Nsryjdtyk

I am trying to find the straight distance from a point C to the beach. Beach line is defined by points A and B and with the Haversine formula I get the distance from C (my marker in Google Maps) to a point D in the AB beach line perpendicular to C.

Everything works fine but the point D is not the right one. I use this code to find D:

function get_perp(C){

        var A = { lat:33.345678, lng:-117.518921 };

        var B = { lat:33.100678, lng:-117.318492 };



        t = ((C.lat-A.lat)*(B.lat-A.lat)+(C.lng-A.lng)*(B.lng-A.lng))/((B.lat-A.lat)*(B.lat-A.lat)+(B.lng-A.lng)*(B.lng-A.lng));



        var D = { lat:0,lng:0};

        D.lat = A.lat + t*(B.lat-A.lat);

        D.lng = A.lng + t*(B.lng-A.lng);



        return D;

}

Returned D point is indeed a point on the line but it is not perpendicular to C. It is when the AB line is horizontal or vertical, but when it is not the angle between AB and CD is not right.

I've tried another functions I've found here but all of them cause the same result.

In this fiddle it is the whole process and if you zoom enough you can see the AB and CD lines are not perpendicular: Shortest distance from AB to C

EDIT: playing with it in geogebra I can see the function is OK in finding the point. The error happens then when google maps api represents the point. Geogebra

As far as I can see, your formula for D is correct, and the linear approximation is justified at such a small scale (deltas about a quarter of a degree; relative errors due to non-linearity should be on the order of 10^-5).

What you see can be due to the fact that the map projection is not conformant (does not preserve angles), so that the angle is not displayed as right. But the point is correct.

Do you know which projection they use ?

Bingo, the angle is right, just a display artifact due to the projection.

You make your calculations using plane geometry approach but they are wrong for spherical geometry. (C.f.: note that you found distance with Haversine formula, not Pythagorean formula).

At this page you can find algorithm and JS code to find cross-track distance and along-track distance (that might be used to find D point using bearing from the first point and this distance)

Cross-track distance

Here’s a new one: I’ve sometimes been asked about distance of a

point from a great-circle path (sometimes called cross track

error).



Formula:    dxt = asin( sin(δ13) ⋅ sin(θ13−θ12) ) ⋅ R

where   δ13 is (angular) distance from start point to third point

θ13 is (initial) bearing from start point to third point

θ12 is (initial) bearing from start point to end point

R is the earth’s radius

JavaScript: 

var δ13 = d13 / R;

var dXt = Math.asin(Math.sin(δ13)*Math.sin(θ13-θ12)) * R;

Here, the great-circle path is identified by a start point and 

an end point – depending on what initial data you’re working from,

you can use the formulæ above to obtain the relevant distance 

and bearings. The sign of dxt tells you which side of the path

the third point is on.



The along-track distance, from the start point to the closest 

point on the path to the third point, is



Formula:    dat = acos( cos(δ13) / cos(δxt) ) ⋅ R

where   δ13 is (angular) distance from start point to third point

δxt is (angular) cross-track distance

R is the earth’s radius

JavaScript: 

var δ13 = d13 / R;

var dAt = Math.acos(Math.cos(δ13)/Math.cos(dXt/R)) * R;