I have a square bitmap of a circle and I want to compute the normals of all the pixels in that circle as if it were a sphere of radius 1:
The sphere/circle is centered in the bitmap.
What is the equation for this?
Don't know much about how people program 3D stuff, so I'll just give the pure math and hope it's useful.
Sphere of radius 1, centered on origin, is the set of points satisfying:
x2 + y2 + z2 = 1
We want the 3D coordinates of a point on the sphere where x and y are known. So, just solve for z:
z = ±sqrt(1 - x2 - y2).
Now, let us consider a unit vector pointing outward from the sphere. It's a unit sphere, so we can just use the vector from the origin to (x, y, z), which is, of course, <x, y, z>.
Now we want the equation of a plane tangent to the sphere at (x, y, z), but this will be using its own x, y, and z variables, so instead I'll make it tangent to the sphere at (x0, y0, z0). This is simply:
x0x + y0y + z0z = 1
Hope this helps.
(OP):
you mean something like:
const int R = 31, SZ = power_of_two(R*2);
std::vector<vec4_t> p;
for(int y=0; y<SZ; y++) {
for(int x=0; x<SZ; x++) {
const float rx = (float)(x-R)/R, ry = (float)(y-R)/R;
if(rx*rx+ry*ry > 1) { // outside sphere
p.push_back(vec4_t(0,0,0,0));
} else {
vec3_t normal(rx,sqrt(1.-rx*rx-ry*ry),ry);
p.push_back(vec4_t(normal,1));
}
}
}
It does make a nice spherical shading-like shading if I treat the normals as colours and blit it; is it right?
(TZ)
Sorry, I'm not familiar with those aspects of C++. Haven't used the language very much, nor recently.
This formula is often used for "fake-envmapping" effect.
double x = 2.0 * pixel_x / bitmap_size - 1.0;
double y = 2.0 * pixel_y / bitmap_size - 1.0;
double r2 = x*x + y*y;
if (r2 < 1)
{
// Inside the circle
double z = sqrt(1 - r2);
.. here the normal is (x, y, z) ...
}
Obviously you're limited to assuming all the points are on one half of the sphere or similar, because of the missing dimension. Past that, it's pretty simple.
The middle of the circle has a normal facing precisely in or out, perpendicular to the plane the circle is drawn on.
Each point on the edge of the circle is facing away from the middle, and thus you can calculate the normal for that.
For any point between the middle and the edge, you use the distance from the middle, and some simple trig (which eludes me at the moment). A lerp is roughly accurate at some points, but not quite what you need, since it's a curve. Simple curve though, and you know the beginning and end values, so figuring them out should only take a simple equation.
I think I get what you're trying to do: generate a grid of depth data for an image. Sort of like ray-tracing a sphere.
In that case, you want a Ray-Sphere Intersection test:
http://www.siggraph.org/education/materials/HyperGraph/raytrace/rtinter1.htm
Your rays will be simple perpendicular rays, based off your U/V coordinates (times two, since your sphere has a diameter of 2). This will give you the front-facing points on the sphere.
From there, calculate normals as below (point - origin, the radius is already 1 unit).
Ripped off from the link above:
You have to combine two equations:
Ray: R(t) = R0 + t * Rd , t > 0 with R0 = [X0, Y0, Z0] and Rd = [Xd, Yd, Zd]
Sphere: S = the set of points[xs, ys, zs], where (xs - xc)2 + (ys - yc)2 + (zs - zc)2 = Sr2
To do this, calculate your ray (x * pixel / width, y * pixel / width, z: 1), then:
A = Xd^2 + Yd^2 + Zd^2
B = 2 * (Xd * (X0 - Xc) + Yd * (Y0 - Yc) + Zd * (Z0 - Zc))
C = (X0 - Xc)^2 + (Y0 - Yc)^2 + (Z0 - Zc)^2 - Sr^2
Plug into quadratic equation:
t0, t1 = (- B + (B^2 - 4*C)^1/2) / 2
Check discriminant (B^2 - 4*C), and if real root, the intersection is:
Ri = [xi, yi, zi] = [x0 + xd * ti , y0 + yd * ti, z0 + zd * ti]
And the surface normal is:
SN = [(xi - xc)/Sr, (yi - yc)/Sr, (zi - zc)/Sr]
Boiling it all down:
So, since we're talking unit values, and rays that point straight at Z (no x or y component), we can boil down these equations greatly:
Ray:
X0 = 2 * pixelX / width
Y0 = 2 * pixelY / height
Z0 = 0
Xd = 0
Yd = 0
Zd = 1
Sphere:
Xc = 1
Yc = 1
Zc = 1
Factors:
A = 1 (unit ray)
B
= 2 * (0 + 0 + (0 - 1))
= -2 (no x/y component)
C
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2 + (0 - 1) ^ 2 - 1
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2
Discriminant
= (-2) ^ 2 - 4 * 1 * C
= 4 - 4 * C
From here:
If discriminant < 0:
Z = ?, Normal = ?
