Fortunately for me I have two telescopes that I can compare directly. One is a 14" SCT using Hyperstar and has a focal length of 711mm. The second is a 4" Refractor with a focal length of 714mm so they are both effectively the same.
I happen to know that one of the scopes requires much less of an exposure to get a high signal shot of a 14th mag asteroid than the other using the same camera. That's simple physics, photons per pixel. I even had a paper published in the Journal for Occultation Astronomy on the subject
Sooo... a C14 gathers more light than a 4" ?
I'd really like to hear from any Ceravolo dual FL owners on whether they see better S/N in the "fast" configuration.
All you will get is a wider field of view over the same sensor. Unless you increase the aperture you will not be collecting more flux.
Google Stan Moore F ratio myth
I guess then the laws of physics somehow do not apply here, and my direct experience with these two f-ratios with my scope n camera, where I could easily see stronger flux per pixel per unit of time at f/4.5 as opposed to f/6 (confirmed by SNR measurements with PI) are either my imagination or camera n PI are “playing Jedi mind tricks’...google does not provide answers to all questions Peter
So if we take a glass window (simulating a very large 1m aperture with loooong focal length), it clearly will be putting photons on a pixel behind the window much more slowly than a 4” strongly converging lens (small aperture fast f-ratio) placed in front of the same pixel.
Let’s create a new myth - “larger aperture is always faster”...
No myth at all Peter. For a given aperture and camera, a fast scope will produce more signal than a slow one. The aperture determines how many photons get through from any point in the sky, but the focal length (and hence the Fno) determines how large an area of sky feeds photons through that aperture and into each pixel - for a given aperture and camera, a faster scope looks at more sky, resulting in more detected signal in each pixel. In more general terms, it is wrong to say that FNo by itself is the only determinant of image SNR quality, but it is equally misleading to contend that aperture by itself is the only determinant of image SNR. As Rick points out, sampling must also be included.
Re Stan Moore's paper, ask why the two images he presents - from f4 and f12 scopes of the same aperture - have the same plate scale. Unfortunately Stan doesn't explain this, but I assume that he got there by binning 3x3 in the f12 configuration. That would give about the same scale and also restore the large SNR loss inherent in operating at f12 vs f4. If he used that approach, the only thing he ended up losing at f12 was ~90% of the field of view, but in the process he did not show the large drop in SNR at f12 vs f4 (with the same camera and no binning).
If anyone needs an illustration that aperture is not all that matters, try imaging with a 2x Barlow in the imaging train
Peter, your observation that you get better SNR results from an f5.5 scope than from an f5 could come about if you use different cameras on the two scopes.
I was also going to suggest making a test with the same scope and camera bin 1x1, but changing f-ratio with a barlow, imaging at native FL and then using a reducer. I tired it and it DOES affect the speed
12" f8 RC with SBIG STXL11000 which is about 0.76" per pixel
12' f4 Newt with QSI683 which is about 0.92" per pixel.
The image scales are not too dissimilar.
Same target to get similar looking results.
12" RC 39 hours
12" Newt 15 hours
Imaging speed does appear make a difference at similar image scales.
Hi Paul
your experience could be explained by:
sampling - the imaging time scales with the square of the sampling, so your 11002 system required 1.5x as long as the 8300 due to this alone QE - the 8300 has about 1.4x the QE of the 11002 at Ha, so there is another 1.4x in favour of the 8300 Read noise -for narrowband imaging, you will almost always be read noise limited and there will be a significant reduction in the required imaging time with the 8300 due to its lower read noise(how much depends on sub length).
So, multiplying these factors, your Newt with 8300 should be considerably more than 2x times faster than the old system was - but much of it is down to the limitations of the 11002, not the FNo of the scopes.
............Peter, your observation that you get better SNR results from an f5.5 scope than from an f5 could come about if you use different cameras on the two scopes..........
That's the rub...I'm slumming it with just the one camera of late.
I have to say I've been tempted to try one of the new Sony chipped SBIG's..but suspect there will be no revelation in doing so with the various 'scope's I'm currently using.
That's correct. You have kept the aperture the same but increased the image scale. You're stuffing more photons into each pixel. Thanks for illustrating my point perfectly, Suavi
there have been lots arguments and it's hard to put it all together (and I hope I didn't miss a contribution already having said the following).
There is one of your statements that might have pushed you in a wrong direction:
Quote:
Originally Posted by Peter Ward
I hear what you are saying Rick...my point however is a 10" can only gather 10" worth of flux.
