After looking at this post I can see that a flat object is more exposed to the sun at the equator. Does this mean that a person walking around (standing up), is much more likely to burn at the poles than at the equator? Because their entire body is facing the sun, not just the top of their head?

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    $\begingroup$ Odds of having much exposed skin at the Poles: small! :-p $\endgroup$ Commented Jun 14, 2018 at 17:22
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    $\begingroup$ I started thinking about doing the math on this, and quickly found myself thinking of the physics stack-exchange. "Assume a spherical human...." $\endgroup$
    – J Thomas
    Commented Jun 15, 2018 at 10:59

1 Answer 1


Assuming that your skin is exposed, you are standing upright are out when the sun is out, yes.

You don't even need to be very polar.

I live near Edmonton Alberta -- latitude 54 degrees.

At the equinox, the sun is at 54 degrees from the zenith. At winter solstice it is 54 + 23 = 77 degrees from the zenith, or 13 degrees off the horizon.

No joke: We use our car's sun-visors driving south at noon.

So if I'm standing up outside, I'm getting roughly the kind of sun intensity you would get lying flat on the beach in more tropical climes. Since you aren't really hot, you can get a burn without your usual warning signs.

Confounding factors:

The sunlight is traveling through more atmosphere which reduces intensity somewhat. At 13 degrees, the light path has about 4 times as much air in the path. Which increases the amount of sunlight that is absorbed -- but not by a factor of 4. The difference between sunlight in space and sunlight at the surface is about 30% overall, but adding to the air column will only increase the absorption of frequencies that are partially absorbed by a more vertical column.

Polar latitudes, and less polar ones in winter when solar angles are low, will tend to have less dust, water and smoke in the air, increasing air's transparency.

You get reflection off the snow which can come close to doubling your exposure. Some of the worst sunburn's I've gotten have been on winter days on my poor nose. (Sticking out of my face it's getting direct sun, reflected off of snow, and reflected off of face.)

One useful measure of the relative intensity is to look at the exposure information for photographs. (You have to know the film speed or digital ISO as well as shutter speed and f/stop) I find that exposures in sunlight at 'low noon' in winter are comparable to summer ones.

Day lengths are short. At winter solstice we theoretically have about 9 hours of daylight. When the sun peaks at 13 degrees however there are a lot of shadows.

The weather tends to discourage large displays of skin, although I've split wood without a shirt on a calm sunny day at -10 C. In general when dressing for outdoors I figure that bright sun in winter is equivalent to 10-15C degrees warmer, since it's hitting more of me.

  • $\begingroup$ I wouldn't call them confounding factors, I'd call them as vital to the question as the sun angle/resultant profile! :) $\endgroup$ Commented Jun 14, 2018 at 18:04
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    $\begingroup$ Another variable -- the thinned ozone layer at the poles means a higher proportion of UV light gets through. Altitude makes a big difference -- easier to sunburn on high mountains than at sea level. $\endgroup$
    – J Thomas
    Commented Jun 15, 2018 at 10:56
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    $\begingroup$ Re So if I'm standing up outside, I'm getting the kind of sun intensity you would get lying flat on the beach in more tropical climes. No, you're not. It's not anywhere close due to the first confounding factor, "the sunlight is traveling through more atmosphere which reduces intensity somewhat." Somewhat is an understatement. The air mass when the sun is 13° above the horizon reduces solar intensity by more than a factor of 2. $\endgroup$ Commented Jun 15, 2018 at 13:09
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    $\begingroup$ While the air mass column is twice the mass, it also has much less water vapour and dust in it. A vertical air column reduces insolation from about 1300 w/m2 at the top of the atmosphere to typically around 1000 w/m2. Doubling the amount absorbed would bring it down to around 700. $\endgroup$ Commented Jun 15, 2018 at 21:45
  • $\begingroup$ This answer explains that a greater area of a person is likely to receive direct sunlight at high latitudes, due to the low angle of the sun. It does not address whether a person is more likely to be sunburnt (which I would imagine depends more on the maximum intensity on the skin that is lit, rather than the area of skin involved) $\endgroup$ Commented Jun 18, 2018 at 15:29

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