I have an array of points, as well as two more points (A and B). The last two points form a line, and I'm trying to find which of the points in my array are furthest from the line. How would I do this in Java?
I'm wondering if it's something along the lines of finding the distance from A and B, but that doesn't sit well in my head.
Additional info:
I think it's a line segment. Given this is QuickHull, I don't know if it makes a difference.
I've never been the greatest when it comes to math and formulas, so more explanation is better. Thanks!
Note that each 3 points [a,b,p] for each p in the array form a trianle, whose area is denoted by: (ab) * h /2 [where h is the distance from p to ab]
You can compute the area these trianles create, and select the minimal. Since ab is constant for all - it guarantees that the trianle with the minimal area will also have the minimal h.
You can find it [the area of each triangle] using
T=(1/2)* abs((x_a - x_p) * (y_b-y_a) - (x_a - x_b)* (y_p - y_a))
[where x_a,x_b,x_p and y_a,y_b,y_p are the x,y coordinates of a,b,p respectively].
Though I find this method very elegant, I believe there are better
ways to do it.
ArrayList<Point> points=new ArrayList();//YOUR POINTS
Point a=new Point(1,1);
Point b=new Point(1,1);
Point ABcenter=new Point((a.x+b.x)/2,(a.y+b.y)/2);//THE CENTER OF THE A AND B POINTS ,USE A OR B IF YOU WANT
int furthestid=0;
float furthestdis=0;
for(int c=0;c<points.size();c++)
{
if(calculate_distance(ABcenter.x,ABcenter.y,points.get(c).x,points.get(c).y)>furthestdis)
{
furthestid=c;
}
}
//closestid now contains the id of the furthest point ,use it like this points.get(c).x ...
public static double calculate_distance (float x1,float y1,float x2 ,float y2){
return Math.sqrt((((x1-x2) * (x1-x2)) + ((y1- y2) * (y1- y2))));
}
I'm assuming you mean the Euclidean distance. If you're working in the plane, then the answer is simple.
First, compute the equation of the line in the form
ax + by + c = 0
In slope-intercept form, this is the same as
y = (-a/b)x + (-c/b)
Now compute the distance from any point (p,q) to the line by
|a*p + b*q + c| / (a^2 + b^2)^(1/2)
For more than 2 dimensions, it's probably easiest to think in terms of parametrized vectors. This means think of points on the line as
p(t) = (A1 + (B1-A1)*t, A2 + (B2-A2)*t, ..., An + (Bn-An)*t)
where the two points are A = (A1,...,An) and B = (B1,...,Bn). Let X = (X1,...,Xn) be any other point. Then the distance between X and p(t), the point on the line corresponding to t, is the square root of
[(A1-X1) + (B1-A1)t]^2 + ... + [(An-Xn) + (Bn-An)t]^2
The distance to the line is the distance to p(t) where t is the unique value minimizing this distance. To compute that, just take the derivative with respect to t and set it to 0. It's a very straightforward problem from here, so I'll leave that bit to you.
If you want a further hint, then check out this link for the 3-dimensional case which reduces nicely.
If the problem is as you have stated it, you can't do much better then computing the distance for each point and choosing the smallest of these.
However you can simplify the calculation of the distance a bit by using a generalized line equation for the line passing through A and B. This would be an equation of the form ax + by + c = 0
You can compute such equation for a line passing through two arbitrary points quite easily:
x * (A.y - B.y) + y * (B.x - A.x) + A.x * B.y - A.y * B.x,
i.e. a = A.y - B.y, b = B.x - A.x and c = A.x * B.y - A.y * B.x
Now that you have computed such equation for the line, you can compute the distance from an arbitrary point P in the plane to the line a * x + b * y + c by substituting x and y with the coordinates of P:
abs(a * P.x + b * P.y + c) / sqrt(a * a + b * b). But as the denominator will be the same for all points, you may igonore it and simply choose the point for which abs(a * P.x + b * P.y + c) is smallest
Here is a link that explains how to compute 2-d distance to a line once you have it's generalized equation.
I'm assuming you talking about line segment not line. First you should find your points distance from your line segment, and you can do it, as the way suggested in this similar question, after that find minimum/maximum distance over all inputs.
