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Astrophotography and Sketching >> DSLR & Digital Camera Astro Imaging & Processing

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Hikari
scholastic sledgehammer
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Reged: 01/05/11

Loc: Maine, USA
Re: Astronomical f-number (N) new [Re: mmalik]
      #5725863 - 03/11/13 01:28 PM

Quote:

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5






+1 Can't be done. However, you can project an image and measure the image circle. You can also use a focusing telescope/microscope and rail as well. You can also measure the illumination cone by looking at the exit pupil and rotating the lens--that will give you an angle and with a focal length, the circle of good illumination can be calculated. All kind of methods, but no real way by simple optical specs.


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mmalik
Postmaster
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Reged: 01/13/12

Loc: USA
Re: Astronomical f-number (N) new [Re: Falcon-]
      #5725946 - 03/11/13 02:10 PM

Quote:

Quote:

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5






Alas it can not be calculated with only those data points. It all depends on the optical design of the system. You could achieve those numbers with an 2 element refractor (archomat), or with a 17 element refractor (telephoto, zoom, image-stabilized camera lens), or with a newtonian (but what size secondary!?) or a cassegrain scope (but what secondary size, what baffles, etc) or a bunch of other potential optical configurations.

To truly *calculate* the image circle you would have to know the details of each element of the optical system that makes up that scope (or camera lens) as well as the physical layout (field stops, baffle tubes, etc) and do the math on exactly what the light paths through the optics will be. This is the domain of optical designers (beyond me!).

Worse once you know the scope's image circle you can still mess it up by using the wrong connections between the camera and the optic. That 88mm image circle you listed for the FSQ-106 is meaningless if you use a 1.25" prime-focus adaptor. In fact I suspect you must use a custom camera adaptor as even a T-ring and 2" adaptor would probably not allow the FULL 88mm image circle to pass.

So we are left with two options:

1) trust the manufacturer's stated specs on image circle
2) do real-world tests to see what the vignetting is (and hopefully report on it here )




This was not a trick question by the way, but a point is made. While dimensionless, f-ratio can be meaningless (the way it is touted by manufactures and how it is comprehended/misunderstood by amateur astro-imagers).


Short of one becoming an optical engineer, practically speaking, most is left to experimentation (the circles, the reducers, the sensors, the pixels, the focuser, the nose piece, etc.). Your thoughts?


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Hikari
scholastic sledgehammer
*****

Reged: 01/05/11

Loc: Maine, USA
Re: Astronomical f-number (N) new [Re: mmalik]
      #5726142 - 03/11/13 03:42 PM

Even if you were an optical engineer, why bother trying to calculate that stuff. Optics is not that hard. Just put the system together and go from there. Even putting it together is not that hard--you should be able to get enough product specifications to know what you need.

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fco_star
Carpal Tunnel
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Reged: 12/13/10

Loc: Texas, Midland
Re: Astronomical f-number (N) new [Re: mpgxsvcd]
      #5726235 - 03/11/13 04:13 PM


Mates, I was following this post very close, but I lack in ability with the numbers, so I ask a friend of mine (Carlos Ortega) that likes everything related to calculations and this type of stuff to help me with this and this is what he came up with.

To my surprise he use words and no formulas, please read it and let me know what do you think?

I consider Carlos a brilliant mind!

“Instead of stars let’s consider objects with a larger angular size, like the Moon, a planet, a galaxy or a nebula. For this example let’s consider we are imaging M31. The following applies basically the same to refractor and reflector telescopes. These basic concepts can be easily understood doing geometric optics.
For a given focal length, the main lens or mirror produces an image on its focal plane of M31 of a given size in millimeters or fractions of an inch. The amount of photons per unit of time that ‘fall’ on the image increases four times if you double the aperture, or changes as the square of the aperture. Therefore, if you have two main telescope lenses or mirrors with the same focal length but one of them has twice the diameter, both of them will make an image of M31 of the same size, but the one with larger diameter will make an image four times brighter. The exposure time will be four times faster with the larger aperture if you want to get similar results with both of them, from the brightness point of view. We are considering here that you keep the rest of the setup the same in both cases, like using the same eyepiece, or imaging directly to your camera sensor with no eyepiece.
Now, if you have two main telescope lenses or mirrors with the same aperture but one of them has a focal length twice that of the other, the one with the larger focal length will make an image twice as big compared to the other. Since both have the same aperture, the number of photons falling on the image per unit of time is the same, but the photons on the larger image have to spread over a larger area, so you have here four times less photons per unit of area, and as a result your image is four times dimmer. The relationship here is quadratic again.
So in summary, if you have two telescopes, one with an aperture ‘x’ times bigger (gaining a factor of x*x in brightness) and a focal length ‘x’ times larger than the other (losing a factor of x*x in brightness), the net result will be that both telescopes will make an image of about the same brightness and will require about the same exposure time for roughly the same result. The main difference will be that the telescope with the larger focal length will make an image ‘x’ times bigger. In other words, if you keep the focal length to aperture ratio constant, you get a similar result from the point of view of exposure time.”


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riverpoet
sage


Reged: 11/03/03

Loc: Slovenia, Europe
Re: Astronomical f-number (N) [Re: fco_star]
      #5726361 - 03/11/13 05:08 PM

We can all be friends:

In real life, you want to image & frame and object or types of objects (big nebulas, small galaxies etc). So you have to decide which FOV to use. FOV is a direct function of focal length & sensor size. Say you already have a camera and can only choose focal length. Now you want to gather as much light as possible (=aperture rules), so you want the largest aperture possible. By doing that, while keeping FL fixed because of FOV, you automatically want a faster f-ratio


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