Velocity Reviews > A Modest Proposal

# A Modest Proposal

Dave Martindale
Guest
Posts: n/a

 06-25-2006
"(E-Mail Removed)" <(E-Mail Removed)> writes:

>Somehow
>some manufacturer introduced the 50 mm as the normal lens and now it is
>general
>accepted as the normal lens. I believe that when people refers to 35
>mm
>equivalent is based on 50 mm. So I used that as the reference.

When you see lenses marked with a "35 mm equivalent" focal length, that
calculation depends only on the sensor dimensions. What's considered a
"normal" lens for the focal length has no part in it.

For example, if you have a digital camera whose sensor is 7.2 mm wide,
that is 1/5 the width of a 35 film frame, and so all of the "35
equivalent focal length" specification are 5 times the true focal
length. If the lens is actually 7-21 mm, it will be quoted as "35-105
equivalent". It simply doesn't matter whether you think "normal" is 40
or 43 or 50 or 55 mm on a 35 camera.

>I am referring to using the digital camera to take a picture through
>the
>eyepiece of a telescope or microscope. If the focal length of the
>eyepiece
>is 32 mm and using the maximum zoom of the CP-995 (which is also 32
>mm),
>the image you got is the same as formed by the objective of the
>telescope,
>i.e., the final magnification is 1. If the focal length of the eyepiece
>is 16, the
>final magnification is 2. Of course you have to make sure that the
>images
>match.

That doesn't make sense to me at all. The magnfication of the
*telescope* is an angular magnification, determined by the ratio of the
objective and eyepiece focal lengths. The resulting virtual image is
projected at infinity (or so we assume, anyway).

Then the camera uses its lens to capture that image. On a camera, the
lens focal length primarily determines angle of view, and you generally
want a rough match between the camera angle of view and the eyepiece
image size. That might mean using a 50 mm lens on a full-frame camera,
or a 10 mm lens on a P&S digicam, but the angle of view is the same for
both of these. And it doesn't matter whether the telescope eyepiece is
a 55 mm or a 9 mm.

If you want to calculate the magnification of the digital image you get,
you do need to know the true focal length of the camera lens as well as
the overall magnification of the telescope. But you'll get the same
final magnification using a telescope with a 2000 mm objective and a 40
mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece.

Having the focal length of the camera lens match the focal length of
the eyepiece has no value at all that I'm aware of. They do two very
different jobs, and have entrance and exit pupils in very different
places.

You *would* like to match the exit pupil of the eyepiece with the entrance
pupil of the camera lens, but that doesn't depend on the focal length,
and can vary greatly between zoom lenses that have the same focal
length.

Dave

Pat
Guest
Posts: n/a

 06-26-2006
One would think that the use of 50 mm instead of 43 mm comes from a
combination of 3 things. First, it's just a heck of a lot easier to
use/remember/etc. For example, would you then got to a 150 mm or a 129
mm. Second, you probably want a little extra light around the outside
of the frame, so you round up a bit. Third, and most likely, when you
are designing a lens, there's a lot of math going into the angles, etc.
Back in the early days when the designers were using slide rules, it
was probably easier to stick 50 into a formula than 43. But these are
just guesses.

Pat.
Dave Martindale wrote:
> "(E-Mail Removed)" <(E-Mail Removed)> writes:
>
>
> >Somehow
> >some manufacturer introduced the 50 mm as the normal lens and now it is
> >general
> >accepted as the normal lens. I believe that when people refers to 35
> >mm
> >equivalent is based on 50 mm. So I used that as the reference.

>
> When you see lenses marked with a "35 mm equivalent" focal length, that
> calculation depends only on the sensor dimensions. What's considered a
> "normal" lens for the focal length has no part in it.
>
> For example, if you have a digital camera whose sensor is 7.2 mm wide,
> that is 1/5 the width of a 35 film frame, and so all of the "35
> equivalent focal length" specification are 5 times the true focal
> length. If the lens is actually 7-21 mm, it will be quoted as "35-105
> equivalent". It simply doesn't matter whether you think "normal" is 40
> or 43 or 50 or 55 mm on a 35 camera.
>
> >I am referring to using the digital camera to take a picture through
> >the
> >eyepiece of a telescope or microscope. If the focal length of the
> >eyepiece
> >is 32 mm and using the maximum zoom of the CP-995 (which is also 32
> >mm),
> >the image you got is the same as formed by the objective of the
> >telescope,
> >i.e., the final magnification is 1. If the focal length of the eyepiece
> >is 16, the
> >final magnification is 2. Of course you have to make sure that the
> >images
> >match.

