Viewfinder and DOF
---- HYPERFOCUSING THE TULIPS II
Mathematical Proof
Martin Tai
May 28 1996
Victoria Day long weekend (May 18-21), took a trip to Ottawa
for the famous annual Tulips Festival.
That trip to Ottawa reaps not only many nice tulip photos,
( 4 x 24 Agfa HDC 100 ) but also allow me to gain a
deeper insight into the relationship of viewfinder and
depth of field
Any one taking photo of a field of flower, will notice that
because your camera is tilting downward, and when focusing the lens,
if the near end is in focus the far end will be out of focus, and if
you make the far end sharp, then the near end will be out of focus.
what is the best point to focus ?
Traditional DOF theory stated that, for a scene with near object
at distance A and far object at distance B, one must focus at a point
X= 2*A*B /(A+B)
To do this you need to measure A, B, then a calculator to
calculate X.
You may think of using your lens and focusing screen, focus
on the near point, read off A, then focus on far point read of B.
Then use a calculator and get the value for X.
But unless you have a measuring tape, how are you
going to measure X ?
Some newer cameras have built in DOF function, it allows you
to meter A then B and compute X; but then, even if the lens
automatically set to X, how do you know where X is in the tulip field ?
Is there any simpler way to determine the optimal point of focus
for a tilted field ?
In a previous article " Hyperfocusing the Tulips "
I stated that when the camera is tilted
down to cover a field ( excluding horizon which is at infinity ),
the point of of sharp focus at the spot encircled by the microprism
ring or split image wedges in the SLR viewfinder is
VERY CLOSE to the correct focusing distance as required by
traditional DOF theory.
The following is a proof that the center spot of
the viewfinder points to a spot PRECISELY required by the
classical DOF theory !
The following diagram illustrate this:
Camera tilt downward
P
\ \
\. \. \.
\. \ \ .
\. \ \
\ \ \
a \ c \ b \
\ \ \
\ \ \
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
S A C B
In the diagram, S is position of the tripod, P the camera,
A is the near end of view, B is the far end, and C is the point
encircled by the microprism ring in the SLR viewfinder ( or the
center spot of any autofocus compacts ).
PS is the height of the camera on tripod ( or handheld )
let it be H
Let c denotes the distance between the camera P and point C
Let a denotes the distance between the camera P and point A
Let b denotes the distance between the camera P and point B
As the camera is pointed towards point C, line PC is the optical
axis, and all distance must measure on this axis.
Let the a' denotes the projection of PA on axis PC
= object distance of point A
Let the b' denotes the projection of PB on axis PC
= object distance of point B
c= object distance of point C.
Let alpha = angle between PA/SA
beta = angle between PB/SA
gamma = angle between PC/SC
delta = angle between PA/PC
= angle between PB/PC
From geometry, we know that gamma = (alpha + beta)/2
delta = (alpha - beta)/2
Objective: to prove that 2* (a' * b')/(a' + b') = c
Proof:
2
2* (a' * b')/(a' + b') = ----------------------------
1/a' + 1/b'
2
= ------------------------------------------
1 1
------------ + --------------
a *cos (delta) b *cos (delta)
2 * cos (delta)
= ------------------------------------------
1 1
-------- + --------
a b
2 H * cos (delta)
= ------------------------------------------
H H
-------- + --------
a b
2 H * cos (delta)
= ------------------------------------------------
sin (alpha ) + sin (beta )
2 H * cos (delta)
= ---------------------------------------------------------
2 * sin [ (alpha +beta )/2] * cos[ (alpha -beta )/2]
2 H * cos (delta)
= ---------------------------------------------------------
2 * sin [ gamma] * cos( delta)
H
= ------------------------
sin [ gamma]
= c.
Q.E.D. !
In summary,
c = PC IS indeed exactly equals to 2* (a' * b')/(a' + b')
as required by classical DOF theory.
In effect, the center of any viewfinder is a built in
DOF computer, it points directly at the point of optimal focus !
Application:
Photographing: fields of tulip, iris, peony ........
railway tracks
highrise building from on end
Setting on banquet table...
etc etc.
copyright
Martin Tai
May 28, 1996
Not so. you are not considering f-stop here, not to mention size of
acceptable circle of confusion, so this premise is not correct, and
leads you to incorrect conclusions with your "proof".
j. albert
The selection of f stop is a different matter.
The Ilford (1939) formula is always true, regardless of f stop.
Can you provide concrete cases to show the Ilford formula is not
true ?
The Ilford (1939) formula is always true, regardless of f stop.
I dont think you are able to provide an example concrete cases to
show the Ilford formula is not true.
If you focus at a distance X, and if the hyperfocal
lenght of your lense at YOUR selected aperture, is H
and X = H/n
then DOF theory tells you that near limit A = H/(n-1)
and far limit B = H/(n+1)
H H
2 * --- * ----
2* A * B n-1 n+1 H
--------------= -------------------------- = --
(A + B) n
2* n * H
-----------------
(n-1) * (n+1)
Whick is X.
You have not a clue what I was talking about.
And I don't think you understand DOF at all.
The Ilford (1939) formula is always true, regardless of f stop.
I dont think you are able to provide one example to
show the Ilford formula is not true.
If you focus at a distance X, and if the hyperfocal
lenght of your lense at YOUR selected aperture, is H
and X = H/n
then DOF theory tells you that near limit A = H/(n-1)
and far limit B = H/(n+1)
H H
2 * --- * ----
2* A * B n-1 n+1 H
--------------= -------------------------- = --
(A + B) n
2* n * H
-----------------
(n-1) * (n+1)
Whick is X.
