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1.6x multiplier and 35mm lens equivalencies

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John Kim

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The question has come up, "I have been using 35mm film cameras and am thinking of getting a D60. What sorts of pictures would a D60 take with my EOS lens collection?"

On the left are the common EOS lenses you're thinking of putting on the D30/D60. On the right is the lens and aperture you'd need to put on your 35mm camera to produce a picture identical in every way except exposure to what a D60 would produce. The only caveat is that the subject must be much further than the focal length of the lens (try a least 10x further to be safe).

Example: Say I wanted to put my 50mm f/1.8 lens on a D60 and take a picture of something 1 meter away. Say I wanted to "preview" what that picture would look like in terms of composition, perspective, and DOF before buying the D60. I look at the chart below and see:

50mm f/1.8 = 80mm f/2.88

So if I position my 35mm camera camera 1 meter away, put an 80mm lens on it and set the aperture to f/2.88, it will yield an image identical to what the D60 would produce with a 50mm at f/1.8 from the same position. Same composition, same perspective, same DOF. The only differrences are exposure (The D60 still gets a f/1.8 exposure, which you can compensate for with a longer shutter speed on the 35mm), and resolution (which will be degraded slightly on the D60 due to the smaller imager, and is dependent on the lens).

The tables assume you're shooting wide open. If you wish to stop down, just multiply the f/ ratio on the D60 by 1.6x to get the equivalent 35mm f/ ratio for DOF equivalence. Remember though that the subject needs to be much further than the focal length of the lens for this conversion to be valid.

Primes
14mm f/2.8 = 22.4mm f/4.48
20mm f/2.8 = 32mm f/4.48
24mm f/1.4 = 38.4mm f/2.24
28mm f/2.8 = 44.8mm f/4.48
28mm f/1.8 = 44.8mm f/2.88
35mm f/1.4 = 56mm f/2.24
35mm f/2.0 = 56mm f/3.2
50mm f/1.0 = 80mm f/1.6
50mm f/1.4 = 80mm f/2.24
50mm f/1.8 = 80mm f/2.88
50mm f/2.5 = 80mm f/4
85mm f/1.2 = 136mm f/1.92
85mm f/1.8 = 136mm f/2.88
100mm f/2 = 160mm f/3.2
100mm f/2.8 =160mm f/4.48
135mm f/2 = 216mm f/3.2
200mm f/1.8 = 320mm f/2.88
200mm f/2.8 = 320mm f/4.48
300mm f/2.8 = 480mm f/4.48
300mm f/4 = 480mm f/6.4
400mm f/2.8 = 640mm f/4.48
400mm f/4 = 640mm f/6.4
400mm f/5.6 = 640mm f/8.96
500mm f/4 = 800mm f/6.4
600mm f/4 = 960mm f/6.4

L zooms
16-35mm f/2.8 = 25.6-56mm f/4.48
17-35mm f/2.8 = 27.2-56mm f/4.48
28-70mm f/2.8 = 44.8-112mm f/4.48
70-200mm f/2.8 = 112-320mm f/4.48
35-350mm f/3.55.6 = 56-560mm f/5.6-8.96
100-400mm f/4.5-5.6 = 640mm f/7.2-8.96

Consumer zooms
20-35mm f/3.54.5 = 56mm f/5.6-7.2
24-85mm f/3.5-4.5 = 38.4-136mm f/5.6-7.2
28-135mm f/3.5-5.6 = 44.8-216mm f/5.6-8.96
75-300mm f/4-5.6 = 120-480mm f/6.4-8.96

If you're wondering why the f/stop changes to maintain DOF, a 50mm lens at f/1.8 has a lens diameter of 27.8mm. An 80mm lens at f/1.8 has a lens diameter of 44.4mm. Since the 80mm lens has a wider diameter, it picks up light from a wider angle than the 50mm. If that light is not focused sharply, it falls onto the sensor at a wider angle, and the blur circle is larger in diameter than what the 50mm can produce. To produce identical diameter blur circles, you need to give the 80mm lens a 27.8mm diameter, which corresponds to f/2.88.
 
John Kim said:
Since the 80mm lens has a wider
diameter, it picks up light from a wider angle than the 50mm.
Click to expand...
Just to clarify, it picks up a wider angle of light coming from a particular target than the 50mm. Just imagine the target is shooting a cone of light at the lens (which the lens focuses into the final image). It's easy to see then that the larger diameter lens picks up a larger angle cone.
 
I can see it now:

"it's not a focal length multiplier it's a FOV crop!"

Jason
John Kim said:
The question has come up, "I have been using 35mm film cameras and
am thinking of getting a D60. What sorts of pictures would a D60
take with my EOS lens collection?"

