Talk:Normal (geometry)

message: "A surface normal holds the three dimensional direction a surface faces in a 3 component vector representation."
Can't this be stated a little more clearly? I don't understand what is meant.
S.
:Probably.  It could also benefit from a simple graphic.  I know what it means—I'll try to get around to it. —Frecklefoot 18:52, 28 Aug 2003 (UTC)
replies: text: Okay, I edited the definition and added a graphic.  I hope it is clearer now.  If not, mention it here and I'll try to clrify it even more. :-) —Frecklefoot 20:34, 28 Aug 2003 (UTC)
replies: text: I took off the stub notice, it looks pretty well fleshed out now. (if someone more mathematical disagrees, they can put it back. --ObscureAuthor 20:37, 28 Aug 2003 (UTC)
thread_title: What does it all mean!?

message: I reverted recent changes to this article as they were not carefully written.
* The picture with an inward pointing normal is not good looking, and a bit confusing. 
* The concepts of inward and outward pointing normals do not make sense unless one defines what is the "inside" and what is the "outside" of a surface, so I guess it does not make sense to all surfaces.
* The wording "Normals do not penetrate the surface." does not make sense. 
As such, the new changes had a point, but were poorly written to the point of being incorrect. Comments? Oleg Alexandrov (talk) 22:34, 20 May 2006 (UTC)
replies: text: Theory says a smooth surface in 3D has a tangent plane at each point. Displacements between points in that plane are perpendicular to exactly two opposite directions. A normal vector is a non-zero vector pointing in one of the two directions; often, but not always, a normal vector has unit length. If a vector is an equivalence class of parallel, equal-length, and consistently directed displacements, which is how a mathematician might view the matter, then it makes no sense to speak of vectors penetrating anything. If a surface is not closed or not orientable (like a Möbius strip), then we cannot speak of inside or outside. We may draw pictures of vectors as physical arrows in space, but we should not confuse the picture with reality. --KSmrqT 01:16, 22 May 2006 (UTC)
replies: text: But how does this relate to an ellipsoidal surface?  For instance, look at Figure 1.4 on the top of pg.8, here (PDF):  Would that nearly horizontal ring connecting points A and B be the curvature of the "normal section"?  If it is, what would the "ring" that connects the two points, cutting through the center of the ellipsoid (i.e., a "great circle"), , be the curvature of?:  The "central section" or "subtended section"?  Orthodromic section? ~Kaimbridge~13:14, 22 May 2006 (UTC)
thread_title:  Outward-pointing normal 

message: Surface normals are defined (and used) for general surfaces in n-dimensional spaces. This article doesn't even mention the possibility of a normal in n-dimensional space. While the 3D examples here are intuitive and may be useful for most people arriving at this page, I think a more general exposition and a section on higher dimensions are necessary. Anyone? --Zvika 06:30, 29 August 2007 (UTC)
replies: text: Well, you are talking about "hypersurface normal", or otherwise about "normal subspace to a surface". If you want a section about this, it should be further down the article, where it won't confuse people who don't know much beyond 3D euclidean geometry. Oleg Alexandrov (talk) 06:43, 29 August 2007 (UTC)
replies: text: I must do a little observation: the word "surface" is universally reserved for two-dimensional manifolds embedded in three-dimensional space, so I am going to change the section name "n-dimensional surface" to the correct and universally accepted "hypersurface". If someone has something to object, He is welcome. Daniele.tampieri (talk) 13:25, 3 September 2009 (UTC)
thread_title:  3D-centrism 

message: It seems to have an 'outer' normal in both directions. But the parameterization is differentiable and the normal vector is a function (which mean it'll only give 1 normal at a point). If we chose a normal in 1 direction at a point A then run across the surface back to that point the normal points in the other direction (since the normal vector is continuous we do that right?), admiting two normals "in the same direction at the same point"... WHY?  —Preceding unsigned comment added by Cod4 better than halo3 (talk • contribs) 07:56, 29 August 2008 (UTC)
replies: text: As you point out, a Mobius strip isn't orientable. As I understand it, a Mobius strip can be thought of not embedded in R3 in which case a "surface normal" isn't meaningful. So in answer to your question of "WHY?": Because a Mobius strip has one side. If you wanted to represent one on a computer embedded in R3, you'd make it out of polygons each of which had a given surface normal, then you'd have a discontinuity somewhere around the loop. —Ben FrantzDale (talk) 21:03, 29 August 2008 (UTC)
thread_title:  the mobius strip 

message: At the moment, the page says
"For a convex polygon (such as a triangle), a surface normal can be calculated as the vector cross product of two (non-parallel) edges of the polygon."
: I'm pretty sure that will also work for concave polygons ...  —Preceding unsigned comment added by 157.193.173.205 (talk) 11:30, 7 December 2009 (UTC)
replies: text:  It will work up to sign. If you want an oriented surface normal (in which the sign matters), then obviously crossing consecutive edges of a concave pair of edges would give you the wrong sign. —Ben FrantzDale (talk) 13:25, 7 December 2009 (UTC)
thread_title:  Only convex polygons? 

