This is sort of a homework question, however no expectations for code or whatever just an idea or hint towards the following problem.
I have a set of cubes in 3D world coordinates and i have to display them using two projections in two separate areas, parallel and perspective. The parallel went fine, no problems there, however displaying the same scene using perspective projection is becoming a nuisance for me.
The world to screen coordinates seemed like a good idea, but i don't know on which coordinates to apply them to, the original real coordinates, the new coordinates.
Thank you for your time.
PS: we are only allowed Java2D Api.
To perform a perspective projection, you need two additional things: the perspective point (where the "eye" is) and the projection plane. With a parallel projection, the perspective point/eye and plane can be any arbitrary distance from the objects (e.g., the cubes). But it is a little more complex with perspective projection.
Once you establish your eye and projection plane, you will need to iterate over your cubes. Ideally, you would iterate over them from the farthest cube to the eye to the nearest - that way the nearer cubes will overwrite the farther ones.
For each cube, determine the distance from the eye for each point. Then for each face (again in order of decreasing distance), calculate the projected points for each vertex. You can skip those faces with occluded points (the farthest vertex for each cube).
To calculate the projected point for a particular vertex, you need to find the point on the projection plane. This point will be where the line from the eye to the vertex intersects the projection plane. This will require some math, but should not be too difficult.
Related
I was looking around a lot but only found how to check collision for 2d objects.. for my current project I want to check colisions in 3d (I'm using obj models) - I could probably figure out something myself the problem is that I only know the center point of each object..
Is there a way to get the boarders of the object so I can check if it touches the boarders of another object? What would be the best way to get this information?
Edit: Some more information that might help:
I'm using lwjgl 2.8,
my objects are obj files,
I can get position, scale and the rotation of an object
EDIT: This is what I found on youtube:
https://www.youtube.com/watch?v=Iu6nAXFm2Wo&list=PLEETnX-uPtBXm1KEr_2zQ6K_0hoGH6JJ0&index=4
What you have is called a triangle soup[1]: you got vertex position (, texture and normal) information couppled with triangle information. What you can do is intersect each triangle from one mesh with another mesh. You can do this either brute-force (by testing all the other triangles each time) or build up a space partitioning data structure to speed up your intersections.
E.g. build an Octree per mesh and iterate over the leaves in one of them: for each of those leaves, test its bounding box against the leaves in the other tree and for each of those collision pairs brute-force test each triangles within either of those pairs against each other (or only test those from mesh A against those in mesh B, if you don't care about self-intersections).
There are libraries for these sorts of algorithms like OpenMesh or Bullet for example. But I only know of one port to Java: JBullet.
EDIT: If you're only interested in approximate collisions you can throw away all information about triangles and build a bounding volume out of your vertices (e.g. an axis aligned box is just the min and the max for all vertex positions, an oriented box is built similarly, but you have to find a good enough orientation first, a sphere is a bit more involved and finally you have a convex mesh, which uses the same sorts of intersection tests as a normal mesh, but is smaller than the original and convex, allowing for some optimisations of the general intersection test).
[1]: there are other kinds of 3D representations that record information about how the triangles are connected. You might imagine finding only two intersecting triangles in mesh A and mesh B and then start searching for intersecting triangles only in the neighbourhood of the initial intersection... these algorithms are much more involved, but for meshes that deform have the advantage, that you don't have to rebuild the space partitioning data structure each time as you have to do with triangle soups.
Say you have a collection of points with coordinates on a Cartesian coordinate system.
You want to plot another point, and you know its coordinates in the same Cartesian coordinate system.
However, the plot you're drawing on is distorted from the original. Imagine taking the original plane, printing it on a rubber sheet, and stretching it in some places and pinching it in others, in an asymmetrical way (no overlapping or anything complex).
(source)
You know the stretched and unstretched coordinates of each of your set of points, but not the underlying stretch function. You know the unstretched coordinates of a new point.
How can you estimate where to plot the new point in the stretched coordinates based on the stretched positions of nearby points? It doesn't need to be exact, since you can't determine the actual stretch function from a set of remapped points unless you have more information.
other possible keywords: warped distorted grid mesh plane coordinate unwarp
Ok, so this sounds like image warping. This is what you should do:
Create a Delaunay triangulation of your unwarped grid and use your knowledge of the correspondences between the warped and unwarped grid to create the triangulation for the warped grid. Now you know the corresponding triangles in each image and since there is no overlapping, you should be able to perform the next step without much difficulty.
