From this article, it seems like the only problem with a purely optical solution (taking lens at flange distance >> relay lens at flange distance >> camera sensor) is the chromatic aberration that must be overcome by CCD sensors.
How terrible would the CA look if you desaturated to B&W?

Of course that kind of sucks, because you'd just have a fairly expensive B&W HD camera. Might as well shoot 16mm B&W.

CA is most noticeable as a colored ring around bright objects next to a dark background, particularly if out of focus.

If you desaturated to black and white, you wouldn't see that distracting color ring except as a slightly less bright area around bright objects.

However, CA also reduces sharpness of in-focus areas. For most lenses, this is far more trivial than the OOFO effect.

But using non-telecentric lenses (shorter than a certain focal length) with 3-chip cameras is a bad idea, even on black and white (though if it's black and white, why bother with 3-chip?). Because the filters that divide the light into the various colors are very, very angle-dependent, you'll actually see bizarre color effects toward the edge of the frame (wavelengths that would normally be reflected by, say, the red filter are transmitted at steeper angles, so that light ends up registering as blue/green and not red). This color shift would affect tonalities even in black and white, and even objects with a constant color would show up as brighter or darker towards the edge of the frame. That shifting would be highly wavelength dependent.

I don't know - maybe it would actually look kind of neat. Anyone with a removable lens 3-chip camera want to try it out? For best effect, use a well-corrected, (super-) wide angle lens at the proper flange focal length, but one designed for use without a beam-splitting prism - removable lenses for digital SLRs or even 35mm stills would be a good choice. It might look very strange if still in color, to maximize the weirdness.

(I'd try it myself, but my camera was recently stolen.)

Now, the real 'loss' is the fact that it only operates at f/3.5ish on the object side - so opening up your taking lens beyond that won't give you more light (at least not in your image; it'll probably give you a little more flare and image defects).
Ryan,

How do you know that it only operates at ƒ/3.5 on the object side? I've searched for info on the object side aperture and can't find any.
After reviewing some of my tests I don't see much of an exposure change for any stops open more than T2.8. There is a definite difference in exposure between T4 and T2.8, (about as much as to be expected) so I would guess that T2.8 is around the max on the object side.

Hi Tim -

Sorry for the confusion - I was thinking of the 35mm to 2/3" chip adapter mentioned in this thread (http://www.cinematography.net/Pages%20GB/%5Bcml-hdtv%5D%20Cine%20lenses%20on%2023%20HD%20CCD%20cameras.htm). Helmut Lenhof says that f/1.4 on the image side corresponds to f/3.5 on the taking side for that adapter (that corresponds to a magnification ratio of 2.7 and an image side speed of f/1.3).

For the Panasonic, I should have looked at the magnification ratio: I think a 1/3" chip corresponds to a 6mm sensor diagonal, and a 16mm film frame corresponds to 12mm. By that token, a taking lens of f/2.8 would correspond to relay optics at f/1.4 (f/# of taking optics divided by magnification ratio), which is within the bounds of reason.

The point stands, though, that if you want to get very shallow DOF on a 1/3" chip camera (without scattering, that is, with a purely optical solution), the relay optics are going to be ridiculously fast, perhaps to the point of impossible.

I presume that a 16mm prime at f/2.8 isn't that shallow a DOF? Take nothing away from the adapter, I think it's a great way to leverage 16mm optics and get cinema-quality footage out of the JVC. As I may have mentioned earlier, there's lots of reasons to use cinema lenses beyond DOF.

And your footage was excellent, thanks for posting.

Ryan,

I've been doing the research to explain my test results. Maybe you can tell me if this makes sense to you?
The JVC adapter has an exit aperture of f/1.4 just like the CLA35. However, as I mentioned earlier I wasn't able to see any exposure or DoF differences after I opened past T2.8 on the attached lens.

It looks like this is the Lagrange Invariant at work.

Taking the following into consideration for the HZ-CA13U I calculated that the maximum usable aperture on any PL lens should be f/2.73, which seems close enough to be consistent with my practical tests.

16mm Horizontal field size: 9.35mm
1/3" Horizontal field size: 4.8mm
Max exit aperture: f/1.4

1.4(9.35/4.8)=2.73

If I plug the numbers in for Super35mm and 2/3" in the case of the CLA35 I get:
1.4(21/9.6)=3.06
Referring to the true F-stop table (http://www.panavision.co.nz/main/kbase/reference/tblefstops.asp), 3.06 would give the CLA35 a maximum aperture of f/2.8 ¼ or rounded to f/3.5.

Someone mentioned the idea of creating a 35mm version for 1/3". The problem is that you would only be able to open to f/5.6 ¼ on the PL lens.
1.4(21/4.8)=6.125 or f5.6 ¼

The point stands, though, that if you want to get very shallow DOF on a 1/3" chip camera (without scattering, that is, with a purely optical solution), the relay optics are going to be ridiculously fast, perhaps to the point of impossible.
I guess until we can change the laws of physics there will be no "perfect" 35mm purely optical solution!


BTW, I've been conducting more tests and the CA almost completely disappeared once the Cooke S4 lenses were mounted. The lens they had for the Sundance test was a Zeiss Super speed. The Cookes looked considerably better for alignment than even the Zeiss Ultra Primes. Both S4 and Ultra Primes were in a class of their own above the 'super' speeds and standard speed Zeiss lense. However, the Cookes do breath quite a bit, and are big and heavy.

Tim,

Right on, I agree completely with your findings. Sounds like an optical-only from 35mm to 1/3" chip is pretty much out of the question.

It also sounds like the Cookes were better matched to the adapter than the Zeiss - that's not a direct knock on the Zeiss lens, of course - the adapter probably has some built-in correction that better matches the Cooke (or even an aberration of the opposite sign, which is basically the same thing).

