Not celestial navigation...but
- Thread starter Paul H
- Start date
- Status
You're right, Ross
The problem stated: "there is a contact bearing 090 @..." Bearing 90 means that it is not off your beam (perpendicular to your own fore and aft) but rather at 90 measured by a hand bearing or other compass. The problem didn't say directly abeam and did say bearing 90 degrees, so your plot makes sense. I figure they're widening the distance between them and are already at their closest point. Evidently, the use of language becomes extremely important in stating, and, hence, resolving the question. Gee, maybe that's why the container ship hit the Oakland - San Francisco Bay Bridge on November 7th. Maybe there's an ezxperienced submariner here who can help out on the terminology, but in all the war movies I've seen, the bearing is NOT from the lubber line of the sub, but rather the (compass or true) bearing to the target.
The problem stated: "there is a contact bearing 090 @..." Bearing 90 means that it is not off your beam (perpendicular to your own fore and aft) but rather at 90 measured by a hand bearing or other compass. The problem didn't say directly abeam and did say bearing 90 degrees, so your plot makes sense. I figure they're widening the distance between them and are already at their closest point. Evidently, the use of language becomes extremely important in stating, and, hence, resolving the question. Gee, maybe that's why the container ship hit the Oakland - San Francisco Bay Bridge on November 7th. Maybe there's an ezxperienced submariner here who can help out on the terminology, but in all the war movies I've seen, the bearing is NOT from the lubber line of the sub, but rather the (compass or true) bearing to the target.
To answer your question Ross
There is no direct equation to do the problem. You have to find out a critical piece of info first. But the answer is: Your boat passsed the nearest approach 0.987nm ago. The nearest approach is 3.01nm. How did I do it?? Drop a tangent line from your fast boat course so it intersects with your current position - that distance is 1.197nm. Find that by sin(20)*3.5 = 1.197. The speed ratio is 14k/6.5k = 2.154 So the equation you have been waiting for is: 2.154a = 1.197 - sin(70)a a=.987. That is the distance your boat has to go(backwards). Hope that helps
There is no direct equation to do the problem. You have to find out a critical piece of info first. But the answer is: Your boat passsed the nearest approach 0.987nm ago. The nearest approach is 3.01nm. How did I do it?? Drop a tangent line from your fast boat course so it intersects with your current position - that distance is 1.197nm. Find that by sin(20)*3.5 = 1.197. The speed ratio is 14k/6.5k = 2.154 So the equation you have been waiting for is: 2.154a = 1.197 - sin(70)a a=.987. That is the distance your boat has to go(backwards). Hope that helps
I laugh now Ross
while you try to figure out my beautiful blue smoke. You'll eventually see that I have it right. Since you are graphical, I might be able to post my solution via a graph tomorrow if you would like. Would you? It has been great fun!
while you try to figure out my beautiful blue smoke. You'll eventually see that I have it right. Since you are graphical, I might be able to post my solution via a graph tomorrow if you would like. Would you? It has been great fun!
One from the peanut gallery
Forget the compass, rotate the problem s1(t) = slow boat distance from initial observation time s2(t) = fast boat distance from initial observation time s1=(-6.5t)i s2=(-14(cos 20)t)i + (3.5 + 14(sin 20)t)j d = s2-s1 = (-6.6t + 14(cos 20)t)i + (3.5 + 14(sin 20)t)j |d|^2 approx 67.29t^2 + 33.52t + 12.25 t* where (|d|^2)'=0 t* = ~ -.25 hrs |d(t*)| = 2.8 nm s1(t*) = 1.62i (ie ~1.6 nm backwards on its course from t=0 position) s2(t*) = 3.28i + 2.3j (ie backward ~3.3 nm along line parallel to s1 course and 2.3nm perpendicular from that course compared to 3.5 at t=0) There's my answer. For comparison, at t = -.7 hr s1(-.7) = 4.55i s2(-.7) = 9.2i + .15j (just about to cross the course of the slow boat, way after the slow boat was there [text edited]) |d(-.7) |= 4.66 nm GuyT, 1) I believe you meant drop a perpendicular. 2) I don't see where Ross was looking for the distance you calculated. 3) The perpendicular you found (1.197) is not the closest approach distance. That would assume that the fast boat was at the other end of that perpendicular at the same time you are at the present location, which is clearly not the case. It does get the distance from the slow boat to the intersection point, which is perhaps all you were trying to do.
Forget the compass, rotate the problem s1(t) = slow boat distance from initial observation time s2(t) = fast boat distance from initial observation time s1=(-6.5t)i s2=(-14(cos 20)t)i + (3.5 + 14(sin 20)t)j d = s2-s1 = (-6.6t + 14(cos 20)t)i + (3.5 + 14(sin 20)t)j |d|^2 approx 67.29t^2 + 33.52t + 12.25 t* where (|d|^2)'=0 t* = ~ -.25 hrs |d(t*)| = 2.8 nm s1(t*) = 1.62i (ie ~1.6 nm backwards on its course from t=0 position) s2(t*) = 3.28i + 2.3j (ie backward ~3.3 nm along line parallel to s1 course and 2.3nm perpendicular from that course compared to 3.5 at t=0) There's my answer. For comparison, at t = -.7 hr s1(-.7) = 4.55i s2(-.7) = 9.2i + .15j (just about to cross the course of the slow boat, way after the slow boat was there [text edited]) |d(-.7) |= 4.66 nm GuyT, 1) I believe you meant drop a perpendicular. 2) I don't see where Ross was looking for the distance you calculated. 3) The perpendicular you found (1.197) is not the closest approach distance. That would assume that the fast boat was at the other end of that perpendicular at the same time you are at the present location, which is clearly not the case. It does get the distance from the slow boat to the intersection point, which is perhaps all you were trying to do.
