We know that in parametric continuity, $C^1$ continuity is two successive curve section $C_1$ and $C_2$ has first parametric derivative is same. That means tangent vector $t_1$ is same for both $C_1$ and $C_2$ and has same direction and magnitudes are same. Like $C_1'(t=1)=C_2'(t=0).$

enter image description here

Confusion:1 My question is the above image is right for $C^1$ continuity w. r. t tangent $t_1$ of $C_1$ and $C_2?$

But geometric continuity $G^1$ continuity is two successive curve section $C_3$ and $C_4$ has first parametric derivative is proportional to each other. That means tangent vector $t_2$ and $t_3$ has same direction for both curve sections $C_3$ and $C_4$ and their magnitudes may or may not be same. Like: $C_3'(t=1)=(a,b,c),C_4'(t=0)=(ka,kb,kc).$

That means two tangents $t_2$ and $t_3$ are parallel to each other. Like: enter image description here

Confusion:2 My question is above image is right, to show the parallelity of the tangents $t_2$ and $t_3$ for curve sections $C_3$ and $C_4?$

We know that $C^1$ and $G^1$ continuity means $C^0$ and $G^0$ respectively. But read from (shown below)internet which showing $C^1$ and $G^1$ continuity, but $C^0$ and $G^0$ continuity not holding because $r(t=1) =(-1,1) {\neq}n(t=0)=(1,1).$

enter image description here

Confusion:3 My question is how can I say $C^1$ or $G^1$ continuity hold inspite of $C^0$ and $G^0$ not holding?

  • $\begingroup$ Closing here since it is a cross-site duplicate: math.stackexchange.com/questions/4302092/… $\endgroup$
    – wychmaster
    Nov 10 at 15:45
  • $\begingroup$ @wychmaster please open the post, in "Math stack exchange" I didn't get any answer till now. $\endgroup$
    – user17488
    Nov 10 at 18:09
  • $\begingroup$ Reopening the question, since the duplication has been removed from Math SE $\endgroup$
    – wychmaster
    Nov 12 at 8:08

I would rather draw the $G_1$ example something like this:

This makes it clear that $t_2$ and $t_3$ are parallel, but have different lengths in general. (They both start at the same point, but $t_3$ is longer.) The way you have drawn it, they have the same length but slightly different directions—they don't quite look parallel.

The standard definition of the $C^k$ smoothness classes implies that any $C^k$ is contained in all the lower-numbered classes $C^{k-1}, C^{k-2}, \ldots, C^0$. So if a curve is in $C^k$ then it is also in all the lower classes down to $C^0$. The book you found is giving a bad example. Its notation also looks inconsistent/wrong from what you posted, so I would not trust that particular book overly much.


Yes, your understanding of $C1$ and $G1$, as shown in your drawings, is roughly correct: $C1$ means equal derivative vectors, and $G1$ means parallel derivative vectors.

I say “roughly” because there are some subtleties when derivative vectors have zero length. But that’s a corner case that doesn’t have much impact on conceptual understanding.

Regarding your last question: people who write books and papers and internet articles are, of course, free to make whatever definitions they want. But I think most people would define a curve joint to be $C1$ if both positions and first derivative vectors match across the join. In other words, most people would define $C1$ to include $C0$.


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