On the left are the common EOS lenses you're thinking of putting on
the D30/D60. On the right is the lens and aperture you'd need to
put on your 35mm camera to produce a picture identical in every way
except exposure to what a D60 would produce. The only caveat is
that the subject must be much further than the focal length of the
lens (try a least 10x further to be safe).

Example: Say I wanted to put my 50mm f/1.8 lens on a D60 and take a
picture of something 1 meter away. Say I wanted to "preview" what
that picture would look like in terms of composition, perspective,
and DOF before buying the D60. I look at the chart below and see:

50mm f/1.8 = 80mm f/2.88

So if I position my 35mm camera camera 1 meter away, put an 80mm
lens on it and set the aperture to f/2.88, it will yield an image
identical to what the D60 would produce with a 50mm at f/1.8 from
the same position. Same composition, same perspective, same DOF.
The only differrences are exposure (The D60 still gets a f/1.8
exposure, which you can compensate for with a longer shutter speed
on the 35mm), and resolution (which will be degraded slightly on
the D60 due to the smaller imager, and is dependent on the lens).

The tables assume you're shooting wide open. If you wish to stop
down, just multiply the f/ ratio on the D60 by 1.6x to get the
equivalent 35mm f/ ratio for DOF equivalence. Remember though that
the subject needs to be much further than the focal length of the
lens for this conversion to be valid.

Primes
14mm f/2.8 = 22.4mm f/4.48
20mm f/2.8 = 32mm f/4.48
24mm f/1.4 = 38.4mm f/2.24
28mm f/2.8 = 44.8mm f/4.48
28mm f/1.8 = 44.8mm f/2.88
35mm f/1.4 = 56mm f/2.24
35mm f/2.0 = 56mm f/3.2
50mm f/1.0 = 80mm f/1.6
50mm f/1.4 = 80mm f/2.24
50mm f/1.8 = 80mm f/2.88
50mm f/2.5 = 80mm f/4
85mm f/1.2 = 136mm f/1.92
85mm f/1.8 = 136mm f/2.88
100mm f/2 = 160mm f/3.2
100mm f/2.8 =160mm f/4.48
135mm f/2 = 216mm f/3.2
200mm f/1.8 = 320mm f/2.88
200mm f/2.8 = 320mm f/4.48
300mm f/2.8 = 480mm f/4.48
300mm f/4 = 480mm f/6.4
400mm f/2.8 = 640mm f/4.48
400mm f/4 = 640mm f/6.4
400mm f/5.6 = 640mm f/8.96
500mm f/4 = 800mm f/6.4
600mm f/4 = 960mm f/6.4

L zooms
16-35mm f/2.8 = 25.6-56mm f/4.48
17-35mm f/2.8 = 27.2-56mm f/4.48
28-70mm f/2.8 = 44.8-112mm f/4.48
70-200mm f/2.8 = 112-320mm f/4.48
35-350mm f/3.55.6 = 56-560mm f/5.6-8.96
100-400mm f/4.5-5.6 = 640mm f/7.2-8.96

Consumer zooms
20-35mm f/3.54.5 = 56mm f/5.6-7.2
24-85mm f/3.5-4.5 = 38.4-136mm f/5.6-7.2
28-135mm f/3.5-5.6 = 44.8-216mm f/5.6-8.96
75-300mm f/4-5.6 = 120-480mm f/6.4-8.96

If you're wondering why the f/stop changes to maintain DOF, a 50mm
lens at f/1.8 has a lens diameter of 27.8mm. An 80mm lens at f/1.8
has a lens diameter of 44.4mm. Since the 80mm lens has a wider
diameter, it picks up light from a wider angle than the 50mm. If
that light is not focused sharply, it falls onto the sensor at a
wider angle, and the blur circle is larger in diameter than what
the 50mm can produce. To produce identical diameter blur circles,
you need to give the 80mm lens a 27.8mm diameter, which corresponds
to f/2.88.
Click to expand...
 
Here's an easier approach:

If I take my 50mm F1.4 lens from my 35mm film camera and put it on a D60 I will have the field of view of 80mm lens but the depth of field of a 50m lens. Take whatever lens you have from your f35mm film camera and simply multiply the focal length by 1.6 for the D60 and you have the new FOV. DOF remains the same.
John Kim said:
The question has come up, "I have been using 35mm film cameras and
am thinking of getting a D60. What sorts of pictures would a D60
take with my EOS lens collection?"

On the left are the common EOS lenses you're thinking of putting on
the D30/D60. On the right is the lens and aperture you'd need to
put on your 35mm camera to produce a picture identical in every way
except exposure to what a D60 would produce. The only caveat is
that the subject must be much further than the focal length of the
lens (try a least 10x further to be safe).

