Front tilt loses middle

wrote:
I don't know the math, but on the ground glass, the middle always goes
out after placing foreground and horizon on the plane that brings both
into sharp focus.

I must say I've never noticed this with any of my lenses. If I choose
the tilt to bring two points, one near and the other far, in focus, then
everything on the line connecting them appears to be in focus. Also, it
is important to realize that you can't ever really see just one plane
when you look at the gg. You are going to see the DOF wedge for the
focusing aperture.

Let me try to make this concrete with an example. Suppose you tilt 5
degrees using a 210 mm lens at f/5.6 to focus. Then the hinge line
will be about 2.4 meters below the lens along a line parallel to the
back. The DOF region starts there and opens up linearly as you move
away from the lens. Using a coc of 0.1 mm, a conservative choice, the
hyperfocal distance is about 75 meters. At that distance, a
cross-section parallel to the back has extent about twice the hinge
distance, or about 4.8 meters. At 1500 meters, which should be
effectively infinity, the cross-section should have extent 1500/75 = 20
times that or almost 100 meters. You can vary the position of that
wedge by focusing, i.e., by changing the distance between the
standards,, but anything outside it will appear to be out of focus on
the gg. However, if you stop down, you decrease the hyperfocal
distance, so the 4.8 meter cross section is pulled closer to the camera.
Thus at f/22, the hyperfocal distance would be at about 5.6/22 = 0.25
times 75 or about 19 meters.

If all the points under consideration are inside the wedge for the
taking aperture, and some in the middle are out of focus, then something
else is going on. It could be field curvature, as Roger suggested, but
as I said, I've never seen anything like that with my lenses.

I think I overlooked the loss of sharpness to the
middle because the image was indistinct and hard to see. Normally I
would have refocused favoring the middle and expecting the small f/stop
to bring it all together. I probably caused more focus loss by tilting
than I would have if I didn't.

What surprises me more is that these distances should have had any
focal variation at all
as they all should have been at infinity. In fact that seems so absurd,
I doubt it can be addressed.

Maybe the standards are not perfectly square, and the ocean scene
demonstrated it.

That seems to me to be more plausible than field curvature as an
explanation for your problem. If the standards are not parallel, that
induces a slight swing. Combined with the tilt, that results in a plane
of exact focus that also tilts from left to right. If your middle
ground points are not in line with your near and far points, then this
left to right tilt could place them outside the DOF wedge. I wouldn't
generally expect this sort of thing to produce a visible effect in 4 x 5
with anything but wide angle lenses, say 90 mm or less.


The camera itself, before any lens tilt, is looking down at the plane
of the ocean from lets say a 30 degree angle. Yet horizon, and even the
middle ground, required a slight difference in focus. In that case just
a little look see on the ground glass would be all that was needed to
reach a perfect squaring of the standards, and all should have focused
at infinity.

I should have tweeked to the right instead of tweeking to the left.



Leonard Evens wrote:
Roger N. Clark (change username to rnclark) wrote:
wrote:

This has been a problem for me in other situations where the foreground
is a few feet or even inches away, however in the seascape shots the
foreground was probably around 300 to 500 yards away.

I was surprised to find any difference in focus given that all areas of
the scene were, I would think, at infinity. There was a few seconds
there where I must have looked like a dog whose bowl was moved. What ?
I have to tilt to focus this??? Well alright....

So I checked to get a sharp foreground and sharp horizon line, thought
the middle was sharp, stepped down to my sharpest f/22, and took a
photograph with a middle section that had about a half inch on a 5x7
neg modestly fuzzed out. With all due respect to the principles of
planes, I'm confused by this.
For a flat plane (e.g. the ocean), the best focus from top
to bottom in the camera's focal plane is not actually a plane
(or a titled plane). The use of tilt in a view camera only
gives an approximation of the focal plane.
The reason is as follows:

Simple lens equation, where f = lens focal length
s= distance to subject, b is distance from the principle
plane of the lens to to focal point behind the lens:

1/b = 1/f + 1/s
Roger,

I have no idea what you are talking about. It is undoubtedly true that
real lenses exhibit field curvature, even without a tilt. But that
doesn't follow from the lens equation. The lens equation applies to an
ideal lens. In that case, the relation between a subject point and the
corresponding image point is

1/u + 1/v = 1/f

where u is the perpendicular distance of the subject point to the lens
plane, v is the perpendicular distance of the image point to the lens
plane, and f is the focal length. You can think of this as a
transformation which maps points in the subject space to points in the
image space. It may seem strange, but that transformation takes planes
into planes, not into curved surfaces. So, if you consider a plane like
the (idealized) surface of the ocean, those subject points create image
points which all lie in a plane, presumably the film plane if you have
positioned everything properly. Field curvature arises because real
lenses are not ideal perfect lenses, not because of the lens equation.
How much there is and how it is distributed depends on the specific
design of the lens and which compromises where made to minimize lens
aberrations, including field curvature.


