I hate to admit but this young lady is ten times the sailor Zac is
Wilbur Hubbard
This young lady, Jessica Watson, appears to be a real sailor. Quite frankly
I must admit I underestimated her.
Reading her blog makes it clear she is everything the wannabe, Zac
Sunderland, was not.
She is brave, she is not a weak-willed, "it-takes-a-village" party animal
who needs constant companionship and a committee of hundreds to shepherd her
around the world.
Jessica isn't planning to motor halfway around the world. And she seems to
know more about motors than Zac did. She even cleans and polishes her
Yanmar.
Jessica is not planning to cheat by using the Panama Canal.
She talks of legs of several thousand miles being routine. And I bet she
doesn't get towed into and out of ports.
My Gosh, but I hope she has fair winds most of the way and I say Godspeed to
you, Jessica.
In closing, it is really pathetic to see how, nowadays, it takes a young
woman to show the world how a man's supposed to do it. What's with all the
wimp men lately?
Wilbur Hubbard
Mike Coon
> http://www.youngestround.blogspot.com/
>
> This young lady, Jessica Watson, appears to be a real sailor. Quite
> frankly I must admit I underestimated her.
But perhaps not a real navigator! Else she would be able to explain how a
great circle route can be shorter than a "straight line" (whatever straight
might mean here). Surely no-one would call a rhumb line "straight"?
Mike.
--
If reply address is invalid, remove spurious "@" and substitute "plus"
where needed.
Wilbur Hubbard
news:xu2dndoOc60UmI3W...@brightview.co.uk...
I thinks she knows how and why but just didn't want to bore her audience
with it or take the time to write about it when few would understand anyway.
She's kewl for a youngster.
Wilbur Hubbard
Duncan Heenan
news:4b0ffc28$0$65851$892e...@auth.newsreader.octanews.com...
> http://www.youngestround.blogspot.com/
> snip most of Wilma's usual crap <
> In closing, it is really pathetic to see how, nowadays, it takes a young
> woman to show the world how a man's supposed to do it. What's with all the
> wimp men lately?
>
>
> Wilbur Hubbard
Quite right Wilma! You show 'em. Get off the internet and in to your boat
yourself. (well....get a boat first of course).
You go ahead and justify your credentials for condemning people who sail
round the world as wimps - you do it! (But please don't take an internet
connection with you).
--
Duncan Heenan
Peter Köhlmann
> Wilbur Hubbard wrote:
>> http://www.youngestround.blogspot.com/
>>
>> This young lady, Jessica Watson, appears to be a real sailor. Quite
>> frankly I must admit I underestimated her.
>> ...
>
> But perhaps not a real navigator! Else she would be able to explain how
> a great circle route can be shorter than a "straight line" (whatever
> straight might mean here). Surely no-one would call a rhumb line
> "straight"?
>
Well, I really hate to destroy your illusions, but a "great circle route"
is always the shortest route to take on a globe
And it *is* a straight line (on a gnomonic chart). It just isn't on a
standard sea chart. Which does not display a globe either
--
What happens if a big asteroid hits Earth? Judging from realistic
simulations involving a sledge hammer and a common laboratory frog,
we can assume it will be pretty bad. --- Dave Barry
Quilljar
each evening, and every day I admire her and her attitude more and more. Her
writing style is also so mature, it is a pleasure to read.
Quilljar
Wilbur Hubbard
news:WKSdnczsFpI1pY3W...@bt.com...
Agreed, and so far she's been dauntless. So refreshing to see it. (And she's
way easier on the eyes than Zac, too.)
Wilbur Hubbard
Stephen Trapani
Dauntless? So far she has had mild weather, no challenges and is barely
one tenth complete. She should be dead right now after her boat collided
with a container ship while she slept on a shakedown trip. You are
judging too soon.
Stephen
Wilbur Hubbard
news:5lXPm.41073$X01....@newsfe07.iad...
