Updated midpoint software - free online tool

[dschmidt]

The Geographic Midpoint Calculator, a free online tool has recently been updated with new features:

http://www.geomidpoint.com

It is used to calculate the midpoint (or central location) for two or more places. You select your places and the latitude and longitude of the midpoint

are calculated. You can view a map with a marker pointing at the exact midpoint for those places. It has been used for setting up and solving puzzle caches

and for caching at midpoints.

New features--it now has a built-in geocoder so that you have millions of world cities, towns and street addresses at your fingertips. You can now import

multiple latitude/longitude data from another source such as Excel or your word processor.

It's fast, accurate and fun to use.

dschmidt

[+TrailGators]

The Geographic Midpoint Calculator, a free online tool has recently been updated with new features:

http://www.geomidpoint.com

It is used to calculate the midpoint (or central location) for two or more places. You select your places and the latitude and longitude of the midpoint

are calculated. You can view a map with a marker pointing at the exact midpoint for those places. It has been used for setting up and solving puzzle caches

and for caching at midpoints.

New features--it now has a built-in geocoder so that you have millions of world cities, towns and street addresses at your fingertips. You can now import

multiple latitude/longitude data from another source such as Excel or your word processor.

It's fast, accurate and fun to use.

dschmidt

Very interesting! Thanks for the link!

[+fizzymagic]

How does it calculate the midpoint for more than 2 points?

[dschmidt]

The midpoint calculation is based on the model of a spherical earth. For more than 2 points each point is converted into cartesian coordinates (x,y,z).

The average x, y and z coordinate is calculated, then that average coordinate is converted back into latitude and longitude.

dschmidt

[+fizzymagic]

The midpoint calculation is based on the model of a spherical earth. For more than 2 points each point is converted into cartesian coordinates (x,y,z).

The average x, y and z coordinate is calculated, then that average coordinate is converted back into latitude and longitude.

 

Hmm.. OK. I am worried that people might try to use it to find a point equidistant from N given points, which it will not do. Instead, it finds the centroid, using what seems a reasonable definition thereof.

[+TrailGators]

The midpoint calculation is based on the model of a spherical earth. For more than 2 points each point is converted into cartesian coordinates (x,y,z).

The average x, y and z coordinate is calculated, then that average coordinate is converted back into latitude and longitude.

 

Hmm.. OK. I am worried that people might try to use it to find a point equidistant from N given points, which it will not do. Instead, it finds the centroid, using what seems a reasonable definition thereof.

That site would have saved me a lot of time solving this puzzle a few years ago. Edited September 3, 2007 by TrailGators

[dschmidt]

You've got a good point. The method uses the center of mass (or center of gravity). For 2 points it works perfectly. It doesn't necessarily find the equidistant point for 3 points; it seems it may be possible to calculate that but I don't know the mathematical procedure. For 4 or more points the center of mass method is by far the best, for example if you made a 3 point triangle then put the 4th point near the center. It would be impossible to find a point with equal distance from all four points.

 

I suppose that if folks understand that this site finds the midpoint based on center of gravity/mass and accept that as their definition of geographic midpoint when setting up and solving their multipoint problems, they should get consistent results when using the calculator. If they want to find a point that is equal distance from other points they should use another method.

dschmidt

[dschmidt]

I've been studying this more today. It turns out that the only triangle on a sphere where the points are equal distance from a center point is a triangle where all 3 sides are the same length, and the calculator handles this ok. The calculated midpoint doesn't give you an overall combined minimum distance from all points, however it gives a reasonable approximation to the location of minimum distance.

The center of mass method that it uses for a problem using 3 points would be like putting a weight at each point on a massless triangle, and the calculated midpoint would be the exact point on the triangle where the triangle balances on a pin.

dschmidt

[CoyoteRed]

Yes, it's good to know the method, reference, scale, or what have you, of various elements of puzzles. The classic is bearing: is that magnetic or true?

 

One can state it outright. The better method, I feel, is have a calibration leg where the seeker figures it out themselves. This inspires confidence they are doing it right. But it just depends on the feel of the hunt you're after. For quick and dirty, I'd rather have it stated.

 

I like to do this when using letterboxing-style clues in calibrating paces. Not only is a "pace" sometimes confusing--is that one or two footfalls--but folks have different length strides.

 

Knowing the method you are calculating the center folks can put that in their clues.

 

I'm wondering how large an area, or how far away the reference points, does it have to be before you start getting significant differences from your method and simply averaging the coordinates.

 

Nice website, BTW, very high on the usability scale, IMHO. My only suggestion would be for folks like myself who use browsers with script blockers is to highlight the parts we might be missing. I floundered around a bit before I checked the script blocking and turned it off for your site.

[dschmidt]

I did some calculations on triangles of different sizes to find the midpoint (center of mass/centroid) method and compared the results with a plain flat map calculated by hand and the Geographic Midpoint Calculator.

For a triangle roughly 100 miles per side at mid US latitudes the flat map and the calculator were 190 feet apart at their calculated centers. For a triangle with 50 mile sides the 2 centers were 53 feet apart. For a triangle with 10 mile sides the 2 centers were 24 feet apart.

I'm the software developer for the site, I'm giving some thought to creating another calculator page that would look pretty much the same as the current page, but would do the centroid/center of mass calculations based on a flat map instead of a spherical earth model. It wouldn't actually take a lot of work.

dschmidt

[+kurchian]

I have been trying to conjure up my school boys of geometry in an attempt to figure out how to solve some puzzles involving both centroids and circumcenters. Would this geographic midpoint calculator assist me in this endeavor?