[MF3D.FORUM:422] Re: fl/30

["David Lee" <koganlee@xxxxxxxxxxxxx>]

> 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