You're right, but I do think she has more than meets the eye about her. I
sense a certain depth and competence beyond her chronological age. I
sincerely hope her mettle isn't tested too soon or too severely. That's the
difference between her and Zac who was a showboater assisted by a huge
committee and a wagon load of contrived hype. Jessica seems to be going
about her business without a lot of hoopla. Notice how hers is not a litany
of one breakdown after another.
Good sailors make voyaging seem uneventful - not some continued comedy of
errors and chaos. She's a good sailor and I'm amazed and appreciative of it.
Would that my heroes could be young men but that seems a thing of the past
with all the emasculated young men these days.
This young woman has what it takes and I wish her all the best. This is just
the opposite of what I felt with the Zac debacle. I wished he'd fail. My
heroes have to be better than mediocre. And, in a way he DID fail because
his so-called record is not recognized by Guinness.
Wilbur Hubbard
Oz...@crackerbox-palace.com
I had the pleasure of talking to her in Sydney a short while before
her departure.
She is an inspiration...and a foxy young thing!
OzOne of the three twins
I welcome you to Crackerbox Palace.
Duncan Heenan
"Wilbur Hubbard" <wilbur...@thefarm.invallid> wrote in message
news:4b10441b$0$65830$892e...@auth.newsreader.octanews.com...
> You're right, but I do think she has more than meets the eye about her. I
> sense a certain depth and competence beyond her chronological age. I
> sincerely hope her mettle isn't tested too soon or too severely. That's
> the difference between her and Zac who was a showboater assisted by a huge
> committee and a wagon load of contrived hype. Jessica seems to be going
> about her business without a lot of hoopla. Notice how hers is not a
> litany of one breakdown after another.
> Wilbur Hubbard
If she's not a 'showboater' too, how come you on the other side of the world
know about her at all?
The sort of brave self effacing sailors you pretend to admire stay anonymous
and just go sailing. They remain unknown to those who have not met them.
When did you last actually go sailing Wilma? Or maybe your many hours on the
internet wouldn't allow you time to do the real thing, even if you had a
boat?
Ian
I just hope she isn't spending all her time writing. On past form, she
really needs to go up on deck and have a look round from time to time.
Of course it may be that her parents are (re)writing the blog for her.
Ian
Mike Coon
> "Mike Coon" <Mike@@mjcoon.+.com> wrote:
>> But perhaps not a real navigator! Else she would be able to explain
>> how a great circle route can be shorter than a "straight line"
>> (whatever straight might mean here). Surely no-one would call a
>> rhumb line "straight"?
>
> Well, I really hate to destroy your illusions, but a "great circle
> route" is always the shortest route to take on a globe
> And it *is* a straight line (on a gnomonic chart). It just isn't on a
> standard sea chart. Which does not display a globe either
I'm sure I have many illusions, but not, I think, in this area. (Although I
have coded up a Molodensky transformation I only understand the purpose, not
the mathematics!)
On the other hand I have not crossed an ocean, or made a passage of a
fracction of the 5000+ nm that Jessica Watson was making. So I don't know if
she would use a gnomonic chart. My cruiser handbook says it "it is only used
for small areas such as harbour plans". On the other hand if it yields
accurate angles over such long passages perhaps that is how one would choose
a compass course for a rhumb line course across an ocean.
The fact that the great circle route is the shortest distance between two
points is precisely the point of my comment. But it may well be that a
gnomonic "straight line" (as just mentioned above) is what she was comparing
it with. The three significant figures that she gave for the distance is
impressive, too! Presumably the great circle route that she says she
travelled was calculated for her by GPS. Otherwise I have no idea how to
work out the varying compass course to steer.
Regards, Mike.
Ronald Raygun
> Peter K�hlmann wrote:
>> "Mike Coon" <Mike@@mjcoon.+.com> wrote:
>>> But perhaps not a real navigator! Else she would be able to explain
>>> how a great circle route can be shorter than a "straight line"
>>> (whatever straight might mean here). Surely no-one would call a
>>> rhumb line "straight"?
I don't see why not, since on "normal" charts (i.e. Mercator charts)
rhumb lines *are* straight!