Because of the inverse relationship, increasing distance
in the scene translates to a curved focal plane that is
tilted. The view camera solves the major problem of the
tilted focal plane, but the curvature of best focus
can only be compensated by using small enough aperture to
increase apparent depth of field. Then add in a 3-dimensional
scene, and the problem just gets worse.

Soft middles are a common problem in view camera images.
That may be true, and I don't claim in can't ever happen because of
field curvature. But usually it is because of the reasons we have been
discussing. Namely, in the example under consideration, the problem
appears to be that the subject points don't in fact lie in a plane to
start with. So the only way to get them in adequate focus is to place
the plane of exact focus appropriately and then stop down sufficiently.
One would do this treating the lens as a perfect lens and then deal
with field curvature as a refinement. Your method may be helpful in
doing that.

Here is what I do to mitigate this problem:
I try and find a local "hyperfocal" distance for the
foreground area and another one for the background area.
I do this with the lens at wide aperture, and experiment
by closing the aperture. If the following illustrates
the focal plane (vertical plane), the film is given by "|"
and the best focus is given by ".":

. |
. |
. |
.|
.| x1
|.
|.
| .
| .
| .
| .
| .
| .
| .
| .
|.
|.
.| x2
.|
. |
. |
. |

The X1 and X2 are the two points in best focus using the
Scheimpflug principle. The actual focal pane has a slight
curve. I set x1 and x2 inside the min and max distances to
minimize depth-of-field requirements. In the field, I use an
8x loupe, and examine the image as I close down the aperture,
examining the entire scene until I reach the desired sharpness.

Roger
Images at: http://www.clarkvision.com

My understanding of DOF follows what Roger is saying. Since few lenses
have flat field focus planes, the DOF isn't a line on both sides of the
focus plane but two shallow curves, concaved to the focus plane and the
narrowest at the middle, wider at the bottom and wider still at the
top. You can get information at:

http://www.trenholm.org/hmmerk/index.html
http://www.trenholm.org/hmmerk/HMArtls.html
http://www.bobwheeler.com/photo/ViewCam.pdf (PDF on the geometry of
View Cameras)

If you want you can calculate the tilt using the Scheimpflug rule. It
takes an angle indicator and calculator, and about 1-2 minutes. If you
do this exercise, you'll discover tilt ranges from 3-5 degrees, with an
maximum of about 8 degrees, so if you see you're tilting a lot, you've
gone too far.

Agreed, but it's a pretty bad lens these days that doesn't focus a very flat
field.

"wsrphoto" wrote in message
oups.com...
My understanding of DOF follows what Roger is saying. Since few lenses
have flat field focus planes, the DOF isn't a line on both sides of the
focus plane but two shallow curves, concaved to the focus plane and the
narrowest at the middle, wider at the bottom and wider still at the
top. You can get information at:

http://www.trenholm.org/hmmerk/index.html
http://www.trenholm.org/hmmerk/HMArtls.html
http://www.bobwheeler.com/photo/ViewCam.pdf (PDF on the geometry of
View Cameras)

If you want you can calculate the tilt using the Scheimpflug rule. It
takes an angle indicator and calculator, and about 1-2 minutes. If you
do this exercise, you'll discover tilt ranges from 3-5 degrees, with an
maximum of about 8 degrees, so if you see you're tilting a lot, you've
gone too far.

Leonard Evens wrote:
Roger N. Clark (change username to rnclark) wrote:

wrote:

This has been a problem for me in other situations where the foreground
is a few feet or even inches away, however in the seascape shots the
foreground was probably around 300 to 500 yards away.

I was surprised to find any difference in focus given that all areas of
the scene were, I would think, at infinity. There was a few seconds
there where I must have looked like a dog whose bowl was moved. What ?
I have to tilt to focus this??? Well alright....

So I checked to get a sharp foreground and sharp horizon line, thought
the middle was sharp, stepped down to my sharpest f/22, and took a
photograph with a middle section that had about a half inch on a 5x7
neg modestly fuzzed out. With all due respect to the principles of
planes, I'm confused by this.


