Re: [math-fun] flat corners
On 8/12/09, rcs@xmission.com <rcs@xmission.com> wrote:
Presumably you've noticed that all your denominators are 2K^2?
Rich
Ouch --- I hadn't --- thanks! But I had noticed that h(q) = h(p), where p = (2-2q)/(2-q) --- from which in particular the maximum h occurs where q = p, that is q = 2-sqrt2 = 0.585786438 ... though we still don't know exactly what h = 1.287188506 ... is here. And now following on from your remark, I notice that denom(h) = 2 (denom(p) denom(q))^2 in all cases [but denominators are always easier than numerators ...]. WFL Latest data: h(-sqrt2) = h(+sqrt2) = 0; h^2(-4/3) = h^2(7/5) = 59/450, h^2(-1) = h^2(4/3) = 11/18, h^2(-2/3) = h^2(5/4) = 287/288, h^2(-1/2) = h^2(6/5) = 231/200, h^2(-1/3) = h^2(8/7) = 1139/882, h^2(0) = h^2(1) = 3/2, h^2(1/4) = h^2(6/7) = 2511/1568, h^2(1/3) = h^2(4/5) = 731/450, h^2(2/5) = h^2(3/4) = 1311/800, h^2(1/2) = h^2(2/3) = 119/72. WFL
Fred Lunnon wrote:
I had noticed that h(q) = h(p), where p = (2-2q)/(2-q) [snip] Latest data:
h(-sqrt2) = h(+sqrt2) = 0; h^2(-4/3) = h^2(7/5) = 59/450, h^2(-1) = h^2(4/3) = 11/18, h^2(-2/3) = h^2(5/4) = 287/288, h^2(-1/2) = h^2(6/5) = 231/200, h^2(-1/3) = h^2(8/7) = 1139/882, h^2(0) = h^2(1) = 3/2, h^2(1/4) = h^2(6/7) = 2511/1568, h^2(1/3) = h^2(4/5) = 731/450, h^2(2/5) = h^2(3/4) = 1311/800, h^2(1/2) = h^2(2/3) = 119/72.
Hmm. 59/450 = 1/9 + 1/50 and 11/18 = 1/2 + 1/9. However, 287/288 = 7*(1/9 + 1/32); the denominators clearly come from squaring q and p, but where does the 7 come from? Aha! 1/9 = 1 - (-4/3)^2/2, 1/50 = 1 - (7/5)^2/2; and then 1 - (-2/3)^2/2 = 7/9 and 1 - (5/4)^2/2 = 7/32. After checking the rest, it seems your data is produced by the formula h^2(q) = 2 - q^2/2 - p^2/2, where p is given by the formula above. [It wasn't as easy as I make it look above; I've omitted a lot of pointless thrashing about.] -- Fred W. Helenius fredh@ix.netcom.com
On 8/12/09, Fred W. Helenius <fredh@ix.netcom.com> wrote:
... After checking the rest, it seems your data is produced by the formula
h^2(q) = 2 - q^2/2 - p^2/2,
where p is given by the formula above.
[It wasn't as easy as I make it look above; I've omitted a lot of pointless thrashing about.]
I already had all the clues, but thrashed about even more woefully. Since h^2(q) = 0 when q = (+/-) sqrt2, it has a factor 2-q^2; since h^2(q) = h^2(p), by symmetry it also has a factor 2-p^2. And since the plot looks suspiciously like a quartic, it must be h^2(q) = c(2-q^2)(2-p^2), where by inspection c = 1. And p = 2(1-q)/(2-q), so this may be recast in various ways, such as Fred's above. Now, what about a civilised proof, I wonder ... WFL
My last posting turns out to have been complete cobblers --- just one of those "senior moments", I guess. Apologies to everybody, and congratulations to Fred. Now, as I was saying, what about a proof ... WFL On 8/13/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 8/12/09, Fred W. Helenius <fredh@ix.netcom.com> wrote:
...
After checking the rest, it seems your data is produced by the formula
h^2(q) = 2 - q^2/2 - p^2/2,
where p is given by the formula above.
[It wasn't as easy as I make it look above; I've omitted a lot of pointless thrashing about.]
I already had all the clues, but thrashed about even more woefully.
Since h^2(q) = 0 when q = (+/-) sqrt2, it has a factor 2-q^2; since h^2(q) = h^2(p), by symmetry it also has a factor 2-p^2. And since the plot looks suspiciously like a quartic, it must be h^2(q) = c(2-q^2)(2-p^2), where by inspection c = 1.
And p = 2(1-q)/(2-q), so this may be recast in various ways, such as Fred's above.
Now, what about a civilised proof, I wonder ... WFL
In an effort to make the torus polyhedron plots a little easier to interpret, the outer cube might be flattened to a unit square cuboid of height c say [so c = 1 earlier]. It apparently remains true that if either type of corner is "flat" (developable) then both types are. The Helenius relation for an origami torus now generalises to 2*h^2 + q^2 + c^2*p^2 = 2*(1+c^2), where 2*(p+q-1) = p*q, recast to show the symmetry between p,q [and this time, properly checked!] As far as proving these conjectures algebraically goes, notice that the squares of the sines (and cosines) of the face angles are rational in the coordinates of the corners. So there's a good chance that FWH's relation can be fed into something like sin^4(A+B) - 2 sin^2(A+B) (1 + sin^2(A) + sin^2(B) + 2 sin^2(A) sin^2(B)) + (1 + sin^2(A) + sin^2(B))^2 = 0 [unchecked!] to show that the U/V corners have sin(face angle-sum) = 0 [two face angles are already known to be \pi/2]; then using continuity from the (improperly flattened) case q = 1 would clinch it for U/V corners. If it could also be proved that both or neither are developable, then we're home and dry. Otherwise, we need some rational addition formula like the above, but for sums of 4 angles --- maybe RWG might know how to do this? Fred Lunnon
On 8/14/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... If it could also be proved that both or neither are developable, then we're home and dry.
Of course they are --- this is simply the Gauss-Bonnet theorem --- as any fule no. What's more, using Gaussian curvature (in the form of a "2-D cosine) gives a far easier way to compute the face angle sums than employed by my existing kludgy program (or my dodgy sine formula). WFL
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Fred lunnon -
Fred W. Helenius