Talk:Chasles' theorem (kinematics)

Requested move 31 December 2017

The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review. No further edits should be made to this section.

The result of the move request was: Not moved. (non-admin closure)  sami  ^talk 23:40, 8 January 2018 (UTC)

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Chasles' theorem (kinematics) → Chasles's theorem (kinematics) – The rule is to use "s's" per WP:MOS as Cherkash seems to insist. The move was reverted by David Eppstein twice. GeoffreyT2000 (talk) 06:06, 31 December 2017 (UTC)

Per WP:COMMONNAME, "Chasles' theorem" has about 1040 results in Google scholar; "Chasles's theorem" has 73, down at the level of spelling mistakes rather than common variants. Also, to me, "s's" implies that both s's should be pronounced separately, which is the incorrect pronounciation in this case. —David Eppstein (talk) 06:43, 31 December 2017 (UTC)

Oppose MOS:POSS says that there is a second choice of Add either 's or just an apostrophe, according to how the possessive is pronounced which can be done as long as it is consistent. Since it is apparently incorrect to pronounce both s's separately, there is no conflict of MOS and commonname - both allow/support this title. (I think allowing that second style is perhaps being debated somewhere; however for now it is allowed). Galobtter (pingó mió) 18:34, 2 January 2018 (UTC)

• Support. Two main reasons: 1) Singular nouns, even the ones that end in s/z, take the " 's " as opposed to a single apostrophe in their possessive form (see here). 2) Singular nouns that terminate with a silent "s/z/x" (as in some French names, an example of which is the name Chasles discussed here), almost uniformly take the " 's " in their possessive form (see here), especially since this " 's " is actually pronounced: "...l" for the name itself, "...lz" for its possessive – which clearly can't be the case if one insists on writing "Chasles' " where the last "s" is silent. So the suggestion above that somehow "Chasles's" forces pronunciation of both last esses doesn't hold any water (one of them is clearly silent). cherkash (talk) 01:32, 4 January 2018 (UTC)
• Oppose per David Eppstein and Galobtter above. Dbfirs 10:41, 4 January 2018 (UTC)

Both of them have used the same argument based on the (incorrect) assumption that both esses are pronounced if spelled s's – which is clearly not the case here. Would you perhaps care to make a better justification than repeating false arguments? cherkash (talk) 22:28, 4 January 2018 (UTC)

No, it's your logic that is based on a false premise. Dbfirs 22:30, 4 January 2018 (UTC)

Which is what? I was very specific above when I outlined why I made my statement and what it's based on. So please be specific and argue your point logically, without leaving out any important details (i.e. which false premise you refer to that you somehow infer from my comment, and how specifically you disagree with the points I made). cherkash (talk) 22:57, 4 January 2018 (UTC)

You seem to believe that omission of the final "s" after "s'" is somehow ungrammatical or is a spelling error. In the UK, it has been taught in schools for the past sixty years, at least. Dbfirs 23:19, 4 January 2018 (UTC)

I believe you may be right about some schools, but definitely not all. The real-world MoS's mostly agree with the " 's " being a more proper / less controversial spelling in this case. cherkash (talk) 02:15, 5 January 2018 (UTC)

• Comment: This theorem is an eponym, and thus does not describe the result. The suggestion puts us in a lose-lose position as either grammar or usage (Google test) is disrespected. The eponym is not particularly direct, as disambiguation lists gravitation and geometry results due also to Chasles, though these instances are weakly referenced. Avoiding the eponym with screw displacement theorem is suggested: the theorem is that any direct Euclidean motion in space may be represented as a screw displacement. This result is important in mechanical engineering and is at the heart of screw theory. As Mozzi (1763) published the result before Chasles (1830), the attribution is subject to challenge. Mathematics, with its abstraction, inevitably relies on eponyms, but this theorem is not so abstract as to call for such attribution. — Rgdboer (talk) 03:32, 5 January 2018 (UTC)
• Instead, drop the possessive and call it the Mozzi–Chasles theorem as many sources do. Dicklyon (talk) 02:24, 8 January 2018 (UTC)
• Oppose per WP:COMMONNAME. Regardless of the pronunciation or grammar, browsing through Google Scholar and Google Books reveals that the current title is the heavily preferred choice among reliable English-language sources. If anything, the 2nd most preferred usage (although still much less than the current title) seems to be Chasles theorem, without any possessive markings at all.--Aervanath (talk) 23:38, 8 January 2018 (UTC)

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The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page or in a move review. No further edits should be made to this section.

Help!

I understand Whittaker's explaination clearly; the article currently says:

In Whittaker's terms, "A rotation about any axis is equivalent to a rotation through the same angle about any axis parallel to it, together with a simple translation in a direction perpendicular to the axis."

