X and the City: Modeling Aspects of Urban Life (14 page)

It’s a lovely weekend and you decide to “catch some rays” in the park. If you’re anything like me, with the fair skin of someone from Northern Europe, you will make certain you lather yourself with sun block. It’s tempting to get the highest possible SPF variety, but is it really necessary?

In fact, SPF 30 does
not
block out twice as much harmful radiation as SPF 15. SPF is not sun-filtering, it is a S(un)P(rotecting)F(actor). The label tells you how much
time
you can spend in the sun before you start to burn (compared with the time for bare skin); 15 times longer for SPF 15, 30 times longer for SPF 30. However, SPF 30 only blocks out about 3% more of the harmful UVA and UVB radiation.

Here’s what happens. Let’s suppose we have SPF 2 (does that exist?). Anyway, that would block out 50% of the UV radiation that causes burning. If you burn after 30 minutes when naked as the day you were born (without any sunscreen), you could stay out for an hour—twice as long—with SPF 2. (Please try any naked sunbathing at home, not in the park.) SPF 4 would block out 75% of the harmful radiation, so you could stay out in the sun four times longer—but it cuts out only 25% more of the incoming UV rays than SPF 2 does, correct? SPF 8 means you can stay out 8 times longer, but only cuts out, well, let’s see how much more.

SPF 2 cuts out 1 −
= 50% of the incoming UV radiation.

SPF 4 cuts out 1 −
= 75% of the incoming UV radiation.

SPF 8 cuts out 1 −
= 87.5% of the incoming UV radiation.

SPF 16 cuts out 1 −
= 93.75% of the incoming UV radiation.

You get the idea. Following this pattern, we see that SPF
X
cuts out a fraction 1 −
of the incoming UV radiation. But we can examine this from another point of view. From the list above we see that the fractions of UV radiation blocked by SPF 2, 4, 8, 16 can be written respectively as

Therefore, for SPF
X
the corresponding fraction of blocked radiation may be written in the form 1 −
, where
X
= 2
ⁿ
. Taking logarithms to base 2 we find that
n
= log
₂
X
, or, using the change of base formula with common logarithms,

Let’s now go back and compare
X
= 15 and 30.

For SPF 15,
so
or 93.3% of the harmful radiation is blocked. Of course, we could have just calculated
to get this result—but where would be the fun in that?

For SPF 30,
, so
or 96.7% of the harmful radiation is blocked. This is a difference of 3.4 percentage points! Again, “we don’t need no logs” to do this, because
.