>> Well, I really hate to destroy your illusions, but a "great circle
>> route" is always the shortest route to take on a globe
>> And it *is* a straight line (on a gnomonic chart). It just isn't on a
>> standard sea chart. Which does not display a globe either
>
> I'm sure I have many illusions, but not, I think, in this area. (Although
> I have coded up a Molodensky transformation I only understand the purpose,
> not the mathematics!)
>
> On the other hand I have not crossed an ocean, or made a passage of a
> fracction of the 5000+ nm that Jessica Watson was making. So I don't know
> if she would use a gnomonic chart. My cruiser handbook says it "it is only
> used for small areas such as harbour plans". On the other hand if it
> yields accurate angles over such long passages perhaps that is how one
> would choose a compass course for a rhumb line course across an ocean.
>
> The fact that the great circle route is the shortest distance between two
> points is precisely the point of my comment. But it may well be that a
> gnomonic "straight line" (as just mentioned above) is what she was
> comparing it with.
I rather think not. The comparison, I understand, was between a great
circle and a "straight line" of unspecified type. No-one would want to
compare anything with itself, and since a great circle *is* a gnomonic
straight line, it should follow that the comparison must have been with
any type of straight line *other than gnomonic*. The most obvious
candidate, I suggest, is the Mercator straight line (i.e. a rhumb line),
the whole point being that a great circle line between any two points
sufficiently far apart is shorter than the rhumb line.
> The three significant figures that she gave for the
> distance is impressive, too! Presumably the great circle route that she
> says she travelled was calculated for her by GPS. Otherwise I have no idea
> how to work out the varying compass course to steer.
In the absence of electronic aids, the usual way to work out what compass
course to steer is to start with the gnomonic chart and draw a straight line
on it. This line will cross several meridians on the way, and will do so at
various angles. These can be measured. The course to steer depends on how
much progress you've made along that line, and a good approximation is to
take the angle at which it crosses the meridian nearest your current
position.
Peter Köhlmann
> Mike Coon wrote:
< snip >
>> The three significant figures that she gave for the
>> distance is impressive, too! Presumably the great circle route that she
>> says she travelled was calculated for her by GPS. Otherwise I have no
>> idea how to work out the varying compass course to steer.
>
> In the absence of electronic aids, the usual way to work out what
> compass course to steer is to start with the gnomonic chart and draw a
> straight line on it. This line will cross several meridians on the way,
> and will do so at various angles. These can be measured.
There exist some relativly simple mathematical formulas to compute the
course from any given point in between. Simple enough to do it without any
electronic calculator in just a few minutes.
> The course to steer depends on
> how much progress you've made along that line, and a good approximation
> is to take the angle at which it crosses the meridian nearest your
> current position.
In reality one will steer nout exactly the great circle but an
approximation of it, with unchanging courses between points. The
difference to the real great circle will be small
--
I doubt, therefore I might be.
Wilbur Hubbard
news:heuduj$p7v$01$1...@news.t-online.com...
What many people don't seem to understand is the "Great Circle" route is
actually a figment of the imagination that is useful when translating a flat
paper chart to the sphere that is the Earth. The angular great circle route
mentioned above is imposed upon the chart in order to try to emulate how it
actually is on a sphere. The fact of the matter is, however, on the sphere
the line is one directly towards the destination. The only circular aspect
of it is traveling straight over the surface (curvature) of the sphere. If
one were using a submarine and the ocean was deep enough one could indeed
proceed on a straight line (as seen in three dimensions) and arrive at the
destination. But, this not possible on the surface of the oceans. One can
travel a straight line in two dimensions but not in three. If viewed from
the side the line that appears straight when viewed from above will appear
as a curved line when viewed from the side because the boat must still
proceed on the surface around the curvature of the Earth. It is this
curvature that the "great circle" route traced upon a flat piece of paper
attempts to describe.
I hope this helps. It takes a Yank to tell a Brit what's going on when it
comes to sailing.
Wilbur Hubbard
Peter Köhlmann
> Peter K�hlmann wrote:
>> "Mike Coon" <Mike@@mjcoon.+.com> wrote:
>>> But perhaps not a real navigator! Else she would be able to explain
>>> how a great circle route can be shorter than a "straight line"
>>> (whatever straight might mean here). Surely no-one would call a
>>> rhumb line "straight"?
