Quote:
Originally Posted by rally
Ray,
I only mention this because your formula includes F No. as one of the primary terms.
But since incoming light based purely on aperture and focal length may not have as much relevance as the percentage incoming illumination of the CCD chip, I think its worth considering.
Say for example we have two OTAs with the same focal length and aperture and therefore F No, but one has an image circle of say 100mm and the other has an image circle of 50mm
Both in theory support a large chipped camera - but one is spreading that light at the focus plane over an area that is 4 times larger than the other the pi.r^2 does this - double the radius - quadruple the area
eg 50mm image circle yields roughly 2000mm2
100mm yields roughly 8000mm
So our efficiency calculation really needs to consider just how the incoming light is being distributed at the focal plane.
In my example that factor is 400% difference - so its not insignificant and if real world examples are used some scopes have image circles well under 50mm - often under 30mm so if for example it was say 25mm then the variation is a factor of 16 times (1600%) difference
So are we are worrying ourselves about small percentage variations yet this is potentially an order of magnitude difference !
I am not sure there is a uniform standard for defining the image circle of a telescope but its usually some notional diameter at which the amount of vignetting falls off below a certain level and or the spot size and chromatic aberration grows to a dimension that makes it unuseable.
But I think manufacturers have left it to their own discretion as to how this is interpreted.
Which makes it very difficult to quantify using a generic formula.
All of this ignores the quality of light that different telescopes are capable of focussing in the first place - strehl ratio, longitudinal plot characteristics in terms of the useable spectral width of a given scope.
Its one thing to illuminate a CCD with incoming photons its another entirely to ensure that only photos that are "supposed" to be received by a given well (pixel) are in fact received at that point and not spread out all over the place.
So the "noise" of the system is entirely relevant to the efficiency since at the end of the day if the signal we are trying to record is lost in the noise of poor optical performance or poor CCD performance (eg low dynamic range, high noise) etc we are losing efficiency in swathes.
Food for thought - its a complex problem methinks.
Rally
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Hi Rally
I think that what you are saying is encapsulated in the formula. by way of explanation, imagine a 2mfl f8 system and a 1mfl f4 system. The f8 system has 9micron pixels and the f4 system has 4.5 micron pixels. Both have the same aperture and angular sampling.
according to the formula, both systems have the same pixel level sensitivity (which is right). However, for the same field of view, the f8 system will need a corrected focal plane of 4x the area of that of the f4 system, because the f8 system pixels are larger - which I think is the point you are getting at - but the larger focal plane does not affect the operation of the formula or change the system sensitivity.
Focal length has as much effect on sensitivity as aperture, since it determines the fraction of the incoming photons that actually gets into a pixel of a given size (the aperture determines how many there are in total). That is why ("myth" papers notwithstanding) FNo rather than aperture is vitally important for fixed pixel size. If you doubt this, try changing your scope FNo with a Barlow and see what happens (it will not change the aperture or the pixel size, but it will drastically reduce the sensitivity).
regards Ray