>
> That doesn't make sense to me at all. The magnfication of the
> *telescope* is an angular magnification, determined by the ratio of the
> objective and eyepiece focal lengths. The resulting virtual image is
> projected at infinity (or so we assume, anyway).
>
> Then the camera uses its lens to capture that image. On a camera, the
> lens focal length primarily determines angle of view, and you generally
> want a rough match between the camera angle of view and the eyepiece
> image size. That might mean using a 50 mm lens on a full-frame camera,
> or a 10 mm lens on a P&S digicam, but the angle of view is the same for
> both of these. And it doesn't matter whether the telescope eyepiece is
> a 55 mm or a 9 mm.
>
> If you want to calculate the magnification of the digital image you get,
> you do need to know the true focal length of the camera lens as well as
> the overall magnification of the telescope. But you'll get the same
> final magnification using a telescope with a 2000 mm objective and a 40
> mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece.
>
> Having the focal length of the camera lens match the focal length of
> the eyepiece has no value at all that I'm aware of. They do two very
> different jobs, and have entrance and exit pupils in very different
> places.
>
> You *would* like to match the exit pupil of the eyepiece with the entrance
> pupil of the camera lens, but that doesn't depend on the focal length,
> and can vary greatly between zoom lenses that have the same focal
> length.
>
> Dave

ih@ece.udel.edu
Guest
Posts: n/a

 06-27-2006

Dave Martindale wrote:
> "(E-Mail Removed)" <(E-Mail Removed)> writes:
>

> When you see lenses marked with a "35 mm equivalent" focal length, that
> calculation depends only on the sensor dimensions. What's considered a
> "normal" lens for the focal length has no part in it.
>
> For example, if you have a digital camera whose sensor is 7.2 mm wide,
> that is 1/5 the width of a 35 film frame, and so all of the "35
> equivalent focal length" specification are 5 times the true focal
> length. If the lens is actually 7-21 mm, it will be quoted as "35-105
> equivalent". It simply doesn't matter whether you think "normal" is 40
> or 43 or 50 or 55 mm on a 35 camera.
>

You have used a legitimate method to calculate the
35 mm equivalent focal lengths by linear proportions.
By using linear proportion, all the "form" factors
are automatically included. This method is good enough
for all practical purposes. However, strickly speaking
this method is correct if the aspect ratio is the same
and the normal focal length is equal to the diagonal.
We know the diagonal the 35 mm (double frame) film format
is 43. Its aspect ratio is 3:2 and that of S&P digicam is 4:3.
50 mm is now generally accepted as the normal length. Because '
of these factors, the calculation may produce slight different
values. However, it is of no important concerns. As I said
above this method is good enough for all purposes.

Let us go through the "formal" exercise in calculating
the 35 mm equivalent. The diogonal of the sensor of 7.2 mm
width is 9 mm. The normal focal length of this sensor is
9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333.
So the 35 mm equivalent is 39 - 117 mm based on 50 mm and 33
- 100 based on 43 (which nobody uses now). You can see that
what normal focal lengths to use does matter if you follow
the "formal" procedure. Is this important? probably not.
I don't know which of the three is more right than others.
I probably chosen either 35 - 105 or 39 - 117. The normal
focal length difinition has been used only as a guideline or reference.

> >I am referring to using the digital camera to take a picture through
> >the
> >eyepiece of a telescope or microscope. If the focal length of the
> >eyepiece
> >is 32 mm and using the maximum zoom of the CP-995 (which is also 32
> >mm),
> >the image you got is the same as formed by the objective of the
> >telescope,
> >i.e., the final magnification is 1. If the focal length of the eyepiece
> >is 16, the
> >final magnification is 2. Of course you have to make sure that the
> >images
> >match.

