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Navigation problem

7K views 42 replies 21 participants last post by  SailNet Archive 
#1 ·
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
 
#2 ·
I should begin by saying that I have never wondered this before. A relatively easy way occurs to me though assuming that you have and use paper charts. Plot your friend's position, and his course. Next mark of where you expect him to be in .5 hours, one hour, 1.5 hours etc, using a compass and your scale from the chart. Plot your position. Then use the compass to draw arc segments to show where you can be in half hour increments. At some point they should come very close to coinciding, and you can plot a course to that point.

Any better ways?
 
#5 ·
I was going to say that if you figure out where his track will put him in 7 hours, he'll have gone 35 miles. If you aim at that point for seven hours, you will have gone 49. 35 and 49 make the 84 miles you said separated you. However, you said your bearing from him was 96ºT, so that's not right. Yotphix seems to have the right idea, but I'm thinking it will take something like 18 to 19 hours to get to see your friend based on the original distance and bearing.
 
#6 · (Edited)
You describe a maneuvering board problem. You use the maneuvering board sheet according to the directions printed on it.

It is a quick graphic method for calculating relative motion between two moving objects. (Your boat and your friend's.)

It is an excellent way of calculating such things as the course and speed of another vessel or closest point of approach to it. Once the technique is learned, it can be used mentally for a rough but quick result when needed. Good stuff to know.

Here are links to places that can supply a pad of them and a manual for using same:

http://www.celestaire.com/catalog/products/5090M.html
http://www.usskawishiwi.org/Crew/Sparkman/Maneuvering-Board.html
 
#7 ·
John,

The maneuvering board solution, as others have mentioned, is the classic way to solve this problem.

However, in these days of computers there are obviously programs which can do the calculations rapidly.

I don't know what the best of these might be, but there is a little program which was designed for gaming (submarines, torpedos, etc.) which can solve such problems. While the calculations are in yards or meters, and there are some limits to distance, etc., with a little ingenuity you can find the solution.

The program is called Intercalc 1.1 Just Google it any you'll find it.

Bill
 
#8 · (Edited)
It's a trig problem. In cartesian coordinates your buddies boat starts at x1=0, y1=0 and at any time will have x1=2.939t, y1=4.045t. Your boat starts off at x2=83.54, y2=-8.78 and will move from there and at any time have a position of x2=7tsin(a)+83.54, y2=7tcos(a)-8.78 where x and y have units of NM, t is in hours and a is the heading in degrees.

Solving this for when x1=x2 and y1=y2, you catch him after ~10.5 hours with your boat's heading of ~314 degrees.
 
#9 ·
CapnHand...I was trying to work that out myself on graph paper. Knew there had to be a trig solution but was at a loss! Are you sure of your answer? At 84 miles away...it would take 12 hours at 7 knots for "my" boat to reach my friends if it were stationary and in a position almost due East from me.

Since the friends boat is moving away from my boat on a course of 36 degrees T ...(Northeast)it does not make intuitive sense that I could close the gap in less time...especially with a heading of 314T or northwest.
It has been a long time since high school...where am I going wrong here?

I am thinking that your starting position for my boat is perhaps different than mine based on our reading of the problem. I am assuming my friends boat is on a bearing of 96T taken from my boat...perhaps you are reading it as a bearing of 96T taken from my friends boat??
 
#11 · (Edited)
camaraderie said:
CapnHand...I was trying to work that out myself on graph paper. Knew there had to be a trig solution but was at a loss! Are you sure of your answer? At 84 miles away...it would take 12 hours at 7 knots for "my" boat to reach my friends if it were stationary and in a position almost due East from me.

Since the friends boat is moving away from my boat on a course of 36 degrees T ...(Northeast)it does not make intuitive sense that I could close the gap in less time...especially with a heading of 314T or northwest.
It has been a long time since high school...where am I going wrong here?