>>
>> Well, I really hate to destroy your illusions, but a "great circle
>> route" is always the shortest route to take on a globe
>> And it *is* a straight line (on a gnomonic chart). It just isn't on a
>> standard sea chart. Which does not display a globe either
>
> I'm sure I have many illusions, but not, I think, in this area.
Well, to call them "illusions" was the friendly way of telling you that
you are utterly wrong
> (Although I have coded up a Molodensky transformation I only understand
> the purpose, not the mathematics!)
Which, even if it were true, has nothing to do with great circle routes
> On the other hand I have not crossed an ocean, or made a passage of a
> fracction of the 5000+ nm that Jessica Watson was making. So I don't
> know if she would use a gnomonic chart.
She quite probably would not. That problem is *easily* solved
mathematically
> My cruiser handbook says it "it
> is only used for small areas such as harbour plans".
Would you please tell us the publisher of that handbook? It is just to
make sure to stay away as far as possible from anything that publisher is
offering.
Because a gnomonic chart is completely unuseable for these things
> On the other hand
> if it yields accurate angles over such long passages perhaps that is how
> one would choose a compass course for a rhumb line course across an
> ocean.
You still have no idea what a "great circle course" or a "rhumb line
course" really is
> The fact that the great circle route is the shortest distance between
> two points is precisely the point of my comment.
No, it was not.
You wrote "Else she would be able to explain how a great circle route can
be shorter than a "straight line""
> But it may well be that
> a gnomonic "straight line" (as just mentioned above) is what she was
> comparing it with.
I rather doubt that. It is easier to figure out mathematically
> The three significant figures that she gave for the
> distance is impressive, too! Presumably the great circle route that she
> says she travelled was calculated for her by GPS.
I doubt that also
But the GPS *would* tell her a near great circle route while underway, as
it would adjust the course to steer accordingly
> Otherwise I have no
> idea how to work out the varying compass course to steer.
Obviously not. Although the formula is relativly simple
--
Klingon function calls do not have 'parameters' -
they have 'arguments' - and they ALWAYS WIN THEM.
Wayne.B
<wilbur...@thefarm.invallid> wrote:
>It is this
>curvature that the "great circle" route traced upon a flat piece of paper
>attempts to describe.
Actually it depends on the chart projection that is used to depict the
great circle route, true enough on the common mercator projection
where longitude lines appear to be parallel, but some chart
projections depict great circle routes as straight lines. An example
of that are gnomonic charts:
Wilbur Hubbard
news:7oq5h5dqpksjt0bup...@4ax.com...
Nevertheless, the gnomic chart is still not a sphere but a sphere
represented on a flat surface so it is less accurate than if one plotted a
course on a globe itself. In effect, one creates a gnomic chart by peeling
the globe. So try peeling a globe and laying the peeled off surface on a
flat table. Doesn't work, does it?
Wilbur Hubbard
Ronald Raygun
> "Mike Coon" <Mike@@mjcoon.+.com> wrote:
>>
>> My cruiser handbook says it "it
>> is only used for small areas such as harbour plans".
>
> Would you please tell us the publisher of that handbook? It is just to
> make sure to stay away as far as possible from anything that publisher is
> offering.
> Because a gnomonic chart is completely unuseable for these things
Rubbish. A gnomonic projection is eminently suitable for such things,
because at such a large scale as harbour plans would use, there is no
noticeable difference between gnomonic, Mercator, and most other
projections. The only thing wrong with "it is only used for small areas
such as harbour plans" is the word "only", since actually the area in
which gnomonic charts are most useful is at small scales where they are
used to discover the great circle route.
Ronald Raygun
> Ronald Raygun wrote:
>
>> Mike Coon wrote:
>
> < snip >
>
>>> The three significant figures that she gave for the
>>> distance is impressive, too! Presumably the great circle route that she
>>> says she travelled was calculated for her by GPS. Otherwise I have no
>>> idea how to work out the varying compass course to steer.
