TACTICAL AND FIRE CONCENTRATION PROBLEMS AND THEIR SOLUTIONS
By Lieut. Commander H. W. Hill, U, S. Navy
Problems
I. When ordered to take position on a given bearing and at a given distance from a moving ship, find:
- Change of course necessary using certain speed.
- Speed necessary if a certain change of course is made.
II. When ordered to change distance from guide to maintain present bearing, find:
- Change of course necessary using a certain speed.
- Speed necessary if a certain change of course is made.
III. In fire concentration, to determine the bearing of the C.-in-C.'s target from ship concerned, having been signaled the bearing of that target from C.-in-C.'s flagship (i.e., the General Bearing Line).
IV. In maneuvers, target practice, etc., to straighten maneuvering ship or unit up on a line which is parallel to, and at a given distance from, course of guide or target.
V. To determine whether an enemy at known range and bearing and approximately known course and speed is within torpedo range, and if so, what the desired gyro setting is.
1. The problems listed above are simple ones, but which in constantly varying form are ever present during all forms of tactical maneuvers. A quick and accurate means of solution is necessary to flag officers, commanding officers, navigators, gunnery officers and officers of the deck.
2. These problems are capable of solution by many methods, but the Battenberg Course Indicator, manufactured by the Washington Navy Yard and issued to all large ships, furnishes, in the minds of those who have tried all methods, the most practical solution. Recently a flag officer commanding a battleship division used the Battenberg Course Indicator entirely, and it was soon a remarked fact that his division, following a fleet tactical or fire concentration signal, always led all other divisions in getting the correct signal hoisted and executed.
3. Small, light, but durable, made of non-corrosive metals, so that it is readily available in all kinds of weather at any part of the bridge, it is in effect a portable, weather-proof mooring board, and requires no instruments to operate it.
4. There are three types of Battenberg Course Indicators now in use in the service. Heretofore their issue has been only to the large ships, but probably in the near future they will be supplied to cruisers and destroyer division leaders. The three types are:
(a) Mark I. Manufactured by Washington Navy Yard in 1913.
(b) Mark II. Manufactured by Washington Navy Yard in 1920.
(c) British type. Manufactured by Elliott Bros., London, Eng.
The Mark II type is shown in Plate I, and may be briefly described as follows:
The Tray (1) is circular, 12 inches in diameter, and its rim is graduated from o to 360 degrees to indicate both bearings and reciprocal bearings. Pivoted centrally in this tray are the disc (2) and three brass arms. The Disc (2) is grooved with parallel lines, marked alternately red and black. The Diameter Groove (3) is of greater width than the other grooves and has an arrow head g,t each end. The unit spaces of all markings on both disc and bars are of equal size.
The two Position Bars (4-5) are graduated from the center in units from 0-30, each unit being subdivided into two parts. These units may be used to represent any desired scale—the larger the scale chosen, of course, the greater the accuracy obtained. On the position bars are Pointers (6) which may be moved to any position and clamped.
One position bar represents the observer's present position relative to the guide, or ship angled and ranged upon, and the other represents the position relative to the guide that the observer is endeavoring to reach. The ends of the position bars override the tray rim scales, so that they may be readily set at the desired angles.
The Guide Bar (7) is marked in the same scale as the position bars and on it is the Graduated Circle (8) which is capable of movement along the bar, and which may be clamped at any point to indicate the speed of the guide. Rigidly attached to this circle, which pivots about the Clamp (9), is the Speed Bar (10), graduated in the same scale as the guide and position bars. The sliding Pointer (11) on this bar may be set directly for any speed of the observer's ship. The guide bar is mounted sufficiently high above the disc to allow the position bars to be moved underneath it.
5. The Battenberg Course Indicator is an instrument for solving graphically the many problems of the mooring board. Two sets of triangles are formed in the solution of any problem. In one of these the sides are made by two position bars and one of the small red or black grooves on the disc, and in the other by the speed bar, its intercept on the guide bar, and its intercept on the diameter groove. A geometrical study of the Battenberg Course Indicator solutions will not be attempted here, but only a description of how the solution is reached.
6. Problem I (a) presents the most common of all tactical situations and the following problem along these lines is solved and discussed in detail. (Plate I shows the Battenberg Course Indicator set for this solution.)
