October 2017 - mathfun@mailman.xmission.com

[math-fun] Frame dragging (was Re: Fabricating a glass/crystal sphere circa 1500)
by Keith F. Lynch 25 Oct '17

25 Oct '17

Brent Meeker <meekerdb(a)verizon.net> wrote: > Keith F. Lynch wrote: >> (Copernicus, Galileo, Kepler et al believed that the celestial >> sphere stood still and Earth rotated. They were mostly right, but >> if they had kept close track, they would have noticed that the >> celestial sphere rotates by about 0.04 seconds of arc per year due >> to frame dragging.) > I don't understand that. The celestial sphere is defined as not > rotating and is empirically fixed by distant galaxies. That certainly wasn't the original definition. Before Copernicus it was generally believed that it rotated about once per day around the stationary Earth. According to relativity, with linear motion it's arbitrary which frame of reference is regarded as stationary. Not so with rotary motion. Numerous experiments can be done to determine whether the Earth turns or whether the celestial sphere, distant galaxies and all, revolve around the Earth. Or some of each. At Earth's north or south pole, set up a frictionless pendulum that's free to swing in any horizontal direction. Set it to swing towards and away from some distant galaxy that's on the horizon. If Earth is rotating, it will continue to swing towards and away from that distant galaxy as the Earth rotates under it. Anywhere on or near Earth, point a gyroscope's axis at some distant galaxy. See if it keeps pointing at that galaxy. Fire a cannonball in the northern hemisphere. See if its trajectory curves to the right, and if so by how much. On the equator, simultaneously launch two satellites into equatorial orbits, identical except that one is eastbound, the other westbound. See whether they're directly overhead when they collide. Send a radio signal in all horizontal directions at once. Measure when it returns to you from each direction. Deduce Earth's rotational rate from the magnitude of its equatorial bulge. The frame dragging effect, also known as Lense-Thirring precession, says that space in and near a rotating mass such as Earth will be dragged along with it somewhat. As such, those experiments ought to show that Earth is rotating slightly less than observations of distant galaxies would imply, and that those distant galaxies are all revolving slowly around the Earth in the opposite direction. Slowly in terms of arc seconds, quite rapidly in meters per second, given how distant those galaxies are. They're going many times faster than light. No, that doesn't contradict general relativity. GR says that that speed limit is strictly local. With the possible exception of Gravity Probe B, Earth frame dragging has never been detected, as the numbers are just too small. But it's certainly been detected around neutron stars and black holes. In _Incandescence_ by Greg Egan, a pre-technological civilization with no knowledge of astronomy (they live inside a hollow asteroid) works out general relativity from a few simple experiments and measurements and some careful reasoning. They have the advantage of being in close orbit of a massive black hole where GR effects are significant. (Well, it's an advantage in discovering GR, but a great disadvantage in all other ways, as it's a rather dangerous neighborhood.)

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[math-fun] Four constants (e, pi, gamma and phi) in a continuous cubic root.
by François Mendzina Essomba 24 Oct '17

24 Oct '17

*Hello, Four constants (e, pi, gamma and phi) in a continuous cubic root. Formula 33 a:=2/((3*sqrt(5))*((1/9)*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(-1/3)-((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2))))) ; b:=(3/sqrt(5))*(((3*sqrt(5)*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2))))-((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(1/3))/(1-9*((exp(-gamma)*Pi)/2+sqrt(27*Pi^2*exp(-2*gamma)-4)/(2*3^(3/2)))^(4/3))); r:=1/((((((((((((exp(-gamma)*Pi+1)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+ …) ; (phi)^(-1)=(1/a)*(b+r); An approximate value will be: (phi)^(-1)=(1/a)*(b+1/(((((((((((exp(-gamma)*Pi+1)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)+exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3) +exp(-gamma)*Pi)^(1/3)); Formula 34 continuous fraction phi:=sqrt(4*exp(-5*Pi)+4106118241)/(sqrt(5)*(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)-sqrt(4*exp(-5*Pi)+4106118241)/(2*sqrt(5))+64079/2)+…) ; An approximate value will be: phi:=sqrt(4*exp(-5*Pi)+4106118241)/(sqrt(5)*(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)/(exp(-5*Pi)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)+64079)-sqrt(4*exp(-5*Pi)+4106118241)/(2*sqrt(5))+64079/2)) ; Best regards...*

