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        riddles >> hard >> Designing a baseball
        
(Message started by: JocK on Aug 25th, 2005, 11:24am)
Title: Designing a baseball
Post by JocK on Aug 25th, 2005, 11:24am
A 3/8 x 1/8 rectangle has unit circumference. Stiching two of these rectangles together yields a cube of size 1/8 x 1/8 x 1/8 with volume 1/512.

Surely we can do better than that!

What is the maximum volume you can obtain by stitching together two identical unit circumference planar shapes?

Can you prove for your solution that planar unit-circumference shapes yielding larger volumes are not possible?




 
Title: Re: Designing a baseball
Post by SMQ on Aug 25th, 2005, 1:07pm
Hmm.  The five platonic solids:

Tetrahedron: a = 1/4; V = sqrt(2)/768 ~ .00184
Cube: a = 1/8; V = 1/512 ~ .00195
Octahedron: a = 1/6; V = 1/648 ~ .00154
Dodecahedron: a = 1/20; V = (15 + 7 sqrt(5)) / 32000 ~ .00096
Icosahedron: a = 1/12; V = 5 (3 + sqrt(5)) / 20736 ~ .00126

Round One to the cube...

--SMQ
Title: Re: Designing a baseball
Post by Barukh on Aug 26th, 2005, 4:10am
Round 2:

2 rectangles with sides 1/3, 1/6 produce a cuboid 1/6 x 1/6 x 1/12 with volume 1/432 = 0.00231...

Two 30-side planar shapes produce icosidodecahedron (http://mathworld.wolfram.com/Icosidodecahedron.html) with volume 0.00307... ~ 1/325.25
Title: Re: Designing a baseball
Post by SMQ on Aug 26th, 2005, 6:05am

on 08/26/05 at 04:10:48, Barukh wrote:
Two 30-side planar shapes produce icosidodecahedron (http://mathworld.wolfram.com/Icosidodecahedron.html) with volume 0.00307... ~ 1/325.25


Eh?  The link gives V = (1/6)(45+17sqrt(5)) for a = 1, so for a = 1/30 that's V = [(1/6)(45+17sqrt(5))] * (1/30)^3 ~ .00051, right?

In general, more closely approaching a sphere seems to require a more complex perimeter, which reduces the scale too much.

My best so far: a bi-cone, where each of the two cones has diameter = height  = 1/(sqrt(5)+pi) ~ .186; V ~ .00337

--SMQ
Title: Re: Designing a baseball
Post by Barukh on Aug 26th, 2005, 6:12am

on 08/26/05 at 06:05:31, SMQ wrote:
Eh?  The link gives V = (1/6)(45+17sqrt(5)) for a = 1, so for a = 1/30 that's V = [(1/6)(45+17sqrt(5))] * (1/30)^3 ~ .00051, right?

Correct. I didn't notice the 1/6 coefficient. Sorry.


Quote:
In general, more closely approaching a sphere seems to require a more complex perimeter, which reduces the scale too much.

This seems to be the right observation.
Title: Re: Designing a baseball
Post by SMQ on Aug 26th, 2005, 6:32am
It just occurred to me that my biconic solution is formed by simply cutting a circle in half.  I think that's a proveably optimum solution for surface area with a fixed perimeter.  If there's a way to show that, for the types of solids we're considering, maximizing surface area also maximizes volume, we may have at least a sketch of the optimality proof...

Or, altermatively, I could be off my nut again. ;D

EDIT: See below; I'm definitely off my nut...

--SMQ
Title: Re: Designing a baseball
Post by JocK on Aug 26th, 2005, 11:09am

on 08/26/05 at 06:05:31, SMQ wrote:
In general, more closely approaching a sphere seems to require a more complex perimeter, which reduces the scale too much.


This is an important observation: you are effectively maximising V/L3 with V the volume and L the total stitch-length (circumference of each of the two shapes). Small sacrifices in 1/L in order to make a gain in V hit back to the third power... So keep it simple! ;)

You guys went a long way: from a cube (1/512) via a block (1/432) to a bi-cone (1/297). But... there are solutions better than a bi-cone...

(Have to admit that I have no idea whether my best solution is optimal. So I am curious to see whether one can find even better solutions...)
Title: Re: Designing a baseball
Post by SMQ on Aug 26th, 2005, 6:06pm
OK, let each planar shape be a circle of radius ~ .1626 with a chord subtending ~85.233 degrees cut out.  These shapes can be folded into truncated cones with a base radius of ~.1241, an end radius of ~.1197, and a height of ~.1100.