Else:
t = (2 + (discriminant) ^ 1 / 2) / 2
If t < 0 (hopefully never or always the case)
t = -t
Then:
Z: t
Nx: Xi - 1
Ny: Yi - 1
Nz: t - 1
Boiled farther still:
Intuitively it looks like C (X^2 + Y^2) and the square-root are the most prominent figures here. If I had a better recollection of my math (in particular, transformations on exponents of sums), then I'd bet I could derive this down to what Tom Zych gave you. Since I can't, I'll just leave it as above.
I have a set of points. I want to separate them into 2 distinct sets. To do this, I choose two points (a and b) and draw an imaginary line between them. Now I want to have all points that are left from this line in one set and those that are right from this line in the other set.
How can I tell for any given point z whether it is in the left or in the right set? I tried to calculate the angle between a-z-b – angles smaller than 180 are on the right hand side, greater than 180 on the left hand side – but because of the definition of ArcCos, the calculated angles are always smaller than 180°. Is there a formula to calculate angles greater than 180° (or any other formula to chose right or left side)?
Try this code which makes use of a cross product:
public bool isLeft(Point a, Point b, Point c){
return ((b.X - a.X)*(c.Y - a.Y) - (b.Y - a.Y)*(c.X - a.X)) > 0;
}
Where a = line point 1; b = line point 2; c = point to check against.
If the formula is equal to 0, the points are colinear.
If the line is horizontal, then this returns true if the point is above the line.
Use the sign of the determinant of vectors (AB,AM), where M(X,Y) is the query point:
position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
It is 0 on the line, and +1 on one side, -1 on the other side.
You look at the sign of the determinant of
| x2-x1 x3-x1 |
| y2-y1 y3-y1 |
It will be positive for points on one side, and negative on the other (and zero for points on the line itself).
The vector (y1 - y2, x2 - x1) is perpendicular to the line, and always pointing right (or always pointing left, if you plane orientation is different from mine).
You can then compute the dot product of that vector and (x3 - x1, y3 - y1) to determine if the point lies on the same side of the line as the perpendicular vector (dot product > 0) or not.
Using the equation of the line ab, get the x-coordinate on the line at the same y-coordinate as the point to be sorted.
If point's x > line's x, the point is to the right of the line.
If point's
x < line's x, the point is to the left of the line.
If point's x == line's x, the point is on the line.
I implemented this in java and ran a unit test (source below). None of the above solutions work. This code passes the unit test. If anyone finds a unit test that does not pass, please let me know.
Code: NOTE: nearlyEqual(double,double) returns true if the two numbers are very close.
