I´m trying to replicate the ego-motion compensation done by the article "Dynamic obstacle avoidance for quadrotors with event cameras". I have the Imu data (only 3 angular velocities rad/sec: gx, gy, gz) and the events data (timestamp, coordinate x, coordinate y, polarity).
As can be seen in the first paragraph of the image, they warped the events by w=(ti-t0). My Question is: What is the equation to warped the events in the image plane?
Thank you!
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I am using a handpose estimation model that uses the webcam to generate (x, y, z) coordinates for each joint from a moving hand (the z is estimated accurately). I also have a .glb character with a full T-bone skeleton structure, with hands, that is made using Blender.
What I cannot figure out is how to use these real-time data points to animate the imported 3D character’s hand in ThreeJS . The (x, y, z) is in 3D cartesian plane and from what I’ve read in the docs, ThreeJS uses Euler/Quaternion angles for rotation (correct me if I’m wrong). I’m at an impasse right now because I am unsure of how to convert this into angular data.
I am fairly new to animation so please do let me know if there are other libraries that can help me do this in an easier fashion.
I think you are looking for inverse kinematics. It calculates variable joint parameters (angles or scales) depending on the end of the chain. The end of the chain i yours xyz position in space. The chain is arm for example. The joints are virtual joints of yours characters rig.
I am trying to solve a question related to transformation of coordinates in 3-D space but not sure how to approach it.
Lets a vertex point named P is drawn at the origin with a 4x4 transformation matrix. It's then views through a camera that's positioned with a model view matrix and then through a simple projective transform matrix.
How do I calculate the new screen coordinates of P' (x,y,z)?
Before explain of pipeline, you need to know is how pipeline do process to draw on screen.
Everything between process is just matrix multiplication with vector
Model - World - Camera - Projection(or Nomalized Coordinate) - Screen
First step, we call it 'Model Space' because of (0,0,0) is based in model.
And we need to move model space to world space because of we are gonna place model to world so
we need to do transform will be (translate, rotation, scale)TRS * Model(Vector4) because definition of world transform will be different
After do it, model place in world.
Thrid, need to render on camrea space because what we see is through the camera. in world, camera also has position, viewport size and
rotation.. It needs to project from the camera. see
General Formula for Perspective Projection Matrix
After this job done, you will get nomalized coordinate which is Techinically 0-1 coordinates.
Finaly, Screen space. suppose that we are gonna make vido game for mobile. mobile has a lot of screen resolution. so how to get it done?
Simple, scale and translate to get result in screen space coordinate. Because of origin and screen size is different.
So what you are tring to do is 'step 4'.
If you want to get screen position P1 from world, formula will be "Screen Matrix * projection matrix * Camera matrix * P1"
If you want to get position from camera space it would be "Screen Matrix * projection matrix * P1".
There are useful links to understand matrix and calculation so see above links.
https://answers.unity.com/questions/1359718/what-do-the-values-in-the-matrix4x4-for-cameraproj.html
https://www.google.com/search?q=unity+camera+to+screen+matrix&newwindow=1&rlz=1C5CHFA_enKR857KR857&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjk5qfQ18DlAhUXfnAKHabECRUQ_AUIEigB&biw=1905&bih=744#imgrc=Y8AkoYg3wS4PeM:
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.
I am trying to build a simple camera matching (or match moving) application. The functionality is the same as that in most 3d applications like 3ds Max or Maya. Given an image of a cube and a 3d model of the cube, the user selects points on the image corresponding to each vertex of the model. The application must then generate a camera view that displays the 3d cube model from the same angle as shown in the image.
Can anyone point me in the direction of an algorithm for that?
PS: The camera is calibrated and the camera calibration matrix is available to the program
You can try with the algorithm illustrated step-by-step on http://www.offbytwo.net/camera-matching/. The Octave source code is provided, too.
As a plus, you don't need to start with a cube, but just with any two edges parallel to the x axis and two in the y direction.
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...