These give a volume of ~.01027 or about 1/97.37

EDIT: On further thought, I don't think you can construct a truncated cone like that...whups.  Back to the drawing board, I guess... :-[

--SMQ
Title: Re: Designing a baseball
Post by Barukh on Aug 28th, 2005, 8:52am
[quote author=SMQ link=board=riddles_hard;num=1124994244;start=0#3 date=08/26/05 at 06:05:31]My best so far: a bi-cone, where each of the two cones has diameter = height  = 1/(sqrt(5)+pi) ~ .186; V ~ .00337[/quote]

on 08/26/05 at 06:32:23, SMQ wrote:
It just occurred to me that my biconic solution is formed by simply cutting a circle in half.

Could you explain how did you get to this result? My derivations for half-circle give 0.00333… ~ 1 / 299.75.
Title: Re: Designing a baseball
Post by Grimbal on Aug 29th, 2005, 5:51am
I would go for a cylinder-cylinder intersection.  The axes being perpendicular, but offset by one radius, i.e. each axis is tangent to one circle around the other cylinder.  The intersection is like a baseball except that both halves are made of a flat roughly elliptic sheet.

I don't know about optimality, I haven't computed the volume or the intersection length for a given radius, but it seems to have all the properties I would expect from an optimal solution.
Title: Re: Designing a baseball
Post by SMQ on Aug 29th, 2005, 6:21am

on 08/28/05 at 08:52:28, Barukh wrote:
Could you explain how did you get to this result? My derivations for half-circle give 0.00333… ~ 1 / 299.75.


My fault; my brain doesn't seem to be doing well at translating planar shapes to 3D in this thread...  The optimal biconic solution has cones with diameter = height; which my brain wrongly jumped to being formed from a half-circle (diameter = slant-height) ... sorry for the confusion.  :-[

--SMQ
Title: Re: Designing a baseball
Post by SMQ on Aug 29th, 2005, 10:27am
For a cylinder-cylinder intersection at right angles, I get an optimum volume of ~.00378 (1/265) when the cylinders are offset by ~.869 of their radius.

That is, of course, assuming that my numerical cylinder intersection solver is accurate, which, given my history in this thread, I wouldn't count on... ::)

--SMQ
Title: Re: Designing a baseball
Post by SWF on Aug 29th, 2005, 7:13pm
I have a way with a ratio of 1/263.6, and expect to be able to improve it a bit, because have not fully optimized.  Will explain the approach later.

Interesting that the Grimbal's cylinder intersection comes so close.  I get 1/266.781 for cylinders offset by one radius.  Haven't tried optmizing the offset a little, as done by SMQ.
Title: Re: Designing a baseball
Post by SMQ on Aug 30th, 2005, 11:03am
Fiddling with the cylinder-cylinder intersection a bit more:

A slightly better result can be obtained by stretching the cylinders (so that their cross sections are elliptical) to make the intersection slightly "deeper".

Stretching by ~12.99% leads to an optimal solution when the axes of the elliptical cylinders are separated by ~.9664 of the major axis.  This appears to be the point where the stretching causes the plane shapes to switch from concave to convex, and leads to a volume of ~.003796 or ~1/263.4.

If I have more than a few minutes free I'll try to solve the maths to get an exact solution for this situation...

--SMQ
Title: Re: Designing a baseball
Post by SWF on Aug 30th, 2005, 6:25pm
263.4 is also what I am getting.  Interesting that deforming intersection cylinders works so well.

The approach I used was to define the shape of the seam in three dimensions, for the shape it takes after the two pieces are combined into the volume.  Define one quarter of the curve parametrically in terms of x with x going from 0 to 'a', assuming some symmetry.  y coordinate is some function of x, f(x).  For the identical pieces to fit together, the z coordinate is given by f(a-x).

The volume is 8 times the integral from 0 to a/2 of f(x)*f(a-x).  Seam length is 8 times integral of sqrt(1+f '(x)^2+f '(a-x)^2 ). Find f(x) to maximize V/L^3.  Perhaps calculus of variations could be used with constraint on length, but I was unsure how to deal with the f(a-x).

