I have the intrisic and extrinsic parameters of the camera.
The extrinsic is a 4 x 4 matrix with rotation and translation.
I have sample data as under, I have this one per camera image taken.
2.11e-001 -3.06e-001 -9.28e-001 7.89e-001
6.62e-001 7.42e-001 -9.47e-002 1.47e-001
7.18e-001 -5.95e-001 3.60e-001 3.26e+000
0.00e+000 0.00e+000 0.00e+000 1.00e+000
I would like to plot the image as given on the Matlab calibration toolkit page or
However I'm unable to figure out the Math of how to plot these 2 images.
The only lead I have is from this page http://en.wikipedia.org/wiki/Camera_resectioning. Which tells me that the camera position can be found by C = − R` . T
Any idea how to achieve this task?
Assume the corners of the plane that you want to draw are 3x1 column vectors, a = [0 0 0]', b = [w 0 0]', c = [w h 0]' and d = [0 h 0]'.
Assume that the calibration matrix that you provide is A and consists of a rotation matrix R = A(1:3, 1:3) and a translation T = A(1:3, 4).
To draw the first view
For every pose A_i with rotation R_i and translation T_i, transform each corner x_w (that is a, b, c or d) of the plane to its coordinates x_c in the camera by
x_c = R_i*x_w + T_i
Then draw the plane with transformed corners.
To draw the camera, its centre of projection in camera coordinates is [0 0 0]' and the camera x axis is [1 0 0]', y axis is [0 1 0]' and z axis is [0 0 1]'.
Note that in the drawing, the camera y-axis is pointing down, so you might want to apply an additional rotation on all the computed coordinates by multiplication with B = [1 0 0; 0 0 1; 0 -1 0].
Draw the second view
Drawing the plane is trivial since we are in world coordinates. Just draw the plane using a, b, c and d.
To draw the cameras, each camera centre is c = -R'*T. The camera axes are the rows of the rotation matrix R, so for instance, in the matrix you provided, the x-axis is
[2.11e-001 -3.06e-001 -9.28e-001]'. You can also draw a camera by transforming each point x_c given in camera coordinates to world coordinates x_w by x_w = R'*(x_c - T) and draw it.
There is now an example in opencv for visualizing the extrinsics generated from their camera calibration example
It outputs something similar to the original questions ask:
Related
I have a system where one axis is moving from [0 -> 2PI]. This movement generates an angled plane. Axis movement.
This yellow plane will be my target plane. I know the normal vector of this yellow plane and its constant. For me to calculate XYZ position on the yellow plane based on the rotation value of the axis (tool). I've come to a "solution" to first calculate what is the XYZ coordinate for a simpler plane vertical plane [1 0 0] as normal vector as I know the sphere origin and also the radius then it is easy to calculate any XYZ position based on the axis angle.
But my probelm is that now that I have the XYZ position on the gray plane: how can get my XYZ position to the corresponding position on the yellow plane? From gray plane to yellow plane Any suggestions would be appreciated.
Solution to this was simple.. I made it more complicated than necessary. There wasn't any need for transforming the points from one plane to another as these values could be calculated easily from the sphere origin and the plane orientation values.
// calculate axis rotation to radians
let radAngle = (angle)*Math.PI/180;
let beeta = (90*Math.PI/180) - radAngle; //rotation value on the circle
let gamma = Math.acos(yellow.normal.x); //plane orientation
// temporary vars
let cb = Math.cos(beeta);
let sb = Math.sin(beeta);
let cg = Math.cos(gamma);
let sg = Math.sin(gamma);
let x = sphere.origin.x + sphere.radius*(cg*sb);
let y = sphere.origin.y + sphere.radius*(sg*sb);
let z = sphere.origin.z + sphere.radius*cb;
Rotation sample
I'm working on a project with two infrared positioning cameras which output the (X,Y) coordinate of any IR source. I'm placing them next to each other and my goal is to measure the 3D coordinate (X,Y,Z) of the IR source, using the same technique our eyes use to measure depth.
I have drawn a (lousy) sketch here
which illustrates what I'm trying to calculate. The red dot is my IR source, which can also be seen on the 'views' of the camera to the right. I am trying to measure the length of the blue line.
I have a few known variables:
The cameras have a resolution of 1024x768 (which also means that this is the maximum of the (X,Y) coordinate mentioned earlier)
Horizontally the field of view is 41deg, vertically 31deg.
I have yet to decide on the distance between cameras (AB), but this will be a known variable. Let's make it 30 cm for now.
