DOF and Sensor Size
Pete
calculations to show the effect of sensor size on depth of field (DOF).
The scenario: capture a studio portrait at a chosen DOF (realistic, not
infinite rear DOF).
The question: what considerations need to be made for the sensor size used.
Totally ignoring practical limitations for the moment and using only
the laws of physics, I found the answer surprising:
The effective f-stop must be scaled with sensor size.
The required shutter speed remains constant.
Lens focal length affects only the angle of view and required object
distance. It would be logical to scale it with sensor size to keep the
required angle of view, but it does not affect DOF.
So, what's the catch? I did say the *effective* f-stop. Macro
photographers will know that the effective f-stop of a lens (Ne) is
slower than its selected mechanical value (N):
Ne = N * (1 + M/P) where
M = image magnification (image size/object size)
P = pupil magnification
Using a small sensor for our portrait, the magnification ratio will be
very low (M/P << 1) so we can safely assume Ne = N.
As sensor size increases, there will be some point at which we will
have to include the term M/P (often called the near-macro effect) and
widen the f-stop accordingly. The exact point at which this is reached
depends on the lens design (which determines the value of P). For
example, a telephoto lens may have P=0.5 and a wide-angle
reverse-telephoto may have P=2.0.
I'll explain why sensor size does not affect shutter speed.
36 mm wide sensor
-----------------
f = 100 mm
Ne = f/4
M = 0.05
v = 2.1 m (object to entrance pupil distance)
C = 30 um (circle of confusion)
DOF= 95 mm
18 mm wide sensor
-----------------
f = 50 mm
Ne = f/2
M = 0.025
v = 2.05 m
C = 15 um
DOF= 95 mm
The area of the 18 mm sensor is only 1/4 that of the 36 mm sensor, but
Ne is 2 stops wider. Using the same shutter speed will yield the same
number of photons hitting each sensor. On sensors with equal photon
efficiency and the same number of pixels, the base ISO of the 18 mm
sensor will be 1/4 that of the 36 mm sensor. With each camera set to
its base ISO the exposure will be the same, as will the signal to
photon noise ratio.
Other Issues
============
Doubling both f and v will give the same magnification and will not
affect the DOF or the required Ne.
The effect of diffraction is exactly the same in both examples
therefore DOF at the diffraction limit is not affected by sensor size.
A 4.5 mm sensor would need a 12.5 mm f/0.5 lens to take the same photo.
Minimum DOF is a practical problem for small sensors.
In macro photography, lens focal length cannot be continually scaled
down with sensor size because, eventually, the front element will touch
a 3-D object! The object would also need to be smaller to keep the same
perspective. While it's an amusing thought, it has practical
implications in my studio portrait example: when the near-macro effect
kicks in (somewhere around medium format size) the lenses must be
close-range corrected and may need to have a slightly longer focal
length than calculated by up-scaling from a smaller sensor.
Thanks for reading. I'm sure there are some typos and technical errors
in the above. It's my first attempt at trying to understand this
complex subject so feedback would be greatly appreciated. I'll be the
first to write a comment "Zzzz."
--
Pete
Gordon Freeman
> The area of the 18 mm sensor is only 1/4 that of the 36 mm sensor, but
> Ne is 2 stops wider. Using the same shutter speed will yield the same
> number of photons hitting each sensor. On sensors with equal photon
> efficiency and the same number of pixels, the base ISO of the 18 mm
> sensor will be 1/4 that of the 36 mm sensor. With each camera set to
> its base ISO the exposure will be the same, as will the signal to
> photon noise ratio.
Someone once did some test shots demonstrating this, unfortunately I don't
have the URL but two sample pictures I have seen were done using a Ricoh
GX100 @ ISO800 and a Canon 5D @ ISO16890 to give the same no of photons
(I'm not sure how they achieved that ISO) and the results were indeed
virtually the same in terms of depth of field and noise level.