>>
>> In the absence of electronic aids, the usual way to work out what
>> compass course to steer is to start with the gnomonic chart and draw a
>> straight line on it. This line will cross several meridians on the way,
>> and will do so at various angles. These can be measured.
>
> There exist some relativly simple mathematical formulas to compute the
> course from any given point in between.
I'd be interested to know what they are. My intuition tells me that they
must be based on something a little more complicated than ordinary spherical
trigonometry, because delta lat is, but delta long is not, a great circle,
so we're not talking about ordinary right angled spherical triangles.
> Simple enough to do it without any
> electronic calculator in just a few minutes.
Had you said seconds, I'd have thought you meant a slide rule.
But minutes? Are you talking pencil and paper and trig tables?
Ronald Raygun
> Nevertheless, the gnomic chart is still not a sphere but a sphere
> represented on a flat surface
That's what "projection" means.
> so it is less accurate than if one plotted a course on a globe itself.
No, there is nothing in principle inaccurate about a projection.
> In effect, one creates a gnomic chart by peeling
> the globe. So try peeling a globe and laying the peeled off surface on a
> flat table. Doesn't work, does it?
No one doesn't. One creates it by making the surface of the globe
transparent and putting a point light source into the centre of the Earth,
and then projecting the surface features onto a flat screen. Usually, but
not necessarily, this screen is taken to be tangential, i.e. touching the
globe's surface at some point. The further away from the touch point you
get, the more distortion you get on the screen, but it happens to be the
case that this type of projection has the property that all great circles
on the globe's surface (not just those which pass through the touch point)
will end up as straight lines (rather than curves) on the screen.
A Mercator projection, on the other hand, is done similarly but not onto
a flat surface but onto a flat sheet which has first been rolled up into
a cylinder around the globe. This also has the property that you get more
distortion the further you get from the touch circle (which becomes a
touch line once the cyclinder is unrolled again), but also that (if the
axis of the cylinder is aligned with the NS axis of the globe) all latitude
lines end up parallel, and so do all longitude lines.
By the way, Wilbur, remind us what the difference is between a great circle
and any old ordinary circle on the surface of a sphere.
Wilbur Hubbard
news:ciEQm.10096$Ym4....@text.news.virginmedia.com...
> Wilbur Hubbard wrote:
>
>> Nevertheless, the gnomic chart is still not a sphere but a sphere
>> represented on a flat surface
>
> That's what "projection" means.
>
>> so it is less accurate than if one plotted a course on a globe itself.
>
> No, there is nothing in principle inaccurate about a projection.
>
>> In effect, one creates a gnomic chart by peeling
>> the globe. So try peeling a globe and laying the peeled off surface on a
>> flat table. Doesn't work, does it?
>
> No one doesn't. One creates it by making the surface of the globe
> transparent and putting a point light source into the centre of the Earth,
> and then projecting the surface features onto a flat screen.
<snip>
And that can't be done without it being distorted to fit the flat surface.
Same thing as peeling the surface and trying to flatten it - distortions.
Wilbur Hubbard
Peter Köhlmann
Yes. And logarithmus tables.
They exist, combined even for all relevant computations.
The german "Fulst" for example enables you to compute your position
astronomical without resorting to HO-tables (249/229/208), for example.
All relevant tables are provided. It really takes just a few minutes to
compute a great circle course with those
--
It's sweet to be remembered, but it's often cheaper to be forgotten.
Ian
wrote:
> Would you please tell us the publisher of that handbook? It is just to
> make sure to stay away as far as possible from anything that publisher is
> offering.
> Because a gnomonic chart is completely unuseable for these things
You should really let the UK Hydrographic Office know, because they
have been publishing large scale Admiralty charts with gnomonic
projections for many, many years. I have several on my boat.
Ian
Wayne.B
<no....@localhost.localdomain> wrote:
>> There exist some relativly simple mathematical formulas to compute the
>> course from any given point in between.