Example: Say I wanted to put my 50mm f/1.8 lens on a D60 and take a
picture of something 1 meter away. Say I wanted to "preview" what
that picture would look like in terms of composition, perspective,
and DOF before buying the D60. I look at the chart below and see:

50mm f/1.8 = 80mm f/2.88

So if I position my 35mm camera camera 1 meter away, put an 80mm
lens on it and set the aperture to f/2.88, it will yield an image
identical to what the D60 would produce with a 50mm at f/1.8 from
the same position. Same composition, same perspective, same DOF.
The only differrences are exposure (The D60 still gets a f/1.8
exposure, which you can compensate for with a longer shutter speed
on the 35mm), and resolution (which will be degraded slightly on
the D60 due to the smaller imager, and is dependent on the lens).

The tables assume you're shooting wide open. If you wish to stop
down, just multiply the f/ ratio on the D60 by 1.6x to get the
equivalent 35mm f/ ratio for DOF equivalence. Remember though that
the subject needs to be much further than the focal length of the
lens for this conversion to be valid.

Primes
14mm f/2.8 = 22.4mm f/4.48
20mm f/2.8 = 32mm f/4.48
24mm f/1.4 = 38.4mm f/2.24
28mm f/2.8 = 44.8mm f/4.48
28mm f/1.8 = 44.8mm f/2.88
35mm f/1.4 = 56mm f/2.24
35mm f/2.0 = 56mm f/3.2
50mm f/1.0 = 80mm f/1.6
50mm f/1.4 = 80mm f/2.24
50mm f/1.8 = 80mm f/2.88
50mm f/2.5 = 80mm f/4
85mm f/1.2 = 136mm f/1.92
85mm f/1.8 = 136mm f/2.88
100mm f/2 = 160mm f/3.2
100mm f/2.8 =160mm f/4.48
135mm f/2 = 216mm f/3.2
200mm f/1.8 = 320mm f/2.88
200mm f/2.8 = 320mm f/4.48
300mm f/2.8 = 480mm f/4.48
300mm f/4 = 480mm f/6.4
400mm f/2.8 = 640mm f/4.48
400mm f/4 = 640mm f/6.4
400mm f/5.6 = 640mm f/8.96
500mm f/4 = 800mm f/6.4
600mm f/4 = 960mm f/6.4

L zooms
16-35mm f/2.8 = 25.6-56mm f/4.48
17-35mm f/2.8 = 27.2-56mm f/4.48
28-70mm f/2.8 = 44.8-112mm f/4.48
70-200mm f/2.8 = 112-320mm f/4.48
35-350mm f/3.55.6 = 56-560mm f/5.6-8.96
100-400mm f/4.5-5.6 = 640mm f/7.2-8.96

Consumer zooms
20-35mm f/3.54.5 = 56mm f/5.6-7.2
24-85mm f/3.5-4.5 = 38.4-136mm f/5.6-7.2
28-135mm f/3.5-5.6 = 44.8-216mm f/5.6-8.96
75-300mm f/4-5.6 = 120-480mm f/6.4-8.96

If you're wondering why the f/stop changes to maintain DOF, a 50mm
lens at f/1.8 has a lens diameter of 27.8mm. An 80mm lens at f/1.8
has a lens diameter of 44.4mm. Since the 80mm lens has a wider
diameter, it picks up light from a wider angle than the 50mm. If
that light is not focused sharply, it falls onto the sensor at a
wider angle, and the blur circle is larger in diameter than what
the 50mm can produce. To produce identical diameter blur circles,
you need to give the 80mm lens a 27.8mm diameter, which corresponds
to f/2.88.
Click to expand...
 
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
The problem with this approach is that you're asking the person to think of a picture taken with a 80mm lens, and project onto it the DOF of a 50mm lens. My approach just lets you take a picture with your 35mm, print it, look at it, and say "so that's what a D60 picture would look like."
 
DOF is produced and freezed by lens, whether you crop big or small, which will affect angle of "view" only. So no need to multiply the f numbers.
Doug Walker said:
John Kim said:
The question has come up, "I have been using 35mm film cameras and
am thinking of getting a D60. What sorts of pictures would a D60
take with my EOS lens collection?"

On the left are the common EOS lenses you're thinking of putting on
the D30/D60. On the right is the lens and aperture you'd need to
put on your 35mm camera to produce a picture identical in every way
except exposure to what a D60 would produce. The only caveat is
that the subject must be much further than the focal length of the
lens (try a least 10x further to be safe).