The apparently poor noise levels of compact cameras is therefore explained
if you think of them as doing the same job as a DSLR set to an absurdly
high ISO so that it can be stopped down to f64 or somesuch tiny aperture to
maximise DoF.
Paul Furman
> I mentioned in another thread that I'd post the results of some
> calculations to show the effect of sensor size on depth of field (DOF).
>
> The scenario: capture a studio portrait at a chosen DOF (realistic, not
> infinite rear DOF).
>
> The question: what considerations need to be made for the sensor size used.
>
> Totally ignoring practical limitations for the moment and using only the
> laws of physics, I found the answer surprising:
>
> The effective f-stop must be scaled with sensor size.
>
> The required shutter speed remains constant.
>
> Lens focal length affects only the angle of view and required object
> distance. It would be logical to scale it with sensor size to keep the
> required angle of view, but it does not affect DOF.
>
> So, what's the catch? I did say the *effective* f-stop.
I have a tough time following the math so I'm going to plod ahead and
summarize in my own words based on what I think you are exploring and I
think my conclusions are a little different or I'm not following you all
the way. I'm pretty sure of that last point <g>.
I see below you are comparing an FX camera with 50mm f/2 lens to
something like a 4/3 camera (such as current Olympus format) which would
need a 12.5mm lens at f/0.5. I think you say the ISO would be 200 for FX
and 50 for 4/3 with the same noise and that the same DOF could be
achieved except that the macro magnification effect kicks in putting the
4/3 camera at a advantage because it's not yet working in it's macro
range and the larger sensor FX camera is. There are losses when working
near/at or beyond macro range and those mean a smaller effective
aperture, which means more depth of field and more diffraction problems.
The diffraction problems mean you can't make as big of a print and you
can't make use of as many megapixels on the sensor. Now I've lost myself
because I know the opposite is true in regards to diffraction. I think
that's right about depth of field though. I suspect your effective
aperture is not my effective aperture, I'm thinking how real aperture
changes at macro distances, your effective aperture refers to effective
depth of field, so lets use e-dof and e-aperture from here on out.
Counting that, all things are equal except the magnification and that
increases e-aperture for the larger sensor camera.
Another way of looking at this is that the FX camera is operating at 1:1
macro conditions when filling the frame with an eye but an 8x10 view
camera will cover a whole face when it's at 1:1 macro conditions. Now
try filling the frame with the eye using the 8x10 camera and you'll need
an enormously long bellows making things quite dark. I'm going to sleep
on this and take a fresh look in the morning... or maybe a few days to
stew on it :-)
Noons
>> So, what's the catch? I did say the *effective* f-stop.
>
> I have a tough time following the math so I'm going to plod ahead and summarize
> in my own words based on what I think you are exploring and I think my
> conclusions are a little different or I'm not following you all the way. I'm
> pretty sure of that last point <g>.
>
Most of that "math" is unverified...
> I see below you are comparing an FX camera with 50mm f/2 lens to something like
> a 4/3 camera (such as current Olympus format) which would need a 12.5mm lens at
> f/0.5.
Hmmmm.... I've got a E-PL1 with that sensor. Using it with a Tamron 17/3.5 I
get excellent DOF control, easily better than the equivalent of a 35/2 in FF.
When I want real shallow DOF, and I mean 50/1.2 style in 35mm, I use the
Voigtlander 25mm/0.95 at 1.8 or less and that gives me as shallow a DOF as
anyone can possibly want. When I want macro, I use it with a Nikon mount Sigma
180/3.5 macro lens. It works a treat and gives me reasonable DOF.
Yet, according to the "math" it all shouldn't happen.
So take it with a grain of salt and experiment rather than calculate.
> are losses when working near/at or beyond macro range and those mean a smaller
> effective aperture, which means more depth of field and more diffraction
> problems. The diffraction problems mean you can't make as big of a print and you
> can't make use of as many megapixels on the sensor. Now I've lost myself because
> I know the opposite is true in regards to diffraction. I think that's right
Hehehe! Dang! :) Precisely what I feel about all these "calculations"...
