Talk:Sphere

Spherical product?

Spherical product redirects to Sphere but it is not mention anywhere in the article, is it? --RokerHRO (talk) 12:20, 31 March 2014 (UTC)

Came here to say the same thing. I still don't know what a spherical product is (I think I've worked it out but that's not what an encyclopedia should be making me do) Andybak (talk) 22:53, 14 July 2021 (UTC)

Is it actually a recognized term? The history is brief with the redirect created by long departed User:Xp54321 who does not look like a mathematician. No obvious mathematical use in a brief search? --Salix alba (talk): 05:49, 15 July 2021 (UTC)

"Spherical product" shows up in a browser search; for example, https://www.cs.mcgill.ca/~dudek/SQ/spherical_products.html, but I don't know how accurate or practical this source is.—Anita5192 (talk) 18:21, 15 July 2021 (UTC)

Thats a useful link, the key paper seems to be Superquadrics and Angle-Preserving Transformations (free copy). We have a paper on Superquadrics which seems like a better target of the redirect. I'll change it, and maybe add a brief section on the spherical product there. --Salix alba (talk): 19:16, 15 July 2021 (UTC)

Dubious definition of axis of rotation

In the Terminology section, it states that you can take any arbitrary point of the sphere as a pole, take its antipodal point as the opposite pole and the line between the two points is the axis of rotation. This is only true in the case where a) The sphere is rotating and b) the two poles are chosen such that their positions on the sphere are unchanged by the rotation - ie. not arbitrary.

The article correctly states (although in a very awkward way) that any diameter of a sphere may be an axis of rotation. This is a mathematical property of spheres and has nothing to do with spinning physical objects. Please sign your comments and put new ones at the end of the page. Thanks. Bill Cherowitzo (talk) 20:32, 27 July 2015 (UTC)

Then they shouldn't use the word 'rotation.' And frankly, mathematicians are totally incompetent in the domain of quality and effective use of language. Which is fine, but they ought to be kept on shorter leashes in regard to terminology and naming conventions. The 'ball/sphere' distinction is another ridiculous example. Although, the latter seems to have been the result of faulty reasoning, while the former seems to be just poor use of language. Firejuggler86 (talk) 22:14, 4 October 2020 (UTC)

numbers of small sphere in a large sphere

I have calculated the the numbers of small sphere in alarge sphere by my formulaدکترغلامعلی نوری Dr Goal.A. Newray 09:05, 12 September 2015 (UTC)

File:Enclosed Spheres.png

prediction of numbers of small sphere enclosed in the larger sphere, by Dr Goal.A.Newray`s formula.

— Preceding unsigned comment added by غلامعلي نوري (talk • contribs) 09:05, 12 September 2015 (UTC)

This looks like Sphere packing in a sphere, your results seem to differ. We cannot publish results here which have not been peer reviewed, please see WP:OR.--Salix alba (talk): 18:01, 12 September 2015 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Sphere/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Seems much too brief, there is plenty more which could be said about the sphere. User:Salix Alba 16:52, 14 February 2007 The language is couched in academic high brow, most people wanting answer to this question will want a far simpler and easier to understand lesson, eg thier are no explanations to formula what it means and so on, formula for pi would be a start. Otherwise good work, somebody might like to add a simpler section but please don't call it PI for idiots. —Preceding unsigned comment added by 78.146.217.198 (talk) 11:00, 16 September 2009 (UTC)

Last edited at 11:02, 16 September 2009 (UTC). Substituted at 02:36, 5 May 2016 (UTC)

ref

(Copied from User:Wcherowi's talk page.)

greetings... You and the other editor are merely asserting "not reliable" source, for that ref, but not offering just how. Also, that source was already discussed and agreed upon, with consensus, and established some time ago. It seems more that you just don't like the ref cuz it doesn't seem to meet up to something like "Britannica" or something...but Wikipedia does not stipulate that it necessarily has to be so well-known to be "reliable". Learning.com is simply a teaching and student and academic website. And WP does not forbid such sites as sources, necessarily. And the statement is unsourced. So what's the big issue here? I always kneel to consensus (even if provably wrong), but let's not let "I don't like" be the real motivation to remove this thing. And forgetting that this is a WIKI...and no one editor "owns" any article...and respect other people's contributions. Regardless of personal tastes or likes. The burden is on YOU to prove that this site is so obviously "unreliable" and just an "advertisement"...and totally not usable. You simply have not done that. Assertions without actual facts and proof don't cut it. Regards. 71.246.98.233 (talk) 18:33, 20 March 2018 (UTC)

that source was already discussed and agreed upon, with consensus, and established some time ago - and exactly where was that. I was certainly part of at least one discussion of this, and there was no consensus, in fact only the editor who put it in had anything positive to say and most of that sounded just like your diatribe above. As for being a reliable source, look at WP:RSVETTING. This will give you an idea of what I look at to determine if a source is reliable. The source you insist on fails in almost all categories. All your huffing and puffing does not change the fact that this is a "teaser ad" for the website and that is unacceptable for Wikipedia. --Bill Cherowitzo (talk) 18:54, 20 March 2018 (UTC)