Given: Ship A is ordered by Flagship B to take station astern of B, distant 1000 yards. B is on course 345°, speed 12. B now bears 260° from A, distant 5000 yards. A has speed for 14 knots.
Required: Course to be steered by A if 14 knots speed is used.
(a) Let the center pivot represent the guide, or ship on which station is to be taken.
The center pivot may be assumed equally well to be the observer's ship, but makes a set of triangles not so clearly visualized.
(b) Using the largest scale available (1 unit=200 yards) set one position bar for the present distance (5000 yards—pointer on 25) and bearing of the observer's ship from guide B. (Reciprocal bearing 260°.)
(c) Using the same scale set the other position bar for the distance (1000 yards—pointer on 5) and bearing of the desired position from guide B, (Reciprocal of B's course or bearing 165°.)
Whenever practicable a multiple of the actual distance should be set on the position bar; the same multiple being set on both bars, of course. This enlarges the triangle formed and so makes for a corresponding increase in accuracy. This may be done without regard to whether any multiple will be used in setting the speeds on the guide and speed bars, for, as noted above, these are arms of an entirely separate triangle, and the new triangle will be similar to that formed by using actual distances.
(d) Rotate the disc till the grooves run parallel to a line joining the pointers on the two position bars. This completes this triangle.
(e) Clamp the guide bar on the course of the guide, or vessel angled and ranged on. (Guide's course 345°.)
If the problem to be solved is one arising from an actual or prospective change of course by the guide, the guide bar should be set for the New Course—i.e., the course that will be taken by the guide before the completion of the maneuver.
(f) Move the graduated circle along the guide bar and clamp it at the speed of the guide. (Clamp at 24—scale one unit equals one-half knot.)
Should the guide change speed during the maneuver, or by zigzagging have a resultant loss in advance on the course set, this must be considered as creating an entirely new problem, and therefore requiring a new solution. In order to obtain greater accuracy, a multiple of the speed of both the guide and observer's ship should be set on the guide and speed bars, but, of course, the same multiple must be set on both.
(g) Set the pointer on the speed bar at the speed of the observer's ship. (Speed 14 knots. Clamp at 28—scale as above.)
Should the speed of the observer's ship be changed during the maneuver, this must be considered as creating a new problem, and requiring a new solution.
(h) Swing the speed bar around till its pointer comes to the diameter groove on the disc. The reading of the graduated circle shows the number of degrees (37) to the left of the guide's course which must be steered to obtain the desired position under assumed conditions of course and speed. The course to be steered, therefore, is 308 degrees. Answer.
It will be noted that there are two solutions to every problem, due to the fact that the speed bar may be swung either to right or left to its position relative to the diameter groove. The problem, when set on the Battenberg Course Indicator, however, becomes self-apparent. A general rule is that the long end of the speed bar must be on the same side of the guide bar as the position bar which represents the present position of the observer's ship. No rule is necessary to determine whether the change of course should be made to the right or left.
7. In most instances changes of course will be made in multiples of 10 or 15 degrees. When the change of course is 10, 15, 20, 30, 45, or 60 degrees, the sines of which are easily kept in mind, the time for completion of a maneuver can be readily computed mentally. The sines of these angles are, roughly:
10 degrees—1/6.
15 degrees—1/4.
20 degrees—1/3.
30 degrees—1/2.
45 degrees—7/10.
60 degrees—7/8.
8. Therefore, if in a maneuver one finds that 1000 yards must be gained to the left and the change of course is 30 degrees, speed 12 knots, the time required to run before straightening up on the base course may be quickly computed as 5 minutes. (12 knots—400 yards per minute. Since 30 degrees = ½, or gain to left in 1 minute is 200 yards. In 5 minutes 1000 yards will be gained to left.)
9. The Battenberg Course Indicator supplies quickly the information needed for the above problem—as to distance to be gained to the flank. After it is set for the data of the problem, rotate the disc till the red and black grooves are parallel to the guide bar, or course of the guide. Now count the number of grooves between the pointers of the two position bars and multiply this by the number of yards per space between grooves, according to the scale used on the position bars.
