Front tilt angle vs. Image Circle

[ralph_barker]

A question for the math-enriched folks in the crowd: is there an easy

way to calculate the amount of front tilt required for a specified

focus plane angle (measured in relation to a vertical back) and the

amount that the tilt will shift the image circle? For example, let's

assume a 150mm SSXL (386mm image circle, flange distance of 157.9mm)

on an 8x10, focused at infinity, and you want to drop the focus plane

almost to the ground - about 85 degress off the vertically plumb back,

but you want to be sure that the image circle of the lens is

sufficient to still cover the negative. How do you calculate the tilt

angle, and how much does that move the image circle?

[john_cook1]

Hello again, Ralph,

 

I have never learned the math behind this because I have never used math to figure it out. I know that the plane of focus, the plane of the front lensboard and the plane of the groundglass are either all parallel or if not, all intersect at some point (Scheimphfug?).

 

Therefore, if the back is vertical and the plane of focus is nearly level, the plane of the lensboard has to be around 45 degrees. I can "eyeball" that and then fine tune in the groundglass.

 

Then I look into the little holes at the corners of the groundglass to see if I can see a perfect circle of light coming out of the back of the lens. If not, and stopping down doesn't fix it, I go to Plan B.

 

Plan B is to straighten up the lensboard a little, then tip the groundglass back until focus is achieved without vignetting. Of course this will distort the shape of vertical subject material, which may or may not matter.

 

Then again, if the subject is in some tourist attraction I can always purchase a color slide at the refreshment stand...

[leonard_evens]

Ralph,

 

The tilt angle doesn't completely determine the position of the subject plane (plane of exact focus). Remember the Scheimpflug Rule. The subject plane, the lens plane, and the film plane intersect in a line. If you keep the subject plane fixed, you can still vary the front tilt and the position of the film plane.

 

Let phi be the tilt angle and alpha the angle made by the subject plane with the film plane. Also, let v be the perpendicular distance from the (rear nodal point of the) lens to the film plane. (If the camera uses center tilt, for most lenses, v will be pretty close to the bellows extension. For base tilt, you will have to add a correction or measure directly from gg to the center of the lens. For telphoto lenses it will be more complicated.)

 

Then

 

alpha = arctan( v sin(phi)/(v cos(phi) - f))

 

Think of the coverage in terms of angular coverage instead of a disc in the film plane. Thus the lens provides a cone with axis the lens axis and the angle beta made between the axis and the side of the cone being that angle. A formula for beta is

 

beta = arctan(R/f)

 

where R is the radius of the circle of coverage and f is the focal length.

 

When you tilt the lens axis, you tilt that entire cone. The projection on the film plane then becomes an ellipse.

 

The longer semiaxis is

 

(v/2)(tan(beta + phi) + tan(beta - phi))

 

where v is the same quantity as above.

 

The shorter semiaxis involves some very complicated calculations which I haven't pushed through yet, but a lower bound will be

 

v sec(phi) tan(beta)

 

I think for a modest tilt, the exact value isn't going to be much different from the untilted radius of the circle of coverage at lens to film distance v, which is

 

v tan(beta) = vR/f.

 

In fact, for modest tilts, the long semiaxis will also not be too much larger than this. Of course, it would be the shorter semiaxis that would be most important as far as estimating the coverage is concerned. If you use v sec(phi) tan(beta), I think you can be reasonably confident you will be within the angular coverage or pretty close to that.

 

Although the mathematics is entirely elementary, I not above making mistakes, so I stand ready to be corrected.

 

There may also be some optical issues arising from lens design. The usual calculations involving lens tilts abstract the lens to a single point, presumably the rear nodal point, but it could be more complicated than that.

[leonard_evens]

Sorry.

 

Let me try again since something I said previously while literally correct is highly misleading. I admit it confused me briefly when I was responding.

 

If you fix the tilt angle, then you fix a certain parameter which Merklinger calls J. J = f/sin(phi) where phi is the tilt angle. If you drop a plumb from the center of the lens and measure the distance J, you reach a line called the hinge line. That stays fixed as long as the tilt angle is fixed, and vice versa, if you fix J, then you fix the tilt angle. Now as you rack the rear standard back and forth, the plane of exact focus pivots on the hinge line. If you fix the plane of exact focus AND the position of the lens, then you do fix the tilt angle. (If you allow the postion of the lens to vary, which is seldom the case, then you can change the tilt angle without changing the subject plane.)

 

The formula I gave before is correct, but there may be a more useful way to state it. Let u be the distance to the subject plane from the lens along a line through the lens and perpendicular to the film plane. (It would be the lens axis of the untilted lens.) If the subject plane makes angle alpha with the film plane, then

 

alpha = arctan(u/J) = arctan(u sin(phi)/f)

 

The problem with this formula is that u is difficult to estimate. For a subject plane which is almost perpendicular to the film plane, u would be 'almost" infinite. So it is easier to use the lens to film distance v which is related to u by

 

1/u + 1/v = cos(phi)/f

 

Substituting that in the previous formula gives the formula I gave previously.

 

As far as the camera is concerned, you set both u and v by focusing until the subject plane comes into focus.

 

If the subject plane is close to horizontal, then a quick way to set the tilt angle is to measure J and then set phi by the formula

 

sin phi = f/J

 

If instead of concentrating on the angle, you measure the distance you should move the top of the front standard forward, it becomes even simpler. If T is the distance from the top of the standard to the tilt axis, then move the top of the standard

 

fT/J

 

units foward.