If one goes by all these statements, m4/3 would never work as a format. Yet,
guess what: those using it are ditching other formats because the blessed
things work and darn well. I'm sure they are all idiots and only the "maths"
operators know how to get DOF isolation and good macro definition...
Pete
Yes.
> There are losses when working near/at or beyond macro range and those
> mean a smaller effective aperture, which means more depth of field and
> more diffraction problems.
Yes, see my new examples below based on an FX camera with 50 mm lens
working at 1:1 macro.
> The diffraction problems mean you can't make as big of a print and you
> can't make use of as many megapixels on the sensor. Now I've lost
> myself because I know the opposite is true in regards to diffraction. I
> think that's right about depth of field though. I suspect your
> effective aperture is not my effective aperture, I'm thinking how real
> aperture changes at macro distances, your effective aperture refers to
> effective depth of field, so lets use e-dof and e-aperture from here on
> out. Counting that, all things are equal except the magnification and
> that increases e-aperture for the larger sensor camera.
That's what had always made it impossible for me to understand this
subject until I did loads of calculations. In my mind, I'd be changing
two parameters at the same time instead of one at a time.
> Another way of looking at this is that the FX camera is operating at
> 1:1 macro conditions when filling the frame with an eye but an 8x10
> view camera will cover a whole face when it's at 1:1 macro conditions.
> Now try filling the frame with the eye using the 8x10 camera and you'll
> need an enormously long bellows making things quite dark. I'm going to
> sleep on this and take a fresh look in the morning... or maybe a few
> days to stew on it :-)
I hope my new examples will make sense of this.
>> the effective f-stop of a lens (Ne) is
>> slower than its selected mechanical value (N):
>>
>> Ne = N * (1 + M/P) where
>>
>> M = image magnification (image size/object size)
>> P = pupil magnification
>>
>> <...> a
>> telephoto lens may have P=0.5 and a wide-angle reverse-telephoto may
>> have P=2.0.
Rearrange the above equation because we know Ne for our given DOF, so
we must calculate the physical f-stop (N) to use:
N = Ne/(1 + M/P)
Let's use a symmetrical lens (P=1), just a single lens, no aperture
iris. At 1:1 macro (M=1):
N = Ne/(1 + 1/1) = Ne/2
We require Ne=2 (an effective f-stop of f/2) for our chosen DOF,
therefore we need a physical f/1 lens. For the purposes of diffraction
and exposure we use f/2 in the calculations not f/1.
If we purchased a telephoto lens with P=0.5:
N = 2/(1 + 1/0.5) = 2/3
we need a physical aperture of f/0.67
which is 3.2 f-stops wider than f/2!
I'll do exactly for the 1/4 size sensor: we require f/0.5 (P=1,
magnification M=0.25):
N = 0.5/(1 + 0.25/1) = 0.4
therefore we need an f/0.4 lens
which is 0.6 f-stops wider than f/0.5
Let's have a look at entrance pupil sizes:
12.5 mm f/0.4 = 31.25 mm
50 mm f/1 = 50 mm (1.6x the diameter, 2.56x the area)
As the near-macro effect kicks in, the larger sensor requires a
substantially bigger chunk of front element glass than the smaller
sensor.
When both sensors are used to image far-field objects at the same DOF,
they require the same size entrance pupil. Therefore all sensors
require the same huge chunk of front element glass to obtain a very
shallow DOF.
In your example of filling an 8x10 frame with an eye we have gone from
macro to micro photography (M >> 1). Let's say P=1 and M=10 then Ne = N
x 11. An f/4 physical lens becomes an f/44. Yikes, the finder image
will be very dim. The diffraction and exposure are calculated using
f/44, not the lens mechanical f-stop.
For macro photography, I would want to balance DOF with diffraction.
