Take the 2-minute tour ×
Geographic Information Systems Stack Exchange is a question and answer site for cartographers, geographers and GIS professionals. It's 100% free, no registration required.

I need Earth radius in pixels for use in Mercator projection. Well let's say that our map's height is 1478 pixels and distance between 0 pixel and 1478 pixel is 2.52 degree. Radius is distance from the equator to pole so the formula for count Earth radius in pixels is R=90/2.52*1478 which is 52785.71429. Well then pixel number for 48.54 degree is 51287.99461 and for 51.06 is pixel number 54885.98194 counted by Mercator so the diference is 3597.987328 which is not correct because it should be those 1478 pixels. I think that Mercator formulas are good because when i make inverses of pixel i get the degrees as in input so mistake can be only in Earth radius. Am I missed something, are my calculations right? Thanks in advance.

share|improve this question
2  
Mercator doesn't go to the pole? How do you get 2.52 degrees for the total distance? –  BradHards Apr 20 '13 at 6:50
1  
I think there is some confusion in terms here as well. You state that 'radius' is the distance from the equator to the pole. The radius is actually the distance from the center of the earth to the pole. Since the equator is in the surface of the earth, what you are looking for is 1/4 of the circumference. –  Get Spatial Apr 20 '13 at 10:26
2  
My previous comment was meant to highlight the need for clarity in your question. It might help to break your question down into sections and paragraphs. You might explain your project and overall goal, then your process and finally, the problem you are running into. It is not clear why you need to work with pixels at all. Is this a web map or a printed map? –  Get Spatial Apr 20 '13 at 10:31
    
Also bear in mind that such a small (vertical) resolution (around a quarter of a degree if I understand you) can cause significant errors when calculating, since in the end pixels are integers. –  lynxlynxlynx Apr 21 '13 at 17:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.