10. The use of the disc as described above is a handy one, and may be used for many problems. For instance, with the fleet in column, suppose a formation signal is made which requires the van division to take station 30 degrees abaft the port beam of the guide (center) division distant 4000 yards. The Battenberg shows a big change necessary, but for sake of example, let us assume that a "Ships left 90 degrees" signal is made. As the ships of the van division run out, successive problems may be made from each range and bearing obtained of the guide, and when the ships are 1000 yards, or the distance of their advance, from the line parallel to the guide's course running through their desired position (shown by the Battenberg by method described in above paragraph), "Ships right 90 degrees" may be executed and the ships straightened up on their base course. They are now "on the line," and distance may be gained or lost on that line to bring them to the required bearing. A person experienced in the use of the Battenberg should be able to straighten up on the base course and pick up speed, within a very few degrees of the required bearing.
11. The following problems are solved and discussed to illustrate more clearly the uses of the instrument. In these solutions each setting is enumerated, making it appear slow and complicated. It is to he emphasized, however, that all hut one or two settings are kept corrected up to date at all times, so that when the signal is made to execute a certain maneuver, the correct course may he almost immediately determined.
Problem I.—Given: Ship A is ordered by Flagship B to take station, astern of B, distant 1000 yards. B is on course 345 degrees, speed 12. B now bears 260 degrees from A, distant 5000 yards. A has speed for 14 knots.
Required: (1) Course to be steered by A if 14 knots speed is used.
(2) Speed required if it is desired to steer course 315 degrees. Solution (1) has already been discussed in Par. 6. Solution (2):
(a) a, b, c, d, e, f, of this solution are the same as in the solution of (1) discussed in Par. 6.
(b) It is desired to steer 315 degrees, which is 30 degrees to the left of B's course. Therefore, swing the speed bar (to the right of the guide bar—or on the same side of it as the position bar
showing the present position of ship) till the graduated circle reads
30 degrees.
(c) Now move the pointer along the speed bar till it is over the diameter groove. Its position on speed bar, 26.2, indicates a speed of 13.1 knots necessary. Answer.
Problem II.—Given: Ship A, now in position 45 degrees on starboard bow of Flagship B, distant 2000 yards, base course 270 degrees, speed 12 knots, is ordered to open distance to 3000 yards, maintaining present bearing. A has boiler power for 17 knots.
Required: (1) To find change of course possible using 14 knots. (2) To find speed necessary changing course 15 degrees to right.
Solution (1):
(a) From an examination of conditions it will be seen that one of the triangles is eliminated in this problem, i.e., the triangle formed by the position bars and diameter groove becomes a straight line. Therefore
(b) Rotate the disc until the diameter groove is coincident with position bars, i.e., to the bearings of Flagship B from ship A (135 degrees). The position bars need not be set on this bearing,
(c) Clamp the guide bar on course 270 degrees.
(d) Clamp the graduated circle at B's speed—12 knots. (Clamp at 24—scale one unit equals one-half knot.)
(e) Set the pointer on the speed bar for A's desired speed—(14 knots). (Pointer at 28—scale one unit equals one-half knot.)
(f) Swing speed bar as directed in par. 6 (h). The graduated circle gives a reading of 9 degrees, which in this case is the desired change of course to be made to the right by ship A (the present base courses of ships A and B are the same). Answer.
Solution (2):
(a) a, b, c, d, of this solution are the same as in Solution (1) above.
(b) It is desired to steer 285 degrees, which is 15 degrees to the right of B's course. Therefore, swing the speed bar until the graduated circle reads 15 degrees.
(c) Now move the pointer along the speed bar until it is over the diameter groove. Its position on the speed bar, at 32.7, indicates a speed of 16.4 necessary.
Problem III.—Given: The fleet deployed in battle formation. Fleet Flagship A bears 15 degrees (true) from ship B, distant 2200 yards. Fleet course is 360 degrees. Enemy is on port beam of fleet. A's corresponding ship in the enemy line is distant 15,000 yards. Flagship A hoists G. B. L. signal "My target bears from me 275 degrees. Concentrate according to plan." In order to make concentration plans the fleet flagship's target must first be identified by ship B.
Required: The bearing of fleet flagship's target from ship B.