I'll choose f/22 on a FF camera. Provided that I open-up the mechanical
f-stop appropriately I will get what I expect. E.g. at 1:1 ratio with a
symmetrical lens (P=1) I would select f/11 to give me an effective f/22.
Notes
-----
Ne = N(1 + M/P) is an approximation for relatively small angles of
thetaE. The correct formula is:
Ne = sqrt((M/2)^2 + (N(1+M/P))^2)
In all of my examples the difference is less than 4%.
To calculate the f-stop difference use:
6.64 x Log10(1 + M/P)
To determine if a lens is approximately symmetrical (P=1) look into
each end: the stopped down aperture opening will look the same size.
--
Pete
Pete
> Paul Furman wrote,on my timestamp of 21/02/2011 5:42 PM:
>
>>> So, what's the catch? I did say the *effective* f-stop.
>>
>> I have a tough time following the math so I'm going to plod ahead and summarize
>> in my own words based on what I think you are exploring and I think my
>> conclusions are a little different or I'm not following you all the way. I'm
>> pretty sure of that last point <g>.
>>
>
> Most of that "math" is unverified...
It took me a long time, but it is verifiable.
>> I see below you are comparing an FX camera with 50mm f/2 lens to something like
>> a 4/3 camera (such as current Olympus format) which would need a 12.5mm lens at
>> f/0.5.
>
> Hmmmm.... I've got a E-PL1 with that sensor. Using it with a Tamron
> 17/3.5 I get excellent DOF control, easily better than the equivalent
> of a 35/2 in FF.
Impossible, as the online DOF calculators will easily verify.
Subject distance = 10 feet, in-focus zone shown:
E-PL1 17 mm f/3.4: 6.5 to 21.2 feet
FF 35 mm f/2.0: 8.7 to 11.7 feet
On the E-PL1 the near-side DOF is 2.7 times greater and the total DOF
is 4.9 times greater than the FF.
Reference:
<http://www.dofmaster.com/dofjs.html>
> When I want real shallow DOF, and I mean 50/1.2 style in 35mm, I use
> the Voigtlander 25mm/0.95 at 1.8 or less and that gives me as shallow a
> DOF as anyone can possibly want. When I want macro, I use it with a
> Nikon mount Sigma 180/3.5 macro lens. It works a treat and gives me
> reasonable DOF.
>
> Yet, according to the "math" it all shouldn't happen.
According to the maths, format size makes no difference to the maximum
DOF available. So, yes, that part *should* happen.
> So take it with a grain of salt and experiment rather than calculate.
>
>> are losses when working near/at or beyond macro range and those mean a smaller
>> effective aperture, which means more depth of field and more diffraction
>> problems. The diffraction problems mean you can't make as big of a
>> print and you
>> can't make use of as many megapixels on the sensor. Now I've lost
>> myself because
>> I know the opposite is true in regards to diffraction. I think that's right
>
> Hehehe! Dang! :) Precisely what I feel about all these "calculations"...
> If one goes by all these statements, m4/3 would never work as a format.
> Yet, guess what: those using it are ditching other formats because
> the blessed things work and darn well. I'm sure they are all idiots
> and only the "maths" operators know how to get DOF isolation and good
> macro definition...
The reason I've attempted to understand this subject is so that I can
sort fact from fiction. Your claim above that your E-PL1 gives better
DOF control is your subjective assessment, however, it is contrary to
the laws of physics.
No offence intended and this reply is certainly not a personal attack.
--
Pete
Noons
wrote:
> > Most of that "math" is unverified...
>
> It took me a long time, but it is verifiable.
I remain sceptic.
> > Hmmmm.... I've got a E-PL1 with that sensor. Using it with a Tamron
> > 17/3.5 I get excellent DOF control, easily better than the equivalent
> > of a 35/2 in FF.
>
> Impossible, as the online DOF calculators will easily verify.
