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I am trying to develop a raster layer with the interference of streetlight in this example: enter image description here

As you can see there are multiple buffer rings around a point layer. I actually want to transform this buffer into raster data that behaves like an inverse square law. the border of the buffer is when the luminosity equals the background light, this constant is given. for example 0.0708 lux.

The raster data in the buffer should behave like: L = L_0/d^2. L is the luminosity of a certain location, L_0 is the luminosity of point source and d is the distance in meters from the point source. Therefore, if the point source is 100 lux, then if you would create cellsize of 1 by 1 meter, 3 cellsizes away the cell value will be: 100/3^2 = 11.11 lux. then the end of the raster data, namely the border of the buffer should have the cell value of 0.0708 lux, the background light lux.

But that is not all, because there are buffers that overlap each other. I actually want to sum up these overlapping values with each other. Let's say the cell value 11.11 overlaps with the other buffer raster of value 20. than this cell value should become 31.11 lux. This value however should continue to behave like the inverse square law. This means that buffers overlapping will not be two identical perfect circles but are distorted since the intersection point would have a bigger distance before the luminosity of the cell equals the background light.

I am working in ArcGIS in ArcMap. I do not have the skill to use a python script for this unfortunately. I hope someone could help me with this problem.

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  • Have a look at the euclidean distance tool that will create your distance pixels that you would feed into your luminosity equation, no python required. – Hornbydd May 12 '20 at 19:23
  • Are values at source points equal each other? If not buffers should be different. – FelixIP May 12 '20 at 20:51
  • @FelixIP Yes the values at the source point are the same and therefore the buffers are identical at first. – Mike May 12 '20 at 20:52
  • There are raster and vector solutions depending on a total number of cells inside buffers. What is your estimate? Are we talking 100s k? – FelixIP May 12 '20 at 20:59
  • So one buffer has an area of more than 5000 m^2. The radius can range between 40 to 70 meters or so. Therefore if you have a cell size of 1 than the amount of cells equal the area. However, if that is too much then the cell size could be enlarged? – Mike May 13 '20 at 7:06

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