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RandallR

A question about the lat/long coordinates system.

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FS2002 gives coordinate info in deg, min, sec format. I found a calculator program on the internet that converts into decimal format (I need decimal format so that can do calculations more easily). How do these values (in decimal format) translate into meters or feet?Here is an example of two points (in decimal format) that I need to know the distance between: Point 1: Long -87.831505, Lat 42.984665Point 2: Long -87.83128045, Lat 42.98464722Now, the difference between the two longitude points is 0.00022455. How do I figure out how this translates into feet (or fraction thereof)? The distances I'm trying to calculate are quite small so I don't need to account for the curving of the earth....I'll just assume a flat plane. I am hoping that it is not necessary to apply Great Circle formulas.I tried converting things and applying different formulas but I only ended up confusing myself. I think there must be something I'm missing. One minute of lat/long = I nautical mile. That

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A better question for me to ask may be: How do I find the distance between two lat/long points in feet, where the coordinates are presented in decimal format?Point 1: Long -87.831505, Lat 42.984665Point 2: Long -87.83128045, Lat 42.98464722Thanks.

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This exact same question was answered on the forum some time ago. Do a search.You have to find the angle between the two points on the great circle and then from 1 minute = 6080 ft calculating the distance is a snack.Calculating the angle requires a bit of maths though. :-)

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By moving the co-ordinates to decimal degrees, you're somewhat complicating the issue. One nautical mile is 1 minute of Longitude at the equator, or 1,852 Kilometers. 1 minute of Longitude at the pole, unfortunately, is zero. The distance betweeen two pairs of points with the same difference of Longitude is therefore dependant on the Latitude of the two points. How accurate do you need to be? Assuming a linear slope for the declining value of Longitude is the easiest, but least accurate, way.One degree of longitude at the equator is 1852 * 60 or 111120 Kilometres. Since you're dealing with small values, we can treat the earth as a perfect sphere, and assume one degree of Latitude to also be 111120 Kilometres.In your example, the difference of Latitude between the two points is 42.984665 - 42.98464722 or 0.00001778 degrees, which is 1.9757 Kilometres.The difference between the Longitudes is 0.00022455 degrees. Assuming a linear change in Longitude tells us that one degree Longitude at 43' Latitude is 43/90 of 111120 Kilometres, or 53091 Kilometres. The difference of Longitude between the two points is thus 11.9215 Kilometres.We now have a right angled triangle, with one side of 1.9757 Kilometres, and the other of 11.9215 Kilometres, and we want to know the length of the hypotenuse. Basic trig tells us that the square of the hypotenuse is the sum of the squares of the other two sides, so the length is the square root of (1.9757*1.9757)+(11.9215*11.9215), or 12.083 Kilometres.Hope this helps.Richard

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Christine,A great tool for this is the Fly! Distance & Bearing Calculator by Colin Sare-Soar. This wonderful little program is something we use in the Fly! world for calculating the position of scenery objects, navigation transmitters, runway endpoints, etc. You can enter a location in deg/min/sec, decimal, or Fly! coordinates, then enter a second location. The calculator instantly gives you the bearing and distance in feet, meters, and nautical miles.You can also enter the first coordinate, then enter the bearing and distance to instantly obtain a new lat/lon. Colin's calculator is in the Avsim File Library here: http://ftp.avsim.com/library/sendfile.php?DownloadID=8474Colin has also just placed a new version of the program on his site: http://www.avsim.com/hangar/fly/colins2 under the "Other Utilities" area. This version resolves a bug for use with Southern Hemisphere coordinates.

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Christine,I tested your two points by entering them into Colin's tool. First, I converted them back to standard lat/long by using the old *60 rule - I did this to make it easier to enter the data into the calculator. This gives us the following points:Point 1: Long 87

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Randall, thank you for taking the time to look this up for me.Christine

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No problem - if you need info like this often, Colin's Calculator is a great thing to have in your toolbox.

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