Given the 3D vector of the direction that the camera is facing and the orientation/direction vector of a 3D object in the 3D space, how can I calculate the 2-dimensional slope that the mouse pointer must follow on the screen in order to visually be moving along the direction of said object?
Basically I'd like to be able to click on an arrow and make it move back and forth by dragging it, but only if the mouse pointer drags (roughly) along the length of the arrow, i.e. in the direction that it's pointing to.
thank you
I'm not sure I 100% understand your question. Would you mind posting a diagram?
You might find these of interest. I answered previous questions to calculate a local X Y Z axis given a camera direction (look at) vector, and also a question to translate an object in a plane parallel to the camera.
Both of these examples use Vector dot product, Vector cross product to compute the required vectors. In your example the vector dot product can be also used to output the angle between two vectors once you have found them.
It depends to an extent on the transformation that you are using to convert your 3d real world coordinates to 2d screen coordinates, e.g. perspective, isometric, etc... You will typically have a forward (3d -> 2d) and backward (2d -> 3d) transformation in play, where the backward transformation loses information. (i.e. going forward each 3d point will map to a unique 2d point, but going back from the point may not yield the same 3d point). You can often project the mouse point onto the object to get the missing dimension.
For mouse dragging, you typically get the user to specify an operation (translation on the plane of projection, zooming in or out, or rotating about an anchor point). Your input is the mouse coordinate at the start and end of the drag, which you transform into your 3d coordinate system to get two 3d coordinates, which will give you dx, dy, dz for dragging / translation etc...
Related
for a personal project, I've created a simple 3D engine in python using as little libraries as possible. I did what I wanted - I am able to render simple polygons, and have a movable camera. However, there is a problem:
I implemented a simple flat shader, but in order for it to work, I need to know the camera location (the camera is my light source). However, the problem is that I have no way of knowing the camera's location in the world space. At any point, I am able to display my view matrix, but I am unsure about how to extract the camera's location from it, especially after I rotate the camera. Here is a screenshot of my engine with the view matrix. The camera has not been rotated yet and it is very simple to extract its location (0, 1, 4).
However, upon moving the camera to a point between the X and Z axes and pointing it upwards (and staying at the same height), the view matrix changes to this:
It is obvious now that the last column cannot be taken directly to determine the camera location (it should be something like (4,1,4) on the last picture).
I have tried a lot of math, but I can't figure out the way to determine the camera x,y,z location from the view matrix. I will appreciate any and all help in solving this, as it seems to be a simple problem, yet whose solution eludes me. Thank you.
EDIT:
I was advised to transform a vertex (0,0,0,1) by my view matrix. This, however, does not work. See the example (the vertex obviously is not located at the printed coordinates):
Just take the transform of the vector (0,0,0,1) with the modelview matrix: Which is simply the rightmost column of the modelview matrix.
EDIT: #ampersander: I wonder why you're trying to work with the camera location in the first place, if you assume the source of illumination to be located at the camera's position. In that case, just be aware, that in OpenGL there is no such thing as a camera, and in fact, what the "view" transform does, is move everything in the world around so that where you assume your camera to be ends up at the coordinate origin (0,0,0).
Or in other words: After the modelview transform, the transformed vertex position is in fact the vector from the camera to the vertex, in view space. Which means that for your assumed illumination calculation the direction toward the light source, is the negative vertex position. Take that, normalize it to unit length and stick it into the illumination term.
I'm working on my raytracer and it seems I can't manage to handle the case where the direction vector of my camera is parallel to the vector (0,1,0).
I think it is linked to my way to compute the vector up and right for camera but I can't manage to find a work around.
Here is how I do it:
cam_up = vector_cross(cam_dir, {0, 1, 0});
camp_right = vector_cross(cam_right, cam_dir);
Can somebody enlighten me?
You have the correct formula for calculation of an orthogonal axis from a single cameraOut vector. However, as has been stated this formula will not account for the camera roll, which could be any direction in the plane perpendicular to the camera direction. This will be apparent when moving a camera across the pole (y-axis) as there will be undesireable behavior (yes it will be correctly aimed, but no doubt the roll won't be desired).
For more information, look into gimbal lock.
The roll itself is not really incorrect, however in reality for this camera transition to be smooth and appear correct (rather than suddenly flip or spin as it's direction becomes 0,1,0), you need to correct any roll incurred. This is a rotation about the cameraOut axis and ideally should be relative to the previous cameraAlong. This means in order to maintain the correct roll (or perceived correct roll) you need to consider the camera POSE (position and orientation) from the previous frame and ensure the roll is mitigated. Of course, if the camera doesn't move (i.e. your rendering a frame with a static camera position) you do not have a previous camera state so the position cannot be calculated and instead must be explicitly defined as part of the scene definition.
Personally I store an entire orthogonal axis for a camera so the orientation and roll is always clearly defined. This is only for completeness, to be honest you don't need to store the entire axis, 2 vectors cameraOut and cameraAlong (the third one being cameraUp) are enough. cameraAlong is dependant on the handed-ness of your coordinate system (e.g. for initial camera position say position (0,0,0) in left hand coordinate system, the cameraAlong direction will be in the right direction in relation to the viewer, for right hand system the cameraAlong would be the other way around. The cameraUp and cameraOut would are the same in both coordinate systems).
Hope this helps.
P.S This isn't ray tracing specific and the same principles apply for OpenGL/DirectX or any 3D representation.
Could someone explain to me how FindPlane works? (I understand the inputs, and the outputs, but not the process.) I am getting random values for the output and therefore I do not understand how it actually functions: does it raycast a normal vector from my camera according to my touch position and gets the depth point that hit the raycast and gets a plane out of that?