To maximize I used Fourier series assuming f '(x) to be 0 at x=a.  Also it seemed reasonable to make f '(x) at 0 be infinite, so fewer terms needed if f(x) is square root of a Fourier series.  The form assumed for f() is:

f(x) = sqrt ( b1*sin(x) + b3*sin(3x) + b5*sin(5x) +...)   with a= pi/2

Find b1, b3, b5, ... that maximize V/L^3 (this was done numerically)

1 term of the series gives approxmately 1/264.921
2 terms in gives 1/263.601
3 terms gives 1/263.454
4 terms gives 1/263.422
Title: Re: Designing a baseball
Post by Barukh on Aug 31st, 2005, 1:06am
This bi-cylinder case is exciting! And raises several questions:

1. What's this "radius offset" parameter?

2. Are all the bodies "unfoldable"?

3. What about cone-cone case?
Title: Re: Designing a baseball
Post by JocK on Aug 31st, 2005, 3:26pm
Well.... with SMQ and SWF reaching values in excess of 1/265, I think it is about time to post my own solution...

I took the same approach as SWF and tried to optimise the 3-D curve that describes the seam of the two planar shapes. However, I was pretty convinced that the optimal curve would lay on a sphere (see the title of this thread!), so I used a parametric representation of the form:


x(phi) = sqrt(R2 - z2(phi)) cos(phi)

y(phi) = sqrt(R2 - z2(phi)) sin(phi)

z(phi) = R(1/sqrt(2) + a1 sin(phi) + a2 sin2(phi)  .. )

for (0 < phi < pi) describing a quarter of the total 3-D curve. (Note that it is crucial that the end-points of the quarter curve end in the coordinates (+/-1/sqrt(2), 0, 1/sqrt(2)) so as to be able to compose the total stitch out of four identically shaped quarter curves.)

For the above 3-D stitch, the enclosed volume V is given by:

V = 4 Integral (o < phi < pi) x(phi) z(phi) dy/dphi dphi

and the 'stitch length' L by:

L = 4 Integral (o < phi < pi) sqrt((dx/dphi)2 + (dy/dphi)2 + (dz/dphi)2) dphi

If all ai's are set to zero, it follows that V/L3 = 1/32 pi2 = 1/315.8.

However, optimising V/L3 by tuning the ai's leads to much improved values. Using Excel's solver I reached values of about 1/265. (Using only the first few ai terms.)

I am not too certain about the numerical accuracy of my exercisez in Excel, but doubt whether I could reach 1/263.4...

Yet, my intuition still tries to tell me the curve should be positioned on a sphere...  ???
Title: Re: Designing a baseball
Post by JocK on Aug 31st, 2005, 3:43pm
Based on SWF's Fourier-parametrisation, I should perhaps have used:

z(phi) = R(1/sqrt(2) + a1 sin(phi) + a2 sin(3.phi) + .. + ak sin((2k-1).phi))

However, a more fundamental question for me is:

Does SWF's more general approach yield a 3-D curve that with suitably chosen z(phi)  (*) can be parametrised as:

x(phi) = sqrt(R2 - z2(phi)) cos(phi)

y(phi) = sqrt(R2 - z2(phi)) sin(phi)


 
(The interesting contributions to this thread leave me with as many questions as Barukh..! ;) )



(*)  z(phi) defined on 0 < phi < pi, z(0) = z(pi) = 1/sqrt(2) and z(pi - phi) = z(phi).
Title: Re: Designing a baseball
Post by markr on Aug 31st, 2005, 9:55pm
This is all quite interesting, but are you tossing out the planar requirement?  You can't wrap a planar figure around a sphere, right?
Title: Re: Designing a baseball
Post by JocK on Sep 1st, 2005, 4:21am

on 08/31/05 at 21:55:06, markr wrote:
[..] are you tossing out the planar requirement?  You can't wrap a planar figure around a sphere, right?


Huh..?  ???

I was discussing two planar shapes - both bended and stitched together with the stitch being located on a sphere. The 3-D object thus created obviously is not a sphere.

The easiest way to visualise the approach is to consider the case z(phi) = constant = 1/sqrt(2).

In that case the object created consists of two half-cylinders, each with height equal to twice the radius, glued together with the flat faces each rotated 90 degrees. The round edges of this object are located on a sphere with radius sqrt(2) times the radius of the cylinders.

For z depending on phi, the object obviously becomes more complicated, but still consists of two bended planar shapes stitched together using a 'spherical 3-D stitch'.
Title: Re: Designing a baseball
Post by JocK on Sep 1st, 2005, 1:14pm

on 08/29/05 at 05:51:00, Grimbal wrote:
IThe intersection is like a baseball except that both halves are made of a flat roughly elliptic sheet.