Example: Say I wanted to put my 50mm f/1.8 lens on a D60 and take a
picture of something 1 meter away. Say I wanted to "preview" what
that picture would look like in terms of composition, perspective,
and DOF before buying the D60. I look at the chart below and see:

50mm f/1.8 = 80mm f/2.88

So if I position my 35mm camera camera 1 meter away, put an 80mm
lens on it and set the aperture to f/2.88, it will yield an image
identical to what the D60 would produce with a 50mm at f/1.8 from
the same position. Same composition, same perspective, same DOF.
The only differrences are exposure (The D60 still gets a f/1.8
exposure, which you can compensate for with a longer shutter speed
on the 35mm), and resolution (which will be degraded slightly on
the D60 due to the smaller imager, and is dependent on the lens).

The tables assume you're shooting wide open. If you wish to stop
down, just multiply the f/ ratio on the D60 by 1.6x to get the
equivalent 35mm f/ ratio for DOF equivalence. Remember though that
the subject needs to be much further than the focal length of the
lens for this conversion to be valid.

Primes
14mm f/2.8 = 22.4mm f/4.48
20mm f/2.8 = 32mm f/4.48
24mm f/1.4 = 38.4mm f/2.24
28mm f/2.8 = 44.8mm f/4.48
28mm f/1.8 = 44.8mm f/2.88
35mm f/1.4 = 56mm f/2.24
35mm f/2.0 = 56mm f/3.2
50mm f/1.0 = 80mm f/1.6
50mm f/1.4 = 80mm f/2.24
50mm f/1.8 = 80mm f/2.88
50mm f/2.5 = 80mm f/4
85mm f/1.2 = 136mm f/1.92
85mm f/1.8 = 136mm f/2.88
100mm f/2 = 160mm f/3.2
100mm f/2.8 =160mm f/4.48
135mm f/2 = 216mm f/3.2
200mm f/1.8 = 320mm f/2.88
200mm f/2.8 = 320mm f/4.48
300mm f/2.8 = 480mm f/4.48
300mm f/4 = 480mm f/6.4
400mm f/2.8 = 640mm f/4.48
400mm f/4 = 640mm f/6.4
400mm f/5.6 = 640mm f/8.96
500mm f/4 = 800mm f/6.4
600mm f/4 = 960mm f/6.4

L zooms
16-35mm f/2.8 = 25.6-56mm f/4.48
17-35mm f/2.8 = 27.2-56mm f/4.48
28-70mm f/2.8 = 44.8-112mm f/4.48
70-200mm f/2.8 = 112-320mm f/4.48
35-350mm f/3.55.6 = 56-560mm f/5.6-8.96
100-400mm f/4.5-5.6 = 640mm f/7.2-8.96

Consumer zooms
20-35mm f/3.54.5 = 56mm f/5.6-7.2
24-85mm f/3.5-4.5 = 38.4-136mm f/5.6-7.2
28-135mm f/3.5-5.6 = 44.8-216mm f/5.6-8.96
75-300mm f/4-5.6 = 120-480mm f/6.4-8.96

If you're wondering why the f/stop changes to maintain DOF, a 50mm
lens at f/1.8 has a lens diameter of 27.8mm. An 80mm lens at f/1.8
has a lens diameter of 44.4mm. Since the 80mm lens has a wider
diameter, it picks up light from a wider angle than the 50mm. If
that light is not focused sharply, it falls onto the sensor at a
wider angle, and the blur circle is larger in diameter than what
the 50mm can produce. To produce identical diameter blur circles,
you need to give the 80mm lens a 27.8mm diameter, which corresponds
to f/2.88.
Click to expand...
Click to expand...
 
Take a picture with your film camera and desired lens combination. Get it developed and crop the print by 1.6.

For example, print a 4x6 and cut it down to 2.5x3.75.

--Ketih
 
Keith Lommel said:
Take a picture with your film camera and desired lens combination.
Get it developed and crop the print by 1.6.

For example, print a 4x6 and cut it down to 2.5x3.75.

--Ketih
Click to expand...
That works too, although I'd suggest printing as 8x10 and cutting it down to 5x7 to compare with other 5x7 prints.

The equivalent FL and f/ numbers are still useful though for comparing different lenses. Most 35mm photographers have already developed a pretty good idea of what sort of picture a certain focal length lens at a certain aperture will produce. The tables provide a way to project that knowledge onto the D60 without having to make prints.
 
The advantage to my approach is that it is not esoteric and is more intuitive. You also don't need reference tables. I'm a big believer in the KISS principle.
John Kim said:
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
The problem with this approach is that you're asking the person to
think of a picture taken with a 80mm lens, and project onto it the
DOF of a 50mm lens. My approach just lets you take a picture with
your 35mm, print it, look at it, and say "so that's what a D60
picture would look like."
Click to expand...
 