>
> Subject distance = 10 feet, in-focus zone shown:
> E-PL1 17 mm f/3.4: 6.5 to 21.2 feet
> FF 35 mm f/2.0: 8.7 to 11.7 feet
>
> On the E-PL1 the near-side DOF is 2.7 times greater and the total DOF
> is 4.9 times greater than the FF.
>
> Reference:
> <http://www.dofmaster.com/dofjs.html>
>
And once again, the nature of the sensor and its imaging properties is
not considered.
> According to the maths, format size makes no difference to the maximum
> DOF available. So, yes, that part *should* happen.
Hmmm.... Disagree. The maths assume only changes to lenses and
sensor area to get at the circle of confusion. They fail to take into
consideration that to blow a FF image to, say, 8X10, takes the same
resulting image magnification than to do the same for a m4/3 image:
the pixel density in a m4/3 sensor is much higher than in an
equivalent FF, to compensate for the smaller imaging surface.
If pixel density was constant then indeed we could see all those
differences. Problem is: a m4/3 image is "blown out" more than a FF
due to the higher pixel density. That causes major changes to circle
of confusion measurements, which are not accounted for in the maths as
presented.
One thing is the theoretical - and correct - circle of confusion which
is only dependent on distance and lens aperture assuming a sensor of
unlmited definition, the other is the practical result of pixel
density and dimensions of any given sensor when applied to an image
taken with any given lens.
> The reason I've attempted to understand this subject is so that I can
> sort fact from fiction. Your claim above that your E-PL1 gives better
> DOF control is your subjective assessment, however, it is contrary to
> the laws of physics.
Not at all. What is contrary is the quantification of those physics
and its laws to the problem in question. I don't dismiss the physics,
I do dismiss the maths that have been used: they make too many
assumptions as to the nature of the imaging sensor and fail to take
into account major fundamental scale differences on how the image is
captured in each case.
> No offence intended and this reply is certainly not a personal attack.
None taken. I wish there was an answer to these questions that kept me
happy. As is, every single time I've gone into the maths of DOF, I
get stumped on simple inconsistencies like failing to take into
account the scale of image capture. It's not the physics, it's the
maths; they are not synonyms.
Pete
Thanks, Noons. I remain sceptical of the maths I've presented; even
more sceptical of most of the websites I've read on the subject - the
authors get lost in places during their explanation. I'm very interest
to demystify it for myself as a hobby. When it come to taking pictures
it makes no difference: I look at the results, learn a little, stick to
what works well and try changing what doesn't (much more fun than
learning maths).
--
Pete
Noons
wrote:
>
> > None taken. I wish there was an answer to these questions that kept me
> > happy. As is, every single time I've gone into the maths of DOF, I
> > get stumped on simple inconsistencies like failing to take into
> > account the scale of image capture. It's not the physics, it's the
> > maths; they are not synonyms.
>
> Thanks, Noons. I remain sceptical of the maths I've presented; even
> more sceptical of most of the websites I've read on the subject - the
> authors get lost in places during their explanation. I'm very interest
> to demystify it for myself as a hobby. When it come to taking pictures
> it makes no difference: I look at the results, learn a little, stick to
> what works well and try changing what doesn't (much more fun than
> learning maths).
Hehehe! I'd raise my glass to that! ;)
(have enough of the stuff in my day to day job...)
Alan Browne
> I mentioned in another thread that I'd post the results of some
Pete. Please start with a very basic point regarding DOF.
It is _essentially_ about magnification ratio from the
negative/positive/sensor size to the size of the print or displayed image.
One more time: It's all about the magnification ratio.
Running the calculations regarding aperture, focal length and subject
distance are of course part of the math.
But the _essential_ thing is the enlargement ratio.
This is reflected in the "magic numbers" in the equation (CoC constant)
which all devolve from the simple act of looking at a print (8x10 for
example) at a given distance, typically 2x the diagonal of the print.
And of course the resolution typically is given at values around 6 or 8
line pairs per mm on said print (about 150 - 200 dpi printed in modern
printer terms).
