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Frank L. Preuss

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How would then the calculation look like for the diameter of the shade?
**

For the calculation of the diameter of the shade I take a line from the edge of the sun also again over the edge of the moon, but this time I take the opposite edge of the moon, and extent this line to the edge of the shade.

The sum of the radius of the sun plus the radius of the moon is 698 088 km.
The inclination of the line is 698 087 / 149 243 590 = 0,0046775074.
And the difference between the radius of the shadow and the radius of the moon I call z.
The inclination of this line is then = z / 350 032 = 0,0046775074.
And then z is = 0,0046775074 x 350 032 = 1 637 km.
And the radius of the shade is then 1 738 km +
1 637 = 3 375 km. And the diameter of the shade is then
**6 750 km**.

When the shade, exactly like the cone shade, has a speed of 28 km/minute, then it wanders through an area of the shade in 241 minutes. When I compare this with the picture from Novosibirsk, then the camera was there 36 x 3 = 108 minutes in the shade. The simplest explanation for this difference would be that the moon did not have its minimum distance from earth, but was further away, and for that reason the diameter of the shade was smaller.

This is the end of "Astronomical question and answer 213"

To the German version of this chapter:
Astronomische Frage und Antwort 213

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