Operation is similar to raycast, but other way round. When you click any point on screen, screen coordinates are recorded. All 3D points in Pointcloud are projected onto image plane using camera intrinsic. Points which are close to screen coordinates are taken. RANSAC method is used to extract plane information from those points. SVD can also be used to extract plane normal from inliers obtained from RANSAC. This method should be used only once per frame transformation operation is applied on all points in pointcloud.
This method gives random values in cases where Sparse Pointcloud, reflections in 3D point cloud, reflective surfaces, Cluttered 3D space, IR from outside etc.,
I'm a fresh in cocos3d, now I have a problem.
In cocos3d, I want to rotate a node. I got the angles in x axis, y axis, z axis, then I used the property:rotation to rotate, like this:
theNodeToBeRotated.rotation = cc3v(x,y,z);
But I found out it didn't rotate as I expected, because the document said the rotate order is y-x-z.
I want to change the order to x-y-z. Can anyone let me know how?
You might need to clarify further regarding the following: "it didn't rotate as I expected"
OpenGL ES (and ergo, cocos3D) uses the y-axis as up so the rotation order is still x-y-z. If you are importing a model, you then need to take into account the 3D editor's co-ordinate system and adapt accordingly.
If you are not used to working with three-dimensional representations, the leap from 2D to 3D can be a significant hurdle. Within Cocos3D:
the x-axis is positive on the right and negative on the left
the y-axis is positive upwards and negative downwards
the z-axis is positive moving towards you and negative moving away from you
Envisage those three lines of axis, or even better, a piece of string.
If you are rotating around the x-axis, hold the string horizontally from left to right: the object would rotating towards you or away from you.
If you are rotating around the y-axis, hold the string vertically from feet to head: the object would rotate as if like a revolving door.
If you are rotating around the z-axis, hold one end close to your chest and the other end as far away as possible: the object would rotate similar to a clock face.
-- Update
I heavily wouldn't recommend changing the rotation order as it is the OpenGL standard to use Y-X-Z. If you wish to modify it, take a look at CC3GLMatrixMath and look for kmMat4RotationYXZ - there is also kmMat4RotationZYX. If you want to have X-Y-Z, you would need to construct your own rotation matrix and update accordingly in CC3GLMatrix and CC3GLMatrixMath.
As a reference, you also have the OpenGL Red book - it should have some suggestions for you.
I'd like to implement a dragging feature where users can drag objects around the workspace. That of course is the easy bit. The hard bit is to try and make it a physically correct drag which incorporates rotation due to torque moments (imagine dragging a book around on a table using only one finger, how does it rotate as you drag?).
Does anyone know where I can find explanations on how to code this (2D only, rectangles only, no friction required)?
Much obliged,
David
EDIT:
I wrote a small app (with clearly erroneous behaviour) that I hope will convey what I'm looking for much better than words could. C# (VS 2008) source and compiled exe here
EDIT 2:
Adjusted the example project to give acceptable behaviour. New source (and compiled exe) is available here. Written in C# 2008. I provide this code free of any copyright, feel free to use/modify/whatever. No need to inform me or mention me.
Torque is just the applied force projected perpendicular to a vector between the point where the force is applied and the centroid of the object. So, if you pull perpendicular to the diameter, the torque is equal to the applied force. If you pull directly away from the centroid, the torque is zero.
You'd typically want to do this by modeling a spring connecting the original mouse-down point to the current position of the mouse (in object-local coordinates). Using a spring and some friction smooths out the motions of the mouse a bit.
I've heard good things about Chipmunk as a 2D physics package:
http://code.google.com/p/chipmunk-physics/
Okay, It's getting late, and I need to sleep. But here are some starting points. You can either do all the calculations in one coordinate space, or you can define a coordinate space per object. In most animation systems, people use coordinate spaces per object, and use transformation matrices to convert, because it makes the math easier.
The basic sequence of calculations is:
On mouse-down, you do your hit-test,
and store the coordinates of the
event (in the object coordinate
space).
When the mouse moves, you create a
vector representing the distance
moved.
The force exterted by the spring is k * M, where M is the amount of distance between that initial mouse-down point from step 1, and the current mouse position. k is the spring constant of the spring.
Project that vector onto two direction vectors, starting from the initial mouse-down point. One direction is towards the center of the object, the other is 90 degrees from that.
The force projected towards the center of the object will move it towards the mouse cursor, and the other force is the torque around the axis. How much the object accelerates is dependent on its mass, and the rotational acceleration is dependent on angular momentum.
The friction and viscosity of the medium the object is moving in causes drag, which simply reduces the motion of the object over time.
Or, maybe you just want to fake it. In that case, just store the (x,y) location of the rectangle, and its current rotation, phi. Then, do this:
Capture the mouse-down location in world coordinates
When the mouse moves, move the box according to the change in mouse position
Calculate the angle between the mouse and the center of the object (atan2 is useful here), and between the center of the object and the initial mouse-down point. Add the difference between the two angles to the rotation of the rectangle.
This would seem to be a basic physics problem.
You would need to know where the click, and that will tell you if they are pushing or pulling, so, though you are doing this in 2D, your calculations will need to be in 3D, and your awareness of where they clicked will be in 3D.
Each item will have properties, such as mass, and perhaps information for air resistance, since the air will help to provide the motion.
You will also need to react differently based on how fast the user is moving the mouse.
So, they may be able to move the 2 ton weight faster than is possible, and you will just need to adapt to that, as the user will not be happy if the object being dragged is slower than the mouse pointer.
Which language?
Here's a bunch of 2d transforms in C