Advanced Pilot Training: TM 1-205 Air Navigation - Section XII Interception Problem

109. Interception for pilot-navigator.-a. Use.-The interception problem can be used for various purposes. As the name implies, it may be the interception of two aircraft or of an aircraft and a surface vessel, etc. This problem is extensively used in the Navy for the interception of the airplane carriers by its airplanes returning to the home carrier. In time of war, bombers intercept the enemy fleet, or pursuit intercept the enemy airplanes.

b. Two kinds.-There are two general classifications of interception problems:

(1) When the aircraft or surface vessel to be intercepted is visible.

(2) When the aircraft or surface vessel is not visible but its position and course and speed are known.

c. Constant bearing.-Interception of visible or invisible craft is possible in the shortest time only when the relative bearing between the approaching craft remains constant. Plotting the relative bearings will immediately indicate whether or not interception is possible.

110. Interception by visual method.-When the craft to be intercepted is visible, it is possible to use a visual check to maintain the direction which will give interception in the shortest possible time, assuming that there is no change in the speed or direction of the craft to be intercepted.

a. In figure 59, assume that A and B are located on the lines CE and DE as shown and that B is directly north of A. Assume that the speed of B is such that he will reach E in exactly 20 minutes, and that A has a speed which will bring him to E at the same time. In this case, if the two airplanes continue, the interception will occur at E, and it will be in the shortest possible time at the speeds being used.

b. Assume that B continues his course and speed and that A is the one to find and hold the course to intercept. The pilot at A knows that if he heads so as to cut in toward the flight path of B too quickly he will finally end up behind B, in a stern chase, and will lose time by doing so. He also knows that if he heads in a direction too nearly parallel to A's course, he will finally have to turn in to keep from getting too far ahead. So he attempts to find the mean of these two headings.

(1) Not knowing that the heading AE will be the correct one, A heads from A toward F. He notes that the original relative bearing of B is such that he can sight B over a portion of the left wing. He holds a steady heading, using the directional gyro, until he notices at point F that it appears as though B is pulling out ahead of him. This shows A that he is turning in too quickly and will have a stern chase if he continues.

(2) At F he swings several degrees to the right, straightens out on a heading with the gyro, and notes the relative position of B. He holds the heading for several minutes and then, at about G, he notices that the position of B seems to be moving rearward as he continues. This shows A that his path is too nearly parallel that of B, and so he turns toward the left again.

(3) When A finally narrows the change in headings down to a heading on which the relative bearing of B remains constant, he will then be on the heading which will give interception in the shortest possible time.

c. In this example, A required a greater air speed than D. Whether or not more speed is required in a given situation is determined by the courses followed, the starting positions of the aircraft, and the wind.

111. Computed interception. - a. General. - A diagrammatic solution of an interception problem made before the target is in sight is frequently of value. Such a problem may be solved when the location, direction of travel, and speed of the target are known, together with the interceptor's air speed and the wind velocity.

b. Example.-M In figure 60, assume that the objective is reported over B at 0830 hour and is making good the course BX at a known ground speed. The interceptor is situated at A and desires to intercept the target. Draw the interceptor's position A on the N-S line of the diagram, then draw the target's course line BX.

(2) Determine the time the interceptor will be ready to take off (0840 hour) and plot C, the location of the target at the time the interceptor takes off.

(3) Draw a line through A and C and label it as the first CB line (first constant bearing line).

(4) The position of point D is determined, so that CD equals 1 hour of the target's travel along his own course. Draw ED through point D and parallel to AC, and label it the second CB line (second constant bearing line).

(5) From A draw AF equal to 1 hour's wind, downwind. From F locate the point G on the second CB line so that FG equals 1 hour of our air speed.

(6) Draw a line from A through G and on until it intercepts the line BCDX at point H.

(7) Point H will be the point of interception; AGH will be the interceptor's course out; and the direction of FG will be the heading out.

(8) There are three methods which may be used to find the time of the interception:

(a) Target speed may be used as follows: CH: CD:: t: 60, or the distance CH is to the distance CD as the unknown time is to 60 minutes,

(h) The interceptor's speed may similarly be used, as AH: AG:: t: 60.

(c) The rate of closure method uses the closing in rate per hour, and it has some advantages over the other two methods. In the above example the distance AC minus the distance GD gives the rate of closure (R/C) ; or R/C = AC - GD. After finding the rate of closure, the distance AC is divided by the rate of closure to give the time AC required to intercept. In figure 60 the time will be AC-GD , and the result will be in hours and fractions of hours. The advantage of using the rate of closure method is shown in figure 61.

1 .     In this example several additional lines of constant bearing show that the interception would occur at about 1230 hour, if the interceptor takes off at 0800 hour. In order to draiv a diagram such as the one shown, on an ordinary sheet of paper, it would be necessary to use a small scale and this would tend to create inaccuracies in the solution.

2.      Use of the rate of closure method will require only the first and second lines of constant bearing to be in the diagram and so a much larger scale can be used in the construction. To find the distance to the point of interception in such a solution the ground speed out is multiplied by the time.

112. Interception before I hour.-a. General.-Positions of the interceptor and the target and the speeds used may allow interception to be made before I hour has elapsed.

b. Example.-In figure 62 assume that conditions are the same as in figure 60, except that the interceptor's air speed is much faster. Interception will occur at point H, where the interceptor course line crosses the course of the target. (Note that the rate of closure in this case is AC+DG.)

113. Interception of aircraft when wind acting on both is the same-a. General.-When the reported data on a target give the objective's heading and air speed, instead of course and ground speed, the diagrammatic solution may be made as follows: (The wind acting on the interceptor and the objective must be the same to use this method.)

b. Example (fig. 63).- (1) Mark the position C of the objective at the instant of the interceptor's proposed departure from base A

(2) Draw the target's heading and air speed as the line BX, with CD equal to his air distance traveled in I hour (air speed). Draw the first and second constant bearing lines as usual.

(3) From A strike an arc with radius equal to air speed of the interceptor. This arc cuts the second bearing line at G.

(4) Draw a line from A through G to H. H is the point of interception in relation to the air; it is the no-wind position of the intercept.

(5) The wind affects both airplanes by the amount of time involved multiplied by the wind velocity. If the wind had blown in the direction from H to O, and the distance HO represents the wind speed times the time interval of the flight, then the actual interception in relation to the ground would be at O. A straight line from A to O would represent the actual course out.

114. Effect of data change.-If a change of data (target course or speed, wind velocity, or air speed of interceptor) occurs after take-off, the problem must be reworked as a new problem. Use the time, the position of the interceptor when he can change his course, and the position of the target at this same time for the first new constant bearing line. Then the second new constant bearing line will be parallel to the first new constant bearing line and will be I hour of the target's travel down his (new) course line. The balance of the problem will then be worked as already described.