For a flat plane (e.g. the ocean), the best focus from top
to bottom in the camera's focal plane is not actually a plane
(or a titled plane). The use of tilt in a view camera only
gives an approximation of the focal plane.
The reason is as follows:

Simple lens equation, where f = lens focal length
s= distance to subject, b is distance from the principle
plane of the lens to to focal point behind the lens:

1/b = 1/f + 1/s


Roger,

I have no idea what you are talking about. It is undoubtedly true that
real lenses exhibit field curvature, even without a tilt. But that
doesn't follow from the lens equation. The lens equation applies to an
ideal lens. In that case, the relation between a subject point and the
corresponding image point is

1/u + 1/v = 1/f

Leonard,

My internet connection has been down.

A correction: I reversed the b and f in my equation; your
equation is correct, and with b and f in my equation reversed,
then our equations are the same.

But the effect I was talking about has nothing to do with field
curvature, and only to do with the above equation.

I wrote an excel spreadsheet to compute the curvature from
the simple lens equation. A plane in object space does not
translate to a plane in the focal plane of the lens unless the
plane is perpendicular to the optical axis.

Imagine the following: a view camera placed 1 meter above
the ground. The ground is flat (e.g. earth's surface, ignore
curvature of the earth). Say the camera views 20 degrees
about the optical axis (40 degrees total field of view).
Point the camera down at a 45 degree angle with a 100 mm
focal length lens. The focal points are illustrated in the
following graph:

http://www.clarkvision.com/tmp/view....l.plane.v1.gif

In the graph, (0, 0) represents the optical axis where the
ground focuses at 107.609 mm behind the lens. Negative
values on the x-axis are looking down closer to the camera,
and positive values are looking up closer to the horizon.

The trend in the arc in the graph represents the tilt needed
for best focus. The best focus of a flat
film plane would be a line intersecting this curve at two
points. But, as you can see, minimizing distance from
the arc to a straight line would result in a focus error
in the middle. That focus error is what causes the
soft focus in many view camera images.

Roger

where u is the perpendicular distance of the subject point to the lens
plane, v is the perpendicular distance of the image point to the lens
plane, and f is the focal length. You can think of this as a
transformation which maps points in the subject space to points in the
image space. It may seem strange, but that transformation takes planes
into planes, not into curved surfaces. So, if you consider a plane like
the (idealized) surface of the ocean, those subject points create image
points which all lie in a plane, presumably the film plane if you have
positioned everything properly. Field curvature arises because real
lenses are not ideal perfect lenses, not because of the lens equation.
How much there is and how it is distributed depends on the specific
design of the lens and which compromises where made to minimize lens
aberrations, including field curvature.



Because of the inverse relationship, increasing distance
in the scene translates to a curved focal plane that is
tilted. The view camera solves the major problem of the
tilted focal plane, but the curvature of best focus
can only be compensated by using small enough aperture to
increase apparent depth of field. Then add in a 3-dimensional
scene, and the problem just gets worse.

Soft middles are a common problem in view camera images.


That may be true, and I don't claim in can't ever happen because of
field curvature. But usually it is because of the reasons we have been
discussing. Namely, in the example under consideration, the problem
appears to be that the subject points don't in fact lie in a plane to
start with. So the only way to get them in adequate focus is to place
the plane of exact focus appropriately and then stop down sufficiently.
One would do this treating the lens as a perfect lens and then deal
with field curvature as a refinement. Your method may be helpful in
doing that.


Here is what I do to mitigate this problem:
I try and find a local "hyperfocal" distance for the
foreground area and another one for the background area.
I do this with the lens at wide aperture, and experiment
by closing the aperture. If the following illustrates
the focal plane (vertical plane), the film is given by "|"
and the best focus is given by ".":

. |
. |
. |
.|
.| x1
|.
|.
| .
| .
| .
| .
| .
| .
| .
| .
|.
|.
.| x2
.|
. |
. |
. |

The X1 and X2 are the two points in best focus using the
Scheimpflug principle. The actual focal pane has a slight
curve. I set x1 and x2 inside the min and max distances to
minimize depth-of-field requirements. In the field, I use an
8x loupe, and examine the image as I close down the aperture,
examining the entire scene until I reach the desired sharpness.

Roger
Images at: http://www.clarkvision.com

wsrphoto wrote:
My understanding of DOF follows what Roger is saying. Since few lenses
have flat field focus planes, the DOF isn't a line

Perhaps you mean a plane. A lot of people get confused because the
usual diagrams are cross-sectional and planes are represented by lines.

on both sides of the
focus plane but two shallow curves, concaved to the focus plane and the
narrowest at the middle, wider at the bottom and wider still at the
top.

I am not an expert in this, but it was my impression that field
curvature depended on the design of the lens and that there was no
simple rule that said it worked the same way for all lenses.

