Are there any navigators here? What is the procedure for plotting an intercept course, as in meeting another boat that is approaching from a different direction. For example, if my friend is on a steady course of 036T and travelling at 5K and I am 84NM away on a bearing of 096T and I make 7K, how do I plot a new course to intersect him so we arrive at the same spot at the same time?(assume a constant track and speed) Anyone? Thanks, John Davidson
Navigation problem
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jrd22
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Ok, so this is simpler than I first thought. This is way I should have done it the first time.
First of all, this is the situation. The distance the friend's boat will travel is 5t (t = hours) NM on a course of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.
It's obvious that the angle between the track of your friend's boat and the bearing from his boat to yours is 96 - 36 = 60 Deg
http://www.sailnet.com/photogallery/bulkupload.php?ppaction=addphotos&do=preview&photopath=131308&upuser=¬ify=no&defcat=500&deftitle=&defdesc=&defdesc=&keywords=&numprocess=10&processall=no&dthumbs=
Remember the law of sines? It's handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.
If A is the angle between the heading you need and the bearing from your boat to your friends,
that means: sin60 / 7t = sinA / 5t
or sin -1 (5/7 * sin 60) = A
A = 38.2 Deg
Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T
How long to get there?
We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 - 38.2 - 60 = 81.8 Deg
From the law of sines, 7t = 84 * sin 60 / sin 81.8
t =10.5 hours
First of all, this is the situation. The distance the friend's boat will travel is 5t (t = hours) NM on a course of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.
It's obvious that the angle between the track of your friend's boat and the bearing from his boat to yours is 96 - 36 = 60 Deg
http://www.sailnet.com/photogallery/bulkupload.php?ppaction=addphotos&do=preview&photopath=131308&upuser=¬ify=no&defcat=500&deftitle=&defdesc=&defdesc=&keywords=&numprocess=10&processall=no&dthumbs=
Remember the law of sines? It's handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.
If A is the angle between the heading you need and the bearing from your boat to your friends,
that means: sin60 / 7t = sinA / 5t
or sin -1 (5/7 * sin 60) = A
A = 38.2 Deg
Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T
How long to get there?
We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 - 38.2 - 60 = 81.8 Deg
From the law of sines, 7t = 84 * sin 60 / sin 81.8
t =10.5 hours
Capt Hand,
Yeah, easy, but I believe confusing on several counts.
First, the friends boat (Vessel #1) is on a COURSE of 036T, not a BEARING. His speed is 5 knots. His distance from your boat (Vessel #2) is 84nm. The BEARING to/from the vessels is ambiguous, the way it was presented. Normally, bearings are taken FROM your vessel, but the wording in the original is ambiguous, as in your solution. Either way, the answers provided do not agree with the computed values. Your boatspeed is 7 knots.
Let's assume that the bearing of 096T was, indeed, from Vessel 1 to Vessel 2. The reciprocal bearing, from your vessel to your friend's vessel, is then 276T.
If we then set up the problem in a conventional way, it looks like this:
Vessel #1 (friends boat): Bearing 276T Distant 84nm Course 036T Speed 5 knots
Vessel #2 (your boat): speed 7 knots
Required: course, time, and distance to intercept
According to the calculator (both of the aforementioned calculators agree):
Course to intercept: 314.21 T
Time to intercept: 10 hours 30 mins
Distance to intercept: 73.5 nm
Bill
Yeah, easy, but I believe confusing on several counts.
First, the friends boat (Vessel #1) is on a COURSE of 036T, not a BEARING. His speed is 5 knots. His distance from your boat (Vessel #2) is 84nm. The BEARING to/from the vessels is ambiguous, the way it was presented. Normally, bearings are taken FROM your vessel, but the wording in the original is ambiguous, as in your solution. Either way, the answers provided do not agree with the computed values. Your boatspeed is 7 knots.
Let's assume that the bearing of 096T was, indeed, from Vessel 1 to Vessel 2. The reciprocal bearing, from your vessel to your friend's vessel, is then 276T.
