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Driving and Sun Damage

2012-06-05 00:00:00 UTC by terrance macgregor

We all do it. We get into our cars and drive. Depending on how long you commute and what time of day you work, you may be exposing yourself to additional doses of radiation.

Imagine you are a giant sponge. Every time you step outside, you are soaking in ultraviolet (UV) rays. The longer your skin is exposed, the more damage is being done. If you don't think this is a real threat, check out this dramatic article. That's quite a visual, isn't it? 

Now, it isn't all that bad. This is one extreme case, obviously. However, the dangers are real. You need sun to make Vitamin D, but you don't need that much. So, the question on everyone's mind is... how much damage is actually being done to your skin? To answer this question correctly, you need to consider the many variables that occur within our atmosphere daily. That is almost impossible. Absorbtion of UVA takes place on skin that is in direct contact with the sun, but it can also occur through clothes that do not block UV rays.

Let's asume, for the sake of argument, that your clothes are a suit of armor. There's nothing getting through that armor, so the only sun exposure that you would get would be on your face, your arms (we're making your armor sleeveless!), neck, and hands. If you are wearing a bikini during your commute, then chances are that you aren't really worried about the sun and probably have one heck of a nice commute. :-) So, we have the parts: face, neck, hands, and arms. How much surface area is that?

After a rather creepy Google search, this is what I found: Anterior Head = 4.5%; Anterior Arm = 4.5 %X 2 (assuming you have 2) = 9%; Posterior Arm = 4.5 %X 2 (assuming you have 2) = 9%. So, this total is about 22.5%. This number is a little high. Let's change this up a little bit.... Let's also assume that since the sun can't hit both the front and back of your arms, we only need to consider one half of them. The new calculation looks like this: Anterior Head = 4.5%; Front or Back of Arm = 4.5 %X 2 (assuming you have 2) = 9%. Total = 13.5%

We're also going to assume that only your left-hand side is really exposed to direct sunlight. Our figure drops to 6.75%. That's not too bad.

Now, what do those percentages actually mean? Some Gross Facts: The average adult has about eight pounds (3.6 kilograms), or about 22 square feet (2 square meters) of skin. It may help to put that in perspective -- a standard doorway is 21 square feet, and the average adult's skin would fill all of that space [Source: National Geographic]. For our calculation, let's stick with the metric system. .0675 X 2 meters = .135 meters squared. So, I know that this is gross, but I think of this as a patch...let's consider this a square of skin, similar to a solar cell. Yeah, this is creeping me out to think about this...but it helps. In fact, there are many great tutorials out there that explain the amount of energy the sun delivers. Check them out here.

Now, all we have to do is figure out how much UV light makes it through the ozone layer, the atmosphere, and anything in between. Once again, we have to ignore the tremendous variability from things like patterns of the stratosphere, ozone thickness in patches, elevation, sea level, dust in the lower air masses, dust in the upper air masses, multiple air masses, forgetting to return Redbox movies and realizing it would have been cheaper to buy the crappy movie, etc. Another important factor to consider is that only of the band within the optical radiation spectrum, broad-spectrum ultraviolet radiation (UVR) is the strongest and most damaging to living things. UVR is divided into wavelength ranges identified as UVA (315 to 400 nm), UVB (280 to 315 nm), and UVC (100 to 280 nm). Of the solar UV energy reaching the equator, 95% is UVA and 5% is UVB. No measurable UVC from solar radiation reaches the earth's surface, because the shortest UV wavelengths are completely absorbed by ozone, molecular oxygen, and water vapor in the upper atmosphere. [Source: http://ntp.niehs.nih.gov/index.cfm?objectid=BD4CD88D-F1F6-975E-792094AC1CE4B062]

Here is where we dive off the deep end of science, making it both fun and confusing (cue maniacal laughter).

We need to understand how irradiance works. Yeah, irradiance. Never heard of it? Neither did I until I had to write this! We are going to get a little messy here. First, we are going to lump both UVA and UVB together. We are going to focus on this. Remember meters in science class? We are going to be dealing with some of these terms: um = micrometer = 10^-6 meters; nm = nanometer = 10 ^-9 meters; Ultraviolet A = 315- 400 nm = .315 - .400 um; Ultraviolet B = 280 - 315 nm = .280 - .315um. Range for our calculations = .280 um - .400 um. [Source: http://en.wikipedia.org/wiki/Ultraviolet

Another factor that we are going to use is Watts. The watt ( /ˈwɒt/ wot; symbol: W) is a derived unit of power in the International System of Units (SI), named after the Scottish engineer James Watt (1736–1819). The unit, defined as one joule per second, measures the rate of energy conversion or transfer. [Source: http://en.wikipedia.org/wiki/Watt] We can calculate the Spectral Irridance using some great tools, one of which can be found here.

Using this tool, we came up with something like this: Preciptable water vapor = 1.42cm; total ozone = .34cm; Turbidity = .5 um: .27; Airmass = 1.5; Ground Albedo: 0.2; Measure Plane to Horizontal Plane: 37 6,363.5 W/m^2/um 6363.5 (spectral irradiation) * .135 (our surface area) = 859 spectral irradiations (W/m^2/um) So, this is per second. Per minute = 859 x 60 = 51,540 (W/m^2/um) http://en.wikipedia.org/wiki/Irradiance

Check out these links to see where the science comes from:

http://www.germar.org/faq.html http://ntp.niehs.nih.gov/index.cfm?objectid=BD4CD88D-F1F6-975E-792094AC1CE4B062