[MF3D.FORUM:423] Re: fl/30

["Bill Glickman" <bglick@xxxxxxxx>]

David

         I ran your formula side by side with John B formula and
surprisingly enough, using a constant deviation in John's formula of 2.7,
your answers were extremely close every time, regardless of the, near
distance , far distance and focus distance.  So I guess you are both on the
same page...and when evaluating a scene, you both would use the same stereo
base.

    But the question that arose earlier.....that is still not clear to me is
this.... if your formula is accurate than I guess the answer to the
deviation question must be..... the deviation does not change even when the
fl of the camera lens is increased or decreased.  A few posts ago, it was
suggested that when you double the camera  fl, you halve the deviation?  But
on the other side of the coin, when you go to shorter fl camera lenses you
should never go below 2.7 deviation. When using longer fl lenses this logic
appealed to me, hence why I am pursuing this one till the bitter end :-)

Here is Johns formula if you do not have it handy...
http://home.mira.net/~kiewavly/bases.html

       David you seem to be very experienced at this, and I guess the bottom
line is this.... if you have very good success with this formula, than maybe
the deviation should not change at all, regardless of the camera fl lens
being used?   Meaning yours or John B's formula (with constant deviation)
would get us to the same stereo base.

      Any one else have experience at this.... following the math is one
thing, knowing what works is something quite different.  Thanks David...
your input is very helpful to us beginners getting a handle on this... it
seems we are now very close... I really look forward to seeing your paper on
this subject.

Regards
Bill G



----- Original Message -----
From: David Lee <koganlee@xxxxxxxxxxxxx>
To: Medium Format 3D Photography <MF3D.Forum@xxxxxxxxxxxxx>
Sent: Thursday, April 06, 2000 9:36 PM
Subject: [MF3D.FORUM:422] Re: fl/30


>
> > David
> >
> >      Thank you for helping out.... so you are saying, in the example you
> > cited, 78mm/150mm, or approx. 1/2, * 2.7 or 1.35, right?  So if I use
this
> > figure (1.35)  in place of 2.7 deviation in John B's formula, you feel
it
> > will provide the best 3d base, right?
> >
> Rather than confuse both of us by talking about John's formula (which I
have
> not analyzed) I will just show you the formula I use (which I suspect
> amounts to the same thing).
>
> (viewer lens/camera lens)  x    (1/30)    x     [(far pt)(near pt)]/(far
> pt - near pt) = separation
>                irst part                  second part
> third part
>
> The first part is the focal length factor and becomes "1" when you are
using
> a normal lens.
> The second part  is the familiar 1/30.
> The third part increases the separation for scenes which have a far point
> closer than infinity. When the far point is at infinity this factor
reduces
> to the near point distance.
> Thus, when using a normal lens with the far point at infinity the formula
is
> simply:
>     1/30 x (near point distance) = separation
> If you're using a non-normal lens you can simply multiply that factor
times
> 1/30 and remember what the new number is. Thus, when using a 150mm lens
> (long lens) the formula becomes:
>     1/2 x 1/30 x (near point distance) = 1/60 x (near point distance) =
> separation
> When using a 35mm to 40mm lens  (wide angle) it would be:
>     2 x 1/30 x (near point distance) = 1/15 x (near point distance) =
> separation
>
> If we want to factor in the third part of the formula it does not really
> have to be as complex as it looks. You never have to do calculations in
the
> field if you remember some simple principles.
>
> 1) If the far point is a really long way off compared to the near point,
> then use:
>     1/30 x (near point distance) = separation
>
> 2) If the far point is equal to or less than twice the distance of the
near
> point use twice the separation:
>     1/30 x (near point distance) x 2 = separation
>         [Amazingly enough, when the far point is twice the near point the
> third part of the formula equals the far point distance. For example, n.p.
=
> 10' and f.p. = 20'.   (fp x np)/(fp-np) = (20x10)/(20-10)=20.       1/30 x
> 20' = 8" which is the same as 1/30 x 10' x 2 = 8"]
> [I never use more than twice the 1/30 separation because I don't want to
> make a flat subject appear to have full depth.]
>
> 3) If the far point is between infinity and twice the distance of the near
> point then simply use a separation somewhere between 1/30 x near point and
> twice that amount. So if the near point was 10' and the far point was 40'
> you would choose a separation somewhere between 4" and 8". (6" would work
> just fine.)(using the formula works out to 5.3".)
>
> If you're using a non-normal lens then just remember to factor in all of
the
> variables. Let's say you are using a 55mm lens and the scene goes from 3'
to
> 6'.
> The factor for the lens is 78mm/55mm = 1.4
> If you multiply 1.4 x 1/30 you get about 1/24, so for that lens you always
> use 1/24 instead of 1/30.
> If the distance is from 3' to 6' feet then we have twice the depth of 3'
to
> infinity:
>     therefore the separation is 1/24 x 3' x  2 = 3"
>     By the way, this separation for this type of scene with a wide angle
> lens gives an awesome feeling of depth without a bit of eyestrain.
>
> I am going to write a paper on these and other principles for determining
> inter-lens separation and present it in a workshop at the Mesa convention.
> Hope to see you all there.
>
> David Lee
>