New DV8 unit!

[Levi]

awww man! Sorry guys, when I clicked "add reply" my cat jumped up on my hand causing me to triple click.

[Actor]

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I was just curious as to how you decided the distance between the gate, lens, and camera? Did you have to play with it for awhile or is there some mathematical way to figure that?
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There are two formulae from Optics 101:
<ul type="square">[*](1/s) + (1/s') = (1/f)[*]m = s'/s[/list]where
<ul type="square">[*]s is the distance from the gate to the center of the lens.[*]s' is the distance from the center of the lens to the CCD.[*]f is the focal length of the lens.[*]m is the linear magnification.[/list]The simplest case is the one where the gate and the CCD are the same size and the same aspect ratio, i.e., m = 1. The second formula then implies that s = s'

The first formula then gives us

(1/s) + (1/s) = (2/s) = (1/f)

Ergo

s = 2f

meaning the gate to lens and lens to CCD distances must both be twice the focal length of the lens.

In the real world the situation where the film image and CCD are the same size and aspect ratio is probably rare. Let's assume the CCD is 1/3 inch diagonally with a width of 0.267" and height of 0.200", an aspect ratio of almost exactly 4:3. Now Super8 is (if memory serves) 0.225" wide and 0.167" high, an aspect ratio of 1.347. If we go for a slight letterbox effect (to preserve all the super8) then the ratio of the widths will determine our choice of magnification...

m = 0.267 / 0.225 = 1.187

Ergo

s' = 1.187s

(1/s) + (1/1.187s) = (1/f)

(1.187/1.187s) + (1/1.187s) = (1/f)

(2.187/1.187s) = (1/f)

f = 1.187s/2.187

s = 2.187s/1.187 = 1.843f

s' = s/1.187 = 1.843f/1.187 = 1.553f

In practice you could fix any one of the three values s, s' and f at any value you want and then determine the magnification and focus by adjusting the other two.

<font color="#a62a2a" size="1">[ April 30, 2003 12:04 AM: Message edited by: Actor ]</font>

[MovieStuff]

Wow! All that math! My head hurts. And to think silly me just fit my lens on some cardboard tubes until the image looked the right size. What was I thinking?...... [img]wink.gif[/img]

Actor, seriously, I wish I had your math skill. I can do the formulas but I always end up choking on them. Fortunately, I sort have a 6th sense regarding optics and distance developed over years of doing this sort of thing so I, literally, just spent about a minute with a cardboard tube until I found the measurement I needed then machined some adjustable extension tubes to match. Wasn't complicated, really.

Roger

[Actor]

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Frankly, short of a flying spot scanner, I truly believe this is about as good as it's going to get.
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Frankly, I don't see why a flying spot scanner should be superior to this approach. Correct me if I'm wrong but a flying spot scanner and a Rank-Cintel are the same thing, right?

Having revealed that much of my ignorance I'll continue. The advantage of a Rank, as opposed to a chain transfer, is that it does a 3:2 pulldown and captures both fields of each frame, all in real time. At least that's my understanding. To my way of thinking the WorkPrinter approach accomplishes the same thing in a more elegant manner. (Albeit not in real time. Big deal.)

I have a feeling that my whole concept of how a Rank works is wrong. If so, would someone please explain it?