I am currently working on a project of drawing thick polylines and I am using interpolation in OpenGL. I managed to calculate all the necessary points but I need to draw two more points. I need to translate one point orthogonally to the line connecting two points. The scatch below shows what are the points. Point L is to be translated for the distance between L and nJ orthogonally to the line AB (B is the central point). Similar thing is with translation to the nK.
I have written the code:
float alpha = atan2(B.y - A.y,B.x - A.x) - deg90;
float alpha2 = atan2(C.y - B.y, C.x - B.x) - deg90;
nJ.x = L.x + w*cos(alpha); // w is distance between A1 and A2
nJ.y = L.y + w*sin(alpha);
nK.x = L.x + w*cos(alpha2);
nK.y = L.y + w*sin(alpha2);
The code works only for some points, not all. I need to fix + sing in above calculations of nJ and nK, but I do not know how. Anyone having suggestion?
First you need the left-hand-side function:
lhs(v) = [-v.y, v.x]
This turns a vector 90 degrees counter-clockwise.
Now you need the turn function:
turn(u, v, w) = sign(lhs(v - u), w - v)
If you have a polyline from u to v to w, turn(u,v,w) tells you whether it's a left turn (counter-clockwise turn) (positive), right turn (clockwise turn) (negative), or colinear (0).
There are four infinite lines in your picture that run parallel to ab and bc, with a distance of w between each pair.
The lines on the lower part are:
f(s) = (a + 0.5 * w * normalize(lhs(b - a))) + (b - a) * s
g(t) = (b + 0.5 * w * normalize(lhs(c - b))) + (c - b) * t
You want to find the intersection of the two lines; i.e., you want to solve for s and t in f(s) = g(t). This is just a system of two linear equations with two unknowns.
The solution is your point L = f(s) = g(t) in the picture.
To compute I you can use the exact same idea:
f(s) = (a - 0.5 * w * normalize(lhs(b - a))) + (b - a) * s
g(t) = (b - 0.5 * w * normalize(lhs(c - b))) + (c - b) * t
Solve for I = f(s) = g(t).
Update
Once you have L you can compute Kn and Jn as follows.
Kn = L - w * normalize(lhs(b - a))
Jn = L - w * normalize(lhs(c - b))
In computational geometry code, trigonometry is usually a code smell - it's not always wrong, but it usually is wrong. Try to stick to linear algebra.
Related
Yes, I know that it is a popular problem. But I found nowhere the full clear implementing code without using OpenGL classes or a lot of headers files.
Okay, the math solution is to transfer ellipsoid to sphere. Then find intersections dots (if they exist of course) and make inverse transformation. Because affine transformation respect intersection.
But I have difficulties when trying to implement this.
I tried something for sphere but it is completely incorrect.
double CountDelta(Point X, Point Y, Sphere S)
{
double a = 0.0;
for(int i = 0; i < 3; i++){
a += (Y._coordinates[i] - X._coordinates[i]) * (Y._coordinates[i] - X._coordinates[i]);
}
double b = 0.0;
for(int i = 0; i < 3; i++)
b += (Y._coordinates[i] - X._coordinates[i]) * (X._coordinates[i] - S._coordinates[i]);
b *= 2;
double c = - S.r * S.r;
for(int i = 0; i < 3; i++)
c += (X._coordinates[i] - S._coordinates[i]) * (X._coordinates[i] - S._coordinates[i]);
return b * b - 4 * a * c;
}
Let I have start point P = (Px, Py, Pz), direction V = (Vx, Vy, Vz), ellipsoid = (Ex, Ey, Ec) and (a, b, c). How to construct clear code?
Let a line from P to P + D intersecting a sphere of center C and radius R.
WLOG, C can be the origin and R unit (otherwise translate by -C and scale by 1/R). Now using the parametric equation of the line and the implicit equation of the sphere,
(Px + t Dx)² + (Py + t Dy)² + (Pz + t Dz)² = 1
or
(Dx² + Dy² + Dz²) t² + 2 (Dx Px + Dy Py + Dz Pz) t + Px² + Py² + Pz² - 1 = 0
(Vectorially, D² t² + 2 D P t + P² - 1 = 0 and t = (- D P ±√((D P)² - D²(P² - 1))) / D².)
