distance denoted...

[9mmCaching]

Is the 1/1000th position of the minutes reading equal to 10 feet? Or what??? If so, then 1/100th is 100 feet? Oh, forget it...

[+Jamie Z]

1/1000 of a minute is roughly six feet, therefore 1/100 of a minute is about sixty feet, 1/10 minute about 600 feet, and a whole minute? About 6000 feet, or just about a nautical mile.

 

Jamie

[+Alan2]

Which explains why sailors use nautical when navigating as it equates to the geo-coordinate system. But the 6 feet, which also represents a fathom. Does fatham come from the 1/1000 or from another measurement standard?

 

Alan

[+bigcall]

I believe fathom goes back to the old way of measuring water depth and/or speed. A sounding line had a series of knots tied at lengths roughly equal a fingertip to fingertip distance (approximately six feet). When entering shallow water harbors this line was cast over the side and could then be used to judge depths and, based on the angle from perpendicular, the speed of the vessel.

 

This is based on my recollection of sea stories so treat it accordingly.

 

Way back in the days when the grass was still green

and the pond was still wet

and the clouds were still clean,

and the song of the Swomee-Swans rang out in space...

one morning, I came to this glorious place.

[Dru Morgan]

1/1000 of a minute is 6 feet only N/S, correct?

For E/W it would get smaller as you get closer to the poles. Does that mean that your GPSr is more accurate as it gets to the poles in the E/W coordinates? Is there an easy way to approximate your 1/1000 distance depending on your location?

 

Everywhere that cache is found,

Bound to Cover Just a Little More Ground.

-Dru Morgan www.theheavenlyhost.com/dru

[+majicman]

It depends, A LOT. North / South and East / West, as mentioned, are not the same.

 

But the answer given that it is "roughly" 6 feet per 1/1000th of a minute "generally" holds true (unless you see polar bears or penguins out your door.)

 

(I hate these kind of generalizations, being so anal-retentive, but I have learned to live with them... I would rather know that it is 5.734533 feet at my "current" location if N/S and 5.9325 if E/W. - Whew, that's better!)

 

majicman

[+The Leprechauns]

For way too much information on antiquated nautical units of measure, see:

Nautical Units and Angles

 

Fathoms as a depth measurement "have been in use since before 1600, and may be derived from "faethm", the Anglo Saxon word for "to embrace" (because it is roughly the distance from one hand to the other if your arms are out-stretched)."

 

Next time, instead of getting married, I think I'll just find a woman I don't like and buy her a house.

[+Waterboy]

1. The definition of a nautical mile is the length of 1 minute of arc at the equator. One minute of arc at the equator is equal to a one minute change in longitude at the equator. One nautical mile is approximately equal to 6,076 feet or 1,852 meters.

2. The earth is not a perfect sphere. Thus the length of one minute of difference in latitude is not the same as the length of one minute change in longitude. The latitude change is slightly smaller. However the difference is so small it should be considered negligible.

3. For all general purposes we may use 6,076 feet or 1,852 meters as the distance of one minute change in latitude. A distance of 608 ft or 185 m is 1/10 of a minute. 60.80 ft or 18.5 m is 1/100, and 6.08 ft or 1.85 m is 1/1000.

4. With a little trigonometry the distances for 1/1000 of a minute change is longitude can be calculated as that distance in latitude times the cosine of the latitude.

5. The table below shows the distance for 1/1000 if a minute in longitude expressed in both feet and meters at various latitudes:

 

At 00° Lat, 1/1000 of a min in Long = 6.08 ft. or 1.85 m.

At 10° Lat, 1/1000 of a min in Long = 5.98 ft. or 1.82 m.

At 20° Lat, 1/1000 of a min in Long = 5.71 ft. or 1.74 m.

At 30° Lat, 1/1000 of a min in Long = 5.26 ft. or 1.60 m.

At 40° Lat, 1/1000 of a min in Long = 4.65 ft. or 1.42 m.

At 50° Lat, 1/1000 of a min in Long = 3.91 ft. or 1.19 m.

