Calculate 'Rectangle' Coordinates Given 2 Points and width [closed]

Question

I want to create a rectangular polygon using two points as guides. So let's say a journey starts in Egypt and ends in London, my polygon should have 4 points:

• 10 miles further from London than Egypt is, following the line between them (roughly south in this example).
• Halfway between the two cities but 50 miles at right angles from the line that joins them.
• Like the above point but 50 miles in the other direction.
• 20 miles further from Egypt than London is, following the line between them (roughly north in this example).

I'll end up with a rough diamond shaped polygon that would completely contain the straight line journey from Egypt to London.

I hope this makes sense; any help for how I can calculate the 4 points is appreciated.

Rough Map Diagram

Answer 1

I will post another answer to presents a new way to do this, since my first way may be useful to other users that uses GIS software.

Well, try this PHP function to help you in the ST_Azimut and ST_Distance_Sphere. This will give you the distance and the azimuth if you pass the two coordinates of source and target. Please check if its in miles, kilometers or nautical miles ( I don't know ). Sorry about some portuguese var names.

PHP:

function CalcDistAndAzimuth($lat1, $lon1, $lat2, $lon2) {
    $dlat = 0.0;
    $dlon = 0.0;
    $apart = 0.0;
    $alfa = 0.0;
    $latm = 0.0;
    $rumo = 0.0;
    $distancia = 0.0;
    if (($lon1 == $lon2) && ($lat1 == $lat2)) {
        $rumo = 0;
        $distancia = 0.0;
    } else {
        $dlat = $lat2 - $lat1;
        $dlon = $lon2 - $lon1;
        $latm = abs(($lat1 + $lat2) / 2.0);
        $apart = $dlon * cos($latm * pi() / 648000); // 180*3600
        // calc distance and azimuth
        if ($dlat == 0) {
            $distancia = abs(($apart) / 60);
            if ($lon2 > $lon1)
                $rumo = 90;
            else
                $rumo = 270;
        } 
        if ($dlon == 0) {
            $distancia = abs(($dlat) / 60.0);
            if ($lat2 > $lat1)
                $rumo = 0;
            else
                $rumo = 180;
        } 
        if (($dlat != 0) && ($dlon != 0)) {
            $alfa = atan(abs($apart / $dlat));
            $distancia = abs(($apart / (sin($alfa))) / 60);
            if ($dlat <= 0) {
                if ($dlon <= 0)
                    $rumo = 180.0 + ($alfa * 180.0 / pi());
                else
                    $rumo = 180.0 - ($alfa * 180.0 / pi());
            } else {
                if ($dlon <= 0)
                    $rumo = 360.0 - + ($alfa * 180.0 / pi());
                else
                    $rumo = ($alfa * 180.0 / pi());
            } 
        }
    } 

    $rumodistancia["distance"] = $distancia;
    $rumodistancia["azimuth"] = $rumo;

    return $rumodistancia;
} 

This is other helpful function. Given a point, an azimuth and a distance, return a new point.

function CalcCinematic($lat1, $lon1, $azimuth, $dist) {
    // $dist is in miles
    // $azimuth is in degrees
    $dlat = 0.0;
    $dlon = 0.0;
    $apart = 0.0;
    $alfa = 0.0;
    $latm = 0.0;
    $delta = 0.0;
    if ($rumo <= 90.0) {
        // convert azimuth from degrees to radians
        $alfa = $rumo * pi() / 180.0;
        $dlat = $dist * cos($alfa);
        $apart = $dist * sin($alfa);
    } else
        if ($rumo <= 180.0) {
            $alfa = (180.0 - $rumo) * pi() / 180.0;
            $dlat = (-1) * $dist * cos($alfa);
            $apart = $dist * sin($alfa);
        } else
            if ($rumo <= 270.0) {
                $alfa = ($rumo -180.0) * pi() / 180.0;
                $dlat = (-1) * $dist * cos($alfa);
                $apart = (-1) * $dist * sin($alfa);
            } else
                if ($rumo <= 360.0) { 
                    $alfa = (360.0 - $rumo) * pi() / 180.0;
                    $dlat = $dist * cos($alfa);
                    $apart = (-1) * $dist * sin($alfa);
                }
    $lat2 = $lat1 +round($dlat * 60);
    if ($lat2 < -324000) {
        $lat2 = (-1) * 648000 + $lat2;
        $delta = 648000;
    } else
        if ($lat2 > 324000) {
            $lat2 = 648000 - $lat2;
            $delta = 648000;
        }

    $latm = abs(($lat1 + $lat2) / 2.0);
    $dlon = ($apart * 60.0) / cos($latm * pi() / 648000);
    $lon2 = $lon1 +round($dlon);

    if ($lon2 > 648000)
        $lon2 = -1296000 + $lon2;
    else
        if ($lon2 < -648000)
            $lon2 = 1296000 - $lon2;

    if ($delta == 648000) {
        $lon2 = $lon2 +round($delta);
        if ($lon2 > 648000) {
            $lon2 = -1296000 + $lon2;
        }
    }

    $newpoint["lat"] = $lat2
    $newpoint["lon"] = $lon2;
    return $newpoint;
}

Answer 2

Do this :

1) Calculate the distance of Hypotenuse between your two points using ST_Distance_Sphere.

2) Divide by four ( or other number of points )

3) Calculate the azimuth between Egypt and London using ST_Azimuth

4) Project the first point from Egypt, using the azimuth and the distance from item 2, using ST_Project.

5) From this point, project the second point, using same way, and so on until you reach London.

Note the use of radians in some functions. You can translate using Degrees() PostGIS function. You can create any pathway using this method.

Answer 3

I posted this question over at math.stackexchange.com after i realised only need to worry about 2D calculations and really wanted a non programming specific solution.

@ju'x answered it brilliantly, so credit should go to him.

Calculate 'Rectangle' Coordinates Given 2 Points and width