RustyBug Offline Upload & Sell: On
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Thank again guys ... I'm still not there (as others groan yet again) @ where I want to be, but as always I appreciate the respectful spirit of help from fellow FM'ers.
I just feel like there is an illusive formulaic Sasquatch out there that provides the missing link that would formulaic connect us to the image circle as an area with an amount of photons falling on it to yield our derived formula from a root formulaic perspective.
Okay, so the light falls on our subject at a given concentration, it is reflected from our subject into a direction that can be captured by our lens. That will put an X amount of photons at our lens (for a given time @ component of flow), with our entrance pupil having a given area. Through the refractive properties of the lens elements, that same amount of photons will be "condensed" to the size of our max aperture which would yield a given amount/area.
After passing through (flow ) the physical (orifice ) aperture (which when stopped down reduces the amount "passing through", i.e. cross-section area iaw the area reduction), that same amount of photons would be "re-expanded" by the refractive properties of the lens elements to the area of the image circle. For a max aperture WO, the amount of light at our entrance pupil should be the same as the amount of light at our image circle
If our lens FL is 50mm at f/1, then our entrance pupil is also 50mm diameter (25mm radius). As such, no orifice is (smaller physical aperture) in place to restrict the amount/flow of photons below that which is available at our entrance pupil. The amount of photons available at our entrance pupil is the same amount available to be projected onto our image circle. If our image circle is the same area as our entrance pupil and our amount of light at our image circle is the same as the amount of light "passing through" our entrance pupil, then the amount/area will be the same for each ... i.e. 1:1 proportion, hence f/1.
If we stop down our physical aperture, we are then reducing the cross-section area of our ability to pass through (flow) photons. This reduction in cross section area, yields a reduction in the amount of photons that can flow (per unit time). Thus a 1/2 reduction in diameter/radius yields a 1/4 reduction in the cross-section area to reduce the amount of photons that can pass through to the image circle area. Now that we have a a reduced amount (without an offsetting change in time) photons reaching our image circle, a 1/4 reduction in amount will yield an amount/area(unit) 1/4 that of our entrance pupil. And, of course, this is why we increase the time by a corresponding two stops (2 x 2 = 4) to allow the total amount of photons reaching our film/sensor to accumulate iaw with Q=Av.
As it turns out, our FF film area is 24X36 for, then the distance to our furthest corner is 43.3 mm, so a "perfectly fitting" image circle would be one of the same size. But for convenience we call 50mm "normal" when it is actually a very, very short tele compared to the format size image circle @ 43.3 mm. Note here that our "normal lens" between formats are correlated to our image circle size (see other sizes in link below).
http://en.wikipedia.org/wiki/Normal_lens
While it is highly doubtful that I've convinced anyone of the proportional relationship to the image circle or to the role of flow relative to the amount/area (unit) ... and have yet to fully derive the formulaic connection ... I'm good.
I'll leave (for now ) the remaining deductions and correlations regarding format size equivalencies for each to resolve in the manner of their choosing. I highly recommend using formulaic, dimensions and proper units to accurately make those equivalencies. However, most photographers are not interested in the fully developed math, but instead lean heavily on the pragmatic reduced/derive equations ... i.e. FL/aperture = f/stop (proportion).
As it turns out, I aspire to understand this proportion in terms of area rather than length (given our significance to exposure @ amount/area(unit). However, since area is derived from length, the proportions of area vs. area can retain the same resultant as length vs. length, and our image circle is defined by length of its radius, similar to how the diameter and area of our entrance pupil are resultant of its radius, thus proportions are retained ... even if not the far easier conventional way that we make the calculation using length to assess the proportion.
But, I would suggest that once you start trying to make certain equivalency comparisons, it is very easy to grab one portion of the underlying math while omitting another aspect of it as we aspire to present our position to one another ... hence the enormity of confusion that abounds on the subject of equivalency as people are mixing/matching area vs. length @ r^2 (i.e. linear vs. exponential).
I still believe that if you consider the amount/area(unit) as key, you must consider the image circle area relative to the entrance pupil area (sans physical aperture restriction) as a proportion, which is what an f-stop is (i.e. proportion) even if we have found an easier way to calculate that proportion via length rather than area ... i.e. saying the same thing in a different way.
Thanks again for the indulgence, assistance and patience ... FM'ers rock !!!
FWIW ... feel free to ignore the ramblings of a brain damaged individual. It's just the way things have to work for me sometimes in a post-TBI reality. In case you didn't already know or forgot, I got whacked. So if you ever thought I was whack (or a little slow sometimes) ...
Thus, my appreciation for the indulgence is genuine.
Edited on Oct 29, 2013 at 10:58 AM · View previous versions
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