October 21st, 2016, 04:48 AM  #1 
Newbie Joined: Oct 2016 From: Amsterdam Posts: 2 Thanks: 0  Lat/lon projection on line
Please see attached picture. I have two lines. One line is between two coordinates P2 and P3. From the second line I know one coordinate P1 and the angle alpha. P1, P2 and P3 are earth coordinates (lat/lon). The angle alpha is in degrees (0360). How to calculate the distances d1, d2 and d3 in meters, as a function of X1,Y1,X2,Y2,X3,Y3 (lat/lon) and alpha (degrees)? Note: The formulas should work for all angles alpha (0360) and also if lines intersect, or are parallel. The distances I have to calculate are max 5000 meters. Therefore the earth's radius can be left out of the equasion (I think??). Last edited by Neo444; October 21st, 2016 at 04:55 AM. 
October 21st, 2016, 06:40 AM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
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Last edited by skaa; October 21st, 2016 at 06:48 AM. 
October 21st, 2016, 06:44 AM  #3 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
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Last edited by skaa; October 21st, 2016 at 06:48 AM. 
October 21st, 2016, 06:47 AM  #4 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
Let's: $\displaystyle \gamma=\left\{\begin{matrix} \arctan{\frac{x3x2}{y3y2}},\ if\ y3\neq y2\\ \frac{\pi}{2},\ if\ y3=y2 \end{matrix}\right.$ Then: $\displaystyle d3=\sqrt{(x3x2)^2+(y3y2)^2}\cdot\cos{(\gamma\alpha)}$ Continue?.. 
October 22nd, 2016, 08:10 AM  #5 
Newbie Joined: Oct 2016 From: Amsterdam Posts: 2 Thanks: 0 
Thank you very much. Could you calculate d1 and d2 for me also? The coordinates I mentioned above are lat/lon (WGS84 system). So first I have to convert the coordinates to X,Y,Z based on WGS84. This is an eliptical system. Do you happen to know the formulas for that also? 
October 24th, 2016, 10:46 AM  #6 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
The ray containing P1: If α≠0 then: $\displaystyle y=\cot\alpha\cdot x+y1\cot\alpha\cdot x1$ (*) else: $\displaystyle x=x1$ The line P2P3: If x3≠x2 then: $\displaystyle y=\frac{y3y2}{x3x2}x\frac{y3y2}{x3x2}x2+y2$ (**) else: $\displaystyle x=x2$ It makes sense to solve this problem if the ray containing P1 intersects P2P3, otherwise d1 and d2 do not exist. Let us designate the point of interception (xi, yi). 1. α≠0. 1a. α≠0 and x3≠x2. $\displaystyle xi=\frac{y1\cot\alpha\cdot x1+\frac{y3y2}{x3x2}x2y2}{\frac{y3y2}{x3x2}\cot\alpha}$, yi can be found from (*). 1b. α≠0 and x3=x2. $\displaystyle xi=x2$, yi can be found from (*). 2. α=0. 2a. α=0 and x3≠x2. $\displaystyle xi=x1$, yi can be found from (**). 2b. α=0 and x3=x2. The ray containing P1 is parallel to P2P3, or is a part of it d1 and d2 do not exist. Continue?.. 
October 24th, 2016, 11:58 AM  #7 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
Just in case if x2≤xi≤x3 and y2≤yi≤y3 or if x3≤xi≤x2 and y3≤yi≤y2 the problem can be solved.


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lat or lon, line, projection 
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