How many degrees in a mile?

[Guest jeremy]

be in degrees? Straight across the longitude or latitude is fine.

 

so, say 47.3952 + .1 is how many miles/meters? + .01? Etc.

 

Jeremy

[Guest bobhogan]

Jeremy - you're making me do math again on my day off!

 

Okay, starting at your coordinates of:

 

Lat: N 47.3952 = 47 23 43

Lon: W 121.4445 = 121 26 40

 

First convert to UTM

 

Zone 10

N 5,250,255

E 617,382

 

Now add one mile in meters to each number:

 

1 mile = 1,609 meters

 

New UTM coordinates:

 

Zone 10

N 5,251,864

E 618,991

 

Now convert back to Lat/Lon:

 

Lat: N 47 24 33.75

Lon: W 121 25 21.91

 

Take the difference between the Lat/Lon pairs:

 

Change Lat: 50.75 seconds = 0.00140972 deg

Change Lon: 78.09 seconds = 0.0216916 deg

 

So, there you have how much moving one mile of latitude or longitude affects your position at your Lat/Lon.

 

These conversions made possible through FREE software that the US Army Corps of Engineers developed. Download your free copy at:

http://crunch.tec.army.mil/software/corpscon/corpscon.html

 

HTH,

 

Bob Hogan

New England Geocaching

[Guest swangner]

Actually, how many miles there are per degree of long/lat (or the other way around, how many degrees/mile) depends on your current location, especially for longitude.

 

Since all lines of longitude converge at both the North and South poles, one mile (along an E-W route) covers more of a degree at 70 degrees latitude than the same mile does at 20 degrees latitude.

 

The same holds true for a much lesser degree for latitude, but since the lines of latitude do not converge (ever, they're always parallel to each other), the difference between one mile at 20 degrees latitude and one mile at 70 degrees latitude is less than the difference in one mile of longitude at the same points.

 

It's actually a fairly complex calculation, which involves calculating the angular distance between the start and end points, then finding the distance along the surface of the earth based on that angular distance and the circumferance (or radius, or diameter, since all of these are directly related) of the earth. Trust me, it's not a calculation you would want to have to do by hand.

 

I did a lot of research on this when I wrote a piece of GPS software for the Palm handheld, so I could calculate distance and bearing from a current position to a given waypoint.

[Guest jeremy]

I really only need an approximate. My goal (which I should have put in my original email ) is to allow people to decide whether they want a huge list of caches or a specific radius. Approximations would work.

 

I definitely know about how much of a pain distance calculations are! I do a pretty complicated one for this site.

 

Jeremy

[Guest jeremy]

I really only need an approximate. My goal (which I should have put in my original email ) is to allow people to decide whether they want a huge list of caches or a specific radius. Approximations would work.

 

I definitely know about how much of a pain distance calculations are! I do a pretty complicated one for this site.

 

Jeremy

[Guest swangner]

Jeremy -

 

Since I have the calculations coded (in BASIC) already, it would be easy enough for me to convert it to an ASP function (I noticed that you're using ASP on the site - I'm assuming you're using VBScript as the language for the ASP pages?).

 

Let me know, I'd be more than happy to send along my code (so far it seems to be pretty accurate) in the form of a VBScript function (i.e., you send it two sets of coordinates as parameters, and you get back the distance).

 

As far as the bearing goes, I'm not quite sure how accurate that part of my code is. I'm calculating the distance difference in both long and lat, then taking the arctangent of that, so it's an approximation since the arctangent function is getting planar geometry coordinates, not spherical geometry. It should be fairly accurate for shorter (less than 500 miles, say) distances, since the difference between using planar geometry and spherical geometry is very small over distances like this (because the distances involved are tiny compared to the curvature of the earth).

[Guest Greg Bernath]

Approximations:

 

1 nautical mile is 1 minute of longitude at the equator, or 1 minute of latitude anywhere.

 

1 nautical mile = 1.15 regular (statute) miles.

 

So one degree of latitude, or longitude at the equator, is 60*1.15=69 statute miles.

 

One degree of longitude anywhere would be 69 miles times the cosine of the latitude.

 

Pushing the invert key on your calculator will give degrees in a mile.

 

For short distances, a flat earth model will give a good approximation of distance.

 

Distance = 69 * sqrt[(change in degrees lat)^2 + {cos(lat)*(change in degrees lon)}^2]

[+GOT GPS?]

I wonder how this eventually was worked out?? How the radius was determined in Alaska versus in Hawaii??

 

[This message was edited by GOT GPS? on May 03, 2003 at 06:39 PM.]

[+phantom4099]

That is a little weird that I happen to see the link to that army software twice in the same day (and I don't think I ever heard of it before today). Whats really strange is that this page and the one I read on Joe and Jacks web page are over two years diffrent.

 

The page on joe and jacks web page describes how to set up state plane coord systems on a magellan (http://gpsinformation.net/state-plane-mag.html), the army software is mentioned toward the bottom.

 

Wyatt W.

 

The probability of someone watching you is directly proportional to the stupidity of your actions.