You can get information at:

http://www.trenholm.org/hmmerk/index.html
http://www.trenholm.org/hmmerk/HMArtls.html
http://www.bobwheeler.com/photo/ViewCam.pdf (PDF on the geometry of
View Cameras)

I am familiar with these web pages, the last of which I have gone
through in detail on several occasions. I also tried searching all
three just now, and I couldn't find any references of the kind you
describe about field curvature. If you can direct me to the appropriate
sections I would appreciate it.

Wheeler does discuss a different issue, namely the shape of the bounding
surfaces for the DOF region after a tilt. We say that these are planes
(actually half planes) emanating from the hinge line. But because of
the tilt, circles of confusion get changed to ellipses of confusion, and
that leads to these bounding surfaces actually being curved surfaces
rather than planes. Wheeler graphs this surface in an example along the
center line and displays the curve. I've investigated this question in
great detail and studied the bounding surface in three dimensions, not
just along the center line. Wheeler found for the center line that in
most practical photography the departure from a plane was negligible.
My analysis confirmed that for the entire surface. I found that it would
only be a significant factor for extreme wide angle lenses and then only
for close-up photography.

The analysis described in the above paragraph is only for an ideal lens.
Actual field curvature, which is a lens aberration, has nothing to do
with it. I am sure that actual field curvature may be a factor in
practical situation and probably more important that the above effect.
but, as I said, I believe it varies from lens to lens. I haven't
seen any formal analysis of how field curvature might affect
calculations about the location of the exact "plane" of focus or DOF
about it. If you know of any such analysis, please give me a reference.

A good source of information about some of these matters is Jeff
Conrad's article "Depth of Field in Depth" at www.lfphoto.info. Let me
quote what he says. (Note that field curvature is a lens aberration.)

"Lens aberrations are ignored. Real lenses have optical defects that
cause the image to be other than predicted by Gaussian optics, and these
aberrations are the primary cause of image degradation at large
apertures. Including the effects of aberrations is nearly impossible,
because doing so requires knowledge of the specific lens design.
Moreover, in well-designed lenses, most aberrations are well corrected,
and at least near the optical axis, often are almost negligible when the
lens is stopped down 2–3 steps from maximum aperture. At this point, a
lens often is described, if somewhat optimistically, as diffraction
limited. Because lenses usually are stopped down at least to this point
when DoF is of interest, treating the lens as diffraction-limited is
reasonable. Some residual aberrations are present even in well-designed
lenses, and their effects usually increase with distance from the lens
axis, so the sharpness realized in practice will be somewhat less than
suggested by this discussion."


If you want you can calculate the tilt using the Scheimpflug rule. It
takes an angle indicator and calculator, and about 1-2 minutes. If you
do this exercise, you'll discover tilt ranges from 3-5 degrees, with an
maximum of about 8 degrees, so if you see you're tilting a lot, you've
gone too far.

I agree that in most practical situation, one seldom uses a tilt of more
than about 8 degrees. But certainly there are situations where one
might want to do that. I've certainly done so on more than one
occasion. What is important is to have some understanding of how the
tilt angle is related to the placement of the subject plane. One simple
rule of thumb would be the following. Try to visualize how far below
the lens the subject plane passes. Estimate its distance to the lens in
mm, and divide the by the focal length by that number. Finally,
multiply by 60. That will give you an estimate in degrees of the
necessary tilt. For example, suppose the camera is about 1.5 meters
above the ground and you want the subject plane to pass at ground level
below that. Suppose you are using a 150 mm lens. 1.5 meters is 1500
mm, so you want a tilt of about 150/1500 x 60 or about 6 degrees.
Notice however that if you were using a 300 mm lens, then you would want
a tilt of about 300/1500 x 60 or about 12 degrees. (This calculation
works for the back level. If the back is tilted, you have to look in a
plane through the lens parallel to the back and measure in that plane.)

Roger N. Clark (change username to rnclark) wrote:
Leonard Evens wrote:
Roger N. Clark (change username to rnclark) wrote:

wrote:

This has been a problem for me in other situations where the foreground
is a few feet or even inches away, however in the seascape shots the
foreground was probably around 300 to 500 yards away.

I was surprised to find any difference in focus given that all areas of
the scene were, I would think, at infinity. There was a few seconds
there where I must have looked like a dog whose bowl was moved. What ?
I have to tilt to focus this??? Well alright....

So I checked to get a sharp foreground and sharp horizon line, thought
the middle was sharp, stepped down to my sharpest f/22, and took a
photograph with a middle section that had about a half inch on a 5x7
neg modestly fuzzed out. With all due respect to the principles of
planes, I'm confused by this.


