I have :
Arc : defined by p1x, p1y - p2x,p2y , radius, center, initial - final ang.
Line : defined by ax,ay - bx,by.
As you can see at the image I want to figure out an arc tangent to arc and linea and passing by the end point of first arc .
I think there is a unique solution. (or maybe two, R + and R - )
I'm trying so see how to implement an algorithm, without results...
Any idea would be appreciated...
What you are trying to do is to find a circle tangent to an infinite line "L", and tangent to a circle at a particular point on the circle. The key observation is that, since the tangent vector to any circle at a given angle is perpendicular to the radius vector for that angle, what we need to find are the point "TC" ("Tangent Center") and distance "d" at which a line offset from the given line by the distance d intersects a normal drawn outward from the circle for the same distance (forgive my bad art):
The easiest way to solve for "d" is as follows:
Construct the normalized normal vector "R" at point "P" by taking P-C and normalizing. (This constructs the outer tangent to the circle. If you want an inner tangent, you can flip.)
Construct the normalized perpendicular vector "N" to the line "L". I'm not really sure what your variables "ax,ay - bx,by" mean, so let's define the line by a start point "A = (ax, ay)" and a direction vector "DA = (dax, day)". In that case the normal is +/- (-day, dax)/sqrt(day*day+dax*dax). (The normalized cross with the (0,0,1) vector in 3d.)
Choose the sign of "N" so that it points away from P, i.e. if the dot product of (P-A) and N is positive, flip N. if the dot product is (nearly) zero, then the tangent circle would have radius (nearly) zero, and so is not defined.
Now consider a point TC defined by P2 = P + d*(R + N) for some d. If P2 lies on the line L, then d is the radius of the tangent circle we seek! P2 lies on the line if and only if the dot product of (P2 - A) and N is zero. This defines a linear equation in one variable -- d -- so you can solve for it. Note that if R + N is (nearly) of length zero, then P is 180 degrees away from the closest point between the circle and the line; you will need to check for this explicitly and handle it.
Once you have d, you can get the center of the circle TC=P + d*R.
This method should have good numerical stability when the tangent to the circle at P is parallel or nearly parallel to the line.
You didn't specify a language, but hopefully this can get you started. My primary language is c#.
Related
My trigonometry needs a little help.
How would I go about calculating the point of the nearest possible intersection with a line along a rounded corner?
Take this image:
What I would like to know is, given that I know point a, and the dimensions of the rectangle, how would I find point b when the edges of the rectangle are curved?
So far, as you can see, I've only managed to calculate the nearest edge of the rectangle as if it had right-angled corners.
If it matters, I'm doing this in ActionScript 3. But example sudo-code will suffice.
Calculate the vector from the midpoint M of the corner to A:
v_x = a_x - m_x
v_y = a_y - m_y
then go radius of the corner r times towards A to get to the intersection point I
i_x = m_x + r*v_x
i_y = m_y + r*v_y
This obviously only works if the nearest intersection is on the rounded corner. Just calculate the other intersections with the edges, too, and then check which has the nearest distance to A.
You need to know the radius R of the circle that generates the round corner and the coordinates (Xr,Yr) of the point where the two sides of a non rounded rectangle cross each other.
Then the coordinates for the center of the circle that generates the round corner are (Xc, Yc) = (Xr-R, Yr-R)
From here, it's a matter of solving the equation of the cross point between the segment line defined by point A=(Xa, Ya) and point (Xc, Yc), whose parametric equation is:
x = Xa + p*(Xc-Xa)
y = Ya + p*(Yc-Ya)
and the circle whose equation is
(x-Xc)^2 + (y-Yc)^2 = R^2
Substitute values for x and y from the parametric euation of the line in the equation of the circle, and you will have an equation with only one unkown: p. Solve the equation, and if there are more than one solution, choose the one that is in the range [0,1]. Substitute the found value of p in the parametric equation of the line to get the point of intersection.
