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POFind nearest position from a fixed position to a line joined by two GPS positions
 

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  1. POFind nearest position from a fixed position to a line joined by two GPS positions
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    <p>I want to find the nearest position from a fixed position to a line joined by two GPS positions. I have tried to illustrate it with the means of a diagram. All positions are in GPS coordinates. I want to find the ´shortest diatance from the position P to the line joined from position A and position B,i.e., the line from P makes a 90 degree on line joining A and B. It would be really great if you know of an existing implementation for this or an algorithm for the same. Many Thanks. <img src="https://i.stack.imgur.com/YazW1.png" alt="enter image description here"></p> <p>The way i was told to calculate is: first convert the GPS positions into cartesian coordinates. For the conversion , it uses a reference position (Q_Ref_longitude, Q_Ref_latitude).</p> <pre><code>x=(longitude-reflong)*π/180*r_e, r_e radius of earth y=artanh(sin⁡(latitude)) Therefore the reference point has the Cartesian coordinates (0,artanh(sin⁡(reflat))). conv(refp)=(0,artanh(sin⁡(reflat))) All positions A,B, P are converted to (x,y) coordinate system Calculation of the intercept point with reference Cartesian coordinates (x,y)∈R^2: In the following context (vp) ⃗∈R^2 is the position vector of the P in reference Cartesian coordinates. &lt;a ⃗,b ⃗&gt; is the dot product of two vectors a ⃗ and b ⃗. (op2) ⃗ refers to position A (op1) ⃗ refers to position B (vp) ⃗ refers to position P (op1) ⃗,(op2) ⃗,(vp) ⃗∈R^2 intPoint=[(&lt;((op2)-(op1)),((vp)-(op1))&gt;)/(&lt;((op2)-(op1)),((op2)-(op1))&gt;)]*((op2)-(op1))+(op1) The intercept point has to be transformed back into GPS coordinates with the following inverse formula: Iconv:R^2→[-180°,180°]×[-90°,90] longitude=x+reflong latitude=arcsin⁡(tanh⁡(y)) </code></pre> <p>Does someone know about this way? I tried to do this way but it gives negative values.</p>
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    1. USjava_learner
    1. USjava_learner
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    1. COcheck http://stackoverflow.com/questions/10358022/find-the-better-intersection-of-two-moving-objects
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    2. COWhat kind of distances are we dealing with? If short enough you can approximate AB as a straight line. If you must account for the curvature of the Earth it gets quite complicated.
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      1. USJoni
    3. CO@Joni: the distances are short enough in the range of less than 50 metres.
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