The Burrage Time Board is designed to assist the officers when maneuvering their ships to determine the time to go from one position to another.
The following explanation of its construction is given to facilitate the understanding of how to use it.
As a ship will steam in three minutes 100 yards at one knot, 200 yards at two knots, 1,200 yards at twelve knots, and so forth, the three-minute line is plotted with these coordinates. See Fig. 1. The scale at the top is hundreds of yards, the scale at the right is in knots.
The distance 1,800 yards steamed in three minutes at eighteen knots is divided at the bottom of the board into minutes and seconds of time, so that the time to steam any distance at eighteen knots up to 1,800 yards is found at the bottom on the scale of time.
By connecting the o by diagonal lines with the time scale at the bottom, the distances steamed in three minutes at the various speeds are divided into the same proportional parts of minutes and seconds; thus, at eight knots, the 800 yards, A-B, steamed in three minutes is divided into minutes and seconds.
If the required distance is 4,000 yards, the time required at eight knots is found by taking one-tenth of 4,000, or 400, which is within the limit of the scale of the board.
Four-hundred-yard vertical line crosses the eight-knot A-B line at C, on the one-minute thirty-seconds diagonal line. Multiplying by ten for 4,000 yards, gives ten minutes 300 seconds, or fifteen minutes, the time to steam 4,000 yards at eight knots.
The problem of going from A to B, Fig. 2, C-L being the course of the guide, requires obliquing at the course and speed of C-D, drawn to scale; then C-G is the horizontal speed made while obliquing on the course and speed C-D.
Having plotted A-B on the maneuvering board, the horizontal distance A-F is determined. Knowing the horizontal distance A-F in yards or miles and with the horizontal speed C-G, the time to go from A to B is found.
Combining Fig. 1 with a quadrant of the maneuvering board gives Fig. 3 on the Time Board.
In any problem of going from one position AJ Fig. 2, to another, B, both positions are plotted on the maneuvering board, the change of course or oblique for the maneuver determined and the horizontal distance A-F found as shown in Fig. 2. Then the time board solves the problem of the time to go from A to B.
How long will it take to go from A to B in a problem requiring, an oblique of 30° at sixteen knots, when the horizontal distance to be gained is 1,200 yards? The solution is in heavy lines, Fig. 3.
Follow the sixteen-knot circle to 30° at E: E-K is the horizontal distance made good in three minutes, or 800 yards by the scale at the top. Follow the vertical line through E to the three-minute diagonal line at F: F-G is the same as E-K, or 800 yards, but F-G is divided into minutes and seconds of time by the diagonal lines.
As the distance E-K steamed in three minutes is 800 yards, or less than the 1,200 yards required, take one third of 1,200, or 400 yards; find the time to go that distance and multiply by three for the 1,200 yards.
The 400-yard vertical line crosses F-G at H, which is on the one-minute-thirty-seconds diagonal line, and is the time to gain 400 yards horizontally. The horizontal distance of 1,200 yards will take three times one minute and thirty seconds, or 4 minutes thirty seconds, which gives the time to gain 1,200 yards horizontally, or the time to go from A to B, as plotted on the maneuvering board.
For long distances, the time scale at the bottom becomes hours and minutes, and the corresponding distances on the scale at the top become miles.
The operation for solving the problem is the same. Example: To reach the new position B with a speed of twenty-four knots and an oblique of 30° requires gaining the horizontal distance of fifteen miles; use half the speed, or twelve knots, to be within the scale of the diagram, and take half the time for the twenty-four knots. Operation is as follows—See Fig. 4. Follow the twelve-knot circle to a 30° oblique at E. Follow the vertical line through E to the three-hour line at F. Follow horizontal line to fifteen-mile vertical line at H. H is on the two-hour-thirty- minute diagonal line, which is the required time for twelve knots; for twenty-four knots take half the time, or one hour fifteen minutes.
Example: The next astern of the guide is directed to be 20,000 yards abeam of the guide at 9:00 a.m., steaming at fourteen knots, standard speed being twelve knots, What is the time to start?
The time board shows that at fourteen knots it requires' an oblique of 30° to make twelve knots advance, the same speed as the guide. Follow the fourteen-knot circle to 30° oblique, then down to three-hour line and across to the ten-mile vertical line gives one hour twenty-five minutes on the diagonal line, which is the time to gain the ten miles, and the start should, therefore, be made by 7:30 a.m.
Construct the diagram on 1" cross-section paper—1" =100 yards or 1 knot.