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2 point question
It could be almost anything depending on your start-latitude. If you
are close to equator it would be very small and if you are close to the pole it could rather large. It could be more than 28 hours x say 6 knot: more than 168 nautical miles, but that would be a very cold trip. Peter S/Y Anicula "Donal" skrev i en meddelelse ... "DSK" wrote in message . .. I was also suprised that nobody caught on to the diff in distance at diff lattitudes... that was the first thing I thought of and was surprised that it was not the point of the question. The original question mentioned a distance of 14 miles. What do yuu think that the maximum offset could be? Regards Donal -- |
2 point question
Correction- I find it interesting that "The Navigator(tm)," who is
definitely Navvie but could possibly be accurately named Navsprit, had nothing to say on the matter other than "Bwahahahahahahaha." DSK Nav wrote: It's a trivial academic exercise. Bwhahahhahahahaha |
2 point question
DSK wrote:
28nm. Is that with a really really bad compass ;) Yes! I'll go for 1.86nm instead. :-) From my limited knowledge of, and ability with, spherical trig, I get an answer of somewhere around 0.8 meters starting at the equator... If one starts from 7nm N of the equator, and goes ESWN, one ends up at the start point. Starting on the equator, I get 0.000116nm - 0.215m. now I have another question, would the offset be constant as you move further north or south? No, it increases. And the change isn't linear. My 1.86nm figure is based on starting at 89deg 57.77 N - the 14nm due E leg is a circle of 14nm circumference. Starting S of this reduces the offset. Starting N means that the 'inner' circle (E leg) is getting smaller much faster than the 'outer' circle (W leg) - when we complete the course and come back to the start latitude, we're on a circle which is getting smaller faster than we can get around the outer circle to try and maximise or distance from the start. (For a given inner circle, the maximum distance one can be from the start is the diameter of the circle - 180 degrees around - but it's impossible to travel 180 degrees around the *outer* circle, such that we'd finish on the inner 180 deg from the start). At 90N, there is no E leg - we go S 14nm, W 14nm (through 57.3 deg longitude - the maximum possible), and N 14nm to arrive back at the start. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Doug wrote:
Is that with a really really bad compass ;) No, not necessarily, but he sure has a slow boat. Peter S/Y Anicula Wally wrote: It specified hours and, later, constant speed. Still, we can assume 1kt and ask ourselves... What do yuu think that the maximum offset could be? 28nm. Is that with a really really bad compass ;) From my limited knowledge of, and ability with, spherical trig, I get an answer of somewhere around 0.8 meters starting at the equator... now I have another question, would the offset be constant as you move further north or south? I also find it curious that "The Navigator(tm)" had nothing to say on this point. Fresh Breezes- Doug King |
2 point question
Don-o,
Go back and read the question again. Miles are never mentioned. The legs were direction (Compass) and time (14 hours) per leg. The only mention of speed was that it was constant. If it was zero and he rotated every 14 hours, 90 degrees on his keel there would be no "Vector" to name. If he sailed at 5 mph each leg would be 70NM. He would have made a 90 digress heading change on his present meridian, which would be more than 90 degrees to the starting meridian. He would have traveled at this angle for 14 hrs then turned to a East west heading,which would have been 90 degrees but greater than 90 with respect to the first 90. Now, he travels 14 hrs due West (70 NM) to a new meridian, short of the starting meridian by how far he set off by the 70miles he traveled on is North - South Leg He now sails North for 14 hours (70NM) and should meet the line of The original leg some where west off the starting position, +or -- any other drift that might have been encountered. A line or vector should be and indication of this "DRIFT" Don-0- I mentioned "DRIFT" in my first reply. Sorry group, I didn't stay out of the discussion Ole Thom |
2 point question
Yep. The answer's in that nasty "book-learning".
Bwhahahahahhahahhahahah Cheers DSK wrote: Correction- I find it interesting that "The Navigator(tm)," who is definitely Navvie but could possibly be accurately named Navsprit, had nothing to say on the matter other than "Bwahahahahahahaha." DSK Nav wrote: It's a trivial academic exercise. Bwhahahhahahahaha |
2 point question
Again I stand corrected. 1 minute=NM A mistype.