Solve this quadratic equation for t and get the two intersections as P + t D. (Don't forget to invert the initial transformations.)
For the ellipsoid, you can either plug the parametric equation of the line directly into the implicit equation of the conic, or reduce the conic (and the points simultaneously) and plug in the reduced equation.
I'm trying to write an algorithm to determine if point is located inside a triangle or on it's edge in 3D coordinate space.
For example, I try to reach such results for different cases
I've figured out how to check if point P inside the triangle, I calculated normal vectors for triangles ABP, BCP, CAP and checked if they are similar.
Can someone explain how to check if a point is on the edge of a triangle (but not outside of a triangle)? You can provide formulas or code as you wish.
Make vectors:
r = p - A (r.x = p.x - A.x, r.y = p.y - A.y, r.z = p.z - A.z)
s = B - A
q = C - A
Calculate normal to ABC plane:
n = s x q (vector product)
Check if p lies in ABC plane using dot product:
dp = n.dot.r
If dp is zero (or has very small value like 1.0e-10 due to the floating point errors, then p is in the plane, and we can continue
Decompose vector p by base vectors s and q. At first check if z-component of normal (n.z) is non-zero. If so, use the next pair of equations (otherwise choose equations for x/z or y/z components):
px = a * sx + b * qx
py = a * sy + b * qy
Solve this system
a = (sy * qx - sx * qy) / (py * qx - px * qy)
b = (px - a * sx) / qx
If resulting coefficients a and b fulfill limits:
a >= 0
b >= 0
a + b <= 1.0
then point p lies in triangle plane inside it.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How do you detect where two line segments intersect?
Given two points a and b plus two vectors v and u I want to find a third point c, which is the point of intersection in the following manner:
vector2 intersection(vector2 a, vector2 v, vector2 b, vector2 u)
{
float r, s;
a + r * v = b + s * u;
r * v - s * u = b - a
r * v.x - s * u.x = b.x - a.x
r * v.y - s * u.y = b.y - a.y
}
Is there any other way than using gaussian elimination to solve this system? Or is this the best (or at least an acceptable) way to handle this?
EDIT:
Definition of vector2
typedef union vector2
{
float v[2];
struct { float x, y; };
} vector2;
a and b are also of type vector2, because the only difference between a point and a vector is in the the way it is transformed by an affine transformation.
Looks like an assignment problem to me. Here is the logic that will help you write the code.
Let us call the first Ray as R0.
Locus of a point on R0 is defined as P:
P = P0 + alpha x V0
For the second ray R1:
P = P1 + beta x V1
Since they should intersect:
P0 + alpha x V0 = P1 + beta x V1
alpha and beta are unknowns and we have two equations in x any y.
Solve for the unknowns and get back the point of intersection.
i.e.,
P0.x + alpha * V0.x = P1.x + beta * V1.x
P0.y + alpha * V0.y = P1.y + beta * V1.y
solve for alpha and beta.
If there is a real positive solution for both alpha and beta, rays intersect.
If there is a real but at least one negative solution for both alpha and beta, extended rays intersect.
It's simple math.
But, first, check that you have intersection. If both vector are parallel you will fail to solve that:
// Edit this line to protect from division by 0
if (Vy == 0 && Uy == 0) || ((Vy != 0 && Uy != 0 && (Vx/Vy == Ux/Uy)) // => Fail.
Then (I won't show the calculation because they are long but the result is):
R = (AxUy - AyUx + ByUx - BxUy) / (VyUx - VxUy)
S = (Ax - Bx + RVx) / Ux
Hope that helped you.
I have two points A (x1,y1) and B (x2,y2) that are given as an input to the program. I have to find a third point C that lies on the line AB and is at a distance 10 away from the point A.
I can easily get the slope of the line but that doesn't give me the full equation for the line. Even if I get the full equation, I am not sure using this equation, how would I find out a point that is x distance away from A.
Any suggestions on how to approach this?
There are always two points on each line:
get the vector from A to B (subtract the coordinates)
normalize the vector (divide by its length; pythagorean theorem)
multiply the vector by 10 or -10
add the vector to A to get C
Note that if A==B, the line is not defined, and this algorithm causes a division by zero. You may want to add a test for equality at the beginning.
