What's the formula

[Guest artmartin]

for converting distance to lat/long amounts? I used to have a link to a post that told me that but I just can't find it anymore. I know that the proportion for latitude will always be the same everywhere on the earth but that distance/longitude value will be larger at the equator. I've decided to write a quick and dirty app that will allow designers to apply an offset to all taxiway signs at once and I need to know if someone wants the signs to move say 3 meters to the east and 5 meters to the north, just how much to add or subtract from the lat/long values before rewriting the file. The app is basically done except for this formula. I could go out and find the darn circumference of the Earth and figure this out but I'm just too darned tired.Art

[Guest archtx]

>for converting distance to lat/long amounts? I used to have>a link to a post that told me that but I just can't find it>anymore. I know that the proportion for latitude will always>be the same everywhere on the earth but that>distance/longitude value will be larger at the equator. I've>decided to write a quick and dirty app that will allow>designers to apply an offset to all taxiway signs at once and>I need to know if someone wants the signs to move say 3 meters>to the east and 5 meters to the north, just how much to add or>subtract from the lat/long values before rewriting the file. >The app is basically done except for this formula. I could go>out and find the darn circumference of the Earth and figure>this out but I'm just too darned tired.>>ArtArt: It involves a little spherical geometry...but basically the circumference of the earth at a given latitude is proportional to the cosine of the angle of latitude. i.e., at the equator the earth has a mean radius of 3963 st. miles,therefore a circumference of approx 24,900 miles. The cosine of 10* is 0.9848, therefore at 10* north the circumference would be 24,900 x 0.9848. Divided by 360, that would be 68.117 miles per degree longitude, or 99.9 feet per second of longitude. At 80* north it is down to about 17.6 feet per second of longitude. As you say, N-S is pretty much a constant of 101.28 feet per sec of latitude. Hope that helps.Metric: Circumference of earth at equator=40,075.16 km. More than you want to know...circumference at poles about 67 km less than at equator.(0.17% difference)

[arno 1]

Hi Art,That all depends on the earth model that you are using of course :). And on the projections used, etc.If you assume the earth is a perfect sphere then for the y direction (north) it would mean that 2*pi*Rearth/360 is the amount of meters covered by one degree. I don't know the radius of the earth by heart (something like 6735 km or so, you would have to check that).For the x direction (east) you would have to multiply with the cosine of the latitude. But as you are also moving in that direction this will give a small error of course. So you might want to take the average of the two latitudes or so.If you take a more complex earth mou can not calculate it that easily. At work I use the Proj4 library to do transformations between different coordinate systems. You might want to take a look at that as well.

[Guest luissa]

Hi Art,I use a Constant and a Function (in Visual Basic):Public Const MeterPerDegLat As Double = 111330# Public Function MeterPerDegLon(ByVal lat As Double) As DoubleMeterPerDegLon = Cos(lat * PI / 180#) * 111330#End FunctionLuis

[Guest luissa]

Hi Art,I use a Constant and a Function (in Visual Basic):Public Const MeterPerDegLat As Double = 111330# Public Function MeterPerDegLon(ByVal lat As Double) As DoubleMeterPerDegLon = Cos(lat * PI / 180#) * 111330#End FunctionLuis

[scott967 62]

Here's a good online calculator, which might give what you want:http://www.wherearewe.co.nz/geodetics.htmlscott s..

[Holger 0]

Howdy,and don't forget that there may be a difference between the real world and what the FS designers used as an earth model. For example, here's what the Creating Terrain.doc from the Terrain SDK says (p.18):In Flight Simulator, the earth is defined as an

[Guest artmartin]

Thanks all. For my purposes ballpark was all I needed since the amounts I'm trying to offset things are a judgment call anyway. I actually came up with the answer on my own after a little research and a brushup on my trig.Art

[scott967 62]

I think SceneGenX build 42 already provides what you need?http://forums.avsim.net/user_files/129210.jpgscott s..

[Guest artmartin]

ah wonderful. That's one product I haven't played with yet.Art