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FORUMS Photo Sharing & Visual Enjoyment Astronomy & Celestial 
Thread started 01 Dec 2009 (Tuesday) 14:58
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Moon from yesterday... can I calculate focal length with it?

 
MintMark
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Dec 01, 2009 14:58 |  #1

Everybody is posting their moon shots, so here is mine :) It's just one picture with a bit of contrast, saturation and sharpening in DPP. This was taken with the 1000D through the Megrez 72 telescope with a 2" 2x barlow, mounted on an astrotrac set for lunar tracking. ISO400, 1/250th second.

I've read that the magnification you get from a barlow depends on the distance between the barlow lens and, in this case, the camera sensor. It's a 2x barlow, but that's if you stick an eyepiece in it.

My moon is 1910 pixels high and 0.5 degrees apparent size. The original image is 2592 pixels high, so roughly 0.68 degrees angle of view. The sensor is 14.8mm high (Canon APS-C), so I can calculate the effective focal length

f = 14.8 / (2*tan(0.68/2))

giving a focal length of 1247mm. The Megrez 72 has focal length 430mm, so the barlow must be providing 2.9x magnification.

Is that right? A 2x barlow in front of a camera gives almost 3x magnification? That means the system was operating at f17... which is why I needed ISO400 I guess.

Anyway, I'm really pleased with my moon picture :)


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Celestron
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Dec 01, 2009 16:27 |  #2

Very nice Mark ! Can you post a brighter shot of this one ?




  
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Bernoulli
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Dec 01, 2009 18:24 |  #3

MintMark wrote in post #9115929 (external link)
The Megrez 72 has focal length 430mm, so the barlow must be providing 2.9x magnification.

Is that right? A 2x barlow in front of a camera gives almost 3x magnification?

MintMark -

I've been down this road myself - allow me to share.

I tried several ways to measure the effective FL of my various possible optical configurations, and I found the best way is to key off the diameter of the moon on the image in pixels. At 900 mm (camera directly into 900 mm scope, no Barlow), the disk is 1740 pixels. Dividing, you get that if you measure the diameter of the moon on the image and multiply by 0.52, you get the effective FL. I found this works great for all my 3 scopes and for my long lenses (300 mm, 100 mm). Now it only works for a given sensor, and I have a 50D. If you want to generalize across other cameras, you could measure the image size in mm on the sensor, but I only have one SLR so I don't bother.

Using this, I measured the EFL of my 2X Barlow and found, you guessed it, it acts like anywhere from 2.5X to almost 3X.

It depends on how you set it up, and I have three configurations:
1. scope, barlow, camera
2. scope, barlow, right-angle diagonal, camera
3. scope, right-angle diagonal, barlow, camera

I use number 1 almost all the time, and I get 4340 pixels which, multiplied by 0.52, gives an EFL of 2260 mm, which means the Barlow is giving 2.5X, not 2.0X.

I asked some of my astronomy friends about it and they tend to just stare off in the distance and mumble something about ". . . Barlows are just funny that way . . ."

So I haven't really tried to figure it out, as long as I get good images I'm happy. But it bugs me. Maybe someone will chime in and help us.

I see what you're doing mathematically, but try the pixel diameter method, it's more straightforward.

And, by the way, nice image!


Rick
rulrich@uark.edu (external link)

  
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Bernoulli
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Dec 01, 2009 18:35 as a reply to  @ Bernoulli's post |  #4

Let me post one more comment that folks might find helpful. When you get to 1650 mm, the full moon will be reaching the edges of your APS-C frame, no matter what your pixel size is. You could measure EFL's from that, I suppose, for any APS-C camera.


Rick
rulrich@uark.edu (external link)

  
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MintMark
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Dec 02, 2009 12:45 |  #5

Celestron wrote in post #9116597 (external link)
Very nice Mark ! Can you post a brighter shot of this one ?

Hello Ron, here is a lighter version. I took many pictures that night because I thought I was going to stack them... this was just one I picked out at random.


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MintMark
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Dec 02, 2009 12:49 |  #6

Bernoulli wrote in post #9117306 (external link)
Let me post one more comment that folks might find helpful. When you get to 1650 mm, the full moon will be reaching the edges of your APS-C frame, no matter what your pixel size is. You could measure EFL's from that, I suppose, for any APS-C camera.

Using this calculation (1910 / 2598 ) * 1650 gives me 1215mm, so roughly the same.

I read in wikipedia that the angular size of the moon varies from 29 to 34 arc minutes, so without knowing the apparent size that night it will never be super accurate.


Mark

  
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Celestron
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Dec 02, 2009 14:06 |  #7

MintMark wrote in post #9121750 (external link)
Hello Ron, here is a lighter version. I took many pictures that night because I thought I was going to stack them... this was just one I picked out at random.

Thanks Mark . I can see it much better ! . I always like brighter moons so i can see better . Much like the ones on my website .




  
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Moon from yesterday... can I calculate focal length with it?
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