For a flat plane (e.g. the ocean), the best focus from top
to bottom in the camera's focal plane is not actually a plane
(or a titled plane). The use of tilt in a view camera only
gives an approximation of the focal plane.
The reason is as follows:

Simple lens equation, where f = lens focal length
s= distance to subject, b is distance from the principle
plane of the lens to to focal point behind the lens:

1/b = 1/f + 1/s


Roger,

I have no idea what you are talking about. It is undoubtedly true
that real lenses exhibit field curvature, even without a tilt. But
that doesn't follow from the lens equation. The lens equation
applies to an ideal lens. In that case, the relation between a
subject point and the corresponding image point is

1/u + 1/v = 1/f

Leonard,

My internet connection has been down.

A correction: I reversed the b and f in my equation; your
equation is correct, and with b and f in my equation reversed,
then our equations are the same.

But the effect I was talking about has nothing to do with field
curvature, and only to do with the above equation.

I wrote an excel spreadsheet to compute the curvature from
the simple lens equation. A plane in object space does not
translate to a plane in the focal plane of the lens unless the
plane is perpendicular to the optical axis.

I'm sorry Roger, but that is just not so, whatever you did with your
spreadsheet. I really believe I understand the mathematics and geometry
behind this. The transformation for an ideal lens from subject space to
image space is what we mathematicians call a projective transformation.
It carries planes into planes.

You should read Bob Wheeler's notes on the subject. He shows how
Scheimpflug's principle follows from a theorem in projective geometry
called Desargue's theorem. He then goes on to show how to derive the
lens equation from Desargues equation. It is conceivable that I am
being stupid here, but I doubt that both I and Bob Wheeler are.


Imagine the following: a view camera placed 1 meter above
the ground. The ground is flat (e.g. earth's surface, ignore
curvature of the earth). Say the camera views 20 degrees
about the optical axis (40 degrees total field of view).
Point the camera down at a 45 degree angle with a 100 mm
focal length lens. The focal points are illustrated in the
following graph:

http://www.clarkvision.com/tmp/view....l.plane.v1.gif

In the graph, (0, 0) represents the optical axis where the
ground focuses at 107.609 mm behind the lens. Negative
values on the x-axis are looking down closer to the camera,
and positive values are looking up closer to the horizon.

The trend in the arc in the graph represents the tilt needed
for best focus. The best focus of a flat
film plane would be a line intersecting this curve at two
points. But, as you can see, minimizing distance from
the arc to a straight line would result in a focus error
in the middle. That focus error is what causes the
soft focus in many view camera images.

Roger

where u is the perpendicular distance of the subject point to the
lens plane, v is the perpendicular distance of the image point to the
lens plane, and f is the focal length. You can think of this as a
transformation which maps points in the subject space to points in the
image space. It may seem strange, but that transformation takes
planes into planes, not into curved surfaces. So, if you consider a
plane like the (idealized) surface of the ocean, those subject points
create image points which all lie in a plane, presumably the film
plane if you have positioned everything properly. Field curvature
arises because real lenses are not ideal perfect lenses, not because
of the lens equation. How much there is and how it is distributed
depends on the specific design of the lens and which compromises where
made to minimize lens aberrations, including field curvature.



Because of the inverse relationship, increasing distance
in the scene translates to a curved focal plane that is
tilted. The view camera solves the major problem of the
tilted focal plane, but the curvature of best focus
can only be compensated by using small enough aperture to
increase apparent depth of field. Then add in a 3-dimensional
scene, and the problem just gets worse.

Soft middles are a common problem in view camera images.


That may be true, and I don't claim in can't ever happen because of
field curvature. But usually it is because of the reasons we have
been discussing. Namely, in the example under consideration, the
problem appears to be that the subject points don't in fact lie in a
plane to start with. So the only way to get them in adequate focus is
to place the plane of exact focus appropriately and then stop down
sufficiently. One would do this treating the lens as a perfect lens
and then deal with field curvature as a refinement. Your method may
be helpful in doing that.


Here is what I do to mitigate this problem:
I try and find a local "hyperfocal" distance for the
foreground area and another one for the background area.
I do this with the lens at wide aperture, and experiment
by closing the aperture. If the following illustrates
the focal plane (vertical plane), the film is given by "|"
and the best focus is given by ".":

. |
. |
. |
.|
.| x1
|.
|.
| .
| .
| .
| .
| .
| .
| .
| .
|.
|.
.| x2
.|
. |
. |
. |

The X1 and X2 are the two points in best focus using the
Scheimpflug principle. The actual focal pane has a slight
curve. I set x1 and x2 inside the min and max distances to
minimize depth-of-field requirements. In the field, I use an
8x loupe, and examine the image as I close down the aperture,
examining the entire scene until I reach the desired sharpness.