/*
* #return integer code for which side of the line ab c is on. 1 means
* left turn, -1 means right turn. Returns
* 0 if all three are on a line
*/
public static int findSide(
double ax, double ay,
double bx, double by,
double cx, double cy) {
if (nearlyEqual(bx-ax,0)) { // vertical line
if (cx < bx) {
return by > ay ? 1 : -1;
}
if (cx > bx) {
return by > ay ? -1 : 1;
}
return 0;
}
if (nearlyEqual(by-ay,0)) { // horizontal line
if (cy < by) {
return bx > ax ? -1 : 1;
}
if (cy > by) {
return bx > ax ? 1 : -1;
}
return 0;
}
double slope = (by - ay) / (bx - ax);
double yIntercept = ay - ax * slope;
double cSolution = (slope*cx) + yIntercept;
if (slope != 0) {
if (cy > cSolution) {
return bx > ax ? 1 : -1;
}
if (cy < cSolution) {
return bx > ax ? -1 : 1;
}
return 0;
}
return 0;
}
Here's the unit test:
#Test public void testFindSide() {
assertTrue("1", 1 == Utility.findSide(1, 0, 0, 0, -1, -1));
assertTrue("1.1", 1 == Utility.findSide(25, 0, 0, 0, -1, -14));
assertTrue("1.2", 1 == Utility.findSide(25, 20, 0, 20, -1, 6));
assertTrue("1.3", 1 == Utility.findSide(24, 20, -1, 20, -2, 6));
assertTrue("-1", -1 == Utility.findSide(1, 0, 0, 0, 1, 1));
assertTrue("-1.1", -1 == Utility.findSide(12, 0, 0, 0, 2, 1));
assertTrue("-1.2", -1 == Utility.findSide(-25, 0, 0, 0, -1, -14));
assertTrue("-1.3", -1 == Utility.findSide(1, 0.5, 0, 0, 1, 1));
assertTrue("2.1", -1 == Utility.findSide(0,5, 1,10, 10,20));
assertTrue("2.2", 1 == Utility.findSide(0,9.1, 1,10, 10,20));
assertTrue("2.3", -1 == Utility.findSide(0,5, 1,10, 20,10));
assertTrue("2.4", -1 == Utility.findSide(0,9.1, 1,10, 20,10));
assertTrue("vertical 1", 1 == Utility.findSide(1,1, 1,10, 0,0));
assertTrue("vertical 2", -1 == Utility.findSide(1,10, 1,1, 0,0));
assertTrue("vertical 3", -1 == Utility.findSide(1,1, 1,10, 5,0));
assertTrue("vertical 3", 1 == Utility.findSide(1,10, 1,1, 5,0));
assertTrue("horizontal 1", 1 == Utility.findSide(1,-1, 10,-1, 0,0));
assertTrue("horizontal 2", -1 == Utility.findSide(10,-1, 1,-1, 0,0));
assertTrue("horizontal 3", -1 == Utility.findSide(1,-1, 10,-1, 0,-9));
assertTrue("horizontal 4", 1 == Utility.findSide(10,-1, 1,-1, 0,-9));
assertTrue("positive slope 1", 1 == Utility.findSide(0,0, 10,10, 1,2));
assertTrue("positive slope 2", -1 == Utility.findSide(10,10, 0,0, 1,2));
assertTrue("positive slope 3", -1 == Utility.findSide(0,0, 10,10, 1,0));
assertTrue("positive slope 4", 1 == Utility.findSide(10,10, 0,0, 1,0));
assertTrue("negative slope 1", -1 == Utility.findSide(0,0, -10,10, 1,2));
assertTrue("negative slope 2", -1 == Utility.findSide(0,0, -10,10, 1,2));
assertTrue("negative slope 3", 1 == Utility.findSide(0,0, -10,10, -1,-2));
assertTrue("negative slope 4", -1 == Utility.findSide(-10,10, 0,0, -1,-2));
assertTrue("0", 0 == Utility.findSide(1, 0, 0, 0, -1, 0));
assertTrue("1", 0 == Utility.findSide(0,0, 0, 0, 0, 0));
assertTrue("2", 0 == Utility.findSide(0,0, 0,1, 0,2));
assertTrue("3", 0 == Utility.findSide(0,0, 2,0, 1,0));
assertTrue("4", 0 == Utility.findSide(1, -2, 0, 0, -1, 2));
}
First check if you have a vertical line:
if (x2-x1) == 0
if x3 < x2
it's on the left
if x3 > x2
it's on the right
else
it's on the line
Then, calculate the slope: m = (y2-y1)/(x2-x1)
Then, create an equation of the line using point slope form: y - y1 = m*(x-x1) + y1. For the sake of my explanation, simplify it to slope-intercept form (not necessary in your algorithm): y = mx+b.