I think some time spend on that concept will be of more potential
benefit for you (others, me) than the basic equations.
In sum (relating to your subject) for a larger sensor, framed the same
way (distance or FL changes), when going to print the enlargement ratio
will be smaller. OTOH, the CoC will also be larger as the magnification
ratio to a "standard print" must increase as the mag ratio goes down.
(Start at the end (the print). That can save a lot of time at the
beginning...)
--
gmail originated posts filtered due to spam.
PeterN
> Paul Furman wrote,on my timestamp of 21/02/2011 5:42 PM:
>
>>> So, what's the catch? I did say the *effective* f-stop.
>>
>> I have a tough time following the math so I'm going to plod ahead and
>> summarize
>> in my own words based on what I think you are exploring and I think my
>> conclusions are a little different or I'm not following you all the
>> way. I'm
>> pretty sure of that last point <g>.
>>
>
> Most of that "math" is unverified...
About three years ago I was asked to comment on a presentation of
quantum computing.
I stated that the subject was chosen because none of us had sufficient
understanding of the subject to know whether there were flaws in his
presentation.
<snip>
--
Peter
Pete
> On 2011.02.16 20:25 , Pete wrote:
>> I mentioned in another thread that I'd post the results of some
>
> Pete. Please start with a very basic point regarding DOF.
>
> It is _essentially_ about magnification ratio from the
> negative/positive/sensor size to the size of the print or displayed
> image.
There is only one starting point from which everything else is derived:
the magnification ratio from the object to the eye of the viewer.
> One more time: It's all about the magnification ratio.
Indeed, it's about correctly combining three magnification ratios
(object:sensor, sensor:print, print:eye) into one ratio (object:eye)
such that the end result has the desired DOF and angle of view
(perspective). From that basic start, we need only one more parameter:
eye resolution. Everything else can then be calculated.
The important thing about the three separate magnification ratios is
that they cannot be multiplied together. The object:sensor
magnification is in 3-D object space; the others are in 2-D image space.
The whole point of my original post was to show that (in theory) sensor
size makes no difference to the final result provided that the
near-macro effect is taken into account. The incredible simplicity of
it all is what surprised me.
> Running the calculations regarding aperture, focal length and subject
> distance are of course part of the math.
>
> But the _essential_ thing is the enlargement ratio.
>
> This is reflected in the "magic numbers" in the equation (CoC constant)
> which all devolve from the simple act of looking at a print (8x10 for
> example) at a given distance, typically 2x the diagonal of the print.
>
> And of course the resolution typically is given at values around 6 or 8
> line pairs per mm on said print (about 150 - 200 dpi printed in modern
> printer terms).
>
> I think some time spend on that concept will be of more potential
> benefit for you (others, me) than the basic equations.
>
> In sum (relating to your subject) for a larger sensor, framed the same
> way (distance or FL changes), when going to print the enlargement ratio
> will be smaller. OTOH, the CoC will also be larger as the
> magnification ratio to a "standard print" must increase as the mag
> ratio goes down.
>
> (Start at the end (the print). That can save a lot of time at the beginning...)
The number of pixels required is determined by the resolution of the
eye (about 32 lp/degree) and the angle subtended at the eye by the
print. Obviously, these same two factors enable us to define DOF in
terms of object space then calculate the f-stop required to produce it.
I have a clunky spreadsheet to do this.
My original post has nothing to do with deriving absolute values; I
assumed the reader would be familiar with it. I am interested only in
the relative effect of sensor size on the the key parameters.
Here are the two examples in my original post. Observe that I took the
CoC into account:
36 mm wide sensor
-----------------
f = 100 mm
Ne = f/4
M = 0.05
v = 2.1 m (object to entrance pupil distance)
C = 30 um (circle of confusion)
DOF= 95 mm
18 mm wide sensor
-----------------
f = 50 mm
Ne = f/2
M = 0.025
v = 2.05 m
C = 15 um <<<======
DOF= 95 mm
The object:sensor magnification ratio is catered for by scaling the
focal length (f).