Optimal sampling will help better detect what's hitting the focal plane...it just won't give you more photons down the pipe
The point is, that only a part of the light hitting your 10" mirror will also hit your sensor. All the light coming from outside your aperture angle will be lost.
Now, if you add a reducer like Suavi did and make your scope 'faster', you will increase the aperture angle and thereby also increase the total amount of light hitting your sensor. I.e. less of the light coming from your mirror will hit the camera housing.
(This is the exactly the same what Rick and others arged, only from a different point of view.)
.........All the light coming from outside your aperture angle will be lost.......
A 10" scope can only ever get 10 inches worth of light...much like a cookie cutter ...light is limited by the aperture of the mirror or lens....i.e. a cookie cutter can't gather pastry beyond its edge...angles simply don't come into it.
Hence I'm not sure what you mean by "aperture angle" as the wavefront from the sky is perfectly parallel to the telescope aperture (or "cutter" )
MBJ did a great talk years ago about this stuff. He used the water buckets and rain analogy. Rain falls down and in straight parallel lines, not in a cone pattern or at an angle so Peter's cookie cutter is on the money. The wider your bucket the more rain falls inside it. The deeper tbe bucket, the longer it takes to fill up. Pretty simple.
Hi Marc. I'd be a bit wary of the rain analogy - like all simplifications, it can be a little bit misleading.
As Erwin suggested, light definitely comes in from a range of angles. The light that has a wavefront parallel to the aperture is focused by the optics into an on-axis point. Anything that is focused off-axis comes in with a tilted wavefront - it comes at a different angle from a different part of the sky and is focused to a point away from the optical axis in the focal plane.
An image with a million pixels samples the light coming from a million different directions - that's how an image is formed. The pixel size and focal length determines how big an area of sky is looked at in each of those directions - for a given aperture and pixel size, the faster the scope, the bigger the solid angle of sky looked at by each pixel. So the faster scope will collect more light in each pixel (for a given aperture and pixel size). Thus, the amount of light getting to each pixel is determined by both the aperture size (how many photons can get through it from any direction) and also the area of sky sampled by each pixel (how many photons you start out with from any direction). The rain analogy completely misses the second bit and wrongly leads to the conclusion that aperture is all that matters - aperture is fundamentally important, but it is definitely not the whole story..
Hence I'm not sure what you mean by "aperture angle" as the wavefront from the sky is perfectly parallel to the telescope aperture (or "cutter" )
The field of view displayed on your image has a diameter also on the sky.
The angle between the outer bounds seen from the point of observation is what I called "aperture angle".
Photons from (a few diameters) aside of this field still hit the mirror, but not the sensor. Adding a reducer (between mirror and sensor) redirects photons from outside the original field to your sensor. The total count of photons you earn is now those from the original field plus the ones from the additional part of the sky your image shows.
(Btw.: Making this angle too wide will lead to shadowing by your tube, but more serious is the aberration introduced by the reducer.)
......The pixel size determines how big an area of sky is looked at in each of those directions .........
This very confusing/misleading. Starlight does not come in at angles. The wave front is perfectly parallel to the aperture...any photons left or right simply don't enter the telescope.
Once the light is past the aperture of the telescope, then the telescope optics do what they do..reflect/refract to form the image.
Sure, faster telescopes put the focused image over a smaller area....nailing down how much a difference that makes is the sticking point.
It does! The angle of the light from a single star is too small to resolve (except Beteigeuze I believe), but the angle between the different stars on our images can be resolved easily.
As a result, the star images will not get brighter with F-number, but extended regions like nebulae will!
"A lens which could only focus light rays striking the glass head-on .... would be fairly useless for astronomy. Fortunately, most lenses can also accept rays which come in at a slight angle to the optical axis, and bring them to a focus as well. This focal point is not the same as the focal point for rays which are parallel to the optical axis; depending on the angle of the incoming rays, their focus lies on one side or the other of the optical axis, as shown in the diagram below. But if the lens is well-made, all these focal points will lie on a plane which is parallel to the face of the lens; this is called the focal plane."
As per Erwin and Ray's comments on the field of view (acceptance angle, solid angle of view or however named), it is like considering a ray diagram for a convex lens with an object at infinity with on-axis parallel light rays(BLUE) and off-axis parallel (GREEN and RED) rays from infinity. The field of view being formed by the angle between the GREEN and RED rays as shown.
This is shown in the attached ray diagram, with the 3D effect forming a cone, or more correctly a truncated cone at the lens surface.