Suppose, I have a projection plane of length 2 like the below image:
If I divide the projection plane into a two dimensional
grid of screen_width x screen_height (720*480), then each pixel will be (2/720) distance apart in the x axis and (2/480) distance apart in the y axis and If I add x+(2/720) with my current co-ordinate, keeping the y constant I will move 1 pixel to the right side. Am I right?
Related
I have a point on a sphere that needs to be rotated. I have 3 different degrees of rotation (roll, pitch, yaw). Are there any formulas I could use to calculate where the point would end up after applying each rotation? For simplicity sake, the sphere can be centered on the origin if that helps.
I've tried looking at different ways of rotation, but nothing quite matches what I am looking for. If I needed to just rotate the sphere, I could do that, but I need to know the position of a point based on the rotation of the sphere.
Using Unity for an example, this is outside of unity in a separate project so using their library is not possible:
If the original point is at (1, 0, 0)
And the sphere then gets rotated by [45, 30, 15]:
What is the new (x, y, z) of the point?
If you have a given rotation as a Quaternion q, then you can rotate your point (Vector3) p like this:
Vector3 pRotated = q * p;
And if you have your rotation in Euler Angles then you can always convert it to a Quaternion like this (where x, y and z are the rotations in degrees around those axes):
Quaternion q = Quaternion.Euler(x,y,z);
Note that Unity's euler angles are defined so that first the object is rotated around the z axis, then around the x axis and finally around the y axis - and that these axes are all the in the space of the parent transform, if any (not the object's local axes, which will move with each rotation).
So I suppose that the z-axis would be roll, the x-axis would be pitch and the y axis would be yaw.You might have to switch the signs on some axes to match the expected result - for example, a positive x rotation will tilt the object downwards (assuming that the object's notion of forward is in its positive z direction and that up is in its positive y direction).
I have three non-colinear 3D points, let's say pt1, pt2, pt3. I've computed the plane P using the sympy.Plane. How can I find the orientation of this plane(P) i.e. RPY(euler angles) or in quaternion?
I never used sympy, but you should be able to find a function to get the angle between 2 vectors (your normal vector and the world Y axis.)
theta = yaxis.angle_between(P.normal_vector)
then get the rotation axis, which is the normalized cross product of those same vectors.
axis = yaxis.cross(P.normal_vector).normal()
Then construct a quaternion from the axis and angle
q = Quaternion.from_axis_angle(axis, theta)
I have a question regarding the projection of an image over a set of 3D points. The image is given to me as a JPG, together with position and attitude information of the camera relative to a cartesian coordinate system (Xc,Yc,Zc and yaw, pitch, roll), as well as the horizontal and vertical field of view (in degrees).
Points are given using solely their 3d position in the same coordinate system (Xp,Yp,Zp).
In my coordinate system, Z is up. To project the image onto the points, I
compute the vector from camera to each point
Vector3 c2p = (Xp,Yp,Zp)-(Xc,Yc,Zc);
rotate c2p according to my camera's attitude (quaternion):
Vector3 c2pCamFrame = getCamQuaternion().conjugate().rotate(c2p);
compute azimuth and elevation from the camera's "center ray" to the point:
float azimuth = atan2(c2pCamFrame.x(),c2pCamFrame.y()));
float elevation = atan2(c2pCamFrame.z(),sqrt(pow(c2pCamFrame.x(),2)+pow(c2pCamFrame.y(),2)));
if azimuth and elevation are within the field of view, I assign the color of the corresponding pixel to the point.
This works almost perfectly, and the "almost" motivates my question. Let me show you:
I cannot figure out why the elevation of the projection is distorted. In the bottom right of the image, you can see that points outside the frustum (exceeding the elevation) actually become colored - and this distortion is null at an azimuth of 0 degrees and peaks at the left and right edges of the image, creating the pillow distortion.
Why does this distortion appear? I'd love to understand this problem both in geometrical as well as mathematical terms. Thank you!
