tvdpid
Member
I saw on google maps that the horizontal distance between the cross on the san marco and the front of the palace is ca. 115 m. I guess you stood ca. 250 m away from that cross.
"The ins and outs of focus" by Harold M. Merklinger
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tvdpid
Member
I will give the formulas a try:
(If you want all the details that take all small deviations like field curvature etc. into account --> see Merklingers book / acticles)
f = focal length
N = aperture number
d = f/N = the effective aperture diameter (e.g. 50 mm lens at N = 10 --> d = 50 mm/10 = 5 mm)
D = focus distance = distance from lens to the point (plane) which is in focus.
In focus means that a point (which has no dimensions) results in a point on your film or sensor.
Every point which is closer to or farther away from the lens will result in a disc with certain dimensions on the film or sensor.
The following formulas will give the size of those discs.
Distances from the lens to D will be indicated by X.
Distances from D to infinity (measured from lens) will be indicated by Y.
Sx = disc size according to distances X
Sy = disc size according to distances Y.
The formulas are really easy:
Sx = d x (D-X)/D
Sy = d x (Y-D)/D
If you substitude d by f/N then the formulas are:
Sx = f/N x (D-X)/D
Sy = f/N x (Y-D)/D
What can we conclude from this formulas:
If X = 0 then Sx = d. So points in front af D will never be depicted on the film or sensor greater than d.
If X (or Y) = D then Sx (or Sy) = 0 mm.
If Y = 2D then Sy = d
(From this you can conclude that the 1/3-2/3 rule = simply not true.)
For Y > 2D, Sy > d
For D < infinity and Y --> infinity: Sy --> infinity.
If D = infinity then for all X's Sx = d.
In the picture taken by Richard G, d must have been 25 mm at f/2. The focus was at infinity. That means that Sx = Always 25 mm. That means that everything smaller than or not much bigger than 25 mm is not recognizable. Everything much bigger (let's say >=5 times) than 25 mm is recognizable.
The hyperfocal distance of a 50 mm lens at f/2 = 41.7 m.
When Richard would have focussed at this distance, Sy at 250 m (cross on the san marco) would have been 25 mm x (250 m - 41,7 m) / 41,7 m = ca. 125 mm. That means that objects smaller than 0.6 m would not have been very recognizable. That also means that the contrast in the San Marco would have been much less.
(If you want all the details that take all small deviations like field curvature etc. into account --> see Merklingers book / acticles)
f = focal length
N = aperture number
d = f/N = the effective aperture diameter (e.g. 50 mm lens at N = 10 --> d = 50 mm/10 = 5 mm)
D = focus distance = distance from lens to the point (plane) which is in focus.
In focus means that a point (which has no dimensions) results in a point on your film or sensor.
Every point which is closer to or farther away from the lens will result in a disc with certain dimensions on the film or sensor.
The following formulas will give the size of those discs.
Distances from the lens to D will be indicated by X.
Distances from D to infinity (measured from lens) will be indicated by Y.
Sx = disc size according to distances X
Sy = disc size according to distances Y.
The formulas are really easy:
Sx = d x (D-X)/D
Sy = d x (Y-D)/D
If you substitude d by f/N then the formulas are:
Sx = f/N x (D-X)/D
Sy = f/N x (Y-D)/D
What can we conclude from this formulas:
If X = 0 then Sx = d. So points in front af D will never be depicted on the film or sensor greater than d.
If X (or Y) = D then Sx (or Sy) = 0 mm.
If Y = 2D then Sy = d
(From this you can conclude that the 1/3-2/3 rule = simply not true.)
For Y > 2D, Sy > d
For D < infinity and Y --> infinity: Sy --> infinity.
If D = infinity then for all X's Sx = d.
In the picture taken by Richard G, d must have been 25 mm at f/2. The focus was at infinity. That means that Sx = Always 25 mm. That means that everything smaller than or not much bigger than 25 mm is not recognizable. Everything much bigger (let's say >=5 times) than 25 mm is recognizable.
The hyperfocal distance of a 50 mm lens at f/2 = 41.7 m.
When Richard would have focussed at this distance, Sy at 250 m (cross on the san marco) would have been 25 mm x (250 m - 41,7 m) / 41,7 m = ca. 125 mm. That means that objects smaller than 0.6 m would not have been very recognizable. That also means that the contrast in the San Marco would have been much less.
tvdpid
Member
You can easily use these formulas to calculate e.g. which aperture you would need to blur out the background completely. The classic DOF scales won't tell you. That blurring out relates to the size of the detail in the background. Big details need more blur than small details ... So it is not just the distance from the background to the subject that matters ...
