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Math Problem
Say you are sailing a course of 190 and making 5 kn. An adverse current
suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
*IF* those are the parameters (in the case I was under we were making something
less than 5 knots and we didn't bother to check how fast we were going east, we just confirmed and tacked) ... .... *THEN* one needs a current componant at 010 at 5 knots and a current componant at 5 knots as well, which would be roughly (would be exact if the original course were at 180 rather than 190) a total current at 045 of just over 7 knots. This would be a pretty strong eddy, though I am not sure that such does not exist. In any event, I believe we were making about 3 knots at 190* (we were sailing, as each of us enjoyed sailing, the owner was exceptionally experienced, we had a long ways to go, and we were not sure we had enough fuel to motor forever), which an eddy of 3 knots (much common in eddies) at 011 would have given us easting enough to show on our gps's. Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
Yeah, sure Jax, you were looking at the GPS and read the COG but not the
speed ... Do you really think everyone here is as ****ed up as you are? Rick JAXAshby wrote: *IF* those are the parameters (in the case I was under we were making something less than 5 knots and we didn't bother to check how fast we were going east, we just confirmed and tacked) ... |
Math Problem
on my gps at that time, speed was displayed on a different screen from course.
rick, I sorry that your lack of knowledge and understanding in this very tiny issue -- and issue known and understood by others who have actually sailed in areas with ocean currents -- so clearly shows you know nothing but what you read on the net on any subject at all. Rick, how was the ag pilot business? The ATP rating? The stolen construction site plywood turned into a fine dinghy? The schooner trip to Mexico that lasted 5 years? The master of a an ocean going freighter for ten years? All before you turned 35? Yeah, sure Jax, you were looking at the GPS and read the COG but not the speed ... Do you really think everyone here is as ****ed up as you are? Rick JAXAshby wrote: *IF* those are the parameters (in the case I was under we were making something less than 5 knots and we didn't bother to check how fast we were going east, we just confirmed and tacked) ... |
Math Problem
JAXAshby wrote:
on my gps at that time, speed was displayed on a different screen from course. You wrote that you were "playing with" the GPS, sounds like a typical Jax backpedal to me. Rick, how was the ag pilot business? The ATP rating? The stolen construction site plywood turned into a fine dinghy? The schooner trip to Mexico that lasted 5 years? The master of a an ocean going freighter for ten years? All before you turned 35? More typical Jax fantasy material ... The only thing you got right is the ATP. The rest must be some envious fantasy you have about some other guy. It looks like they let you out a bit early this time or they haven't got your meds quite right yet. The speedos a bit too tight maybe? Rick |
Math Problem
ricky, I know you don't understand this, but you do NOT navigate minute by
minute by minute by minute when you are well offshore outside the 100 fathom line and not another vessel in sight. "playing" is the right word for the situation. as for the rest, I see you have conveniently forgotten all that you posted while your were still "finding" your screen personna. on my gps at that time, speed was displayed on a different screen from course. You wrote that you were "playing with" the GPS, sounds like a typical Jax backpedal to me. Rick, how was the ag pilot business? The ATP rating? The stolen construction site plywood turned into a fine dinghy? The schooner trip to Mexico that lasted 5 years? The master of a an ocean going freighter for ten years? All before you turned 35? More typical Jax fantasy material ... The only thing you got right is the ATP. The rest must be some envious fantasy you have about some other guy. It looks like they let you out a bit early this time or they haven't got your meds quite right yet. The speedos a bit too tight maybe? Rick |
Math Problem
Ball is in your park, Jax. Show the posts. Put up or show yourself to be