All recent (close to optimal?) approaches to this problem seem to lead to (roughly) elliptical planar shapes.

Could it be that two identical ellipses - with a suitable eccentricity - when bended and stitched together would produce an optimal result?

I have to think a bit about how to calculate the stitch length and the volume. (Doesn't seem to be easy..)
Title: Re: Designing a baseball
Post by SMQ on Sep 1st, 2005, 1:56pm
Hmm, I've now tried playing with other distortions of the cylinder intersection (specifically, representing the cross-section as (x/a)i+yj=1) and made only extremely minimal gains.

The same thought had occurred to me about ellipses, but all the solutions I'm finding from pseudo-cylinder intersections look far more like ovals (two semicircles joined by flat sides) than ellipses.

--SMQ
Title: Re: Designing a baseball
Post by JocK on Sep 1st, 2005, 2:17pm

on 08/31/05 at 15:43:22, JocK wrote:
[..] a more fundamental question for me is:

Does SWF's more general approach yield a 3-D curve that with suitably chosen z(phi)  (*) can be parametrised as:

x(phi) = sqrt(R2 - z2(phi)) cos(phi)

y(phi) = sqrt(R2 - z2(phi)) sin(phi)


Well.. not sure about SWF's solution, but SMQ's modification of Barukh's intersecting cylinder model does come close...

The seam curve for the Barukh/SMQ model for elliptical cylinders (with general offset 2r) satisfies:

x2/a2 + (z - r)2/b2 = 1

y2/a2  + (z + r)2/b2 = 1

which can be re-expressed as an equation for a spheroid plus an expression that describes a curve traced out on that spheroid:

(x2 + y2)/a2 + 2 z2/b2 = 1 - r2/b2

y2 - x2 + 4 (a/b)2 z r = 0

So over to curves on spheroids...  :-/  (I dare to conjecture that SWF is effectively also creating seams on a spheroid..)

(No, I am not going to rename this thread to "Designing a rugby ball"...  ;D   ... B.t.w.: no means for me to check, but I assume we are dealing here with oblate spheroids? SMQ, what is the value of a/b that led to your optimum result?)
Title: Re: Designing a baseball
Post by SMQ on Sep 1st, 2005, 5:10pm

on 09/01/05 at 14:17:17, JocK wrote:
B.t.w.: no means for me to check, but I assume we are dealing here with oblate spheroids? SMQ, what is the value of a/b that led to your optimum result?)


Um, if I'm translating my coordinate system to yours correctly in my head, then no, it's a prolate spheroid.  I believe I have a = 1, b = 1.1299 for my best ellipsoidal solution.

I also, as mentioned above, have an ever-so-slightly better non-ellipsoidal solution given by pseudo-cylinders with the following equations: (where the all-positive solutions are mirrored across the appropriate axes)

(x/1.22185)2.22878 + y2.00202 = 1

(x/1.22185 - 1.0504)2.22878 + z2.00202 = 1

--SMQ
Title: Re: Designing a baseball
Post by JocK on Sep 3rd, 2005, 6:38am

on 09/01/05 at 17:10:56, SMQ wrote:
I also, as mentioned above, have an ever-so-slightly better non-ellipsoidal solution given by pseudo-cylinders


Yes, that proves that the optimal planar shape leads to a seam that can not be located on a spheroid.

Whilst the deviations from a spheroid seem to be very small, and the improvements in terms of V/L3 miniscule, it does draw the attention to SWF's  more general approach, which I transcribe as follows:

For two optimal identical planar shapes 'stitched together', there is a function f(t) such that the resulting 3-D volume is described by:

-f(a+z) < x < f(a+z)

-f(a-z) < y < f(a-z)

with -a < z < a.

Here, the function f(..) maps the non-negative reals onto themselves, and satisfies f(0) = 0.

The volume of the above shape is given by:

V = 4 [Integral  0 < z < a] f(a+z) f(a-z) dz

and the 'seam length' by:

L = 8 [Integral  0 < z < a] sqrt(1 +f'(a+z)^2 + f'(a-z)^2) dz

where f'(t) denotes the derivative df/dt.

As suggested by SWF, we should now apply calculus of variations. We don't need to optimise V whilst constraining L (or the other way around) using Lagrange multiplyers, but we can in principle also optimise V/L^3 directly. Either way, the expressions seem to get rather ugly.

