On Wrapping Spheres and Cubes with Rectangular Paper - Discrete and Computational Geometry and Graphs - page 40

Start with an initialization rectangle of width x with the diagonal 2 ˀR
wrapped exactly around an equator. This rectangle is indicated by the dotted
region in Fig. 6 . The top edge forms a line segment with dʸ

dˆ held constant.
Now extend the initialization rectangle, spiraling up the sphere continuing
to hold

dʸ

dˆ constant. Terminate when the center of the ends reaches the poles.
A “cut and paste” argument, similar to that used for Lower Bound 5 ,
rearranges the ends of the strip to ensure the poles are covered. This is visualized
in Fig. 6 by moving the paper above ˆ = 0 into the gap at the top and doing the
same for ˆ = ˀ .
The length of the strip, 1 /x , is the length of the initialization rectangle plus
the amount needed to spiral in each hemisphere: 1 /x = L init +2 L spiral . In cal-
culating the lengths, care must be taken to only look at fully stretched paths
because the rest of the strip is being contracted. These paths are bold in Fig. 6 .
The initializati on rectangle has diagonal 2 ˀR and height x , so its length is

given by L init = (2 ˀR ) 2 x 2 .

Consider the upper hemisphere. By construction, in the upper hemisphere,
the bottom of the strip incurs no contraction. Using the spherical arc length
formula:

−

sin 2 ( ˆ ) dʸ
dˆ

2

dl 2 =( R sin( ˆ ) dʸ ) 2 +( Rdˆ ) 2 + dR 2 = Rdˆ +1

Integrating over ˆ gives the length:

⊛ ⊞

sin 2 ( ˆ ) dʸ
dˆ

2

L spiral = dl = R

π
2

⊝ ⊠ dˆ

+1

ˆ 0

where ˆ 0 is the value of ˆ such that when the bottom of the strip is at an angle
ˆ 0 , the middle of the strip reaches the pole.
The slope dʸ/dˆ is constant and thus equal to the ratio of the total change
in ʸ to the total change in ˆ over the initialization rectangl e. ʸ ranges from 0

to 2 ˀ . Similar triangles demonstrate that ˆ changes by x/ R 2

− ( x/ (2 ˀ )) 2 ,so

dividing gives

dˆ =2 ˀ R 2

= (2 ˀR/x ) 2

dʸ − ( x/ (2 ˀ )) 2

x

− 1 .

The angular distance between the middle and the bottom of the strip is
constant, so reasonin g with similar r ight triangles in the initialization rectangle

yields ˆ 0 = x/ (2 R ) 1 − ( x/ (2 ˀR )) 2 .

Lower Bound 8. An x × 1 /x paper wraps an R -sphere if R satisfies

sin 2 ( ˆ ) (2 ˀR/x ) 2

1 +1 dˆ

+2 R

π
2

x = (2 ˀR ) 2

1

− x 2 − 1

− ( x
2 πR

) 2

x
2 R

init each hemisphere