Low sun in winter

During frosty days, just before the winter solstice, I came along this place.

Frosty meadow in England. The parts in the shade from the hedge have three days of frost to them, the other parts just one day.
Frosty meadow in England. The parts in the shade from the hedge have three days of frost to them, the other parts just one day.

The sun was shining the days before, and cleared the frost where it reached.

I was surprised, how wide the shady area was. My instinct would have been, that the shadow would just have been a bit wider than the height of the hedge. But it is much wider. In the picture I measured the width of the frosty area and the height of the hedge at a few places and calculated the angle of the suns maximum elevation using

alpha = arctan(height/width)
Height [pixel]Width [pixel]Angle alpha [degree]
344112017.1
656208017.5
1012361215.7
Measured values and calculated angle of the sun

In the table above one has to take into account that it’s impossible to measure with one pixel precision for a hedge on uneven ground. Another person might measure the heights 30 pixel higher, or lower, and widths 80 pixel wider or narrower (for the second row), which could vary the angle by about 1.5 degrees. That was the reason, why I did three measurements of different areas in the picture.

So I get about 16 to 18 degrees, which corresponds with the real value of 90 – 51 – 23.4 = 15.6 degrees. The values in the formula are:
90: mid day height of the sun over the equator
51: my geographical latitude
23.4: angle between equator and ecliptic.
At the winter solstice the latter value needs to be subtracted, at the summer solstice it has to be added, giving a height of 62 degrees for the sun during mid day

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