I have scaled the effective f-stop (Ne), which automatically scales the
CoC (C) appropriately for the different sensor:print magnification
ratio (30 um down to 15 um).
Digital makes it much easier because the sensor:print magnification
ratio becomes irrelevant when we compare different sized sensors having
the same number of pixels. The half-size sensor needs twice the
magnification, but its pixels have half the spacing. If we halve the
CoC used to determine the f-stop, we should get identical prints.
If the number of pixels start appearing in DOF calculations then it's
time to buy more pixels. DOF calculations based on a CoC approaching
pixel spacing is a sure indication that the system is already being
pushed beyond its resolution limit.
There is something I missed in my original post. The focal length does
not scale exactly with image sensor size as some websites would have us
believe. Notice the slightly longer object to entrance pupil distance
(v) required by the longer lens. The effect becomes more pronounced as
magnification increases.
--
Pete
Charles E Hardwidge
"Pete" <pete3....@rediculous.ntlworld.com> wrote in message
news:2011022415232498144-pete3attkins@rediculousntlworldcom...
> The whole point of my original post was to show that (in theory) sensor
> size makes no difference to the final result provided that the near-macro
> effect is taken into account. The incredible simplicity of it all is what
> surprised me.
I don't have a fucking clue what you're talking about. Maybe you can take a
step back and think about whatever the question is and provide a less
engineered explanation. Simple is good but whatever you're trying to
communicate isn't carrying over like that.
--
Charles E Hardwidge
Pete
I was doing ok with my original post. Actually posting it was the error
I hadn't spotted :-(
--
Pete
Charles E Hardwidge
I've no real clue what your saying apart from a few bits clicking with the
crap I've read elsewhere about lenses and printing. Call me dense but fewer
numbers and more pictures might help. Failing that a readable narrative that
isn't peppered with jargon.
--
Charles E Hardwidge
Paul Furman
> The whole point of my original post was to show that (in theory) sensor
> size makes no difference to the final result provided that the
> near-macro effect is taken into account. The incredible simplicity of it
> all is what surprised me.
And the key is "in theory".
I believe you were just having fun with the numbers, not even intending
to imply any practical use for the intellectual exercise.
In practice, for the purpose of portraits, smaller sensors can produce
softer backgrounds because there are longish/super-fast lenses available
for 35mm FX which; when mounted on DX or 4/3, provide more extreme bokeh
than anything available for 4/3 like 85mm f/1.2 or 135mm f/1.8 or even
200mm f/2. I actually don't know that for a fact but would be very
surprised if more extreme bokeh could be achieved with lenses designed
for 4/3. Perhaps that's true for 300mm f/2.8 or 400mm f/2.8 but there is
little advantage to designing such a lens for 4/3. There is no such
thing as a DX 300mm f/2.8 for example - I think Olympus may have such a
beast but not much smaller or less expensive.
Now, let's say you want wide angle with shallow DOF, the opposite is
true. FX can use 24mm f.1.4 or even 35mm f/0.95 I think, on a short
mount like Leica with some weird adapters. Larger formats probably can
do even better. Now, there are some video lenses like 12mm f/1.4 as I
recall, which are meant for small sensors but those aren't going to
match 24mm f/1.4 because 12mm on 4/3 looks like 24mm f/2.8 (quick guess)
on FX.
Here's one odd toy of mine that might be able to provide really shallow
DOF on m4/3, it's a 20mm f/2.8 with a very large flange mount distance
and I think a focal reducer such as used with telescopes for widefield
work could be put between the camera and lens to make it wider and
faster while reducing the mount distance:
http://www.flickr.com/photos/edgehill/5129933204/
I haven't really studied that possibility indepth but even if possible,
it would be ridiculous :-)
Oh, here's a new one, Voigtlander 25mm f/0.95 for 4/3:
http://www.luminous-landscape.com/reviews/lenses/voigtlander_f095_25mm_micro_43_nocton.shtml
But that's only equivalent to a 50mm (f/2?) on FX.
