Why for perfect reflections a surface must have G2 continuity (class A surface)?

I would like a mathematical answer.

  • 3
    $\begingroup$ Any context? Or reference where you have read that? Because it does not make sense to me. Plus, if I'm not mistaken, Gn continuity is defined only for piece-wise polynomial surfaces, there is no reason for a surface to be polynomial and in practice most of the surfaces are piece-wise linear. $\endgroup$
    – tom
    Oct 30, 2015 at 14:29
  • 2
    $\begingroup$ G2 just mentions the geometric n-derivability, independently of any parameterisation. $\endgroup$ Oct 30, 2015 at 20:06
  • $\begingroup$ @tom He is talking of general surfece design like in CAD. No they dont need to be polynomials, but in practice they often are (except for circular arcs and conics) $\endgroup$
    – joojaa
    Nov 1, 2015 at 8:00
  • $\begingroup$ @joojaa Than I'm still puzzled why the use of special notation Gn. In mathematics there is standard notion of Cn differentiable manifold. So is Gn and Cn the same? I thought that Gn manifold is piece-wise polynomial, so it is C-infty manifold except at the patch seams. $\endgroup$
    – tom
    Nov 1, 2015 at 16:48
  • $\begingroup$ @tom C continuity is the parametric continuity and G is the geonetric continuity and in this case the continuity over 2 separate geometries. $\endgroup$
    – joojaa
    Nov 3, 2015 at 5:19

2 Answers 2


What you see on reflects is the n-continuity of normals, which are the derivative of positions. -> a G1-only surface would have G0-only normal field, i.e., with sudden change of gradient in the normals (and thus, the reflects), that the eyes can notice. G2 surfaces have G1 normals fields, which is smooth enough for your eyes.

  • G0 Continuity means that the separate surfaces meet,
  • G1 Continuity that the surfaces meet at same angle,
  • G2 Continuity means that the change in angle matches in point of contact.

The G2 requirement does not mean that the surface is good quality. Just means that without this the surface is not going to have a continuous reflection flow so humans can see the difference. That may or may not be a good thing depends on what you want.

Mathematically the surface normal is:

$f(u,v) \frac{\partial}{\partial u} \times f(u,v) \frac{\partial}{\partial v}$

Since both sides are derived that means that the function field of the surface normal has one degree less than the original surface. So for the reflection to be first degree continuous it has to have a second degree continuity.

So far we have established the relationship between the continuity of the surface and the continuity of the reflection. Nothing thus far proves that the surface reflection needs to be first degree continuous. To understand why we must exit the realm of mathematics and enter the realm of biology.

The eye is equipped with an edge detection algorithm on a structural level right on the retina. This edge detection algorithm in essence works as a discrete derivative of the input signal. So, if your surface is not G2 continuous then the human edge detection kicks in and shows itself up. For references read on Mach Bands and so forth.

Since the edge detection is discrete G2 continuity is not enough. The change has to not only be locally satisfied but also satisfied on the retina. So the change should still be shallow enough not to cause problems.

  • $\begingroup$ What does, "The change has to not only be locally satisfied but also satisfied on the retina," mean? $\endgroup$
    – Dan Hulme
    Nov 19, 2015 at 13:05
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    $\begingroup$ The eye is not recording a continious signal. Its discrete, so even if your surface might technically meet the condition presented on a mathematical level. It might not be enough if the dicrete sample spacing does not see the change. So the slope still has to be big enough for human eye to notice. $\endgroup$
    – joojaa
    Nov 19, 2015 at 13:38
  • $\begingroup$ It sounds like you're saying the derivative (of the normal) doesn't just have to be continuous, but its derivative has to be below some limit. If that's what you mean, I think that last paragraph of your answer could be clearer. $\endgroup$
    – Dan Hulme
    Nov 19, 2015 at 13:45
  • $\begingroup$ @DanHulme its not a limit the derivate, its not a question of slope, only, but the interwall of the slope. So it is about a discrete sampling. So a very sharp angle but small difference in slope might seem continious. Likewise continious changes under a short interwall might seem sharp. Its not about mathematics its about sampling. Its just hard to qantify as its a biological system. $\endgroup$
    – joojaa
    Nov 19, 2015 at 14:41

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