OT |
2 point question
Wally wrote: DSK wrote: now I have another question, would the offset be constant as you move further north or south? No, it increases. And the change isn't linear. sin(theta) - theta at small theta. By small, I mean theta should be 0.2 rad for a 1% error. Cheers |
2 point question
1.86nm is not far enough for me: For one thing, I can sail much faster
than your 1 knot. If I can do 6 knots, the figure would be 11.16 nm, assuming your math is correct, but you can do better than that, even in your slow boat. Why don't you take a trip to the South Pole ? Peter S/Y Anicula "Wally" skrev i en meddelelse ... DSK wrote: 28nm. Is that with a really really bad compass ;) Yes! I'll go for 1.86nm instead. :-) From my limited knowledge of, and ability with, spherical trig, I get an answer of somewhere around 0.8 meters starting at the equator... If one starts from 7nm N of the equator, and goes ESWN, one ends up at the start point. Starting on the equator, I get 0.000116nm - 0.215m. now I have another question, would the offset be constant as you move further north or south? No, it increases. And the change isn't linear. My 1.86nm figure is based on starting at 89deg 57.77 N - the 14nm due E leg is a circle of 14nm circumference. Starting S of this reduces the offset. Starting N means that the 'inner' circle (E leg) is getting smaller much faster than the 'outer' circle (W leg) - when we complete the course and come back to the start latitude, we're on a circle which is getting smaller faster than we can get around the outer circle to try and maximise or distance from the start. (For a given inner circle, the maximum distance one can be from the start is the diameter of the circle - 180 degrees around - but it's impossible to travel 180 degrees around the *outer* circle, such that we'd finish on the inner 180 deg from the start). At 90N, there is no E leg - we go S 14nm, W 14nm (through 57.3 deg longitude - the maximum possible), and N 14nm to arrive back at the start. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Peter S/Y Anicula wrote:
1.86nm is not far enough for me: For one thing, I can sail much faster than your 1 knot. If I can do 6 knots, the figure would be 11.16 nm, assuming your math is correct, but you can do better than that, even in your slow boat. The misapprehension of the question that led to this discourse was concerned with distances of 14nm. You might sail the course six times faster, but you'll still be 1.86nm from the start. Why don't you take a trip to the South Pole ? Because my ice-breaker would founder on the mountains. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
In most of my posts on this subject I have mentioned a trip going east
north west and south. In this post I referred to the original question, and in the question the trip goes east south west and north, so in this post the destination would be east of the start-point. Sorry if anyone got confused. Peter S/Y Anicula "Peter S/Y Anicula" skrev i en meddelelse ... I was taught that: Speed is the forward motion through the water Leeway is the sideways movement of the boat through the water Current is the movement of the water over the ground Steered course is the way you point Sailed course is the direction that you move through the water Course over the ground (is that called course made good ?) is the direction you move over the ground etc. If one sails as you described in the question (on the northern hemisphere), one sails toward a destination B that is west of the starting-point A (if there were no current). The position after 56 hours is point C. So, If your teacher taught you that when you sail from A toward B and arrive at C then AC is the "current" then he must have been an electrician and not a sailor. You could justify to call BC for current, though the distance might be a result of more than just the surface-waters movement over the ground. Peter A/Y Anicula "Bart Senior" skrev i en meddelelse et... In most places where people are sailing, current would be the greater effect. Other factors, can all be lumped into something that for lack of a better word is called current. That is what I was taught. DSK wrote Yes it is, but a minute of longitude differs in length as you go north and/or south. |
2 point question
On the south pole, even if you insist on going only 14 nm on every
leg, you could reach the 28 nm that you mentioned in your earlier post, so you wouldn't have to admit that you have a really bad compass. You might have to bring an iceboat though. Peter S/Y Anicula "Wally" skrev i en meddelelse ... Peter S/Y Anicula wrote: 1.86nm is not far enough for me: For one thing, I can sail much faster than your 1 knot. If I can do 6 knots, the figure would be 11.16 nm, assuming your math is correct, but you can do better than that, even in your slow boat. The misapprehension of the question that led to this discourse was concerned with distances of 14nm. You might sail the course six times faster, but you'll still be 1.86nm from the start. Why don't you take a trip to the South Pole ? Because my ice-breaker would founder on the mountains. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Peter S/Y Anicula wrote:
On the south pole, even if you insist on going only 14 nm on every leg, you could reach the 28 nm that you mentioned in your earlier post, ... How? -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
You would start with the outer circle going east then go closer to the