If we then set up the problem in a conventional way, it looks like this:
Vessel #1 (friends boat): Bearing 276T Distant 84nm Course 036T Speed 5 knots
Vessel #2 (your boat): speed 7 knots
Required: course, time, and distance to intercept
According to the calculator (both of the aforementioned calculators agree):
Course to intercept: 314.21 T
Time to intercept: 10 hours 30 mins
Distance to intercept: 73.5 nm
Bill
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While I applaud your attempt to answer the question, I see it as seriously flawed.CapnHand said:
Given that your distance from his boat is 84NM to start with, and his is moving away from you, and you only move at 7 knots... I don't see how the answer could be 10.5 hours.
Draw the vectors... In 10.5 hours you have gone only 73.5nm on a heading of 73.8ËšT. His position has now become 52.5nm @ 36ËšT + the original 84NM @ 96ËšT in the same 10.5 hours-119NM from your original position, or 45.5NM further along the same heading.
If you assume you were at coordinate (0,0) and have moved to (70,20.5)... boat B's position has gone from (83, -8.5) to (114, 33). You're still nowhere near them...
Your solution assumes an interior angle of 60 degrees, but it has an interior angle of 120 degrees (96 -> 36). However, you are on a direct vector to their current position.dog,
I think his time to get there (10.5 hours) is correct, that is, if you assume that the ambiguous bearing info given is actually the bearing FROM the friend's boat TO your boat, as calculated in my response.
This confusion would not have existed if the original post had been unambiguous and/or phrased in the conventional way (i.e., FROM the maneuvering vessel TO the target vessel).
Bill
I think his time to get there (10.5 hours) is correct, that is, if you assume that the ambiguous bearing info given is actually the bearing FROM the friend's boat TO your boat, as calculated in my response.
This confusion would not have existed if the original post had been unambiguous and/or phrased in the conventional way (i.e., FROM the maneuvering vessel TO the target vessel).
Bill
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True... the wording is somewhat unusual...and makes a huge difference in the calculations.
On a paper chart, plot your friend's DR track out past the time you want to meet, then set a compass to the distance you can make during that same period. Set the compass point on your current position, and strike an arc through your friend's DR track; this is point where and when the meeting will take place, if you steer the course between your current position and there. If the arc does not intersect your friends track, one of your assumptions is unrealistic (speed, etc.). You can adjust for set and drift as you near each other. That's the way I did it thirty years ago, in the Navy, on the bridge.
tmhdgpth,
That's a useful technique, but doesn't respond directly to the problem as stated.
This is because there is no fixed time (time desired) for meeting. Rather, the desired meeting time is the EARLIEST time possible, given the relative positions of the vessels, the (fixed) course of one of them, and their respective speeds.
It's a classic intersect problem used in war gaming and targeting, as well as air and maritime navigation. Maneuvering Boards were the classic way to solve the problem, and a few brave souls used trig, but now calculators and computers make it easy.
Regards,
Bill
That's a useful technique, but doesn't respond directly to the problem as stated.
This is because there is no fixed time (time desired) for meeting. Rather, the desired meeting time is the EARLIEST time possible, given the relative positions of the vessels, the (fixed) course of one of them, and their respective speeds.
It's a classic intersect problem used in war gaming and targeting, as well as air and maritime navigation. Maneuvering Boards were the classic way to solve the problem, and a few brave souls used trig, but now calculators and computers make it easy.
Regards,
Bill
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I'm impressed with all the effort that has gone into solving this. I apologize for the ambiguity in my original post. I meant the relative bearing TO the other boat (friend)was 096T, but did not make that clear. The actual USCG question uses different courses, I was looking for the method, not an answer, and if I understood trig better (at all?) I think you guys have done it. The testing is multiple choice with four possible answers given. The answers only vary by 2-3 degrees each so you have to be right on. In the actual test nothing except a simple calculator, plotter, parallel rule, dividers and a pencil are allowed. By the way, the CG has a data bank of 14,000 total questions, (Navigation General, Rules of the Road, Deck General, Navigation problems) and I am working my way through all of them(lot's of duplicates). Thanks again for all the help and ideas. If I come across something else that stumps me I'll post it. Now, back to how many round turns to put on a bitt before figure eights using nylon line) John
) John
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The more I read, the easier the maneuvering board solution seems to be.
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Oops, my bad, thanks for catching that. I've edited the post to correct it.btrayfors said:
As for the question of the initial bearing from one boat to the other, I responded to the post the way it was asked. This was explained several posts back and a link to a diagram of the vectors was included.