>
>I'd be interested to know what they are. My intuition tells me that they
>must be based on something a little more complicated than ordinary spherical
>trigonometry, because delta lat is, but delta long is not, a great circle,
>so we're not talking about ordinary right angled spherical triangles.
>
>> Simple enough to do it without any
>> electronic calculator in just a few minutes.
"American Practical Navigator" by Nathaniel Bowditch, Chapter 24,
sections 2404 to 2410, will tell you just about everything there is to
know about great circle navigation:
<http://www.nga.mil/MSISiteContent/StaticFiles/NAV_PUBS/APN/Chapt-24.pdf>
or
Calculating the initial bearing and distance of a great circle route
is a spherical triangle problem and can be solved using the same
tables or calculations used for celestial navigation. By repeating
the calculation every 100 miles or so, the great circle is broken up
into straight line segments, each of which can be sailed in
conventional fashion.
Edgar
> Calculating the initial bearing and distance of a great circle route
> is a spherical triangle problem and can be solved using the same
> tables or calculations used for celestial navigation. By repeating
> the calculation every 100 miles or so, the great circle is broken up
> into straight line segments, each of which can be sailed in
> conventional fashion.
>
Why not just insert a waypoint off Cape Horn and set your autopilot to steer
to the waypoint. That should automatically give you the great circle
course.
Peter Köhlmann
Still bullshit to use for harbours.
Apart from the fact that you would not see any real difference on such a
scale
Gnomonic charts are for *large* scale charts, when you want to do a great
circle course
--
Linux is simply a fad that has been generated by the media
Ian
> Ian wrote:
> > You should really let the UK Hydrographic Office know, because they
> > have been publishing large scale Admiralty charts with gnomonic
> > projections for many, many years. I have several on my boat.
> Still bullshit to use for harbours.
> Apart from the fact that you would not see any real difference on such a
> scale
Hmm. See a contradiction here?
Ian
Peter Köhlmann
You are aware that the computational model for doing gnomonic charts is
much more involved than doing mercator charts?
Why would someone in his right mind compute curves which are not even
visible on the scale projected?
Bloody Horvath
wrote this crap:
Good point. But does your autopilot steer from your GPS or magnetic?
My old tillermaster steered from magnetic settings so this would not
work. My newer autopilot intergrates with the Raytheon instruments
and a laptop computer with a GPS. No problem.
Vote for Palin-Arnold in 2012.
Hor...@Horvath.net
I'm Horvath and I approve of this post.
Bloody Horvath
>I'd be interested to know what they are. My intuition tells me that they
>must be based on something a little more complicated than ordinary spherical
>trigonometry, because delta lat is, but delta long is not, a great circle,
>so we're not talking about ordinary right angled spherical triangles.
Technically you can't have a great circle route on the Earth, because
the Earth is not a perfect sphere. (Sorry, I've always been an
abstractist.) However, compensating for the irregularities will still
get you there by the shortest route.
>> Simple enough to do it without any
>> electronic calculator in just a few minutes.
>
>Had you said seconds, I'd have thought you meant a slide rule.
>But minutes? Are you talking pencil and paper and trig tables?
I just put the destination in my GPS and it does all the computations.
(Too bad I get people to follow orders the same way.)
Ronald Raygun
> Technically you can't have a great circle route on the Earth, because
> the Earth is not a perfect sphere. (Sorry, I've always been an
> abstractist.) However, compensating for the irregularities will still
> get you there by the shortest route.
Suppose the Earth were a perfect spheroid, so that all the density-related
irregularities could be ignored. Then there would exist a simple linear
mapping by which any point P on the spheroid would correspond with a point
P' on a sphere.
What I have in mind is that the spheroid is described by an ellipse rotating
about the Earth's axis, and the mapping is that which transforms said
ellipse into a circle, so that the orthogonal distances of points P and P'
from the equatorial plane are the same, but the distance of point P' from
the axis is obtained by multiplying that of P by a constant K.
Use that mapping on two spheroid points A and B to get sphere points
A' and B'. Find the great circle route from A' to B', and let C' be
any point on it.