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Re: [math-fun] Random slice of a cube
by Keith F. Lynch 24 Oct '17

24 Oct '17

Allan Wechsler <acwacw(a)gmail.com> wrote: > Your lowbrow procedure excludes some planes that do intersect the > cube. Pick a point Q just "above" the center of one of the cube's > faces, (so it would have been excluded by your procedure) and erect > the plane through Q perpendicular to OQ. Now tilt the plane just a > little toward one of the four vertices of the chosen face, until the > plane nicks that vertex. There is a point Q' in this tilted plane > such that the tilted plane is perpendicular to OQ'; I contend that > this can be done so that Q' is still outside the cube, and hence the > tilted plane would have been excluded even though it intersects the > cube. My method was to pick a point Q at a distance less than the cube root of 3 from the cube's center, i.e. a point inside the sphere that circumscribes the cube. But now I'm wondering what *is* the locus of points such that a plane including Q and perpendicular to the line connecting Q to the center would intersect the cube. The locus obviously includes everything in the cube and nothing outside the circumscribing sphere. But it's not coincident with either the cube or the sphere. What is its shape, its volume, and its surface area? I'm assuming the cube has a side length of 2 and is centered on the origin. I'm pretty sure that parts of the shape are concave. A simpler problem is to reduce it by a dimension. What's the shape, area, and perimeter of the locus of points such that a line through the point and perpendicular to the line connecting the point to the center intersects the square with side length 2 and centered on the origin? Again, I'm pretty sure that parts of the shape are concave.

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Bill Gosper 23 Oct '17

23 Oct '17

Interesting 16/sec ball bearing production process: https://www.youtube.com/watch?v=19duYMdiXi0 Falling raindrops form cabochons: https://www.youtube.com/watch?v=a9CRrGvQwe0 1961|2 freshman chemistry lecturer demoed Prince Rupert's Drops, and then something more interesting. (Prince <othername>'s Drops? Google fails me.) Hollow, thick-walled glass blobs, open at one end. He hammered nails with one, then wrapped it in a towel, dropped in some carborundum crumbs, BAM! --rwg On 2017-10-22 11:57, Dan Asimov wrote: > Henry got that right — in fact, something quite interesting happens, > as Wikipedia says: > > ----- > Prince Rupert's Drops (also known as Dutch or Batavian tears) are > toughened glass beads created by dripping molten glass into cold > water, which causes it to solidify into a tadpole-shaped droplet with > a long, thin tail. These droplets are characterized internally by very > high residual stresses, which give rise to counter-intuitive > properties, such as the ability to withstand a blow from a hammer or a > bullet on the bulbous end without breaking, while exhibiting explosive > disintegration if the tail end is even slightly damaged. In nature, > similar structures are produced under certain conditions in volcanic > lava. > ----- > > more at https://en.wikipedia.org/wiki/Prince_Rupert's_Drop. > > —Dan > > > Henry Baker wrote: > ----- > If you don't remove the air in such a column, you're likely to get > shot that is more raindrop-shaped than sphere-shaped. > ----- > > I wrote: > ----- > ... I'd try dropping molten glass some distance through the air to > make it form into a sphere, and have it land in water to make it > rigid. > > —Dan > > > Henry Baker wrote: > ----- > It's the year 1500, and some wealthy patron wants a transparent glass > and/or rock crystal sphere perhaps 6-8" in diameter. > > How would you fabricate it? > ... > ... > ----- > ----- > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Getting dense in a sphere
by Dan Asimov 23 Oct '17

