Re: [photo-3d] RE: Depth ranges

["Dennis M. Hanser" <dmhanser@xxxxxxxxxxx>]

"David W. Kesner" wrote:

> Is there anyone out there that can do the math and tell me
> what the maximum near and farpoints are for this setup
> assuming a 1.2mm mofd?

I'm no math major, so forgive the somewhat crude use of fractions and
decimals in these equations.

You could try a variation of the following formula used for finding
base  separation for macro subjects;

b=((p*1/m)d1)/d2

b=base
p=deviation 
m=magnification factor in decimal
d1=film plane to near point in subject
d2=distance from near point to far point

for your question, you need to solve for d2

d2=((p*1/m)d1)/b

example

p=1.2

m=1  (life size)

d1=420mm (a 105mm lens would be equidistant from the film and the
subject at 1:1 magnification. To get life size images on film a simple
lens is placed twice it's focal length from the film and will be twice
it's focal length from the subject. Thus 420mm from film to near point.)

b=11mm(your stated fixed base)

1/m cancels as 1/1 in this case

p*d1 converts to 1.2*420 and equals 504
504 is then divided by your stated base of 11 to make d2 equal to 45.8 
So, with a 105mm lens and the bellows racked out to give 1:1(life size)
on film, with the near point at the focus point and the far point at
45.8mm behind the near point, you would get 1.2mm deviation on film.
Now granted, you would want to also consider the hyperfocal and maximum
depth of field equations before focusing on the near point, but I think
you can get the idea for maximum distance from near point to far point
using this equation. 
I used 1:1 because it simplifies the example.
To solve for other magnifications you need to also know how to find the
total distance from film to near point.

The second equation will find the lens to subject distance, and then add
that to the lens to film distance to get total distance from film to
subject(near point)

1/f - 1/n1 = 1/n2 (these should be d1&d2 but substitute n to avoid
confusion with the first equation in which we are using d1 and d2.)

1 over the focal length minus 1 over the film to lens distance will be
equal to 1 over the distance from lens to subject(in this case near
point)

So-- one more example;

Find the near point to far point distance that will give 1.2mm deviation
on film for a 1/2 life size image using a 105mm lens.

First find the total distance from film plane to object(near point)

To get 1/2 life size on film a lens needs to extended 50% of it's focal
length beyond infinity focus, so 105+52.5mm = 157.5 from film to lens.

1/f-1/n1=n2

1/105 - 1/157.5 = 52.5/16537.5 ---- 315mm from lens to subject
157.5 from film to lens and 315 from lens to subject = 472.5film to near
point

now we plug that into the first equation as d1

1/m converts to 1/.5(magnification in decimal)and =2
1.2*2=2.4
2.4*472.5=1134
1134/11=103.09

near point to far point distance of 103mm will give 1.2mm of deviation
on film with a 105mm lens and subject at 1/2 life size on film. 

You can probably can round a bunch of these numbers and get close enuff
results, but I don't have enuff of a feel for this stuff to know what
can be gotten away with. Anyhow hope this (finally) answers your
original question.
Bear in mind that you do not necessarily want to use the MAXIMUM and
also need to consider depth of field and hyperfocal to optimize results.
It will often be necessary to block detail that falls outside the
maximum. This can be accomplished with plain background materials, far
enuff back to be out of focus. If flash is used, care must be taken to
keep shadows from falling on the background.

Dennis Hanser
member Detroit Stereographic Society
dmhanser@xxxxxxxxxxx


 

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