I'm trying to convert from world space coordinates to screen coordinates with a perspective camera. Basically I have this situation:
where I have the near plane at z=0 and the far plane at z=1, now I'm trying to get x' y' and z' coordinates from x y z coordinates. Since I can use similar triangles properties, I can write
The red squares are the things I don't understand. How come z' is encoded with such an expression?
And how can the transformation written above be written in the matrix below? I see a z-scale term and a z-translation term but I don't understand them
Simply solving the matrix with the link indicated by Oli solves the problem. Sorry for the trouble!
Related
I have a question about the feasability of an idea.
I have a surface (which can be parametrized or defined implicitly by an equation F(x,y,z) = 0) and I want to draw some helix that fit the surface, litterally helix on the surface.
What would be the process to achieve that ?
I have a basic idea ,which is inspired from ray marching methods, : I have my surface (which have a finite area) then I 'draw' a big helix curve around it and I decrease the radius of the helix. If the helix intersect the surface then I save that point, and finally I would obtain the set of points that draw an helix on the surface...
Feel free to ask me questions about the problem.
Thank you for your attention.
thomas
There are different ways to understand "drawing a helix curve" on a surface. By the way, I am not sure that you use the proper term, as a helix is a spring-like curve and is not flat at all. I will instead assume some planar curve described by an implicit C(x,y)=0 or parametric x=Xc(t),y=Yc(t) representation.
One approach is by using a (u,v) parameterization of the surface, as used in texture mapping, x=Xs(u,v), y=Ys(u, v), z=Zs(u, v). For example, for a sphere (u,v) could correspond to the angle coordinates in the spherical system. In this case, it suffices to map x=u, y=v and there will be a direct correspondence between the points of the curve and the points of the surface.
Another approach is by "extruding" the curve in the z direction to form a cylindric surface, and intersect the cylindre with the surface. In this case, you form the system
S(x, y, z)= 0
C(x, y)= 0
where z is free, and solve for (x, y) as a function of z. (You can also use the parametric equations, there are different combinations.) In fact, you perform a parallel projection of the curve.
Instead of a cylindre, you can also think of a conic surface, by choosing a vertex point, say the origin, and considering the points (zx, zy, az) where a is an "aperture" coefficient and z is free. This idea is vey close to your "radius decrease" method and corresponds to a central projection.
I'm trying to code the Ritter's bounding sphere algorithm in arbitrary dimensions, and I'm stuck on the part of creating a sphere which would have 3 given points on it's edge, or in other words, a sphere which would be defined by 3 points in N-dimensional space.
That sphere's center would be the minimal-distance equidistant point from the (defining) 3 points.
I know how to solve it in 2-D (circumcenter of a triangle defined by 3 points), and I've seen some vector calculations for 3D, but I don't know what the best method would be for N-D, and if it's even possible.
(I'd also appreciate any other advice about the smallest bounding sphere calculations in ND, in case I'm going in the wrong direction.)
so if I get it right:
Wanted point p is intersection between 3 hyper-spheres of the same radius r where the centers of hyper-spheres are your points p0,p1,p2 and radius r is minimum of all possible solutions. In n-D is arbitrary point defined as (x1,x2,x3,...xn)
so solve following equations:
|p-p0|=r
|p-p1|=r
|p-p2|=r
where p,r are unknowns and p0,p1,p2 are knowns. This lead to 3*n equations and n+1 unknowns. So get all the nonzero r solutions and select the minimal. To compute correctly chose some non trivial equation (0=r) from each sphere to form system of n+1 =equations and n+1 unknowns and solve it.
