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Old 06-02-2013, 22:19   #16
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Re: Coordinate Math Question

whoa.
I think daddle gave the right answer. to the simple question.

Spherical trig isn't required until the distances are ...global?
A "Mercator solution" should work pretty well, won't it?

IF it is not so easy as the "too-easy!" cardinal direction example above, just use the distance along the "bearing" as hypotenuse and angle in a simple right angle equation: latitude being one side, longitude the other. Apply his coefficient to the longitude side as he said...paraphrased: "the cos of lat is the ratio of long". This should work well enough wherever a mercator chart does.
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Old 06-02-2013, 22:33   #17
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Re: Coordinate Math Question

The OP didn't say anything about accuracy or limited distance although the example he gave was for a short distance. If he wanted to check the distance, for instance, from NY to Moscow then he would need to calculate the GC. So I guess it depends on what he is doing.
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Old 07-02-2013, 01:11   #18
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Re: Coordinate Math Question

They're short distances. Very short.

I'm calculating the lat-long of waypoints, relative to navigation aids.

My current example (but there are others) is a point 2,000 yards east of the Tyee Shoal Light, 200 yards south, and 200 yards west.

Tyee Shoal is at the entrance to Eagle Harbor, in Puget Sound.

I'm plotting a course around it, into the harbor.

There is another situation, in Eagle Harbor, where I want the longitude and latitude of a point halfway between two daymarks, but I think that's going to be easier.
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Old 07-02-2013, 01:18   #19
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Re: Coordinate Math Question

Whatever happened to:...sooner or later, I'll get there eventually...not in a hurry to do anything...retired...don't ask me to do a d*** thing! Mauritz
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Old 07-02-2013, 01:29   #20
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Re: Coordinate Math Question

Quote:
Originally Posted by Jammer Six View Post
They're short distances. Very short.

I'm calculating the lat-long of waypoints, relative to navigation aids.

My current example (but there are others) is a point 2,000 yards east of the Tyee Shoal Light, 200 yards south, and 200 yards west.

Tyee Shoal is at the entrance to Eagle Harbor, in Puget Sound.

I'm plotting a course around it, into the harbor.

There is another situation, in Eagle Harbor, where I want the longitude and latitude of a point halfway between two daymarks, but I think that's going to be easier.
Not sure if you are doing this manually (or just want to do it manually) but the few GPS units I'm familiar with will create a lat lon waypoint from an existing waypoint if you enter the range and bearing from the existing waypoint.
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Old 07-02-2013, 15:45   #21
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Re: Coordinate Math Question

I would call it calculation not logical deduction. You're not deducing anything, you're just running numbers.

Technically, simple math, or trigonometry, or spherical trig if you really want to get precise and start covering larger distances. If it isn't covered in Bowditch, there's enough basic navigation there to show you what to look for.

daddle-
"A nautical mile and therefore a minute of longitude is about 6000 feet." I'd rough ti the same way, but a minute of longitude will vary dramatically in distance, from the pole to the equator. A minute of latitude will be a constant distance. So..."good enough for government work" or a fast estimate, but that 6000 feet can be measured only woth a rubber ruler.
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Old 07-02-2013, 16:17   #22
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Re: Coordinate Math Question

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Originally Posted by hellosailor View Post
I would call it calculation not logical deduction. You're not deducing anything, you're just running numbers.

Technically, simple math, or trigonometry, or spherical trig if you really want to get precise and start covering larger distances. If it isn't covered in Bowditch, there's enough basic navigation there to show you what to look for.

daddle-
"A nautical mile and therefore a minute of longitude is about 6000 feet." I'd rough ti the same way, but a minute of longitude will vary dramatically in distance, from the pole to the equator. A minute of latitude will be a constant distance. So..."good enough for government work" or a fast estimate, but that 6000 feet can be measured only woth a rubber ruler.
Agree with hellosailor, disagree with daddle:

a minute of LATITUDE is about 6080 feet

A minute of longitude diminishes to zero feet, at the poles.
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Old 07-02-2013, 16:23   #23
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Re: Coordinate Math Question

To answer Jammer's original question: sure you can do it mathematically, but the usual way is simply to plot it on the chart or chartplotter, then run perpendiculars back to the edge for lat/long.

For a distance scale on a chart,
be sure to use the (vertical) 'scale' of latitude up the side:

one degree of latitude = 60 nautical miles,

hence one minute of latitude = 1 nautical mile,

and one tenth of a minute (usual minor graduation)
is a tenth of a nautical mile or 608 feet.
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Old 09-02-2013, 20:02   #24
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Agree with hellosailor, disagree with daddle:

a minute of LATITUDE is about 6080 feet

A minute of longitude diminishes to zero feet, at the poles.
You disagree with my post? That was grade school trig?? You do know what a cosine is? That's the diminish to zero part. And I said about 6000 feet ... Which is close enough for sailboat work.
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Old 09-02-2013, 20:42   #25
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Re: Coordinate Math Question

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Ya. Too easy. 200 yards is 600 feet. A nautical mile and therefore a minute of longitude is about 6000 feet. So that's 0.1 minute of position west.