Edit: Also from this top coder article you can find distance simply:
//Compute the dot product AB ⋅ BC
int dot(int[] A, int[] B, int[] C){
AB = new int[2];
BC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
BC[0] = C[0]-B[0];
BC[1] = C[1]-B[1];
int dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
int cross(int[] A, int[] B, int[] C){
AB = new int[2];
AC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
AC[0] = C[0]-A[0];
AC[1] = C[1]-A[1];
int cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double distance(int[] A, int[] B){
int d1 = A[0] - B[0];
int d2 = A[1] - B[1];
return sqrt(d1*d1+d2*d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double linePointDist(int[] A, int[] B, int[] C, boolean isSegment){
double dist = cross(A,B,C) / distance(A,B);
if(isSegment){
int dot1 = dot(A,B,C);
if(dot1 > 0)return distance(B,C);
int dot2 = dot(B,A,C);
if(dot2 > 0)return distance(A,C);
}
return abs(dist);
}
I think code has self explanation, if you are familiar with basic geometry, but if you aren't familiar, you should to read the article, if there is any problem for you we can help you.
Related
This question is a bit hard to formulate so I'll start by showing this image:
I want to test if points (such as p1 and p2 on the image) are within the dotted lines that are perpendicular to the line at its limits. I know the coordinates of the points to test and the line's limits.
So for p1 it would be false, and for p2 it would be true.
What would be the most computationaly efficient way to calculate this?
I'm working with floats in Java.
This can be very efficiently done with the dot product:
This is positive if A has a component parallel to B, and negative if anti-parallel.
Therefore if you have a line segment defined by points A and B, and a test point P, you only need two dot product operations to test:
dot(A - B, P - B) >= 0 && dot(B - A, P - A) >= 0
EDIT: a graphical explanation:
The dot product can be shown to be:
Thus if θ > 90 then dot(A, B) < 0, and vice versa. Now for your problem:
In case 1, when dot(A - B, P - B) > 0 we say that P is on the correct side of the dotted line at B, and vice versa in case 2. By symmetry we can then do the same operation at A, by swapping A and B.
If the point to test (tp), together with start point (sp) and end point (ep) of the range, form an obtuse triangle and you can conclude that it is out of range.
https://en.wikipedia.org/wiki/Acute_and_obtuse_triangles
A special case would be tp is same slope as sp and ep, then you calculate the 2 distances:
distSpAndTp - distance between sp and tp
distEpAndTp - distance between ep and tp
If the sum of these 2 distances = distance between sp and ep, then tp is in range, otherwise it is out of range.
The best approach would be to first create a rectangle Rect (of android.graphics package), with x and width being the beginning and end of line, and y and height being the beginning and end of screen height.
Then use the Rect method Rect.contains(int x, int y) to check if the point coordinates lie within.
For floats, use RectF class instead.
This approach involves using 2D vectors (which you should have been using anyway, makes working in any coordinate system a lot easier) and the dot (scalar) product. It does not require any relatively expensive operations like the square root or trignometic functions, and therefore is very performant.
Let A and B be the start and end points (by point I mean a vector representing a position in 2D space) of the line segment, in any order. If P is the point to test, then:
If dot(PA, AB) and dot(PB, AB) have the same sign, the point lies outside the region (like p1 in your example).
If the dot products above have opposite signs, the point lies inside the region (like p2).
Where PA = A - P, PB = B - P and AB = B - A.
This condition can be surmised as follows:
if dot(PA, AB) * dot(PB, AB) <= 0 {
// Opposite signs, inside region
// If the product is equal to zero, the point is on one of the dotted lines
}
else {
// Same signs, outside region
}
Let's say, x1 is the beginning of the line and x2 is the end. Then ax is the dot in an horizontal plan and; y1, y2 and ay in vertical plan.
if((ax >= x1 && ax <= x2) && (ay >= y1 && ay <= y2)){ // do your work
}
There are a few answers already but there is another solution that will give you the unit distance on the line segment that the points is perpendicular to.
