Time for some maths...
Here the the question:
What focal length of lens is required to produce a full frame image of the moon?
I want to produce am image like this:
Where the diameter of the moon projected on my full frame sensor is 24mm.
Clearly, we need to know the anglular size of the moon. The Wiki tells us...
http://en.wikipedia.org/wiki/Angular_diameter
It varies. 29.3′ – 34.1'
That's 29.3 to 34.1 minutes. Sixty minutes in a degree.
We'll make it easy and call it half a degree.
Next, we need to be able to calculate the vertical field of view from any given lens.
For this purpose, I think that we can consider a pin-hole lens. I'll draw a diagram for a 100mm pin hole:
The 7.1 degrees shown is for half the image. The full (vertical) angle of view would be double that, 14.2 degrees.
The angle (x) is calculated with trig:
tan(x) = 12.5/100
x = atan(12.5/100)
x = 7.125 degrees
(don't forget to double it...)
Now, lets check that against Canon's reference:
http://web.canon.jp/imaging/l-lens/spec/lens_spec.html
The vertical angle of view on the 100-400 starts at 14 degrees, against the 14.25 that I predict.
Let check again on the 800mmm
tan(x) = 12.5/800
x = atan(0.015625)
x = 0.89 degrees
Double it to 1.79 degrees.
Which is 1 degree 47 minutes. Canon say 1 degree 40 minutes. Close enough.
So, we weem to be on the right track.
Turn it round and ask what is the focal length for 0.5 degrees?
tan(0.5) = 12.5/x
tan(0.5) * x = 12.5
x = 12.5/tan(0.5)
x = 1432mm
So, I need a 1432mm lens.
An 800 with 1.4x come close at 1120mm and would autofocus on a 1D body.
An 800 with a 2x would crop the moon.
Can someone give me an 800mm f5.6 so that I can check?
Andrew
Here the the question:
What focal length of lens is required to produce a full frame image of the moon?
I want to produce am image like this:
Where the diameter of the moon projected on my full frame sensor is 24mm.
Clearly, we need to know the anglular size of the moon. The Wiki tells us...
http://en.wikipedia.org/wiki/Angular_diameter
It varies. 29.3′ – 34.1'
That's 29.3 to 34.1 minutes. Sixty minutes in a degree.
We'll make it easy and call it half a degree.
Next, we need to be able to calculate the vertical field of view from any given lens.
For this purpose, I think that we can consider a pin-hole lens. I'll draw a diagram for a 100mm pin hole:
The 7.1 degrees shown is for half the image. The full (vertical) angle of view would be double that, 14.2 degrees.
The angle (x) is calculated with trig:
tan(x) = 12.5/100
x = atan(12.5/100)
x = 7.125 degrees
(don't forget to double it...)
Now, lets check that against Canon's reference:
http://web.canon.jp/imaging/l-lens/spec/lens_spec.html
The vertical angle of view on the 100-400 starts at 14 degrees, against the 14.25 that I predict.
Let check again on the 800mmm
tan(x) = 12.5/800
x = atan(0.015625)
x = 0.89 degrees
Double it to 1.79 degrees.
Which is 1 degree 47 minutes. Canon say 1 degree 40 minutes. Close enough.
So, we weem to be on the right track.
Turn it round and ask what is the focal length for 0.5 degrees?
tan(0.5) = 12.5/x
tan(0.5) * x = 12.5
x = 12.5/tan(0.5)
x = 1432mm
So, I need a 1432mm lens.
An 800 with 1.4x come close at 1120mm and would autofocus on a 1D body.
An 800 with a 2x would crop the moon.
Can someone give me an 800mm f5.6 so that I can check?
Andrew