>
> That doesn't make sense to me at all. The magnfication of the
> *telescope* is an angular magnification, determined by the ratio of the
> objective and eyepiece focal lengths. The resulting virtual image is
> projected at infinity (or so we assume, anyway).
>
> Then the camera uses its lens to capture that image. On a camera, the
> lens focal length primarily determines angle of view, and you generally
> want a rough match between the camera angle of view and the eyepiece
> image size. That might mean using a 50 mm lens on a full-frame camera,
> or a 10 mm lens on a P&S digicam, but the angle of view is the same for
> both of these. And it doesn't matter whether the telescope eyepiece is
> a 55 mm or a 9 mm.
>
> If you want to calculate the magnification of the digital image you get,
> you do need to know the true focal length of the camera lens as well as
> the overall magnification of the telescope. But you'll get the same
> final magnification using a telescope with a 2000 mm objective and a 40
> mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece.
>
> Having the focal length of the camera lens match the focal length of
> the eyepiece has no value at all that I'm aware of. They do two very
> different jobs, and have entrance and exit pupils in very different
> places.

I never said that.

>
> You *would* like to match the exit pupil of the eyepiece with the entrance
> pupil of the camera lens, but that doesn't depend on the focal length,
> and can vary greatly between zoom lenses that have the same focal
> length.
>
> Dave

What you said is generally correct and you try to give an overall
description
of the whole operation. In my post I only try to describe the image
"relay"
operation. Two things can be discussed separately. You said the image
is projected at infinite. That means the image is formed right at the
focal plane
of the eyepiece. Then you said that the camera uses its to capture
that image.
The lens must be set at infinite in order to capture an image at
infinite. This
means that the image is formed at the focal plane of the camera. So the

real operation is that the image is formed at focal plane of the
eyepiece and
transferred (or "relay") to the focal plane of the camera. This is
the relay
operation I was talking about. Please see below for a more detailed
explanation. I also generally agree with you about the pupils match.
Since I was only talking about the relay, I don't need to discuss that.

The best way to record the telescope image is to directly
expose it on the sensor of a DSLR. Since we cannot remove
the lense of a P&S digicam, one way is to "relay" the
image to the sensor by using a "relay lense" formed by the
eyspiece and the lens of the camera. This is normally done.
In doing so there is a "final" magnification/demagnification
depending on the focal ratio of the two. It is best to see this
by the following experiment (theoretical or physical). Align
two lenses of equal focal length on their optical axis. If you
put an image at the first lens and an image of equal size
(inverted) is formed at the focal plane of the second. If you
double (or halve) the focal length of the second lens, the image
is doubled (or halved). I did not discuss the many technical
details which you discussed some. The operation the relay lens
is just like this.

Your description of the visual operation
of the telescaope and many technical aspects of the operations
are correct. We are in general agreement except that you
are looking at the big picture and I was talking only about a small
part of it.

Charles S. Ih

Dave Martindale
Guest
Posts: n/a

 06-28-2006
"(E-Mail Removed)" <(E-Mail Removed)> writes:

>You have used a legitimate method to calculate the
>35 mm equivalent focal lengths by linear proportions.
>By using linear proportion, all the "form" factors
>are automatically included. This method is good enough
>for all practical purposes. However, strickly speaking
>this method is correct if the aspect ratio is the same
>and the normal focal length is equal to the diagonal.

No, it doesn't matter what the normal focal length is considered to be,
since that doesn't need to enter the calculations at all.

If the aspect ratios of the two formats are different, then you have to
make it clear whether you are calculating "equivalent" focal lengths by
making the horizontal FOV, the vertical FOV, or the diagonal FOV the
same. There are 3 different answers depending on which you choose to
make equal.

>Let us go through the "formal" exercise in calculating
>the 35 mm equivalent. The diogonal of the sensor of 7.2 mm
>width is 9 mm. The normal focal length of this sensor is
>9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333.
>So the 35 mm equivalent is 39 - 117 mm based on 50 mm and 33
>- 100 based on 43 (which nobody uses now). You can see that
>what normal focal lengths to use does matter if you follow
>the "formal" procedure.

I've never seen that particular procedure used before. You're
converting every focal length into a ratio against the normal lens for
the one format, then multiplying that by the normal lens for the other
format. That does give you a sort of "comparative" focal length, but it
doesn't give you the same field of view unless the two normal FLs are
carefully matched!

Instead, here is the usual method of focal length conversion: scale
focal lengths proportional to sensor dimensions to preserve FOV.
For example, suppose we want to preserve horizontal FOV. The sensor
width is 7.2 mm, while full frame width is 36 mm. So focal lengths of 7
and 21 mm on the digital camera are equivalent to 7 * 36/7.2 = 35 and
21 * 36/7.2 = 105 mm. These are exactly equivalent, in terms of
matching horizontal FOV.

