A Photography forum. PhotoBanter.com

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below.

Go Back   Home » PhotoBanter.com forum » Photo Equipment » Large Format Photography Equipment
Site Map Home Register Authors List Search Today's Posts Mark Forums Read Web Partners

Image circle versus stopping down?



 
 
Thread Tools Display Modes
  #1  
Old June 27th 04, 09:05 PM
Nick Zentena
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?


The way I understood it stopping down increased the image circle
because it improved the quality of the image. But then I found this quote

"4. Why does the size of the image circle in a Large Format lens change as
the lens is stopped down?

Well . . . it doesn't actually. What actually happens is that when the
lens is wide open, the center of the image is very bright and the corners of
the image are VERY dark. So dark in fact that they do not expose the film
(When the film is properly exposed for the center of the image). As the lens
is stopped down, uniformity of illumination improves and therefore the
circle of usable illumination increases."

That's from here

http://www.schneideroptics.com/info/...mat_lenses/#q3

So if I understand this. You could use a lens wide open and get the same
image circle if you could use something like a centre filter? In other words
the edges shouldn't be soft. Does this make sense to anybody?

Nick

  #2  
Old June 28th 04, 11:08 AM
Richard Knoppow
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?

Nick Zentena wrote in message ...
The way I understood it stopping down increased the image circle
because it improved the quality of the image. But then I found this quote

"4. Why does the size of the image circle in a Large Format lens change as
the lens is stopped down?

Well . . . it doesn't actually. What actually happens is that when the
lens is wide open, the center of the image is very bright and the corners of
the image are VERY dark. So dark in fact that they do not expose the film
(When the film is properly exposed for the center of the image). As the lens
is stopped down, uniformity of illumination improves and therefore the
circle of usable illumination increases."

That's from here

http://www.schneideroptics.com/info/...mat_lenses/#q3

So if I understand this. You could use a lens wide open and get the same
image circle if you could use something like a centre filter? In other words
the edges shouldn't be soft. Does this make sense to anybody?

Nick




A center filter does not perform the same function as the stop.
The explanation on the Schndeider site is incomplete and
oversimplified. The main aberrations which vary with the stop are
spherical aberration and coma. Spherical exists all over the image,
coma is not present at the center but increases with the image angle.
Spherical aberration causes a hazy surround to highlights. Enough
spherical causes an overall haze and a blury image, it is the cause of
the softness of soft focus lenses. Coma is related to spherical but is
asymmetrical. It causes smeared, somewhat tear-drop-shaped blur spots
along radiuses toward the center. They can point toward or away from
the center of the image depending on where they mainly occur in the
lens. Coma varies rapidly with the stop.
Both of these aberrations vary with the angle the light rays take
through the lens. Other factors being equal they increase with the
speed of the lens, and coma especially increases with the angle of
coverage, i.e., its harder to correct for wide angle lenses.
These aberrations become smaller as the light going through the
lens is concentrated toward the center of the lens. This is what
stopping down accomplishes.
Uniformity of illumination is not affected by the stop except
where there is some mechanical vignetting from the lens mount. For
most lenses this is gone when stopping down about two stops. Some wide
angle lenses are designed to reduce this mechanical vignetting (the
Angulon is an example).
One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. A non-orthograpic lens, like a fish-eye
lens, can have better illumination because the light is more
concentrated near the margins.
A center filter acts to compensate, at least in part, for the fall
off by letting more light in at the margins of the image. The filter
is not exactly a stop but rather a tapered obstruction. The lens
projects a very blury, out of focus, image of the center filter onto
the film. One reason a center filter must be used with the lens
stopped down is that the range of angles the light going through the
filter must be limited for it work.
It is possible that a center filter _may_ act, to some extent, as
a obstructive stop. An obstructive stop limits the rays of light to
the margins of the lens rather than the center, as does a normal stop.
An obstructive filter changes the balance of all the aberrations and
has an effect on the MTF curves. In comparison to a standard stop the
obstrucive stop can raise the edge contrast at the price of lowering
the resolution.
The Schneider explanation about the balance of aberrations to
diffraction is correct. As the stop size is reduced the aberrations
are reduced but the diffraction is increased. At some point the two
curves cross. That is the "optimum stop". Actually, there isn't a
single optimum stop. Since coma and sphrical abberation play a part in
the image quality, and, since both vary with image angle, the optmum
stop will vary with the image angle wanted from the lens. For
instance, a Dagor, which has relatively low coma (as do all
symmetrical lenses) has a circle of illumination of nearly 90 degrees.
At about f/11 it has a coverage with good image quality of around 60
degrees. To get the maximum coverage angle it must be stopped down to
f/45. The image quality at the margins of the image at 87 degrees will
be best at f/45 but not at the center, there it will be best at around
f/11 or f/16.