2) You have not a clue what I was talking about.
People who understand more about DOF, wrote me that
my message is "extremely interesting and useful"
and X = H/n
Whick is X.
*******************************
Martin, this was extremely interesting and useful! Thanks. Did you
discover this yourself? If so, congratulations.
Being of a heavy mathematics background myself, I can appreciate the
formulas behind this observation.
SLMR 2.1a
* Origin: Winnipeg PC User Group BBS (1:348/204)
********************************
hyperfocal distance H is dependent on f-stop. above you are doing the
inverse problem, choosing X based on H, which already has loaded
in it the f-stop selected, then computing A and B. that is fine.
but when you try to compute X from A and B without having selected an
f-stop (or other parameters such as H that are dependent on f-stop)
you make the false claim that choosing X this way is sufficient
for everything from A to be to be in acceptably sharp focus/
j. albert
it is the essence of the matter, actually.
> The Ilford (1939) formula is always true, regardless of f stop.
>
> Can you provide concrete cases to show the Ilford formula is not
>true ?
sure. focus at some point Y, and set the f-stop to some value N
such that DOF at a particular size of acceptable circle of confusion
runs from near point C to distant point D, where D is not infinity.
now take A = C/2 (ie A is half as far away as C) and
take B = 2*D (ie B is twice as far away as C).
now, without changing f-stop or focal length or camera position,
tell me the point between A and B where you would focus so that
everything from A to B is in focus within acceptable size of
circles of confusion?
the formula is correct. it is your interpretation and application of
it that is wrong. at the given f-stop, there is no point where you
can focus and have everything from A to B in sharp focus at the
same size of circle of confusion. if you compute X as above, I
believe it will give you the point where everything from A to
B is in as sharp a focus as possible, but not acceptable within
the desired circle of confusion. you still have to determine
the f-stop that brings everything from A to B into acceptably sharp
focus.
hence,. the conclusion of your "proof" that simply by choosing X
according to the formula, everything from A to B is in acceptably
sharp focus is not correct. and the point to focus on
isn't always the center of the viewfinder either. consider
when the camera is perfectly level, and the center of the
viewfinder is on the horizon line, and you can see that this
isn't correct
J. Albert
What he is saying is that if you want those two objects equally sharp
(i.e. have same circle of confusion) on the film, the focusing distance
has to be X given above. It is independent on f-stop or focal length.
Hugh
jal...@nyx.cs.du.edu (Joseph Albert) wrote:
>In article <8C1E170.2401...@westonia.com>,
>MARTIN TAI <marti...@westonia.com> wrote:
>>
>>
>> Viewfinder and DOF
>>
>>
>> ---- HYPERFOCUSING THE TULIPS II
>>
>> Mathematical Proof
>>
>> Martin Tai
>> May 28 1996
>>
>>
>>
>> deeper insight into the relationship of viewfinder and
>> depth of field
>>
>> Any one taking photo of a field of flower, will notice that
>>because your camera is tilting downward, and when focusing the lens,
>>if the near end is in focus the far end will be out of focus, and if
>>you make the far end sharp, then the near end will be out of focus.
>>what is the best point to focus ?
>>
>> Traditional DOF theory stated that, for a scene with near object
>>at distance A and far object at distance B, one must focus at a point
>>
>>
>> X= 2*A*B /(A+B)
>Not so. you are not considering f-stop here, not to mention size of
[ a bunch of math deleted ]
> Is there any simpler way to determine the optimal point of focus
>for a tilted field ?
>
> In a previous article " Hyperfocusing the Tulips "
> I stated that when the camera is tilted
>down to cover a field ( excluding horizon which is at infinity ),
>the point of of sharp focus at the spot encircled by the microprism
> ring or split image wedges in the SLR viewfinder is
>VERY CLOSE to the correct focusing distance as required by
> traditional DOF theory.
Very good work, Martin. The only error you made was in waffling. In
fact, it is *exactly* the correct distance. I was unaware of this
result. Thank you for a good posting.
-- David Jacobson
Something was not quite right, that I suspect was just an oversight on
Martin's part in leaving out some information. When it is said that to
bring everything form near point A to far point B into acceptably sharp
focus, one can focus at point X, computed from only A and B and
independent of aperture, something is not correct, since for wide enough
apertures, it won't be possible to get everything from A to B in sharp
enough focus at any point in between.
I suspect that what Martin meant to say was that if an f-stop is chosen
so that near point A is the hyperfocal distance, then point of focus X
can be computed from A and B using the formula given, which will be the
center of the viewfinder for the geometry of the view, but I didn't do
the calculations to see if that suspicion was correct.
J. Albert
No. What Martin was saying was that if you focus at an angle to a
plane (i.e. the tulips) and focus perfectly on the point that is in
the center of the image the circles of confusion from points at the
top of the image and points at the bottom of the image will be the
same size. Of course, to get those circles of confusion down to an
acceptable size you have to choose a suitable f-stop. He doesn't
address how to do that.
[snip]
>J. Albert
-- David Jacobson
You still did'nt get the point.
The post was not about hyperfocal distance.
It was about the center of the viewfinder and Ilford formula.
As for the relationship of hyperfocal distance with aperture
and circle of confusion, David Jacobson explained them clearly
in his excellent article "LENS FAQ", and I suggest you read his
article.