Wow, lotta incorrect information here. I've got a US Navy maneuvering board textbook that I found online dated 1941.
Page 26:
(copy pasted)
Case XI
TO DETERMINE COURSE, SPEED, AND RELATIVE POSITION OF A TARGET BY BEARINGS ALONE
GIVEN: TWO SETS OF THREE OR MORE TIMED BEARINGS, EACH TAKEN ON A TARGET BY AN OBSERVING
UNIT WHICH CHANGES ITS OWN KNOWN COURSE OR SPEED OR BOTH BETWEEN SETS OF
BEARINGS.
TO DETERMINE: COURSE, SPEED, AND RELATIVE POSITION OF THE TARGET.
Example.—An observing vessel G, while on course 110°, speed 15.0 knots, obtains radio bearings on target vessel Mas
follows:
Time 0800. Bearing of M 000°.
Time 0900. Bearing of M 349°.
Time 1000. Bearing of M 336°.
At 1015 the observing vessel changes course to 045° and increases speed to 20.0 knots. Bearings are next received as
follows:
Time 1030. Bearing of M 327°.
Time 1130. Bearing of M 302°.
Time 1230. Bearing of M 273°.
Required.—(a) Course and speed of M. (b) Relative position of M at 1230. (See fig. 15.)
Procedure.—Plot position of observing ship at any convenient point G, and lay out the 0800 bearing line as G . . . . bu
the 0900 bearing as G .... 62, and the 1000 bearing as G .... 63. By any of the methods shown for case IX, draw a slope
P .... Q .... R across these bearing lines so inclined that the intercepts P .... Q and Q . . . . R are proportional to the
time intervals between bearings.
In a similar manner lay out the second set of bearings, G . . . . bit G . . . . b5 , and G . . . . &6 - Draw the
slope T .... U .... V so inclined that the intercepts T . . . . U and 17 .... V are proportional to the time between
these bearings.
From any point e, lay out the first vector of G as e . . . . gu and transfer the slope P .... Q .... R to gi. Draw the
second vector of G, e . . . . g2 , and transfer the slope T . . . . U .... V to g^. This intercepts the slope from gi at zn.
e . . . . m represents the course and speed of M.
Determine the Relative Speed g2 . . . . m and by means of the Logarithmic Scale find the Relative Distance travelled
by M between the 1030 and the 1230 bearings. Lay off this distance, T . . . . W, and by completion of the parallelogram
locate the position M at 1230. T' .... M is equal to T .... W and is the Line of Relative Movement between 1030 and
1230. M's bearing and distance from G at 1230 is G .... M.
Answer.—(a) Course 164°, speed 12.6 knots, (b) 57.5 miles bearing 273° from observing vessel.
NOTE.—Solution by this method will not be obtained if the second vector of G should lie along the transferred slope P .... Q .... R.
For this reason this slope is transferred to g\ before G changes either course or speed or both.
If the bearing does not change in either set, a solution is still obtainable. The slope in this case is a constant bearing and is laid off in both
directions from the head of the vector concerned.
In case G makes a change of course when the last bearing of the first set is obtained, this bearing may be used as the first bearing of the
second set.
When bearings alone are available, the results should be considered as approximations only. This is occasioned by the fact that a small error
in one or more bearings will change the inclination of the slopes to be transferred, and this in turn will change the final results. If sufficient time
is available, a third set of bearings will act to check the course and speed of the target.
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