Richard Knoppow
Los Angeles, CA, USA

  #3  
Old July 1st 04, 03:24 PM
jjs
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?


"Richard Knoppow" wrote in message
om...

[...]
One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. [...]


Care to help an innumerate? Taking the 38mm Biogon as an example, what's the
light fall-off in terms of F-stops?


  #4  
Old July 2nd 04, 06:33 PM
f/256
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?


"jjs" wrote in message
...

"Richard Knoppow" wrote in message
om...

[...]
One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. [...]


Care to help an innumerate? Taking the 38mm Biogon as an example, what's

the
light fall-off in terms of F-stops?


If you consider the light at the center of the image circle to have a
magnitude of 1 (units do not matter because we will be dealing with ratios),
the number of stops of fall off at an angle Theta degrees off of the center
of the center of the image would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 of the angle Theta)

If you assume the fall off of the lens design is Cos^3 instead, the formula
would obviously be:

Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta)

For instance, the fall off at a point 30 degrees off of the center,
considering a lens with Cos^4 fall off, would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 ( 30 ))
Fall-off in Stops = 3.322 x Log(1 / (0.866^4))
Fall-off in Stops = 3.322 x Log(1 / 0.5625 )
Fall-off in Stops = 3.322 x Log(1.77777)
Fall-off in Stops = 3.322 x 0.2498
Fall-off in Stops = 0.83 stops

Guillermo




  #5  
Old July 2nd 04, 07:10 PM
jjs
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?


"f/256" wrote in message
ogers.com...

"jjs" wrote:


Care to help an innumerate? Taking the 38mm Biogon as an example, what's

the
light fall-off in terms of F-stops?


If you consider the light at the center of the image circle to have a
magnitude of 1 (units do not matter because we will be dealing with

ratios),
the number of stops of fall off at an angle Theta degrees off of the

center
of the center of the image would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 of the angle Theta)

If you assume the fall off of the lens design is Cos^3 instead, the

formula
would obviously be:

Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta)

For instance, the fall off at a point 30 degrees off of the center,
considering a lens with Cos^4 fall off, would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 ( 30 ))
Fall-off in Stops = 3.322 x Log(1 / (0.866^4))
Fall-off in Stops = 3.322 x Log(1 / 0.5625 )
Fall-off in Stops = 3.322 x Log(1.77777)
Fall-off in Stops = 3.322 x 0.2498
Fall-off in Stops = 0.83 stops


Very helpful, Guillermo. I'm a bit closer to understanding.

So, finding the Cos power of the lens-design remains problematic. Here's the
lens I'm working with:
http://course1.winona.edu/jstafford/...s1/index2.html There is very
little light fall of even in the corners. In fact there is so little I can
hardly find any. Note the construction of the lens: it covers 4x5" (actually
more than 5x5") and the rear lens is 4.5" in diameter.


  #6  
Old July 2nd 04, 08:06 PM
f/256
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?



So, finding the Cos power of the lens-design remains problematic. Here's

the
lens I'm working with:
http://course1.winona.edu/jstafford/...s1/index2.html There is very
little light fall of even in the corners. In fact there is so little I can
hardly find any. Note the construction of the lens: it covers 4x5"

(actually
more than 5x5") and the rear lens is 4.5" in diameter.


That's a big piece of glass!

As Richard said, and who am I to argue with him!, designs give you not
better than Cos^3, that lens 3" FL give you and angle of 46 degrees at the
corner of a 4x5 negative, still 1.5 stops according to Cos^3 fall off, so it
may not look like there is any fall off but likely there is.

Here is an explanation to the Cosine^4 and how to find the fstops of fall
off, in case it is helpful for you to understand what's going on:

The way Cosine law works is like this: because a point off the center is
farther away by 1/Cosine of the angle, light at that point is dimmer than at
the center, because the InverseSquareLaw light is dimmer by 1/Cos^2 at that
point (that account for 2 times Cos in the cos^4 law), because light falls
on an angle on the film plane, it covers an area 1/cos bigger that at the
center of the film, more coverage with the same light means dimmer light per
unit of area (another Cos in the Cos^4 law) and finally because the aperture
is no seen as a circle but as an oval that has Cosine of the angle less
area, it allows less light thru by a factor of -you guessed it!- cosine of
the angle (yet another Cosine in the cos^4 law), the whole thing combines to
Cosine^4.

