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Math/Exposure challenge


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#1 Gustav Lassen

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Posted 28 January 2011 - 09:44 AM

Hey all,

I was given this challenge by a 1.AC and I must admit I have absolutely no idea how to answer it, can anyone give me some hints.

"If I have T-stop 5,6 as a base and wants to run at 0,15fps with the same aperture what will be my smallest exposuretime and how much ND will be needed on a 435 Extreme?"

Anyone have some help?
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#2 K Borowski

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Posted 28 January 2011 - 10:16 AM

Do you want some help or do you just want the answer?

I can walk you through the steps you'd need to take if you want to fully understand the concept of exposure as it relates to motion picture camera film exposure (shutter angle, shutter speed), filtration, and transmission-stop.

Edited by K Borowski, 28 January 2011 - 10:17 AM.

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#3 Gustav Lassen

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Posted 28 January 2011 - 10:22 AM

Well I would like some help really, the answer just seems like the easy way.

I have calculated the exposure time using the simple formula 180/(360x0.15) which is 3,3333333. Am I on the right track?

I would love some help on how to tackle this.
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#4 David Mullen ASC

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Posted 28 January 2011 - 10:31 AM

Exposure time is easy -- 15 fps with a 180 degree shutter (1/2 open) is 1/30th of a second.

So if you are comparing that to 24 fps with a 180 degree shutter, which is 1/48th of a second, you just have to figure out how many stops to adjust to lose the difference between 1/30th and 1/48th.

Of course the math is really easy if it is 12 fps instead of 15 fps... Since 12 fps is half of 24 fps, that's a gain of one-stop more exposure if you don't change the shutter angle, so an ND.30 would compensate (or a 90 degree shutter angle, again, half of 180 degrees). So since you said 15 fps, I assume compared to 24 fps, the difference will be less than a full stop. So you probably have to change the shutter angle to compensate, basically calculate what shutter angle at 15 fps gives you an exposure time of 1/48th.
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#5 Gustav Lassen

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Posted 28 January 2011 - 10:34 AM

Exposure time is easy -- 15 fps with a 180 degree shutter (1/2 open) is 1/30th of a second.

So if you are comparing that to 24 fps with a 180 degree shutter, which is 1/48th of a second, you just have to figure out how many stops to adjust to lose the difference between 1/30th and 1/48th.

Of course the math is really easy if it is 12 fps instead of 15 fps... Since 12 fps is half of 24 fps, that's a gain of one-stop more exposure if you don't change the shutter angle, so an ND.30 would compensate (or a 90 degree shutter angle, again, half of 180 degrees). So since you said 15 fps, I assume compared to 24 fps, the difference will be less than a full stop. So you probably have to change the shutter angle to compensate, basically calculate what shutter angle at 15 fps gives you an exposure time of 1/48th.


Actually the exposure time is 0.15fps, not 15fps.
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#6 K Borowski

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Posted 28 January 2011 - 10:40 AM

David, I think he wants 0.15 (15/100) frames per second. 3-1/3 seconds per frame is what I get too, Gustav. You can use that formula, but what I did was compare 1/48 sec. to 0.15 per second.

You are also assuming a 180° shutter angle, correct?



Gustav: In English usage, commas are used to separate thousands, and periods are used for decimals. I.E. $1,549,999.99.


Are you sure that is the right number? That is a VERY long exposure time, where you'd run into reciprocity on almost all films. From an academic standpoint, you'd need to stack a lot of ND.

Edited by K Borowski, 28 January 2011 - 10:43 AM.

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#7 Gustav Lassen

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Posted 28 January 2011 - 10:42 AM

David, I think he wants 0.15 (15/100) frames per second. 3-1/3 seconds per frame is what I get too, Gustav. You can use that formula, but what I did was compare 1/48 sec. to 0.15 per second.

You are also assuming a 180° shutter angle, correct?


Yes, he didn't state what shutter angle it was, but I assumed 180 degrees, yes.
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#8 K Borowski

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Posted 28 January 2011 - 10:44 AM

Another missing part to the problem is the exposure.

Did the AC tell you that the base T-5.6 aperture is the proper exposure at a 180° shutter angle?


If not you need to know footcandles and film speed ISO to determine exposure.
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#9 Gustav Lassen

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Posted 28 January 2011 - 02:35 PM

Another missing part to the problem is the exposure.

Did the AC tell you that the base T-5.6 aperture is the proper exposure at a 180° shutter angle?


If not you need to know footcandles and film speed ISO to determine exposure.


The AC said only what I posted in my original post. I can just answer that I take it that 5.6 is the proper exposure, and then I can say that in order to determine correct exposure I would have to know ISO, etc.

So if we say that exposure time is 3-1/3 then how would I get the ND's needed in front of the lens? I would have to figure how many stops of light is between 1/0.30 and 1/48? Or am I wrong? And how would I go about doing that?
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#10 Gustav Lassen

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Posted 28 January 2011 - 02:52 PM

Does anyone know why I cannot edit my posts? Or am I missing some button somewhere? :-)

Am I wrong when I say that the difference between 1/0.30 and 1/48 are 7+1/4 stops?