Keith,

right, but also wrong!

you are capturing the image with a smaller FoV. But that FoV is recorded edge-to-edge on a medium with a totally different design (than film). That image, printed to 4 x 5 will produce a given quality. To produce the same size print from a 35mm, taken at the same distance from the subject, you would have to enlarge a crop from the negative. Thus you are losing some of the recording capability of the film since the image is not coming from across the entire meduim.

the comparison should be based on two shots of total edge-to-edge coverage, based on an identical FoV, not on a shot from the same spot.

Frank
Keith Lommel said:
Take a picture with your film camera and desired lens combination.
Get it developed and crop the print by 1.6.

For example, print a 4x6 and cut it down to 2.5x3.75.

--Ketih
Click to expand...
 
Different things work for different people. For me, trying to combine two different images produced by two different lenses in my head to predict the final output is not intuitive nor simple. If that is easier for you, then all the more power to you.

The reference tables are just to save people from doing the math. All it is is the f/ratio multiplied by 1.6x. Any grade school student should be able to do that if pressed.
Doug Walker said:
John Kim said:
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
The problem with this approach is that you're asking the person to
think of a picture taken with a 80mm lens, and project onto it the
DOF of a 50mm lens. My approach just lets you take a picture with
your 35mm, print it, look at it, and say "so that's what a D60
picture would look like."
Click to expand...
Click to expand...
 
I am a little confused as to the reason for this; doesn’t the DOF Preview button give you a pretty good idea of this?

Not meant to be sarcastic... I am just missing something here
John Kim said:
The question has come up, "I have been using 35mm film cameras and
am thinking of getting a D60. What sorts of pictures would a D60
take with my EOS lens collection?"

On the left are the common EOS lenses you're thinking of putting on
the D30/D60. On the right is the lens and aperture you'd need to
put on your 35mm camera to produce a picture identical in every way
except exposure to what a D60 would produce. The only caveat is
that the subject must be much further than the focal length of the
lens (try a least 10x further to be safe).

Example: Say I wanted to put my 50mm f/1.8 lens on a D60 and take a
picture of something 1 meter away. Say I wanted to "preview" what
that picture would look like in terms of composition, perspective,
and DOF before buying the D60. I look at the chart below and see:

50mm f/1.8 = 80mm f/2.88

So if I position my 35mm camera camera 1 meter away, put an 80mm
lens on it and set the aperture to f/2.88, it will yield an image
identical to what the D60 would produce with a 50mm at f/1.8 from
the same position. Same composition, same perspective, same DOF.
The only differrences are exposure (The D60 still gets a f/1.8
exposure, which you can compensate for with a longer shutter speed
on the 35mm), and resolution (which will be degraded slightly on
the D60 due to the smaller imager, and is dependent on the lens).

The tables assume you're shooting wide open. If you wish to stop
down, just multiply the f/ ratio on the D60 by 1.6x to get the
equivalent 35mm f/ ratio for DOF equivalence. Remember though that
the subject needs to be much further than the focal length of the
lens for this conversion to be valid.

Primes
14mm f/2.8 = 22.4mm f/4.48
20mm f/2.8 = 32mm f/4.48
24mm f/1.4 = 38.4mm f/2.24
28mm f/2.8 = 44.8mm f/4.48
28mm f/1.8 = 44.8mm f/2.88
35mm f/1.4 = 56mm f/2.24
35mm f/2.0 = 56mm f/3.2
50mm f/1.0 = 80mm f/1.6
50mm f/1.4 = 80mm f/2.24
50mm f/1.8 = 80mm f/2.88
50mm f/2.5 = 80mm f/4
85mm f/1.2 = 136mm f/1.92
85mm f/1.8 = 136mm f/2.88
100mm f/2 = 160mm f/3.2
100mm f/2.8 =160mm f/4.48
135mm f/2 = 216mm f/3.2
200mm f/1.8 = 320mm f/2.88
200mm f/2.8 = 320mm f/4.48
300mm f/2.8 = 480mm f/4.48
300mm f/4 = 480mm f/6.4
400mm f/2.8 = 640mm f/4.48
400mm f/4 = 640mm f/6.4
400mm f/5.6 = 640mm f/8.96
500mm f/4 = 800mm f/6.4
600mm f/4 = 960mm f/6.4

L zooms
16-35mm f/2.8 = 25.6-56mm f/4.48
17-35mm f/2.8 = 27.2-56mm f/4.48
28-70mm f/2.8 = 44.8-112mm f/4.48
70-200mm f/2.8 = 112-320mm f/4.48
35-350mm f/3.55.6 = 56-560mm f/5.6-8.96
100-400mm f/4.5-5.6 = 640mm f/7.2-8.96