John McWilliams
> On Feb 22, 2:20 am, Pete<pete3.attk...@rediculous.ntlworld.com> wrote:
>> It took me a long time, but it is verifiable.
>
> I remain sceptic.
You mean skeptical, or a skeptic.
I mean septic.
--
lsmft
Pete
That's the best reply I've seen so far :-) Unfortunately, that makes it
harder to resist starting another topic I'll regret posting...
--
Pete
Charles E Hardwidge
Oh, be-have. *wet slap*
I feel an urge to wallow in orgasmic creativity. Reading someone else's shit
makes a change to listening to the bollocks in my own head. Kirk Tuck's on a
semi-official gotta-finish-the-book break and I need my fix.
--
Charles E Hardwidge
Noons
http://www.thefreedictionary.com/sceptic
I meant exactly what I said.
You on the other hand - as usual - can't make up your mind...
But what else can be expected of trolls such as yourself?
Noons
>
> I feel an urge to wallow in orgasmic creativity. Reading someone else's shit
> makes a change to listening to the bollocks in my own head. Kirk Tuck's on a
> semi-official gotta-finish-the-book break and I need my fix.
>
He's back...
Charles E Hardwidge
news:ikl54g$m3$2...@news.eternal-september.org...
Yeah, I noticed that. Dude can't keep his trap shut.
Golly, I hope it's not contagious. ;-)
--
Charles E Hardwidge
Pete
> "Pete" <pete3....@rediculous.ntlworld.com> wrote in message
> news:20110301215355245-pete3attkins@rediculousntlworldcom...
>> On 2011-03-01 16:49:24 +0000, John McWilliams said:
>>
>>> On 2/21/11 PDT 1:05 PM, Noons wrote:
>>>> On Feb 22, 2:20 am, Pete<pete3.attk...@rediculous.ntlworld.com> wrote:
>>>
>>>>> It took me a long time, but it is verifiable.
>>>>
>>>> I remain sceptic.
>>>
>>> You mean skeptical, or a skeptic.
>>> I mean septic.
>>
>> That's the best reply I've seen so far :-) Unfortunately, that makes it
>> harder to resist starting another topic I'll regret posting...
>
> Oh, be-have. *wet slap*
I've been trying to figure out why you said that.
I thought John was indicating that my post deserved a septic reply. I'd
agree with that and thought it was very amusing.
If I'm mistaken then my apologies to Noons.
--
Pete
tony cooper
For those that don't catch the reference, "septic" is UK rhyming slang
for an American. Septic tank = Yank = septic. Used on the old "Red
Dwarf" TV show.
Like, "Merkin" (a pubic wig), a derogatory term for Americans.
--
Tony Cooper - Orlando, Florida
Charles E Hardwidge
> On 2011-03-01 22:11:38 +0000, Charles E Hardwidge said:
>
>> Oh, be-have. *wet slap*
>
> I've been trying to figure out why you said that.
I was just camping around.
Carry on.
--
Charles E Hardwidge
Pete
Ah, I wasn't mistaken; I was hopelessly lost. Strangely enough, it's
not the first time :-)
--
Pete
John McWilliams
Not to worry. It's way easier than all this. I meant simply that the
little man from down under is often "septic" in the original meaning of
"septic tank" and the contents thereof.
--
Over and out.
Peter N
Not sure if u meant aseptic, antiseptic, or sepsis.
--
from my Droid
Noons
> Not to worry. It's way easier than all this. I meant simply that the
> little man from down under is often "septic" in the original meaning of
> "septic tank" and the contents thereof.
and once again, for sheer ignorance, small-mindedness and utter
stupidity, you stuck your foot in your mouth.
Wanna compare my height to yours? Anytime.