23 Oct '17

Henry's question suggests the following math problem in the context of polishing a (large) BB: Suppose you are given a continuous curve A: [0,L] —> S^2 parametrized by arclength for some L > 0, where S^2 is the unit sphere. Let D(r, p) denote the closed disk of radius r about the point p of S^2, where r is measured along S^2. Then define gap_L(A) = sup{ r > 0 | For some p in S^2, D(r, p) does not intersect A([0,L])} For any L > 0, define mingap(L) = inf { gap(A) | A: [0,L] —> S^2 is parametrized by arclength }. Then what can be said about the values of mingap(L) ? What is its asymptotic behavior as L —> oo ??? —Dan Henry Baker wrote: ----- It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. How would you fabricate it? ... ... -----

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Re: [math-fun] Random slice of a cube
by Keith F. Lynch 23 Oct '17

23 Oct '17

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron. I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes. > Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons). I meant regular polyhedrons, not polygons. > * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides. I meant the maximum distance from the center of the cube to the closest point on the plane. > Unfortunately, my program cannot easily be adapted to other regular > polygons. Again, I meant polyhedrons, not polygons. > I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested. In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices. > The average number of sides is 4.0004. I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4. This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square. I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4. I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried. Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.

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Re: [math-fun] Fabricating a glass/crystal sphere circa 1500
by Dan Asimov 23 Oct '17

23 Oct '17

Henry got that right — in fact, something quite interesting happens, as Wikipedia says: ----- Prince Rupert's Drops (also known as Dutch or Batavian tears) are toughened glass beads created by dripping molten glass into cold water, which causes it to solidify into a tadpole-shaped droplet with a long, thin tail. These droplets are characterized internally by very high residual stresses, which give rise to counter-intuitive properties, such as the ability to withstand a blow from a hammer or a bullet on the bulbous end without breaking, while exhibiting explosive disintegration if the tail end is even slightly damaged. In nature, similar structures are produced under certain conditions in volcanic lava. ----- more at https://en.wikipedia.org/wiki/Prince_Rupert's_Drop. —Dan Henry Baker wrote: ----- If you don't remove the air in such a column, you're likely to get shot that is more raindrop-shaped than sphere-shaped. ----- I wrote: ----- ... I'd try dropping molten glass some distance through the air to make it form into a sphere, and have it land in water to make it rigid. —Dan Henry Baker wrote: ----- It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. How would you fabricate it? ... ... ----- -----

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[math-fun] accelerated convergence
by François Mendzina Essomba 23 Oct '17

23 Oct '17

Hello, After reading some articles on the Brent-Salamin algorithm based on the AGM and algorithms of the Borweins brothers arising from the famous Ramanujan formulas on the Pi number. I wondered if it was possible to find an algorithm for any convergence speed m considered. Finally I came to establish the following general formula which is of convergence (2 * m): x(k+1)=x(k)+sum(2*((m!)^2)*sin(n*x(k))/(n*((m+n)!)*((m-n)!)),n=1..m); x(0) = 3 ; k --- infinity ; x(k) ---- Pi ; If we take m = 5, the convergence is (2 * m) = 10; we are getting: x(k+1)= x(k)+(5/3)*sin(x(k))+(10/21)*sin(2*x(k))+(5/42)*sin(3*x(k))+(5/252)*sin(4*x(k))+(1/630)*sin(5*x(k)); x(0)=3; x(1)=3.141592653589628958434997482316410876703079409530919198836399516902375075829274562410138150533960715 ; x(2)=3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068 ; So if I want to write a 2000000 convergence formula, then I would simply have: x(k+1)=x(k)+sum(2*((1000000!)^2)*sin(n*x(k))/(n*((1000000+n)!)*((1000000-n)!)),n=1..1000000); Unfortunately, this formula can not be useful for the calculation of Pi given the difficulty related to the calculation of sinuses. Best regards...