Now, to find the corresponding point A, in the warped image:
Find the triangle A lies in and use the transformation between the triangle in the unwarped grid and the warped grid to figure out the new position.
This is explained explicitly in detail here.
Another (much more complicated) method is the Thin Plate Spline (which is also explained in the slides above).
I understood that you have one-to-one correspondence between the wrapped and unwrapped grid points. And I assume that the deformation is not so extreme that you might have intersecting grid lines (like the image you show).
The strategy is exactly what Jacob suggests: Triangulate the two grids such that there is a one-to-one correspondence between triangles, locate the point to be mapped in the triangulation and then use barycentric coordinates in the corresponding triangle to compute the new point location.
Preprocess
Generate the Delaunay triangulation of the points of the wrapped grid, let's call it WT.
For every triangle in WT add a triangle between the corresponding vertices in the unwrapped grid. This gives a triangulation UWT of the unwrapped points.
Map a point p into the wrapped grid
Find the triangle T(p1,p2,p3) in the UWT which contains p.
Compute the barycentric coordinates (b1,b2,b3) of p in T(p1,p2,p3)
Let Tw(q1,q2,q3) be the triangle in WT corresponding to T(p1,p2,p3). The new position is b1 * q1 + b2 * q2 + b3 * q3.
Remarks
This gives a deformation function as a linear spline. For smoother behavior one could use the same triangulation but do higher order approximation which would lead to a bit more complicated computation instead of the barycentric coordinates.
The other answers are great. The only thing I'd add is that you might want to take a look at Free form deformation as a way of describing the deformations.
If that's useful, then it's quite possible to fit a deformation grid/lattice to your known pairs, and then you have a very fast method of deforming future points.
A lot depends on how many existing points you have. If you have only one, there's not really much you can do with it -- you can offset the second point by the same amount in the same direction, but you don't have enough data to really do any better than that.
If you have a fair number of existing points, you can do a surface fit through those points, and use that to approximate the proper position of the new point. Given N points, you can always get a perfect fit using an order N polynomial, but you rarely want to do that -- instead, you usually guess that the stretch function is a fairly low-order function (e.g. quadratic or cubic) and fit a surface to the points on that basis. You then place your new point based on the function for your fitted surface.
i want to find a circular object(Iris of eye, i have used Haar Cascase with viola Jones algorithm). so i found that hough circle would be the correct way to do it. can anybody explain me how to implement Hough circle in Java or any other easy implementation to find iris with Java.
Thanks,
Duda and Hart (1971) has a pretty clear explanation of the Hough transform and a worked example. It's not difficult to produce an implementation directly from that paper, so it's a good place for you to start.
ImageJ provides a Hough Circle plugin. I've been playing around with it several times in the past.
You could take a look at the source code if you want or need to modify it.
If you want to find an iris you should be straightforward about this. The part of the iris you are after is actually called a limbus. Also note that the contrast of the limbus is much lower than the one of the pupil so if image resolution permits pupil is a better target. Java is not a good option as programming language here since 1. It is slow while processing is intense; 2. Since classic Hough circle requires 3D accumulator and Java probably means using a cell phone the memory requirements will be tough.
What you can do is to use a fact that there is probably a single (or only a few) Limbuses in the image. First thing to do is to reduce the dimensionality of the problem from 3 to 2 by using oriented edges: extract horizontal and vertical edges that together represent edge orientation (they can be considered as horizontal and vertical components of edge vector). The simple idea is that the dominant intersection of edge vectors is the center of your limbus. To find the intersection you only need two oriented edges instead of three points that define a circle. Hence dimensionality reduction from 3 to 2.
You also don’t need to use a classical Hough circle transform with a huge accumulator and numerous calculations to find this intersection. A Randomized Hough will be much faster. Here is how it works (~ to RANSAC): you select a minimum number of oriented edges at random (in your case 2), find the intersection, then find all the edges that intersect at approximately the same location. These are inliers. You just iterate 10-30 times choosing a different random sample of 2 edges to settle in a set with maximum number of inliers. Hopefully, these inliers lie on the limbus. The median of inlier ray intersections will give you the center of the circle and the median distance to the inliers from the center is the radius.