Consumer zooms
20-35mm f/3.54.5 = 56mm f/5.6-7.2
24-85mm f/3.5-4.5 = 38.4-136mm f/5.6-7.2
28-135mm f/3.5-5.6 = 44.8-216mm f/5.6-8.96
75-300mm f/4-5.6 = 120-480mm f/6.4-8.96

If you're wondering why the f/stop changes to maintain DOF, a 50mm
lens at f/1.8 has a lens diameter of 27.8mm. An 80mm lens at f/1.8
has a lens diameter of 44.4mm. Since the 80mm lens has a wider
diameter, it picks up light from a wider angle than the 50mm. If
that light is not focused sharply, it falls onto the sensor at a
wider angle, and the blur circle is larger in diameter than what
the 50mm can produce. To produce identical diameter blur circles,
you need to give the 80mm lens a 27.8mm diameter, which corresponds
to f/2.88.
Click to expand...
 
Doug,

Unfortunately you are not using the "standard" definition of DoF. Everything I have read on the subject points to the fact that the DoF also has to factor in the magnification of the image output. This "factor is known as the CoC -- Have you heard of it? The CoC is set based on the magnfication of the negative/crop to final output and viewing distance although it is commonly assumed to be a 8x12 output viewed at about 2 feet. Below are a couple of ready references on the Internet for those that are interested. Do you have some other explaination? How would you set the CoC based on Film and/or sensor size?

In similar threads you keep stating the same thing (wrongly based on the references below) and have not offered ANY proof or references other than stating that you are right.

Perhaps you should try READING a bit and see what you think. There is a chance that you got it wrong.

http://www.tangentsoft.net/fcalc/help/CoC.htm

Quoting from the above site (f/calc) on Circle of Confusion:

Imagine a perfect white point in an empty black room. The point has no height, and no width. If you focus an optically perfect lens on that point, it forms a perfect point on the film as well. If, however, you focus slightly in front of or behind the point, the point will image on the film as a small blurry circle. If that circle is small enough, it will still look like a point when enlarged for printing. The "circle of confusion" is typically calculated as the largest on-film circle that you see as a point when you make an 8 × 12 print and view it from a "normal" viewing distance, typically 2-3 feet. Anything larger is seen as a small circle, and is therefore perceived as out of focus.

f/Calc calculates the CoC using the "Zeiss formula": d/1730, where d is the diagonal measure of the film, in millimeters. This formula yields acceptable values for most uses.

http://www.cs.mtu.edu/ ... ... User-Guide/990/EXPOSURE/EV-depth-of-field.html

http://www.outsight.com/hyperfocal.html

Books by Kodak also say about the same thing, namely that DoF is a function of the film format size. If LMC54 has some references for some other definition of DoF, that would be interesting.

Karl
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
 
Answer this question karl:

If you scan a 35mm negative and crop out and enlarge the center section equivalent to the size of a D30 sensor, did you change the DOF?
Karlg said:
Unfortunately you are not using the "standard" definition of DoF.
Everything I have read on the subject points to the fact that the
DoF also has to factor in the magnification of the image output.
This "factor is known as the CoC -- Have you heard of it? The CoC
is set based on the magnfication of the negative/crop to final
output and viewing distance although it is commonly assumed to be a
8x12 output viewed at about 2 feet. Below are a couple of ready
references on the Internet for those that are interested. Do you
have some other explaination? How would you set the CoC based on
Film and/or sensor size?

In similar threads you keep stating the same thing (wrongly based
on the references below) and have not offered ANY proof or
references other than stating that you are right.

Perhaps you should try READING a bit and see what you think. There
is a chance that you got it wrong.

http://www.tangentsoft.net/fcalc/help/CoC.htm

Quoting from the above site (f/calc) on Circle of Confusion:

Imagine a perfect white point in an empty black room. The point has
no height, and no width. If you focus an optically perfect lens on
that point, it forms a perfect point on the film as well. If,
however, you focus slightly in front of or behind the point, the
point will image on the film as a small blurry circle. If that
circle is small enough, it will still look like a point when
enlarged for printing. The "circle of confusion" is typically
calculated as the largest on-film circle that you see as a point
when you make an 8 × 12 print and view it from a "normal" viewing
distance, typically 2-3 feet. Anything larger is seen as a small
circle, and is therefore perceived as out of focus.

f/Calc calculates the CoC using the "Zeiss formula": d/1730, where
d is the diagonal measure of the film, in millimeters. This formula
yields acceptable values for most uses.

http://www.cs.mtu.edu/ ...
... User-Guide/990/EXPOSURE/EV-depth-of-field.html

http://www.outsight.com/hyperfocal.html

Books by Kodak also say about the same thing, namely that DoF is a
function of the film format size. If LMC54 has some references for
some other definition of DoF, that would be interesting.