You can use the sine or the cosine (times 10) of the angle of the line to get the horizontal or vertical distance of the point that is a distance of 10 from a given point. A shortcut is to use the horizontal or vertical distance divided by the direct distance between the points to get the sine or cosine.
You can do it using vectors like this:
Let D = the difference between B and A (D = B - A)
Then any point on the line can be described by this formula:
point = A + Dt
where t is a real number.
So just plug in any value for t to get another point. For example if you let t == 1 then the equation above reduces to point = B. If you let t = 0 then it reduces to point = A. So you can see that you can use this to find a point between A and B simply by let t range from 0 to 1. Additionally if you let t > 1, you will find a point past B.
You can see from the image that your given points are x1,y1 and x2,y2. You need to find an intermediate point at a distance 'R' from point x1,y1.
All you need to do is to find θ using
Tan θ = (y2-y1)/(x2-x1)
Then you can get the intermediate point as (R * cos θ),(R * Sin θ)
I have drawn this assuming positive slope.
Going on similar lines you can seek a solution for other special cases lile:
i. Horizontal line
ii. Vertical line
iii. Negative slope
Hope it clarifies.
I have done the calculation in Andengine using a Sprite object. I have two Array List x coordinates and y coordinates. Here i am just calculating using the last two values from these two array list to calculate the third point 800 pixel distant from Your point B. you can modify it using different values other than 800. Hope it will work.The coordinate system here is a little different where (0,0) on the top left corner of the screen. Thanks
private void addExtraCoordinate(CarSprite s) {
int x0, y0, x1, y1;
float x = 0f, y = 0f;
x0 = Math.round(xCoordinates.get(xCoordinates.size() - 2));
x1 = Math.round(xCoordinates.get(xCoordinates.size() - 1));
y0 = Math.round(yCoordinates.get(yCoordinates.size() - 2)) * (-1);
y1 = Math.round(yCoordinates.get(yCoordinates.size() - 1)) * (-1);
if (x1 == x0 && y1 == y0) {
return;
} else if (y1 == y0 && x1 != x0) {
if (x1 > x0) {
x = (float) x1 + 800f;
} else
x = (float) x1 - 800f;
y = Math.round(yCoordinates.get(yCoordinates.size() - 1));
} else if (y1 != y0 && x1 == x0) {
if (y1 > y0) {
y = (float) Math.abs(y1) - 800f;
} else
y = (float) Math.abs(y1) + 800f;
x = Math.round(xCoordinates.get(xCoordinates.size() - 1));
} else {
float m = (float) (yCoordinates.get(yCoordinates.size() - 1) * (-1) - yCoordinates
.get(yCoordinates.size() - 2) * (-1))
/ (float) (xCoordinates.get(xCoordinates.size() - 1) - xCoordinates
.get(xCoordinates.size() - 2));
if (x1 > x0) {
x = (float) ((float) x1 + 800f / (float) Math
.sqrt((double) ((double) 1f + (double) (m * m))));
} else
x = (float) ((float) x1 - 800f / (float) Math
.sqrt((double) ((double) 1f + (double) (m * m))));
if (y0 > y1) {
y = (float) ((float) Math.abs(y1) + 800f / (float) Math
.sqrt((double) (((double) 1f / (double) (m * m)) + (double) 1f)));
} else
y = (float) ((float) Math.abs(y1) - 800f / (float) Math
.sqrt((double) (((double) 1f / (double) (m * m)) + (double) 1f)));
}
xCoordinates.add(x);
yCoordinates.add(y);
}
I need a method that allows me to find the Y-coordinate on a Cubic Bezier Curve, given an x-coordinate.
I've come across lots of places telling me to treat it as a cubic function then attempt to find the roots, which I understand. HOWEVER the equation for a Cubic Bezier curve is (for x-coords):
X(t) = (1-t)^3 * X0 + 3*(1-t)^2 * t * X1 + 3*(1-t) * t^2 * X2 + t^3 * X3
What confuses me is the addition of the (1-t) values. For instance, if I fill in the X values with some random numbers:
400 = (1-t)^3 * 100 + 3*(1-t)^2 * t * 600 + 3*(1-t) * t^2 * 800 + t^3 * 800
then rearrange it:
800t^3 + 3*(1-t)*800t^2 + 3*(1-t)^2*600t + (1-t)^3*100 -400 = 0
I still don't know the value of the (1-t) coefficients. How I am I supposed to solve the equation when (1-t) is still unknown?