So like SWF, I turned to a numerical approach.  Unlike SWF, I used a simple Taylor expansion for the square of f(t):

f(t)2 =  a1 t + a2 t2 + a3 t3 + ...

Note that although the expression is different from SWF's Fourier expansion, also here I expanded the square of f(t) rather than f(t) itself so as to allow lim t -> 0 1/f'(t) = 0.

Retaining only two terms, in the above expression:

f(t)2 = 3.030 t - 0.783 t2

we obtain a V/L3 approximately equal to the often quoted value 263.4. The benefit of adding more terms quickly drops, and using above expression, one can only make improvements in the less significant digits.

This suggests that we have reached the optimal value. To prove this in a rigorous way needs a brave person who derives the optimal curve from variational principles.

Any takers?  ;D
Title: Re: Designing a baseball
Post by SMQ on Sep 4th, 2005, 5:49am

on 09/03/05 at 06:38:05, JocK wrote:
Yes, that proves that the optimal planar shape leads to a seam that can not be located on a spheroid.

Hmm, the differences are so small I don't feel confident they're not an unintended artifact of my hand-coded numeric solver...  or even just a floating-point artifact.  I 'suppose I could run the integrals through Mathematica or Maple to be more sure.


EDIT: OK, after crashing my newly-downloaded-and-registered trial Mathematica kernel several times, I finally obtained a convincing result:

In[1] := y[x_, a_, d_, i_, j_] := (1 - Abs[x/a]^i)^(1/j)

In[2] := z[x_, a_, d_, i_, j_] := (1 - Abs[-d + x/a]^i)^(1/j)

In[3] := V[a_, d_, i_, j_] := 4Integrate[y[x, a, d, i, j]z[x, a, d, i, j], {x, a(-1 + d), a}]

In[4] := L[a_, d_, i_, j_] := 4Integrate[Sqrt[1 + D[y[x, a, d, i, j], x]^2 + D[z[x, a, d, i, j], x]^2], {x, a(-1 + d), a}]

In[5] := W[a_, d_, i_, j_] := V[a, d, i, j]/L[a, d, i, j]^3

In[6] := W[a, d, i, j]
Out[6] := Integrate[((1 - Abs[x/a]^i)^(1/j))((1 - Abs[-d + x/a]^i)^(1/j)), {x, a(-1 + d), a}] /
   (16Integrate[Sqrt[1 + ((i^2)(Abs[x/a]^(-2 + 2i))((1 - Abs[x/a]^i)^(-2 + 2/j))(Abs'[x/a]^2))/((a^2)(j^2)) + ((i^2)(Abs[-d + x/a]^(-2 + 2i))((1 - Abs[-d + x/a]^i)^(-2 + 2/j))(Abs'[-d + x/a]^2))/((a^2)(j^2))], {x, a(-1 + d), a}]^3)

In[7] := WN[a_?NumberQ, d_?NumberQ, i_?NumberQ, j_?NumberQ] :=
 Re[NIntegrate[Re[((1 - Abs[x/a]^i)^(1/j))((1 - Abs[-d + x/a]^i)^(1/j))], {x, a(-1 + d), a}] /
   (16NIntegrate[Re[Sqrt[1 + ((i^2)(Abs[x/a]^(-2 + 2i))Re[(1 - Abs[x/a]^i)^(-2 + 2/j)])/((a^2)(j^2)) + ((i^2)(Abs[-d + x/a]^(-2 + 2i))Re[(1 - Abs[-d + x/a]^i)^(-2 + 2/j)])/((a^2)(j^2))], {x, a(-1 + d), a}]^3)]

In[8] := NMaximize[{WN[a, d, 2, 2], .5 <= a <= 2, 0 <= d < 2}, {a, d}]
Out[8] := {0.00379605, {a -> 1.12984, d -> 0.966366]

In[9] := NMaximize[{WN[a, d, i, j], .5 <= a <= 2, 0 <= d < 2, 1 < i <= 3, 1 < j <= 3}, {a, d, i, j}]
 NIntegrate::pwncvb: NIntegrate failed to converge to prescribed accuracy after 7 recursive bisections.
Out[9] := {0.00379652, {a -> 1.18269, d -> 1.01795, i -> 2.12355, j -> 2.02463]

So I guess that convinces me that no, the optimum solution does not lie on the surface of an ellipsoid.


--SMQ


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