pole moving south... If the inner circle has an arc of 180 degrees plus the arc of the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Peter S/Y Anicula "Wally" skrev i en meddelelse ... Peter S/Y Anicula wrote: On the south pole, even if you insist on going only 14 nm on every leg, you could reach the 28 nm that you mentioned in your earlier post, ... How? -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
The terminology is not quite right, substitute inner circle with the
leg along the inner circle... So it should be: You would start with the leg along the outer circle going east then go closer to the pole moving south... If the leg along the inner circle has an arc of 180 degrees plus the arc of the leg along the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Peter S/Y Anicula "Peter S/Y Anicula" skrev i en meddelelse ... You would start with the outer circle going east then go closer to the pole moving south... If the inner circle has an arc of 180 degrees plus the arc of the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Peter S/Y Anicula "Wally" skrev i en meddelelse ... Peter S/Y Anicula wrote: On the south pole, even if you insist on going only 14 nm on every leg, you could reach the 28 nm that you mentioned in your earlier post, ... How? -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Both Peter and Wally deserve 5 points for taking this
silly question so far. Bart Peter S/Y Anicula wrote The terminology is not quite right, substitute inner circle with the leg along the inner circle... So it should be: You would start with the leg along the outer circle going east then go closer to the pole moving south... If the leg along the inner circle has an arc of 180 degrees plus the arc of the leg along the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Peter S/Y Anicula "Peter S/Y Anicula" skrev i en meddelelse ... You would start with the outer circle going east then go closer to the pole moving south... If the inner circle has an arc of 180 degrees plus the arc of the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Peter S/Y Anicula "Wally" skrev i en meddelelse ... Peter S/Y Anicula wrote: On the south pole, even if you insist on going only 14 nm on every leg, you could reach the 28 nm that you mentioned in your earlier post, ... How? -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Peter S/Y Anicula wrote:
The terminology is not quite right, substitute inner circle with the leg along the inner circle... So it should be: You would start with the leg along the outer circle going east then go closer to the pole moving south... I hadn't realised it would work differently at the south pole... I'll do some numbers later and see what I come up with. If the leg along the inner circle has an arc of 180 degrees plus the arc of the leg along the outer circle, then the end-point should be opposite the start-point, That doesn't quite sound right - if the leg along the outer circle, plus the leg along the inner circle, add up to a total of 180 degrees longitude, then the end point would be opposite the start point. and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Yup. Proving that my compass does indeed work, in some places at least. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Bart Senior wrote:
Both Peter and Wally deserve 5 points for taking this silly question so far. Ah, bonus points for waffling... (The apple's in the post, teach!) -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Wally wrote:
If one starts from 7nm N of the equator, and goes ESWN, one ends up at the start point. Right! I didn't think of that angle... ... Starting on the equator, I get 0.000116nm - 0.215m. Now that I have my referance library and a spreadsheet instead of a pocket calculator and faulty memory, I'll try again. It's a more interesting problem than I thought it'd be.... now I have another question, would the offset be constant as you move further north or south? No, it increases. And the change isn't linear. Yes, I realized this thinking about it on the drive home. So what is it proportional to, the arcsine? You and Peter deserve your points. Maybe one day Taddy will write a poem about you... now there's something to brag about to your grandkids... Fresh Breezes- Doug King |
2 point question
Wally wrote:
If the leg along the inner circle has an arc of 180 degrees plus the arc of the leg along the outer circle, then the end-point should be opposite the start-point, That doesn't quite sound right - if the leg along the outer circle, plus the leg along the inner circle, add up to a total of 180 degrees longitude, then the end point would be opposite the start point. If you count the arc with sign, whitch is probably the right thing to do, you are right. I just counted the size - independent of direction. Peter S/Y Anicula "Wally" skrev i en meddelelse ... Peter S/Y Anicula wrote: The terminology is not quite right, substitute inner circle with the leg along the inner circle... So it should be: You would start with the leg along the outer circle going east then go closer to the pole moving south... I hadn't realised it would work differently at the south pole... I'll do some numbers later and see what I come up with. If the leg along the inner circle has an arc of 180 degrees plus the arc of the leg along the outer circle, then the end-point should be opposite the start-point, That doesn't quite sound right - if the leg along the outer circle, plus the leg along the inner circle, add up to a total of 180 degrees longitude, then the end point would be opposite the start point. and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Yup. Proving that my compass does indeed work, in some places at least. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
