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Name: Donna Schakelaar
Status: other
Age: 30s
Location: N/A
Country: N/A
Date: 1999 - 2000

What are the relative intensities at sunrise, sunrise + 10 minutes, sunrise + 30 minutes, and noon, and sunset? How would altitude and geographical location affect this e.g. between Darwin, Australia and Melbourne,Australia? Are there any other factors that affect the intensity of the sunlight?

Unfortunately, I don't know a simple mathematical formula to get these numbers. So, I can't answer your question quantitatively, as you would want. I can only repond in generalizations.

The intensity of sunlight during any given day is pretty close to constant: as long as the sun is up, its intensity depends mostly on the day of the year. It rises sharply at sunrise and drops sharply at sunset, and in between it curves gently. A plot of the solar intensity versus time looks a bit like an upside-down "U", somewhat squared off. The main factor influencing solar intensity is the length of atmosphere through which the sunlight has to travel. If the sun were directly overhead, such as at noon on an equinox at the equator, the sunlight is at its most intense, as it only needs to penetrate one atmospheric thickness. At latitudes away from the equator, the sunlight reaches the earth's surface at a more grazing angle, so that it needs to penetrate a greater slice of atmosphere. The angle depends on the latitude, the time of year (more overhead in summer, closer to the horizon in winter), and the time of day (more overhead at noon, right at the horizon at sunrise and sunset). The lower the angle (closer to the horizon), the more atmosphere stands between you and the sun, and the lower the intensity of the sunlight that arrives.

Richard Barrans Jr., Ph.D.

I don't know the answer, but I know how I'd try to calculate it. You need to make some simplifying assumptions and do a little trigonometry. Let's say the difference in intensity is entirely attributable to the greater distance in the atmosphere the morning and evening sunlight must travel to get to you. I think this is a very good assumption. You can solve for the distance variation by drawing a circle of radius 4000 miles (the Earth's surface) and a concentric circle of 4010 miles (the top of the atmosphere) and doing the trig. The varying air density with height is ignorable because you're interested in relative intensity, and because we'll assume different air densities affect all light wavelengths in the same way (not too bad an assumption if we're only interested in light that actually makes it down to the surface). A quick a dirty calculation says morning sunlight travels through 283 miles of air while noon sun travels through 10, if you're at the equator in midsummer.

So now you can calculate the distance variation. You can calibrate this scale to an intensity-variation scale with two numbers: the intensity at noon at the equator in midsummer, and the intensity at the pole in midsummer. If you can't find these numbers, you could make a reasonable stab at them by comparing the temperatures at those places and times (in Kelvins: Fahrenheit and Centigrade have arbitrary zeros so they're no good for this purpose).

Altitude makes no difference under the assumptions I've made. Geographical location makes a big difference, of course. (If you're at the pole, there's no difference between morning and noon. I'd guess the dependence on latitude involves nothing worse than a cosine.) Other factors that affect the relative intensity include clouds and mountains.

Tim Mooney

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