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)

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)

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)

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)

Retrieved from "http://en.wikipedia.org/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)

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

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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 ...