If you want to match diagonal FOV instead, the diagonal of a 7.2 mm wide
1.333 aspect sensor is 9 mm, the diagonal of a 36 mm wide 1.5 aspect
ratio 35 frame is 43.27 mm, and the multiplier to convert from true FL
to "35 equivalent" FL is 4.8 instead of 5.0.

But none of these calculations depend on what is considered "normal" for
either format.

>The best way to record the telescope image is to directly
>expose it on the sensor of a DSLR. Since we cannot remove
>the lense of a P&S digicam, one way is to "relay" the
>image to the sensor by using a "relay lense" formed by the
>eyspiece and the lens of the camera. This is normally done.

>In doing so there is a "final" magnification/demagnification
>depending on the focal ratio of the two. It is best to see this
>by the following experiment (theoretical or physical). Align
>two lenses of equal focal length on their optical axis. If you
>put an image at the first lens and an image of equal size
>(inverted) is formed at the focal plane of the second. If you
>double (or halve) the focal length of the second lens, the image
>is doubled (or halved). I did not discuss the many technical
>details which you discussed some. The operation the relay lens
>is just like this.

Ok. But, in general, it's not desirable to make the FL of the camera
lens equal to the FL of the telescope eyepiece. That *is* true if
you're trying to simulate the effect of removing the lens from the
camera, removing the eyepiece from the telescope, mounting the camera
directly on the telescope, and doing "prime focus" photography.
But if you leave the eyepiece and camera lens in place, you generally do
not want 1:1 relaying. You want to relay most of the image that the
eyepiece "sees", whose real size is set by the field stop inside the
eyepiece, onto the sensor in the camera. For a given eyepiece and field
diameter, this means that the camera lens FL should be proportional to
the size of the camera sensor. Put another way, you want the camera to
capture a particular field of view, no matter whether that's a 14 mm FL
on a small digicam or a 70 mm FL on a full-frame SLR.

So you're right: the magnification of the "relay optics" is determined
by the ratio of their focal lengths. Doing calculations with these, you
need to determine how the objective's FL converts an angle at infinity
into a distance in the primary focal plane, and then use the relay
optics magnification to convert that to a distance on the sensor, which
ultimately becomes a distance on the print given the printing
magnification.

In contrast, I was keeping things in the angular domain as long as
possible. The telescope objective and eyepiece together provide an
angular magnification, and this angle is a certain portion of the FOV of
the camera. The two methods are equivalent and should give the same

Dave

ih@ece.udel.edu
Guest
Posts: n/a

 06-29-2006

Dave Martindale wrote:
> "(E-Mail Removed)" <(E-Mail Removed)> writes:
>
> >You have used a legitimate method to calculate the
> >35 mm equivalent focal lengths by linear proportions.
> >By using linear proportion, all the "form" factors
> >are automatically included. This method is good enough
> >for all practical purposes. However, strickly speaking
> >this method is correct if the aspect ratio is the same
> >and the normal focal length is equal to the diagonal.

>
> No, it doesn't matter what the normal focal length is considered to be,
> since that doesn't need to enter the calculations at all.
>
> If the aspect ratios of the two formats are different, then you have to
> make it clear whether you are calculating "equivalent" focal lengths by
> making the horizontal FOV, the vertical FOV, or the diagonal FOV the
> same. There are 3 different answers depending on which you choose to
> make equal.
>
> >Let us go through the "formal" exercise in calculating
> >the 35 mm equivalent. The diogonal of the sensor of 7.2 mm
> >width is 9 mm. The normal focal length of this sensor is
> >9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333.
> >So the 35 mm equivalent is 39 - 117 mm based on 50 mm and 33
> >- 100 based on 43 (which nobody uses now). You can see that
> >what normal focal lengths to use does matter if you follow
> >the "formal" procedure.

