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Re: [Digital BW] Digest Number 1589

Re: [Digital BW] Digest Number 1589

2003-06-16 by Bruce

on 6/16/2003 3:46 AM, DigitalBlackandWhiteThePrint@yahoogroups.com at
DigitalBlackandWhiteThePrint@yahoogroups.com wrote:

> Message: 17
> Date: Mon, 16 Jun 2003 04:03:53 -0000
> From: "Tim Timmermans" <zenphoto7@...>
> Subject: Taking the plunge?
> 
> Well after holding out, partly because of money, I'm considering
> jumping into the digital camera realm. I still have budget
> considerations and since my current set up is a Canon EOS 1N with
> some good lenses it makes sense to look at the Canons. Used D60's and
> new D10's are not that far off in price so I'm focusing on the D10's
> which I see for around $1400 on ebay.
> 

> 
> I ask because Burning Man is coming up again soon and I'd like to
> bring a digi out there this year.



Wow, you're going to bring your brand new digi to the sands storms of the
Black Rock Desert? That alkali dust gets into everything. Bring lots and
lots of baggies!



-Bruce

Visit my website at:
http://home.earthlink.net/~smthopr

Re: [Digital BW] Digest Number 1589

2003-06-17 by claudej1@aol.com

In a message dated 6/16/2003 3:48:46 AM Pacific Daylight Time, 
DigitalBlackandWhiteThePrint@yahoogroups.com writes:

> Yes, electronically, 100 ohms in parallel with 100 ohms does give you 50
> ohms, but electronics are not the same as optics and film.  Film records the
> data just like a CCD array does, using discrete elements.  If the photons
> being "measured" happen to fall right on the sensing area, it will record at
> that resolution, if it happens to straddle two sensing areas, it'll record

> half that resolution.  What you will get are values from %50 gray on two
> adjacent sensing areas, to one %100 white and one %100 black (if you are
> using a test pattern that is).  Of course, film grain size is random, but
> still within a rather narrow size band.
> 
> It's more than a formula, it's an understanding.
> 
> Austin
> 

The electronic analogy was merely to point out the classic formula used to 
figure out system MTF between lenses and film. The "understanding" that you 
outline concerning "straddling" two sensing areas is precisely what is taken into 
account with the formula.

The classic U.S. Air Force LPPMM test target contains hard edged lines on 
Kodalith to represent, as best as possible 0% and 100% modulation (which would 
averge out to 50 mathematically). This is not what happens in a continuous tone 
image, but gives an extreme tonal contrast with which to efficiently judge an 
optical system. So when we make a contact print of the test target using an 
enlarger as a collimated light source onto a particular film, we remove the 
influence of optics and their aberrations from the light path. This give us pure 
film MTF to work with.  Now if we test a lens separately in the same manner we 
also isolate the performance of the lens without the influence of film 
anomalies. If the best we get by looking at the arial image from the Lens under test 
with high power optics is 100 LPPMM also, we need to combine the two.

The accepted formula for doing so, in all literature I have ever come across 
is to invert the sum of their inverses.  So, 1/100 is 0.01 for the film and 
lens. If we add 0.01 + 0.01 = 0.02. Now taking the inverse of 0.02 (1/.02), we 
get 50. This is the BEST possible performance we can expect with a high 
contrast subject, with the lens and film working together. If either one of them is 
slightly lower (usually the lens), then the system drops to less than 50 LPPMM. 
If we only happen to get 60 LPPMM from the lens at a certain aperture, then 
the system MTF drops to 37.5 LPPMM.

This is why, historially,  larger formats have always won the sharpness 
contest, as the gains are much greater by adding more Millimeters to throw the 
image onto rather than trying to cram more line pairs into the same millimeters 
(which solid state imagers do better than color film for a given number of 
millimeters). But that is beyond the scope of this text.

Your reasoning sounds good, but it doesn't appear to be congruous with the 
traditional formulas cited above.

Claude


[Non-text portions of this message have been removed]

RE: [Digital BW] Digest Number 1589

2003-06-17 by Austin Franklin

Claude,

> > It's more than a formula, it's an understanding.
> >
> > Austin
> >
>
> The electronic analogy was merely to point out the classic
> formula used to
> figure out system MTF between lenses and film. The
> "understanding" that you
> outline concerning "straddling" two sensing areas is precisely
> what is taken into
> account with the formula.

Some things ARE 1/x + 1/x etc (like the resistors), and some things are not,
and understanding enough to know when some are and some aren't is the
understanding I was mentioning you apparently don't have.  Straddling gives
you the WORST case, and aligned gives you the best case.  There is no such
thing as "straddling" with respect to resistors, but there IS when respect
to sampling.  Resistors are not sampling, but imaging is.

> The accepted formula for doing so, in all literature I have ever
> come across
> is to invert the sum of their inverses.  So, 1/100 is 0.01 for
> the film and
> lens. If we add 0.01 + 0.01 = 0.02. Now taking the inverse of
> 0.02 (1/.02), we
> get 50. This is the BEST possible performance we can expect with a high
> contrast subject, with the lens and film working together. If
> either one of them is
> slightly lower (usually the lens), then the system drops to less
> than 50 LPPMM.

Your statement that started this was about DIGITAL sensors, BTW, which is an
easier example to show you're incorrect.  If you have a lense that gives 100
lp/mm and a sensor that is also, say, 100 lp/mm, then you CAN IN FACT get
100 lp/mm at the sensor if the.  You can only RELIABLY get 50, but in
reality you get a range of between 50 and 100.  This is just plain and
simple fact of sample theory, and if you just don't understand it, well,
then you just don't understand it.

> Your reasoning sounds good, but it doesn't appear to be congruous
> with the
> traditional formulas cited above.

Since you obviously don't actually understand how the system works, or the
actual dynamics of "the formula", or how it actually applies, you're going
to draw erroneous conclusions.

Austin

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