A technical question on MM calculations

I have a few lenses that overlap in mm settings, however, the images from them
are not the same size. I would accept slight differences in size, but one lens
produces images at least 50% bigger than the other lenses!

I have a big zoom lens that also does 300mm, although I couldn't set it to 300,
it either says 270 or 320... but anyway, it is close to one of the lenses, but
not the other

I believe my Nikkor 70-300 is actually 100-375!!! The other lenses make smaller
images, one is a Nikkor 18-70 and is much smaller at 70.

I compared shots from a 28-300, an 18-70, and a 50-500, with the 70-300, and the
70-300 is out in every case. (I set the lens to either 70 or 300 to compare with
similar settings.)

weird...

Anyway, I need a way to calculate the mm of an 'unknown' lens. There must be a
way to take a picture of a ruler placed at a set distance, and calculate by the
size on the negative. (ok its digital but the same thing applies!)

I'm sure I saw some formulas for this type of thing but maybe someone here has
them?

Thanks!

wrote:

Anyway, I need a way to calculate the mm of an 'unknown' lens. There must be a
way to take a picture of a ruler placed at a set distance, and calculate by the
size on the negative. (ok its digital but the same thing applies!)

I'm sure I saw some formulas for this type of thing but maybe someone here has
them?

1 ) Position the ruler and adjust the focus of the camera for
a 5:1 reduction.

2) Move the ruler further away and note the amount that you have
moved the ruler

3) Refocus in such a way that the front of the lens is in exactly
the same position as before. (this is easier to do with a view
camera with back focussing movement, but as long as you can somehow
make sure that the front of the lens is in the same position it will
work.)

4) Determine the reduction at the new distance

5) Subtract 5 (the original reduction) from the new reduction.
The distance that you moved the ruler is that many times
the focal length of the lens.

Another method of determining focal length is to focus the camera
on a distant narrow object. Rotate the camera a carefully measured
number of degrees, making sure that the object appears in the field
of view both times. Measure the distance the image moves on the groundglass.

Then solve:

(d/2)/f = tan (a/2)

where d = the distance between the two marks on the groundglass
f= the focal length of the lens
a= the angle that you rotated the camera.

I hope this helps.

Peter.
--

On Tue, 18 Nov 2008 03:39:02 +0000 (UTC), Peter Irwin wrote:

wrote:

Anyway, I need a way to calculate the mm of an 'unknown' lens. There must be a
way to take a picture of a ruler placed at a set distance, and calculate by the
size on the negative. (ok its digital but the same thing applies!)

I'm sure I saw some formulas for this type of thing but maybe someone here has
them?

1 ) Position the ruler and adjust the focus of the camera for
a 5:1 reduction.

Do you mean that I want the image on the negative to be 5 times smaller than the
object? I'm not sure what you mean by 'reduction'.

2) Move the ruler further away and note the amount that you have
moved the ruler

3) Refocus in such a way that the front of the lens is in exactly
the same position as before. (this is easier to do with a view
camera with back focussing movement, but as long as you can somehow
make sure that the front of the lens is in the same position it will
work.)

When calculating focal length is the front of the lens always used? I know that
telephoto lenses don't follow any rules, the point of measure could be anywhere.

4) Determine the reduction at the new distance

5) Subtract 5 (the original reduction) from the new reduction.
The distance that you moved the ruler is that many times
the focal length of the lens.

Another method of determining focal length is to focus the camera
on a distant narrow object. Rotate the camera a carefully measured
number of degrees, making sure that the object appears in the field
of view both times. Measure the distance the image moves on the groundglass.

Then solve:

(d/2)/f = tan (a/2)

where d = the distance between the two marks on the groundglass
f= the focal length of the lens
a= the angle that you rotated the camera.

I hope this helps.

Peter.

I found a site that mentioned that the ratio of lens magnification is the same
for distance as it is for height... or Di/Do = Hi/Ho for object and image. I
figure that if I put a meter stick at a distance where it fills the sensor,
(24mm) then I will have a ratio of 24/1000, then I just need to measure the
distances between lens center point and film plane and object, and once the
ratio is verified, then using 1/f = 1/Do + 1/Di , I can calculate the f in mm.

What you think?

wrote:

Anyway, I need a way to calculate the mm of an 'unknown' lens. There must be a
way to take a picture of a ruler placed at a set distance, and calculate by the
size on the negative. (ok its digital but the same thing applies!)


I think it's figured at infinity. Closeups are likely some other focal
length (think extension tubes). Maybe tape measure a building for the tests.

Also there is a lot of rounding of numbers involved.

--
Paul Furman
www.edgehill.net
www.baynatives.com

all google groups messages filtered due to spam

sorry about the header, my mistake, filter mixups...

wrote:

Do you mean that I want the image on the negative to be 5 times smaller than the
object? I'm not sure what you mean by 'reduction'.

Yes. - One fifth size.