Karl
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
Click to expand...
 
Keith????

What is more simple than looking through the viewfinder? So who needs to print anything to see the relationship? Just look...

And why do you insist on "only" referring to cropping a 35mm negative to get the same results? Why not just refer to moving in tighter when taking the 35mm shot for equivilency with the digital image - so that the image is spread from edge to edge across the entire available medium (film)?

-or- move back with the digital body to make it equivilent in content to the full 35mm frame ?

It's like zooming a 70-200 from 100mm (35mm) to 160mm (digital 1.6) only you don't have to zoom.... In both cases you are capturing the same field of view onto the full recording medium... All that's left to argue then is the production capabilities of the film medium vs. the sensor medium.... Then it becomes which film you are talking about.... or what ISO the digital was set toooooo..... and the argument never ends.

And at anything over about 50 feet for the focal point in the image, few if any educated judges will see any difference in DoF at the same stop. At anything under 50 ft (just for example) the difference in visible DoF will increase, but marginally on a foot-by-foot basis until the focal point is a lot closer.

The significant issue here is the FoV on the near side of the lens (zoom). To get under the FoV of a 28mm you need a 16 or 17mm lens or a zoom with that kind of "bottom".

My old....

28-70mm L (35mm frame fill)
70-200mm L (35mm frame fill)
100-300mm (35mm frame fill)

Total range of 35mm frame fill:

28-300mm

My anticipated (with D60)
16-28mm L -or- 17-28mm L ---> 26-45mm Medium Frame Fill
28-70mm L ---> 45-112mm MFF
70-200mm L ---> 112-320mm MFF

Total range of MFF (Medium Frame Fill):

26-320mm

Frank
(the "frame fill" specialist)
John Kim said:
Keith Lommel said:
Take a picture with your film camera and desired lens combination.
Get it developed and crop the print by 1.6.

For example, print a 4x6 and cut it down to 2.5x3.75.

--Ketih
Click to expand...
That works too, although I'd suggest printing as 8x10 and cutting
it down to 5x7 to compare with other 5x7 prints.

The equivalent FL and f/ numbers are still useful though for
comparing different lenses. Most 35mm photographers have already
developed a pretty good idea of what sort of picture a certain
focal length lens at a certain aperture will produce. The tables
provide a way to project that knowledge onto the D60 without having
to make prints.
Click to expand...
 
The answer is that the DoF has NO MEANING except in relationship to an output size and viewing distance, in . Yes, the depth with focusing effects are "frozen" on the film, but this is NOT DoF. DoF is defined in terms of when the image will become "unacceptably" blurred. What is unacceptably blurry is affected by the viewing distance and image size. If you print an image at 4x6 it may seem very sharp, if you blow it up to 20x30 and view it from 2 feet away, it may seem very blurry.

In part this is an argument over the definition of DoF. I have quoted a few references on the subject (do you also want me to scan a book from Kodak?) and yet I have seen nothing from you or anyone else that DoF is not a function of the output size. The other part is that you do not seem to accept that the sharpness required on the negative/sensor varies with output size, because the more you blow up the image the more blur you will see.

There are other factors like film grain and pixel size that affect sharpness, but they are NOT the only reason that you cannot blow up an image without it appearing blurry.

Why don't you look at the STANDARD DoF equations, AND look at how DoF is Defined, and how CoC is defined. I have given you the references, why don't you take a few minutes and read them. If you have a counter reference that defines DoF some other way, then please give it.
Doug Walker said:
Karlg said:
Unfortunately you are not using the "standard" definition of DoF.
Everything I have read on the subject points to the fact that the
DoF also has to factor in the magnification of the image output.
This "factor is known as the CoC -- Have you heard of it? The CoC
is set based on the magnfication of the negative/crop to final
output and viewing distance although it is commonly assumed to be a
8x12 output viewed at about 2 feet. Below are a couple of ready
references on the Internet for those that are interested. Do you
have some other explaination? How would you set the CoC based on
Film and/or sensor size?

In similar threads you keep stating the same thing (wrongly based
on the references below) and have not offered ANY proof or
references other than stating that you are right.