[notes]
To ease up the processing you can have your equations in this form:
(p.xi-p0.xi)^2=r^2
and use sqrt(r^2) only after solution is found (ignoring negative radius).
there is also another simpler approach possible:
You can compute the plane in which the points p0,p1,p2 lies so just find u,v coordinates of these points inside this plane. Then solve your problem in 2D on (u,v) coordinates and after that convert found solution form (u,v) back to your n-D space.
n=(p1-p0)x(p2-p0); // x is cross product
u=(p1-p0); u/=|u|;
v=u x n; v/=|v|; // x is cross product
if memory of mine serves me well then conversion n-D -> u,v is done like this:
P0=(0,0);
P1=(|p1-p0|,0);
P2=(dot(p2-p0,u),dot(p2-p0,v));
where P0,P1,P2 are 2D points in (u,v) coordinate system of the plane corresponding to points p0,p1,p2 in n-D space.
conversion back is done like this:
p=(P.u*u)+(P.v*v);
My Bounding Sphere algorithm only calculates a near-optimal sphere, in 3 dimensions.
Fischer has an exact, minimal bounding hyper-sphere (N dimensions.) See his paper: http://people.inf.ethz.ch/gaertner/texts/own_work/seb.pdf.
His (C++/Java)code: https://github.com/hbf/miniball.
Jack Ritter
jack#houseofwords.com
What is the relationship between the depth of the target 3D object that you wish to draw perception on, and the perceptive depth that is visible.
E.g. if I know a square is x by y by z and I wish to draw it in perspective at distance d hwo do I know how long the z (depth) line should be drawn in relation to the vanishing point (should I draw it 2/3's of the way or 1/5th of the way).
Is there a relation between the two, like; multiply the depth by DepthObject / distanceToVanishingPoint = perceptiveDepth.
THanks.
With perspective projection, your view frustum appears like a triangle from a top (x, z) view. You have a nice theorem which explains the relation between sides of triangles and parallel lines.
For a more practical solution, you should have a closer look at projection matrices.
After a 3d point has been transformed by a perspective projection matrix, what do the Z coordinate stand for? Distance from the 'eye'? Distance from the near clip plane? None of this?
Edit
I've set up a matrix using the glFustrum definition found here.
Then, I transform a 3D point with that matrix.
I'm left with a 3D point where X and Y are the coordinates of the point on the near frustum face, and Z, supposedly a depth information that I have a hard time exploiting.
Thanks!
The transformed z (that you use as denominator for dividing x and y) is the distance from the eye in the perpendicular direction to the projection plane, scaled so that the projection plane is at distance 1.
It's been a while for me, but if you have the X and Y coordinates on the frustrum, does not the Z control the layer order?
I've spent a good amount of time getting intersections working correctly between various 2D shapes (circle-circle, circle-tri, circle-rect, rect-rect - a huge thanks to those who've solved such problems from which I drew my solutions from) for a simple project and am now in the process of trying to implement an triangle-AABB intersection test.
I'm a bit stuck however. I've tried searching online and thinking it through however I've been unable to get any ideas. The thing that's given me the biggest issue at the moment is checking whether the edges of triangle (which is an isosceles btw) intersect the rectangle when no vertexes lie within the rectangle.
Any ideas how I could get this working?
EDIT: To give a bit more insight as to stages as I think they should occur:
1 - Check to see if any vertexes lie with in the rectangle (this part is easy). If yes, collision, otherwise continue.
2 - Check to see if any edges are intersecting the rectangle. This is where I'm stuck. I have little idea how to implement this.
I'd calculate a collection of equations which define the 4 lines of the rectangle, and then solve against a collection of equations which define lines of the triangle.
For example, gievn a rectangle with lowest point (x1, y1) and one side having a gradient of g, one of the lines of the rectangle will be y = gx + y1. Find equations to represent the other 3 sides of the rectangle as well.
The lines which form the sides of the triangle will be calculated similarly. The equation for a line given two points is
y - y1 = (x - x1) * (y2 - y1)/(x2 - x1)
If there are any possible x & y values that satisy all 7 equations then you have an intersection.
edit: I realise that although this is a simple algorithm it might be tricky to code; another option is to calculate formulae for the intervals that form each edge (essentially lines with a min and max value) and solve these.