It is also easy to do the calculations by heart ...
d = f/N = easy...
(D-X)/D or (Y-D)/D = also easy and gives you the factor with which you reduce or blow up d ...
It is also easy to do the calculations by heart ...
d = f/N = easy...
(D-X)/D or (Y-D)/D = also easy and gives you the factor with which you reduce or blow up d ...
tvdpid
Member
Note that we never talked about film or sensor size nor print size ... you don't need to...
We think of disks of confusion in "object space" instead of circles of confusion in "image space".
We think of disks of confusion in "object space" instead of circles of confusion in "image space".
Richard G
Veteran
Impressive calculations. I have been thinking this morning about Leitz's infinity locks. I have been using my Elmar 50 3.5 and the 35 3.5 Summaron. Barack and Berek must have thought all these things through very carefully. Why have an infinity lock? They must have considered that many focussing situations would be ideally served by just leaving the lens focussed at infinity. And this is in a camera with a wonderful and highly accurate rangefinder.
I have been making great use of the infinity locks, but not to lock the focus at infinity, but just short of that by using the lock as an infinity stop, the point at which the focus stops before the resistance that needs to be overcome to lock it properly. This sets the focus at just beyond 20m with the 50 and just short of 20m with the 35. These are very useful focus points for streets and buildings that aren't so far off as being practically at infinity, but they are still a long way farther than any traditional hyperlocal focussing distance except, conveniently, for the maximum aperture of f3.5 with these lenses.
I have been making great use of the infinity locks, but not to lock the focus at infinity, but just short of that by using the lock as an infinity stop, the point at which the focus stops before the resistance that needs to be overcome to lock it properly. This sets the focus at just beyond 20m with the 50 and just short of 20m with the 35. These are very useful focus points for streets and buildings that aren't so far off as being practically at infinity, but they are still a long way farther than any traditional hyperlocal focussing distance except, conveniently, for the maximum aperture of f3.5 with these lenses.
Dektol Dan
Well-known
Jeezus!
Jeezus!
If 10 angels can stand on the head of a pin which will be in focus and which will be consumed by bokeh?
Jeezus!
If 10 angels can stand on the head of a pin which will be in focus and which will be consumed by bokeh?
Richard G
Veteran
Consumed by bokeh, when desired, is indeed one of the spin-offs of the OP's pursuit. Mastering photography is like many things a matter of the detail. I appreciate the link to Merklinger and there are some important lessons in his way of looking at things, even if you aren't up to the maths. Unlike your analogy, this is practical not philosphical and certainly not theological at all.
hap
Well-known
I think most of us, shooting film understand the issues regarding focus and analog display. Lloyd Chambers of Diglloyd fame has posted a blog item on how focus issues will become more problematic as digital display resolution increases. Since 8K displays are not far off even 24MP images will not fill the screen and every little pixel will get more than it is due. No smearing or interpolation to adversely affect the zone of critical sharpness. Even the current mac 5K display is close to this point with some digicams. DOF will not necessarily compensate for expected sharpness. He discusses focus stacking as one proposed solution.
Richard G
Veteran
At the bottom of this thread on the Leitz infinity lock is a further amusing reference to the Angels. http://www.l-camera-forum.com/topic/32162-why-infinity-focus-lock/
tvdpid
Member
When I unlock the infinity lock of my Summaron 35 mm f/2.8. I can turn a little further than infinity (as described by Merklinger). This could also explain why there is a lock.
tvdpid
Member
Since the head of a pin has a diameter of 0,03 mm, I'd advice you to put 1/3th of the angels in the first 0,01 mm and the rest in the following 0,02 mm. I think you will be able to get them all in focus if you use a lens with an appropriate focal length combined with an appropriate aperture. Focus at the front of the pin + 0,01 mm.
Looking forward to seeing the results,
Tom
Roger Hicks
Veteran
Dear Richard,
Not so much as "not up to the math" as "can't be arsed to look for more precision than exists in the system, let alone waste good picture-taking and processing time in piddling about with pointless calculations".
Cheers,
R.
David Hughes
David Hughes
Exactly, it takes all the fun and skill out of it.
Once you reach a certain age, amount of experience you 'know' it and don't bother with the mechanics. 'though it's probably nearer the truth to say your fingers know it...
Regards, David
Once you reach a certain age, amount of experience you 'know' it and don't bother with the mechanics. 'though it's probably nearer the truth to say your fingers know it...
Regards, David
charjohncarter
Veteran
bmattock, thanks for the link and your summary.
