Not celestial navigation...but

T37 Phil · Dec 3, 2007

Clarification

An object bearing 090 from a boat on a course of 290 would be at a compass heading of 020 (on the slow boat's compass, ignoring any variation&deviation). If that object is a boat steering a course (heading) of 310 on its compass (tracking 310??), then the two paths definitely intersect. If so, then I agree with Ross' first description, but not his numeric values. The faster boat reaches the intersection point first and the closest approach occurs at a bearing of less than 90 degrees.

Ross · Dec 3, 2007

The faster boat would cross the course of the

slower boat in 45 minutes, if they were on a converging course. In that time the slower boat would travel .75 times 6.5 knots.

Roger (Ninette Manitoba) · Dec 3, 2007

the PLOT thickens

(Couldn't resist that pun!)Okay, I plotted it out with my new assumptions, verified by Ross that indeed the contact was 090 (from north ref). Given that, and that they are on divergent courses, the closest he will ever be is when directly abeam. He is travelling at 14 kts so, with me at my 6.5 he is gaining on my by 7.5 kts/hr, ie he will be 7.5 nm ahead of his present positon on his track so I advance his position 7.5 nm downrange from his present position. I then draw an intersecting line directly abeam from my track to intercept that point (where he will be 7.5 nm from his original position). I then measure on my track from my current position to the line that intersects my track when he will be directly abeam, and measure that distance to determine that I have travelled a distance of 4 nm at 6.5kts. (divide 4 by 6.5) and I get .615 of an hour (60 min x .615 = about 37 minutes until I get to his nearest position. When I measure the bearing it is 020 north ref when we are closest and we will be separated by 4.75 nm.From my present position he we will be closest at a point that bears 331(north ref), about 5.3 nm on that bearing line

Roger (Ninette Manitoba) · Dec 3, 2007

frivolous answer

Of course, if he has beer, the obvious answer is to power up, bear down on him using as acute an intercept angle as possible, take his beer, and the problem becomes moot!

Ross · Dec 3, 2007

I have been troubled by this problem and just

reworked it. This is what I did : On a paper I drew an east-west line and a north-south line. Through this intersection I drew a line at 290 degrees and called this line "my course" I also lightly drew a line at 310 degrees and another at 90 degrees to "my course" on this line I marked off 3.5 miles and with my parallel rules I moved the 310 degree line to the 3.5 mile mark and drew that line and marked it "bogey". The lines extend forward and back from the intersect and they converge prior to the time stated in the problem.Please try this and check my work I think it is correct. My original numbers are reasonable but closest approach and convergence I believe are prior to the time of the problem not after as I stated originally.

T37 Phil · Dec 3, 2007

Ross

I see your picture now. I don't have my napkin from last night, so I'm not sure what I was thinking. I think I may have had the 290 and 310 reversed (too much salsa). Anyway, I agree the intersection is behind them. I also calculate about 45 minutes to the intersection. I guess we'll have to wait to see what Paul actually had in mind.

GuyT · Dec 4, 2007

Intersection only occurs..

If the faster boat changes direction.Its quite easy to see that you will never intersect with those headings.An interesting problem would be to show how they could intersect.If the faster boat changed course to 228 degrees, he would reach you after you travelled 1.84 KM. He would have travelled 3.95 KM.

Ross · Dec 4, 2007

Paul It is called analytical geometery.

GuyT, Unless two lines are parallel they will intersect somewhere if they are in the same plane. In this case the courses intersected before the time in the problem.

FrankR3878 · Dec 4, 2007

Actually there are two points of path intersection

Since the earth is a sphere any two great circles will intersect twice- one on either side of the earth. This assumes that continential drift will occur more rapidly, the ice caps will melt and both boats continue on their present course for at least 3323 more hours. As far as collision occurring - very remote.

