Giles Rais

Or, one could simply play the footage back faster than normal until the motion feels correct, and work out how much faster than normal it was shot at that way. I needed something more official for my purposes, but thank you for your idea.

I *gasp* don't have this book...I just ordered it. Thanks for the suggestion and for the info from Photosonics.

Yes, neither did I. Another mystery from this magnificent film. Thank you so much for looking!

Right, he would be no stranger to modifying a camera to fit his purposes (Barry Lyndon). I just find it strange that I can find no information about this very important shot on all the plethora of work written about this film.

Don't think so; wouldn't step printing add stuttering? Footage is completely smooth.

That's what I was afraid of, because the shots look faster than 72 fps to my eye: https://youtu.be/ypEaGQb6dJk?t=420 Thank you so much for your comment, Mr Mullen!

Hello, I'm trying to find out the frame rate used during the slow motion shots of Moonwatcher striking the skull with the bone in the "Dawn of Man" sequence from 2001. It looks like it might be 120 fps, but I can't find a quote anywhere (checked the SK archives, the new Taschen book on 2001, and several biographies and studies). Does anybody know for sure, or perhaps has a more educated guess based on the cameras used during production? Thank you.

sorry, double post. Please delete

Dear John: Mr. Viera unfortunately passed away. Mr. Brown I have not contacted yet. I've seen this information repeated in many other books as well, so I thought there must be something I am not getting. I will try asking Brown himself. Thanks again for all your help. Giles

Dear John: But they are not saying that the 48:1 tonal range would equal 5.5 stops. If they did, I wouldn't be scratching my head trying to understand this. They list the following reflectances as equaling the 9 or 10 zones in a grayscale, which makes no sense to me because of the reasons I explained before. What you are saying makes complete sense to me. a 48:1 ratio should only equal about 5.5 stops. Take a look at what is included in their books: This is what Viera writes: black velvet = 2% black face = 10% green leaves= 14% brown face = 16% midgray = 18% caucasian face= 36% light grays = 70% off-whites = 80% white chalk = 96% "The typical range of natural reflectances-from whitest white to blackest black-is about 96% to 2%, usually expressed as the ratio 48:1" As you can see, he equates the steps in the grayscale to these reflectance values, and goes on to say that the "typical range" is about a 48:1 ratio. What I am continuing to ask, is how can a 48:1 ratio equal 9 steps in the grayscale when doubling a value every steps would make for at least a 256:1 ratio? (2,4,8,16,32,64,128,256=8 stops not copunting the one we began with). I am sorry to keep asking the same question, but I don't think my particular point has been adressed. In fact, this is the breakdown Brown uses: Zone 0= 3.5% Zone 1= 4.5% Zone 2= 6% Zone 3= 9% Zone 4= 12.5% Zone 5= 17.5% Zone 6= 25% Zone 7= 25% Zone 8= 50% Zone 9=70% Zone 10= 100% and he writes: "A very shiny surface can reflect up to 98% of the light that falls on it. Black velvet reflects about 2% of the light that fall son it. Thsi is a brightness ratio (BR) of 1:48" If I shine 100 footcandles FLATLY on a grayscale with these reflectances, I would get a ratio of only 28:1 (the brightest value divided by the darkest value to obtain a luminance range). I am not asking about real life values here, or the overall luminance range, but just the rationale they are using in their books. Since every zone equlas an f/stop, a 28:1 ratio makes no sense whatsoever according to the reflectances given. If somebody understands my question and can point out how to reconcile what these books are showing (or where my error lies), please let me know. Thanks. Giles

Thank you John. However, I am not referring to the overall luminance range, which, as you point out, can easily exceed the 9 f/stop range of the grayscale. What I am trying to find out is why several books equate a ratio of 48:1 (obtained by going from 2% to 96% luminance) to all the 9 steps in a grayscale, whereas that ratio could not be lower than at least a 256:1 ratio. It seems there is somehting fundamentally wrong if a 48:1 is presented as the whole range and yet it does not match a 256:1 ratio (which should be since zones equals f/stops equals doubling/halving of brightness values, therefore 2, 4, 8, 16, 32, 64, 128, 256...or 256:1 ratio equals a range of 8 zones or f/stops). Both Brown and Viera quote this 48:1 ratio but I am trying to see why it does not fit with their own interpretation of a ratio for a grayscale with 9 steps. Hopefully my question is clearer now. Thanks. Giles

Dear friends: Perhaps you can help me wrap my head around this. I have read in a couple of books (Brown's "Cinematography" and Viera's "Lighting for Film" that the grayscale goes from about 2% reflectance from the lower black to 96% to the brighter white. If you shine 100 fc, you then would get 2 footlamberts from the black and 96 footlamberts from the white. If you divide these figures to get the luminance ratio, you get a 48:1 (96 divided by 2). This is what I don't get: a 48:1 brightness ratio does not equal the 8 zones of brightness that go from Zone 1 to Zone 9...the difference in brightness between zones equal those between f/stops, meaning that 8 zones would equal a ratio of 256:1, not 48:1. Yet those books show a grayscale with reflectance values that go from 2% to 96% across the entire 9 steps of the graycale...how can a 48:1 ratio accomodate for 8 steps in the scale, when 8 steps equal 256:1 ratio? (2, 4, 8, 16, 32, 64, 128, 256). PLease point out where I am making a mistake here. Thanks. Giles.

Dear Mr. Case: "I think you will be looking for a long time to find more of an answer than you have been given here. Perhaps you need to realise that the reason you didn't get a simple answer to your question is that it is unanswerable: because the original premise is flawed." The premise of my question is simply to find out the premise used by Kodak to place the 0.0 camera stop along the Log E axis. If there is no quantifiable relationship between these three scales (Density, Log E and camera stops), the flaw would be Kodak's by including them in the same graph. "A sensitometric curve is designed to show you the curve shape for a typical emulsion and process. As John has said, the exposure in a sensitometer for a particular emulsion is designed so that the 21 steps of the strip show the entire range of response, rather than with any precise mathematical relationship to the exposure of any other emulsion. As I said, the recommended speed rating of a stock is based more on "what exposure gives a good-looking result" than the original scientific graph-plotting method that placed a certain exposure a certain density above d-min at a certain gamma. I'm not sure if the 0 point on the published curves has a precise relationship to the published EI rating - but John's post leads me to suppose not." It sure wouldn't hurt to find out IF there is a precise relationship, wouldn't it? If I didn't know better, I would think that people somehow are trying to discourage me from digging deeper...is this a world wide cinematographic conspiracy?? : ) Maybe I can force Kodak to remove their camera stops scale from appearing integrated with their characteristic curves if they are not correlated at all (as well as asking them nicely to please make sure their curves match [nobody said anything about the discrepancies I found and posted here]). Again, thanks for the time and effort of putting up with me. See you soon!

No, it isn't, because the information I'm looking for is not going to be used on a particular film, but a paper that requires a generalization on the concept and not the specific response a batch may give the filmmaker. Besides, I thought it was a simple matter, but given the fact that the more experienced members of this forum could not confirm or deny the exact points of my question, now I just "gots to know"!! Thanks anyway!

"The "0" point on the Camera Stop scale may not represent what you consider normal exposure for the "look" you want" OK, thanks to everyone who tried answering my question. I'll keep looking until I find the answer and post my findings back here. Thanks again! Dear Patrick: in answer to your question, it's for something very important. ; )