I am thinking that your starting position for my boat is perhaps different than mine based on our reading of the problem. I am assuming my friends boat is on a bearing of 96T taken from my boat...perhaps you are reading it as a bearing of 96T taken from my friends boat??
The post read "I am 84NM away on a bearing of 096T" so I took both the distance and bearing to be relative to the friend's boat. If it stated that "HE is 84NM away on a bearing of 096T", then I'd have taken the distance and bearing relative to his boat.

So from the perspective I took, yes, I'm sure of my answer. Does my answer make more sense now? The fact that you disagreed indicates that you were thinking about it too and that's good.

Last time I was in high school was to pick up my son. This was a good mental excercise, jrd. Thanks for posting the question.

There's another way to figure this out using vector math instead of cartesian coordinates. It's simpler and more elegant, but I couldn't remember how to do it last night, so I just went ahead with the trig. It's coming back to me, I'll post it when I get it straight. If someone else knows, have a go.

Funny thing is, there's probably a math forum out there somewhere with people discussing roller furlings.
 
#12 ·
The maneuvering board is just a piece of paper with a polar plotting grid printed on it. For the problem you describe, all you need do is to put your vessel at the center and plot your friends range and bearing from you. That wil establish the initial relative positions of both boats.

Now, plot a vector representing the course and speed of your friend's boat, originating at the center of the plot. Draw a line parallel to the his relative bearing from you and long enough to intersect the circle that represents your vessel's speed. The intersections thus drawn will be at the courses that will give (1) best closing speed, and (2) best opening speed relative to your friend.

There are also other techniques that allow you, for example, to estimate course and speed of another vessel from its observed relative motion.

The principles are especially useful when using radar because of its ranging capability.
 
#14 ·
#15 ·
uh, you'll see him tuesday... or weds, or about an hour before your beer runs out.
 
#16 ·
This could be a CPA problem using real time vector (radar) Or using "Dutton's" and follow their directions on using the maneuvering board. Either way will solve your intercept problem.
Mark
 
#17 ·
CapnHand...yup that was the problem...180 degrees out of "phase". You answer makes perfect sense now. Thanks.

Funny thing is, there's probably a math forum out there somewhere with people discussing roller furlings.
The really funny thing is that with our penchant for thread drift (mea culpa too!) ....we'll end up talking about roller furling on this thread!!
 
#18 ·
I have to agree with Camaraderie, it would have to be an Easterly heading. I think the trig approach is correct, if only I knew how to do it. I am home studying for my 100 ton cert. and have not been able to find an explanation of how to solve this in any of the huge stack of books and reference material I have. Thanks for all the ideas, this one makes set and drift problems look easy. John
 
#19 ·
For some reason when I checked in I didn't get the last replies, CapnHand you must be a genius, your answer makes perfect sense from the reference you used. If you think of the trig solution I'd really appreciate you posting it. That question has made me question my sanity trying to "go it alone" for my captains license instead of taking a class. Thanks, John
 
#20 ·
Problem Solved, I Think

OK, John...

I think this will do what you want.

There's a shareware/freeware set of nav programs called EZ-NAV93.EXE which contains a number of useful navigation programs, including one called INTERCPT.EXE which will solve the problem you posed.

You can find EZ-NAV93.EXE at: http://ftp.linux.org.uk/pub/Navigation/

Just download it, save it to your computer in its own folder, and run it. It will automatically generate the 12 nav programs in the same folder, together with documentation. Run the INTERCPT.EXE program, and away you go.

By the way, the answer to your problem seems to be:

28 hours to intercept on a course of 57.8T and a distance of 196 nm.

This assumes: (1) that the program is accurate; and (2) no effect of current or leeway.

Cheers,

Bill
 
#21 · (Edited)
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=&notify=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
 
#23 · (Edited)
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
 
#24 · (Edited)
CapnHand said:
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 bearing 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=&notify=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
While I applaud your attempt to answer the question, I see it as seriously flawed.

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 73.8˚T. His position has now become 52.5nm @ 36˚T + the original 84NM @ 96˚T 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... :D 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.
 
#25 ·
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
 
#26 ·
True... the wording is somewhat unusual...and makes a huge difference in the calculations. :D
 
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