Determining the angular size of celestial bodies and accurately portraying them in the sky
No.13805889 ViewReplyOriginalReport
Quoted By: >>13805911 >>13806074
So I've been working out how I can portray different celestial objects in the sky such as Jupiter and Neptune if they had the same distance between us as the Moon. Pic related. First step I needed to do was calculate the angular width that my camera captures photos with, to do this I used the Sun as a reference. I took a Sun photo at sunset so it appeared as almost as a perfect circle with no distortions. I then used the sun's 0.5 degree known angular size, and its size in pixels as a conversion ratio to find the angular width of the photo as a whole which turned out to be ~60 degrees. To angular size I used this formula:
delta = 2arctan(d/2D)
Where:
delta: Angular size/diameter
d: Actual diameter of the object
D: Distance between objects
Pic related is what Jupiter would look like if it was as close to us as the Moon. Seen from an observer on the ground (angular size ~ 21 deg).
My main issue right now is how I can reliably get the angular width of different horizon images like in the OP's background. I assume that the only way is if there's an object in the image you can use as reference, where you have to know both its real dimensions and its distance from where the image was taken. Thoughts?
delta = 2arctan(d/2D)
Where:
delta: Angular size/diameter
d: Actual diameter of the object
D: Distance between objects
Pic related is what Jupiter would look like if it was as close to us as the Moon. Seen from an observer on the ground (angular size ~ 21 deg).
My main issue right now is how I can reliably get the angular width of different horizon images like in the OP's background. I assume that the only way is if there's an object in the image you can use as reference, where you have to know both its real dimensions and its distance from where the image was taken. Thoughts?