Also, I don't mind lens breathing at all. I play around with large format optics all the time, and they're inherently pretty breath-y... I kind of like the look, though it definitely calls attention to any focus pulls.

And just out of curiosity, how did you figure out the exit aperture size? Was it in the documentation, or did you do some crazy measurement? Also, it seems the optical invariant could be a help, not a hindrance, at least in one regard: did you notice if the illumination level was fairly high? It seems to me that the exit aperture should be indicative of the exposure level... so you would basically get a F/2.8 image with a F/1.4 illumination... or am I missing something?


Cheers,


Ryan

And just out of curiosity, how did you figure out the exit aperture size? Was it in the documentation, or did you do some crazy measurement?
f/1.4 is indicated on the instruction sheet. It is the only documentation I have on the adapter.


Also, it seems the optical invariant could be a help, not a hindrance, at least in one regard: did you notice if the illumination level was fairly high? It seems to me that the exit aperture should be indicative of the exposure level... so you would basically get a F/2.8 image with a F/1.4 illumination... or am I missing something?

I'm glad you mentioned this because I've been trying to figure out why there is a 1½ stop discrepancy between the T stop of a cine lens, and the corresponding F stop on a 1/3" video zoom directly connected to the camera.
I expected there to be a slight difference, but 1½ stops would explain your theory. T2.8 on the cine lens was the same light level as approx f/1.7 on the 1/3" zoom lens. I've heard (but it is not documented) that there is a ½ stop loss due to absorbtion/scattering.
So if there is a 2 stop increase in illumination, minus a ½ stop loss for absorption, 1½ stops would make perfect sense.

Can you show me the proper math with the Lagrange invariant applied to illumination? I don't know the algebraic symbol for illumination, so I'll call it "I".
I'm guessing (Iexit)=(Ientry)(ƒ stop)/1.4 , therefore if we use a value of f/2.8 and "1" for entry illumination, the exit illumination will be 2? I'm not sure if I've got it right because to me "two times" would mean 1 stop.

I agree that it's definitely the Lagrange invariant causing the apparent light gain. I
think I can explain in more detail why I think that...

First, the background you probably already know: the optical invariant
is, roughly speaking, a quantity that does not vary no matter where you
are in an optical system - anywhere from the object itself, through the
lens, and on to the image at the CCD/film plane. This quanitity pertains
to lots of different, related variables, including f-stop, illumination,
and magnification factor, such that when one value changes, the others
change as well but the optical invariant remains the same. It's
essentially a mathematical shorthand that makes the relationship between
these values easier to compute (it's usually calculated and referenced for
individual rays, but the general points are the same for ray bundles and
images).

The extreme shorthand: when you crunch all that light down into a smaller
area, the brightness must go up (by the inverse of the magnification
ratio, less transmission losses). This is not so much 'because of' the
optical invariant, as the things which make the optical invariant true
(mostly trigonometric relationships and geometry) also dictate the above
relationship.

So when you were shooting side by side with the JVC video lens, it may
have been open to f/1.7, but next to it the PL lens at f/2.8 was
collecting even more light, but that light is distributed across a much
larger image plane (the 16mm sized aerial image plane). The adapter then
takes (most) of that light and squishes it down to the final size at the
JVC's chips.

It's important to note that the adapter is actually working at an
image-side speed of f/1.4 (or whatever) that more closely corresponds to
the f-stop on the video lens; one thing the optical invariant guarantees
is there's no free lunch in optics. When you squeeze the image into the
smaller format size, you must do it with proportionally faster optics - so
you're not really getting more light than the *entire system* takes in,
just what is printed on the taking lens.

Put it another way: the f-stop ratings on the taking lens's barrel are
strictly for DOF calculation, the entire system actually performs at the
back (image-side) f-stop. In other words, because there's no scattering,
you have to look at the lens system from front element to back element to
determine the actual f-stop. And since f-stop is related to pupil
diameter, you sort of have to conclude that the entire system is actually
performing massively faster than the front lens is rated at (because the
relay lens magnifies the exit pupil, but also shrinks its magnification -
these two phenomena are related). I think you calculated that ratio to be
about 2 (or two stops).

So... now to the math, which is actually pretty straightforward. You
mentioned in the article that the frame sizes were 9.35mm and 4.8mm for
16mm and 1/3", respectively. That makes a linear relationship of 9.35/4.8
= 1.94, which is why the magnification factor is about 2:1. The way we
calculate DOF is basically linear - we compare the diameter of a circle of
confusion to determine whether or not something is in focus... In
contrast, illumination is a function of area (exit aperture size is the
easiest way to think about it). Since area is a squared relationship, the
total illumination difference should be 1.97^2, which is 3.76 or about 2
stops.

The other way to think of it is that there's a certain amount of flux (or
simply, 'light') that goes through the 16mm 'virtual' frame of the aerial
image. Flux is area*illumination, when you shrink the area by a factor of
4 the illumination goes up by an equal factor. (That's just the
definition of flux; you can set the two equal to each other due to -
drumroll please - the optical invariant.) Getting this to more
photographic-looking equations, you could simply propose the squared
relationship (with or without associated hand-waving) and state that flux
is constant (minus transmission losses), and flux equals brightness times
area (F=B*A). This is a shade disingenuous, if only because the units
involved are pretty counter-intuitive: flux is in lumens, Area is in cm^2,
and Illumination is in lumens/cm^2.

By the way, I'm working off of Warren J. Smith's "Modern Optical
Engineering," specifically the discussions of photometry and the optical
invariant (different sections). Anything specific to photography
(f-stops, etc) is my math and hand-waving.


Cheers!