The maneuvering and mooring board described below will be found useful in making approaches at target practice, changing position in formation and working out mooring problems. It is believed that most maneuvering and mooring problems can be worked out in much less time and with less labor than when using the regulation mooring board.
The board consists of a movable compass rose a, graduated from 0° to 360°, so arranged as to move about the center and along the edge of the board. The compass rose can be clamped in any convenient position at the top of the board, its position depending on the conditions of the problem.
Under the compass rose a is another rose b, graduated on either side of d from 0° to 90° and from 90° to 0°. b is arranged to move along the board with a, but does not turn about the center, a key preventing the movement. Pivoting about the center of both compass roses is an arm c, graduated in yards to the same scale as the plotting sheet. Any number of scales can be used on the same arm. In this case ¾ inch equals 100 yards and 200 yards, respectively. The arm is also graduated in inches from zero to 30 inches. A pointer is attached to the arm, so that either the direct or reverse bearing can be read off the compass rose.
The protractor e consists of a compass rose having two arms, f and g, pivoting about the center. The arm f can be clamped on any course or bearing desired, while g is free to move about the center of the rose. Both have pointers, so that either the direct or reverse bearing can be read. Both arms are graduated in yards to the same scale as the arm c, and are also graduated in knots. In this case, one-half inch equals 1 knot, although any arbitrary scale may be used provided both arms are graduated the same. A pair of cross wires locates the center of the compass rose.
The plotting sheet, mounted on the board, is of cross-section paper, four squares to the inch. In this case one side of a square equals 25 yards or 50 yards, depending on the scale used. Every fourth line is drawn in red ink to facilitate plotting.
The diagram near the lower right-hand corner will be found useful in determining the time required to perform the evolution and will be explained in connection with examples of problems given below.
In the sketches of problems, the cross-section is not drawn. The black horizontal and vertical lines shown are those nearest the plotted positions on the plotting sheet. The full lines in red ink are the only lines required to be drawn in working out the problems. In some cases it is not necessary to draw any lines, as shown in Examples 3 and 5. In Example 5 the turning circle need not be plotted.
In some cases where it is desired to moor on a given bearing from another ship's anchor, the compass rose a cannot be used. In such cases, however, all the plotting can be done at any convenient position on the board, using the protractor. One arm can be clamped on north and laid parallel to the vertical lines, or on east, laying the edge parallel to the horizontal lines. By using the protractor the various bearings and distances entering into the problem can be laid off much more quickly than with dividers and parallel rulers as in the case when using the regulation mooring board.
EXAMPLE 1
Forming on a moving ship, preserving the bearing. Speeds unequal.
C, standing on course 50° magnetic, speed 12 knots, signals A, who is bearing 290° magnetic, distant 1200 yards, to close in to 800 yards and maintain bearing. A's available speed is 16 knots. Find A's course, distance and time required to perform the evolution.
To Find the Course.—Clamp the compass roses a and b near the left-hand side of the board, with the zero line of b on any vertical line and with C's course at j. Plot the position, A 1200 yards from the center of the compass rose on bearing 290°. Plot B 800 yards on same bearing, using compass rose a and arm c. Connect AB. Clamp the arm f of the protractor on C's course and lay the edge on or parallel to the horizontal line passing through B with the 12-knot graduation at B. Swing the arm g around so that the 16-knot graduation intersects AB. A's course can then be read off the compass rose of the protractor.
To Find the Distance.—Draw the line xy paralled to g, passing it through B and intersecting the horizontal line passing through A. Bt’ is the distance steamed in performing the evolution.
To Find the Time.—The diagram near the lower right-hand corner gives the distance steamed in 1 minute, at various speeds on the scale of 1 inch= 200 yards. By using this diagram the time required to perform the evolution can be stepped off with the dividers. If another scale is used, the distance run in 1 minute can be increased or decreased proportionately. For instance, if the scale used is 400 yards to the inch and the speed is 16 knots, step off the distance run in i minute on the 8-knot line.
