**The Pattern for the Ball Shaped Test Envelopes**

Given the basic dimensions x and y, first one cuts out four circular sectors and one long rectangle as shown (of course allowing some margins for sewing):

(where the mathematically exact value of the sector angle is easily determined if desired, since each of these sectors is sqrt(3)/4 of a complete circle). Then one sews together the pieces in the obvious manner into an envelope of the type shown in the photographs, of which the cone angle turns out to be 60 degrees, while the height of each of its conical portions is x and the height of its cylindrical portion is y, so that the total height point-to-point is 2x+y. It is easily shown that the volume and area of the resulting solid are:

Volume V = pi.x^{2}(2x+3y)

Area A = 2.pi.x.sqrt(3)(2x+y)

In the case of the actual envelopes we made, the values were:

Envelope (A): x=49 cm, y=68 cm, V=2.28 m^{3}, A=8.85 m^{2}

Envelope (B): x=49 cm, y=76 cm, V=2.46 m^{3}, A=9.28 m^{2}

Obviously it is much easier to cut out and sew together these five pieces (or six, if the long rectangle is made from two pieces as is usually convenient) than it would be to make a sphere-and-cone shape from twelve or more gores. It is not only a matter of having less pieces to sew together. The distortion of the sewn-together mass as it grows by accretion, little by little, into a three-dimensional shape, which is really what determines the difficulty of sewing on the later pieces, is very much less troublesome with this pattern.

If one is to cut out an envelope of dimensions like the envelopes (A) and (B) above from a run of material of width 150 cm, a single piece about 8 meters run is quite sufficient.

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