Where then line is x1,y1 to x2,y2 and the point is px,py
Find the vector from line start to end
vx = x2 - x1;
vy = y2 - y1;
And the vector from the line start (or end) to the point
vpx = px - x1;
vpy = py - y1;
And then divide the dot product of the two vectors by the length squared of the line.
unitDist = (vx * vpx + vy * vpy) / (vx * vx + vy * vy);
If the perpendicular intercept is on the line segment then unitDist will be 0 <= unitDist <= 1
In shortened form
x2 -= x1;
y2 -= y1;
unitDist = (x2 * (px - x1) + y2 * (py - y1)) / (x2^2 + y2^2);
if (unitDist >= 0 && unitDist <= 1) {
// point px,py perpendicular to line segment x1,y1,x2,y2
}
You also get the benefit of being able to get the point on the line where the intercept is.
p2x = vx * unitDist + x1;
p2y = vy * unitDist + y1;
And thus the distance that the point is from the line (note not line segment but line)
dist = hypot(p2x - px, p2y - py);
Which is handy if you are using a point to select from many lines by using the minimum distance to determine the selected line.
This can be solved nicely using complex numbers.
Let a and b the endpoints of the segment. We use the transformation that maps the origin 0 to a and the point 1 to b. This transformation is a similarity and its expresion is simply
p = (b - a) q + a
as you can verify by substituting q=0 or q=1. It maps the whole stripe perpendicular to the given segment to the stripe perpendicular to the segment 0-1.
Now the inverse transform is obviously
q = (p - a) / (b - a),
and the desired condition is
0 <= Re((p - a) / (b - a)) <= 1.
To avoid the division, you can rewrite
0 <= Re((p - a) (b - a)*) <= |b-a|²
or
0 <= (px - ax)(bx - ax) + (py - ay)(by - ay) <= (bx - ax)² + (by - ay)².
I'm programming a software renderer in Java, and am trying to use Z-buffering for the depth calculation of each pixel. However, it appears to work inconsistently. For example, with the Utah teapot example model, the handle will draw perhaps half depending on how I rotate it.
My z-buffer algorithm:
for(int i = 0; i < m_triangles.size(); i++)
{
if(triangleIsBackfacing(m_triangles.get(i))) continue; //Backface culling
for(int y = minY(m_triangles.get(i)); y < maxY(m_triangles.get(i)); y++)
{
if((y + getHeight()/2 < 0) || (y + getHeight()/2 >= getHeight())) continue; //getHeight/2 and getWidth/2 is for moving the model to the centre of the screen
for(int x = minX(m_triangles.get(i)); x < maxX(m_triangles.get(i)); x++)
{
if((x + getWidth()/2 < 0) || (x + getWidth()/2 >= getWidth())) continue;
rayOrigin = new Point2D(x, y);
if(pointWithinTriangle(m_triangles.get(i), rayOrigin))
{
zDepth = zValueOfPoint(m_triangles.get(i), rayOrigin);
if(zDepth > zbuffer[x + getWidth()/2][y + getHeight()/2])
{
zbuffer[x + getWidth()/2][y + getHeight()/2] = zDepth;
colour[x + getWidth()/2][y + getHeight()/2] = m_triangles.get(i).getColour();
g2.setColor(m_triangles.get(i).getColour());
drawDot(g2, rayOrigin);
}
}
}
}
}
Method for calculating the z value of a point, given a triangle and the ray origin:
private double zValueOfPoint(Triangle triangle, Point2D rayOrigin)
{
Vector3D surfaceNormal = getNormal(triangle);
double A = surfaceNormal.x;
double B = surfaceNormal.y;
double C = surfaceNormal.z;
double d = -(A * triangle.getV1().x + B * triangle.getV1().y + C * triangle.getV1().z);
double rayZ = -(A * rayOrigin.x + B * rayOrigin.y + d) / C;
return rayZ;
}
Method for calculating if the ray origin is within a projected triangle:
private boolean pointWithinTriangle(Triangle triangle, Point2D rayOrigin)
{
Vector2D v0 = new Vector2D(triangle.getV3().projectPoint(modelViewer), triangle.getV1().projectPoint(modelViewer));
Vector2D v1 = new Vector2D(triangle.getV2().projectPoint(modelViewer), triangle.getV1().projectPoint(modelViewer));
Vector2D v2 = new Vector2D(rayOrigin, triangle.getV1().projectPoint(modelViewer));
double d00 = v0.dotProduct(v0);
double d01 = v0.dotProduct(v1);
double d02 = v0.dotProduct(v2);
double d11 = v1.dotProduct(v1);
double d12 = v1.dotProduct(v2);
double invDenom = 1.0 / (d00 * d11 - d01 * d01);
double u = (d11 * d02 - d01 * d12) * invDenom;
double v = (d00 * d12 - d01 * d02) * invDenom;
// Check if point is in triangle
if((u >= 0) && (v >= 0) && ((u + v) <= 1))
{
return true;
}
return false;
}
Method for calculating surface normal of a triangle:
private Vector3D getNormal(Triangle triangle)
{
Vector3D v1 = new Vector3D(triangle.getV1(), triangle.getV2());
Vector3D v2 = new Vector3D(triangle.getV3(), triangle.getV2());
return v1.crossProduct(v2);
}
Example of the incorrectly drawn teapot:
What am I doing wrong? I feel like it must be some small thing. Given that the triangles draw at all, I doubt it's the pointWithinTriangle method. Backface culling also appears to work correctly, so I doubt it's that. The most likely culprit to me is the zValueOfPoint method, but I don't know enough to know what's wrong with it.