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Re: [math-fun] math-fun Digest, Vol 176, Issue 31
by françois mendzina essomba2 23 Oct '17

23 Oct '17

Gosper's identities (Bill Gosper); Hello, Thank you dear Professor Bill Gosper, for the details related to my recurrence, I did not verify it experimentally. The negative terms (-2) ^ n or (-3) ^ n of this identities seem to me so rich and so profound ... It took me longer to understand Best regards. Le Samedi 21 octobre 2017 20h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. 31 lines of zero (mod 15) puzzle. (Ed Pegg Jr) 2. Gosper's identities (Bill Gosper) 3. Gosper's nonidentities (Bill Gosper) ---------------------------------------------------------------------- Message: 1 Date: Fri, 20 Oct 2017 23:00:36 -0500 From: Ed Pegg Jr <ed(a)mathpuzzle.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: [math-fun] 31 lines of zero (mod 15) puzzle. Message-ID: <CAGog0+Y90bRpf2mWS9eZ_Ho3DbAdt6UHVWvNfGCRev2CbLqm7g(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" Here's an orchard of 31 lines of three on 15 points. That might be maximal according to A003035. https://i.imgur.com/w18qs7R.jpg Arrange the fifteen numbers -7 to 7 in the disks so that the (mod 15) sum of all lines is zero. --Ed Pegg Jr ------------------------------ Message: 2 Date: Sat, 21 Oct 2017 02:22:35 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Gosper's identities Message-ID: <CAA-4O0E0cmKVasfuY65rGjf496FDtY8anutimpgecOeLyngZgw(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" Dear Dr. Essomba, Thank you for catching my D111 mistake. Your claim is immediately substantiated by the limit ??0, which gives 1 = ?. And confirmed by rederivation, simply multiplying D110 by itself with ???3?. There are several errors on that page (which needs to be done over in Mathematica) due to faulty conversion to html, (e.g., p for ?, f for ?,...) but this isn't one of them, since the error is repeated in C111. Puzzling. Also, here is experimental confirmation of your 2017-10-13 07:51 recurrence: In[502]:= NestList[(Print[??(2^(4 #[[1]] + 5) (3 - 2 ?(2 + 2 #[[2]]) + #[[2]]))]; {#[[1]] + 1, ?(2 + 2 #[[2]])/2}) &, {0, -1.}, 9] During evaluation of In[502]:= 2.82842712474619 During evaluation of In[502]:= 3.061467458920717 During evaluation of In[502]:= 3.12144515225806 During evaluation of In[502]:= 3.136548490545999 During evaluation of In[502]:= 3.140331156951562 During evaluation of In[502]:= 3.141277250930965 During evaluation of In[502]:= 3.141513800680598 During evaluation of In[502]:= 3.141572952943204 During evaluation of In[502]:= 3.141587832038716 Out[502]= {{0, -1.}, {1, 0.}, {2, 0.7071067811865476}, {3, 0.9238795325112867}, {4, 0.9807852804032304}, {5, 0.9951847266721969}, {6, 0.9987954562051724}, {7, 0.9996988186962042}, {8, 0.9999247018391445}, {9, 0.9999811752826011}} where u_n (= #[[2]]) converges to 1 while a_n converges to ??. --rwg Date: 2017-10-16 05:35 From: Fran?ois Mendzina Essomba via math-fun <math-fun(a)mailman.xmission.com> To: math-fun(a)mailman.xmission.com, m_essob(a)yahoo.fr Reply-To: Fran?ois Mendzina Essomba <m_essob(a)yahoo.fr>, math-fun < math-fun(a)mailman.xmission.com> Hello, I found in the page http://www.tweedledum.com/rwg/idents.htm strange identities, as incredible as each other ... I have observed at length some of the identities ranging from C105 to C115. And this is what I deduced from it: (9) D106 product(2*cos(theta/3^(n/m))-1,n=1..infinity)=product(cos(3^(i/m)*theta/2),i=0...m-1) ; (10) D105 product(2*cos(theta/2^(n/m))-1,n=1..infinity)=product((1+2*cos(2^(i/m)*theta))/3,i=0..m-1); (11) D110 product((2*cos(theta/3^(n/m))+1)/3,n=1..infinity)=(2^m/(theta)^m)*product(sin(3^(i/m)*theta/2)/3^(i/m),i=0..m-1); However