This makes complete sense! However, the lede to the article currently states:

Chasles' theorem says that the most general rigid body displacement can be produced by a translation along a line followed (or preceded) by a rotation about an axis parallel to that line.

I cannot understand that at all. The drawing of the screw does not help. Consider the following example: I have a book lying flat on a table, set to be read. Oriented conventionally: title facing up, spine to the left. Now I rotate it so that it is standing upright: the title now faces me; the spine is still to the left. This is a 90 degree rotation along the left-right axis. Next, I grasp the book, and raise it vertically. This is a movement perpendicular to the axis of rotation. The title still faces me, the spine is still to the left, but it has been raised off the table-- say, to eye height.

The lede, as written, implies that it is possible to find some other rotational axis, and a translation parallel to that axis, that will result in the same initial and final orientation+displacement. For the life of me, I cannot imagine what might possibly be. Perhaps the correct wording was supposed to be:

... translation parallel to the plane of rotation.

which makes sense, since the plane of rotation is perpendicular to the axis of rotation (and thus agrees with Whittaker.) I'ma gonna change the lede, unless someone can show up and explain what I got wrong above. 67.198.37.16 (talk) 18:28, 20 May 2024 (UTC)

When you grasp and raise the book it moves in a plane parallel to or containing the axis for which the book was rotated upright. Parallel is correct in the text so your edit is reverted. These 3D motions can be confusing. The rigid body motion includes a general rotation and a general translation. This theorem’s content is that the translation can be decomposed into two: one along the axis of rotation and the other perpendicular to the axis, and that the latter can be made part of the rotation by selecting an appropriate axis parallel to the axis of rotation in the original rigid body motion. A proof using linear fractional transformations of quaternions has been given in Wikibooks: b:Associative Composition Algebra/Quaternions#Screw displacement. Continue to ask questions if this is still opaque. — Rgdboer (talk) 23:16, 20 May 2024 (UTC)

The rotation axis runs from left to right. The translation is from low to high. These are perpendicular, not parallel. There is no such thing as a "plane of translation", there is only an affine direction of translation. The reason I provided an explicit book example was as a challenge: If you believe that the "parallel" formulation is correct, then please exhibit the rotation and translation that results in the same initial and final book positions. Simply asserting that "they exist" and "its confusing" and saying it has something to do with "screw motions", doesn't resolve the issue. Worse: screw motions are neither rotations, nor are they translations, and thus aren't even in-bounds for the theorem, anyway: the screw motion is not one of the two allowed motions. It seems wiser to tag you with a @Rgdboer: than to enter into revert warfare. Oh, please tag me back; I'd rather not check this page three times a day to see if there's a response. 67.198.37.16 (talk) 04:22, 21 May 2024 (UTC)

In your example your composite motion (book rotation upward about the proximate edge composed with translation vertically off the table) can be expressed as a simple rotation with no translation at all (this time a rotation about an axis along the length of the table sitting somewhere closer to you than the book and above the level of the tabletop). You'll need something slightly trickier if you want it to be expressible as a screw motion. –jacobolus (t) 06:03, 21 May 2024 (UTC)

Ah. I slept on it, and now I see. The axis is still left-right, the angle of rotation is still 90 degrees. The axis is located vertically exactly half-way up between the initial and final locations of the book, and an equal amount in front of the vertical drop from final to initial positions. Looked at from the side, this forms a right triangle, with the hypotenuse exactly vertical, the bottom vertex of hypotenuse being the original book position, and the top of the hypotenuse being the final position. Indeed, very clever. A diagram is in order, but alas. I now see that the thm is true as worded. Which then suggests that the Whittaker quote is faulty: it currently says perpendicular; it too should say parallel. I'll change this now, and leave you to revert, if this is again a misunderstanding? Oh, I see what Whittaker was saying as well, now. He's explaining how to get rid of the perpendicular component. Phew.

A conventional proof, instead of that "geometric algebra" thing would be nice, as well; it is inscrutable. "four reflections" holds as a theorem in an arbitrary number of Euclidean dimensions, and I think it holds in pseudo-Euclidean as well(??). It is not clear that the rest of the theorem holds in anything but 3D Euclidean space. 67.198.37.16 (talk) 15:22, 21 May 2024 (UTC)

In n-dimensional pseudo-Euclidean space, every isometry which preserves a fixed point is a composition of (up to) n reflections in hyperplanes passing through that point. That's the Cartan–Dieudonné theorem. You can also represent any translation as a composition of two reflections in parallel hyperplanes. I think the most general isometry may need up to n+2 reflections (Edit: thinking about it a bit more n+1 reflections is clearly sufficient for some values of n.... or maybe in general). But Chasles' theorem is just about orientation-preserving isometries of Euclidean 3-space, which are always screw motions (that title should really be a separate article). –jacobolus (t) 15:57, 21 May 2024 (UTC)