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