what we all know you to be anyway, a jealous, impotent, incompetent wannabe with a vivid fantasy life who thinks a flamefest means someone cares. Have a nice day, Jax. You aren't worth the bandwidth. Rick JAXAshby wrote: ricky, I know you don't understand this, but you do NOT navigate minute by minute by minute by minute when you are well offshore outside the 100 fathom line and not another vessel in sight. "playing" is the right word for the situation. as for the rest, I see you have conveniently forgotten all that you posted while your were still "finding" your screen personna. on my gps at that time, speed was displayed on a different screen from course. You wrote that you were "playing with" the GPS, sounds like a typical Jax backpedal to me. Rick, how was the ag pilot business? The ATP rating? The stolen construction site plywood turned into a fine dinghy? The schooner trip to Mexico that lasted 5 years? The master of a an ocean going freighter for ten years? All before you turned 35? More typical Jax fantasy material ... The only thing you got right is the ATP. The rest must be some envious fantasy you have about some other guy. It looks like they let you out a bit early this time or they haven't got your meds quite right yet. The speedos a bit too tight maybe? Rick |
Math Problem
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Math Problem
"SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) Current = 50*, 8kts and again at 2.5 kn. 36* 6.2 kts. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? I don't know the Gulf stream, but I am fairly sure that I could find it! Regards Donal -- |
Hunter for sale
On 28 Feb 2004 14:52:54 -0800, (Horvath) wrote
this crap: " wrote in message om... My fine hunter sailboat is up for sale. Last night as I stumbled back to my boat I fell off the dock, hitting my head on the cabin top. When I woke up this morning I found a big crack in the fibergkass. As I've told Douglass, when my boat has a crack, I buy a new one. Of course I'll buy another Hunter. or maybe a Magregor? BB Dude, your mixed up. This signature is now the ultimate power in the universe If you're going to forge posts from me, at least learn to spell, dumbass. This signature is now the ultimate power in the universe |
Math Problem
Skitch,
You said it better than me, Ole Thom |
Math Problem
donny, your math wouldn't get you promoted from 10th grade to 11th.
(what the hell do *you* think the sq rt of 50 is, for the krist sakes?) "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) Current = 50*, 8kts and again at 2.5 kn. 36* 6.2 kts. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? I don't know the Gulf stream, but I am fairly sure that I could find it! Regards Donal -- |
Math Problem
maybe ole fart, but donny don't even know what the sq rt of 50 is.
do you? [it *IS* important, in this case] Skitch, You said it better than me, Ole Thom |
Hunter for sale
Why would they need to do that?
"Horvath" wrote in message ... On 28 Feb 2004 14:52:54 -0800, (Horvath) wrote this crap: " wrote in message om... My fine hunter sailboat is up for sale. Last night as I stumbled back to my boat I fell off the dock, hitting my head on the cabin top. When I woke up this morning I found a big crack in the fibergkass. As I've told Douglass, when my boat has a crack, I buy a new one. Of course I'll buy another Hunter. or maybe a Magregor? BB Dude, your mixed up. This signature is now the ultimate power in the universe If you're going to forge posts from me, at least learn to spell, dumbass. This signature is now the ultimate power in the universe |
Math Problem
7.071067812
AXAshby wrote: maybe ole fart, but donny don't even know what the sq rt of 50 is. do you? [it *IS* important, in this case] Skitch, You said it better than me, Ole Thom |
Math Problem
7.071067812
donny don't even know what the sq rt of 50 is. actually, the square root of 50 is an irrational number (meaning never end, never repeating) that is just a rough estimate. a more accurate answer to 1,000 decimal places is: 5 * 1.4142135623 73095048801688724209 69807856967187537694807317667973799073247846210703 88503875343276415727 350138462309122970249248360558 507372126441214970999358314132 226659275055927557999505011527 820605714701095599716059702745 345968620147285174186408891986 095523292304843087143214508397 626036279952514079896872533965 463318088296406206152583523950 547457502877599617298355752203 375318570113543746034084988471 603868999706990048150305440277 903164542478230684929369186215 805784631115966687130130156185 689872372352885092648612494977 154218334204285686060146824720 771435854874155657069677653720 226485447015858801620758474922 657226002085584466521458398893 944370926591800311388246468157 082630100594858704003186480342 194897278290641045072636881313 739855256117322040245091227700 226941127573627280495738108967 504018369868368450725799364729 060762996941380475654823728997 180326802474420629269124859052 181004459842150591120249441341 728531478105803603371077309182 869314710171111683916581726889 419758716582152128229518488472 08969 Note: if a thousand digits is not accurate enough, let me know and I can do the math quickly up to maybe 5 million decimal places. More than 5 mil will take me a bit longer. |
Hunter for sale
"Jonathan Ganz" wrote in message ...
Why would they need to do that? Because that is his sig. BB |
Hunter for sale
If you're going to forge posts from me, at least learn to spell dumbass. " D U M B A S S " This signature is now the ultimate power in the universe |
Math Problem
Wrong again.
JAXAshby wrote: 7.071067812 donny don't even know what the sq rt of 50 is. actually, the square root of 50 is an irrational number (meaning never end, never repeating) that is just a rough estimate. a more accurate answer to 1,000 decimal places is: 5 * 1.4142135623 73095048801688724209 69807856967187537694807317667973799073247846210703 88503875343276415727 350138462309122970249248360558 507372126441214970999358314132 226659275055927557999505011527 820605714701095599716059702745 345968620147285174186408891986 095523292304843087143214508397 626036279952514079896872533965 463318088296406206152583523950 547457502877599617298355752203 375318570113543746034084988471 603868999706990048150305440277 903164542478230684929369186215 805784631115966687130130156185 689872372352885092648612494977 154218334204285686060146824720 771435854874155657069677653720 226485447015858801620758474922 657226002085584466521458398893 944370926591800311388246468157 082630100594858704003186480342 194897278290641045072636881313 739855256117322040245091227700 226941127573627280495738108967 504018369868368450725799364729 060762996941380475654823728997 180326802474420629269124859052 181004459842150591120249441341 728531478105803603371077309182 869314710171111683916581726889 419758716582152128229518488472 08969 Note: if a thousand digits is not accurate enough, let me know and I can do the math quickly up to maybe 5 million decimal places. More than 5 mil will take me a bit longer. |
Math Problem
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Hunter for sale
We know you're a bumpass. What's your point?
"Horvath" wrote in message om... If you're going to forge posts from me, at least learn to spell dumbass. " D U M B A S S " This signature is now the ultimate power in the universe |
Math Problem
You missed a digit.
"JAXAshby" wrote in message ... 7.071067812 donny don't even know what the sq rt of 50 is. actually, the square root of 50 is an irrational number (meaning never end, never repeating) that is just a rough estimate. a more accurate answer to 1,000 decimal places is: 5 * 1.4142135623 73095048801688724209 69807856967187537694807317667973799073247846210703 88503875343276415727 350138462309122970249248360558 507372126441214970999358314132 226659275055927557999505011527 820605714701095599716059702745 345968620147285174186408891986 095523292304843087143214508397 626036279952514079896872533965 463318088296406206152583523950 547457502877599617298355752203 375318570113543746034084988471 603868999706990048150305440277 903164542478230684929369186215 805784631115966687130130156185 689872372352885092648612494977 154218334204285686060146824720 771435854874155657069677653720 226485447015858801620758474922 657226002085584466521458398893 944370926591800311388246468157 082630100594858704003186480342 194897278290641045072636881313 739855256117322040245091227700 226941127573627280495738108967 504018369868368450725799364729 060762996941380475654823728997 180326802474420629269124859052 181004459842150591120249441341 728531478105803603371077309182 869314710171111683916581726889 419758716582152128229518488472 08969 Note: if a thousand digits is not accurate enough, let me know and I can do the math quickly up to maybe 5 million decimal places. More than 5 mil will take me a bit longer. |
Math Problem
"JAXAshby" wrote in message ... donny, your math wouldn't get you promoted from 10th grade to 11th. (what the hell do *you* think the sq rt of 50 is, for the krist sakes?) As it happens, I don't have an opinion about the square root of 50. Now, perhaps you could tell us why you think that my mathematical skills are lacking?? While you are at it, you can also tell us about the malicious code that I sent you in an e-mail? I haven't forgotten about your lying accusation! I think that you should admit that you might have made a mistake Think about it! Regards Donal -- |
Math Problem
yeah, maybe. but what the hey. one digit in a thousand.
You missed a digit. "JAXAshby" wrote in message ... 7.071067812 donny don't even know what the sq rt of 50 is. actually, the square root of 50 is an irrational number (meaning never end, never repeating) that is just a rough estimate. a more accurate answer to 1,000 decimal places is: 5 * 1.4142135623 73095048801688724209 69807856967187537694807317667973799073247846210703 88503875343276415727 350138462309122970249248360558 507372126441214970999358314132 226659275055927557999505011527 820605714701095599716059702745 345968620147285174186408891986 095523292304843087143214508397 626036279952514079896872533965 463318088296406206152583523950 547457502877599617298355752203 375318570113543746034084988471 603868999706990048150305440277 903164542478230684929369186215 805784631115966687130130156185 689872372352885092648612494977 154218334204285686060146824720 771435854874155657069677653720 226485447015858801620758474922 657226002085584466521458398893 944370926591800311388246468157 082630100594858704003186480342 194897278290641045072636881313 739855256117322040245091227700 226941127573627280495738108967 504018369868368450725799364729 060762996941380475654823728997 180326802474420629269124859052 181004459842150591120249441341 728531478105803603371077309182 869314710171111683916581726889 419758716582152128229518488472 08969 Note: if a thousand digits is not accurate enough, let me know and I can do the math quickly up to maybe 5 million decimal places. More than 5 mil will take me a bit longer. |
Math Problem
As it happens, I don't have an opinion about the square root of 50.
opinion? how the hell can one had an **opinion** re sq rt 50???? Now, perhaps you could tell us why you think that my mathematical skills are lacking?? ah ..... maybe see above??? |
Math Problem
Since no one seems willing to do this probably, here are the answers:
Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
jeffie, the answer was posted three days ago. And it was posted out to 1,000
decimal places. Earth to jeff, Earth to jeff ... Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
Why give an answer to 1000 places when it wrong in the second place? The point
of nav problems is to be able to get the correct answer, not something within 10%. "JAXAshby" wrote in message ... jeffie, the answer was posted three days ago. And it was posted out to 1,000 decimal places. Earth to jeff, Earth to jeff ... Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
ah, it seems you didn't understand the ramifications of the question.
not surprising. Why give an answer to 1000 places when it wrong in the second place? The point of nav problems is to be able to get the correct answer, not something within 10%. "JAXAshby" wrote in message ... jeffie, the answer was posted three days ago. And it was posted out to 1,000 decimal places. Earth to jeff, Earth to jeff ... Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
You've only answered one part of the question.
Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
go back and re-read the answer, skitch, and you find that the question was more
comprehensively answered than you asked for. If you don't understand that, well you don't understand that. You've only answered one part of the question. Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
I think I gave the answer to the "math problem."
As to whether this current can exist in the Gulf Stream or an Eddy, I'm skeptical. However, I really don't know, and there are certainly many cases of current running substantially higher than what is "advertised." Since Jax claims to have only been making 3 knots, and doesn't specify the final SOG, its actually possible that they were not in an eddy, but in the Gulf Stream itself. Given the obviously poor navigational skill onboard, I wouldn't doubt it! "SkitchNYC" wrote in message ... You've only answered one part of the question. Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
I know that *you* don't understand, jeffies, but the problem really was
answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can exist in the Gulf Stream or an Eddy, I'm skeptical. However, I really don't know, and there are certainly many cases of current running substantially higher than what is "advertised." Since Jax claims to have only been making 3 knots, and doesn't specify the final SOG, its actually possible that they were not in an eddy, but in the Gulf Stream itself. Given the obviously poor navigational skill onboard, I wouldn't doubt it! "SkitchNYC" wrote in message ... You've only answered one part of the question. Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
Yes I know you "think" you answered the question, just like you "think" you're a
member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can exist in the Gulf Stream or an Eddy, I'm skeptical. However, I really don't know, and there are certainly many cases of current running substantially higher than what is "advertised." Since Jax claims to have only been making 3 knots, and doesn't specify the final SOG, its actually possible that they were not in an eddy, but in the Gulf Stream itself. Given the obviously poor navigational skill onboard, I wouldn't doubt it! "SkitchNYC" wrote in message ... You've only answered one part of the question. Since no one seems willing to do this probably, here are the answers: Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 5, current must be 50 degrees, speed of 7.66 Starting with heading of 190 and speed through water 5, to have a COG of 90 and SOG of 2.5, current must be 34.37 degrees, speed of 5.96 For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c then: the law of sines says: a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle and the law of cosines says: c^2 = a^2 + b^2 - 2*a*b* cos C "SkitchNYC" wrote in message ... Say you are sailing a course of 190 and making 5 kn. An adverse current suddenly gets you and you are now making a COG of 90. Assume the new COG is at the same speed (5kn) and again at 2.5 kn. What direction and speed must the current be to produce either of these results? Can such a current exist in a Gulf Stream eddie? |
Math Problem
jeffies, it is near impossible to have a rational discussion with you, as you
once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e |
Math Problem
Hey Jax, how come you didn't get the right answer to the question? I did. I
then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e |
Math Problem
the square root of 50 was in fact important to the discussion, but it seems you
still don't know why. go back and re-read the post again and again and again and again until you see why. (hint: 5^2 + 5^2 = 50) Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e |
Math Problem
Yes, if you can't solve the real problem you can make an approximation. What's
your point? You don't know how to solve the real problem? We already know that. "JAXAshby" wrote in message ... the square root of 50 was in fact important to the discussion, but it seems you still don't know why. go back and re-read the post again and again and again and again until you see why. (hint: 5^2 + 5^2 = 50) Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e |
Math Problem
jeff the *real* problem was what kind of counter-current was needed to change a
boat's course by a large amount. That question was answered within minutes by the very first repsonder (me). After that, the posts got playful as to what precision -- not accuracy -- that a "little over 7 knots" could be. you, on the other hand, concerned yourself with precision calculations on imprecise -- and unknowably precise -- assumptions. in other words, you didn't understand the question, only the need for complexity. The simple and original answer was understood by everyone -- but you -- without regard to their understanding of, let alone the capability of calculating, the lengths of non-square triangles. Yes, if you can't solve the real problem you can make an approximation. What's your point? You don't know how to solve the real problem? We already know that. "JAXAshby" wrote in message ... the square root of 50 was in fact important to the discussion, but it seems you still don't know why. go back and re-read the post again and again and again and again until you see why. (hint: 5^2 + 5^2 = 50) Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e Hey Jax, how come you didn't get the right answer to the question? I did. I then provided the appropriate formulae. All you did was babble about "components" and then give the wrong answer. You went on for 4 posts about the "square root of 50" which isn't relevant to the problem. You're right, you can't have a "rational discussion" with me, or with anyone else. BTW, one doesn't describe a current by which direction is comes from, one describes it by which direction its going. Take the Power Squadron course, really. "JAXAshby" wrote in message ... jeffies, it is near impossible to have a rational discussion with you, as you once AGAIN show you have zero understanding of either the issue or the resultant answer. you really, really, really don't understand the question, i.e. how much current and from which direction would be needed to cause a boat to change course by some large angle. you, you blithering idiot, are rambling on and on and on and on hoping against all hope that newbies will forget that you motor training wheels while Walter Mittying yourself to be Shackleton. Yes I know you "think" you answered the question, just like you "think" you're a member of Mensa, and you "think" you graduated high school. But saying that the answer is something that has a collection of components is NOT showing that you actually know how to compute such a collection. The fact that you keep insisting you solved it proves you don't really know how to do it. And being able to guess within 15% isn't bad, but doesn't show you know how to navigate. "JAXAshby" wrote in message ... I know that *you* don't understand, jeffies, but the problem really was answered comprehensively within a few minutes of the question being posted. jeffie, rational discussion with you is near to impossible, for you are not only much lacking in mental candlepower you are also so lacking in such you are not even capable of understanding just how lacking. don't buy an EPIRB, jeffie. Let Darwin help you sail, if you ever do decide to sail in lieu motoring. I think I gave the answer to the "math problem." As to whether this current can e |
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