Graphically:
If you know the radius and center of the corner as R and C=(Xc, Yc), then the nearest point on the corner to the given point A=(Xa, Ya) is the intersection point of the corner and the line defined by the given point and the center. This point can be directly expressed as
X = Xc + R*(Xa-Xc)/|AC|
Y = Yc + R*(Ya-Yc)/|AC|
where |AC| = Sqrt((Xa-Xc)^2 + (Ya-Yc)^2)
I have the plane equation describing the points belonging to a plane in 3D and the origin of the normal X, Y, Z. This should be enough to be able to generate something like a 3D arrow. In pcl this is possible via the viewer but I would like to actually store those 3D points inside the cloud. How to generate them then ? A cylinder with a cone on top ?
To generate a line perpendicular to the plane:
You have the plane equation. This gives you the direction of the normal to the plane. If you used PCL to get the plane, this is in ModelCoefficients. See the details here: SampleConsensusModelPerpendicularPlane
The first step is to make a line perpendicular to the normal at the point you mention (X,Y,Z). Let (NORMAL_X,NORMAL_Y,NORMAL_Z) be the normal you got from your plane equation. Something like.
pcl::PointXYZ pnt_on_line;
for(double distfromstart=0.0;distfromstart<LINE_LENGTH;distfromstart+=DISTANCE_INCREMENT){
pnt_on_line.x = X + distfromstart*NORMAL_X;
pnt_on_line.y = Y + distfromstart*NORMAL_Y;
pnt_on_line.z = Z + distfromstart*NORMAL_Z;
my_cloud.points.push_back(pnt_on_line);
}
Now you want to put a hat on your arrow and now pnt_on_line contains the end of the line exactly where you want to put it. To make the cone you could loop over angle and distance along the arrow, calculate a local x and y and z from that and convert them to points in point cloud space: the z part would be converted into your point cloud's frame of reference by multiplying with the normal vector as with above, the x and y would be multiplied into vectors perpendicular to this normal vectorE. To get these, choose an arbitrary unit vector perpendicular to the normal vector (for your x axis) and cross product it with the normal vector to find the y axis.
The second part of this explanation is fairly terse but the first part may be the more important.
Update
So possibly the best way to describe how to do the cone is to start with a cylinder, which is an extension of the line described above. In the case of the line, there is (part of) a one dimensional manifold embedded in 3D space. That is we have one variable that we loop over adding points. The cylinder is a two dimensional object so we have to loop over two dimensions: the angle and the distance. In the case of the line we already have the distance. So the above loop would now look like:
for(double distfromstart=0.0;distfromstart<LINE_LENGTH;distfromstart+=DISTANCE_INCREMENT){
for(double angle=0.0;angle<2*M_PI;angle+=M_PI/8){
//calculate coordinates of point and add to cloud
}
}
Now in order to calculate the coordinates of the new point, well we already have the point on the line, now we just need to add it to a vector to move it away from the line in the appropriate direction of the angle. Let's say the radius of our cylinder will be 0.1, and let's say an orthonormal basis that we have already calculated perpendicular to the normal of the plane (which we will see how to calculate later) is perpendicular_1 and perpendicular_2 (that is, two vectors perpendicular to each other, of length 1, also perpendicular to the vector (NORMAL_X,NORMAL_Y,NORMAL_Z)):
//calculate coordinates of point and add to cloud
pnt_on_cylinder.x = pnt_on_line.x + 0.1 * perpendicular_1.x * 0.1 * cos(angle) + perpendicular_2.x * sin(angle)
pnt_on_cylinder.y = pnt_on_line.y + perpendicular_1.y * 0.1 * cos(angle) + perpendicular_2.y * 0.1 * sin(angle)
pnt_on_cylinder.z = pnt_on_line.z + perpendicular_1.z * 0.1 * cos(angle) + perpendicular_2.z * 0.1 * sin(angle)
my_cloud.points.push_back(pnt_on_cylinder);
Actually, this is a vector summation and if we were to write the operation as vectors it would look like:
pnt_on_line+perpendicular_1*cos(angle)+perpendicular_2*sin(angle)
Now I said I would talk about how to calculate perpendicular_1 and perpendicular_2. Let K be any unit vector that is not parallel to (NORMAL_X,NORMAL_Y,NORMAL_Z) (this can be found by trying e.g. (1,0,0) then (0,1,0)).
