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Astronomical question and answer 231


Frank L. Preuss


What does the following picture tell us?



"series of short exposures at 15 minute intervals of solar eclipse at Chicago in 1954."

The first question could be, is this in the morning or is it in the evening? When on sees the picture only, without the description, then one could also assume that it was taken in a place somewhere in the southern hemisphere and was therefore in the evening.

So only the description indicates that the picture was taken in the northern hemisphere and that it was daybreak - and that it was taken looking towards the east, not towards the west.

The second question could be about the distance between two exposures.

The sun needs about 2.13 minutes to rise or to set. We had seen this in answer 26. And in 15 minutes it would cover about a distance of 7 diameters of the sun. The distance in the picture seems to be less than that, but the picture is not very clear.

A third question would be the angle of the course of the eclipse and therefore of the course of the sun and of the moon.

The angle against the vertical on the picture is about 47 degrees. And when one then has a look on a map where Chicago is shown, then one finds that Chicago is about on the 42th parallel. So that would agree only more or less, because these two angles should be about the same.

A fourth detail is that the angle of the course of the moon should be steeper than that of the course of the sun and that is so.

Now after I saw this picture I made some inquiries about this eclipse and found out that it occurred on June 30, 1954 and that it was in the morning, at 07:25.

The astronomical tables for Chicago give the time of 07:25 for new moon on 30 June 1954. So that agrees completely.

For moonrise they give 05:15 and a distance of the moon of 372 308 km.

The sunrise is given as 05:17 and as distance of the sun is given 152.096 million km.

This information therefore confirms that the moon went ahead of the sun and rose first. The moon rose two minutes before the sun.

Then the information is given: "Max. width of band 153 km (95 mi)":


I have used my spreadsheet calculation and calculated the diameter of the cone shadow with these distances of the moon and of the sun and got 125 km, so there is a difference.

The duration is given as 155 sec (2m 35 s). When I take a speed of 28 km/min, then I get a travelling distance of 28 x 155 /60 = 72 km. So the speed has been a different one, because the 28 km/min is valid for the equator and increases the further away it is from the equator.

Now a last image:



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To the German version of this chapter: Astronomische Frage und Antwort 231



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