Now plug in (x3, y3) for x and y. Here is some pseudocode detailing what should happen:
if m > 0
if y3 > m*x3 + b
it's on the left
else if y3 < m*x3 + b
it's on the right
else
it's on the line
else if m < 0
if y3 < m*x3 + b
it's on the left
if y3 > m*x3+b
it's on the right
else
it's on the line
else
horizontal line; up to you what you do
I wanted to provide with a solution inspired by physics.
Imagine a force applied along the line and you are measuring the torque of the force about the point. If the torque is positive (counterclockwise) then the point is to the "left" of the line, but if the torque is negative the point is the "right" of the line.
So if the force vector equals the span of the two points defining the line
fx = x_2 - x_1
fy = y_2 - y_1
you test for the side of a point (px,py) based on the sign of the following test
var torque = fx*(py-y_1)-fy*(px-x_1)
if torque>0 then
"point on left side"
else if torque <0 then
"point on right side"
else
"point on line"
end if
Assuming the points are (Ax,Ay) (Bx,By) and (Cx,Cy), you need to compute:
(Bx - Ax) * (Cy - Ay) - (By - Ay) * (Cx - Ax)
This will equal zero if the point C is on the line formed by points A and B, and will have a different sign depending on the side. Which side this is depends on the orientation of your (x,y) coordinates, but you can plug test values for A,B and C into this formula to determine whether negative values are to the left or to the right.
basically, I think that there is a solution which is much easier and straight forward, for any given polygon, lets say consist of four vertices(p1,p2,p3,p4), find the two extreme opposite vertices in the polygon, in another words, find the for example the most top left vertex (lets say p1) and the opposite vertex which is located at most bottom right (lets say ). Hence, given your testing point C(x,y), now you have to make double check between C and p1 and C and p4:
if cx > p1x AND cy > p1y ==> means that C is lower and to right of p1
next
if cx < p2x AND cy < p2y ==> means that C is upper and to left of p4
conclusion, C is inside the rectangle.
Thanks :)
#AVB's answer in ruby
det = Matrix[
[(x2 - x1), (x3 - x1)],
[(y2 - y1), (y3 - y1)]
].determinant
If det is positive its above, if negative its below. If 0, its on the line.
Here's a version, again using the cross product logic, written in Clojure.
(defn is-left? [line point]
(let [[[x1 y1] [x2 y2]] (sort line)
[x-pt y-pt] point]
(> (* (- x2 x1) (- y-pt y1)) (* (- y2 y1) (- x-pt x1)))))
Example usage:
(is-left? [[-3 -1] [3 1]] [0 10])
true
Which is to say that the point (0, 10) is to the left of the line determined by (-3, -1) and (3, 1).
NOTE: This implementation solves a problem that none of the others (so far) does! Order matters when giving the points that determine the line. I.e., it's a "directed line", in a certain sense. So with the above code, this invocation also produces the result of true:
(is-left? [[3 1] [-3 -1]] [0 10])
true
That's because of this snippet of code:
(sort line)
Finally, as with the other cross product based solutions, this solution returns a boolean, and does not give a third result for collinearity. But it will give a result that makes sense, e.g.:
(is-left? [[1 1] [3 1]] [10 1])
false
Issues with the existing solution:
While I found Eric Bainville's answer to be correct, I found it entirely inadequate to comprehend:
How can two vectors have a determinant? I thought that applied to matrices?
What is sign?
How do I convert two vectors into a matrix?
position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
What is Bx?
What is Y? Isn't Y meant to be a Vector, rather than a scalar?
Why is the solution correct - what is the reasoning behind it?
Moreover, my use case involved complex curves rather than a simple line, hence it requires a little re-jigging:
Reconstituted Answer
Point a = new Point3d(ax, ay, az); // point on line
Point b = new Point3d(bx, by, bz); // point on line
If you want to see whether your points are above/below a curve, then you would need to get the first derivative of the particular curve you are interested in - also known as the tangent to the point on the curve. If you can do so, then you can highlight your points of interest. Of course, if your curve is a line, then you just need the point of interest without the tangent. The tangent IS the line.
Vector3d lineVector = curve.GetFirstDerivative(a); // where "a" is a point on the curve. You may derive point b with a simple displacement calculation:
Point3d b = new Point3d(a.X, a.Y, a.Z).TransformBy(
Matrix3d.Displacement(curve.GetFirstDerivative(a))
);
Point m = new Point3d(mx, my, mz) // the point you are interested in.
