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Frank L. Preuss

**
How could one calculate the diameter of the cone shadow?
**

I take a line from the edge of the sun over the edge of the moon to the edge of the cone shadow on the surface of the earth. This line has an inclination to the line through the three centres of these three circles.

149 600 000 km is the sun away from earth.

356 410 km is the closest distance of the moon from the earth.

149 243 590 km is then the sun away from the moon.

696 350 km is the radius of the sun.

1 738 km is the radius of the moon.

694 612 km is the difference.

The inclination of the line is 694 612 / 149 243 590 = 0,0046542166.

The difference between the radius of the moon and the radius of the cone shadow I call y. The radius of the earth is 6 378 km. The distance of the moon from the surface of the earth = 356 410 - 6 378 = 350 032 km.

The inclination of the line is then = y / 350 032 = 0,0046542166.
And then is y = 0,0046542166 x 350 032 = 1 629 km.
And the radius of the cone shadow is then the radius of the moon minus y and that is

1 737 km minus

1 629 km and that is

109 km. And the diameter of the cone shadow is then

**218 km**.

That would be so the maximum diameter of the cone shadow, because I have taken the minimum distance of the moon from the earth, 356 410 km.

This is the end of "Astronomical question and answer 212"

To the German version of this chapter:
Astronomische Frage und Antwort 212

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