My zValueOfPoint method was not working correctly. I'm unsure why :( however, I changed to a slightly different method of calculating the value of a point in a plane, found here: http://forum.devmaster.net/t/interpolation-on-a-3d-triangle-using-normals/20610/5
To make the answer here complete, we have the equation of a plane:
A * x + B * y + C * z + D = 0
Where A, B, and C are the surface normal x/y/z values, and D is -(Ax0 + By0 + Cz0).
x0, y0, and z0 are taken from one of the vertices of the triangle. x, y, and z are the coordinates of the point where the ray intersects the plane. x and y are known values (rayOrigin.x, rayOrigin.y) but z is the depth which we need to calculate. From the above equation we derive:
z = -A / C * x - B / C * y - D
Then, copied from the above link, we do:
"Note that for every step in the x-direction, z increments by -A / C, and likewise it increments by -B / C for every step in the y-direction.
So these are the gradients we're looking for to perform linear interpolation. In the plane equation (A, B, C) is the normal vector of the plane.
It can easily be computed with a cross product.
Now that we have the gradients, let's call them dz/dx (which is -A / C) and dz/dy (which is -B / C), we can easily compute z everywhere on the triangle.
We know the z value in all three vertex positions.
Let's call the one of the first vertex z0, and it's position coordinates (x0, y0). Then a generic z value of a point (x, y) can be computed as:"
z = z0 + dz/dx * (x - x0) + dz/dy * (y - y0)
This found the Z value correctly and fixed my code. The new zValueOfPoint method is:
private double zValueOfPoint(Triangle triangle, Point2D rayOrigin)
{
Vector3D surfaceNormal = getNormal(triangle);
double A = surfaceNormal.x;
double B = surfaceNormal.y;
double C = surfaceNormal.z;
double dzdx = -A / C;
double dzdy = -B / C;
double rayZ = triangle.getV1().z * modelViewer.getModelScale() + dzdx * (rayOrigin.x - triangle.getV1().projectPoint(modelViewer).x) + dzdy * (rayOrigin.y - triangle.getV1().projectPoint(modelViewer).y);
return rayZ;
}
We can optimize this by only calculating most of it once, and then adding dz/dx to get the z value for the next pixel, or dz/dy for the pixel below (with the y-axis going down). This means that we cut down on calculations per polygon significantly.
this must be really slow
so much redundant computations per iteration/pixel just to iterate its coordinates. You should compute the 3 projected vertexes and iterate between them instead look here:
triangle/convex polygon rasterization
I dislike your zValueOfPoint function
can not find any use of x,y coordinates from the main loops in it so how it can compute the Z value correctly ?
Or it just computes the average Z value per whole triangle ? or am I missing something? (not a JAVA coder myself) in anyway it seems that this is your main problem.
if you Z-value is wrongly computed then Z-Buffer can not work properly. To test that look at the depth buffer as image after rendering if it is not shaded teapot but some incoherent or constant mess instead then it is clear ...
Z buffer implementation
That looks OK
[Hints]
You have too much times terms like x + getWidth()/2 why not compute them just once to some variable? I know modern compilers should do it anyway but the code would be also more readable and shorter... at least for me
I'm trying to get random point inside triangle in Java.
I have three points with x, y coordinates and trying to use this formula.
P = (1 - sqrt(r1)) * A + (sqrt(r1) * (1 - r2)) * B + (sqrt(r1) * r2) * C
Where r1 and r2 are random double from 0 to 1.