Now perform the inverse mapping (which will involve dividing axis distance
by K) to find the spheroid point C which corresponds to sphere point C'.
As C' moves along the great circle route, C will move along some route on
the spheroid.
Is that route the shortest from A to B? Discuss.
Ignatios Souvatzis
> Peter K�hlmann wrote:
>
>> course from any given point in between.
>
> I'd be interested to know what they are. My intuition tells me that they
> must be based on something a little more complicated than ordinary spherical
> trigonometry, because delta lat is, but delta long is not, a great circle,
> so we're not talking about ordinary right angled spherical triangles.
-- snippets from http://en.wikipedia.org/wiki/spherical_trigonometry --
Spherical triangles satisfy a spherical law of cosines
\cos c= \cos a \cos b + \sin a \sin b \cos C \!
The identity may be derived by considering the triangles formed by
the tangent lines to the spherical triangle subtending angle C and
using the plane law of cosines. Moreover, it reduces to the plane
law in the small area limit.
They also satisfy an analogue of the law of sines
\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}.
Here, a, b, and c are the angles at the centre of the sphere
subtended by the three sides of the triangle, and A, B, and C are
the angles between the sides, where angle A is opposite the side
which subtends angle a, etc.
------- end of snippets ---------------------------------------------------
I've used one of them (or both) to compute the direction to point a
directional antenna to in my youth, so it's not too hard.
You only have to compute A, B and C out of the lat/long pairs,
after setting e.g. A to the north pole, B and C to the two positions
you're considering.
Details left as an exercise to the reader.
-is
Edgar
"Bloody Horvath" <Blo...@Horvath.net> wrote in message
news:hkc7h5t5ug1rjej0t...@4ax.com...
Yes, it must of course be linked to GPS. Then all you have to do is make
sure you avoid islands and of course other ships.
Ronald Raygun
> Ronald Raygun wrote:
>> Peter K�hlmann wrote:
>>
>>> There exist some relativly simple mathematical formulas to compute the
>>> course from any given point in between.
>>
>> I'd be interested to know what they are. My intuition tells me that they
>> must be based on something a little more complicated than ordinary
>> spherical trigonometry, because delta lat is, but delta long is not, a
>> great circle, so we're not talking about ordinary right angled spherical
>> triangles.
>
> -- snippets from http://en.wikipedia.org/wiki/spherical_trigonometry --
>
> Spherical triangles satisfy a spherical law of cosines
>
> ------- end of snippets
Yes I know about all that, but it fails to address my specific point that
the "triangle" formed by the departure and destination points and a third
point chosen in such a way that the latitude and longitude differences
make up the two sides (the third side being the great circle route) is
not actually a proper spherical triangle (and the spherical cosine rule
therefore does not apply), because the longitude difference "side" is
not a great circle.
> I've used one of them (or both) to compute the direction to point a
> directional antenna to in my youth, so it's not too hard.
Easier in your youth than now, I expect! :-)
> You only have to compute A, B and C out of the lat/long pairs,
> after setting e.g. A to the north pole, B and C to the two positions
> you're considering.
Indeed. I had meanwhile remembered that the triangle you want to use
should have the North (or South if preferred) Pole as one of its corners,
and then we can work with three sides, all great circles: the GC route, and
the two other points' co-latitudes (you follow their meridians to the pole
instead of to the equator). The three angles in the triangle are the
longitude difference (at the pole corner) and at the other two corners they
are the bearings of each other (one applied clockwise from North, the other
anticlockwise).
> Details left as an exercise to the reader.
The fortunate thing is that provided we have the technology to compute
sines and cosines (and inverse cosines) and to multiply and divide,
we don't really need specialised navigation tables.
Also, it's convenient that because co-latitude of a point is equal to
90 degrees minus its latitude, and vice versa, and cos(x)=sin(90-x),
and vice versa, we can replace cos(colatitude) by sin(latitude) etc to
simplify the solution procedure. WE can work directly in terms of
latitude angles without needing to bothering to subtract them from
90 degrees first.
We don't need the sine rule, we need the cosine rule, twice. Once to
solve for the third side (this will give us the great circle distance).