In some cases, where the distance run is great, the line xv will run off the board before intersecting the horizontal line passing through A. In such cases, part of the distance can be measured from B to t and the remainder from t" to u.
The line xy does not necessarily have to pass through the point B or A. It may be drawn on any part of the board, the only condition being that it must intersect the horizontal lines passing through A and B.
EXAMPLE 2 (CASE 1)
Forming on a given bearing and distance from a moving ship, speeds unequal.
C, standing on course 200° magnetic, speed 14 knots, signals A, who is bearing 355° magnetic, distant 1400 yards, to take position one point forward of C's starboard beam, distant 1600 yards. A decides to change course 18° to the right. Find A's speed, distance run and time required to perform the evolution.
To Find A's Speed.—Clamp the compass rose a on the right side of the board with the zero mark of b on any vertical line and with C's course at i. Plot positions A and B on the proper bearings and distances as shown in sketch, using compass roses a and b and arm 0°. Connect AB. Clamp the arm f of the protractor on C's course as shown and lay the edge on or parallel to the line passing through B with 14-knot graduation at B. Place the pointer of arm g on course 218°. A's speed can then be read off the arm g where it intersects AB.
To Find the Distance.—In this case the line xy runs off the board before intersecting the horizontal line passing through B. Draw another line uv through B, parallel to xy, so that uv and xv will intersect the same horizontal line. Part of the distance run can be measured from A to y and the remainder from u to B.
The time is found as explained in Example 1.
EXAMPLE 2 (CASE 2)
C is standing on course 180° magnetic, speed to knots, signals A, who is bearing 340° magnetic, distant 1200 yards, to take position on bearing 40° magnetic from C, distant 1000 yards. A's speed is 22 knots.
In this case A is on C's starboard quarter and must take position on the port quarter. For this reason the compass rose is clamped on C's course at k and the arm f of the protractor is on or parallel to the nearest vertical line passing through B. The course, distance and time are found in the same manner as described in other problems.
EXAMPLE 2 (CASE 3)
C, standing on course 48° magnetic, speed 11 knots, signals A, who is 60° on her starboard bow, distant 950 yards, to take position 1000 yards ahead of C. A's speed, is 18 knots. Find distance, course and time.
In this case A is on C's starboard bow, making it necessary to work down the board. The compass rose is clamped on C's course at l. The course is found the same as in the other problems. In this case xy runs off the bottom of the board before intersecting the vertical line passing through B. Draw another line, uv, parallel to xy so that it intersects the vertical line passing through B and any other vertical line intersecting xy. Part of the distance can be measured from A to t and the remainder from t' to t".
EXAMPLE 3
Finding the amount to change course to pass a stationary object abeam at a given distance.
A stationary subcaliber target C is 10° on A's starboard bow, distant 2900 yards. A wishes to pass the target 1600 yards on starboard beam. Find how much course must be changed to pass C abeam on a 1600-yard line.
In this case A may pass C on any course, the only condition being that C must be 1600 yards on A's beam. Plot A's position on relative bearing 10°, 2900 yards from C. Swing the arm c around so that the 2900-yard graduation intersects the 1600-yard line, as at A2, and read the relative bearing off the rose b. This is how the target should bear if A were on the 1600-yard line. The difference between the two relative bearings is the amount to change course to the left to get on the 1600-yard line.
EXAMPLE 4
To determine the point where a ship, sent out on a given bearing, should return.
A destroyer A is ordered at 10 a. m. to scout on course 135° magnetic, speed 20 knots, and rejoin fleet C at 8 p. m. The fleet continues at 16 knots on course 150° magnetic. At what point should A change course to intercept C at 8 p. in. and what should be A's course?
Let 1inch equal 10 miles. Clamp the compass more on C's course at l, and locate C 150 miles from center of compass rose. Locate A 200 miles on 135°. Clamp arm f of protractor on any course and lay the edge on line AA2 in such a position that the distances AA2 and A2C are the same on both legs of the protractor. The point A2 is the point where A must change course to intercept C at 8 p. m.