Solution:
(a) After deploying into battle formation, individual ship or division maneuvering practically ceases, and therefore the Battenberg Course Indicator may be kept ready for this problem, anti its setting kept corrected, during the battle approach, for the range and bearing of the fleet flagship, and also, if visibility is good, for the approximate range and bearing of B's corresponding ship in the enemy line.
(b) Assume the fleet flagship A to be at the center pivot.
(c) Set one position bar for range 2200 yards (pointer on 4.4—scale one unit equals 500 yards), and reciprocal bearing of Flagship A—195 degrees.
(d) Set the other position bar for range of B's corresponding ship in the enemy line (pointer on 30—scale one unit equals 500 yards), which is also the approximate range of the flagship's target from the flagship.
(e) When the signal is made, set position bar (step d) on signaled bearing (G. B. L.)—275 degrees.
(f) Rotate the disc until the grooves are parallel to the line adjoining the pointers of the two position bars.
(g) The reading of the rim scale of the tray marked by the arrow head on the end of the diameter groove is the bearing of the fleet flagship's target ship B, in this problem 283 ½ degrees.
Note.—Having this bearing of the flagship's target, it may be readily identified by glancing over an azimuth circle on a pelorus and its numerical position in the enemy line fixed. Then considering the number of enemy ships ahead or astern of this target, relative to our own battle line, the concentration plan may be quickly decided upon.
Problem IV.—Given: Ship A is making a run on target B, and wishes to reach a point where she may parallel target B's course, when latter bears 65 degrees to the right of this base course, on a line which will be 10,000 yards abeam of target B. Speed of B is estimated at 7 knots, course 360 degrees, and A's speed is 20 knots. Target B now bears from 72 degrees, distant 22,000 yards.
Required: (1) To find the course which must be set to allow straightening up on the proper line when the target bears 65 degrees true (i.e., 65 degrees to the right of the base course)
(2) To find the approximate time required to reach the turning point, so that the gunnery officer may be notified.
(3) Assuming A's advance for a 6o-degree turn to be 800 yards, what will be the range of the target at the proper movement to put the rudder over when straightening to the base course?
Solution (1):
(a) Set guide bar on course (360 degrees) of target B.
(b) Set one position bar for position abeam of target B or bearing 270 degrees, distance 10,000 yards. (Pointer on 10—scale one unit equals 1000 yards.)
(c) Rotate the disc until the grooves are parallel to the guide bar.
(d) Set the other position bar for the bearing of the desired position (245 degrees, or the reciprocal of 65 degrees) and note where the grooves passing through the pointer of the first position bar cut this bar. In this example 11. 10—11,100 yards. (Scale one unit equals 1000 yards.)
This is the range of position desired.
(e) Having found the range and bearing of this desired position, this now becomes a simple problem and may be solved as described above in par. 6. The correct solution shows the desired course to be 60 ¼ degrees. For sake of example, let us use 60 degrees. Answer.
Solution (2):
(a) The Battenberg Course Indicator is set for solution (1). Now rotate the disc until the red and black grooves are parallel to the guide bar, or base course of target B.
(b) Count the number of spaces between the grooves running through the pointers of the two position bars. (In this case 10.8.)
(c) Multiply this number by the yards per-space (in this case 1000 yards) or 10.8 X 1000=10,800 yards.
(d) Following the method described in par. 8. Twenty knots—666 yards per minute. Cos. 60 degrees = 7/8. Ship A will dose at the rate of 666 cos. 60 degrees = approximately 580 yards per minute. The time required in problem will therefore be
10,800/ 580 = 18 ½ minutes. Answer.
Note.—For this approximate solution it is not necessary to correct for the time of turning.
Solution (3):
(a) Let us assume that the approach has been made correctly and that when about to turn the target B bears 66 degrees from A.
(b) Set both position bars on bearing 245 degrees (reciprocal of 65 degrees) with one pointer marking the distance from the target that ship A will be when on her line at that bearing. (In this case, as found in solution (1), 11,100 yards.)
(c) Add 800 yards (assumed advance) to this distance which is the point at which the rudder should be put over, or range 11,900 yards.