DSK wrote:
Now that I have my referance library and a spreadsheet instead of a pocket calculator and faulty memory, I'll try again. It's a more interesting problem than I thought it'd be.... Yup, the south pole behaves differently - gonna to play with that later. No, it increases. And the change isn't linear. Yes, I realized this thinking about it on the drive home. So what is it proportional to, the arcsine? The function I used is: cos(lat) x 60 Since the cosine produces the same curve as a sine, but 90deg out of phase [it starts high instead of low - cos(0)=1, sin(0)=0], for the first 0-90deg, we're seeing the second half of the bendy top of the curve, followed by the first half of the straightish negative-going part of the curve. The most linear part starts at about 50-60 degrees and continues to the pole. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Well thank you Bart for the points, that's all I wanted in the first
place. Peter S/Y Anicula "Bart Senior" skrev i en meddelelse et... Both Peter and Wally deserve 5 points for taking this silly question so far. Bart |
2 point question
Wally wrote:
That doesn't quite sound right - if the leg along the outer circle, plus the leg along the inner circle, add up to a total of 180 degrees longitude, then the end point would be opposite the start point. This is wrong - it should be the *difference* between E leg on outer circle and W leg on inner circle that gives 180 degrees... Start at 89d 42.447m S (roughly)... The 14nm E leg traverses 45.7d longitude. Go 14nm S to 89d 56.447m S... The W leg covers 225.7d longitude. Do 225.7 - 45.7 to get 180, and then go N 14nm to complete the course. You are now on the original lat, but 180 degrees away from your start point. The distance from the start is twice the minutes from the start latitude to the pole... 90 - StartLat = 17.553 minutes from the pole ....which gives a distance of 35.11nm from the start to the finish. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
A few comments below:
Peter S/Y Anicula "Wally" skrev i en meddelelse ... Wally wrote: That doesn't quite sound right - if the leg along the outer circle, plus the leg along the inner circle, add up to a total of 180 degrees longitude, then the end point would be opposite the start point. This is wrong - it should be the *difference* between E leg on outer circle and W leg on inner circle that gives 180 degrees... Now you confuse me, isen't that what I said in the first place? Start at 89d 42.447m S (roughly)... The 14nm E leg traverses 45.7d longitude. Go 14nm S to 89d 56.447m S... The W leg covers 225.7d longitude. Do 225.7 - 45.7 to get 180, and then go N 14nm to complete the course. You are now on the original lat, but 180 degrees away from your start point. The distance from the start is twice the minutes from the start latitude to the pole... 90 - StartLat = 17.553 minutes from the pole ...which gives a distance of 35.11nm from the start to the finish. So that would be around 210 nautical miles in my boat, assuming it would keep a speed of 6 knot even with skies mounted under it - that is well over 168 nm, and all that without any help from current. Good work Wally! I will top Barts 5 points with a couple of Anicula-points. -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Peter S/Y Anicula wrote:
Now you confuse me, isen't that what I said in the first place? Yes. I had read yours as an addition (big arc = small arc + 180), but somehow thought it was the wrong addition. I then realised that it was a subtraction (big arc - small arc = 180), which it was what you had said, but in a different way. ...which gives a distance of 35.11nm from the start to the finish. So that would be around 210 nautical miles in my boat, assuming it would keep a speed of 6 knot even with skies mounted under it - that is well over 168 nm, and all that without any help from current. If the legs were 14x6=84nm, there would be a factor involving the diameter of the inner circle, which would increase to maintan the big-small=180 relationship between degrees covered on each arc. I think that means it would be greater than 210. Good work Wally! I will top Barts 5 points with a couple of Anicula-points. A good haul, considering I don't think I answered the original question... -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
"Wally" wrote in message ... Donal wrote: The original question mentioned a distance of 14 miles. It specified hours and, later, constant speed. Still, we can assume 1kt and ask ourselves... What do yuu think that the maximum offset could be? 28nm. Do you mean 28 nautical miles??? Perhaps you meant 28 nautical metres!!! Regards Donal -- |
2 point question
"Peter S/Y Anicula" wrote in message ... You would start with the outer circle going east then go closer to the pole moving south... If the inner circle has an arc of 180 degrees plus the arc of the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Could you please repost that nonsense when I'm sober. Thanks. Regards Donal -- |
2 point question
"Thom Stewart" wrote in message ... Don-o, Go back and read the question again. Miles are never mentioned. The legs were direction (Compass) and time (14 hours) per leg Well, OT, I take my hat off to you! I was wrong. You seem to be the only one who is paying attention. Regards Donal -- |
