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Why is G3 continuity only achievable between two edges?

For example, curve 1/2/3, 4/5/6, 7/8/9, 10/11/12 are all G3 continuous. The center surface is built with G3 constraint on edge 5/2. Since curve 1/2/3, 4/5/6 are G3 already, how come edge 8/11 can only achieve G1 tangent?

Can't I just build a function to use the first & second & third derivatives at each u/v's edge control point to compute the 3 control points adjacent to this edge control point to achieve G3 on all 4 edges?

enter image description here

yacc
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June Wang
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  • This is not a programming question so you might get a better reception at the [math.se] site. – Guy Incognito May 24 '20 at 15:46
  • @GuyIncognito CAD programmers use SO to ask/answer questions like this. – June Wang May 24 '20 at 15:50
  • You are specifically asking for a mathematical reasoning, not about CAD programming. Just a suggestion, depends on whether you want an answer or not. – Guy Incognito May 24 '20 at 15:52
  • @GuyIncognito Mathematical reasoning and programming are bundled in such topics. I've invited SO users to answer this, so don't worry if it'll have an answer or not. – June Wang May 24 '20 at 15:54
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    @JuneWang The two are as you say bundled in those topics, however they don't seem to be bundled in your question. It seems that any explanation would require mathematical notations, and as SO doesn't even have latex support it would be much better suited to math.stackexchange. Also, I think you're much more likely to stumble upon someone that will be able to answer your question on the math site. – Ash May 24 '20 at 16:21
  • From my simple mathematical view, in order to fit curve 7 and 9 G3-compliant, curve 8 must match the 1st, 2nd and 3rd degree derivation at the junction points. This leaves no degree of freedom to create a G3 junction to the 1/4 shape. – yacc May 24 '20 at 16:32
  • And by the way LaTeX is a type-setting language, I doubt you'll reach the right audience there. – yacc May 24 '20 at 16:34
  • @yacc Yes, but 1/4 shouldn't be influenced by 7/8/9's relationship as 1/8 & 4/8 are G0. We need a stack3D. – June Wang May 24 '20 at 16:43
  • So shape 1/4 is not given/constant? Might be a limitation of the application. I'd add the app name to the tags. – yacc May 24 '20 at 17:09
  • @yacc Shape 1/4 are given/constant, but they are G0 position to 8. i.e curve 2 fit curve 1 and 3 G3 continuous, curve 5 fit curve 4 and 6 G3 continuous and so on. Well the name of the app isn't listed here, it's Autodesk Alias. It should be quite powerful in surface continuity as it's specifically designed for Class A modeling. – June Wang May 24 '20 at 17:13
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    well I've set the tag at least, I'm out of clues but good luck – yacc May 24 '20 at 17:17

1 Answers1

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The reason that the center surface cannot be built to be G3 to all 4 surfaces is because the 4 surfaces might not meet at the 4 corner points with G3 continuity. In fact, for the given condition that the curves 1/2/3, 4/5/6, 7/8/9, 10/11/12 are all G3 continuous only ensures that the 4 surfaces meet at the 4 corner points with G1 continuity.

The following is a bit more details per OP's request.

Let's denote two of the 4 surfaces as Surface A and B with coincident control points P and Q as shown in the following picture.

enter image description here

The normal vector of Surface A at point P is obtained by taking cross product of vector(P,P1) and vector(P,P2) and that for surface B at point Q is the cross product of vector(Q,Q1) and vector(Q,Q2). Since curve 1 and 2 are connected with G3 continuity, it means that vector(P,P1) is parallel to vector(Q,Q1). Similarly we have vector(P,P2) parallel to vector(Q,Q2). Therefore, we can conclude that surface A and B have the same unit normal vector at point P (or Q), which means the two surface meet with G1 continuity.

In order for surfaces A and B to meet with G2 at point P, 3 more control points from each surface will get involved (shown in the picture as green dots P3, P4 and P5 for surface A). All these 12 control points (6 from each surface) need to form a specific relationship in order for the two surfaces to meet with G2 continuity. The fact that curve 1/2 and 8/9 are connected with G3 continuity only affects the location of P3 and P5, and not the location of P4. Therefore, it does not ensure the two surfaces meet with G2 continuity at point A, let alone G3 continuity.

fang
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  • Could you please elaborate more on why ‘1/2/3, ...G3 continuous only ensures that the 4 surfaces meet at the 4 corner points with G1 continuity’? What condition can ensure the 4 corner points with G3 then? – June Wang May 25 '20 at 00:43
  • More details are added. – fang May 25 '20 at 23:45
  • Thanks, I get it now. Is there a way to compute the location of P4 to ensure Surface A and B meet with G2 ? – June Wang May 26 '20 at 03:30
  • There is no unique solution for the location of P4 to make the two surfaces meet with G2. Furthermore, the command you are using is supposed to create the center surface from the 4 surfaces, therefore, it is not supposed to modify the input. – fang May 26 '20 at 05:48
  • I’m trying to write an algorithm to compute the two surfaces meet with G2, using fixed Surface 1( Qi are given). So I can modify input from surface 1. Not sure what’s the general solution for the location of P4. – June Wang May 26 '20 at 15:58