GuyT · Dec 4, 2007

When BOATS intersect - not lines

I thought my post was pretty clear that I meant when the BOATS would intersect - not their course.I thought Paul was trying to get to his contact by changing his approach.If that is the case - he cannot.His contact needs to change their approach.Maybe I am reading that wrong??Ross, I agree with you that their intersection of their courses happened behind them and I'm pretty sure that thing you said about lines is correct

Ross · Dec 4, 2007

By calculator our boat crossed the course of the

"bogey" 1 hour and 23 minutes ago. The "bogey" crossed our course 44 minutes ago. At that time we were 5.3 miles ahead of the "bogey". My math isn't good enough to determine closest approach. The courses crossed 9.6 miles astern our boat, 10.23 miles astern the "bogey"

Ross · Dec 4, 2007

Brian, it is time to buy some of the tools of the

trade. Dividers, protractor, parallel rules and pencil are all you need. Paper is optional but helpful if drawing on the walls upsets anyone.

GuyT · Dec 4, 2007

By your terms and definitions of what the

closest approach is Ross, the nearest approach would be when the faster boat has the slower boat directly off its port beam.You know the speeds, angles and the distance relationships, all that is left to do is some trigonometry.

Ross · Dec 4, 2007

Please do it Guy and show us the math.

I haven't been able to work it out.

Rick Webb · Dec 4, 2007

Why not Just Use the MARPA Button?

That is much easier than getting out the maneuvering boards and doing it by hand

sailaway78 · Dec 4, 2007

Simple Logic

Lots of very interesting answers to this little problem. Mine is this. Simple logic tells me that if he is 90degrees off my beam, 3.5 miles away and heading 20degrees off my base course, that we are as close as we are going to be. This should require no paper, calculator, divider or rulers. This problem is a good example of why we, including myself, many times make a much more complicated deal out of something than it really is.

T37 Phil · Dec 4, 2007

Lemonade

I suspect that Paul misunderstood/miscommunicated the original problem, believing that there "will be" a point of closest approach, much as you are focusing on close as we're "going to be", not close as "we were". From there on out it's lemonade, which I happen to favor. I particularly like Frank's great circles. Now there's a sailor looking ahead! And I was going to go for non-Euclidean remarks!

Brian M H23 · Dec 4, 2007

I'm going to do this mathematically

However, I think my first try (i posted and then deleted) was wrong, because I put the "bogey" to the east of me, not off my beam.PS: Ross, I didn't have them to whip out during class...but (unfortunately for me, I guess) my first thought it to solve it like this:** x goes west...y goes north**Boat 1heading of 290. tan 290 ~ slope of 2.75Postion of 0,0Boat 2Boat 2 heading of 310. tan 310 ~ slope of 1.20Postion of 3.5 nm at 20..sin 20 = .34...cos 20 = .94 ... position of -1.2 , 3.3This boat is traveling 2.15x faster so I need to scalesqrt(2.75^2 + 1^2) = 2.93...multiply by 2.15 i get 6.3sqrt(1.2^2 + 1^2) = 1.56...6.3 / 1.56 gives me a velocity scalar of 4So my boat travels 2.75 west and 1 north in every timestep.Their boat travels 4.8 west and 4 north in every timestep.Every timestep is equal to the 14 knot boat going 6.25 knots, or 26.7 minutes. I could correct for this if I wanted to make 1 timestep 30 mins, or whatever, but i just need to know the conversion.Parametrics:1x = 2.75t1y = t2x = 4.8t-1.22y = 4t+3.3Distance between boats ==sqrt[(x1-x2)^2 + (y1-y2)^2]=sqrt[ (-2.05t+1.2) ^2 + (-3t-3.3)^2 ]=sqrt[ 4.2t^2 - 4.9t + 1.44 + 9t^2 +19.8t +10]Note: because sqrt operates evenly over all times, i can remove it to calculate nearest approach. but, I can tell you that the closest approach is just before time 0 just from looking at this. The distance at t0 is sqrt 11.44, or about 3.4 nm, so i obviously rounded a little bit more than I would have using a calculator.Does anyone see a mistake here?edit/ disclaimer : i used a trig lookup: is this allowed. Do navigators carry these with their other tools?