Distance from reference line at right angle

Question

A few weeks back I received a script that, I think, calculates the distance of a point from a reference line (at right angle) and returns the distance along the reference li

Answer 1

My opinion on the first function depends on whether the points described by arrays x,y lie on a line (as you say) or not (as the second function suggests; in the second script they are obtained from spline interpolation.

If x,y are consecutive points on the line

Then the computation of sina, cosa within the loop is redundant since the angle is always the same. For each point in the arrays xp,yp the function calculates its orthogonal projection onto the line. It returns the position along the line, measured from the first point x(1),y(1), and signed normal distance. All of this happens in a rather inefficient way for this task.

If x,y do not lie on a line

Then the computation of sina, cosa is incorrect. These are meant to be components of a unit vector in the direction from (x(i),y(i)) to (x(i+2),y(i+2)). But the denominator is s(i+2)-s(i), which [looking at the second function] will generally be greater than the straight-line distance from (x(i),y(i)) to (x(i+2),y(i+2)).

The following would make more sense:

cosa = (x(i+2)-x(i))/sqrt((x(i+2)-x(i))^2 + (y(i+2)-y(i))^2);
sina = (y(i+2)-y(i))/sqrt((x(i+2)-x(i))^2 + (y(i+2)-y(i))^2);

The confusion of straight-line distances and along-the-curve distances persists through the rest of the function: the comparisons of sproj with s(i+2) and especially with s(ns) are not geometrically motivated.

Anyway, the function attempts to locate a point near the orthogonal projection of xp,yp onto the curve (which is not unique to begin with); depending on how wiggly the curve is, it may have moderate success or fail badly.

Answer 2

If I understand what you're looking for, this is it.

This function takes in three points, the first two as a vector of x values and a vector of y values (I'll call them p1 and p2) and a third point (p3). The function returns the minimum distance from p3 to the line intersecting p1&p2 and the minimum of the distance from (p1 to the intersection point) or (p2 to the intersection point).

Example:

The two points to define the line are (1,2) and (5,10). The line defining them would be y = 2x. If the third point were (1,0). The perpendicular line is y = -(x-1)/2. The intersection point of the two lines would be (2/5, 1/5). The closest point defining the reference line would be (1,2) and the distance from (2/5, 1/5) to (1,2) is 1.7 and the distance from (2/5, 1/5) to (1,0) is 0.89. The function would return [1.7, 0.89].

Please let me know if this is what you want.

function [ dRef, dInt ] = StackExchange( x, y, xP, yP )
%StackExchange: Calculate line intersecting (x(1), y(1)) & (x(2), y(2))
%finds perpendicular line which intersects (xP, yP) which occurs as
%(IP(1), IP(2)) then calculated the minimum distance from the reference points
%(x(1), y(1)) or (x(2), y(2)) to (IP(1), IP(2))
%and from (IP(1), IP(2)) to (xP, yP)

%Calculate  slope of reference line, the slope of the perpendicular line
%will be -1/m
m = diff(y) ./ diff(x) ;

%IP(1) is the x-coordinate of the intersection point
%This formula is solving the equation
% m * (x - x(1)) + y(1) = -(1/m) * (x - xP) + yP
%for x
IP(1) = (m^2 * x(1) + xP + m * (yP - y(1))) ./ (m^2 + 1);
%IP(2) is the y-coordinate of the intersection point
IP(2) = m * (IP(1) - x(1)) + y(1);

%Minimum distance from the reference points to the intersection point
dRef = sqrt(min((IP(1) - x).^2 + (IP(2) - y).^2));

%Distance from the intersection point to  point of interest
dInt = sqrt(sum(([xP, yP] - IP).^2));

end