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Sekonic light meters


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#1 Bon Sawyer

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Posted 18 July 2005 - 11:50 PM

Hi all,

Hopefully this has not already been asked... I searched the forum for "sekonic," but did not find what I was after.

I have a Sekonic L-558 (non-Cine) light meter. As many of you know, it has cine frame rate settings in addition to standard shutter speeds. There is no mention in the manual of what shutter speed each frame rate setting corresponds to.

Does anyone know where I would find this information? Or do all cine cameras conform to the same shutter speed at any given frame rate/shutter angle?

To be specific, I am using a Krasnogorsk-3 16mm camera. According to its manual, it has a shutter speed of 1/60 (when using a framerate of 24fps), and a shutter angle of 150 degrees. The Sekonic manual indicates that the meter assumes a shutter angle of 180 degrees, and to compensate using the ISO setting for other angles. I am using 250 ASA film (Kodak 7205), so have been using an ISO setting of 200 on the light meter, and a shutter speed setting of 24 fps.

Do I have my meter correctly set up for my camera, or are there other factors I have not considered?

Thanks,

-Bon

Edited by Bon Sawyer, 18 July 2005 - 11:51 PM.

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#2 John Hall

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Posted 19 July 2005 - 09:04 AM

I own the same meter (non cine). Quite happy with it (except for the lack of footcandle reading, but I can do my own math.

Does anyone know where I would find this information? Or do all cine cameras conform to the same shutter speed at any given frame rate/shutter angle?


The shutter speed is determined by the shutter opening and the frame rate you select. As you said, the meter assumes a shutter opening of 180 degrees (open half the time). At 24 fps, the shutter speed will be 1/48 of a second.

If your camera has a 150 deg shutter opening, then it has an effective shutter speed of 1/60 of a second when shooting at 24 fps.
On the L-558, I would just select 1/60 shutter speed. If you want to use the fps settings, you could compensate simply by setting the frame rate on the meter to 30fps, which, assumining as the meter does, a 180 degree opening, will be a shutter speed of 1/60 second.
A compensation in the ISO speed is unnecessary, though it's good to understand their relationship to each other.
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#3 Bon Sawyer

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Posted 19 July 2005 - 11:06 AM

John,

Thanks very much for the info!

-Bon
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#4 scorsesebull

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Posted 19 July 2005 - 11:18 AM

If your camera has a 150 deg shutter opening, then it has an effective shutter speed of 1/60 of a second when shooting at 24 fps.

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Hi John,

Could you explain for a thirsty student in a little bit more detail the math and relationship of "effective" shutter speed and how you can calculate shutter speed based on knowing the degrees of your shutter opening? I understand your logic using the 1/60 and by switching the meter to 30fps but the figuring 150 into 1/60 I can't quite picture. Thanks!

--Luke Kalteux
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#5 Dominic Jones

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Posted 19 July 2005 - 01:58 PM

I'm not John, but I can explain shutter angles to you, if that's ok?!

The shutter angle determines the percentage of the shutter's spinnning cycle it is open (and therefore passing light through to the film) for - so the shutter speed for a film camera is a factor of the framerate (or how many cycles the shutter does per second) and the shutter angle (or how long the shutter is open for per cycle).

A 180 degree shutter angle, which is about normal, gives half the time per cycle for exposure - so 1/48th for 24fps as previously mentioned.

A 90 degree shutter angle will give you a shorter exposure - 1/96th at 24 fps whereas a 270 degree angle will give you a longer exposure period of 1/32nd...

Many cameras (the Bolex H16 is an example) use a shutter angle of 172 degrees to give the easier to work with shutter speed of 1/50th of a second.

The mathematical equation for calculating exposure time for a given shutter angle and framerate is:

1 / (<framerate> * (360 / <shutter angle>))

So, for the example of a 150 degree shutter angle whilst shooting at 24fps you get:

1 / (24 * (360/150)) = 1/57.6, to be precise!

I hope that's fairly lucid and helps rather than hinders...

Edited by Dominic Jones, 19 July 2005 - 02:00 PM.

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#6 scorsesebull

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Posted 19 July 2005 - 04:00 PM

Thank you so much Dominic! That was very helpful. I've always understood the 1/48, and I've worked with oooollld bolex cameras at school and I know those are closer to 1/50, but I needed to SEE the math. Thanks. Unfortunately and reluctantly I must admit math is not my strongest suit in the realm of cinematography. So my follow up question would be more trivial: In movies like Saving Private Ryan, or Gladiator (seemed to make this popular) when you get more clarity and less motion blur, for instance in Gladiator when you can see the individual specs of dirt flying through the air with what looks like unusual clarity, is that accomplished with a longer or shorter exposure? (does that make sense?)

--Luke
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#7 Dominic Jones

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Posted 19 July 2005 - 07:07 PM

Those strobing high detail effects ala Saving Private Ryan are accomplished with a short exposure time - that's part of a constant across all forms of photography:

The longer the exposure time, the more blurred motion will become, and vice-versa.

I don't know what the shutter angle was for SPR (I'm sure someone here will, it's probably fairly common knowledge), but I'd guess the exposure time was around 1/500th - 1/1000th, giving a shutter angle determined but the following (just to provide a demonstration of the maths the other way around):

The reverse equation, which is actually often more useful is:

Shutter Angle = <exposure time> * <framerate> * 360

So the shutter angle for SPR was probably between (1/500th sec):
1/500 * 24 * 360 = 17.2 degrees

And (1/1000th):
1/1000 * 24 * 360 = 8.6 degrees

It's worth noting that the maths here is pretty secondary. The trick is to learn the creative effects of shorter and longer exposure times and decide what you want to shoot for a particular scene or effect - then all you've got to do is plug the numbers into a calculator and hey presto!
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#8 scorsesebull

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Posted 19 July 2005 - 07:57 PM

Again, thank you! As you can probably see, I'm pretty new to the website, but it's great getting fast and very efficient advice from all the other people who truly enjoy this field. Thanks a lot!

--Luke K.
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