2 point question
Donal wrote:
What do yuu think that the maximum offset could be? 28nm. Do you mean 28 nautical miles??? I actually meant 35.11 nautical miles. (Eventually...) Perhaps you meant 28 nautical metres!!! I don't think so... -- Wally www.artbywally.com www.wally.myby.co.uk |
2 point question
Donal wrote: "Wally" wrote in message ... Donal wrote: The original question mentioned a distance of 14 miles. It specified hours and, later, constant speed. Still, we can assume 1kt and ask ourselves... What do yuu think that the maximum offset could be? 28nm. Do you mean 28 nautical miles??? Perhaps you meant 28 nautical metres!!! What happened to yards and cables -don't tell the the EU has changed that as well! Cheers |
2 point question
Donald wrote:
Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. Peter S/Y Anicula "Donal" skrev i en meddelelse ... "Peter S/Y Anicula" wrote in message ... You would start with the outer circle going east then go closer to the pole moving south... If the inner circle has an arc of 180 degrees plus the arc of the outer circle, then the end-point should be opposite the start-point, and the distance over the pole would be 2 x length of legs plus diameter of inner circle: more than 28 nm in at 1 knot and more than 168 nm at 6 knot. Could you please repost that nonsense when I'm sober. Thanks. Regards Donal -- |
2 point question
"Peter S/Y Anicula" wrote in message ... Donald wrote: Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. I've had a (sobering)cup of cofee in anticipation of your reply. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. You're absolutely correct. It didn't help. Care to elucidate(sp?)? Regards Donal -- |
2 point question
Have a look at this diagram
http://www.geocities.com/aniculapeter/Diagram.htm Does it help ? Peter S/Y Anicula "Donal" skrev i en meddelelse ... "Peter S/Y Anicula" wrote in message ... Donald wrote: Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. I've had a (sobering)cup of cofee in anticipation of your reply. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. You're absolutely correct. It didn't help. Care to elucidate(sp?)? Regards Donal -- |
2 point question
then have a look at this diagram
Scout http://www.ebaumsworld.com/flashloops-booty.html "Peter S/Y Anicula" wrote in message ... Have a look at this diagram http://www.geocities.com/aniculapeter/Diagram.htm Does it help ? Peter S/Y Anicula "Donal" skrev i en meddelelse ... "Peter S/Y Anicula" wrote in message ... Donald wrote: Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. I've had a (sobering)cup of cofee in anticipation of your reply. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. You're absolutely correct. It didn't help. Care to elucidate(sp?)? Regards Donal -- |
2 point question
I am very impressed with your ability Scout, but how do you do that ?
Peter S/Y Anicula "Scout" skrev i en meddelelse ... then have a look at this diagram Scout http://www.ebaumsworld.com/flashloops-booty.html "Peter S/Y Anicula" wrote in message ... Have a look at this diagram http://www.geocities.com/aniculapeter/Diagram.htm Does it help ? Peter S/Y Anicula "Donal" skrev i en meddelelse ... "Peter S/Y Anicula" wrote in message ... Donald wrote: Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. I've had a (sobering)cup of cofee in anticipation of your reply. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. You're absolutely correct. It didn't help. Care to elucidate(sp?)? Regards Donal -- |
2 point question
Diet, exercise, and the Sunshine Band.
Scout "Peter S/Y Anicula" wrote in message ... I am very impressed with your ability Scout, but how do you do that ? Peter S/Y Anicula "Scout" skrev i en meddelelse ... then have a look at this diagram Scout http://www.ebaumsworld.com/flashloops-booty.html "Peter S/Y Anicula" wrote in message ... Have a look at this diagram http://www.geocities.com/aniculapeter/Diagram.htm Does it help ? Peter S/Y Anicula "Donal" skrev i en meddelelse ... "Peter S/Y Anicula" wrote in message ... Donald wrote: Could you please repost that nonsense when I'm sober. Thanks. Ok, I'll keep a copy of the post ready, should that day arrive. Just say when. I've had a (sobering)cup of cofee in anticipation of your reply. But I'm not convinced it would help. One of the reasons being that it is rather incoherent. The correction post, that followed it, is slightly more understandable though. You're absolutely correct. It didn't help. Care to elucidate(sp?)? Regards Donal -- |
2 point question
Peter?
How in the name of all that's Holy do you hold a compass heading for 14 hours, while going in a circle? Please hide your Diagram. It's stupidity! OT |
2 point question
Peter?
How in the name of all that's Holy do you hold a compass heading for 14 hours, while going in a circle? Please hide your Diagram. It's stupidity! OT |
2 point question
Thom wrote:
How in the name of all that's Holy do you hold a compass heading for 14 hours, while going in a circle? I will admit that it is not easy. Every time I get to the point in the circle where I have my back toward the compass I tend to get a little bit off course. Peter S/Y Anicula "Thom Stewart" skrev i en meddelelse ... Peter? How in the name of all that's Holy do you hold a compass heading for 14 hours, while going in a circle? Please hide your Diagram. It's stupidity! OT |
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