Sadly I cannot seem to find the focal length of the camera.
Ultimately I'm hoping for an (X,Y,Z) coordinate relative to the middle point of AB. How would I go about measuring (Z)?
I am not sure how well aligned your cameras are, but from your pictures I am beginning to assume that the camera A and camera B are so well aligned that the rectangle representing the camera B's screen is simply horizontal translation of the rectangle representing the camera A's screen. What I mean by that is that the corresponding edges of the screens' rectangles are parallel to each other and the two rectangular screens lie in a common vertical plane perpendicular to the ground. Now, consider the plane parallel to the vertical plane that contains the two camera screens and passing through the focal points A and B of the two cameras. Call this latter plane the screen_plane. Also, the focal points A and B are at an equal height from the ground. If that is the case, and if I assume that c = |AB| is the distance between the focal points of the two cameras, and if I put a coordinate system at A, so that the x axis is horizontal to the ground, the y axis is perpendicular to the ground, and the z axis is parallel to the ground but perpendicular to the screen, then the focal point of camera B would have coordinates ( c, 0, 0 ). As an example, you have given c = 30 cm. Also the screen_plane is spanned by the x and y axes described above and the z axis is perpendicular to the screen_plane.
If that is the setting you want to work with, then the red point P will appear on both screens with the same coordinate Y_A = Y_B but different coordinates X_A and X_B.
Then let us denote by theta the horizontal field of view angle, which you have determined as theta = 41 deg. Just to be clear, I am assuming the angle between the leftmost side to the rightmost side of view is 2 * theta = 82 deg.
If I understand correctly, you are trying to calculate the distance Z between the vertical plane screen_plane that contains both camera focal points and the plane parallel to screen_plane and passing through the red point P, i.e. you are trying to calculate the distance from P to the vertical plane screen_plane.
Then, here is how you calculate Z:
Step 1: From the image of point P on screen A calculate the distances (e.g. the number of pixels) from P to the vertical edges of the screen. Say they are dist_P_to_left_edge and dist_P_to_right_edge. Set
a_A = dist_P_to_left_edge / (dist_P_to_left_edge + dist_P_to_right_edge) (this one is not really necessary)
b_A = dist_P_to_right_edge / (dist_P_to_left_edge + dist_P_to_right_edge)
Step 2: Do the same with the image of point P on screen B:
a_B = dist_P_to_left_edge / (dist_P_to_left_edge + dist_P_to_right_edge)
b_B = dist_P_to_right_edge / (dist_P_to_left_edge + dist_P_to_right_edge) (this one is not really necessary)
Step 3: Apply the formula:
Z = c * cot(theta) / (2 * (1 - b_A - a_B) )
So for example, from the pictures of the screens of camera A and B you have provided, I measured with a ruler, that
b_A = 4/38
a_B = 12.5/38
and from the data you have included
theta = 41 deg
c = 30 cm
so I have calculated that the length of the blue segment on your picture is
Z = 30 * cos(41*pi/180) / (sin(41*pi/180) * (1 - 4/38 - 12.5/38))
= 60.99628 cm
I have 4 points, point A(x1, y1), point B(x2, y1), and point C(x2, y2), point D(x2, y3) creating two scales: Scale1 yRange(B, D), Scale2 yRange(B, C).
I want to scale the y axis only, so that point D is scaled down to point C, so that the angle between CAB is a certain degree. Then I want to rotate the scaled point C around point B a certain degree, obtaining point E. Then I want to find the real value of point E on the un-scaled coordinate grid where original point D is located. I think I need to use affine transformations, but all examples are rotate first, then scale. But I need to scale first, then rotate. How do I find this new value? I know how to perform rotations alone and a bit of scaling alone, but not together.
Maybe I am confused, after I perform the scaling and rotation, I would not need to scale back, because the new value of E equates to F? I know there are plenty of examples, but I can not wrap my head around this...
Here is my objective. I have the 3 points A, B and D. I want to scale so that DAB equates to let's say 60 degrees, creating point C, then perform several point rotations on C inside the scaled grid. I ultimately want to find the value of F, and the other rotated points, which lies in geometrical positions inside the original unscaled grid that I can not calculate unless I scale the 3 original points to said degree first. The scaled grid contains the correct ratio I need in order to rotate my points. I need to do all my rotations inside the scaled grid, without losing values, but changing aspect ratio, and then take those new points inside the scaled grid and plot them in the unscaled grid, which will then lose correct aspect ratios, which is fine.
Coding in python.
