Re: Re:Superimposing 2 images taken with 2 FL lenses

----- Original Message -----
From: "John Fletcher" <jfl10593@bigpond.net.au>

> Dude I told you how to do it...
>
> 50% of 127mm = 63.5
>
> 31.75/50 = tan(angle) = 0.635
>
> Width/1.27 = Distance
>
> If subject is 2m wide it must be 78 cm from the camera to occupy 50% of
> 5"

If the camera side's half cone of light has tangent = 31.75/50 then the
same angle on the subject side should be 1m/distance, then the distance
subject to pinhole lens for a subject 2m wide should be 50/31.75 meters or
1.5748meters, about twice what you say. Nevertheless, I don't think that
answers Colin's question completely. I'll give it a try bellow.

----- Original Message -----
From: "Colin Talcroft" <ctalcroft@yahoo.com>

> Sorry for the slow response, but thanks for the
> thoughts about superimposing images. I understand the
> way that perspective will alter the images. I don't
> mind that. I'm not looking for a perfect fit, just a
> reasonable correspondence. I'm just looking for a rule
> of thumb. Is it possible to say, for example, that a
> subject occupying 50% of the width of a 4X5 negative
> using a 50mm pinhole at a given distance from the film
> and at X feet will also occupy 50% of the width of the
> negative using a 210mm lens on a 4x5 view camera if
> the film-to-subject distance is some multiple of X, or
> something like that? Seems like it should be a fairly
> simple relationship.

> I just don't know what it is.

Neither do I, but after drawing couple of triangles and writing 2
relationships between their sides I can tell you that an approximation
answer to your hypothetical case above is given by multiplying the distance
from the subject being photographed to the pinhole lens by the ratio or
factor given dividing the glass lens focal lens by the pinhole focal length,
like this:

Factor = 210 / 50 = 4.2

So assuming that when you took the pinhole camera exposure, the pinhole lens
was 1000mm from the subject, to get the same size subject on film when using
your 210mm glass lens, that glass lens should be placed:

Distance glass lens to subject = distance subject to pinhole lens * Factor

Distance glass lens to subject = 1000mm * 4.2

Distance glass lens to subject = 4200mm or 4.2 meters

The answer should stop right here, but.....

For a more exact answer and also for more on the above approximation, read
bellow:

as per http://photo.net/photo/optics/lensFAQ

The lens equation is:

1/Si + 1/So = 1/f See above link for meaning of Si, So and f

Using that equation and a little algebra we end up with the following:

If:

Y = distance from subject to glass lens (lens focal length 210mm in your
example)
X = distance from subject to pinhole lens (lens focal length 50mm in your
example)

Then:

Y = ( 210 * X / 50 ) + 210 (a)

and

X = ( 50 * Y / 210 ) - 50 (b)

In general, if GFL is the glass lens focal length and PFL is the pinhole
lens focal length:

Y = ( GFL * X / PFL ) + GFL (c)

X = ( PFL * Y / GFL ) - PFL (d)

For instance, if your subject was 1000mm from your 50mm pinhole lens camera
when photographed ( X = 1000mm), when making the 210mm glass lens exposure
the distance from subject to lens should be:

Y = ( 210 * 1000 / 50 ) + 210

Y = 4410mm or 4.41 meters

A more simple approximation could be made if we forget that a glass lens
"actual" focal length distance increases as we focus on subjects closer to
the lens than infinity, in such case:

Y = X * (210 / 50) (e)

X = Y * (50 / 210) (f)

Again, in general:

Y = X * (GFL / PFL) (g)

X = Y * (PFL / GFL) (h)

Using the same example above, subject 1000mm from your 50mm pinhole lens
camera when photographed ( X = 1000mm), when making the 210mm glass lens
exposure the distance from subject to lens should be:

Y = 1000 * (210 / 50)

Y = 4200mm or 4.2 meters

Bear in mind that formulas (a) to (d) must have all values expressed in the
same units of measure. In the case of the approximation equations (e) to
(h), both focal length have to be in the same units, but the Y and X
distances can be in the same or any other units of measure.

If you were looking for the simple relationship, all the above should do it,
whether exact or approximated results are wanted, it doesn't get simpler
than that, I am pretty sure.

Corrections welcomed.

Guillermo
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Received on Wed Jan 3 22:20:06 2007