 

This has all been discussed in some detail by Merklinger and Wheeler and also in

 

math.northwestern.edu/~len/photos/pages/dof_essay.pdf

[leonard_evens]

I notice I didn't explicitly answer one of your questions.

 

When you tilt the lens, the point where the lens axis intersects the film plane will move up on the gg

 

v sin(phi)

 

units, using the same notation as before. But this won't be the center of the ellipse of coverage, which will be a bit further up. If you really need to know where that is, I can find it for you, but note that for most tilts, sin(phi) is pretty small, so the shift is not very large.

[ralph_barker]

Thanks to both for your responses.

 

John - I like the simplicity of Plan C (buying the slide at the refreshment stand), but doesn't that negate the wisdom of having purchased all the expensive equipment? (lol)

 

Leonard - your explanation is pretty clear, but being a liberal arts type, I have to confess that I remain b(fuddled), a state which a Baptist minister might explain by cos(u sin). ;-)

[leonard_evens]

Let me try it with your specific numbers. Maybe that will help. So suppose the subject plane makes an 85 degree angle with the vertical. Let's also assume the lens is J = 68 inches or about 1727 mm above the subject palne. (That would be a little high for me, but I assume you are taller.) Then

 

sin(phi) = 150/1727 = .086846

 

and hitting the inv sin keys on your scientific calculator should yield

 

phi = 4.98 degrees.

 

Putting alpha = 85 deg and J = 1727 mm in

 

tan(alpha) = u/J will yield

 

11.43 = u/1727 or u ~ 19740 mm or 19.74 meters

 

Now putting this value of u and phi = 4.98 deg in

 

1/u + 1/v = cos(phi)/f

 

yields

 

1/19740 + 1/v = cos(4.98)/150

 

and some algebra gives v ~ 151.73 mm.

 

I assume you mean the image circle has DIAMETER 386 mm so its radius is 193 mm. That means the angle beta I described (which is actually half the angular coverage) is

 

beta = arctan(193/150) ~ 52 degrees.

 

The quantity I described as a rough estimate of the radius of coverage was

 

v sec(phi) tan(beta) ~ 152 x sec(4.98) 193/150 =

 

152 x 1/cos(4.98) x 193/150 ~ 196 mm

 

In other words the coverage would not be materially affected by such a small tilt. Again, although I haven't yet done the exact computation, I doubt it you have to worry about how far up on the gg the region of coverage will be shifted.

 

As I long ago learned, everything becomes clearer if you work out an example, but I'm afraid I got tied up in the theory when originally answering. The reason for that is that I've spent quite a lot of time working out that theory for a variety of reasons, and I got carried away. ;-)

[ralph_barker]

<p>Thanks, Leonard, for walking through the calculations. That certainly clarified matters, and I <i>really</i> appreciate your effort, which has answered my question.</p>

 

<p>I suppose part of my confusion stems from the variable names commonly used in such problems. Walking through the calculations made it clear, however. Now, all we need to do is to come up with a graphical version of Bob Wheeler's Vade Mecum Palm Pilot program for guys like me who never grasped the magic of mathematics. My excuse is having had a rather "unphotogenic" algebra teacher in high school - not a great motivation for staying after class for additional mathematical inspiration. ;-) </p>

[leonard_evens]

How about just using arithmetic?

 

If you use Merklinger's method, you can simplify it as follows. Measure the distance from the lens to the point in the subject plane below it. Divide the focal length by that distance. Multiply the result by 60, and that will give you a reasonable estimate of the tilt in degrees. (The 60 is to convert from radians to degrees. 57.3 would be more accurate. And there are some further approximations I won't go into.)

 

Better yet, calculate that same ratio, but instead of multiplying by 60 do the following. Measure the distance in mm from a reference point on the front board, say its top, to the axis on which it tilts. Multiply that distance by the ratio. Then move the reference point that distance foward. An easy way to do that is to take a ruler with a fairly wide base, place it against the rear standard and line it up with the refererence mark in front. Then move forward the requisite distance. You will need a fairly long ruler for 8 x 10 and most typical lenses you would use. If you use a metric ruler, you can measure the distance to within a mm.

 

After you have the tilt angle established, you can position the subject plane visually by what you see on the gg. It hinges on the line below the lens which stays fixed.

 

The trouble with Merklinger's method as I see it is that often one may not want to fix the hinge line, and it may be difficult to specify just where it should be, hence the first distance may be in question. But in any event it gives you a reasonable starting point, and you can make further adjustments visually.

 

Personally, I use Wheeler's Method which uses instead the ratio of focus spread on the gg between near and far points to the distance on the gg between those points. I get very close that way, but if necessary I make further adjustments visually.

[ralph_barker]

As a "word person" once said, "There are multiple methodologies by which one can divest the feline of its epidural enclosure." (more than one way to skin a cat)

 

Although I have Bob Wheeler's program loaded on my Palm Pilot, I use it mostly for DOF calculations when in the field. Using it for tilt and such in the field is something I've not found personally appealing due to the other tools needed to measure distances and angles. (In addition to not remembering what all of the variable names mean, the extra paraphernalia makes picture taking more like an engineering project.) So, even though it's not very efficient, doing multiple iterations of focus-for-the-far/tilt-for-the-near keeps me thinking in visual terms when I have my head up my darkcloth. ;-)

 

Having a program with a graphical interface (a diagram where you could click on what you want to calculate) that would do the calculations when at home considering whether to buy a particular lens (will it do what I want it to?) would be handy, though.