I'm working with libGDX particle effect system, and I have a following need: to reverse an effect in time. These effects look like scattering particles, appearing in a single spot, and what I need is exactly the opposite - particles appearing in random places around the central spot and gathering in the central spot. It all is the sprite animation, and it could be reversed, but how?
I think you can because the particle editor allows negative velocity plus I've done it in the particle editor.
1 set the velocity to -300 low to +300 high (or whichever)
2 particle size is zero to begin with (invisible), this allows the particle to move away from origin at appear at a radius
3 at the desired radiues, when you arrive, restore the size (become visible).
4 this point is when the velocity reverses heading back to the origin.
The point is that negative or positive velocity both head away from the origin, but as you interpolate between the two you reverse the course. As long as you are invisible (make the particle 0 size) then you appear at the radius and travel inwards.
Related
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.
I'm wanting to create a physics engine within Java. However it's not the code I'm bothered about. It's simply the math of rigid body physics, specifically forces and how they affect the rotation of an object.
Let's say for example that I have a square with same length sides. The square will be accelerating towards ground level due to gravity (no air resistance). This would mean that there would be a vector force of (0,-9.8)m/s on every point in the square.
Now let's say that this square is rotated slightly. When this rotated square comes into contact with the ground (a flat surface) there will be an impulse velocity vector at the point of contact (most likely a corner of the square). However, what happens to the forces of the other corners on the square? From the original force of gravity, how are they affected?
I apologize if my question isn't detailed enough. I'd love to upload a diagram but I don't yet have the reputation.
rotation is form of kinetic energy
first the analogy to movement
alpha - angular position [rad]
omega - angular speed [rad/s]
epsilon - angular acceleration [rad/s^2]
alpha(t)/(dt^2)=omega(t)/dt=epsilon(t)
now the inertia
I - quadratic rotation mass inertia [kg.m^2]
m - mass [kg]
M - torque [N.m]
and some equations to be exploited
M=epsilon*I - torque needed to achieve acceleration or vice versa [N.m]
acc=epsilon*radius - perimeter acceleration [m/s^2]
vel=omega*radius - perimeter speed [m/s^2]
equation #1 can be used to directly compute the force. Equations #2,#3 can be used to calculate friction based forces like wheels grip/drag. Do not forget about the kinetic energy Ek=0.5*m*vel^2+0.5*I*omega^2 so you can exploit the law of preserving energy.
During continuous contact of object1 with object2 in rotation happens this
Perimeter speed/acceleration create interaction force, this is slowing down the rotation of object2 creating drag force on the object2 and reacting force on the object1.
if object1 is not fixed then this force also create torque and rotates the object1
If the rotation is forced to stop suddenly then all rotational part of kinetic energy is moved to the collision reaction Force impulse.
If object is in more complicated rotation motion then you should compute the actual rotation axis and alpha,omega,epsilon and use that because object can rotate with more rotations each with different center of rotation.
Also if object is rotating and another rotation is applied in different axis then this creates gyroscopic torque creating also rotation in the third axis perpendicular to both.
So when yo put all these together you have a idea of what structures you need. Sorry can not be more specific than this without further info about the structures and properties of your simulation ...
Applied forces do not play a role in the calculation of contact impulses because the impulses are said to occur on a time scale much smaller than the simulation time step. Basically the change is velocity during an impact because of gravity or other forces is negligible.
If I understand correctly, you worry about the different corners of the square - one with an impact, three without.
However, since you want to do rigid body dynamics, it is more helpful to think about the rigid body as having a center of mass (in this case, the square's center), a position, a rotation, and a geometry (in this case the square, but it could be anything).
The corners of the vertices are in constant position and rotation with regards to the center of mass - it's only the rigid body's position and rotation which change all four corners position in the world at once. An advantage of this view is that it is independent of the geometry - you could have 10 or 20 corners, and the approach would be the same.
With regard to computing the rotation:
Gravity is working as before. However, you now have another force (from the impulse over the time it acts) - and you have to add the effects of the two in order to get the complete outcome of the system.
The impulse will be due to one of the corners being in collision in the case you describe. It has to be computed at the contact point, with a contact normal - in this case the normal of the flat surface.
If the normal points in a different direction than the center of mass, this will lead to a rotation (as well as a position change).
The amount of the position change is due to how you model the contact computation and resolution, material properties, numerical stepper, impact velocity, time step, ...
As others mentioned, reading up on physics (rigid body dynamics) and physics simulations might be a good starting point to understand the concepts better.