Once to solve for the bearing (of destination from departure point, i.e.
the initial true course to steer).
R E C I P E :
Begin by computing sin and cos of the two latitudes and writing down
the answers. Three of these four quantities will be used twice, the
other once. Also compute the cos of the longitude difference:
S = sin(Lat1)
T = sin(Lat2)
C = cos(Lat1)
D = cos(Lat2)
L = cos(Long2-Long1)
Now comes the first use of the cosine rule, to compute the cos of the
great circle distance.
X = S*T + C*D*L
Then arccos(X) is the great circle distance (as an angle, if this is in
degrees, multiply by 60 to get miles).
As a convenience for what follows, work out Y = sin(arccos(X)). This
should be easy enough given that we've already calculated arccos(X).
The second use of the cosine rule solves for the bearing angle:
Z = (T - X*S) / (Y*C)
Then arccos(Z) gives us the bearing of point 2 from point 1. We need
to decide separately whether this bearing is OK as it is, or needs to
be subtracted from 360 degrees.
Example:
Depart from the junction of the Equator and the Greenwich Meridian
(Lat1=0, Long1=0) and navigate to a point at 75N 90W, somewhere in
the North of Canada (Lat2=75, Long2=-90). We get:
S=0, T=0.9659, C=1, D=0.2588, L=0,
X = 0, Y = 1, Z = 0.9659;
distance = 90 degrees or 5400 miles; bearing = 15 degrees, though
the required course is obviously 270.
Different example, same points, but going the other way:
S=0.9659, T=0 C=0.2588, D=1, L=0; X=0, Y=1, Z=0;
distance same as above, bearing 90 degrees.
A little thought should explain why the initially counterintuitive
result, that you should head Due East in order to get to the Equator,
is in fact correct.
cavelamb
Bloody Horvath
wrote this crap:
>>
>> Good point. But does your autopilot steer from your GPS or magnetic?
>> My old tillermaster steered from magnetic settings so this would not
>> work. My newer autopilot intergrates with the Raytheon instruments
>> and a laptop computer with a GPS. No problem.
>
>Yes, it must of course be linked to GPS. Then all you have to do is make
>sure you avoid islands and of course other ships.
>
Excellent! My newer autopilot connects with the Raytheon instruments
and a laptop computer. The computer shows wind direction, wind speed,
true windage and apparent. Water temp, depth, GPS coordinates, and
true vs. magnetic compass headings. (The Raytheon instruments are
connected to a magnetic flux sensor.) The computer is connected to
all of these and the GPS, it also runs a chart of the area we are in
showing all nav points and fixed obstructions and WILL steer around
any fixed obstructions such as islands.
The navigation software will not show other ships, (unless I had the
Raytheon radar to connect to it.)
Vote for Palin-Ahhnold in 2012.
Ian
> You are aware that the computational model for doing gnomonic charts is
> much more involved than doing mercator charts?
> Why would someone in his right mind compute curves which are not even
> visible on the scale projected?
You claimed that large scale gnomonic charts would be "completely
unusable". Then you said that "you would not see any real difference".
When you've made up your mind about what you're claiming, I'll
consider a response.
Ian
Ignatios Souvatzis
> Yes I know about all that, but it fails to address my specific point that
> the "triangle" formed by the departure and destination points and a third
> point chosen in such a way that the latitude and longitude differences
> make up the two sides (the third side being the great circle route) is
> not actually a proper spherical triangle (and the spherical cosine rule
> therefore does not apply), because the longitude difference "side" is
> not a great circle.
Sorry, I misread your original post and didn't notice what triangle
you had in mind.
>> I've used one of them (or both) to compute the direction to point a
>> directional antenna to in my youth, so it's not too hard.
>
> Easier in your youth than now, I expect! :-)
Just not any longer in memory, but nothing fancy. What I still didn't
do is to derive the sine and cosine rule from a series of rotations,
which would be a way for me to really remember the stuff. But I didn't
do that in my youth, either.
Regards,
-is
--
seal your e-mail: http://www.gnupg.org/