But, how to define A, B, C? Because now A have x and y coordinates.
P(x) = (1 - sqrt(r1)) * A(x) + (sqrt(r1) * (1 - r2)) * B(x) + (sqrt(r1) * r2) * C(x)
P(y) = (1 - sqrt(r1)) * A(y) + (sqrt(r1) * (1 - r2)) * B(y) + (sqrt(r1) * r2) * C(y)
More information can be found here math.stackexchange and this papaer
I would rather not use a formula wich involves square roots and thus, floating point errors + computation time. The following approach only uses multiplication and addition, wich makes it efficient, and more float-friendly. It is also quite easy to implement/understand :
Generating randomly uniformly a point in ABC :
The idea is to generate a point in a parallelogram ABCD, and project the obtained point inside ABC.
pick a point p inside the parallelogram ABCD (D is the translation of A by vector AB + AC)
two cases :
p is inside ABC, keep it
p is outside ABC, pick p', its symetrical according to the middle of [BC]
Few additional details
Checking if a point is inside a triangle : How to determine if a point is in a 2D triangle?
(in fact you only need to check on wich side of bc it is)
Random point p in parallelogram ABCD :
let V1 (resp. V2) the vector from A to B (resp A to C).
Point p is given by the translation of A by (r1 * V1 + r2 * V2) where r1 and r2 are two random double between 0 and 1.
Uniformity : The choosen point in the parallelogram is obviously uniformly choosen. Moreover every point in ABC can be obtained from two points in ABCD, except for the points lying on BC, which are twice less likely, however this does not harm unifromity as BC is of null area comparing to ABC.
This approach can easily be generalized to n-dimensional simplices
Here's another method to achieve this goal which is also introduced in Graphics Gems (Turk).
if (r1 + r2 > 1) {
r1 = 1 - r1;
r2 = 1 - r2;
}
a = 1 - r1 - r2;
b = r1;
c = r2;
Q = a*A + b*B + c*C
This method cannot be extended to a higher dimensional space. If that is the case, you need to use your formula which is essentially Barycentric coordinates.
Since the question was for Java code and not pseudo code or mathematical notation, here's Vaibhav's solution in Java:
public class Point{
public double x;
public double y;
public Point(double x, double y){
this.x = x;
this.y = y;
}
}
public class Triangle {
Point A;
Point B;
Point C;
public Point getRandomPoint(){
double r1 = Math.random();
double r2 = Math.random();
double sqrtR1 = Math.sqrt(r1);
double x = (1 - sqrtR1) * A.x + (sqrtR1 * (1 - r2)) * B.x + (sqrtR1 * r2) * C.x;
double y = (1 - sqrtR1) * A.y + (sqrtR1 * (1 - r2)) * B.y + (sqrtR1 * r2) * C.y;
return new Point(x, y);
}
}
Further optimization is possible, the code just becomes less readable.
For school, I have recently started creating my own raytracer. However, I've hit a snag with either computing the viewing rays, or checking for an intersection between a triangle and a ray. As far as I can tell, the computations seem to be executed correctly, as I place my camera in the origin and have it face the -z axis towards an object right in front of it, allowing for simple vector maths by hand. Everything seems to check out, but nothing gets painted on the screen.
I will post the code I am using the calculate the viewing rays.
public Ray generateRay(float nX, float nY , Point2f coordinates)
{
// Compute l, r, b and t.
Vector3f temp = VectorHelper.multiply(u, nX/2.0f);
float r = temp.x + Position.x;
temp = VectorHelper.multiply(u, -nX/2.0f);
float l = temp.x + Position.x;
temp = VectorHelper.multiply(v, nY/2.0f);
float t = temp.y + Position.y;
temp = VectorHelper.multiply(v, -nY/2.0f);
float b = temp.y + Position.y;
// Compute the u and v coordinates.
float uCo = (l + (r - l) * (coordinates.x + 0.5f)/nX);
float vCo = (b + (t - b) * (coordinates.y + 0.5f)/nY);
// Compute the ray's direction.