message: Not entirely sure, but why does one need to solve:
:\left( A{{A}^{*}}+b{{b}^{*}} \right)n=b
I mean, I know it's originated from a constrained LS, but it seems to be an insane move to solve this system in order to obtain n.
The condition number of A{{A}^{*}}+b{{b}^{*}} might be horrid, since there are no promises about A,
and this is just the start of it.
Wouldn't it be much simpler to generate a vector b which isn't in the span of A and then just solve:
:n^*\left[ A,b \right]=\left[ {{0}_{N-1}},1 \right]
Here there is no need to multiply matrices, generate unneeded matrices (bb^*) if one doesn't know how to bypass this, and most importantly the condition number should be low enough for randomly selected b. 79.182.0.184 (talk) 12:52, 29 March 2010 (UTC)
thread_title:  Unnecessary complication 

message: "If a surface S is given implicitly as the set of points (x,y,z) satisfying F(x,y,z) = 0, then, a normal at a point (x,y,z) on the surface is given by the gradient
since the gradient at any point is perpendicular to the level set, and F(x,y,z) = 0 (the surface) is a level set of F."
I have not been able to find "how to find the gradient or partial derivatives of an implicitly defined function" anywhere in wikipdia.  --    Here is a short example: "it is true that the gradient of Ax^2 + By^2 + Cz^2 + D = 0 is (2Ax, 2By, 2Cz)." - source: my instructor.   -- this should be added to the articles for Gradient, Partial Derivatives, and maybe other articles as well.  BriEnBest (talk) 00:10, 16 November 2010 (UTC)
thread_title:  Partial Derivatives of Implicitly Defined Functions 

message: When describing how planes can be calculated from normals, the text says "For a plane given by the equation r = a + αb + βc, where a is a vector to get onto the plane". I have NO IDEA what 'vector to get onto the plane' is supposed to mean, and I am not unfamiliar with the material. swestrup (talk) 20:31, 29 March 2011 (UTC)
replies: text: You are right; it's not clear. What it is trying to do is parametrize a plane by defining all points, r, on that plane as a function of α and β. That is, the point r(0,0) is on the plane at position a. r(0,1) is on the plane at a+c, etc. That is, a is a vector from the origin to some point on the plane and c and d define directions on the surface. Does that make sense? —Ben FrantzDale (talk) 23:25, 29 March 2011 (UTC)
thread_title: Terminology is still confusing

message: While it is IMO important to provide a generalization, is the generalization of the vector space orthogonal to the tangent space of the manifold valid usage, and not pushing the generalization too far?  For example, do we speak of the normal surface to a curve in 3-space?  I would simply call this the space orthogonal to the tangent space; the term "normal" is already not great because of its double-use to mean "normalized". The generalization I would expect would be a normal vector or line to an (n−1)-manifold in a Euclidean n-space. But then, I'm not a mathematician; input gladly accepted. — Quondumtc 18:48, 19 December 2011 (UTC)
replies: text: The general notion of normal (vector) space is an important mathematical notion which deserves to be fully treated in Wikipedia, which is presently not. For example, in Differential geometry of curves, an orthogonal basis of the normal plane is defined which consists in the curvature vector and the binormal vector. Another example: as the normal space is the orthogonal space of the tangent space, the normal space plays a role in the theory of duality of manifolds and varieties. However I am unable to be more precise, being not a specialist of this theory and being unable to follow the Wikipedia pages on duality theory, which are written in a style which is much too abstract for me. D.Lazard (talk) 20:08, 19 December 2011 (UTC)
replies: text: I am merely concerned about encroaching on the perfectly good term orthogonal, which does cover spaces orthogonal to linear vector subspaces. The only added concept is the idea of a tangent space.  I guess part of my reticence stems from the other use of the term "normal" to mean "of norm 1", which occurs in the same contexts as "orthogonal". However,  the criterion is how the term gets used in the literature, not my preferences. I was only wanting to be sure that the term "normal" does get used rather than "orthogonal" in the more abstract context. — Quondumtc 21:42, 19 December 2011 (UTC)
thread_title:  Generalization to manifolds 

message: I'm an artist, who dabbles in the intersection of math and art. Normal maps are frequently called bump maps in game dev. Pretty obvious reason: because it encodes bumps, right?
The more correct and general term is "normal map" because that's what it's a map of, the normal of the surface. But WHY is it called a normal? What is the history?
I think it's important for wikipedia to include historical context for mathematical ideas. Mathematical terms are not pre-existing platonic Forms, they have historical and cultural context that helps ground a student's engagement with the abstract. 
Does anyone know why Normal has become the standard terminology for this? What is the history for it? Who first coined the term, and why did they choose Normal?
This is especially important considering how Normal is used in many other aspects of math for different things. Having an understanding of the history of the idea helps distinguish those different uses of the term. Thedonquixotic (talk) 19:55, 24 March 2023 (UTC)
thread_title:  Why is it called a Normal? 

message: The redirect Unit normal has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at  until a consensus is reached.  1234qwer1234qwer4 12:37, 12 November 2023 (UTC)
thread_title:  "[[:Unit normal]]" listed at [[Wikipedia:Redirects for discussion|Redirects for discussion]] 

message: Do you have a reference to the claim: "Since a surface does not have a tangent plane at a singular point, it has no well-defined normal at that point."נצה מ.ה. (talk) 14:14, 27 December 2023 (UTC)
thread_title:  Looking for a reference to one sentence