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[sorry for the bad english] I am fond of astronomy and environment. I wanna try to make a "light pollution map" but I haven't my satellites... so I use as approximation of light pollution the cities' population. Let say we have for each city C citizens, each one spreads an average of X watt of electricity for lightning ( I have these data ). Skip the units ( I need just a rough dimensionless "light power" ) : city light power = C * X

I have a map, with many cities. I know light power is inverse proportional to the distance^2. I don't know about sky, air diffraction, clouds reflections.

Start from the simplest model. A flat terrain map. N light sources, every one at position X(n), Y(n) has a specific "total light power" = C(n) * X(n)

At a specific point of coordinates x,y wich is the light power, sum of all the cities light ?

I tried to calculate and plot, but it seems weird ( too far from some real satellite night shot ) and too slow to calculate.

Please help !

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In a sufficiently detailed model the concept of albedo might come into play: en.wikipedia.org/wiki/Albedo –  Joseph Malkevitch Dec 30 '10 at 18:20
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2 Answers

You may be confusing the energy distribution received by a satellite with the energy present at a place on the ground. The satellite is above the ground and records the amount of energy received as a function of direction (equivalently as a function of location on the ground) but doesn't add all the cities together. So it would make a map with light at each city location measured as C*X/r^2. This should match roughly with the photos you see, though different cities will have very different light/person.

At a location on the ground (in your model) you receive light from all other cities, again in the amount C*X/r^2 from each city, but there is no particular reason to expect this sum to match the satellite photos as the light is traveling along the ground, not up into space. If you imagine a ring of cities in a circle, there would be a lot of light at the center, but the satellite would see nothing coming from there.

It is surprising the calculation takes a long time. How many locations are you calculating the power at? Even if you have a few hundred cities in the list and are calculating at thousands of locations, that is a few million. Did you get trapped by the curse of dimension and take a thousand points in each of the x and y directions, which would make hundreds of millions of calculations?

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I'm not sure what you mean by "light power," but intensity seems to be the measure you want (and is inversely proportional to the square of the distance.)

The intensity at a point $(x,y)$ of a spherical light source originating at, say, $(x_0, y_0)$ is given by

$I = \mathcal{P} / 4 \pi r^2$,

where $\mathcal{P}$ is the power of the light source (what I think you've called $X$), and $r$ is the distance between $(x,y)$ and $(x_0, y_0)$. This gives the final expression, for all light sources $(x_n, y_n)$ with power $\mathcal{P}_n$:

$I = \displaystyle \sum_{n} \frac{\mathcal{P}_n}{4 \pi ((x-x_n)^2 + (y-y_n)^2)}$

(Unless I've forgotten some physics and intensities don't sum like that). For the third dimension, just add the $z$ terms in the distance formula. Hope this helps.

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Since the intensity is proportional to $1/r^2$, and not to $1/r$, you shouldn't have the square root in your second formula. –  Rahul Dec 30 '10 at 22:40
    
Ha, right you are. Thanks. –  Jay Kopper Dec 31 '10 at 18:21
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