The Solution:
return (b.X - a.X) * (m.Y - a.Y) - (b.Y - a.Y) * (m.X - a.X) < 0; // the answer
Works for me! See the proof in the photo above. Green bricks satisfy the condition, but the bricks outside were filtered out! In my use case - I only want the bricks that are touching the circle.
Theory behind the answer
I will return to explain this. Someday. Somehow...
An alternative way of getting a feel of solutions provided by netters is to understand a little geometry implications.
Let pqr=[P,Q,R] are points that forms a plane that is divided into 2 sides by line [P,R]. We are to find out if two points on pqr plane, A,B, are on the same side.
Any point T on pqr plane can be represented with 2 vectors: v = P-Q and u = R-Q, as:
T' = T-Q = i * v + j * u
Now the geometry implications:
i+j =1: T on pr line
i+j <1: T on Sq
i+j >1: T on Snq
i+j =0: T = Q
i+j <0: T on Sq and beyond Q.
i+j: <0 0 <1 =1 >1
---------Q------[PR]--------- <== this is PQR plane
^
pr line
In general,
i+j is a measure of how far T is away from Q or line [P,R], and
the sign of i+j-1 implicates T's sideness.
The other geometry significances of i and j (not related to this solution) are:
i,j are the scalars for T in a new coordinate system where v,u are the new axes and Q is the new origin;
i, j can be seen as pulling force for P,R, respectively. The larger i, the farther T is away from R (larger pull from P).
The value of i,j can be obtained by solving the equations:
i*vx + j*ux = T'x
i*vy + j*uy = T'y
i*vz + j*uz = T'z
So we are given 2 points, A,B on the plane:
A = a1 * v + a2 * u
B = b1 * v + b2 * u
If A,B are on the same side, this will be true:
sign(a1+a2-1) = sign(b1+b2-1)
Note that this applies also to the question: Are A,B in the same side of plane [P,Q,R], in which:
T = i * P + j * Q + k * R
and i+j+k=1 implies that T is on the plane [P,Q,R] and the sign of i+j+k-1 implies its sideness. From this we have:
A = a1 * P + a2 * Q + a3 * R
B = b1 * P + b2 * Q + b3 * R
and A,B are on the same side of plane [P,Q,R] if
sign(a1+a2+a3-1) = sign(b1+b2+b3-1)
equation of line is y-y1 = m(x-x1)
here m is y2-y1 / x2-x1
now put m in equation and put condition on y < m(x-x1) + y1 then it is left side point
eg.
for i in rows:
for j in cols:
if j>m(i-a)+b:
image[i][j]=0
A(x1,y1) B(x2,y2) a line segment with length L=sqrt( (y2-y1)^2 + (x2-x1)^2 )
and a point M(x,y)
making a transformation of coordinates in order to be the point A the new start and B a point of the new X axis
we have the new coordinates of the point M
which are
newX = ((x-x1)(x2-x1)+(y-y1)(y2-y1)) / L
from (x-x1)*cos(t)+(y-y1)*sin(t) where cos(t)=(x2-x1)/L, sin(t)=(y2-y1)/L
newY = ((y-y1)(x2-x1)-(x-x1)(y2-y1)) / L
from (y-y1)*cos(t)-(x-x1)*sin(t)
because "left" is the side of axis X where the Y is positive, if the newY (which is the distance of M from AB) is positive, then it is on the left side of AB (the new X axis)
You may omit the division by L (allways positive), if you only want the sign
How can I draw a perpendicular on a line segment from a given point? My line segment is defined as (x1, y1), (x2, y2), If I draw a perpendicular from a point (x3,y3) and it meets to line on point (x4,y4). I want to find out this (x4,y4).