Then
perpendicular_1 = K X (NORMAL_X,NORMAL_Y,NORMAL_Z)
perpendicular_2 = perpendicular_1 X (NORMAL_X,NORMAL_Y,NORMAL_Z)
Here X is the vector cross product and the above are vector equations. Note also that the original calculation of pnt_on_line involved a vector dot product and a vector summation (I am just writing this for completeness of the exposition).
If you can manage this then the cone is easy just by changing a couple of things in the double loop: the radius just changes along its length until it is zero at the end of the loop and in the loop distfromstart will not start at 0.
Given a description of an arc which has a startpoint and endpoint (both in Cartesian x,y coordinates), radius and direction (clockwise or counter-clockwise), I need to convert the arc to one with a start-angle, end-angle, center, and radius.
Is there known algorithm or pseudo code that allows me to do this? Also, is there any specific term to describe these kinds of transformations?
You can find a center solving this equation system:
(sx-cx)^2 + (sy-cy)^2=R^2
(ex-cx)^2 + (ey-cy)^2=R^2
where (sx,sy) are coordinates of starting point, (ex,ey) for ending point, unknowns cx, cy for center.
This system has two solutions. Then it is possible to find angles as
StartAngle = ArcTan2(sy-cy, sx-cx)
EndAngle = ArcTan2(ey-cy, ex-cx)
Note that known direction doesn't allow to select one from two possible solutions without additional limitations. For example, start=(0,1), end=(1,0), R=1 and Dir = clockwise give us both Pi/2 arc with center (0,0) and 3*Pi/2 arc with center (1,1)
I'd propose a different approach than MBo to obtain the centers of the two circles, which have the given radius and pass to both start and end point.
If P and Q are start and end point of the arc, the center of each of the two circles lies on the line L which is orthogonal to PQ, the line from P to Q, and which bisects PQ. The distance d from the centers to L is easily obtained by Pythagoras theorem. If e is the length of PQ, then d^2 + (e/2)^2 = r^2. This way you avoid to solve that system of equations you get from MBo's approach.
Note that, in case you have a semicircle, any approach will become numerically unstable because there is only one circle of the given radius with P and Q on it. (I guess I recall the correct term is 'the problem is ill posed' in that case. It happens when P and Q are precisely 2r apart, and to figure out whether this actually true you need to check for equality of two doubles, which is always a bit problematic. If, for some reason, you know you have a semicircle you are better of to just calculate the center of PQ).
I'm trying to find the best way to get the most distant point of a circle from a specified point in 2D space. What I have found so far, is how to get the distance between the point and the circle position, but I'm not entirely sure how to expand this to find the most distant point of the circle.
The known variables are:
Point a
Point b (circle position)
Radius r (circle radius)
To find the distance between the point and the circle position, I have found this:
xd = x2 - x1
yd = y2 - y1
Distance = SquareRoot(xd * xd + yd * yd)
It seems to me, this is part of the solution. How would this be expanded to get the position of Point x in the below image?
As an additional but optional part of the question: I have read in some places that it would be possible to get the distance portion without using the Square Root, which is very performance intensive and should be avoided if fast code is necessary. In my case, I would be doing this calculation quite often; Any comments on this within the context of the main question would be welcome too.
What about this?
Calculate A-B.
We now have a vector pointing from the center of the circle towards A (if B is the origin, skip this and just consider point A a vector).
Normalize.
Now we have a well defined length (the length is 1)
If the circle is not of unit radius, multiply by radius. If it is unit radius, skip this.
Now we have the correct length.
Invert sign (can be done in one step with 3., just multiply with the negative radius)
Now our vector points in the correct direction.
Add B (if B is the origin, skip this).
Now our vector is offset correctly so its endpoint is the point we want.
(Alternatively, you could calculate B-A to save the negation, but then you have to do one more operation to offset the origin correctly.)
By the way, it works the same in 3D, except the circle would be a sphere, and the vectors would have 3 components (or 4, if you use homogenous coords, in this case remember -- for correctness -- setting w to 0 when "turning points into vectors" and to 1 at the end when making a point from the vector).
