[QUOTE=Beetle179;53168568]I'm taking a course on differential geometry, and while most of the time I feel like I'm understanding what's going on, the assignments have been owning me big-time
[img]https://i.imgur.com/Jt2CPY1.png[/img]
Where a surface normal [B]n[/B] is given by
[img]https://i.imgur.com/p098m6u.png[/img]
This (part a) seems pretty straightforward, and it would be, if the algebra didn't make it a complete disaster. For example, for sigma(0, 0), sigma_u = [B]b_01[/B] - [B]b_00[/B] and sigma_v = [B]b_10[/B] - [B]b_00[/B]. The cross product of these is absolutely disgusting, much less its norm, and it makes me think I'm going about this entirely wrong. I feel like I must be missing something pretty obvious but I've been kinda stuck just thinking about it for a liiiiittle bit too long and if anyone has any insight I would actually love you so much[/QUOTE]
Sigma_u and Sigma_v are partial derivatives. Cross product of those derivatives is normal.
Anyway I think you must calculate normal as a function, so normal(u,v).
Then with this function you calculate normals on 4 control points (n_00 = normal(0,0), n_10 = normal(1,0)...). Results of function will still be algebraic (not a numbers), and you should simplify those.
In the end you make this patch n(u,v), which should look something like this:
n(u,v) = (1-u)*(1-v)*n_00 + (1-u)*v*n_10 + u*(1-v)*n_01 + u*v*n_11
Weirdly though, end results of n(u,b) shouldn't be normalized normal, just interpolated between control points (which are normalized normals).
[QUOTE=JohnnyMo1;53169298]I've taken differential topology and I'm in a differential geometry reading group now and I have no idea what any of that means. Are you using a textbook?[/QUOTE]
We're using Andrew Pressley's [url=http://www.springer.com/us/book/9781848828902]Elementary Differential Geometry[/url], but only loosely -- the book doesn't discuss Bezier surfaces (the subject of this question), or even Bezier curves, although we've spent a lot of time studying them in lecture applying the concepts from the book.
[QUOTE=Fourier;53169322]Sigma_u and Sigma_v are partial derivatives. Cross product of those derivatives is normal.
Anyway I think you must calculate normal as a function, so normal(u,v).
Then with this function you calculate normals on 4 control points (n_00 = normal(0,0), n_10 = normal(1,0)...). Results of function will still be algebraic (not a numbers), and you should simplify those.
In the end you make this patch n(u,v), which should look something like this:
n(u,v) = (1-u)*(1-v)*n_00 + (1-u)*v*n_10 + u*(1-v)*n_01 + u*v*n_11
Weirdly though, end results of n(u,b) shouldn't be normalized normal, just interpolated between control points (which are normalized normals).[/QUOTE]
Right, I think I understand the process here, but when trying to apply it things just do not end up clean or even workable in any reasonable manner. Here's as far as I've gotten:
The partials of sigma(u, v) are pretty straightforward:
sig_u = (1-v)(b_01 - b_00) + v(b_11-b_10)
sig_v = (1-u)(b_10 - b_00) + u(b_11-b_01)
At (u, v) = (0, 0), these become:
sig_u(0, 0) = b_01 - b_00
sig_v(0, 0) = b_10 - b_00
The problem, though, is that I can't see how or where this is supposed to simplify when taking their cross product to calculate the actual surface norm at (0, 0):
(b_01 - b_00) cross (b_10 - b_00)
is not readily simplifiable (as far as I can tell), and that's only the numerator -- the denominator is the norm of that!
So my only guess is to leave it in that form and not actually expand the cross product or norm, and hope that things collect down to the form you've mentioned -- I think this is what you're suggesting, yes? -- but anyway, I'll definitely give it a try later when I have the time. Thanks :happy:
-snip- found a solution.
First post on Newpunch!
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