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Peter Smith (Smith_Peter@EMC.COM) wrote-

>Pierre Boucher [mailto:pboucher@LAVOILE.COM] said:
>> Here is the problem (Radar) - from Ocean Navigator
>>
>> Own course and speed:
>> 150d   12.0 kt
>>
>> On the radar screen (relative bearing mode):
>> at 0448 a "spot" at 7.0 nm  322d   on the screen
>> at 0554 a "spot" at 5.4 nm  322d   on the screen
>>
>> Request :
>> Other ship's course
>> Other ship's speed
>...
>> The reason for my posting is:
>>
>> Solving with the maneuvering board (relative mouvement NIMA Pub 217
>method)
>> give answers that somehow do not agree with the geographical plot solution
>???
>
>I don't have pub 217 handy, but if it gives a similar example to that
>in the on-line Radar Navigation and Maneuvering Board Manual
>(http://pollux.nss.nima.mil/pubs/pubs_j_show_sections.html?dpath=RNM&ptid=10
>&rid=244)
>you have to be careful. The example in the on-line manual is worked from
>a PPI display -- i.e., north-is-up, as opposed to ship's-head-is-up. For
>the vector addition to work, all vectors must be relative to the same
>reference, either North or ship's heading.
>
>To solve the problem on a maneuvering board:
>
> -- Mark the two target positions: angle 322d, distance 7.0 and
>    angle 322d, distance 5.4. Draw a vector from the first to the
>    second. This is the target's relative motion vector. By my
>    reckoning it is 142d, distance 1.6 .
>
> -- Draw your own motion vector from the origin: the direction of
>    the vector is 000d (your own heading relative to your own heading),
>    length 13.2 (12 knots for 1h 06m).
>
> -- Transfer the target's vector (142d/1.6) to the end of the vector
>    just drawn, giving the vector sum of the two.
>
> -- Draw a vector from the origin to the end of vector sum. This will
>    be the target's course and distance run RELATIVE TO YOUR OWN COURSE.
>    By my reckoning, it is 004.7d, distance=12.0 (which over 1h 06m
>    equals a speed of 10.9 knots)
>
> -- To convert the relative course to true, add 150d (your course) and
>    wah la! Target's course is 154.7d, speed 10.9 .
>
> -- Peter (who also get the wrong answer the first time by applying
>           the example in the manual)

==================

Comment from George Huxtable.

I don't think Peter Smith's solution can possibly be correct: commonsense
tells me so. We have two ships travelling in nearly the same direction, and
the second ship is slowly overtaking the first. In that case, the speed of
the overtaking ship must be greater than that of the ship being overtaken.

I have sent my solution privately to Pierre Boucher, and copy it below.

Velocity of ownship relative to the world is 12 knots at 150 deg

Velocity of othership with respect to ownship is-
1.45 knots ( relative distance change 1.6 miles in 1.1 hours), in a
constant relative direction of 142 degrees (reciprocal of 322), straight
towards ownship. Or if you prefer you can think of it as a relative
velocity of (-1.45 knots), because the distance is decreasing, in direction
322 deg: it's the same thing.

This is an overtaking situation which will end with a collision if no
action is taken.

Velocity of othership with respect to the world is velocity of ownship with
respect to the world plus (vectorially) velocity of othership with respect
to ownship.

You can guess at an approximate solution by observing that the two courses
have to be very nearly parallel so you can simply add the velocity of
ownship to the relative velocity in nearly the same direction and end up
with velocity of othership being about 13.45 knots, direction nearly 150
deg.

If you add the vectors 12 knots at 150 and 1.45 knots at 142, using trig or
drawing, you end up with 13.44 knots at 149.1.

To me, this looks a more reasonable answer than that given by Peter Smith.

George Huxtable.

------------------------------

george@huxtable.u-net.com
George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
Tel. 01865 820222 or (int.) +44 1865 820222.
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