In the picture below bright colors correspond to inliers and orientation is shown with little line segment. The set of original edges is shown in the middle (horizontal only). While original edges lie along an ellipse, Hough edges were transformed by an Affine transform to make those belonging to limbus to lie on a circle. Also note that edge orientations are pretty noisy.
I'm working with a mapping application (worldwind) and trying to determine the minimum and maximum latitudes and longitudes that are currently displayed. I have access to the frustum, but I'm not sure how to account for the fact that the globe can have its heading and/or pitch changed. Any help on this problem would be appreciated.
thanks,
Jeff
Actually the Frustum isn't the most useful thing in this context. What you need to do is reverse the ModelView-Project transform.
If you can retrieve your projection (frustum) and model-view matrices, then you can invert them. If you project a ray from your camera position along projection space, then you can use those inversions to find that ray in world space. From there, you can intersect that ray with your world to find the exact point where that ray hits the globe.
Do this for the four corners of the screen and then calculate your 2D bounding box based upon those intersection coordinates.
What exactly are you trying to calculate ? Corner points of your window or that of your globe. Remember a circle does not have corners (nor does a sphere). Have you seen the minimap tool in worldwind (WorldMapLayer) . It shows the currently visible extent as an overlay in the minimap, this can give you an idea to work out what exactly is being displayed.
Good to see worldwind questions flowing everywhere. We mustn't be doing a good enough job at the forums.
The corner cases (1) of my previous proposal are making the algorithm difficult for you (I've never had to zoom out from the globe and still don't understand why you still don't just use the hemisphere in that case), so here is another method that can work.
Take your projection frustum as a 3D object and intersect it with the globe (as a 3D object). There are a variety CSG algorithms that can give you union, intersection, and difference. Use intersection. This will result in a 3D mesh of the pieces of the globe that intersect with the frustum. The extents of that mesh are the extents of the bounding box. (Project along the major axis of the frustum.)
This is horribly less efficient than my previous proposal. :-)
(1) Pun completely intended.
I'm currently writing an application that actually acts as a "cut" tool for 3D meshes. Well, I had some problems with it now which I am clueless on how to solve, since it is my first application.
I have loaded a model from an object file onto the canvas, then on the same canvas, I use the mouse drag event to draw lines to define the cutting point.
Let us say I want to cut a ball into half and I draw the line in the middle. How do I detect the vertices of the ball under the line.
Secondly, if I rotate/translate the ball, would all the the vertices information change?
Think of what you'd do in the real world: You can't cut a ball with a line, you must use a knife (a line has no volume). To cut the ball, you must move the knife through the ball.
So what you're looking after is a plane, not a line. To get such a plane, you must use some 3D math. What you have is the canvas orientation and the "side view" of the plane (which looks like a line).
So the plane you're looking for is perpendicular to the canvas. A simple way to get such a plane is to take the canvas orientation and create a plane which has the same orientation and then rotate the plane around the line by 90°.
After that, you can visit all edges of your model and determine on which side of the plane they are. For this, determine on which side of the plane the end points of the edge are. Use the cross product. If they are on the same side (both results of the cross products will have the same sign), you can ignore the edge. Otherwise, you need to determine the intersection point of the edge and plane. Create new edges and connect them accordingly.
See this page for some background on the math. But you should find some helper methods for all this in your opengl library.
if I rotate / translate the ball, would all the the vertices information change
Of course.
It's not going to be that easy.
I assume the line you are drawing induces a plane which then cuts the sphere.
To do so, you have to calculate the intersecting area of the sphere and the plane.
This is not a trivial task and I suggest using an existing framework for this or if you really want to do this yourself, read about basic intersection problems to get a feeling for this kind of problem. This paper offers a good introduction to various intersection tests.
In general boundary represended volumes, as in your case, are difficult to handle when it comes to more advanced manipulations. Cutting a sphere in half is easy compared to burring a small hole into it. Sometimes it's better to use a volume representation, like tetrahedral meshes or CSG.
Regarding your second question, you shouldn't rotate or translate the sphere, rotate and translate the camera.