Here's a Python3 program that I think does what you want:
import math
# Use a negative angle since we're rotating clockwise
rotate_C_angle = math.radians(-60)
A = [0, 0]
B = [100, 0]
D = [100, 250]
# Create point D such that angle CAB = target_angle_CAB
target_angle_CAB = math.radians(60)
C = [B[0], B[0] * math.tan(target_angle_CAB)]
# What do we scale B by in order to get C
scale_factor = C[1] / D[1]
# Rotate C around B by rotate_C_angle
# (translate C by -Bx, then rotate, then translate back)
E = [
C[0] - C[1] * math.sin(rotate_C_angle),
C[1] * math.cos(rotate_C_angle)
]
# Scale E back
F = [E[0], E[1] / scale_factor]
This is assuming that A is at the origin. If it isn't then subtract Ax from all the x coordinates and Ay from the y coordinates, then do the calculations before translating back again.
Whether you need to scale back E to F depends on what you're doing. What happens here assumes that the scaling transformation along y does not get rotated.
I have three points a,b,c whose x,y,z coordinates are
a=[ -0.3519052 0 0];
b=[ 0 -0.674984 0];
c=[ 0 0 -0.6485047];
how do plot a plane(triangle) using these three points in scilab
plot3d and plo3d1 are not giving the form i am looking for.
I figured out the problem!
plot3d1 needs column vectors.
plot3d1(a',b',c')
produced the plot
how to plot triangle in scilab
Plot a triangle using its sides
Consider the fist vertex lies at origin(0,0)
Second vertex lies on X-Axis at (a,0)
From distance formula of triangle
ie.
length of side = sqrt( (x2−x1)^2+(y2−y1)^2 )
Program to Plot a Triangle when sides are given is as below in scilab :
clf()
//Length of the sides
a = 10;
b = 10
c = 10
//Vertex of third point
xc=(a^2-(b^2-c^2))/(2*a)
yc=sqrt(c^2-xc^2);
clf()
x=[0,0,xc]
y=[0,yc,0]
plot2d(0,0,-1,"010","",[0,0,0,0]);
xpoly(x,y,"lines",1)
I have two squares, S1 = (x1,y1,x2,y2) and S2 = (a1,b1,a2,b2)
I'm looking for the A transformation matrix with which
A * S1 = S2
As far as I see, A is an affine 3x3 matrix, so I have 9 unknown values.
How can I calculate these values?
thanks and best,
Viktor
There are really only four unknown values here. A rotation angle, a scale factor and an x and y translation. Of your three by three matrix the bottom row is always 0,0,1 which reduces you to six unknowns. The right hand column will be Tx,Ty,1 which are your translations (and the 1 we already know about).
The two by two "matrix" left will be your rotation and scaling. This will (off the top of my head) be something like:
ACos(B), -Asin(B)
ASin(B), aCos(B)
So in total:
ACos(B), -Asin(B), Tx
ASin(B), ACos(B), Ty
0 , 0 , 1
You extend your co-ordinate matrices with the 1 on the end of each co-ordinate to give 2x3 matrices and they then multiply to give you the four equations you need to solve for the four variables. That is left as an exercise for the reader.
A transformation matrix is a factor of scaling matrix Ss, transition matrix St and rotation matrix Sr.
Assume the old point is Po is (Xo,Yo) and as vector will be represented as (Xo Yo 1)' same for the new point Pn
Then Pnv =SsStSrPov
Where Sx is
Sx 0 0
0 Sy 0
0 0 1
St is
1 0 Tx
0 1 Ty
0 0 1
Sr is
Cos(th) -Sin(th) 0
Sin(th) Cos(th) 0
0 0 1
Now back to your question. if two point are giving to represent a rectangle we can just find the parameter of two matrix and the third one will be an identity matrix.
Rect1 is represented as Top-Left point P11 and Bottom-Right Point P12
Rect2 is represented as Top-Left point P21 and Bottom-Right Point P22
S=Ss*St
Sx 0 Tx
0 Sy Ty
0 0 1
Now you have 4 missing parameters and 4 set of equations
P21=S*P11
P22=S*P12
X[P21] =Sx*X[P11]+Tx
Y[P21] =Sy*Y[P11]+Ty
X[P22] =Sx*X[P12]+Tx
Y[P22] =Sy*Y[P12]+Ty
Solve it and you'll get your answer.
and if you have transition and rotation then
S=Sr*St.