Ross, re your #45
Yes, you're right, I was wrong in my #43. If the bearing of the bogey was 90 degrees it would be off the starboard quarter. Paul, I just don't understand your "solution" and explanation, because nowhere in there is the 90 degree bearing to the bogey, that I can see or understand. HAAAALP!
Yes, you're right, I was wrong in my #43. If the bearing of the bogey was 90 degrees it would be off the starboard quarter. Paul, I just don't understand your "solution" and explanation, because nowhere in there is the 90 degree bearing to the bogey, that I can see or understand. HAAAALP!
Backwards
Paul, Either you have a backward arrow somewhere or you have violated the assumptions required for using that board. It's -.25 hr, 2.8nm, well within the error of your graphical method. Your relative is the vector difference in velocities and indicates an increase in distance from the time the bearing was taken. [Text added] You've given the bogey bearing as 090 true as perhaps a radarman would instead of as a watchman would report it, 90 deg off the starboard bow or on the starboard beam (as others have pointed out). The funny thing is that the results are related due to the symmetry of the problem.
Paul, Either you have a backward arrow somewhere or you have violated the assumptions required for using that board. It's -.25 hr, 2.8nm, well within the error of your graphical method. Your relative is the vector difference in velocities and indicates an increase in distance from the time the bearing was taken. [Text added] You've given the bogey bearing as 090 true as perhaps a radarman would instead of as a watchman would report it, 90 deg off the starboard bow or on the starboard beam (as others have pointed out). The funny thing is that the results are related due to the symmetry of the problem.
darn traffic is a pain
i guess ill just have to be ready to report a collison or change course parallel to the fast mover and wait till he goes by! You guys are way to good at this!!
i guess ill just have to be ready to report a collison or change course parallel to the fast mover and wait till he goes by! You guys are way to good at this!!
Inverse stealth
As it happens, I've never had my own radar to play these games. Maybe I'll work up the explicit relationship between the two solutions (true brng versus brng from course line), to satisfy my own curiousity anyway.
As it happens, I've never had my own radar to play these games. Maybe I'll work up the explicit relationship between the two solutions (true brng versus brng from course line), to satisfy my own curiousity anyway.
Phil, I was a radar operator in the Navy
for 16 years...so yes, I guess I automatically assume true bearings. The method that was used to calculate on the maneuvering board is the same way that it's calculated in radar navigation courses that I've taken... The rules of the arrows are...you (single arrow) chase the relative (circled arrow), and the relative and the bogey (double arrow)converge. I don't think I violated the rules of the arrows. There is some error...for these types of problems...instructors usually give +/- 3 degrees, +/- 3 minutes...called using the rule of 3 to calculate CPA... Did you guys like this problem?...I have a tonne of them...also, opening and closing on a steady bearing...useful for buoy approaches.
for 16 years...so yes, I guess I automatically assume true bearings. The method that was used to calculate on the maneuvering board is the same way that it's calculated in radar navigation courses that I've taken... The rules of the arrows are...you (single arrow) chase the relative (circled arrow), and the relative and the bogey (double arrow)converge. I don't think I violated the rules of the arrows. There is some error...for these types of problems...instructors usually give +/- 3 degrees, +/- 3 minutes...called using the rule of 3 to calculate CPA... Did you guys like this problem?...I have a tonne of them...also, opening and closing on a steady bearing...useful for buoy approaches.
PaulH
No, Paul, I agree. The arrows are correct. I finally saw on your plot that the difference in solutions was due to using true brng versus brng from course line. There is a nice symmetry in the problem, which is why one answer appears to be the negative of the other (from the proper perspective). Busy day today, but I can email you the analysis when I do it, if you'd like. It was a fun problem, even more so due to the confusion. I like puzzles.
No, Paul, I agree. The arrows are correct. I finally saw on your plot that the difference in solutions was due to using true brng versus brng from course line. There is a nice symmetry in the problem, which is why one answer appears to be the negative of the other (from the proper perspective). Busy day today, but I can email you the analysis when I do it, if you'd like. It was a fun problem, even more so due to the confusion. I like puzzles.
Paul Show Us Some More
The maneuvering board is the way to go on these things way too many ways to screw it up running the numbers. Paul, in my navigation class there was no mention of setting a course and speed for intercepting the target. I am sure it is a simple variation on the CPA. Give us a rundown on that one.
The maneuvering board is the way to go on these things way too many ways to screw it up running the numbers. Paul, in my navigation class there was no mention of setting a course and speed for intercepting the target. I am sure it is a simple variation on the CPA. Give us a rundown on that one.
I'm stumped Paul H
I re-did based on your numbers of 90 degrees true instead of 90 degrees relative. In your boat going 6.5 knots, I cannot find a solution where you can converge. You are just going too slow. Even if you took a heading of 20 degrees immediately(hard starboard), your bogey would still have blown by you. You are too far away or are too slow for you to intercept your bogey. Are you sure this is a workable problem? Am I missing something here?
I re-did based on your numbers of 90 degrees true instead of 90 degrees relative. In your boat going 6.5 knots, I cannot find a solution where you can converge. You are just going too slow. Even if you took a heading of 20 degrees immediately(hard starboard), your bogey would still have blown by you. You are too far away or are too slow for you to intercept your bogey. Are you sure this is a workable problem? Am I missing something here?
Still before time zero
Intersection is still backwards in time. Travel on the red dashed line from current position to CPA is opposite direction of relative vector (red arrow with circle). The intersection is still prior to current time. It's the same course tracks as before, just a different relative position along them at t=0.
Intersection is still backwards in time. Travel on the red dashed line from current position to CPA is opposite direction of relative vector (red arrow with circle). The intersection is still prior to current time. It's the same course tracks as before, just a different relative position along them at t=0.
- Status