Currently, I'm taking each corner of my object's bounding box and converting it to Normalized Device Coordinates (NDC) and I keep track of the maximum and minimum NDC. I then calculate the middle of the NDC, find it in the world and have my camera look at it.
<Determine max and minimum NDCs>
centerX = (maxX + minX) / 2;
centerY = (maxY + minY) / 2;
point.set(centerX, centerY, 0);
projector.unprojectVector(point, camera);
direction = point.sub(camera.position).normalize();
point = camera.position.clone().add(direction.multiplyScalar(distance));
camera.lookAt(point);
camera.updateMatrixWorld();
This is an approximate method correct? I have seen it suggested in a few places. I ask because every time I center my object the min and max NDCs should be equal when their are calculated again (before any other change is made) but they are not. I get close but not equal numbers (ignoring the negative sign) and as I step closer and closer the 'error' between the numbers grows bigger and bigger. IE the error for the first few centers are: 0.0022566539084770687, 0.00541687811360958, 0.011035676399427596, 0.025670088917273515, 0.06396864345885889, and so on.
Is there a step I'm missing that would cause this?
I'm using this code as part of a while loop to maximize and center the object on screen. (I'm programing it so that the user can enter a heading an elevation and the camera will be positioned so that it's viewing the object at that heading and elevation. After a few weeks I've determined that (for now) it's easier to do it this way.)
However, this seems to start falling apart the closer I move the camera to my object. For example, after a few iterations my max X NDC is 0.9989318709122867 and my min X NDC is -0.9552042384799428. When I look at the calculated point though, I look too far right and on my next iteration my max X NDC is 0.9420058636660581 and my min X NDC is 1.0128126740876888.
Your approach to this problem is incorrect. Rather than thinking about this in terms of screen coordinates, think about it terms of the scene.
You need to work out how much the camera needs to move so that a ray from it hits the centre of the object. Imagine you are standing in a field and opposite you are two people Alex and Burt, Burt is standing 2 meters to the right of Alex. You are currently looking directly at Alex but want to look at Burt without turning. If you know the distance and direction between them, 2 meters and to the right. You merely need to move that distance and direction, i.e. right and 2 meters.
In a mathematical context you need to do the following:
Get the centre of the object you are focusing on in 3d space, and then project a plane parallel to your camera, i.e. a tangent to the direction the camera is facing, which sits on that point.
Next from your camera raycast to the plane in the direction the camera is facing, the resultant difference between the centre point of the object and the point you hit the plane from the camera is the amount you need to move the camera. This should work irrespective of the direction or position of the camera and object.
You are playing the what came first problem. The chicken or the egg. Every time you change the camera attributes you are effectively changing where your object is projected in NDC space. So even though you think you are getting close, you will never get there.
Look at the problem from a different angle. Place your camera somewhere and try to make it as canonical as possible (ie give it a 1 aspect ratio) and place your object around the cameras z-axis. Is this not possible?
I would like to draw an artificial horizon. The center of the view would represent perfectly horizontal view with roll rotating the horizontal line and pitch moving it up or down.
The question is: what is the correct calculation to translate the horizon line up or down (pitch) given the pitch angle.
My guess is that this would probably depend on the FOV angle that one would assume for an assumed camera, so this angle would need to be a factor in the algorithm sought. Ideally I would figure out this angle for the iPhone/iPad camera so that the artificial horizon would line up with the actual horizon if you hold the device in front of you and look towards the horizon.
Until now I've been guesstimating the offset, but I would like to have the exact formula.
Try horizon_offset/(screen_height/2)=tan(pitch)/tan(vertical_FOV/2).
Look at the picture, and the formula derives itself.
(source: zwibbler.com)
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Update I have two angles mixed up. One is the FOV angle of the camera, the other is the viewing angle of the screen. These are two different things. The latter depends on the viewing distance. You probably have to estimate this distance, and adjust magnification and/or focal distance such that objects visible on the screen are the same angular size as the same objects visible with the naked eye. (With my particular phone, you would need to magnify the image by an additional factor of about 3 after the 5x zoom, if the user stretches his hand with the phone all the way forward). Then the two angles are the same, and the formula works.
If you want to introduce magnification (i.e. objects on the screen have different sizes from their real-life counterparts), multiply the horizon offset by the magnification factor.
Update 2 When taking the viewing distance into account, the screen size cancels out, and the offset simply becomes viewing_distance*tan(pitch_angle) (with unit magnification).
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.