Roger
Images at: http://www.clarkvision.com

babelfish wrote:
Agreed, but it's a pretty bad lens these days that doesn't focus a very flat
field.

I agree with you, but way back in the 60s, I discovered that the lens in
my Rolleiflex had detectable field curvature. If I remember correctly,
when focusing at 6 feet from the lens, the field in the corners was a
few inches in front of the theoretical image plane. But you could only
see this under extreme magnification. So, while in most practical
situations field curvature may not be that serious an issue, it is
certainly possible for obsessive types like me to find it. :-;


"wsrphoto" wrote in message
oups.com...
My understanding of DOF follows what Roger is saying. Since few lenses
have flat field focus planes, the DOF isn't a line on both sides of the
focus plane but two shallow curves, concaved to the focus plane and the
narrowest at the middle, wider at the bottom and wider still at the
top. You can get information at:

http://www.trenholm.org/hmmerk/index.html
http://www.trenholm.org/hmmerk/HMArtls.html
http://www.bobwheeler.com/photo/ViewCam.pdf (PDF on the geometry of
View Cameras)

If you want you can calculate the tilt using the Scheimpflug rule. It
takes an angle indicator and calculator, and about 1-2 minutes. If you
do this exercise, you'll discover tilt ranges from 3-5 degrees, with an
maximum of about 8 degrees, so if you see you're tilting a lot, you've
gone too far.

Leonard Evens wrote:
Roger N. Clark (change username to rnclark) wrote:

Leonard Evens wrote:

Roger N. Clark (change username to rnclark) wrote:

wrote:

This has been a problem for me in other situations where the
foreground
is a few feet or even inches away, however in the seascape shots the
foreground was probably around 300 to 500 yards away.

I was surprised to find any difference in focus given that all
areas of
the scene were, I would think, at infinity. There was a few seconds
there where I must have looked like a dog whose bowl was moved. What ?
I have to tilt to focus this??? Well alright....

So I checked to get a sharp foreground and sharp horizon line, thought
the middle was sharp, stepped down to my sharpest f/22, and took a
photograph with a middle section that had about a half inch on a 5x7
neg modestly fuzzed out. With all due respect to the principles of
planes, I'm confused by this.



For a flat plane (e.g. the ocean), the best focus from top
to bottom in the camera's focal plane is not actually a plane
(or a titled plane). The use of tilt in a view camera only
gives an approximation of the focal plane.
The reason is as follows:

Simple lens equation, where f = lens focal length
s= distance to subject, b is distance from the principle
plane of the lens to to focal point behind the lens:

1/b = 1/f + 1/s



Roger,

I have no idea what you are talking about. It is undoubtedly true
that real lenses exhibit field curvature, even without a tilt. But
that doesn't follow from the lens equation. The lens equation
applies to an ideal lens. In that case, the relation between a
subject point and the corresponding image point is

1/u + 1/v = 1/f


Leonard,

My internet connection has been down.

A correction: I reversed the b and f in my equation; your
equation is correct, and with b and f in my equation reversed,
then our equations are the same.

But the effect I was talking about has nothing to do with field
curvature, and only to do with the above equation.

I wrote an excel spreadsheet to compute the curvature from
the simple lens equation. A plane in object space does not
translate to a plane in the focal plane of the lens unless the
plane is perpendicular to the optical axis.


I'm sorry Roger, but that is just not so, whatever you did with your
spreadsheet. I really believe I understand the mathematics and geometry
behind this. The transformation for an ideal lens from subject space to
image space is what we mathematicians call a projective transformation.
It carries planes into planes.

Hmmm. Let's step back and look at the lens equation,
1/u + 1/v = 1/f (your symbols). Assume an object at infinity
approaches the lens on the optical axis with linear velocity.
Describe the focus position of that object with time.
Is it linear? No. As the object approaches the lens,
the focus position accelerates. A plot of focus
position versus time is not a straight line.

It is the same with different objects in the field of view
of the camera: they come to different focus positions
behind the lens, and those distances are not linear
with distance to the subject. If they were, the
hyperfocal distance would always be half way from close
to far. But closer objects have an inverse relationship
increasingly greater focus distance behind the lens.


You should read Bob Wheeler's notes on the subject. He shows how
Scheimpflug's principle follows from a theorem in projective geometry
called Desargue's theorem. He then goes on to show how to derive the
lens equation from Desargues equation. It is conceivable that I am
being stupid here, but I doubt that both I and Bob Wheeler are.