The calculation:

0.30*2 1 stop
0.60*2 1 stop
1.2*2 1 stop
2.4*2 1 stop
4.8*2 1 stop
9.6*2 1 stop
19,2*2 1 stop
38,4*1.25 0.25 stops = 48

Which would mean that an ND9+ND9+ND3 would take seven stops of light away, the remaining 0.25 would have to be by changing the shutter, right?
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#11 Mark Dunn

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Posted 28 January 2011 - 02:59 PM

I think you'd want to get the exposure down with the shutter angle first and then use ND for the rest.
The 435 goes down to 11.2deg. which gets you 4 stops down from 180 for a start. That leaves you with a factor of 10 which is near enough ND9 .
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#12 John Sprung

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Posted 28 January 2011 - 03:31 PM

If you really want 0.15 frames per second, a continuous motion camera is the wrong choice. 60 seconds per minute, so 0.15 x 60 = 9 frames per minute. What you want is time lapse.




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#13 Gustav Lassen

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Posted 29 January 2011 - 03:09 PM

If you really want 0.15 frames per second, a continuous motion camera is the wrong choice. 60 seconds per minute, so 0.15 x 60 = 9 frames per minute. What you want is time lapse.




-- J.S.


I am just trying to answer his question, I am not debating whether or not it is an efficient way of shooting it.
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#14 Gustav Lassen

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Posted 29 January 2011 - 03:20 PM

I think you'd want to get the exposure down with the shutter angle first and then use ND for the rest.
The 435 goes down to 11.2deg. which gets you 4 stops down from 180 for a start. That leaves you with a factor of 10 which is near enough ND9 .


Wouldn't that affect the motion blur of the image too much, you know, too much "gladiator-effect"? :-)
Is there a disadvantage of stacking ND's?
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#15 K Borowski

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Posted 29 January 2011 - 04:20 PM

NDs wouldn't, but I wonder how a rotary shutter is set up at that speed. . . Does it stop intermittently from rotating all-together?

I agree with John here: A modified still camera for timelapse would be a better choice here, something with a focal plane or a leaf shutter would have far less artifact issues than a movie shutter.


You can run into issues of having the top of the frame having a noticeably different moment in time than the bottom of the frame with this type of shutter and timelapse exposure.




The "Gladiator effect" was a SMALL shutter angle, yes, but the film was being esedxpo at a frame rate at or near the standard 24fps. You get the effect with a very SHORT exposure, so short that there is no motion blur to fool the viewer's eyes into thinking that they're seeing continuous motion. They then can see the discrete frame exposures and notice the differences in each one, not seeing it completely as continuous movement.

At a very low shutter speed, there is TOO MUCH motion blur. Something like a car or a human will appear as a blur or a streak of light at night with the case of a car and it's lights. If this shutter speed and exposure were used, I'd be concered with the movie shutter. I've never shot this slow. I've done TIMELAPSE, but in that case the shutter was held open for a discrete amount of time and then suddenly flipped back. It didn't slowly cover and uncover the frame.

At 11.2 degrees, does the shutter even have enough angle to cover an entire frame of film? I guess it must.
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#16 Gustav Lassen

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Posted 29 January 2011 - 06:54 PM

NDs wouldn't, but I wonder how a rotary shutter is set up at that speed. . . Does it stop intermittently from rotating all-together?

I agree with John here: A modified still camera for timelapse would be a better choice here, something with a focal plane or a leaf shutter would have far less artifact issues than a movie shutter.


You can run into issues of having the top of the frame having a noticeably different moment in time than the bottom of the frame with this type of shutter and timelapse exposure.




The "Gladiator effect" was a SMALL shutter angle, yes, but the film was being esedxpo at a frame rate at or near the standard 24fps. You get the effect with a very SHORT exposure, so short that there is no motion blur to fool the viewer's eyes into thinking that they're seeing continuous motion. They then can see the discrete frame exposures and notice the differences in each one, not seeing it completely as continuous movement.

At a very low shutter speed, there is TOO MUCH motion blur. Something like a car or a human will appear as a blur or a streak of light at night with the case of a car and it's lights. If this shutter speed and exposure were used, I'd be concered with the movie shutter. I've never shot this slow. I've done TIMELAPSE, but in that case the shutter was held open for a discrete amount of time and then suddenly flipped back. It didn't slowly cover and uncover the frame.

At 11.2 degrees, does the shutter even have enough angle to cover an entire frame of film? I guess it must.


Of course, makes sense. So if we were to assume that he wants to use the 435 Extreme as the camera anyways, would it be correct to say that a shutter of 11.2 plus an ND9 would give the correct exposure?
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#17 Chris Millar

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Posted 29 January 2011 - 09:52 PM

Does anyone know why I cannot edit my posts? Or am I missing some button somewhere? :-)

Am I wrong when I say that the difference between 1/0.30 and 1/48 are 7+1/4 stops?

The calculation:

0.30*2 1 stop
0.60*2 1 stop
1.2*2 1 stop
2.4*2 1 stop
4.8*2 1 stop
9.6*2 1 stop
19,2*2 1 stop
38,4*1.25 0.25 stops = 48

Which would mean that an ND9+ND9+ND3 would take seven stops of light away, the remaining 0.25 would have to be by changing the shutter, right?


A more accurate way to figure it out:

LOGbase2 (48/0.3) = 7.321928094887363 stops
:lol:
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