Film-Tech Forum ARCHIVE: What's the most film you can put on a Potts Platter?

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Posted by Chase Taylor (Member # 778) on 07-22-2002, 10:28 AM:

With "Gods and Generals" coming out that is 330 Min. (5hr 30Min) what is the largest film you can put on a Potts platter. I know you can't fit that movie on there, It just made me wonder. My butt is already tired and I have not even watched the movie yet!

Posted by John Pytlak (Member # 331) on 07-22-2002, 11:18 AM:

Why wouldn't you run it with an intermission, so you can use two platter levels?

You can calculate roll diameter if you know the hub diameter, film thickness, and length of the print:

____________________________________________________________________

The roll diameter can be calculated from the following equation:

D = Squareroot [ ( (48 x L x t) / Pi) + (C x C) ]

Where:
D is roll diameter in inches
L is length of film roll in feet
t is film thickness in inches (about 0.0053 for Kodak VISION Color Print film)
C is core/hub diameter in inches

For example, for a shipping reel with a 4-inch core, and a film thickness of 0.0053 inches, 2000 feet of film will have the following diameter:

D = Squareroot [ ( (48 x 2000 x 0.0053) / Pi) + (4 x 4) ]

= 13.3 inches

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John P. Pytlak, Senior Technical Specialist
Worldwide Technical Services, Entertainment Imaging
Research Labs, Building 69, Room 7525A
Rochester, New York, 14650-1922 USA
Tel: +1 585 477 5325 Cell: +1 585 781 4036 Fax: +1 585 722 7243
e-mail: john.pytlak@kodak.com
Web site: http://www.kodak.com/go/motion

Posted by William T. Parr (Member # 677) on 07-22-2002, 11:54 AM:

Whew! Having flashbacks of College Algebra here. I vote we change John's Title from Human Search Engine to just plain Genius.

Posted by John Pytlak (Member # 331) on 07-22-2002, 12:05 PM:

That roll diameter equation was derived decades ago, likely by someone at Kodak. Someone much more expert in math than me (Evans?) may be able to detail the theory involved in calculating the diameter of a given length of a spiral made of a material of constant finite thickness wound around a hub of known diameter.

Posted by Tim Sherman (Member # 569) on 07-22-2002, 12:33 PM:

From what i have been told the potts will handle about 4.5 hours. I have never filled mine up all the way. But i have doubled two prints on one platter many times and always had room left. Amazing how slow both the take-up and payout platters go when you get to the outside of the platter.

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Posted by Steve Guttag (Member # 268) on 07-22-2002, 12:45 PM:

This would be a rate of change problem...as such Calculus would be the tool. It would be a matter of expressing the diameter increasing in size as a function of film length. So you have Pi*Dmin=L. D is the variable since on the next revolution, D = Dmin + (2*Film thickness). Also this problem presumes that the film thickness is << D (which it is). This continues until D = Dmax. Now you are at a point where you can integrate D as it goes from min to max.

Or...you can just use the formula John provided!

Steve

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"Old projectionists never die, they just changeover!"

Posted by Timothy Eiler (Member # 383) on 07-23-2002, 06:39 AM:

15minutes worth of intermission stuff. The Sum of all fears and windtalkers. All on one deck. it does come out near the edge of the deck. But it runs fine

tim