I solved the equations for you:
k = ((y2-y1) * (x3-x1) - (x2-x1) * (y3-y1)) / ((y2-y1)^2 + (x2-x1)^2)
x4 = x3 - k * (y2-y1)
y4 = y3 + k * (x2-x1)
Where ^2 means squared
From wiki:
In algebra, for any linear equation
y=mx + b, the perpendiculars will all
have a slope of (-1/m), the opposite
reciprocal of the original slope. It
is helpful to memorize the slogan "to
find the slope of the perpendicular
line, flip the fraction and change the
sign." Recall that any whole number a
is itself over one, and can be written
as (a/1)
To find the perpendicular of a given
line which also passes through a
particular point (x, y), solve the
equation y = (-1/m)x + b, substituting
in the known values of m, x, and y to
solve for b.
The slope of the line, m, through (x1, y1) and (x2, y2) is m = (y1 - y2) / (x1 - x2)
I agree with peter.murray.rust, vectors make the solution clearer:
// first convert line to normalized unit vector
double dx = x2 - x1;
double dy = y2 - y1;
double mag = sqrt(dx*dx + dy*dy);
dx /= mag;
dy /= mag;
// translate the point and get the dot product
double lambda = (dx * (x3 - x1)) + (dy * (y3 - y1));
x4 = (dx * lambda) + x1;
y4 = (dy * lambda) + y1;
You know both the point and the slope, so the equation for the new line is:
y-y3=m*(x-x3)
Since the line is perpendicular, the slope is the negative reciprocal. You now have two equations and can solve for their intersection.
y-y3=-(1/m)*(x-x3)
y-y1=m*(x-x1)
You will often find that using vectors makes the solution clearer...
Here is a routine from my own library:
public class Line2 {
Real2 from;
Real2 to;
Vector2 vector;
Vector2 unitVector = null;
public Real2 getNearestPointOnLine(Real2 point) {
unitVector = to.subtract(from).getUnitVector();
Vector2 lp = new Vector2(point.subtract(this.from));
double lambda = unitVector.dotProduct(lp);
Real2 vv = unitVector.multiplyBy(lambda);
return from.plus(vv);
}
}
You will have to implement Real2 (a point) and Vector2 and dotProduct() but these should be simple:
The code then looks something like:
Point2 p1 = new Point2(x1, y1);
Point2 p2 = new Point2(x2, y2);
Point2 p3 = new Point2(x3, y3);
Line2 line = new Line2(p1, p2);
Point2 p4 = getNearestPointOnLine(p3);
The library (org.xmlcml.euclid) is at:
http://sourceforge.net/projects/cml/
and there are unit tests which will exercise this method and show you how to use it.
#Test
public final void testGetNearestPointOnLine() {
Real2 p = l1112.getNearestPointOnLine(new Real2(0., 0.));
Real2Test.assertEquals("point", new Real2(0.4, -0.2), p, 0.0000001);
}
Compute the slope of the line joining points (x1,y1) and (x2,y2) as m=(y2-y1)/(x2-x1)
Equation of the line joining (x1,y1) and (x2,y2) using point-slope form of line equation, would be y-y2 = m(x-x2)
Slope of the line joining (x3,y3) and (x4,y4) would be -(1/m)
Again, equation of the line joining (x3,y3) and (x4,y4) using point-slope form of line equation, would be y-y3 = -(1/m)(x-x3)
Solve these two line equations as you solve a linear equation in two variables and the values of x and y you get would be your (x4,y4)
I hope this helps.
cheers
Find out the slopes for both the
lines, say slopes are m1 and m2 then
m1*m2=-1 is the condition for
perpendicularity.
Matlab function code for the following problem
function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);
px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;
u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;
Pr=[x,y];
end
Mathematica introduced the function RegionNearest[] in version 10, 2014. This function could be used to return an answer to this question:
{x4,y4} = RegionNearest[Line[{{x1,y1},{x2,y2}}],{x3,y3}]
This is mostly a duplicate of Arnkrishn's answer. I just wanted to complete his section with a complete Mathematica code snippet:
m = (y2 - y1)/(x2 - x1)
eqn1 = y - y3 == -(1/m)*(x - x3)
eqn2 = y - y1 == m*(x - x1)
Solve[eqn1 && eqn2, {x, y}]
This is a C# implementation of the accepted answer. It's also using ArcGis to return a MapPoint as that's what we're using for this project.
private MapPoint GenerateLinePoint(double startPointX, double startPointY, double endPointX, double endPointY, double pointX, double pointY)
{
double k = ((endPointY - startPointY) * (pointX - startPointX) - (endPointX - startPointX) * (pointY - startPointY)) / (Math.Pow(endPointY - startPointY, 2)
+ Math.Pow(endPointX - startPointX, 2));
double resultX = pointX - k * (endPointY - startPointY);
double resultY = pointY + k * (endPointX - startPointX);
return new MapPoint(resultX, resultY, 0, SpatialReferences.Wgs84);
}
Thanks to Ray as this worked perfectly for me.
c#arcgis
Just for the sake of completeness, here is a solution using homogeneous coordinates.