I am a little perplexed as to the last formula that differs from Gosper's D111 identity by a factor of 4. The result I get is this: (12) D111 product((2*cos(theta/3^(n/2))+1)/3,n=1..infinity)=(2^2/(theta)^2)*product(sin(3^(i/2)*theta/2)/3^(i/2),i=0..2-1); I suppose it must be the same for this small series, I have yet to check. Best regards ------------------------------ Message: 3 Date: Sat, 21 Oct 2017 10:15:19 -0700 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Gosper's nonidentities Message-ID: <CAA-4O0ER2aMjEHX2+SdqNzGc0rpkfXApmnGrx4jX9mWAzuNT0g(a)mail.gmail.com> Content-Type: text/plain; charset="UTF-8" While collecting reasons why Mathematica's StirlingS1 and StirlingS2 should, like Macsyma, admit negative and fractional arguments, I found, to my discomfort, >From rwg(a)NEWTON.Macsyma.COM Wed Jul 24 03:00:00 1996 Received: from optima.CS.Arizona.EDU by cheltenham.cs.arizona.edu; Wed, 24 Jul 1996 03:03:19 MST Received: from NEWTON.Macsyma.COM by optima.cs.arizona.edu (5.65c/15) via SMTP id AA07538; Wed, 24 Jul 1996 03:03:16 MST Received: from SWEATHOUSE.macsyma.com ([192.233.166.105]) by NEWTON.Macsyma.COM via INTERNET with SMTP id 363715; 24 Jul 1996 06:01:03-0400 Date: Wed, 24 Jul 1996 03:00-0700 From: Bill Gosper <rwg(a)NEWTON.macsyma.com> Reply-To: rwg(a)NEWTON.macsyma.com Subject: funny-looking sum To: math-fun(a)cs.arizona.edu Message-Id: <19960724100013.6.RWG(a)SWEATHOUSE.macsyma.com> (Using Knuthian (nonnegative) Stirlings) inf n - 2 ==== ==== x k + 1 \ \ (%e - x - 1) stirling_s1(n, n - k) - x n > ( > ----------------------------------------) %e = log(x + 1) / / (k + 1)! ==== ==== n = 1 k = 0 apparently for x>=0, anyway. (Convergence rate is useless.) ----------------------------------- It doesn't even slightly work, presumably mistranscribed. To my active distress, Eric Weisstein has faithfully quoted this bogon as eqn (26) in http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html, (except my S is his |S|). I am so far unable to guess its correct form, which presumably existed because I would have Taylor-expanded the ???? out of anything this weird. Help? --rwg ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 31 *****************************************

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[math-fun] Fabricating a glass/crystal sphere circa 1500
by Henry Baker 22 Oct '17

22 Oct '17

It's the year 1500, and some wealthy patron wants a transparent glass and/or rock crystal sphere perhaps 6-8" in diameter. How would you fabricate it? I don't know how they actually did it, but here are some of my speculations: 1. To make a solid glass sphere, one might make a spherical mold, melt glass into it, and then polish the result. 2. To make a rock crystal sphere, one might build a lathe, and then utilize a semicircle guide to cut a rough crystal block into nearly spherical shape, and then polish the result. One might even take the glass from the mold & mount it onto a lathe for the polishing step. Clearly (!), one could continually reposition the sphere in the lathe to work on different axes to attempt to make the sphere as precisely spherical as possible. Are there more elegant methods? Using just the technology available in 1500, what kind of precision could be achieved? [BTW, a similar problem arose in the 19th Century when *ball bearings* became important. Apparently, the longevity of various kinds of engines depended critically on making the bearings as spherical as possible. I don't know enough about how this precision was achieved, but during WWII, bearing manufacturing plants were heavily targeted for bombing.]

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