What? You are stuck in a cold latitude and not here on the equator where minutes of longitude are not 6000 feet. Multiply by the cosine of your unfortunate latitude.

It's all done for you with the MEASURE tool in OpenCPN.
Now that I've decoded your post, I realise that it was muddled and ambiguous rather than strictly incorrect.

"A nautical mile and therefore a minute of longitude is about 60x0 feet." is not a correct statement.

It would have been, if you had said "a minute of latitude"


The word "therefore" suggests that the two items are the same.


You do go on to qualify it, to explain that it has limited validity, by making a puzzlingly cryptic statement* in the next paragraph

I'm not sure that turns it into suitable advice for someone who did not already know what you're trying to tell them.

* containing a mind-bending double negative and a misplaced phrase -- as a result of which you seem to be saying "minutes of longitude are not 6000 feet here on the equator" which is the converse to what you mean
... then you don't specify what to multiply

I was not poking holes in your trig, but in the misleading totality of what you ended up writing, particularly given the guy's level of experience.

I don't think on this occasion I'm making a mere pedantic quibble: it's essential that sailors know from day 1 never to refer to the top or bottom of the chart for a distance scale, even if they do happen to be sailing at the equator.

It needs to be something they never do, like crossing the street without looking both ways.

I'll never forget Naomi James, telling her story of being the first woman to sail solo around the world.

She confessed that it was only when she was well into the trip, and about to approach land for the first time, (well south of the equator) before her husband back in England realised that she had missed this essential point...

... in her hurried attempts to learn the rudiments of navigation from him, a highly experienced racing navigator, while preparing for the trip.
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Old 10-02-2013, 06:27   #26
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Re: Coordinate Math Question

You got the answer you needed with Cos(lat) and the differences of given location . . . pretty accurate for nm.

Simple calculations for a simpler time...

A sailboat(simpler times), makes 50 nm or so in a travel day, maybe 100nm in a 24 hour period.

From a crow's nest, a person should be able to see 15 or so miles.

In all fairness, these mathematics are not concise enough to put spit in a spitoon, but when used with a compass, these mathematics should get a navigator from point A, to someplace near point B.

Even if you have a know distance and and a know direction, your speed over ground is almost an educated estimate.

If Captain Ahab wants to get from point A to point Z in one fell swoop, at a given time of arrival and with a single calculation, he needed an accurate timepiece, needed to know the constellations and how to properly use a good sextant . . . oh and he also needed a crew member with a good eye on the crow's nest.

If we want to do the same today, we get out our GPS.
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Old 10-02-2013, 08:59   #27
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Re: Coordinate Math Question

" And I said about 6000 feet ... Which is close enough for sailboat work. "
Actually, at 45 degrees, the median latitude, one minute of longitude is down to about 4311 feet. Which is off by about 1/3 of a statute mile, and considerably off, even for a sailboat trying to rockhop in a channel.

Close enough if you're clear at sea, but the OP might be surprised by that.

And if they're plotting on one of those consarned Mercator projected maps...<G>...
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Old 11-02-2013, 05:08   #28
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Re: Coordinate Math Question

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Originally Posted by Jammer Six View Post
Is there a recognized way to deduce one position from another?

That it, given a bouy, and the coordinates for it in degrees, minutes and seconds, is it possible to determine mathematically the coordinates for a point 200 yards west of it?
have you ever heard about great circle sailing

yes of course there is "a recognized way". . people used to tho that when they use to do great circle sailing, literary until mid 80-s (no GPS). only instead 200 yards they used 2000, or 3000 miles etc, etc but procedure is same

take dividers mesure your position, mesure lat and long of said buoy, take nories tables or nautical tables and you have it

and jus tell me where you are sailing. after questions like this I do not wonnt to come close to that area.
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Old 11-02-2013, 07:28   #29
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Re: Coordinate Math Question

sorry I forgot it

here it is...

Haversine formula:
R = earthís radius (mean radius = 6,371km)
Δlat = lat2− lat1
Δlong = long2− long1
a = sin≤(Δlat/2) + cos(lat1).cos(lat2).sin≤(Δlong/2)
c = 2.atan2(√a, √(1−a))
d = R.c
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Old 11-02-2013, 07:47   #30
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Re: Coordinate Math Question

OK, now I'm confused. The distance scale on charts is normally in 10ths for lat, so I plot using degrees and minutes and decimals, rather than seconds, and for longitude, isn't the scale sized to accomodate the particular location the chart is based?
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