Super-Zoom P&S Camera Beats DSLR (again) - Film at 11

________ wrote:

Shorter focal-lengths create smaller airy-disks when using
diffraction limited glass.

Yes, smaller cameras create smaller airy discs *and* dimmer fainter airy
discs, and they have to cram them onto smaller, less sensitive sensors
with more noise using 20x super-zoom lenses stretched to their limit. I
do think you are beginning to understand this.

AndyTalberts wrote:

You can't obtain diffraction-limited effects in non-diffraction limited optics.

You are talking gibberish. Of course you can see diffraction in any lens
that can be stopped down. What the hell are you talking about???

On Wed, 03 Dec 2008 19:03:57 -0800, Paul Furman wrote:

________ wrote:

Shorter focal-lengths create smaller airy-disks when using
diffraction limited glass.

Yes, smaller cameras create smaller airy discs *and* dimmer fainter airy
discs, and they have to cram them onto smaller, less sensitive sensors
with more noise using 20x super-zoom lenses stretched to their limit. I
do think you are beginning to understand this.

The light is spread over a larger area on a larger sensor to try to illuminate
larger photosites, there is no net gain to luminosity level at identical
f/ratios between lenses on their correspondingly matched sensors.

I do know (which is quite different than just thinking or believing something)
that you're still just as ****ingly ignorant as you've always been and continue
to prove to be with every sentence that you attempt to post.

_________ wrote:
Paul Furman wrote:
________ wrote:

Shorter focal-lengths create smaller airy-disks when using
diffraction limited glass.
Yes, smaller cameras create smaller airy discs *and* dimmer fainter airy
discs, and they have to cram them onto smaller, less sensitive sensors
with more noise using 20x super-zoom lenses stretched to their limit. I
do think you are beginning to understand this.

The light

More light, due to larger lenses...

is spread over a larger area on a larger sensor to try to illuminate
larger photosites, there is no net gain to luminosity level at identical
f/ratios between lenses on their correspondingly matched sensors.

Right, it's a wash at the same aperture and ISO on that basis. You are
getting it...


Except that the DSLR is gathering more light so it has less noise even
at base ISO, and works in lower light when you need to crank up the ISO.

On Mon, 01 Dec 2008 22:21:57 GMT, "Andrew Koenig" wrote:

"John O'Flaherty" wrote in message
news
The *absolute* size of a lens' entrance pupil (measured, for example, in
inches or millimeters) determines the smallest detail in the *subject*
that
can be resolved (i.e. the angle of resolution).

So, it determines the angle of resolution, rather than the absolute
size of something in the subject, right? Also, I wondered about
absolute size of the aperture - it seems to be a function of the
wavelength of light, but that fixes it within a factor of two for
visible light. This was interesting on that subject:
http://labbey.com/Articles/Limits/Limits.html

Right, on both counts.

The *relative* size of a lens' entrance pupil (i.e. its f-stop) determines
the smallest detail in the *image* that can be resolved.

I'm still trying to get a comprehensive view of this, relating pixel
physical size, sensor size, and resolution. What you say seems to
imply a property of a sensor that could be seen as independent of the
lens, that is, you could assume a perfect lens of any description and
for a given sensor, the sensor's own minimum f number limit would
still apply. Is that right?

Pretty much. What I'm saying is that for a given f-number, there is a
corresponding absolute size (i.e. a specific number of mm, or microns, or
whatever) that is the smallest detail that can be resolved *on the sensor*
(or on the film), and that that f-number is the same for any lens focal
length. So making the pixels smaller than that size will not result in any
additional information in the digital image, even if it results in more
pixels.

Putting it another way, if you give me a pixel pitch, I can tell you the
smallest relative aperture (i.e. largest f-number) that that pixel pitch
will tolerate.

You've inspired me to get specific. Note that the following numbers are
only approximations, because the notion of resolving power is approximate.
Still, it should give you the idea.

According to Dawes' formula, the resolving power of a lens with diameter D
millimeters is given by R = 116/D arcseconds. So, for example, a 100mm
telescope should have a resolution of 1.15 arcseconds, which squares well
with my personal experience.

Now, suppose we have a pixel pitch of P millimeters and a lens with a focal
length of F millimeters. That's equivalent to an angle of P/F radians. 1
radian = 57.3 degrees (more or less) and one degree = 3600 arcseconds, so 1
radian is 2.06E5 arcseconds.

So a pixel pitch of P millimeters with a focal length of F millimeters is
equivalent to an angle of 2.06E5*P/F arcseconds.

What lens diameter do you need to resolve that angle? Remember that R =
116/D; that equation is equivalent to D = 116/R. So the diameter we seek is
116/(2.06E5*P/F) millimeters, which in turn is equivalent to
(116*F)/(2.06E5*P), which in turn is approximately equal to F/(1778*P).

Now, of course, when we express a number as F/n, we're talking about a lens
with an f-stop of n, irrespective of the value of F.

So we have concluded that for a pixel pitch of P millimeters, we need a lens
at least as fast as f/(1778*P) to deal with it.

Let's see how that works out in practice. A Nikon D700 has 4,256 pixels
across the long dimension of its images, which is 36mm. So its pixel pitch
is 36mm/4256, or 8.45E-3 mm. 8.45E-3 * 1778 is 15.03, which means that the
D700's pixel pitch is equivalent to an f/15 lens, regardless of focal
length.

