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Making Hemispheres out of Paper

Date: 07/28/99 at 16:23:08
From: Matt Osborn
Subject: Making spheres from paper

I'm a high school math teacher, and I've been trying to figure out how 
you would make a sphere from a piece of paper, for example, if I 
wanted my geometry students to be take a piece of paper and fold/cut 
it so that it would fit perfectly over the northern hemisphere of a 

I first decided that I would focus on how to get a piece of paper to 
fold into a hemisphere. I think the best way would be to make the 
North Pole the center of the figure. I decided to use triangular 
wedges of 30 degrees. These triangular wedges are actually triangles 
I studied in non-Euclidean geometry because their angle measure is 
above 180 degrees, where the two congruent sides extending from the 
North Pole to the Equator are both perpendicular to the equator, but 
are 30 degrees apart in terms of longitudinal degrees.

I can figure out the length of the base of this unique isosceles 
triangle, and I can also figure out the lengths of the congruent legs 
(since I can determine the radius). Where I have trouble is in how to 
convert this figure to a piece of paper. I know all three sides of 
this triangle are curved. I also know they are all parts of great 
circles. I don't know how to actually draw these curves or what the 
actual angle is for the vertex angle of this isosceles triangle. What 
type of curves are the sides? I know the vertex angle would be less 
than 30 degrees when drawn on a flat piece of paper, but I don't know 
what it would be in this case. Please help!

What I have envisioned for the finished product (a pattern on a flat 
piece of paper that can be cut/folded so that it would become a 
hemisphere) of this figure would be similar to a flower, with the 
North Pole being the center of the flower, and each petal representing 
one of the non-Euclidean triangles. The empty spaces between these 
"petals" would be the unnecessary part of the hemisphere that would be 
folded under or cut off. Thanks for any help that you can give.


Date: 07/29/99 at 16:46:25
From: Doctor Rick
Subject: Re: Making spheres from paper

Hi, Matt.

I think you realize that any construction of cut and/or folded paper 
will not fit perfectly over a sphere. The narrower you make your 
wedges, the closer it will come, but you will never make a perfect 
sphere. To be more specific, with 30-degree wedges, any cross-section 
parallel to the equator will have the shape of a dodecagon rather than 
a circle.

The method you describe, however, happens to be just what globe makers 
do. They make pieces of map of this shape (which they call "gores") 
and glue them on a sphere.

Now let's work out the shape of the gores. We'll make it so the center 
line of each gore lies exactly on a sphere of radius r. Call the 
angular width of each gore (the number of degrees of longitude it 
covers) phi.

Look at an arbitrary cross-section of the globe parallel to the 
equator. The plane of the cross-section meets the surface of the 
sphere at latitude L. I will work with the co-latitude (angle measured 
from the pole), in radians, which I will call theta. 

     theta = (90 - L) * pi/180

The radius of the circle that is the cross-section of the sphere is

     r_c = r*sin(theta)

You can see this if you take another cross-section perpendicular to 
this one, containing the axis of the sphere. There will be a right 
triangle whose hypotenuse is r, one leg is r_c, and the opposite angle 
is theta.

The cross-section of one gore is a line segment tangent to the circle 
at its center. We want to find the width W of this line segment. Draw 
a line from the center of the circle to the point of tangency, and 
another from the center to the end of the tangent segment (the gore). 
You get a right triangle with legs of length r and W/2; the angle 
opposite W/2 is phi/2. Simple trigonometry gives us the width of the 
gore as

     W = 2*r_c*tan(phi/2)
       = 2r*tan(phi/2)*sin(theta)

The distance (along the arc of the globe) from the point of the gore 
to the line whose width we have just measured is r*theta. (Remember, 
theta is measured in radians.) The full length of a gore, from tip to 
tip, is pi*r. (The gore for a hemisphere will be half this, or 
pi*r/2.) If you lay it out on graph paper so that the tip is at the 
origin and the centerline is along the x-axis, then the two sides of 
the gore will lie along the lines

     y = +/- r*tan(phi/2)*sin(x/r)

So that's the shape of the sides of the "triangle." It's a sine curve 
whose amplitude is determined by the angle phi. 

As for the angle at the poles, it _will_ be exactly the angle phi. 
However, the sides begin curving inward quickly, so we are talking 
about the angle between the _tangents_ to two curved lines. The 
equator end of the "spherical triangle" is a vertical line at
x = pi*r/2, and the sine curves are horizontal at x = pi*r/2 so the 
angles are both right angles.

I haven't tried making one of these, so I don't know how easy or hard 
it might be to assemble. I would try using a card stock rather than 
thin paper so it's easier to handle; I would cut out the gores right 
along their edges and tape them together on the inside of the 
hemisphere. If that worked, I would think about fancier ways to do it.

Have fun!

- Doctor Rick, The Math Forum   
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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