(d) Due to many causes the approach, as assumed in (a), will not be perfect. If not, the same method of locating the desired line must be used as described in solution (1) of this problem, and the advance added to the range of the line at the actual bearing.
Problem V.—Given: Ship A steaming on course 350 degrees, speed 18 knots, is engaging enemy ship B, bearing 95 degrees, distant 13,000 yards, on estimated course 330 degrees, and speed 17 knots. A's torpedoes are good for range 10,000 yards at 30 knots, 12,000 yards at 26 knots, and 15,000 yards at 22 knots. Ship Q of A's fleet, bears 150 degrees from A, distant 2000 yards, course and speed the same.
Required: (1) To find if the enemy is within torpedo range, and if so what is the highest of the three speed settings that can be used.
(2) To find with what gyro angle setting the torpedo must be fixed.
(3) When fired with this setting, will the torpedo pass 500 yards ahead of ship C, assuming it runs true?
Solutions (1) and (2):
(a) Set the guide bar for the estimated course 330 degrees and speed 17 knots of enemy ship B.
(b) Set one position bar for the range 13,000 yards and reciprocal bearing 275 degrees of ship B.
(c) Set the pointer on the speed bar for the highest speed, 30 knots, of torpedo.
(d) Swing the speed bar till its pointer—set as in (c)—reaches the line between the center pivot and the pointer on the position bar, (It will be noted that the center of the pointer on the position bar is about 1/16 of an inch from the edge of the bar.) This completes the speed triangle.
(e) Note the reading of the graduated circle. (In this case 97 degrees.)
(f) Rotate the disc until its grooves are parallel to the speed bar.
(g) The reading of the scale on the tray rim opposite the arrow head on the diameter groove, in this case 67 ½ degrees, shows the true course which must be made by the torpedo in order to hit, at 30 knots speed, an enemy vessel on course 330, speed 17, now bearing from A 95 degrees true. This hit will be made irrespective of the range, provided that the range, measured along the torpedo track, is not in excess of 10,000 yards, the assumed maximum.
(h) Now consider the speed bar as a range bar (the scale is the same as that of the position bars) and set its pointer for the maximum range for 30 knots, i.e., 10,000 yards.
(i) Keeping the speed bar at the angle obtained above, in (e) and (f), slide it along the guide bar until its pointer again reaches the line between the center pivot and the pointer on the position bar. This completes the range triangle, maintaining the angle between the course of the torpedo and the present bearing of enemy ship B.
(j) If the pointer on the position bar is closer to the center pivot than the pointer on the speed bar set, as in (i), the target is within range for this speed of the torpedo, otherwise not.
(k) This problem, when solved for a torpedo range of 10,000 yards, speed 30, shows the target to be about 1000 yards out of range.
(l) The solution, by the same method, for a torpedo range of 12,000 yards, speed 26, shows the target to be well within range, and the true course for the torpedo to be 62 ½ degrees. As A's course is 350 degrees, this is a course 72 ½ degrees to right of A's course and will require a gyro setting of 17 ½ degrees left on the torpedo. Answer.
Solution (3):
(a) This is a problem similar to others previously described.Assume ship C to be at the center pivot.
(b) Set one position bar for range 2000 yards and reciprocal bearing of C from A (330 degrees).
(c) Set the other position bar on C's course 350 degrees, and the distance 500 yards; it is desired that the torpedo pass ahead of C.
(d) Rotate the disc until the grooves are parallel to the line joining the pointers on the two position bars.
(e) Set the guide bar for C's course 350 degrees, and speed 18 knots.
(f) Set the speed bar for the torpedo's speed 26 knots, and swing it until it cuts the diameter groove, extended. The graduated circle on the guide bar reads 131 degrees. This shows that the torpedo could be fired on course 350 degrees plus 131 degrees—121 degrees and still pass 500 yards ahead of C. Looking at it in another way, it may be figured that the torpedo may average as little as 10 ½ knots on its course 62 ½ degrees true and still pass 500 yards ahead of C. Answer.
13. It will be seen from the above problems that the Battenberg Course Indicator is. an instrument capable of solving almost any type of problem which may confront an officer of the deck, navigator, captain, or flag officer during maneuvers and battle, and it is sincerely hoped that its value will be fully utilized by the units of the fleet.