When calculating focal length is the front of the lens always used? I know that
telephoto lenses don't follow any rules, the point of measure could be anywhere.

When photographing one fifth life size, the object is five focal
lengths in front of the anterior focal point. In this method you
do not need to know where the anterior focal point is, only that
its position is fixed relative to the lens. But it is relative to
the lens and not the film plane - so you need to move the film plane
and not the lens when you change focus. This is easy on a view camera
which allows back focussing, but a bit tricky in practice with a
typical 35mm camera.


a
I found a site that mentioned that the ratio of lens magnification is the same
for distance as it is for height.

That is true.

I
figure that if I put a meter stick at a distance where it fills the sensor,
(24mm) then I will have a ratio of 24/1000, then I just need to measure the
distances between lens center point and film plane and object, and once the
ratio is verified, then using 1/f = 1/Do + 1/Di , I can calculate the f in mm.

What you think?

If you serious about obtaining a more accurate figure than
the one marked on the front of your lens barrel then you can't
ignore lens thickness (the distance between the front and
rear principal planes) as your method seems to do.

It is easier to pick an object at infinity and rotate the
camera. You can improvise a big protractor to make an accurate
angle adjustment fairly easily.

Peter.
--

On Tue, 18 Nov 2008 06:34:22 +0000 (UTC), Peter Irwin wrote:

wrote:

Do you mean that I want the image on the negative to be 5 times smaller than the
object? I'm not sure what you mean by 'reduction'.

Yes. - One fifth size.


When calculating focal length is the front of the lens always used? I know that
telephoto lenses don't follow any rules, the point of measure could be anywhere.

When photographing one fifth life size, the object is five focal
lengths in front of the anterior focal point. In this method you
do not need to know where the anterior focal point is, only that
its position is fixed relative to the lens. But it is relative to
the lens and not the film plane - so you need to move the film plane
and not the lens when you change focus. This is easy on a view camera
which allows back focussing, but a bit tricky in practice with a
typical 35mm camera.


a
I found a site that mentioned that the ratio of lens magnification is the same
for distance as it is for height.

That is true.

I
figure that if I put a meter stick at a distance where it fills the sensor,
(24mm) then I will have a ratio of 24/1000, then I just need to measure the
distances between lens center point and film plane and object, and once the
ratio is verified, then using 1/f = 1/Do + 1/Di , I can calculate the f in mm.

What you think?

If you serious about obtaining a more accurate figure than
the one marked on the front of your lens barrel then you can't
ignore lens thickness (the distance between the front and
rear principal planes) as your method seems to do.

It is easier to pick an object at infinity and rotate the
camera. You can improvise a big protractor to make an accurate
angle adjustment fairly easily.

Peter.

OK I'll try that then, I didn't get too far yet... Thanks for all the info!

On Mon, 17 Nov 2008 22:00:29 -0500, wrote:


I have a few lenses that overlap in mm settings, however, the images from them
are not the same size.

I'll post an example on alt.binaries.photos.original

3 pictures taken with 3 different lenses, all set to 70mm, on a tripod 57" from
film plane to object.

What a difference!

Must be a conspiracy...

wrote:

3 pictures taken with 3 different lenses, all set to 70mm, on a tripod 57"
from
film plane to object.

What a difference!

Must be a conspiracy...

Yep. It's a conspiracy to make smaller and easier to use lenses.

Most modern SLR lenses change focal length with focus distance.
The exception are lenses which focus as a unit by extension. All
IF lenses fall into the "change" category, as do most zooms, as
do other prime lenses with "floating" elements to improve close
focus performance.

Even worse, the magnitude of the change isn't constant from lens
to lens, even within a manufacturer's line.

There's also some "rounding" or "slop" in the advertised focal
lengths. For example, Nikon and Leica's 5cm/50mm rangefinder
lenses were actually 51.6mm in focal length. A 5% variation
is not unusual.

--
Michael Benveniste -- (Clarification required)
Legalize Updoc.

Michael Benveniste wrote:

Most modern SLR lenses change focal length with focus distance.
The exception are lenses which focus as a unit by extension. All
IF lenses fall into the "change" category, as do most zooms, as
do other prime lenses with "floating" elements to improve close
focus performance.

Even worse, the magnitude of the change isn't constant from lens
to lens, even within a manufacturer's line.

Correct me if I've got this wrong, but it seems to me that if
a lens is designed to focus by adjusting its focal length
instead of moving, it must be one diopter stronger to focus
at 1 metre than to focus at infinity. Thus a 50mm (20 diopter)
lens would have to become a 47.62 (21 diopter) lens in order to
focus at 1 metre without moving.


There's also some "rounding" or "slop" in the advertised focal
lengths. For example, Nikon and Leica's 5cm/50mm rangefinder
lenses were actually 51.6mm in focal length. A 5% variation
is not unusual.

It seems to me that if you want to measure the focal length of
a lens to have more accurate knowledge than a within 5% lens
marking will give you - you have to really do the measurements
right.


Peter.
--