>
> I've never seen that particular procedure used before. You're
> converting every focal length into a ratio against the normal lens for
> the one format, then multiplying that by the normal lens for the other
> format. That does give you a sort of "comparative" focal length, but it
> doesn't give you the same field of view unless the two normal FLs are
> carefully matched!
>
> Instead, here is the usual method of focal length conversion: scale
> focal lengths proportional to sensor dimensions to preserve FOV.
> For example, suppose we want to preserve horizontal FOV. The sensor
> width is 7.2 mm, while full frame width is 36 mm. So focal lengths of 7
> and 21 mm on the digital camera are equivalent to 7 * 36/7.2 = 35 and
> 21 * 36/7.2 = 105 mm. These are exactly equivalent, in terms of
> matching horizontal FOV.
>
> If you want to match diagonal FOV instead, the diagonal of a 7.2 mm wide
> 1.333 aspect sensor is 9 mm, the diagonal of a 36 mm wide 1.5 aspect
> ratio 35 frame is 43.27 mm, and the multiplier to convert from true FL
> to "35 equivalent" FL is 4.8 instead of 5.0.
>
> But none of these calculations depend on what is considered "normal" for
> either format.

The way I described is also OK. Let me give you a real
example, Canon S400. The specifications are the following.

Focal lelngth: 7.4 - 22.2 (35mm film equivalent 36 - 108 mm)
CCD Size: 1/1.8"

The CCD size follows the TV Vidicon convention. The diagonal
of the image on the 1" tube is 16 mm. Thus the diagonal of the
1/1.8" CCD is 16/1.8 = 8.88889.

7.4/8.8889 = .8325
22.2/8.8889 = 2.4975

The true diagonal of the 35 mm film is 43.2666153.
If you multiple this number to the above number, you get

36 - 108 mm exactly as given by the manufacturer.

If you use 50 mm, you get 41 - 125 mm. Use your method,
you get 37.46 - 112.38 mm.

Apparently the manufacturers use the true diaponal to
calculate the equivalent. So the 35 mm equivalent given
by the manufacturers is as good you can get. From the
very beginning I said your method is legitimate. The
slightly different values are due to the different aspect
ratio of the two format and I also mentioned that. If the
aspect ratio is the same, the two methods will give
the same results. Conventionally people try to match
the diagonal as normally given by the manufacturers.
If you are interested in matching horizonal FOV,
your calculation is right because you started by matching
the horizonal width. If you started by matching
the diagonal, you wil get the same answer as given by the
manufacturer. Both method are right and legitimate. It
happens that one method needs the diagonal value and the
other doesn't. There is nothing wrong with this.

>
>
> Ok. But, in general, it's not desirable to make the FL of the camera
> lens equal to the FL of the telescope eyepiece. That *is* true if
> you're trying to simulate the effect of removing the lens from the
> camera, removing the eyepiece from the telescope, mounting the camera
> directly on the telescope, and doing "prime focus" photography.
> But if you leave the eyepiece and camera lens in place, you generally do
> not want 1:1 relaying. You want to relay most of the image that the
>

> So you're right: the magnification of the "relay optics" is determined
> by the ratio of their focal lengths.

That's all I said and never mentioned to match the
focal lengths or making them 1:1. I never intended to discuss all the
technical details.

Charles S. Ih

>

Dave Martindale
Guest
Posts: n/a

 06-29-2006
"(E-Mail Removed)" <(E-Mail Removed)> writes:

>The way I described is also OK. Let me give you a real
>example, Canon S400. The specifications are the following.

>Focal lelngth: 7.4 - 22.2 (35mm film equivalent 36 - 108 mm)
>CCD Size: 1/1.8"

>The CCD size follows the TV Vidicon convention. The diagonal
>of the image on the 1" tube is 16 mm. Thus the diagonal of the
>1/1.8" CCD is 16/1.8 = 8.88889.

>7.4/8.8889 = .8325
>22.2/8.8889 = 2.4975

>The true diagonal of the 35 mm film is 43.2666153.
>If you multiple this number to the above number, you get

>36 - 108 mm exactly as given by the manufacturer.

Right. That's perfectly legitimate, when you use the true diagonal.
Note that you didn't need to assume a "normal" focal length for either
format.

>If you use 50 mm, you get 41 - 125 mm.

Which I argue is the wrong answer, since it doesn't match *any* field of
view - neither horizontal, vertical, or diagonal.

>you get 37.46 - 112.38 mm.

The method I described gives 3 different answers, depending on whether
you decide you care most about horizontal, vertical, or diagonal FOV.
If you pick diagonal, it is exactly the method you used above.

>Both method are right and legitimate. It
>happens that one method needs the diagonal value and the
>other doesn't. There is nothing wrong with this.

But notice that neither method needs to make any assumption about the
normal lens focal length. That was my original point.

Dave