If the bearing of the intial positions is interpreted in the conventional way, from your boat to the friend's boat, then the internal angle between the initial bearing to the friends boat and his course will be 120 deg. The solution is to head on a course of 57.8 deg. The boats will intercept in 28 hours.
Is this horse dead?
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Okay guys, I have a really easy fix to this problem with the least amount of errors that could occur.
One guy heaves-to and reports their position
the other guy sails to his position.
Done!
Take drift and current into account of the other boat, and it's flawless!
One guy heaves-to and reports their position
the other guy sails to his position.
Done!
Take drift and current into account of the other boat, and it's flawless!
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To the best of my knowledge, you are allowed to use manoeuvering board sheets in the USCG exam and, as previously mentioned, they come complete with the directions on them. It really is the easiest, most fool-proof way to do the problem. They can also be used for small area plotting sheet construction used in celestial navigation. A word to the wise on USCG exam questions (I've passed four, not counting renewals). The multiple choice answers have accounted for the, "oh good, it's multiple choice if I'm close the answer will be right there" factor. The incorrect answers are the result of the most common errors made in answering the question and, as such, will be exactly wrong. Clever people those CGs. Good luck and remember to read the question. I taught the navigation portion of the license prep course at the USMMA and one of the most common complaints was, "oh, I thought they meant something different". The wording must be read and interpreted exactly as written and not as to what you "think" it's trying to say. (I know that doesn't make sense to those who haven't been there-just think in terms of a double trick question!) If you are in doubt about the accuracy of the wording of the question, you can protest it in writing, but the protest writing comes out of your exam time. Don't waste time asking the proctor for an explanation of the question, he is generally a petty officer with little or no knowledge of the exam itself-he's just there to pass out the papers and monitor.
Just nit-picking, but relative bearings are not marked True, Magnetic, Compass or otherwise. Relative bearings are only related to where the bow is and the bow doesn't change regardless of what choose to call the course you are on.
As btrayfors said, it is of utmost importance to use these navigational terms with rigorous exactitude. Mis-speaking can lead to mis-construing, and that can ruin your day at sea. As Robert Ganier, who has also taught navigation, will tell you-the devil is in the details. Small math mistakes ruin the other 99% done correctly. A "passing" grade on the navigation portion of the exam is 90% on the principle that the correct answer is right where the ship is located and an incorrect answer is, well.....aground.
Allowing for set and drift does not detract from the relevance of the question. We always have to factor that in, usually as we go, and it doesn't change the fact that we do have to set an initial course. I suppose one could intercept Bishop Rock starting with even a westerly heading, but something in the northeastern quadrant might be less time consuming.
Just nit-picking, but relative bearings are not marked True, Magnetic, Compass or otherwise. Relative bearings are only related to where the bow is and the bow doesn't change regardless of what choose to call the course you are on.
As btrayfors said, it is of utmost importance to use these navigational terms with rigorous exactitude. Mis-speaking can lead to mis-construing, and that can ruin your day at sea. As Robert Ganier, who has also taught navigation, will tell you-the devil is in the details. Small math mistakes ruin the other 99% done correctly. A "passing" grade on the navigation portion of the exam is 90% on the principle that the correct answer is right where the ship is located and an incorrect answer is, well.....aground.
Allowing for set and drift does not detract from the relevance of the question. We always have to factor that in, usually as we go, and it doesn't change the fact that we do have to set an initial course. I suppose one could intercept Bishop Rock starting with even a westerly heading, but something in the northeastern quadrant might be less time consuming.
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camaraderie: CapnHand's solution is indeed assuming that my friend's boat is to the west of my boat.
Given that interpretation of the problem, CapnHand's solution is correct.
Kind of a fun little algebra problem!
Given that interpretation of the problem, CapnHand's solution is correct.
Kind of a fun little algebra problem!
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Yep..Epic, That WAS fun...now my brain is full...may I be excused? <g>
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Did anyone suggest that the two vessels sail reciprocal courses based on their relative positions? (i.e. toward each other.) It would be the quickest way to close on each other.
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Looks like the wind is about 10 deg to the reciprocal course. The optimal courses will depend on how well the boat sailing upwind will point.Goodnewsboy said:
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