I agree with that analysis. The talk page discussion was most definitely not in favour of including this ref, this IP does sound much like that previous supporter (and somewhat suspiciously added the ref with the old access date still intact), and the ref is not reliable. Meters (talk) 05:17, 21 March 2018 (UTC)

since when is a learning academic website, with credentialed teachers on the site, forbidden outright on Wikipedia, and dogmatically considered an "unreliable source"? 71.246.98.233 (talk) 16:47, 21 March 2018 (UTC)

This is really getting tedious, but I will take advantage of the respite, due to your being blocked, to lay out what is wrong with your source and your attempts to include it.

Study.com is a commercial site. It sells courses to help people prepare to take exams (given by accredited institutions or the GED exams). The site is itself not accredited (it can not give credit for its courses, only certify that they have been taken) and has no authority to which it is accountable. As you have stated many times now, there is nothing that prohibits Wikipedia from using a commercial site as a source, but this does raise some red flags and that means that the source needs to be evaluated more carefully than would normally be the case. The reason for this is that a for profit site has, as its first priority, the making of money rather than the scholarly integrity of its content. Editors are put in the awkward situation of having to decide if the profit motive has led to compromises in the content. Under the general philosophy of Wikipedia, this is not something that editors should be asked to do, so such sites should normally be avoided. I have nothing against this site. They are providing a service that some individuals will find useful and I applaud them for this effort. However, to use the site as a source, I must determine (at least for myself) whether or not it meets Wikipedia's reliability standards. The author of the notes under consideration is Jennifer Beddoe, who has been described as a math teacher and instructor. These are rather vague qualifications; where is she employed? why is she an authority on this subject? if she is certified, who has done the certification? What has she used as the source material for her lessons? Who has vetted this material? If I can't find answers to these questions, I can't determine how reliable her lessons are and the default position is to assume unreliability until there is contrary evidence.

Moving on to other issues. After a few pages in this lesson, you come to a message that says you can not continue reading the lesson unless you register($) for the course. In the advertising world, this is called a "teaser ad"; you are shown just enough of the product to make you want to purchase the whole thing. As the purpose of an ad is to make you want to purchase the product, the veracity of the content of any ad is always suspect. Finally, you claim that this source is cited to support a statement. But this statement is a vague generality and the very next sentence is a more precise statement of the same thing with a valid supporting citation. So I find your claim to be very suspect and it appears that the only reason you insist on putting this in is to sell this course to Wikipedia readers, which runs counter to Wikipedia policies. --Bill Cherowitzo (talk) 19:13, 22 March 2018 (UTC)

Formulae for surface area and volume, n?

I'm not 100% sure, but it looks like the n in the surface are and volume formulae refers to the lowest-dimensional Euclidean space in which the sphere can be embedded, not the dimension of the sphere itself as it's used elsewhere. In particular, it seems to me that the surface area should be proportional to r^n (e.g., that of a circle proportional to the radius), and the volume of the ball to r^(n+1). 50.252.247.245 (talk) 19:11, 15 July 2020 (UTC)

Lead v2

As previous editors commented, it's circular for this page to define a sphere as the surface of a ball and for Ball (mathematics) to define a ball as the region enclosed by a sphere. The ball-sphere distinction is important but not so important that it needs to be in the lead, as another commenter complained. Definition as the set of points at a given radius from a center seems more common: see [1], [2], [3]. Given previous lengthy battles over the lead I thought I'd leave a note just in case. Feel free to disagree with me as long as you don't start a multi-page flame war like in § Lead.

In general I'd like to rearrange this page so that the grade school material comes first and the grad school material comes later. I will work on combing out the knots in this hairy ball as I have time. -Apocheir (talk) 04:28, 29 December 2021 (UTC)

I'll make a suggestion for the lead; this suggestion is intended to be consistent with the above remarks from Apocheir: "A sphere (from Greek σφαῖρα—sphaira, "globe, ball") is a geometrical object consisting of the set of points that are all at the same distance r from a given point in three-dimensional space. That given point is the center of the sphere, and r is the sphere's radius. The earliest known mentions of spheres appear in the work of ancient Greek mathematicians." For legibility I have not included citations and hyperlinks. --Jzimba (talk) 20:49, 11 February 2022 (UTC)

Inconsistent claim about umbilical points

In the section on Eleven Properties of the Sphere, it is claimed that "Any closed surface will have at least four points called umbilical points." However, if we click to read the hyperlinked article about umbilical points, we read that "A torus can have no umbilics"; this contradicts the claim that any closed surface has at least four umbilical points. Moreover, again according to the article about umbilical points, "An unproven conjecture of Constantin Carathéodory states that every smooth topological sphere in Euclidean space has at least two umbilics" - and it is strange that a claim in one article should be stronger than an unproven conjecture in another article.

