Having recently become familiar with large format and movements, it struck me that I hadn't seen anywhere some guidance on what angles are good to use for tilt on a TS lens and under what circumstances. It's hard enough on a LF camera to get the tilt right and I found a calculation that is useful to provide a good starting point. With only the viewfinder on an SLR it's even harder, especially if it's not a 1 series body ;-)

The following picture therefore shows the simple situation of : camera back vertical (therefore optical axis is horizontal) and a landscape, whose plane you want to be in focus, going off horizontally in front of the camera (e.g. football field or roadway as examples of the concept) Here, the little diagram shows that the Scheimpflug point ( really a line coming out of the picture) is beneath the camera at the intersection of the lens plane, ground (focus) plane and sensor plane.

Calculating the tilt is quite simple and the table shows the values (in degrees of tilt) for a 24, 45 and 90mm lens at different heights above the ground (in either inches or mm) where the Scheimpflug condition is met.
Starting at these values for tilt should save some time and frustration with focus evaluation during the set-up of a shot. Even if the camera isn't perfectly horizontal or the groundplane rises or falls a bit as you move away from the camera, these should still be good reference values until the requisite empirical experience is gained. You still need to stop down to increase the depth of the DoF wedge and confirm (angle finder?) that it has worked as desired.

I don't have my TS lens to confirm these specific numbers but the calculations really helped with some of my 4x5 shot setups. However, we can probably trust the basic equation ;-) This should also work well if vertical shift (rise/fall) is used if the shift is very small compared to the height off the ground (usually true unless you're very low).

Note that the main issue is how high off the ground the camera is!


Comments/corrections welcome and I hope this is helpful, particularly for those contemplating purchase of or starting out with such a lens.

Cheers, Andy. I'm experimenting with making one of these myself (MF lens, old enlarger bellows, etc.) so these angles should come in very handy.

KRs
Chris

Chris - good luck with the project!

I should have added in the text as well as the picture that if you focus closer than infinity you should use lens to sensor distance (which will now be greater than the FL) to calculate the tilt and that it will be larger than the infinity value. If someone were to see what extension is needed for the hyperfocal distances at f/11, f/16 etc, I could the calculate "exact" tilt values ...

Andy

Andy,

When using tilt, I don't shift the lens this way. I don't see much point in doing so either.When I tilt, I rotate about an axis running vertically, which I would imagine is the more usual case, especially for landscape.

But even in your case, why does the Scheimpflug point have to be on the ground plane? And where is the plane of focus? This is the thing that matters. If the plane of focus does not form part of your equation, then I don't see what the point of it all is.


Stevve

skefford wrote:
Andy,

When using tilt, I don't shift the lens this way. I don't see much point in doing so either.When I tilt, I rotate about an axis running vertically, which I would imagine is the more usual case, especially for landscape.
Stevve


The diagram is a side view so the optical axis that is tilted would be vertical in a non-TS lens. If you want to keep the sensor vertical to avoid converging verticals, then shift (fall) is a good way to alter the composition to reduce e.g., the amount of sky and increase the amount of foreground in the picture. Not always needed, as you point out.
skefford wrote:

But even in your case, why does the Scheimpflug point have to be on the ground plane? And where is the plane of focus? This is the thing that matters. If the plane of focus does not form part of your equation, then I don't see what the point of it all is.

Stevve


The example is, as explained, specific for the case where the ground plane IS the plane of focus - thus the Scheimpflug line must be in that plane, as must the plane of the lens's vertical axis and the sensor plane - therefore , with the camera vertical, the line is under the camera, on the ground. If you don't use fall, but rather lower the camera to get the foreground to be more prominent, then you'll need more tilt - this table helps you find that angle if you measure (guess) the height, so you'll be close when you start to fine-tune.

In principle, you would only need to look through the viewfinder to compose. Then set the focus to infinity (or hyperfocal or user selected - based on the distance scale n the lens), measure the height, dial in the tilt and set the aperture - ideally you would have it all in focus (of course you would check).

I hope that clarifies what the point is

Andy

AJSJones wrote:
...The example is, as explained, specific for the case where the ground plane IS the plane of focus - ...

I hope that clarifies what the point is

Andy

Yes, I see now. The plane of focus is horizontal, and anything above it is out of focus, depending upon dof of course. I had never thought of shooting like that. I am still trying to imagine the sort of image where it could be used, but it's probably my imagination that's lacking.

Steve

Sorry if the description wasn't sufficiently clear first time around. Fields of flowers, receeding tidepools, beach-scapes, Death Valley Racetrack (classic example), anything where the objects of interest are largely in a plane that is (close to) horizontal is where this approach (using the TS and these tilts) is useful, if simply stopping down doesn't provide enough DoF to do the job.

Andy

Thanks for the great info, Andy!!! Now I can revert to this post to all my friends here that ask about the tilt function. I am going to try it out tomorrow and see for myself.

Jimmy, Hope it "pans" out OK (Mike has my TS lens to play with!)

For completeness, the generic LF workflow,( as Guy Tal pointed out at LL) is
Focus with focus ring on near point
Tilt to get far point in focus
Refocus on near point
Adjust tilt for far point
Repeat until desired objects are in focus as much as possible
Stop down to include more if needed...

Meter with tilt in neutral position

The table is for convenience in a defined situation (I don't memorize it, I have it in the back of my little field notes book) - rather like a table of hyperfocal distances is a useful thing but not a substitute for focusing ;-)

Andy

Couple of things

-Andy, did you do this in excel?
If so could I have a xls file. I could then import it into my palm (My field book...)

-Interesting proceedure from Guy. I found this one on the web:
(source: http://eosdoc.com/manuals/notes/tilt-shift_desc)

A. Select a near point and a far point through which the plane of focus is to pass. The two points should contain sufficient detail to enable determination of sharp focus. Ideally, the two points would as far apart as possible in the viewfinder, but they need not be the same distances from the center of the viewfinder. The near and far points in the procedure below can be interchanged if desired.
B. Choose an initial value for the tilt, and repeat until the sharpness of the near point does not change:
1. Focus on the far point.
2. Slowly decrease the focus distance (i.e., focus closer); then
* If the near point becomes sharper, increase the tilt; or
* If the near point becomes less sharp, decrease the tilt; or,
* If the change in the sharpness of the near point is difficult to determine,
a. Refocus on the far point.
b. Slowly increase the focus distance; then
o If the near point becomes sharper, decrease the tilt; or,
o If the near point becomes less sharp, increase the tilt.

Paul, I couldn't figure out how to attach anything so hope this helps - if not PM me your email and I'll send the .xls

In Excel (or similar) the formula is to paste into column D for example is " =DEGREES(ATAN(B5/C5)) " where column B has the focal lengths and column C has the heights of the optical axis from the ground plane
Both these are in mm. For heights in inches it's =DEGREES(ATAN(B5/(C5*25.4)))

Andy