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mortensen
26-Oct-2009, 15:37
Hi everyone

Because of my lack of patience searching this fantastic forum (sorry), I dare asking this - probably very simple - question:

What is the relationship between focal lenght and the amount of tilt/swing needed?

I am quite new to LF and I use a chamonix 45n-1 with a Rodenstock 90mm and a Caltar 210mm. I find, that I need to swing the lens more with the 210 than the 90 in order to obtain the 'same' bending of the focus plane. We are probably in league of 'basics I ought to know', but if any of you know of an old thread or have the time to explain, I would greatly appreciate it.

Thanks :)

/lars

Jim Michael
26-Oct-2009, 16:18
Well, for a 1:1 reproduction of an object on the ground glass you'll have the lens about 2x the focal length of the lens. An object at great distance ("infinity") will be focussed at the lens' focal length. That should give you an intuitive feel for what you're observing.

Gem Singer
26-Oct-2009, 16:35
You answered your own question.

It seems that you have already discovered that you need to use a greater degree of swing (tilt) with a longer focal length (210) lens than you need to use with a shorter focal length (90) lens in order to extend the plane of focus.

Wide angle lenses need less swing or tilt in order to achieve near-far focusing.

Swing is merely the horizontal version of tilt.

mortensen
26-Oct-2009, 23:37
thanks, both of you. Your explanation, Jim, really gives the point - it must be an almost exponential relationship.

What confused me, was the fact that I only need to tilt the 90 a couple of degrees to give some amount of near-far-focusing. Anyway, I'd better trust my eyes...

/lars

Jeff Keller
26-Oct-2009, 23:47
Adding to Jim's comment, think of the Scheimpflug principle. If the only change is greater focal length, there has to be greater tilt to keep the intersection of the three planes.
http://en.wikipedia.org/wiki/Scheimpflug_principle
Jeff Keller

Jeff Conrad
27-Oct-2009, 00:39
The dependence is linear with focal length. When the lens is tilted, the plane of focus rotates about an axis below the lens (for simplicity, let's assume the camera back is vertical) as the focus is changed. If f is the focal length and θ is the tilt, the distance of that axis from the lens is


J = f/sin θ .

This is shown in the diagrams in the linked article.

In practice, I've never found the calculation useful in the field, except perhaps as a sanity check: with a tripod height of 1.5 m and 90 mm lens, a 3.4 tilt puts the axis on the ground. In most cases, at least for what I do, the axis should be at or below ground. But I find it easier to set tilt visually, using Howard Bond's Focus/Check method (a transcript is available here (http://www.largeformatphotography.info/articles/bond-checklist.html#fc)).

Setting the tilt is easy; deciding where the plane of focus should be is the challenging part.

mortensen
27-Oct-2009, 13:09
you're right, 2xJeff and, well, my math is seriously rusty (as pointed out, haha)
... will check out the howard bond link shortly.

Lynn Jones
27-Oct-2009, 13:28
You may have to use "Swing Opposite" in order to get the necessary correction.

Lynn

Jeff Conrad
28-Oct-2009, 01:39
As Leonard Evens reminded me, Bond describes the plane of focus as moving parallel to itself when focus is adjusted, which isn't quite right, but this doesn't affect the usability of the method (which you may or may not like). Another must read, if you haven't already found it, is QT's How to focus the view camera (http://www.largeformatphotography.info/how-to-focus.html).

Leonard Evens
28-Oct-2009, 08:18
It can get a little complicated.

If you know just where you want to put the exact subject plane---which, as Jeff points ou t, may be questionable--- then that determines where it crosses below the lens, i.e., the parameter J. That then determines the tilt angle through

sin theta = f/J

If the tilt angle is small, then theta is approximately equal to sin theta, when the former is measured in radians. To get degrees, you need to multiply by 57.3 ( but 60 is usually close enough).

I sometimes use this information in order to determine an initial guess for the tilt angle. (Actually, I use the shift in mm of the top of the lens board from vertical instead, but that involves another formula.) But often you can skip that step and just choose some value like 5 degrees for the initial guess.

You can then proceed by using a near point/far point method based on two such points which you know to be in the exact subject plane to refine tilt until you get the desired exact subject plane.

One more point may be of interest. Where you put the exact subject plane also depends on the focal length. For a shorter focal length it will tend to be closer to being parallel to the film plane. So, for example, it may cross the horizontal line of sight further in than would be the case for a longer focal length. Also, you may want to let it cross the line below the lens further down, which may mean a larger J, which might partially cancel out the decrease in theta due to smaller f.

Finally, with smaller focal length, you generally get larger depth of field about the desired exact subject plane, which may make it harder to decide vidually just where you have placed it.