Assuming I'm not missing something here, I recommend deleting the sentence "Any closed surface will have at least four points called umbilical points." The paragraph also needs some copy editing (example: "...because the sphere these are the lines..."), and it could be briefer. Proposal:

At any point on a surface, the normal direction is at right angles to the surface. The intersection of a plane that contains the normal with the surface is a curve that is called a normal section; the curvature of this curve is called the normal curvature. For a typical point on a typical surface, different normal sections at the point will have different curvatures; but there may exist points on the surface at which all normal curvatures are equal. Such points are called umbilical points. Umbilical points can be thought of as the points where the surface is closely approximated by a sphere. For any point of the sphere, the curvatures of all normal sections are equal, so every point is an umbilical point. The sphere and plane are the only surfaces with this property. --Jzimba (talk) 20:21, 11 February 2022 (UTC)

pi squared radius squared

$\pi ^{2}r^{2}$ is the surface area of a sphere, not $4\pi r^{2}.$

It's interesting how long this equation has been wrong, potentially thousands of years if Archimedes was heralded from antiquity. The real mind numbing part about it is everyone else before us took it for granted, just like we apparently had. No one just sat down and tried to do the math themselves, or make real measurements. Just a bunch of assumptions. How much do you assume?

By the way, the previously embedded "proof through calculus" for $A=4\pi r^{2}$ is false and self-referential. It is derived from the volume of a sphere equation, which itself assumes $4\pi r^{2}$ is true for the surface area. It is cyclically self-fulfilling and not a real proof. As such, the volume of a sphere equation is also wrong and is instead $V={\frac {\pi ^{2}r^{3}}{3}}$

The following are some calculus sum proofs which can be solved through wolfram|alpha or equivalent computer aided calculation.

1) $2\times \sum _{x=1}^{R}2\pi {\sqrt {R^{2}-x^{2}}}$

1) When R is equal to a large value (i.e. 10,000), the sum for larger the value of R begins to converge to the same R value plugged into $\pi ^{2}R^{2}$. The method implied through this sum is to take the total area defined by circumferences of equidistant cylinder cuts of a half sphere, multiplied by two for the full sphere. As the lateral surface area for a cylinder is $2\pi r\times height$, when the equidistant cuts are based upon the integer increase $x=1\rightarrow R$, this means the height/thickness of each cylinder is simply 1, so the $2\pi r$ circumference is all that is necessary to sum. Understanding that sine and cosine trace a circular pattern through the plotting of their ratios, this logic is fed back into the formula via Pythagorean theorem to determine the radius values which are used at every step of the sum of circumferences of $\sum _{1}^{radius}2\pi (y)$; where here $(y)={\sqrt {R^{2}-x^{2}}}$.

$(a^{2}+b^{2}=c^{2})$

$(x^{2}+r^{2}=R^{2})\rightarrow (r^{2}=R^{2}-x^{2})\rightarrow (r={\sqrt {R^{2}-x^{2}}})$

The value of $x$ is incremented by 1 towards $R$ as per the instruction of the sum, defining the partial sums using $x$ to be different lengths of a line increasingly reaching from the outside of the half sphere towards the center, eventually covering the full radius length $R$. Little $r$ in the equation defines the height of a right triangle always with hypotenuse $R$ using the base $x$, and the logic of sine and cosine automatically keep these ratios bound through a circular plot of equidistant slices from $x\rightarrow R$. The height of each right triangle then becomes the radius for the sum of $\sum 2\pi (y)$, which attempts to converge to a limit of $\pi ^{2}R^{2}$. Although it appears at first glance that $r={\sqrt {R^{2}-x^{2}}}$ is not a linear function, but possibly an exponential function, it is important to remember that the Pythagorean theorem's structure insures that the base of the triangle can be linearly known and linearly sequential to define the non-linear sine relationship to the height of the triangle when the hypotenuse is known, providing a strict relationship between polar coordinates $r={\sqrt {R^{2}-x^{2}}}$ and linear coordinates $R-x$. It is indeed knowable that each cylinder height/thickness is only 1 because of this relationship, and guaranteed that each circumference towards the sum is equidistant, unnecessary to accommodate the point distance between the maximums of each triangular-height-defined radii. This unfortunately provides a measure of inaccuracy and difference from $\pi ^{2}r^{2}$ which only becomes somewhat negligible with larger values for $R$, of which the following sums will attempt to remedy.