To Find the Course.—Plot C's position at C2. This is C's position when A changes course. Clamp the arm f of the protractor on C's course with the 16-knot graduation at C2. Swing the arm g around until it intersects the continuation of C2A2, at the 20-knot graduation. The course to steer can be read off the compass rose of the protractor as before.
EXAMPLE 5
To determine how to allow for turning when taking station on another moving ship.
A destroyer A is scouting ahead of the flagship C and is ordered to take position on C's starboard bow, distant 800 yards. C is heading 250°, speed to knots. A's speed is 18 knots, tactical diameter 800 yards, advance 800 yards, transfer 400 yards. Speed reduced in turning 20 per cent. What should be the bearing and distance of C when A's helm is put over?
Solution.—Set compass rose on C's course at i. Locate A on C's starboard bow, distance 800 yards. On the vertical line passing through A locate A2 800 yards (tactical diameter) from A. A has to run 1457 yards at 14.4 knots, performing the evolution in 3 minutes. During this time C runs 1000 yards. On the horizontal line passing through A2 locate A3, 1200 yards from A2.
The distance A2A3 is equal to the distance C runs in the time it takes A to perform the evolution plus the difference between the advance and transfer. (1200 = 1000 + (600 – 400).)
EXAMPLE 6
Anchoring on a given bearing and distance from another ship's anchor.
A vessel A, approaching an anchorage on course 5° magnetic, is ordered to anchor on bearing 95° magnetic from C's anchor, distant 600 yards. C signals bearing of her anchor 50° magnetic, distant 200 yards from foremast. The distance of observer's position from hawse-pipe on A is 30 yards. Find the bearing and distance of C's foremast when A's anchor should be let go.
Clamp the compass rose a on 50° at i. With the arm c and the compass rose a locate the position of A's anchor at A1. Move the compass rose 200 yards to the left and clamp in position.
Set the arm f of the protractor on 50° and lay the edge parallel to the horizontal line passing through A1. Set the pointer of arm g on A's course and draw A1A2. The bearing and distance of C's foremast when A's anchor is let go can be read off the compass rose and arm, respectively.
C's bearing and distance can be checked frequently along the line A2A1 as at A2.
The red line A2A1 is the only line necessary to be drawn in working the problem.
EXAMPLE 7
Mooring on a given bearing from a ship already moored.
A vessel C, moored with anchors on line of bearing 25° to 205° magnetic, signals A, who is bearing 65° magnetic, distant 2425 yards to moor with 45 fathoms of chain on each anchor on bearing 60° magnetic from C, distance 800 yards. Line of bearing of A's anchors 25° to 205°. C's heading is 325 magnetic. The distance of C's foremast from a point midway between her anchors is 75 yards. The position of observer on A is 50 yards abaft the hawse-pipe. A's advance is 400 yards.
Find the bearing and distance of C's foremast when each of A's anchors should be let go.
Find the bearing and distance of C's foremast when A's helm should be put over in making the turn to come on line of bearing of anchorage.
Solution.—Clamp the compass a on C's heading 325° at i. Set the arm c on line of bearing of C's anchors 25° to 205° and draw xy. Plot the point l on bearing 60° from C, distance 800 yards. A is the point where A's stem will be when moored. Transfer xy parallel to itself through A. x'y' is the line of be aring of A's anchors and also the course on which A must approach the anchorage. Lay off t and t' 45 fathoms on each side of A. t and t' are the positions of A's anchors. Lay off u and u' 50 yards from t and t', respectively. u and u' are the positions of observer on A when anchors are let go. Move the compass rose a 75 yards to the left, keeping C's heading at i. The center of the compass rose is the position of C's foremast. The bearing and distance of C's foremast from u and u' can be read off the compass rose and arm, respectively.
Plot A's position at A1, using compass rose and arm. Draw vw at right angles to x'y’, passing the line through A1. Lay off 400 yards, A's advance, to e. The bearing and distance of C's foremast when helm should be put over .can be read off the compass rose and arm as before.