At 60° Lat, 1/1000 of a min in Long = 3.04 ft. or 0.93 m.

At 70° Lat, 1/1000 of a min in Long = 2.08 ft. or 0.63 m.

At 80° Lat, 1/1000 of a min in Long = 1.06 ft. or 0.32 m.

At 90° Lat, 1/1000 of a min in Long = 0.00 ft. or 0.00 m.

[+Team Hoijong]

This is way to much information for me. my brain can't handle al these things..

 

I just stick to the things my GPS tells me..

I'm not going to figure out to much..

 

Irresisti

N12º 55.475

E100º 52.865

[Dru Morgan]

Does that mean that if my GPSr displays 1/1000 of a minute, and that is about 5 feet at my house. If I go to the poles, am I more accurate? Can I approach 0 as an accuracy? If I am off by 2/1000 where I live, that could be 10 feet, but near the poles, it could be 2 feet. Or, is it still as inaccurate regardless of your location?

 

Everywhere that cache is found,

Bound to Cover Just a Little More Ground.

-Dru Morgan www.theheavenlyhost.com/dru

[+ClayJar]

quote:

---

Originally posted by The Heavenly Host:

Does that mean that if my GPSr displays 1/1000 of a minute, and that is about 5 feet at my house. If I go to the poles, am I more accurate? Can I approach 0 as an accuracy? If I am off by 2/1000 where I live, that could be 10 feet, but near the poles, it could be 2 feet. Or, is it still as inaccurate regardless of your location?

---

The "hdd mm.mmm" coordinate space gets more accurate as you approad the poles. One thousandth of a minute of east or west even drops to identically zero when you're *at* the poles. So, the "rectangle" of the coordinates plus or minus half a thousandth of a minute will shrink as your latitude increases.

 

Now, the accuracy of your GPS receiver will not be intrinsically any better. You should have a really good view of the sky, which would improve reception, but you almost certainly be out of WAAS reception (and even if you weren't, there are likely no ground stations in the area to compute the errors anyway), which means you wouldn't have that helping.

 

So, basically, if you could know *exactly* where you were, as you go nearer the poles, you can express that more and more precisely in a given format of lat-long coordinates; however, that does not mean that your GPS receiver will in fact be working better (and it may even be that you need a block heater for it just to keep it warm, depending on season ).

[mwelch8404]

I like the depth answer - and the good ol'

 

"By the mark - twain"

 

Two fathoms... (IIRC min for a "riverboat")

 

Once you are over six fathoms it's:

 

"By the deep six." This is 36 feet and really ment "at sea." This is where the navy term to

"Deep Six" something (ie toss or throw away) came from.

 

"Archaeology is... the search for fact. Not truth. If it's truth you're interested in, Doctor Tyree's Philosophy class is right down the hall." Indy- Last Crusade

[matjaz]

Hi!

 

I'm a newbie, with 0/0 caching score, but I think I understand the works of the GPS system...

 

> poles, it could be 2 feet. Or, is it still as

> inaccurate regardless of your location?

 

That's right. See, the GPS gadget just measures distance to the visible satellites, calculates your position, and only then (for your convenience) converts it to LAT/LONG. As far as the position fix accuracy, the earth axis orientation (i.e. where the poles are) is completely irrelevant.

 

It follows from that, that no better accuracy should be expected at the poles, since their position is just a matter of convention, nothing fundamental. (Well, yes, so they are where the earth bearings are, but with GPS, it doesn't enter the story at all.)

 

(The earth axis actually goes throuh Slovenia, as anyone from Europe can tell you, that is, 46degN,15degE.

 

So, if you stand right on the pole or within a few feet of it, you'll still get those 20 feet or so of accuracy, but the longitude reading will spin around in despair...

 

Hope that helped..

take care,

Matjaz

[+Alan2]

From a practical standpoint it would seem that geocachers in Alaska will suffer more from "loose bearings" than those caching at the equator as the coordinates displayed will tend to bounce around more. As you move closer to the poles,each slightly different caculation from the satellites will result in a larger difference in the possible final coordinates displayed for the same amount of "error".

 

Yes? No?

 

Alan