Richard Baker
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Second order Barycentric interpolation
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
Bump. Instead of interpolating the normal factors I could interpolate a function of the normal vectors such as diffuse and specular lighting which is still a function of secondorder partial derivatives but the question is still the same. If I have a property that is a function of the u, v parameters of a surface, and I also have the rate of change of the function I am trying to interpolate then how do I factor in the rate of change at the vertices of a triangle as well as the properties of the triangle at the vertices? 
Second order Barycentric interpolation
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
Bump. Please help poster doesn't have access to internet. Tried to interpolate u and v values then .  multiplied the change in u and v by the gradient of the normal vector with respect to u and v. But it didn't work. 
I already know that if I have a triangle with vertices v0, v1, v2 in property p0, p1, p2 then the property linearly interpolated at location l = a*p0 + b*p1+(1ab)*p2 a=(area of triangle formed by the points v1, v2 and l) / (area of the triangle formed by points v0, v1, v2) b= (area of triangle formed by the points v0, v2, and l) / (area of the triangle formed by points v0, v1, v2) My question is how do I generalize this to a higher order interpolation method? In my problem the property p = the normal vector to a surface and I have already calculated the Gaussian curvature of the surface to decide whether to subdivide the triangle into more triangles, but I don't want the secondorder partial derivatives to go to waste.

implicitization of NURBS surface
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
Perhaps I illphrased my question. The problem is not the divide by zero error, it is a fact that when I apply Kramer's rule the determinant of the denominator matrix is zero. From my understanding this means the system of the equations is not defined. My question is why is it not defined and how do I make it defined? To summarize, I have nine points that I want to interpolate using an implicit quadric surface which may be a sphere, parabolic, hyperboloid, or cylinder etc. This quadric surface has nine unknowns, the coefficients a  i assuming j equals one. So why is the surface not defined? 
For simplicity, I am only using second order NURBS. I have a NURBS surface which essentially takes the form of x(u,v) and y(u,v) and z(u,v) are all quadratic functions of u and v. Would I be correct to say that there exists a quadric of the form 0=ax2 + by2 + cz2 + dxy + eyz + fzx + gx + hy + iz + j which is equivalent to the parametric representation? I tried applying Kramer's rule to interpolate the test points along the parametric representation but the determinant of the 9x9 matrix = 0 and I get a divide by 0 error.

smoothness of Bezier patches
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
Just finished talking with him. I think he can get a good idea of what you are showing him from zoom visits. He is now using NURBS. He showed me an ouroboros he had made, so he thinks he has figured it out. Thanks. More questions coming with next audio call. 
smoothness of Bezier patches
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
Acting as his scribe is not working. I am going to have a zoom call with him on Wednesday. I will show him your illustrations then. We have to work out something more direct. 
smoothness of Bezier patches
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
I had to explain what you graphed, so it is iffy whether he has the mental picture of what you graphed. I took the tensor product of two bezier curves where U and V are the parameters and now I have a bezier surface that gives the X,Y,and Z coordinates in terms of U and V. Now that gives me (for the cubic case) four control points times four control points which is sixteen control points where you have four control points going in the V direction and four going in the U direction. I am already aware of the case where we are dealing with a two dimensional curve. But how do I generalize these results to the three dimensional case of the surface? 
smoothness of Bezier patches
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
I added weights to my Bernstein polynomials and now I believe I will have more freedom in the shapes I can create. But the original question still stands; given a NURBS surface defined by control points and weights under what conditions will an adjacent NURBS surface have g1 continuity? 
smoothness of Bezier patches
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
If it is not possible to readily enforce continuity, then let me know and I will have to learn about NURBS. Thank you. 
My son asks How do I enforce continuity in a grid of Bezier surfaces? I know that the two adjacent control points at the border of a Bezier curve have to be colinear in the onedimensional case. But how do I extend this to surfaces in three dimensional space? Which points would have to be coplanar?

Suppose I have that hyperbolic paraboloid rotated in space so the saddle is pointed to the right or left or any other direction. I would have to parameterize that in terms of hyperbolic sine and hyperbolic cosine, correct? How would I go about doing that? I worked out the implicit expression by rotating the coordinate system but whenever I use hyperbolic sine or cosine in my parametrization it only captures the upper half of the saddle. (he had to leave the phone, says you can figure out the rest of the question.)

Possible typo in a PDF I am reading
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
How do you get the specialized symbols? I will relay this to him. Thank you. 
https://www.researchgate.net/publication/242501702_The_Mathematical_Description_of_Shape_and_Form On page 16, there appears to be a typo on formula (2.13). Should there be a dot above psi? So curvature should equal a derivative of the angle with respect to arc length. Thank you.

Need help understanding Laplacian
Richard Baker replied to Richard Baker's topic in Analysis and Calculus
" Informally, the Laplacian Δf(p) of a function f at a point p measures by how much the average value of f over small spheres or balls centered at p deviates from f(p)." Please explain this sentence from the Wikipedia entry for laplacian. https://en.wikipedia.org/wiki/Laplace_operator