Vector3f rayDirection = VectorHelper.multiply(w, -FocalLength);
temp = VectorHelper.add(VectorHelper.multiply(u, uCo), VectorHelper.multiply(v, vCo));
rayDirection = VectorHelper.add(rayDirection, temp);
rayDirection = VectorHelper.add(rayDirection, Position);
rayDirection = VectorHelper.normalize(VectorHelper.add(rayDirection, temp));
// Create and return the ray.
return new Ray(Position, rayDirection);
}
The following code is what I use to calculate an intersection. It uses Cramer's Rule to solve the matrix equation.
public static Point3f rayTriangleIntersection(
Ray ray, Point3f vertexA, Point3f vertexB, Point3f vertexC)
{
// Solve the linear system formed by the ray and the parametric surface
// formed by the points of the triangle.
// | a d g | | B | | j |
// | b e h | * | Y | = | k |
// | c f i | * | t | = | l |
// The following uses Cramer's rule to that effect.
float a = vertexA.x - vertexB.x; float d = vertexA.x - vertexC.x; float g = ray.getDirection().x;
float b = vertexA.y - vertexB.y; float e = vertexA.y - vertexC.y; float h = ray.getDirection().y;
float c = vertexA.z - vertexB.z; float f = vertexA.z - vertexC.z; float i = ray.getDirection().z;
float j = vertexA.x - ray.getOrigin().x;
float k = vertexA.y - ray.getOrigin().y;
float l = vertexA.z - ray.getOrigin().z;
// Compute some subterms in advance.
float eihf = (e * i) - (h * f);
float gfdi = (g * f) - (d * i);
float dheg = (d * h) - (e * g);
float akjb = (a * k) - (j * b);
float jcal = (j * c) - (a * l);
float blkc = (b * l) - (k * c);
// Compute common division number.
float m = (a * eihf) + (b * gfdi) + (c * dheg);
// Compute unknown t and check whether the point is within the given
// depth interval.
float t = -((f * akjb) + (e * jcal) + (d * blkc)) / m;
if (t < 0)
return null;
// Compute unknown gamma and check whether the point intersects the
// triangle.
float gamma = ((i * akjb) + (h * jcal) + (g * blkc)) / m;
if (gamma < 0 || gamma > 1)
return null;
// Compute unknown beta and check whether the point intersects the
// triangle.
float beta = ((j * eihf) + (k * gfdi) + (l * dheg)) / m;
if (beta < 0 || beta > (1 - gamma))
return null;
// Else, compute the intersection point and return it.
Point3f result = new Point3f();
result.x = ray.getOrigin().x + t * ray.getDirection().x;
result.y = ray.getOrigin().y + t * ray.getDirection().y;
result.z = ray.getOrigin().z + t * ray.getDirection().z;
return result;
}
My question is rather simple. What am I doing wrong? I've looked and debugged this code to death and cannot single out the errors, google offers little more than the theory I already have in the book I am using. Also, the code is still rather rough as I'm just focusing on getting it to work before cleaning it up.
Thanks in advance,
Kevin
Hard to say precisely what is going wrong. Especially since you aren't using descriptive variable names (what are nX, nY etc.??)
Here are some tips:
First make sure it's not a bug in your display code. Fake an intersection to prove that you get visible output when you hit something, e.g. make all rays in the bottom right of the screen hit an axis-aligned plane or something similar so that you can easily verify the co-ordinates
Try a ray/sphere intersection first. It's easier than a ray/triangle intersection.
Consider using vector/matrix operations rather than computing all the components by hand. It's too easy to make a mistake in many lines of jumbled letters.
If you have a scale problem (e.g. the object is too small) then double-check your conversions between world and screen co-ordinates. World co-ordinates will be in the range of small double numbers (0.2 ....5.0 for example) while screen co-ordinates should be pixel locations according to your view size (0 .. 1024 for example). You should be doing most of your maths in world co-ordinates, only converting from/to screen co-ordinates at the beginning and the end of your rendering code.
Step through the top level raytracing code in a debugger and make sure that you are producing rays in sensible directions for each screen co-ordinate(especially the corners of the screen)
Check that your camera direction is pointing towards the target object. It is quite an easy mistake to have it looking in exactly the opposite direction!
Example set-up that should work:
Camera position [0 0 4]
Object position [0 0 0]
Camera DIRECTION [0 0 -1] (note the minus if you want it to look towards the origin!)