There are three common ways of expressing a cubic bezier curve.
First x as a function of t
x(t) = sum( f_i(t) x_i )
= (1-t)^3 * x0 + 3*(1-t)^2 * t * x1 + 3*(1-t) * t^2 * x2 + t^3 * x3
Secondly y as a function of x
y(x) = sum( f_i(x) a_i )
= (1-x)^3 * y0 + 3*(1-x)^2 * x * y1 + 3*(1-x) * x^2 * y2 + x^3 * y3
These first two are mathematically the same, just using different names for the variables.
Judging by your description "find the Y-coordinate on a Cubic Bezier Curve, given an x-coordinate on it." I'm guessing that you've got a question using the second equation are are trying to rearrange the first equation to help you solve it, where as you should be using the second equation. If thats the case, then no rearranging or solving is required - just plug your x value in and you have the solution.
Its possible that you have an equation of the third kind case, which is the ugly and hard case.
This is both the x and y parameters are cubic Beziers of a third variable t.
x(t) = sum( f_i(t) x_i )
y(t) = sum( f_i(t) y_i )
If this is your case. Let me know and I can detail what you need to do to solve it.
I think this is a fair CS question, so I'm going to attempt to show how I solved this. Note that a given x may have more than 1 y value associated with it. In the case where I needed this, that was guaranteed not to be the case, so you'll have to figure out how to determine which one you want.
I iterated over t generating an array of x and y values. I did it at a fairly high resolution for my purposes. (I was looking to generate an 8-bit look-up table, so I used ~1000 points.) I just plugged t into the bezier equation for the next x and the next y coordinates to store in the array. Once I had the entire thing generated, I scanned through the array to find the 2 nearest x values. (Or if there was an exact match, used that.) I then did a linear interpolation on that very small line segment to get the y-value I needed.
Developing the expression further should get you rid of the (1 - t) factors
If you run:
expand(800*t^3 + 3*(1-t)*800*t^2 + 3*(1-t)^2*600*t + (1-t)^3*100 -400 = 0);
In either wxMaxima or Maple (you have to add the parameter t in this one though), you get:
100*t^3 - 900*t^2 + 1500*t - 300 = 0
Solve the new cubic equation for t (you can use the cubic equation formula for that), after you got t, you can find x doing:
x = (x4 - x0) * t (asuming x4 > x0)
Equation for Bezier curve (getting x value):
Bx = (-t^3 + 3*t^2 - 3*t + 1) * P0x +
(3*t^3 - 6*t^2 + 3*t) * P1x +
(-3*t^3 + 3*t^2) * P2x +
(t^3) * P3x
Rearrange in the form of a cubic of t
0 = (-P0x + 3*P1x - 3*P2x + P3x) * t^3+
(3*P0x - 6*P1x + 3*P2x) * t^2 +
(-3*P0x + 3*P1x) * t +
(P0x) * P3x - Bx
Solve this using the cubic formula to find values for t. There may be multiple real values of t (if your curve crosses the same x point twice). In my case I was dealing with a situation where there was only ever a single y value for any value of x. So I was able to just take the only real root as the value of t.
a = -P0x + 3.0 * P1x - 3.0 * P2x + P3x;
b = 3.0 * P0x - 6.0 * P1x + 3.0 * P2x;
c = -3.0 * P0x + 3.0 * P1x;
d = P0x;
t = CubicFormula(a, b, c, d);
Next put the value of t back into the Bezier curve for y
By = (1-t)^3 * P0x +
3t(1-t)^2 * P1x +
3t^2(1-t) * P2x +
t^3 * P3x
So I've been looking around for some sort of method to allow me to find the Y-coordinate on a Cubic Bezier Curve, given an x-coordinate on it.
Consider a cubic bezier curve between points (0, 0) and (0, 100), with control points at (0, 33) and (0, 66). There are an infinite number of Y's there for a given X. So there's no equation that's going to solve Y given X for an arbitrary cubic bezier.
For a robust solution, you'll likely want to start with De Casteljau's algorithm
Split the curve recursively until individual segments approximate a straight line. You can then detect whether and where these various line segments intercept your x or whether they are vertical line segments whose x corresponds to the x you're looking for (my example above).