I'm working with SMFL/C++ to make a 2D isometric game engine, i got this when i do the isometric calculations :
Here is my formula to calculate isometric coordinates in my 2D engine :
For I-J coordinates i have :
x = (I - J) * (tileWidth / 2);
y = (J + I) * (tileHeight / 2);
//Totally working with classics tiles
EDIT: My problem is due to my tiles' shape wich is a cube, but i don't have a clue about how to fix it. Did i really have to do somes complicated maths to handle 3D objetcs(i would rather avoid this) or i can just change the formula a little bit ?
EDIT 2: Solution : int isoY = (x + y) * (height / 4);
First if it is a 2D engine I wonder why there are 3 dimensions, why and how you use z in your engine.
Assuming you want to have a plan of tiles in isometric projection ((x,y) in pixels) given the coordinates (I,J) in number of tiles in orthographic projection.
In that case your formula for x and y are fine by me given tileWidth and tileHeight are correct (i.e. value in isometric projection). And you shouldn't have to use any z.
On the other hand if your problem is to get (x,y) pixels coordinates of a 3D object given (x,y,z) cartesian coordinates i suggest you read this: Computing the Pixel Coordinates of a 3D Point
In case i assumed wrong I'll edit or delete.
Related
I gone through some similar post for this problem like this,
#RafazZ
OpenCV unproject 2D points to 3D with known depth `Z`
2D Coordinate to 3D world coordinate
as doing 2d to 3d conversion in camera_frame they have removed R,T matrix from calculation and only using like this,
x = (u-cx)/fx
y= (v-cy)/fy
actual (X,Y,Z)
X= x * depth.at(u,v)
Y= y * depth.at(u,v)
Z = depth.at(u,v)
so, I want to understand why we can remove R,t from calculation like this when doing 2d-->3d in camera_frame?
I render a 3D mesh model using OpenGL with perspective camera – gluPerspective(fov, aspect, near, far).
Then I use rendered image in a computer vision algorithm.
At some point that algorithm requires camera matrix K (along with several vertices on the model and their corresponding projections) in order to estimate camera position: rotation matrix R and translation vector t. I can estimate R and t by using any algorithm which solves Perspective-n-Point problem.
I construct K from the OpenGL projection matrix (see how here)
K = [fX, 0, pX | 0, fY, pY | 0, 0, 1]
If I want to project a model point 'by hand' I can compute:
X_proj = K*(R*X_model + t)
x_pixel = X_proj[1] / X_proj[3]
y_pixel = X_proj[2] / X_proj[3]
Anyway, I pass this camera matrix in a PnP algorithm and it works just fine.
But then I had to change perspective projection to orthographic one.
As far as I understand when using orthographic projection the camera matrix becomes:
K = [1, 0, 0 | 0, 1, 0 | 0, 0, 0]
So I changed gluPerspective to glOrtho. Following the same way I constructed K from OpenGL projection matrix, and it turned out that fX and fY are not ones but 0.0037371. Is this a scaled orthographic projection or what?
Moreover, in order to project model vertices 'by hand' I managed to do the following:
X_proj = K*(R*X_model + t)
x_pixel = X_proj[1] + width / 2
y_pixel = X_proj[2] + height / 2
Which is not what I expected (that plus width and hight divided by 2 seems strange...). I tried to pass this camera matrix to POSIT algorithm to estimate R and t, and it doesn't converge. :(
So here are my questions:
How to get orthographic camera matrix from OpenGL?
If the way I did it is correct then is it true orthographic? Why POSIT doesn't work?
Orthographic projection will not use the depth to scale down farther points. Though, it will scale the points to fit inside the NDC which means it will scale the values to fit inside the range [-1,1].
This matrix from Wikipedia shows what this means:
So, it is correct to have numbers other than 1.
For your way of computing by hand, I believe it's not scaling back to screen coordinates and that makes it wrong. As I said, the output of projection matrices will be in the range [-1,1], and if you want to get the pixel coordinates, I believe you should do something similar to this:
X_proj = K*(R*X_model + t)
x_pixel = X_proj[1]*width/2 + width / 2
y_pixel = X_proj[2]*height/2 + height / 2
Anyway, I think you'd be better if you used modern OpenGL with libraries like GLM. In this case, you have the exact projection matrices used at hand.