2) $2\times {\big (}\sum _{x=1}^{R}2\pi {\sqrt {R^{2}-(x-1)^{2}}}+{\frac {1}{2}}{\Big (}2\pi {\sqrt {R^{2}-x^{2}}}-2\pi {\sqrt {R^{2}-(x-1)^{2}}}{\Big )}{\big )}$

2) Similar to the first equation with the same caveat of a large number $R$, this sum attempts to evaluate the area through conical slices rather than cylindrical slices. $x$ and $x-1$ are used to define the upper and lower circumferences of each slice. The slices are implied to be cut and unraveled to present a rectangle with sides defined by the circumference, and the additional area comes from a right triangle area using the leftover length of $x$. This formula is not perfect since it assumes the height of each conical cylinder as a rectangle is $1$, while the heights should be different based on the lateral slants connecting each radius as a cone. Nonetheless, this sum converges to the limit of $\pi ^{2}R^{2}$ even quicker than the first sum.

47.224.167.12 (talk) 11:24, 27 September 2022 (UTC)

Do you have any reliable sources that make this claim? ... discospinster _talk 17:28, 2 October 2022 (UTC)

if you cannot do math yourself then you cannot be or suggest to be an arbiter of truth relative to mathematics. The reliable source is math. 47.224.167.12 (talk) 18:07, 2 October 2022 (UTC)

No the reliable source is a scholarly journal article. ... discospinster _talk 18:10, 2 October 2022 (UTC)

you're making excruciatingly ridiculous assumptions about the validity of human intelligence if you need a scholastic journal. We're talking about a simple math equation that has been wrong for 2 millennia. If academia was worth anything, $A=\pi ^{2}r^{2}$ and $V={\frac {\pi ^{2}r^{3}}{3}}$ could have, would have, and should have previously been discovered. Instead we live in a world presumably largely populated by minds of similar effect as your own, whom negligently and recklessly consume scholastic knowledge as a measure of kissing up to the professor for a social welfare credit boost, a pretend way to feel smart and accomplished; a yes-man. The professors didn't understand the value of what they taught to perpetuate into future generations, and the students didn't practically learn how to use Archimedes' equations nor attempt to sieve novel solutions on their own. Academia is a goddamned joke, evidently.

You don't need a scholastic article. You need a reality check, a hard drink, and a wake up call. 47.224.167.12 (talk) 19:10, 2 October 2022 (UTC)

Wikipedia is not the place to promote your equation. Please see WP:No original research and while you're at it WP:No personal attacks. ... discospinster _talk 19:40, 2 October 2022 (UTC)

Learn how to do math. It's only a strong suggestion. The world does not revolve around you. You are nowhere near just enough to claim personal attacks, nor are you anywhere near correct enough to claim basic math is original research. You don't even recognize the irony of your actions and understanding as WP:Fringe. You're the one who made it a personal attack when you defaulted to the abhorrent suggestion that something cannot be truthful unless discovered by academics. You were told that you cannot be an arbiter regarding this because of your own absence of knowledge on the topic. Math is not a democracy. — Preceding unsigned comment added by 47.224.167.12 (talk) 02:08, 3 October 2022 (UTC)

There is an easy way to verify which is the correct formula, take a polygonization of a sphere and calculate its surface area. I hope you agree that the area of the polyhedron will be less than that of the circumscribing sphere. Try the Icosidodecahedron, the surface area can be easily calculated. Still better try a polygonization with many more sides. --Salix alba (talk): 17:50, 3 October 2022 (UTC)

Actually, a much easier way to verify is using a cubic unit 1 to contain the shapes. Diameter 1; Radius 0.5 gives us

$4\pi r^{2}==4\pi 0.5^{2}==4\pi 0.25==({\frac {4}{1}}\times {\frac {1}{4}}\times \pi )=={\frac {4}{4}}\pi ==\pi$

Archimedes formula puts the surface area of the sphere as the same as the circumference; $2\pi r=({\frac {2}{1}}\times {\frac {1}{2}}\times \pi )={\frac {2}{2}}\pi =\pi$

The lateral surface area of the cylinder containing the sphere would be $2\pi 0.5\times height$, or circumference times height, where height is 1; gives that the lateral surface area of the cylinder containing the sphere has $LA=\pi$.

Now use those wonderful eyeballs of yours and tell me with a straight face that it makes sense for a sphere to have the same surface area as the cylinder containing the sphere. The cylinder is obviously a larger shape. No icosidodecahedrons necessary. 47.224.167.12 (talk) 20:25, 3 October 2022 (UTC)