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AR15.COM
9/15/2005 10:47:04 AM EDT
Is there any software/hardware tool available that calculates the distance between two coordinates?

I need a simple calculation, 'as the crow flies' rather than following roads.

Software that can be loaded on a palm would be a plus.
9/15/2005 10:48:10 AM EDT
[#1]
a GPS would
9/15/2005 10:49:27 AM EDT
[#2]
Do you have the coordinates in a spreadsheet or database type file?

What system are they in?


IM them to me and I'll do it for you.

9/15/2005 11:00:17 AM EDT
[#3]
I have an Excel spreadhsheet that will do it (and do various conversions). Email me and I will sned it to you.
9/15/2005 11:02:22 AM EDT
[#4]
If it's UTM or State Plane or something else on a grid you could just compare the X coords and the Y coords, get the difference between each and use Pythagoras.

9/15/2005 11:02:41 AM EDT
[#5]
Any decent mapping software shuld be able to do that. DeLorme does it, MS Streets does it, the software that comes with your GPS does it...
9/15/2005 11:03:43 AM EDT
[#6]
Dist = ((XCoord2-XCoord1)^2+(YCoord2-YCoord1)^2)^(1/2)

Or, the distance is the square root of the sum of the squares of the differences of the coordinates. Pythagorean theorem and all that.
9/15/2005 11:05:19 AM EDT
[#7]
Hardware:  Math Book

Software:  Brain

You do the math. <rimshot>
Sorry, bad pun, and I'm too tired to pass it up.
9/15/2005 11:06:43 AM EDT
[#8]

Quoted:
Dist = ((XCoord2-XCoord1)^2+(YCoord2-YCoord1)^2)^(1/2)

Or, the distance is the square root of the sum of the squares of the differences of the coordinates. Pythagorean theorem and all that.



As long as it's on a grid.  If the coordinates are geodetic (lat long) it gets messed up because the further from the equator, the bigger the difference between the X and Y axis (in terms of how far, in feet or miles or whatever, one degree is).
9/15/2005 11:10:13 AM EDT
[#9]

Quoted:

Quoted:
Dist = ((XCoord2-XCoord1)^2+(YCoord2-YCoord1)^2)^(1/2)

Or, the distance is the square root of the sum of the squares of the differences of the coordinates. Pythagorean theorem and all that.



As long as it's on a grid.  If the coordinates are geodetic (lat long) it gets messed up because the further from the equator, the bigger the difference between the X and Y axis (in terms of how far, in feet or miles or whatever, one degree is).

I'm presuming that the poster has a coordinate system on a rectangular plane. If he's got a more sophisticated coordinate system, then he will need mapping software to account for the datum projections.
9/15/2005 11:14:08 AM EDT
[#10]

Quoted:

Quoted:

Quoted:
Dist = ((XCoord2-XCoord1)^2+(YCoord2-YCoord1)^2)^(1/2)

Or, the distance is the square root of the sum of the squares of the differences of the coordinates. Pythagorean theorem and all that.



As long as it's on a grid.  If the coordinates are geodetic (lat long) it gets messed up because the further from the equator, the bigger the difference between the X and Y axis (in terms of how far, in feet or miles or whatever, one degree is).

I'm presuming that the poster has a coordinate system on a rectangular plane. If he's got a more sophisticated coordinate system, then he will need mapping software to account for the datum projections.




I sorta figure the poster doesn't even know what a coordinate system is - most people don't - so he PROBABLY has lat/long coordinates.

Chances are, if he had coordinates in one of the various state planes, he waouldn't be asking for our help anyway.






Either way, it amazes me how many arfcommers understand his problem well enough to help him.


We should form a corporation.


'Hive Mind All-Purpose Tech Support'.
9/16/2005 10:16:46 AM EDT
[#11]
Thanks everyone for all the replies.  ARFcom beats any tech forum for quick answers

I have lat/longs of various sites in the metro area.  They are located along two rivers and I want to determine the distance between the sites to calculate an elevation offset for our calculations.

I used MapPoint but it is a pain to do over 160 sites manually.