Bob Wheeler's notes:
http://www.bobwheeler.com/photo/ViewCam.pdf

I see nothing there that contradicts what I've said. Most
of the discussion is with regard to near and far
focus points and not the shape over the whole plane
(unless I missed a section).

It seems everyone could easily check this by experiment
with their view cameras: set it up looking at a flat
plane (e.g. table top or floor) and put in some tilts,
getting close to the subject plane to exaggerate the effect.
Can you get the entire plane in perfect focus wide open?

Roger


Imagine the following: a view camera placed 1 meter above
the ground. The ground is flat (e.g. earth's surface, ignore
curvature of the earth). Say the camera views 20 degrees
about the optical axis (40 degrees total field of view).
Point the camera down at a 45 degree angle with a 100 mm
focal length lens. The focal points are illustrated in the
following graph:

http://www.clarkvision.com/tmp/view....l.plane.v1.gif

In the graph, (0, 0) represents the optical axis where the
ground focuses at 107.609 mm behind the lens. Negative
values on the x-axis are looking down closer to the camera,
and positive values are looking up closer to the horizon.

The trend in the arc in the graph represents the tilt needed
for best focus. The best focus of a flat
film plane would be a line intersecting this curve at two
points. But, as you can see, minimizing distance from
the arc to a straight line would result in a focus error
in the middle. That focus error is what causes the
soft focus in many view camera images.

Roger

Roger N. Clark (change username to rnclark) wrote:
Leonard Evens wrote:
Roger N. Clark (change username to rnclark) wrote:

Leonard Evens wrote:

Roger N. Clark (change username to rnclark) wrote:

wrote:

This has been a problem for me in other situations where the
foreground
is a few feet or even inches away, however in the seascape shots the
foreground was probably around 300 to 500 yards away.

I was surprised to find any difference in focus given that all
areas of
the scene were, I would think, at infinity. There was a few seconds
there where I must have looked like a dog whose bowl was moved.
What ?
I have to tilt to focus this??? Well alright....

So I checked to get a sharp foreground and sharp horizon line,
thought
the middle was sharp, stepped down to my sharpest f/22, and took a
photograph with a middle section that had about a half inch on a 5x7
neg modestly fuzzed out. With all due respect to the principles of
planes, I'm confused by this.



For a flat plane (e.g. the ocean), the best focus from top
to bottom in the camera's focal plane is not actually a plane
(or a titled plane). The use of tilt in a view camera only
gives an approximation of the focal plane.
The reason is as follows:

Simple lens equation, where f = lens focal length
s= distance to subject, b is distance from the principle
plane of the lens to to focal point behind the lens:

1/b = 1/f + 1/s



Roger,

I have no idea what you are talking about. It is undoubtedly true
that real lenses exhibit field curvature, even without a tilt. But
that doesn't follow from the lens equation. The lens equation
applies to an ideal lens. In that case, the relation between a
subject point and the corresponding image point is

1/u + 1/v = 1/f


Leonard,

My internet connection has been down.

A correction: I reversed the b and f in my equation; your
equation is correct, and with b and f in my equation reversed,
then our equations are the same.

But the effect I was talking about has nothing to do with field
curvature, and only to do with the above equation.

I wrote an excel spreadsheet to compute the curvature from
the simple lens equation. A plane in object space does not
translate to a plane in the focal plane of the lens unless the
plane is perpendicular to the optical axis.


I'm sorry Roger, but that is just not so, whatever you did with your
spreadsheet. I really believe I understand the mathematics and
geometry behind this. The transformation for an ideal lens from
subject space to image space is what we mathematicians call a
projective transformation. It carries planes into planes.

Hmmm. Let's step back and look at the lens equation,
1/u + 1/v = 1/f (your symbols). Assume an object at infinity
approaches the lens on the optical axis with linear velocity.
Describe the focus position of that object with time.
Is it linear? No. As the object approaches the lens,
the focus position accelerates. A plot of focus
position versus time is not a straight line.

That is correct so far.


It is the same with different objects in the field of view
of the camera: they come to different focus positions
behind the lens, and those distances are not linear
with distance to the subject.

That is also correct, but I don't see what difference it makes.
Remember what u and v are. They are respectively the perpendicular
distance of a subject point to the lens plane and the corresponding
image point to the lens plane. So, for example, all subject points in
a plane perpendicular to the lens axis come to focus in the same image
plane perpendicular to the lens axis. All such subject points share the
same value of u and all the corresponding image points points in the
corresponding image plane share the same value of v.

If they were, the
hyperfocal distance would always be half way from close
to far. But closer objects have an inverse relationship
increasingly greater focus distance behind the lens.