EDIT:
(in reply of pseudocode)
Assuming you have a vec2 class which is a struct of two float numbers with operators for vector subtraction and scalar multiplicaion (pretty trivial, around a dozen lines of code) and a function normalize which needs to be no more than a shorthand for multiplying with inv_sqrt(x*x+y*y), the pseudocode (my pseudocode here is something like a C++/GLSL mix) could look something like this:
vec2 most_distant_on_circle(vec2 const& B, float r, vec2 const& A)
{
vec2 P(A - B);
normalize(P);
return -r * P + B;
}
Most math libraries that you'd use should have all of these functions and types built-in. HLSL and GLSL have them as first type primitives and intrinsic functions. Some GPUs even have a dedicated normalize instruction.
Given two image buffers (assume it's an array of ints of size width * height, with each element a color value), how can I map an area defined by a quadrilateral from one image buffer into the other (always square) image buffer? I'm led to understand this is called "projective transformation".
I'm also looking for a general (not language- or library-specific) way of doing this, such that it could be reasonably applied in any language without relying on "magic function X that does all the work for me".
An example: I've written a short program in Java using the Processing library (processing.org) that captures video from a camera. During an initial "calibrating" step, the captured video is output directly into a window. The user then clicks on four points to define an area of the video that will be transformed, then mapped into the square window during subsequent operation of the program. If the user were to click on the four points defining the corners of a door visible at an angle in the camera's output, then this transformation would cause the subsequent video to map the transformed image of the door to the entire area of the window, albeit somewhat distorted.
Using linear algebra is much easier than all that geometry! Plus you won't need to use sine, cosine, etc, so you can store each number as a rational fraction and get the exact numerical result if you need it.
What you want is a mapping from your old (x,y) co-ordinates to your new (x',y') co-ordinates. You can do it with matrices. You need to find the 2-by-4 projection matrix P such that P times the old coordinates equals the new co-ordinates. We'll assume that you're mapping lines to lines (not, for instance, straight lines to parabolas). Because you have a projection (parallel lines don't stay parallel) and translation (sliding), you need a factor of (xy) and (1), too. Drawn as matrices:
[x ]
[a b c d]*[y ] = [x']
[e f g h] [x*y] [y']
[1 ]
You need to know a through h so solve these equations:
a*x_0 + b*y_0 + c*x_0*y_0 + d = i_0
a*x_1 + b*y_1 + c*x_1*y_1 + d = i_1
a*x_2 + b*y_2 + c*x_2*y_2 + d = i_2
a*x_3 + b*y_3 + c*x_3*y_3 + d = i_3
e*x_0 + f*y_0 + g*x_0*y_0 + h = j_0
e*x_1 + f*y_1 + g*x_1*y_1 + h = j_1
e*x_2 + f*y_2 + g*x_2*y_2 + h = j_2
e*x_3 + f*y_3 + g*x_3*y_3 + h = j_3
Again, you can use linear algebra:
[x_0 y_0 x_0*y_0 1] [a e] [i_0 j_0]
[x_1 y_1 x_1*y_1 1] * [b f] = [i_1 j_1]
[x_2 y_2 x_2*y_2 1] [c g] [i_2 j_2]
[x_3 y_3 x_3*y_3 1] [d h] [i_3 j_3]
Plug in your corners for x_n,y_n,i_n,j_n. (Corners work best because they are far apart to decrease the error if you're picking the points from, say, user-clicks.) Take the inverse of the 4x4 matrix and multiply it by the right side of the equation. The transpose of that matrix is P. You should be able to find functions to compute a matrix inverse and multiply online.
Where you'll probably have bugs:
When computing, remember to check for division by zero. That's a sign that your matrix is not invertible. That might happen if you try to map one (x,y) co-ordinate to two different points.
If you write your own matrix math, remember that matrices are usually specified row,column (vertical,horizontal) and screen graphics are x,y (horizontal,vertical). You're bound to get something wrong the first time.
EDIT
The assumption below of the invariance of angle ratios is incorrect. Projective transformations instead preserve cross-ratios and incidence. A solution then is:
Find the point C' at the intersection of the lines defined by the segments AD and CP.
Find the point B' at the intersection of the lines defined by the segments AD and BP.
Determine the cross-ratio of B'DAC', i.e. r = (BA' * DC') / (DA * B'C').
Construct the projected line F'HEG'. The cross-ratio of these points is equal to r, i.e. r = (F'E * HG') / (HE * F'G').
