Intercepting course - anyone know how to plot/calculate one?

[shmoo]

The problem is this:

Boat A (me) is at a known postition. Boat B is also at a known, and different, postition and is moving. Its course and speed are known. How does boat A select a course and speed to intercept boat B. There are clearly many solutions and if boat A's speed is greater than boat B's top speed then it may be impossible. (There are lots of other impossible ones too, like boat A in the English Channel and boat B in Bristol Channel, but let's ignore those for now)

How do I go about this? A reference will do if its too complicated to explain in forum-speak.

Edit: should have made it clear the aim is to meet the other boat. I am not planning to get in to piracy.

 

[Searush]

You need boat B's position, speed and course. Plot them on a chart & estimate times along the line. Then, using your own speed, set a compass to show where you will intercept boat B's course. It may not fall into place first time with a given speed, but you can adjust it & try again.

Alternatively, select a point on B's course where you wish to meet & plot your course & speed to get there at the same tiime as B. It's simple geometry really, provided you know the other guy's current position plus proposed course & speed.

 

[Guest]

How about drawing Boat B's course on the chart as a line and then a line at right angles to it through your point. i.e Closest approach to your starting point. That is the course. The speed is then set by finding the time Boat B will take to get to the point and applying it to Boat A distance to the point.

This will minimize Boat A travel distance. But assumes Boat A is ahead.

However, there are many possibilities. The hardest one would be finding the quickest time to intercept. Which I would guess would be an iterative solution. i.e. keep picking times and plotting where Boat B would be and drawing a position circle for Boat A at its maximum speed. If the circle crosses ahead of where boat B would be, then pick a smaller time. If behind pick a longer time.

 

[CalmSkipper]

You could always phone the guy on Boat B and arrange what pub you're going to meet him at????????/

 

[Powersalt]

HMM seem to remember the u boats did a similar thing< saw it on the movies

 

[philmarks]

Subtraction of vectors! (I think). If you do intercept, then your angle of relative velocity will have been a straight line between the two vessels - ie as if he is stationary and you have travelled straight towards him. BUT, is the interception possible (your max speed may be too slow to achieve it). Need to do this with diagrams really. It's just like plotting course to steer to allow for tide (he is tide) to reach point of intersection, using pair of compasses as other mailer here suggests.

 

[bilbobaggins]

This is one of a class of plotting problems with which all navies and air forces are familiar - 'Tactical Problems - Interception'.....

....Ship meeting ship for a RAS; aircraft meeting aircraft carrier to land on it; aircraft meeting aircraft to intercept or refuel from it....

The graphical solution posted below ( language largely retained ) comes from the RAF's 'Manual of Air Navigation AP 3456G, written largely by Sir Francis Chichester, who was a famous air pioneer before his 'wild streak' got him into hot water with the authorities of the day. Then he took up ocean racing.....

"In the figure a typical interception is shown. The target B is on a ground track of 050º at a known Speed Over the Ground of 400kts. The interceptor A has an known indicated speed of 440kts. The wind velocity ( substitute tidal vector ) is 260º/80kts. ( Substitute as appropriate )

The initial Line of Constant Bearing ( LCB ), marked 'X', is drawn joining the simultaneous position of A and B at 1440hrs.

The heading ( Course to Steer ) required to maintain the LCB to P, the Point of Interception, is determined by the following construction:

a. The LCB BA is transferred through C, the predicted position of target B after 1 hour, to form CY.

b. The wind ( tidal ) vector AD is drawn for the hour 1400-1500.

c. With cebtre D and radius A's indicated speed ( TAS, or speed through the water ) an arc is struck on CY at E. The vector triangle ADE is now completed, and E represents the position of A after 1 hour.

d. The track (COG ) to intercept, AE, is produced to P, the Point of Interception.

e. The Time to Interception can be calculated in several ways: by calculating the time to cover A-P at Speed Over Ground AE, or the time to cover BP at Speed Over Ground BC....."


OK?


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[shmoo]

Thanks folks - the solutions offered are embarrassingly simple now you have kindly pointed them out!

We have managed to meet other boats at sea, with both of us keeping to our orginal plans as far as possible, but its been a bit hit and miss.

I assumed that this problem would be essential part of "naval tactics 101" courses and thus be all over the internet but not the case.

 

[Danny Jo]

I suggest using the "constant bearing" principle used to avoid collisions, only in this case you are trying to arrange the quickest possible "collision" (i.e. rendezvous).