The field of view angles are only valid on the principal axes. But you can do it the other way around. I.e. calculate the x/y bounds from the angles:
maxX = tan(horizontal_fov / 2)
maxY = tan(vertical_fov / 2)
And check
if(abs(c2pCamFrame.x() / c2pCamFrame.z()) <= maxX
&& abs(c2pCamFrame.y() / c2pCamFrame.z()) <= maxY)
Additionally, you might want to check if the points are in front of the camera:
... && c2pCamFrame.z() > 0
This assumes a left-handed coordinate system.
Sorry.I have a homework but I cannot do.I have to write code for between cube and ellipse'volume with Monte Marlo Method.Can you help me? Dimension for cube is 10 and radius for ellipse is 5 and height is 3.
Thank you
I wrote this codes for the intersection of a cylinder and a sphere but I cannot translate for ellipse and cube
%clear the workspace and command window
clc
clear
% find the volume of the intersection of a cylinder and a sphere
RunLength=10000; % run length
Count=0;
%initialize the empty arrays
px=[];
py=[];
pz=[];
qx=[];
qy=[];
qz=[];
d=0.2
%try 10000 points
for i=1:RunLength
x=2*rand()-1; %x coordinate random number between (-1,1)
y=2*rand()-1; %y coordinate random number between (-1,1)
z=2*rand()-1; %z coordinate random number between (-1,1)
if(x^2+y^2+z^2<=1); %if the point is in the big sphere (x^2+y^2+z^2<=1^2)
if (y^2+(z-(1-0.5)^2)>=(0.5)^2);
qx(end+1)=x; %add this point's x coordinate to the qx array
qy(end+1)=y; %add this point's y coordinate to the qy array
qz(end+1)=z; %add this point's z coordinate to the qz array
Count=Count+1; %increase the number of points in the shape by 1
end
else
px(end+1)=x; %add this point's x coordinate to the px array
py(end+1)=y; %add this point's y coordinate to the py array
pz(end+1)=z; %add this point's z coordinate to the pz array
end
end
%ratio is the number of points in the shape/total number of points
disp(8*Count/RunLength); %multiply the ratio with 8 and display
%use plot3 command because it is a 3D plot and use ‘ro’ for showing points as
%red circles in the plot
%use plot3 command because it is a 3D plot and use ‘ro’ for showing points as
red circles in the plot
plot3(qx,qy,qz,'ro');
I have two normalized vectors:
A) 0,0,-1
B) 0.559055,0.503937,0.653543
I want to know, what rotations about the axes would it take to take the vector at 0,0,-1 to 0.559055,0.503937,0.653543?
How would I calculate this? Something like, rotate over X axis 40 degrees and Y axis 220 (that's just example, but I don't know how to do it).
Check this out. (google is a good thing)
This calculates the angle between two vectors.
If Vector A is (ax, ay, az) and
Vector B is (bx, by, bz), then
The cos of angle between them is:
(ax*bx + ay*by + az*bz)
--------------------------------------------------------
sqrt(ax*ax + ay*ay + az*az) * sqrt(bx*bx + by*by + bz*bz)
To calculate the angle between the two vectors as projected onto the x-y plane, just ignore the z-coordinates.
Cosine of Angle in x-y plane =
(ax*bx + ay*by)
--------------------------------------
sqrt(ax*ax + ay*ay) * sqrt(bx*bx + by*by
Similarly, to calculate the angle between the projections of the two vectors in the x-z plane, ignore the y-coordinates.
It sounds like you're trying convert from Cartesian coordinates (x,y,z) into spherical coordinates (rho,theta,psi).
Since they're both unit vectors, rho, the radius, will be 1. This means your magnitudes will also be 1 and you can skip the whole denominator and just use the dot-product.
Rotating in the X/Y plane (about the Z axis) will be very difficult with your first example (0,0,-1) because it has no extension in X or Y. So there's nothing to rotate.
(0,0,-1) is 90 degrees from (1,0,0) or (0,1,0). If you take the x-axis to be the 0-angle for theta, then you calculate the phi (rotation off of the X/Y plane) by applying the inverse cos upon (x,y,z) and (x,y,0), then you can skip dot-products and get theta (the x/y rotation) with atan2(y,x).
Beware of gimbal lock which may cause problems.