Perhaps you should try READING a bit and see what you think. There
is a chance that you got it wrong.

http://www.tangentsoft.net/fcalc/help/CoC.htm

Quoting from the above site (f/calc) on Circle of Confusion:

Imagine a perfect white point in an empty black room. The point has
no height, and no width. If you focus an optically perfect lens on
that point, it forms a perfect point on the film as well. If,
however, you focus slightly in front of or behind the point, the
point will image on the film as a small blurry circle. If that
circle is small enough, it will still look like a point when
enlarged for printing. The "circle of confusion" is typically
calculated as the largest on-film circle that you see as a point
when you make an 8 × 12 print and view it from a "normal" viewing
distance, typically 2-3 feet. Anything larger is seen as a small
circle, and is therefore perceived as out of focus.

f/Calc calculates the CoC using the "Zeiss formula": d/1730, where
d is the diagonal measure of the film, in millimeters. This formula
yields acceptable values for most uses.

http://www.cs.mtu.edu/ ...
... User-Guide/990/EXPOSURE/EV-depth-of-field.html

http://www.outsight.com/hyperfocal.html

Books by Kodak also say about the same thing, namely that DoF is a
function of the film format size. If LMC54 has some references for
some other definition of DoF, that would be interesting.

Karl
Doug Walker said:
Here's an easier approach:
If I take my 50mm F1.4 lens from my 35mm film camera and put it on
a D60 I will have the field of view of 80mm lens but the depth of
field of a 50m lens. Take whatever lens you have from your f35mm
film camera and simply multiply the focal length by 1.6 for the D60
and you have the new FOV. DOF remains the same.
Click to expand...
Click to expand...
Click to expand...
 
Karlg said:
Doug,

Unfortunately you are not using the "standard" definition of DoF.
Everything I have read on the subject points to the fact that the
DoF also has to factor in the magnification of the image output.
Click to expand...
The link you posted says nothing about DOF and is instead refering to the sharpness as it relates to CoC, the size of the media it was recorded on and the perceived sharpness when enlarged. Maybe you could post a link that directly backs up your assertion.
You have somehow tried to twist this into some definition of changed DOF.

The fact remains that DOF is a function of the lens not the body. A 100mm lens on a D30 will have the same DOF as on a 35mm film body given the same subject to camera distance and same aperture. You cannot change the DOF by enlargement unless you also want to change the laws of optics. Enlargement simply increases the perceived size of the CoC if viewed at the same distance from the image but that's not the same as increasing DOF.
 
I think Karl has it right. Although the DOF as you are defining it does not change when you crop the film the CoC does if viewed at the same enlargment size. Basically even if the DOF is greater, the print will look like it has the same DOF you are accustomed to because you enlarged the "negative" more.

Don't know how to better explain the relationship.

It would be nice if someone proved the point, but I am too lazy to do the math right now.

Ian
 
Hmmm. By your own admission DOF is "frozen" on the film. Hung by your own petard Karl!

I'm afraid you are suffering from the "can't see the forest for the trees" syndrome.

You seem to be stuck on the idea that "enlarging" increases DOF as if we stopped down the lens when in fact "enlarging" simply reduces apparent sharpness. This is not the same as increasing DOF. If we took your assertion to its extreme and enlarged a 35mm negative to the size of a billboard and viewed it closely we would see infinite DOF. This is blatantly false. All we would see is an unsharp image. DOF was not increased by enlargement just the lack of sharpness was increased.
Time to do a little "outside the box" thinking Karl.
 
Doug,
Doug Walker said:
Hmmm. By your own admission DOF is "frozen" on the film. Hung by
your own petard Karl!
Click to expand...
That is not what I said and what is "frozen" is NOT the same as DoF. So apparently it is you inability to read that is causing you problem with reading the references on the subject. Otherwise, why can't you read the references? Or is is because you are so high and mightly all knowing that you don't have to read anything else?

You are the one who one who has learned it WRONG. You are apparently unwilling or incapable or reading anything outside of your very narrow veiw of the world. You are the one that is only expressing YOUR opinion without a shread of evidence.
Doug Walker said:
You seem to be stuck on the idea that "enlarging" increases DOF as
if we stopped down the lens when in fact "enlarging" simply reduces
apparent sharpness. This is not the same as increasing DOF.
Click to expand...
First, enlarging the negative/senor image when viewed from the same distance tends to decrease the DoF -- oh, I forgot you don't understand DoF.
Doug Walker said:
If we
took your assertion to its extreme and enlarged a 35mm negative to
the size of a billboard and viewed it closely we would see infinite
DOF.
Click to expand...
No, it is the other way around. As you enlarge the image, the blur becomes more visible (viewed from the same place) and the original image would have to be sharper, have a smaller CoC, to have the same DoF.
Doug Walker said:
Time to do a little "outside the box" thinking Karl.
Click to expand...
I'm pretty good, outside the box and have over 120 U.S. patents. As an engineer, I learned to READ definitions of technical terms and understand them.

You appear to be incapable of bringing yourself to read anything that even MIGHT not agree with your wrongly held opinion.

Once again, read the references, if you like please give us some counter references. Otherwise please stop confusing the situations with you uninformed opinion.

Karl
 
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