UP vector [0 0.75 0]
RIGHT vector [+/-1 0 0]
Then your ray direction should be something like (for a pixel [screenX, screenY]):
ray = DIRECTION + (2*(screenX / screenWidth)-1)*RIGHT + (1-2*(screenY/screenHeight))*UP
ray = normalize(ray)
I'm completely stuck.
I have an ellipse, and a line. Line is set by two points, ellipse - by bottom-left and top-right corners. I have to find their points of intersection, using java.
I tried to solve an equation system:
(1) y = kx + m;
x^2/a^2 + y^2/b^2 = 1;
but I could't make things work properly. I assume it's because of java's coordinate system, but it also may be my own mistake somewherem beacuse I'm confused with it.
Is there any better way to find points of intersection and, if not, how can I get them properly?
Thank you in advance.
Code:
double r1 = in_y2-in_y;
double r2 = in_x2-in_x;
double k = r1/r2;
double m = in_y2 - k*in_x2;
double a = Math.abs((double)x2 - (double)x)/2;
double b = Math.abs((double)y2 - (double)y)/2;
double A1 = 1/(a*a) + (k*k)/(b*b);
double B1 = (2*k*m)/b*b;
double C1 = (m*m)/(b*b);
double D = Math.sqrt(B1*B1 - 4*A1*C1);
double ex1 = (-B1 + D/(2*A1));
double ey1 = k*ex1 + m;
double ex2 = (-B1 - D/(2*A1));
double ey2 = k*ex2 + m;
This is probably no longer relevant to the original problem owner, but since I encountered the same question, let me present my answer.
There are three mistakes in the original computation that I can see: (i) the one pointed out by #mprivat, (ii) the bracketing in the assignment to B1 which should instead be
double B1 = (2*k*m)/(b*b);
and (iii) a more fundamental one: the presented computation does not correct for the origin of the ellipse. Since the ellipse is defined by itse circumscribing bounds, there is no guarantee that it is centered on (0,0).
Let's call the center (cx,cy), then the equation for the ellipse becomes
(x-cx)^2/a^2 + (y-cy)^2/b^2 = 1
rather than the original
x^2/a^2 + y^2/b^2 = 1
The simple repair, I think, is to translate the line wrt (cx,cy) and translate the results back, like so:
...
double m = (in_y2-cy) - k*(in_x2-cx);
...
double ex1 = (-B1 + D/(2*A1)) + cx;
double ey1 = k*(ex1-cx) + m + cy;
double ex2 = (-B1 - D/(2*A1)) + cx;
double ey2 = k*(ex2-cx) + m + cy;
The more elegant repair is to solve the correct equation for the ellipse instead, but this results in even more impenetrable formulas for B1 and C1:
double B1 = (2*k*(m-cy))/(b*b) - (2*cx)/(a*a);
double C1 = (m-cy)*(m-cy)/(b*b) - 1 + (cx*cx)/(a*a);
As a final remark, note that this breaks down for vertical lines, as then r2 = 0 so k isn't defined.
Java can't solve the algebra problem, but it can compute the solution once you tell it what to compute.
Sounds like you just need to replace your y in the ellipse's equation with kx+m then solve for x. Looks like it's a simply binomial equation. Write a program that computes x=... based on k, m, a and b. Java can help you compute the roots if you tell it what to compute and what the values of k, m, a and b are.
In your particular case, you want to use Java as a simple calculator...
can you please put your code so that we could see if it's correct?
anyway , here's an algorithm:
http://mathworld.wolfram.com/Ellipse-LineIntersection.html
note that since it has a square root , you might get a solution that is not precise.
Line2D.Double line = new Line2D.Double(x1,y1,x2,y2);
Ellipse2D.Double ellipse = new Ellipse2D.Double(x,y,width,height);
int resolution = 1000;
int x_distance = ellipse.getWidth()/2;
int y_distance = ellipse.getHeight()/2;
double angle = 360.0/(double)resolution;
Point center = new Point(width/2,height/2);
Point point = new Point();
for (int index = 0; index < resolution; index++)
{
int x = (center.x+x_distance)*Math.sin(Math.toRadians(angle*index)));
int y = (center.y+y_distance)*Math.cos(Math.toRadians(angle*index)));
Ellipse2D.Double dot = new Ellipse2D.Double(x,y,1,1);
if (line.intersects(dot.getBounds()))
{
point.setLocation(x,y);
index = resolution;
}
}