m with a radius equal to V we make a mark on the line of relative movement S (straight line MoM). Connecting points b and a, we obtain the speed vector of the maneuvering ship m. The direction of this vector also determines the desired course of the maneuvering ship for a change in position. m' From point Mo we draw the line of course of the maneuvering ship. The intersection of this line with a line parallel to the course of the guide, drawn from point M1, will yield point M1-the position of the maneuvering ship at the completion of the maneuver. Drawing from point M1 a straight line, parallel to designated bearing B1 until it intersects the line of course of the guide, we obtain the position of the guide at the end of the maneuver K1. Time required for the maneuver Assume that guide K is at the center of the maneuvering board (Fig. 108). From the initial and designated bearings and distances BoDo and B1D1, from the center of the maneuvering board we plot the initial and designated relative positions of the maneuvering ship Mo and M. The straight line connecting points M and M' will yield the relative movement vector Sp In the center of the maneuvering board we construct a speed triangle using the direct construction method. From the line of course of the guide we plot vector V. From the end of this vector we draw a line parallel to S,, to its intersection with a circle the radius of which is equal to the speed of the maneuvering ship, and obtain the speed vector of relative motion V. Connecting the center of the maneuvering board with the end of vector V we obtain the speed vector of the maneuvering ship m, and its direction will yield the desired course for taking up station Km. Time required for the maneuver (164) 7. Keeping Station When the Guide Changes Course Solving the Problem on a Chart ρ From bearing Bo and distance Do we plot the old position of the maneuvering ship Mo with respect to the guide Ko (Fig. 109). Then, after swinging the ship around, at the same angle on the bow with respect to the new course of the guide, and at the same distance, we plot the new relative position of the maneuvering ship M1. Straight line MoM will be the relative movement vector 5. From point M1 we draw a line parallel to the new course of the guide. Using the inverse construction method, we construct a speed triangle at point M1 whereby the direction of vector V will indicate the desired course which must be set for the maneuvering ship in order to take up the previous station relative to the guide. Time required for the maneuver m t = Sm Sk Fig. 109. Keeping station with a (165) Solving the Problem on a Maneuvering Board Let us assume that guide K is at the center of the maneuvering board. From the bearing and distance we plot on the maneuvering board the positions of the maneuvering ship relative to the old and new courses of the guide-points Mo and M1 (Fig. 110). Connecting these points, we obtain the relative movement of the maneuvering ship S,. In the center of the maneuvering board we construct a speed triangle, for which from point K along the course of the guide we plot its speed vector V. From the end of vector V we draw a line parallel to vector 5, until it intersects a circle the radius of which is equal to the speed of the maneuvering ship Vm, and obtain the speed vector of relative motion V. (166) Fig. 110. Keeping station with a change in the course of the guide. Connecting point K with the end of vector V, we obtain the maneuvering ship m, the direction of which will yield the desired course for taking up station Km. The time required to execute the maneuver: ρ t = 8. Closing at the Shortest Distance Solving the Problem on a Chart From bearing Bo and distance Do we plot the position of the reference ship: point Ko; through it we construct critical angle Q and draw its boundary (Fig. 111). From initial position of the maneuvering ship Mo we plot the closing course, perpendicular to the boundary of critical course angle Q. At its point of intersection with the line of course of the reference ship we obtain the position of the reference ship at the moment of completion of the maneuver K1. Then from point Mo we draw a line parallel to the boundary of the critical course angle, which is relative course K.. Dropping from point Ko a perpendicular to line K,, we obtain KOM1-the shortest distance at which it is possible to close with the guide. Plotting from point K1 segment KOM1 along the line of course of the maneuvering ship, we obtain its position at the moment of completion of the maneuver M1. Time required for the maneuver Solving the Problem on a Maneuvering Board Assume that reference ship K is at the center of the maneuvering board. From bearing Bo and distance Do we plot the position of the maneuvering ship Mo. At the center of the maneuvering board we construct a speed triangle, for which from the course of the reference ship we plot speed vector V (Fig. 112). From the end of vector we draw a tangent to a circle the radius of which is equal to the speed vector of the maneuvering ship and in the opposite direction from the maneuvering ship Mo. The direction obtained will be relative course K. Using a parallel ruler we shift it to point Mo, from which we plot relative course K. Then through the center of the maneuvering board we draw line MK, perpendicular to relative course K. Distance KM' will be Dsh m We continue line MK until it intersects the outermost circle of the maneuvering board. Closing course Km can be read from the scale. Time required for the maneuver 9. Crossing Ahead of the Reference Ship at Maximum Distance Solving the Problem on a Chart From bearing Bo, and distance Do we plot position of the maneuvering ship Mo. At point Ko we construct critical angle Q (Fig. 113). Then from point Mo we construct a line parallel to the boundary of critical angle Q. This is relative course K,, which at its intersection with the course of the reference ship will yield point M. Here distance KoM is the maximum distance at which the maneuvering ship can cross ahead of the reference ship at maximum distance. A line drawn perpendicular to the relative course (boundary of critical course angle 2) is the course of the maneuvering ship for execution of the maneuver. At its intersection with the line of course of the reference ship we locate the position of the maneuvering ship at the moment of completion of the maneuver M1. |