how does the angular size tool work? how can we aplicate it to our pictures?

Haha!  You beat me to the punch on this -- I had some time last week so I came up with the idea of adding things like special calculators and Photoshop plugins to the site, so I changed the "Tutorials" section to "Tutorials & Tools" and started adding the Angular Size calculator, but got sidetracked on some other work before I could finish it up.

I'll try to get the calculator finished up today and will add some instructions and examples.

ok thanks, good ideas!

The tool is now ready, with instructions -- For those of you who haven't seen it yet, go to the Tutorials & Tools page and select the Body Size Calculator.

This is a pretty handy tool, inspired by a conversation with Fahad at THIS TOPIC regarding how to compute how large a celestial body would be in an image.

Some fun ones:

Earth as viewed from the Moon:
Object Diameter (earth): 12700 km
Distance: 384403 km
Angular size: 1.89 degrees

The sun as viewed from Mercury:
Object Diameter (sun): 1392000 km
Distance: 46001200 km
Angular size: 1.73 degrees
(I thought it would be bigger!)

If you have any questions, let me know.

Very interesting! Thanks!! 

Administrator wrote: The sun as viewed from Mercury: Object Diameter (sun): 1392000 km Distance: 46001200 km Angular size: 1.73 degrees (I thought it would be bigger!) If you have any questions, let me know.

Nice tool that, especially with the calculating on the number of pixels!

As for the Sun from Mercury, you've actually calculated its perihelion distance (0.31 AU). That's 3.26 times closer to the sun than we are, so the sun should be 3.26 times bigger. The sun's angular diameter from Earth is 0.53 degrees, and 3.26 times bigger than that is 1.73 degrees. So your calculation is correct

you should also explain how the image width and the DOV work

EDG1 wrote: Nice tool that, especially with the calculating on the number of pixels! As for the Sun from Mercury, you've actually calculated its perihelion distance (0.31 AU). That's 3.26 times closer to the sun than we are, so the sun should be 3.26 times bigger. The sun's angular diameter from Earth is 0.53 degrees, and 3.26 times bigger than that is 1.73 degrees. So your calculation is correct

It's also kind of cool running a range of values where you view the sun and moon from Earth, varying the lunar distance from earth and the Earth's distance from the sun based on their orbital ellipse parameters -- You can see how you have solar eclipses where the moon completely covers the sun as well as "annular" eclipses where the moon has a ring of sun around it!

pogona wrote: you should also explain how the image width and the DOV work

Taking the example there - the field of view is 70 degrees, and the image width is 1000 pixels.

That means if Jupiter is about 19 degrees across as seen from Io, then it should take up 19/70ths of the view, which is 0.27 of the width. 1000 * 0.27 = 270 pixels, so Jupiter should be about 270 pixels across in your view.


Of course, to flip the whole thing around, you could easily just draw a picture and then say what the field of view is afterwards - if Jupiter takes up 540 pixels in your field of view then you can still say that it's a view from Io, but the FOV is only 35 degrees instead of 70 degrees. That said, how you render the foreground could change as a result of this (e.g. a narrow field of view would be more "zoomed in")

Thank you very much Tom! I want you to know that I appreciate it very much, and you have made things better for me, with this tool.