The homogeneous points are:
p1 = (x1,y1,1), p2 = (x2,y2,1), p3 = (x3,y3,1)
a line through two points is their cross-product
l_12 := p1 x p2 = (y1-y2, x2-x1, x1*y2 - x2*y1)
The (signed) distance of a point to a line is their dot product.
d := l_12 * p3 = x3*(y1-y2) + y3*(x2-x1) + x1*y2 - x2*y1
The vector from p4 to p3 is d times the normal vector of l_12 divided by the squared length of the normal vector.
n2 := (y1-y2)^2 + (x2-x1)^2
p4 := p3 + d/n2*(y1-y2, x2-x1, 0)
Note: if you divide l_12 by the length of the normal vector
l_12 := l_12 / sqrt((y1-y2)^2 + (x2-x1)^2)
the distance d will be the euclidean distance.
First, calculate the linear function determined by the points
(x1,y2),(x2,y2).
We get:
y1 = mx+b1 where m and b1 are constants.
This step is easy to calculate by the formula of linear function between two points.
Then, calculate the linear function y that goes through (x3,y3).
The function slope is -m, where m is the slope of y1.
Then calculate the const b2 by the coordinates of the point (x3,y3).
We get y2 = -mx+b2 where m and b2 are constants.
The last thing to do is to find the intersection of y1, y2.
You can find x by solving the equation: -mx+b2 = mx+b1, then place x in one of the equations to find y.
This is a vectorized Matlab function for finding pairwise projections of m points onto n line segments. Here xp and yp are m by 1 vectors holding coordinates of m different points, and x1, y1, x2 and y2 are n by 1 vectors holding coordinates of start and end points of n different line segments.
It returns m by n matrices, x and y, where x(i, j) and y(i, j) are coordinates of projection of i-th point onto j-th line.
The actual work is done in first few lines and the rest of the function runs a self-test demo, just in case where it is called with no parameters. It's relatively fast, I managed to find projections of 2k points onto 2k line segments in less than 0.05s.
function [x, y] = projectPointLine(xp, yp, x1, y1, x2, y2)
if nargin > 0
xd = (x2-x1)';
yd = (y2-y1)';
dAB = xd.*xd + yd.*yd;
u = bsxfun(#rdivide, bsxfun(#times, bsxfun(#minus, xp, x1'), xd) + ...
bsxfun(#times, bsxfun(#minus, yp, y1'), yd), dAB);
x = bsxfun(#plus, x1', bsxfun(#times, u, xd));
y = bsxfun(#plus, y1', bsxfun(#times, u, yd));
else
nLine = 3;
nPoint = 2;
xp = rand(nPoint, 1) * 2 -1;
yp = rand(nPoint, 1) * 2 -1;
x1 = rand(nLine, 1) * 2 -1;
y1 = rand(nLine, 1) * 2 -1;
x2 = rand(nLine, 1) * 2 -1;
y2 = rand(nLine, 1) * 2 -1;
tic;
[x, y] = projectPointLine(xp, yp, x1, y1, x2, y2);
toc
close all;
plot([x1'; x2'], [y1'; y2'], '.-', 'linewidth', 2, 'markersize', 20);
axis equal;
hold on
C = lines(nPoint + nLine);
for i=1:nPoint
scatter(x(i, :), y(i, :), 100, C(i+nLine, :), 'x', 'linewidth', 2);
scatter(xp(i), yp(i), 100, C(i+nLine, :), 'x', 'linewidth', 2);
end
for i=1:nLine
scatter(x(:, i)', y(:, i)', 100, C(i, :), 'o', 'linewidth', 2);
end
end
end