focusing a self-build 6x17 camera

[flip bastiaan]

I’ve build my own 6x17camera with a canham back, bellows and a Schneider Super Angulon XL 90 mm 5,6. Foccusing is possible with a grounglass, but very inacurate. So I would like to know the exact figures.

What I do know:

-normaly the distance from film to the centre of the lens equals the lens-length. So it would be 90 mm with this lens (with the focus on infinity).
-macro-foccusing (1:1) needs a distance of twice the lens-length. So it would be 180 mm.
-all focusing distances have to ly in between.

From the Schneider datasheet I know that the distance form the center of the lens to the film has to be 103,5 mm (instead of the theoretical 90 mm). Maybe the lens isn’t symetrical....

So my question is: does anybody know a formula or a table to focus my camera. Schneider didn’t answer my mailings till now.....

Thanks for any help at all!

Flip

[Bob Salomon]

Why not just buy the helical focus mount for the lens? It is calibrated for both distance as well as DOF. Both Sschneider and Rodenstock make focusing helicoids for their lenses.

[flip bastiaan]

the mount is not big enough for the XL lens.....

thanks for your respons

Flip

[Emmanuel BIGLER]

The distance between the reference plane and the film plane is standard for cut film holders ; my understanding is that the Canham 6x17 rollfim back is interchangeable with a regular 5"x7" cut film holder therefore the distance between film and holder reference plane is 0.228"+-0.010" = 5.79+-0.025mm (for 5"x7" and 13x18) as listed here


http://home.earthlink.net/~eahoo/filmhold.html

So if you know where the image on a simple ground glass is located, in principle you should know where to place your rollfilm back. Do note however that a precision in the micrometer range is an illusion if you are not in a temperature-controlled environment ;-) a few hundredths of a millimetre are already a challenge with amateur tools ;-)

For an amateur approach I do not see any reason why putting a ground glass at the right place determined mechanically would not work (cut film holder reference plane 5.79 m ahead of the ground glass plane) ; focusing on a far distant object to fix the lens in place if you want to use an helical mount. If you accept to focus on the ground glass you do not need any measurement except the 5.79 offset value between the film plane and the reference plane where you push the film back againts rails or precise stops.
A real professional approach would involve an auto-collimator. Optically this is how you reach the ultimate precision, and even if the ground glass is sandwiched with a Fresnel lens, but an auto-collimator is not an affordable piece of optical equipment except as a military surplus ;-)

[flip bastiaan]

I think I didn't explain my problem very good. The right place for the filmholder or the groundglass was not the problem. I just want to know the distance between the shutterplane (the plane were the lens rests) and the film in relation to the distances'

Something like:

infinity - 103.5 mm (this is the only distance I know for sure)

10 m - ?

8 m - ?

6 m - ?

4 m - ?

2 m - ?

1 m - 180 mm?

This because I can not focus on the groundglass very well and I want to use the camera without replacing the filmback versus the groundglass for each photo. The focusing helicoil they make is not big enough for this lens. And besides I did make a camera with bellows because I want to have the possibilty to use several lenses.

[Bob Salomon]

Flip,

"the mount is not big enough for the XL lens....."

You appear to be looking at te wrong focus mount.

Item # 08-025624 is the Schneider Helical Focus Mount for the 90mm 5.6 XL SA lens.

This is the same helical that is on the Technorama 617S II (special version) and the current SIII model.

Why re-invent the wheel?

[Emmanuel BIGLER]

Flip. What you need is simple

1/ You need to know the exact focal length "f" of your lens. This exact focal length may differ from what is engraved and listed in the manufacturer's catalogue by a few millimetres ;
2/ then you have to apply Newton's formula. If you know where the focal plane is located, and you seem to know this quite precisely, the additional extension "e" needed to move the lens with respect to the focal plane (object at infinity) is related to the object distance "D" measured from the entrance principal plane H by :
D = f + f*f/e. (Newton's formula)
or : e = f*f/(D-f)

The total distance "D_t" between the object and film plane is given by :
D_t = f*f/e + f + HH' +f + e
where HH' is the spacing between principal planes.
Traditional engravings refer to the total distance D_t, not D. The relation giving e as a function of D_t is more complicated, it is simply one of the roots of a second degree equation ; but simply use the approximate value : D instead as a starting point.

"e" is the additional displacement required to focus on an object located at a finite distance. For example e=0 in infinity-focus,
for D=1 metre , e = (90*90)/(1000-90) = 8.9 mm
for D=2 metres, e = (90*90)/(2000-90) = 4.2 mm
for D=5 metres, e = (90*90)/(5000-90) = 1.65 mm

This formula is valid for any lens even a telephoto or a very asymmetrical lens, but you need to know where the entrance principal plane is located.
The position of the entrance principal plane 'H' with respect to the first lens vertex and the distance HH' are given in your optical manufacturer's data sheet.
With an helical mount the lens travel is proportional to the rotation angle of the helical. the additional travel "e" is simply proportional to the rotation angle of the helical measured with respect to the initial position in infinity-focus setting. You just have to write down a non linear scale D_t or D as a function of "e" for each position of the helical. Or with a simple translation device for focusing, simply write a non linear scale for various focusing positions.
So your are all set now.

[flip bastiaan]

Thanks Emmanuel,

Schneider gives a value of 103,5 mm (focus at infinity) so I have to add 13,5 mm to all the values I get using this formula. It is clear although not functioning for values smaller than one meter because the distance between the two planes is neglected in this formula when using the approximate value.

Maybe it is just the mathematics I'm having difficultys with....

[Emmanuel BIGLER]

See detailed specs here.


http://www.schneideroptics.com/photography/large_format_lenses/super-angulon/pdf/super-angulon_xl_56_90.pdf

103.5 is not the focal length, it is the flange focal distance i.e. the distance between the back of the shutter (= front of the lensboard) to the focal plane. The focal length is 90.69mm. At least at room temperature ;-);-) principal plane separation is HH'=36.23 mm. H' is located exactly one focal length = 90.69mm ahead from the focal plane. So the lensboard is approximately located in the middle of the lens but the distance HH' is non negligible !!
90.69 is the focal length, it is the exact value tobe inserted into Newton's formula. Now you are really all set. It is not a question of mathematics it is a question of understanding the datasheet of an optics and understanding the meaning of principal planes and focal distances in a thick comound lens, I agree this is cryptic and not only because it is sometimes written in German ;-)