Note that the critical aperture for a 4/3 camera is about f/8; for a
typical
P&S camera it's about f/2.8.

I just did that experiment with a Canon 40D and a 50 mm f/1.8 lens.
The f/8 is quite a bit sharper than the f/22.

Well, let's see. The 40D produces an image of 3,888 x 2,592 pixels. The
long dimension of the sensor is 22.2 mm. So the pixel pitch is 5.7E-3 mm.
Multiply that by 1778 and you get 10.15. So I would expect that the 40D
will start losing sharpness to diffraction by f/11.

Of course 1778 is way too precise for these rough calculations. It would be
quite good enough to compute the pixel pitch in microns and multiply by 18.

I see you have the correct 1.8 in your other replies. This is very
interesting. For a Canon G10 P&S, with a horizontal resolution of 4416
and 7.6 mm, that would give 1.72 um and f/3.1 for the crossover. The
lens has a maximum aperture that varies from 2.8 to 4.5, so ignoring
any other consideration, sharpness is lost at the long end even if the
lens is wide open. It looks as if any narrowing of the aperture to get
greater depth of field would give more blurring due to diffraction.

--
John

On Wed, 03 Dec 2008 19:18:36 -0800, Paul Furman wrote:

AndyTalberts wrote:

You can't obtain diffraction-limited effects in non-diffraction limited optics.

You are talking gibberish. Of course you can see diffraction in any lens
that can be stopped down. What the hell are you talking about???

Do everyone a favor. Go to your local high-school. See what courses they offer
for idiots in their physics department. Go play with a wave-tank (a.k.a.
ripple-tank) and 2-speaker sound systems that explain to you the most simplistic
principles of waves, wave reinforcement, wave-propagation, diffraction, and
optics.

Come back after you've completed at least 5 of those full courses. Why 5? I know
that you'll flunk at least the first 4 attempts of trying to comprehend things
this simple.

Here's a test you can do to prove your own ignorance to yourself. Got a laser
pointer handy? Shine that light through a soft open-weave cloth mesh where the
mesh openings are smaller than the width of the laser beam. Look at the pattern
that the laser beam forms on a distant wall about 30ft. away. See anything odd
around the main beam? No, just a bunch of random-dot blurriness. Now shine that
laser through a piece of metal window screen where the right-angle weave now has
sharply defined edges (edges sharp and cleanly defined enough to cause
diffraction). Look at the pattern that laser-beam forms on that wall 30ft. away.
Wow, what's that? All those equidistantly spaced dots in 2 directions! What's
causing that?!?

DSLR Glass = Soft-edged cloth of unevenly defined fibers. P&S Glass =
Clean-edged window-screen wires capable of causing diffraction patterns.

You'll probably figure it out what that all means after you come back from your
5th year of having to re-take same remedial-level high-school courses in
science.

What the hell are you talking about?

Of course you can see diffraction in any lens that can be stopped down.

What the hell are you planning to do with diffraction rings? Maybe
there's some fourier whateverthe**** trick used in astro imaging for
extracting information from echos of lost data but that's irrelevant to
pictorial photography.

On Wed, 03 Dec 2008 19:28:25 -0800, Paul Furman wrote:


Except that the DSLR is gathering more light so it has less noise even
at base ISO, and works in lower light when you need to crank up the ISO.

The one and only minor advantage to a larger sensor, which has been proven
countless times that high ISOs aren't a requirement of a real professional. Then
their archaic OVF system doesn't allow you to focus in that lower-light to which
you claim it has an advantage. Advantage null.

P&S cameras perform better for many areas which a DSLR can't even attempt to
duplicate. You can't obtain the same light-gathering ability at longer focal
lengths. You can't obtain the same light-gathering ability with deep
depths-of-field required for macro photography. You need to read that 25-point
list again that totally disproves any large sensor's meager and unnecessary (or
outright useless) advantages.

Simply put: Total amateurs need high ISOs to compensate for their lack of skill
and talent as photographers. Just like they need faster auto-focus. Those are
the arguments of a totally amateur snapshooter who doesn't know any better.

On Wed, 03 Dec 2008 20:04:45 -0800, Paul Furman wrote:

What the hell are you talking about?

Of course you can see diffraction in any lens that can be stopped down.

What the hell are you planning to do with diffraction rings? Maybe
there's some fourier whateverthe**** trick used in astro imaging for
extracting information from echos of lost data but that's irrelevant to
pictorial photography.

The only appropriate response: an eye-roll and sigh done when in the presence
of a total moron

"John O'Flaherty" wrote in message
...

Of course 1778 is way too precise for these rough calculations. It would
be
quite good enough to compute the pixel pitch in microns and multiply by
18.

I see you have the correct 1.8 in your other replies.

Yes, I dropped a character in typing.

This is very
interesting. For a Canon G10 P&S, with a horizontal resolution of 4416
and 7.6 mm, that would give 1.72 um and f/3.1 for the crossover. The
lens has a maximum aperture that varies from 2.8 to 4.5, so ignoring
any other consideration, sharpness is lost at the long end even if the
lens is wide open. It looks as if any narrowing of the aperture to get
greater depth of field would give more blurring due to diffraction.

Right! Of course, these numbers are all approximate because we're dealing
with amounts of blur :-)

But still, these computations sure suggest (and strongly!) that if you have
a camera such as a G10 and stop it down to, say, f/11, then you're wasting
lots of pixels. And it doesn't matter how good the lens is.