Cos(th) -Sin(th) Tx
Sin(th) Cos(th) Ty
0 0 1
Now you have 3 missing parameters and 4 set of equations
P21=S*P11
P22=S*P12
X[P21] =Cos(th)*X[P11]-Sin(th)*Y[P11]+Tx
Y[P21] =Sin(th)*X[P11]+Cos(th)*Y[P11]+Ty
X[P22] =Cos(th)*X[P11]-Sin(th)*Y[P12]+Tx
Y[P22] =Sin(th)*X[P11]+Cos(th)*Y[P12]+Ty
Replace Cos(th) with A and Sin(th) With B and solve the equations.
X[P21] =A*X[P11]-B*Y[P11]+Tx
Y[P21] =B*X[P11]+A*Y[P11]+Ty
X[P22] =A*X[P11]-B*Y[P12]+Tx
Y[P22] =B*X[P11]+A*Y[P12]+Ty
Check if its correct A^2+B^2 =? 1 if is true then th = aCos(A)
The last part of the solution, if you'll have all three matrixes, then S=SrStSs is
Sx*sin(th) -Sx*cos(th) Tx
Sy*cos(th) Sy*sin(th) Ty
0 0 1
Now we have 5 missing variables and we need 6 different set of equations to solve it. which is mean 3 points from each rectangle.
You shouldn't have a 3x3 matrix if you're just looking to transform a 2D object. What you're looking for is a 2x2 matrix that solves A*S1=S2. This can be done in many different ways; in MATLAB, you'd do a S2/S1 (right matrix division), and generally this performs some kind of Gaussian elimination.
How can I calculate these values?
When applied to 2d/3d transformations, matrix can be represented a coordinate system, unless we are talking about projections.
Matrix rows (or columns, depending on notation) form axes of a new coordinate system, in which object will be placed placed if every object vertex is multiplied by the matrix. Last row (or columne, depending on notation) points to the center of the new coordinate system.
Standard OpenGL/DirectX transformation matrix (NOT a projection matrix):
class Matrix{//C++ code
public:
union{
float f[16];
float m[4][4];
};
};
Can be represented as combination of 4 vectors vx (x axis of the new coordinate system), vy(y axis of a new coordinate system), vz(z axis of a new coordinate system), and vp (center of the new system). Like this:
vx.x vx.y vx.z 0
vy.x vy.y vy.z 0
vz.x vz.y vz.z 0
vp.x vp.y vp.z 1
All "calculate rotation matrix", "calculate scale matrix", etc go down to this idea.
Thus, for 2d matrix, you'll have 3x3 matrix that consists of 3 vectors - vx, vy, vp, because there is no z vector in 2d. I.e.:
vx.x vx.y 0
vy.x vy.y 0
vp.x vp.y 1
To find a transform that would transform quad A into quad B, you need to find two transforms:
Transform that will move quad A into origin (i.e. at point zero), and convert it into quad of fixed size. Say, quad (rectangle) whose one vertex x = 0, y = 0, and whose vertices are located at (0, 1), (1, 0), (1, 1).
Transform that turns quad of fixed size into quad B.
You CANNOT do that it this way if opposite edges of quad are not parallel. I.e. parallelograms are fine, but random 4-sided polygons are not.
A quad can be represented by base point (vp) which can be any vertex of the quad and two vectors that define quad sizes (direction of the edge multiplied by edge's length). I.e. "up" vector and "side" vector. Which makes it a matrix:
side.x side.y 0
up.x up.y 0
vp.x vp.y 1
So, multiplying a quad (vp.x = 0, vp.y = 0, side.x = 1, side.y = 0, up.x = 0, up.y = 1) by this matrix will turn original quad into your quad. Which means, that in order to transform
quad A into quad B, you need to do this:
1) make a matrix that would transform "base 1unit quad" into quad A. Let's call it matA.
2) make a matrix that would transform "base 1 unit quad" into quad B. let's call it matB.
3) invert matA and store result into invMatA.
4) the result matrix is invMatA * matB.
Done. If you multiply quad A by result matrix, you'll get quad B. This won't work if quads have zero widths or heights, and it won't work if quads are not parallelograms.
This is hard to understand, but I cannot to make it simpler.
What do you mean by S1 = (x1,y1,x2,y2)?
Do they represent the top-left and bottom-right corners of the square?
Also, can you guarantee there's only rotation between the squares or do you need a full affine transformation which allows for scaling, skewing, and translation?
Or do you also need a perspective transformation?
Only if it's a perspective transformation, will you need 3x3 matrix with 8 dof as you've mentioned in your post.