mean that the total change will be a maximum on north and south and a minimum on east and west, however, for this would only be the case when the effects of H and Z both act in the same direction, and in other cases they would counteract each other, and the maximum combined effect would be on east and west headings.
The following table shows the practical results of change of trim in a supposed case:
Lat. 30° ∞’ North. ? = .8.
Long. 40° ∞’ West. a = +.02.
? = 60°. k = —.03.
Z = 2.25. c = +.09.
g - +.02.
R = +.02.
Then we have: ___1_____
G = 1.6 —.11t
M = +.071.
The following table gives the results for the above ship, in that position on the earth's surface, for various headings and angles of change of trim. In giving the amount of submergence of the bow for any given angle of inclination, the ship is supposed to pivot in inclining about a point 300 feet from her bow. Her length is supposed to be 450 feet. For these conditions we have that, to get the required angle of inclination, the bow would have to sink as follows :
For t = 1° the bow goes down 5.2 feet.
For t = 2° the bow goes down 10.4 feet.
For t = 3° the bow goes down 15.7 feet.
For t = 5° the bow goes down 26.1 feet.
For t = 10° the bow goes down 52.1 feet.
In the case under consideration it therefore appears that no probable ordinary change of trim would introduce any material error. It is believed that a similar investigation for each ship of the fleet, using formulæ (56), (57) and (58) would be of sufficient interest to pay for the time it would take, and perhaps in some cases values might be obtained of sufficient magnitude to make it necessary to consider them in actual practice. It will be noted that, the larger the horizontal deviations, the greater will be the actual error in degrees introduced by change in trim.