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Solid angle in lat-long


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#1 juanMorata   Members   -  Reputation: 135

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Posted 19 August 2014 - 04:56 AM

I have an application for which I want to normalize a lat-long map by the solid angle subtended by each pixel, but I'm a bit lost with this. I did read that to have an equal area I have to multiply the values in the map by the vertical cosine fallof (cos phi), however as I'm curious to understand better, do you know a good source to fully understand why is this or can you explain to me why is this? I mean I know that the mapping is not equi-area, but why is the correction by cos phi (if it is cos phi)? 

 

Thank you! 



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#2 Álvaro   Crossbones+   -  Reputation: 13914

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Posted 19 August 2014 - 05:10 AM

I have better intuitions about geometry on the sphere if I imagine it's the Earth. Let's see how much land is covered by a little "rectangle" of latitude and longitude. Near the equator, a 1 degree x 1 degree region is roughly 111 Km x 111 Km. Latitude is well behaved, with 1 degree corresponding to 111 Km everywhere. But if you move North to New York (latitude 40.7 degrees), 1 degree of longitude only covers about 84 Km.

In order to compute how much "shrinkage" happens as you move away from the equator, you can look at the length of the parallels. As you move closer to a pole, parallels get smaller and smaller, following the formula

parallel_length = 40000 Km * cos(latitude)

Is that enough of an explanation?

#3 juanMorata   Members   -  Reputation: 135

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Posted 19 August 2014 - 09:28 AM

I have better intuitions about geometry on the sphere if I imagine it's the Earth. Let's see how much land is covered by a little "rectangle" of latitude and longitude. Near the equator, a 1 degree x 1 degree region is roughly 111 Km x 111 Km. Latitude is well behaved, with 1 degree corresponding to 111 Km everywhere. But if you move North to New York (latitude 40.7 degrees), 1 degree of longitude only covers about 84 Km.

In order to compute how much "shrinkage" happens as you move away from the equator, you can look at the length of the parallels. As you move closer to a pole, parallels get smaller and smaller, following the formula

parallel_length = 40000 Km * cos(latitude)

Is that enough of an explanation?

 

Intuitively yes, thanks!

As I'm actually scaling pixel intensity values I should scale them by the absolute value of the cosine right? Otherwise as the latitude varies between 0 and pi I may end up with negative values. 



#4 Álvaro   Crossbones+   -  Reputation: 13914

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Posted 19 August 2014 - 12:00 PM

Latitude varies between -pi/2 and pi/2 and the cosine is always positive.




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