It goes like this. Plot B's current position and course, and mark off B's position in one hour's time (or choose shorter period if expected meeting is sooner). Now plot your current position and determine B's current bearing in relation to you. Using a parallel ruler, draw a line with the same bearing through B's position in one hour's time. Call that line B2. Set your dividers to the distance you can travel in one hour, put one point on your current position and the other point on B2. If it won't reach B2, you won't meet unless he slows down or changes course, or you speed up. If it only just reaches B2, you'll be sailing on parallel courses and will meet at the restaurant at the end of the universe. If there are two points at which your dividers can touch B2, the closer one defines the collision course. Make a mark, join your present position to that mark, and read off your course.

Edit: damn! Beaten to it by one minute. But it's good to know that the principle is already well established.

 

[shmoo]

There were enough key words in bilbo's piece for me to track it down on the internet. The US SAR handbook at www.njwg.cap.gov/docs/jp3_50_1.pdf has a method spookiely like bilbo's quote from Chichester, and it has a diagram.

This forum really is a valuable resource.

Edit: and of course, thanks!

 

[Danny Jo]

I've just being listening to the Who broadcast live from Glastonbury, but my attention keeps drifting back to this question. There must be a simpler way than reading a 140 page manual or getting one's head around complex phraseology. This one doesn't claim to be new, but call it the relativity method.

The objective is B, so treat her current position as the reference point. She's not moving through the water, the water is moving under her, so to speak, and is effectively a current that A has to take into account in calculating a course to steer, in the same way as any other current. The speed of the current is B's speed, and it's direction is the reciprocal of B's course over ground. So just plot your "course to steer" using the conventional vector triangle with A as the starting point, A-B as the desired course over ground, and B's speed and reciprocal of course over ground as the speed and direction of the current.

 

[Bergman]

"aircraft meeting aircraft to intercept or refuel from it...."

And try to get it right this time!

 

[shmoo]

I am quite happy with line of constant bearing method, now I have got my head round it. My first attempts all involved iterative trig and made my head ache. Made the mistake of trying to find best solution (ie shortest time to intercept) rather than simple one!

Of course all this is quite insignificant compared with getting boat B to hold a course for 6 hours!

 

[jimmynoboat]

This is gonna sound really patronising but, ... what a brilliant question!
I really enjoy these exercises but can't understand why I've never thought to try hitting a moving target. And Freestyle's solution seems delightfully simple.

 

[shmoo]

[ QUOTE ]

This is gonna sound really patronising but, ... what a brilliant question!


[/ QUOTE ]

II take praise where I can find it. And as for patronizing, I must point out that since I am married I get matronized regularly and that's much worse.

 

[Guest]

Draw his course and speed over several hours if necessary on any chart or graph paper - making sure that orientation is related to reference lines on paper or long / lat on chart.

Now you draw a line with course you intend to make ... where they cross is the intercept point and distnace divided by your expected speed = time to Intercept.

If you want to meet at a given time ..... then mark of on his course line the location at that time ... now draw from your location to that spot - gives you distnace and course to make ... knowing time needed to get there - gives speed you need to make ...

If you are not fussed about location of intercept - then decide time to meet ... this gives you distnace to run at expected speed for both boats ..... set that on a pair of compass dividers his distance to make ... draw an arc from his location for his distance ... now set to your distance and draw arc from your location ..... the two will meet if its possible and that is the location ... for time and speed ....

Simple Triangles ...

 

[Guest]

The constant bearing method fails if boat A is either nearly behind or ahead of the boat B.

The tiny angles between the lines allow a lot of inaccuracies. Take the limiting case of boat A leaving the same port and boat B, but some time later, and being a faster boat. The extrapolation of the course line after the 1 hour point, to Boat B's course line does not yield a single point (instead a line). Even when there is a small offset, the position accuracy error would be large.

Vastly different relative speeds would also lead to inaccuracies. The iterative algorithm always yields an answer and only requires a sensible guess to start it. The initial guess is the same complexity as deciding the first constant bearing point (1 hour). Choose a bad first point (say 1 minute) and the accuracy of the extrapolation to B's course is inexact also.

 

[Guest]

But now you are being over-calculating and ignoring the practical side.

The intercept if on shallow angle will be a long coincidental line cross - but both will see each other at sufficient distance to basically null and void the error ... In fact it would have to be pretty bad conditons and an extreme case to be really that worthy of worry.

The line cross drawing method is sufficient in 99.999% of cases. Otherwise we would not have been able to manually plot radar targets, sights, intercepts of vessels in war and peace !!

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[Danny Jo]

[ QUOTE ]
The speed of the current is B's speed, and it's (sic) direction is the reciprocal of B's course over ground.

[/ QUOTE ] Aaarghhh! The curse of the pedant - to be guilty of the crime he so despises, the inappropriate apostrophe.

 

[bilbobaggins]

[ QUOTE ]
The Inappropriate Apostrophe......

[/ QUOTE ]

......sounds like a Fleet Street wine bar! Or the pub round the corner from IPC Towers.....

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