I think you may be confusing the distance from the point to the lens
with the perpendicular distance from the point to the lens plane. The
lens equation uses the perpendicular distances to the lens plane.



You should read Bob Wheeler's notes on the subject. He shows how
Scheimpflug's principle follows from a theorem in projective geometry
called Desargue's theorem. He then goes on to show how to derive the
lens equation from Desargues equation. It is conceivable that I am
being stupid here, but I doubt that both I and Bob Wheeler are.

Bob Wheeler's notes:
http://www.bobwheeler.com/photo/ViewCam.pdf

I see nothing there that contradicts what I've said. Most
of the discussion is with regard to near and far
focus points and not the shape over the whole plane
(unless I missed a section).

The very first section derives the fact that the subject PLANE, the lens
PLANE, and the image PLANE intersect in a common line. These are all
planes. In the next section, he derives the lens equation from what he
did in the first section, and it is clear from his discussion that the
quantities in that equation are the perpendicular distances from the
relevant points to the lens plane.

You can also work it back in the other direction. Start with the lens
equation. Built into its very structure is the assertion that a subject
plane perpendicular to the lens axis is carried into an image plane also
perpendicular to the lens axis. But you can now prove that for any
subject plane, not necessarily perpendicular to the lens axis, the
corresponding set of image points form a plane, and moreover, the
subject plane, the lens plane, and that image plane intersect in a line.
That is, Scheimplug's Rule may be deduced from the lens equation.


It seems everyone could easily check this by experiment
with their view cameras: set it up looking at a flat
plane (e.g. table top or floor) and put in some tilts,
getting close to the subject plane to exaggerate the effect.
Can you get the entire plane in perfect focus wide open?

Yes. That is more or less what happens, at least with my lenses.
Remember, however, that when you look at the gg, you can't every see the
exact plane of focus. That is because implicit in what you see is a
circle of confusion which depends on the aperture, the fineness of the
gg, and how strong a loupe you use. Because of that, what you will see
will involve some fuzziness, described by the DOF region. In case of a
tilt, it will be the familiar wedge hinged on the hinge line.

Of course, all this is supposed only to apply to ideal lenses. What I'm
saying is that if you see anything like this, it is not because it
follows from the geometric optics of ideal lenses. That theory is
quite clear in saying that the image of a plane is a plane, not a curved
surface. But of course, real lenses are far from ideal, and the further
out in the field you go the more they will depart from what you would
expect from an ideal lens. It has been my experience that in practical
situations, the ideal theory comes pretty close to predicting what you
see, but I suppose that depends on how closely you look. But, in any
event, to predict in detail what happens with any specific lens, you
have to know something about its design, and even then, the analysis
would be quite difficult. Nothing as simple as the lens equation will
tell you what to expect.


Roger


Imagine the following: a view camera placed 1 meter above
the ground. The ground is flat (e.g. earth's surface, ignore
curvature of the earth). Say the camera views 20 degrees
about the optical axis (40 degrees total field of view).
Point the camera down at a 45 degree angle with a 100 mm
focal length lens. The focal points are illustrated in the
following graph:

http://www.clarkvision.com/tmp/view....l.plane.v1.gif

In the graph, (0, 0) represents the optical axis where the
ground focuses at 107.609 mm behind the lens. Negative
values on the x-axis are looking down closer to the camera,
and positive values are looking up closer to the horizon.

The trend in the arc in the graph represents the tilt needed
for best focus. The best focus of a flat
film plane would be a line intersecting this curve at two
points. But, as you can see, minimizing distance from
the arc to a straight line would result in a focus error
in the middle. That focus error is what causes the
soft focus in many view camera images.

Roger

In article ,
says...
babelfish wrote:
Agreed, but it's a pretty bad lens these days that doesn't focus a very flat
field.

I agree with you, but way back in the 60s, I discovered that the lens in
my Rolleiflex had detectable field curvature. If I remember correctly,
when focusing at 6 feet from the lens, the field in the corners was a
few inches in front of the theoretical image plane. But you could only
see this under extreme magnification. So, while in most practical
situations field curvature may not be that serious an issue, it is
certainly possible for obsessive types like me to find it. :-;

Just to throw another red herring into the mix, I've had problems with
film flatness. The old Graphic roll holders I used to own were terrible
over the 6x7 field. Since 120 film is thinner that sheet film the
problem was exaggerated. Even sheet film can "pop" in humid weather and
one never knows from one shot to another how flat the film will be.

--
Robert D Feinman
Landscapes, Cityscapes and Panoramic Photographs
http://robertdfeinman.com
mail: