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