If we perform calculations with respect to point M2, then in order to execute the closing maneuver at D3 considerably more time is required than with respect to point M1. Therefore, this method is unacceptable.

Solving the Problem on a Maneuvering Board

Assuming that the reference ship K is at the center of the maneuvering board, we shall plot the initial position of the maneuvering ship Mo (Fig. 102). At point K we plot the speed vector V. After determining the length of segment x = D3(Vx/Vm) we plot it in the opposite direction from vector V and obtain point C. Connecting points Mo and C, we obtain the direction of relative movement or relative course K..

In order to determine the course of the maneuvering ship Km with initial position of the reference ship K, we construct a speed triangle using the direct construction method. For this purpose, through the end of the speed vector of the reference ship Vk we draw a line parallel to relative course K, until it intersects a circle with a radius equal to the speed of the maneuvering ship. Direction determines the desired course of the maneuvering ship required in order to execute the maneuver. At the moment of completion of the maneuver, the reference ship must be right ahead of the maneuvering ship.

m

The time required for execution of the maneuver

t =

S

(158)

4. Closing at the Designated Distance on a Close Approach Course

Solving the Problem on a Chart

From initial bearing Bo and distance Do we plot the positions of the reference ship Ko and meaneuvering ship Mo (Fig. 103). From point Ko we plot the course of the reference ship, then from bearing Bo we plot a segment equal to D3 and obtain point M1.

In order to close on the target at the designated distance D3, we must solve the close approach problem from point M1. For this purpose, a speed triangle is constructed at point M1. Plotting the speed vector of the reference ship V from point M1, we obtain point a. After making from it a mark with a radius equal to the speed of the maneuvering ship Vm on line of bearing Bo, we obtain point b.

A straight line, drawn from the initial position of the maneuvering ship Mo parallel to speed vector m, will yield the desired course for closing on the reference ship at the designated distance. In order to determine the position of the maneuvering ship at the moment the maneuver ends, we must draw from point M1 a straight line parallel to the line of course of the reference ship until it intersects the line of course of the maneuvering ship. If from point M1 we draw a straight line parallel to the line of intiial bearing, then at its point of intersection with the line of course of the reference ship it will yield its position at the moment of completion of the maneuver. The time required for the maneuver

Let us assume that the reference ship K is at the center of the maneuvering board (Fig. 104). From initial bearing Bo and distance Do we plot the initial relative position of the maneuvering ship Mo. Then from the initial position of

the reference ship K, along the bearing on the maneuvering ship, we plot the distance D3 at which we must close, and obtain point M. Line MoM is the relative movement.

In the center of the maneuvering board, at point K, we construct a speed triangle. After plotting the speed vector of the reference ship Vk, we draw a straight line from its end, parallel to vector 5, until it intersects a circle with a radius equal to the speed of the maneuvering ship Vm, and obtain point b. Connecting the center of the maneuvering board with point b, we read the course for execution of the maneuver Km on the outermost circle of the maneuvering board. Time required for the maneuver

t =

S

=

Do - D3
V

(160)

5. Taking up Station in the Shortest Time With Respect to

a Guide on a Steady Course and at a Constant Speed

Solving the Problem on a Chart

We plot on a chart the initial position of the guide Ko and maneuvering ship Mo (Fig. 105). Then, according to the designated B, and D, from initial position Ko we plot designated relative position of the maneuvering ship M1 with respect to the guide. Connecting with a straight line the old point Mo and M1, we obtain relative movement vector S..

From point M1 we draw a line parallel to the course of the guide. At point M1 we construct a speed triangle. The direction of vector m will yield the desired course for execution of the maneuver Km. Drawing from point M1 a straight line parallel to the line of course of the guide to its intersection with the line of course of the maneuvering ship, we determine its position at the end of the maneuver M1. Drawing from this point a straight line parallel to designated bearing B1, at its intersection with the line of course of the guide, we locate its position at the moment of completion of the maneuver. Time required for the

Solving the Problem on a Maneuvering Board

Let us assume that guide K and the initial relative position of the maneuvering ship Mo (Fig. 106) are at the center of the maneuvering board. From designated bearing B1 and distance D1 we plot the new relative position of the maneuvering ship M1. Connecting the old and new relative positions of the maneuvering ship with a straight line, we obtain the relative movement vector

At point K we construct a speed triangle, draw from the end of speed vector of the guide a straight line parallel to the line of relative movement S, to its intersection with a circle with a radius equal to vector V, and obtain point b. Segment ab is the relative speed vector V. Connecting point K in 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 Km› required in order to alter position. Time required for the maneuver

6. Changing Position With Respect to a Guide Proceeding on a

Steady Course and at a Constant Speed

Solving the Problem on a Chart

We plot on a chart the initial positions of the guide Ko and maneuvering ship Mo (Fig. 107).

Ко

B. D

Fig. 107. Change in position with respect to a guide pro-
ceeding on a steady course and at a constant speed.

Then from designated bearing B1 at designated distance D1 we plot the new relative position of the maneuvering ship M1 and draw from it a line parallel to the line of course of the guide.

We construct a speed triangle at point M. For this purpose, from point M1 we construct the speed vector of the guide . From its end, as from the center,