F'F and G'G will intersect at the projected point Q so equating the cross-ratios and knowing the length of the side of the square you can determine the position of Q with some arithmetic gymnastics.
Hmmmm....I'll take a stab at this one. This solution relies on the assumption that ratios of angles are preserved in the transformation. See the image for guidance (sorry for the poor image quality...it's REALLY late). The algorithm only provides the mapping of a point in the quadrilateral to a point in the square. You would still need to implement dealing with multiple quad points being mapped to the same square point.
Let ABCD be a quadrilateral where A is the top-left vertex, B is the top-right vertex, C is the bottom-right vertex and D is the bottom-left vertex. The pair (xA, yA) represent the x and y coordinates of the vertex A. We are mapping points in this quadrilateral to the square EFGH whose side has length equal to m.
Compute the lengths AD, CD, AC, BD and BC:
AD = sqrt((xA-xD)^2 + (yA-yD)^2)
CD = sqrt((xC-xD)^2 + (yC-yD)^2)
AC = sqrt((xA-xC)^2 + (yA-yC)^2)
BD = sqrt((xB-xD)^2 + (yB-yD)^2)
BC = sqrt((xB-xC)^2 + (yB-yC)^2)
Let thetaD be the angle at the vertex D and thetaC be the angle at the vertex C. Compute these angles using the cosine law:
thetaD = arccos((AD^2 + CD^2 - AC^2) / (2*AD*CD))
thetaC = arccos((BC^2 + CD^2 - BD^2) / (2*BC*CD))
We map each point P in the quadrilateral to a point Q in the square. For each point P in the quadrilateral, do the following:
Find the distance DP:
DP = sqrt((xP-xD)^2 + (yP-yD)^2)
Find the distance CP:
CP = sqrt((xP-xC)^2 + (yP-yC)^2)
Find the angle thetaP1 between CD and DP:
thetaP1 = arccos((DP^2 + CD^2 - CP^2) / (2*DP*CD))
Find the angle thetaP2 between CD and CP:
thetaP2 = arccos((CP^2 + CD^2 - DP^2) / (2*CP*CD))
The ratio of thetaP1 to thetaD should be the ratio of thetaQ1 to 90. Therefore, calculate thetaQ1:
thetaQ1 = thetaP1 * 90 / thetaD
Similarly, calculate thetaQ2:
thetaQ2 = thetaP2 * 90 / thetaC
Find the distance HQ:
HQ = m * sin(thetaQ2) / sin(180-thetaQ1-thetaQ2)
Finally, the x and y position of Q relative to the bottom-left corner of EFGH is:
x = HQ * cos(thetaQ1)
y = HQ * sin(thetaQ1)
You would have to keep track of how many colour values get mapped to each point in the square so that you can calculate an average colour for each of those points.
I think what you're after is a planar homography, have a look at these lecture notes:
http://www.cs.utoronto.ca/~strider/vis-notes/tutHomography04.pdf
If you scroll down to the end you'll see an example of just what you're describing. I expect there's a function in the Intel OpenCV library which will do just this.
There is a C++ project on CodeProject that includes source for projective transformations of bitmaps. The maths are on Wikipedia here. Note that so far as i know, a projective transformation will not map any arbitrary quadrilateral onto another, but will do so for triangles, you may also want to look up skewing transforms.
If this transformation has to look good (as opposed to the way a bitmap looks if you resize it in Paint), you can't just create a formula that maps destination pixels to source pixels. Values in the destination buffer have to be based on a complex averaging of nearby source pixels or else the results will be highly pixelated.
So unless you want to get into some complex coding, use someone else's magic function, as smacl and Ian have suggested.
Here's how would do it in principle:
map the origin of A to the origin of B via a traslation vector t.
take unit vectors of A (1,0) and (0,1) and calculate how they would be mapped onto the unit vectors of B.
this gives you a transformation matrix M so that every vector a in A maps to M a + t
invert the matrix and negate the traslation vector so for every vector b in B you have the inverse mapping b -> M-1 (b - t)
once you have this transformation, for each point in the target area in B, find the corresponding in A and copy.
The advantage of this mapping is that you only calculate the points you need, i.e. you loop on the target points, not the source points. It was a widely used technique in the "demo coding" scene a few years back.