I am writing software to determine the viewable locations of a camera in 3D. I have currently implement parts to find the minimum and maximum length of view based on the camera and lenses intrinsic characteristics.
I now need to work out that if the camera is placed at X,Y,Z and is pointing in a direction (two angles, one around the horizontal and one around the vertical axis) what the boundaries the camera can see at are (knowing the viewing angle). The output I would like is 4 3D locations, making a rectangle that show the minimum position, top left, top right, bottom left and bottom right. The same is also required for the maximum positions.
Can anyone help with the geometry to find these points?
Some code I have:
QVector3D CameraPerspective::GetUnitVectorOfCameraAngle()
{
QVector3D inital(0, 1, 0);
QMatrix4x4 rotation_matrix;
// rotate around z axis
rotation_matrix.rotate(_angle_around_z, 0, 0, 1);
//rotate around y axis
rotation_matrix.rotate(_angle_around_x, 1, 0, 0);
inital = inital * rotation_matrix;
return inital;
}
Coordinate CameraPerspective::GetFurthestPointInFront()
{
QVector3D camera_angle_vector = GetUnitVectorOfCameraAngle();
camera_angle_vector.normalize();
QVector3D furthest_point_infront = camera_angle_vector * _camera_information._maximum_distance_mm;
return Coordinate(furthest_point_infront + _position_of_this);
}
Thanks
A complete answer with code will be probably way too long for SO, I hope that this will be enough. In the following we work in homogeneous coordinates.
I have currently implement parts to find the minimum and maximum length of view based on the camera and lenses intrinsic characteristics.
That isn't enough to fully define your camera. You also need a field of view angle and the width/height ratio.
With all these information (near plane + far plane + fov + ratio), you can build a 4x4 matrix known as perspective matrix. Google for it or check here for some references. This matrix maps the pyramidal region of the space which your camera "sees" (usually simply called frustrum) to the [-1,1]x[-1,1]x[-1,1] cube. Call it P.
Now you need a 4x4 camera matrix which transform points in world space to points in camera space. Since you know the camera position and the camera orientation this can be constructed easily (there is no room here to full explain how transformation matrices in homogeneous coordinates work, google for it). Call this matrix C.
Now consider the matrix A = P * C.
This matrix transforms points in world coordinates to points in the perspective space. Your camera will "see" those points if they are inside the [-1,1]x[-1,1]x[-1,1] cube. But you can invert this matrix in order to map points inside the cube to points in world space. So in order to obtain the 8 points you need in world space you can simply do:
y = A^(-1) * x
Where x =
[-1,-1,-1, 1] left - bottom - near
[-1,-1, 1, 1] left - bottom - far
etc.
I wish to generate rays from the camera through the viewing plane. In order to do this, I need my camera position ("eye"), the up, right, and towards vectors (where towards is the vector from the camera in the direction of the object that the camera is looking at) and P, the point on the viewing plane. Once I have these, the ray that's generated is:
ray = camera_eye + t*(P-camera_eye);
where t is the distance along the ray (assume t = 1 for now).
My question is, how do I obtain the 3D coordinates of point P given that it is located at position (i,j) on the viewing plane? Assume that the upper left and lower right corners of the viewing plane are given.
NOTE: The viewing plane is not actually a plane in the sense that it doesn't extend infinitely in all directions. Rather, one may think of this plane as a widthxheight image. In the x direction, the range is 0-->width and in the y direction the range is 0-->height. I wish to find the 3D coordinate of the (i,j)th element, 0
General solution of the itnersection of a line and a plane see http://local.wasp.uwa.edu.au/~pbourke/geometry/planeline/
Your particular graphics lib (OpenGL/DirectcX etc) may have an standard way to do this
edit: You are trying to find the 3d intersection of a screen point (eg a mouse cursor) with a 3d object in you scene?
To work out P, you need the distance from the camera to the near clipping plane (the screen), the size of the window on the near clipping plane (or the view angle, you can work out the window size from the view angle) and the size of the rendered window.