If we assume a light intensity of 1 at the center, the fall off 60 degrees
off of the center would be Cos^4 (60 degrees) = 0.0625, in other words if
you have 100 units of light at the center, at the corners you would have
only 6.25 units of light !! How do you calculate the numbers of stops
between the 6.25 and 100 ? You double 6.25 until you get 100, like this:

6.25 x 2 = 12.5 one stop
12.5 x 2 = 25 two stops
25 x 2 = 50 three stops
50 x 4 = 100 four stops

There are 4 stops fall off with respect to the center at a point 60 degrees
from the center.

Guillermo



  #7  
Old July 2nd 04, 07:28 PM
f/256
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?

Here is an example based on the Biogon 38mm:

Assuming it has a fall off as good as Cos^3

The size of a 6x6 negative is actually 55mmx55mm, its diagonal is 77.77mm
and there is arcTan(77.77/(2x38)) degrees from the center to the corners of
the film, that is 45.66 degrees or a total angle of view of 91.32 degrees
from corner to corner.

The fall off at the corners would be:

Theta = 45.66
Cos(45.66) = 0.6989

Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta)

Fall-off in Stops = 3.322 x Log(1 / (0.6989)^3)
Fall-off in Stops = 3.322 x Log(1 / 0.3414)
Fall-off in Stops = 3.322 x Log(2.929)
Fall-off in Stops = 3.322 x 0.46672

Fall-off in Stops = 1.55

There are ONE and a HALF stops of fall off at the corners of the film with
respect to the center of the film!!

Guillermo






  #8  
Old July 2nd 04, 09:02 PM
jjs
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?


"f/256" wrote in message
. rogers.com...
Here is an example based on the Biogon 38mm:

Assuming it has a fall off as good as Cos^3

[...]
There are ONE and a HALF stops of fall off at the corners of the film with
respect to the center of the film!!


Funny, but I don't find the 38mm Biogon that bad in real life.


  #9  
Old July 3rd 04, 07:40 AM
Stacey
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?

Richard Knoppow wrote:

One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. A non-orthograpic lens, like a fish-eye
lens, can have better illumination because the light is more
concentrated near the margins.


What about retro focus wide angle lenses? While not common on LF cameras,
how does that design compare to "normal" wide angle lenses as far as light
fall off? I have noticed almost no light fall off on the 30mm medformat
fisheye and even the regular wide angle SLR lenses don't seem to be as
affected by fall off as some of the wide LF lenses I've used.
--

Stacey
  #10  
Old July 3rd 04, 02:12 PM
brian
external usenet poster
 
Posts: n/a
Default Image circle versus stopping down?

Stacey wrote in message ...
Richard Knoppow wrote:

One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. A non-orthograpic lens, like a fish-eye
lens, can have better illumination because the light is more
concentrated near the margins.


What about retro focus wide angle lenses? While not common on LF cameras,
how does that design compare to "normal" wide angle lenses as far as light
fall off? I have noticed almost no light fall off on the 30mm medformat
fisheye and even the regular wide angle SLR lenses don't seem to be as
affected by fall off as some of the wide LF lenses I've used.


Extreme examples of reversed telephotos can have zero falloff *and*
zero distortion. The problem is that they are really really big. Way
too big for large format. See the simple two-element example I posted
above. Systems like this are actually used for high performance DMD
and LCD projections systems which require perfect telecentricity,
extremely large working distance, and zero distortion. Zero falloff
means cos^0. Actually, its possible for the corner illumination to be
slightly *greater* than the center illumination.

Brian
www.caldwellphotographic.com
 




Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is Off
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
[SI] XXXI Critique street shooter 35mm Photo Equipment 18 July 5th 04 04:04 PM
[SI] I miss... The Dave© 35mm Photo Equipment 55 July 5th 04 03:45 PM
difficulty drum scanning negatives Jytzel Film & Labs 51 April 10th 04 08:56 PM
IMAGE MANAGEMENT AND CREATIVITY HAS NEVER BEEN EASIER so get it for just £2 Amethyst Film & Labs 0 December 14th 03 02:28 PM


All times are GMT +1. The time now is 07:10 AM.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.
Copyright ©2004-2022 PhotoBanter.com.
The comments are property of their posters.