Roger Hicks
Veteran
Dear David,
You also come to realize that a lot of the theorizing, especially of the Merklinger variety in this book, is wildly overblown. I'm all in favour of understanding the theory behind things, when necessary -- look at my book on exposure -- but equally, theory is something that normally needs to be in the back of your mind, not the forefront.
Cheers,
R.
tvdpid
Member
For me math IS fun. Even slide rules are fun.
Being able to calculate and explain why some settings didn't give you the picture you had in mind creates precisely the skills you're lacking.
For me it is (partially) math. For others it is getting feedback from another skillfull photographer that creates the skills.
mrmeadows
Established
Merklinger's book is an excellent exposition of its subject, but it is not true that DOF is all fixed at the time of exposure by 1) focal length; 2) f-stop; 3) focus distance. If you read again the first two paragraphs of Merklinger's Chapter 3, you will find that, when one views a negative or a print or electronic display made from a negative or digital sensor image, the DOF also depends on 4) the print magnification; 5) the viewing distance of that print; 6) the visual acuity of the viewer. These dependencies are all eliminated from Merlinger's mathematics and his applications of it by a (somewhat hidden) assumption that he makes throughout his book: the circle-of-confusion must be no larger than one-thirtieth of a millimeter in diameter. He mentions this at the beginning of Chapter 3 and then moves on without further elaboration. If one does not make this assumption, then the mathematics must must include the three factors involving the magnification and viewing of prints, and the math becomes a bit more complex. A simple way of understanding this further complication is to state that "the acceptable circle-of-confusion depends on factors 4, 5 and 6." The extra math is hidden in the formula arising from this statement. If one is content to consider only 8x10 enlargements viewed from 250mm, then Merklinger's mathematics and the advice he derives from it are excellent. If one intends, on the other hand, to make bigger enlargements or to view enlargements from a distance that differs much from about one diagonal of the enlargement, then Merklinger's mathematics are incomplete and his advice mostly likely will not be directly applicable.
As a practical matter, I agree with the opinion that it is important to understand the physical basis and consequences of imaging by lenses, of focus and of DOF. But I use my understanding as background information, and I don't recall ever making any calculations when actually making photographs. I do, however, believe that people with other goals than mine for their photography may find calculations essential to getting the images they want.
--- Mike
As a practical matter, I agree with the opinion that it is important to understand the physical basis and consequences of imaging by lenses, of focus and of DOF. But I use my understanding as background information, and I don't recall ever making any calculations when actually making photographs. I do, however, believe that people with other goals than mine for their photography may find calculations essential to getting the images they want.
--- Mike
tvdpid
Member
@mrmeadows
Maybe I should read Mr. Merklingers book again, but what I understood is that he completely leaves the theories and ideas about any circle of confusion. In Chapter 3 he illustrates its impractical complexities. That's why he deliberately uses the term Disc of confusion (DoC). This term is used, like he says, in 'object space'. (CoC is used in 'image space').
If we calculate a DoC of 5 mm at a given distance. This either can mean that an object photographed at that distance will appear sharp, recognizable or completely blurred out in the print (depending on several things ...). This has (implicitly nor explicitly) nothing to do with CoCs greater or smaller than 0,03 mm.
Maybe I should read Mr. Merklingers book again, but what I understood is that he completely leaves the theories and ideas about any circle of confusion. In Chapter 3 he illustrates its impractical complexities. That's why he deliberately uses the term Disc of confusion (DoC). This term is used, like he says, in 'object space'. (CoC is used in 'image space').
If we calculate a DoC of 5 mm at a given distance. This either can mean that an object photographed at that distance will appear sharp, recognizable or completely blurred out in the print (depending on several things ...). This has (implicitly nor explicitly) nothing to do with CoCs greater or smaller than 0,03 mm.
David Hughes
David Hughes
True but they (slide rules) weren't called guessing sticks for nothing. I guess I just see them as a useful tool for some things but I'm not that serious as I turned down an antique one the other day because I figured some collector would appreciate it more. FWIW, it was in a charity shop for a pound or two.
Regards, David
PS I thought that forums were circles of confusion...
Richard G
Veteran
Dear Roger
I was trying to defend the OP for his effort. I will not be using any of the formulae or maths in the field, but some of the principles inspiring the Merklinger approach I find very useful indeed. For him, to merely have put forward his conclusions would have been useless. In my discipline something new would not convince some colleagues even after the careful laying out of evidence. The more we understand something the better it is. In many situations it won't matter much if our theory or understanding are slightly wrong, but there are some times when results will surprise us such as the extreme case when the old assumption falls down.
Regards
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