Scale the screen position to the range -1 < x < +1 and -1 < y < +1 where +1 is the top/right and -1 is the bottom/left
Scale normalised x,y by the view window size
Scale by the right and up vectors of the camera and sum the results
Add the look at vector scaled by the clipping plane distance
In effect, you get:
p = at * near_clip_dist + x * right + y * up
where x and y are:
x = (screen_x - screen_centre_x) / (width / 2) * view_width
y = (screen_y - screen_centre_y) / (height / 2) * view_height
When I directly plugged in suggested formulas into my program, I didn't obtain correct results (maybe some debugging needed to be done). My initial problem seemed to be in the misunderstanding of the (x,y,z) coordinates of the interpolating corner points. I was treating x,y,z-coordinates separately, where I should not (and this may be specific to the application, since the camera can be oriented in any direction). Instead, the solution turned out to be a simple interpolation of the corner points of the viewing plane:
interpolate the bottom corner points in the i direction to get P1
interpolate the top corner points in the i direction to get P2
interpolate P1 and P2 in the j direction to get the world coordinates of the final point
(This is all in ortho mode, origin is in the top left corner, x is positive to the right, y is positive down the y axis)
I have a rectangle in world space, which can have a rotation m_rotation (in degrees).
I can work with the rectangle fine, it rotates, scales, everything you could want it to do.
The part that I am getting really confused on is calculating the rectangles world coordinates from its local coordinates.
I've been trying to use the formula:
x' = x*cos(t) - y*sin(t)
y' = x*sin(t) + y*cos(t)
where (x, y) are the original points,
(x', y') are the rotated coordinates,
and t is the angle measured in radians
from the x-axis. The rotation is
counter-clockwise as written.
-credits duffymo
I tried implementing the formula like this:
//GLfloat Ax = getLocalVertices()[BOTTOM_LEFT].x * cosf(DEG_TO_RAD( m_orientation )) - getLocalVertices()[BOTTOM_LEFT].y * sinf(DEG_TO_RAD( m_orientation ));
//GLfloat Ay = getLocalVertices()[BOTTOM_LEFT].x * sinf(DEG_TO_RAD( m_orientation )) + getLocalVertices()[BOTTOM_LEFT].y * cosf(DEG_TO_RAD( m_orientation ));
//Vector3D BL = Vector3D(Ax,Ay,0);
I create a vector to the translated point, store it in the rectangles world_vertice member variable. That's fine. However, in my main draw loop, I draw a line from (0,0,0) to the vector BL, and it seems as if the line is going in a circle from the point on the rectangle (the rectangles bottom left corner) around the origin of the world coordinates.
Basically, as m_orientation gets bigger it draws a huge circle around the (0,0,0) world coordinate system origin. edit: when m_orientation = 360, it gets set back to 0.
I feel like I am doing this part wrong:
and t is the angle measured in radians
from the x-axis.
Possibly I am not supposed to use m_orientation (the rectangles rotation angle) in this formula?
Thanks!
edit: the reason I am doing this is for collision detection. I need to know where the coordinates of the rectangles (soon to be rigid bodies) lie in the world coordinate place for collision detection.
What you do is rotation [ special linear transformation] of a vector with angle Q on 2d.It keeps vector length and change its direction around the origin.
[linear transformation : additive L(m + n) = L(m) + L(n) where {m, n} € vector , homogeneous L(k.m) = k.L(m) where m € vector and k € scalar ] So:
You divide your vector into two pieces. Like m[1, 0] + n[0, 1] = your vector.
Then as you see in the image, rotation is made on these two pieces, after that your vector take
the form:
m[cosQ, sinQ] + n[-sinQ, cosQ] = [mcosQ - nsinQ, msinQ + ncosQ]
you can also look at Wiki Rotation
If you try to obtain eye coordinates corresponding to your object coordinates, you should multiply your object coordinates by model-view matrix in opengl.
For M => model view matrix and transpose of [x y z w] is your object coordinates you do:
M[x y z w]T = Eye Coordinate of [x y z w]T
This seems to be overcomplicating things somewhat: typically you would store an object's world position and orientation separately from its set of own local coordinates. Rotating the object is done in model space and therefore the position is unchanged. The world position of each coordinate is the same whether you do a rotation or not - add the world position to the local position to translate the local coordinates to world space.
Any rotation occurs around a specific origin, and the typical sin/cos formula presumes (0,0) is your origin. If the coordinate system in use doesn't currently have (0,0) as the origin, you must translate it to one that does, perform the rotation, then transform back. Usually model space is defined so that (0,0) is the origin for the model, making this step trivial.