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Who are the math nerds then ?


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#1 Nick Mulder

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Posted 25 May 2007 - 10:21 PM

Hiya,

Just want to confirm I'm going about my mafamatatics correctly setting up my new light meter for the light loss in my reflex prism Bolexes ...

From measurement I know the prism soaks up %17 of the light coming through it to the film ...
ie. I need to increase exposure (time or aperture) by:

1 / 0.83 = 1.2 (0.83 = 1 - 0.17)

So thats a %120 increase or an equivalent filter factor of 1.2

The compensation in the light meter works in EV which works in base 2 - ie. a difference of 1 between two readings is a double or halving of light ... (like Bels, of which decibels are derived)
so using a nifty base2 log calculator I work out the following:

log2 (1.2) = 0.263

bummer is the light meter only has a resolution of 1/10 increments - so I choose 0.3 and work from there (ass!) ...

Basically - my question is: have I got everything right ?

and also when 'ND3' is referred to I'm assuming its an error and they really mean ND0.3 which in fact has a filter factor of 2 and could also be called ND2 which only has a reduction of 1 stop ie. could potentially be referred to in error as ND1 ...
~
Spoiler
~

Anyway, I know how it works and at last resort would just explain "I want a ND filter that leaves only a quarter of the light" (ND0.6/ND4) - but as I usually work alone I'm wondering how often communication breakdowns happen within camera depts ?

To me filter factors make more sense than base2/EV they are simply the reciprocal of the percentage transmission - something most humans have had experience with in conceptual terms since childhood ...
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#2 chuck colburn

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Posted 25 May 2007 - 10:27 PM

Nick,

I believe a filter factor of 1.2 is 4 stops.
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#3 Jonathan Benny

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Posted 25 May 2007 - 10:49 PM

Nick,

I believe a filter factor of 1.2 is 4 stops.



That is not correct.

A filter factor of 16 is 4 stops.

A filter factor of 2 is 1 stop: (you need twice as much light, so to speak, to compensate - or you open up 1 stop, allowing twice as much light in).

4 is 2 stops, 8 is 3 stops, 16 is 4 stops.

I believe you're referring to an ND1.2 (which operates on a different scale related to 1/3rds of a stop) which has a fliter factor of 16: ie 4 stops.

AJB
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#4 Nick Mulder

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Posted 25 May 2007 - 10:55 PM

I believe a filter factor of 1.2 is 4 stops.

In writing the post I came to the conclusion I was %96 sure I was correct in my math, but went ahead with it anyway as I thought it would be an interesting discussion for the nerds in the crew ...

That being said it will make a bit more sense (yet still like I'm a smarty-pants) when I say I think you're wrong there chuck ... a 4 stop filter would have a filter factor of 16

What/where lead you to think that 1.2 was the factor ?

have a quick look here (same link as from my original post)
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#5 Nick Mulder

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Posted 25 May 2007 - 11:05 PM

I believe you're referring to an ND1.2 (which operates on a different scale related to 1/3rds of a stop) which has a fliter factor of 16: ie 4 stops.


So Chuck is correct but just got the term mixed up ?

This is the confusion that I was referring to.

We all (well, most of us) understand how ND works and affects the exposure and why we might want it for whatever reason - but find it hard to communicate in direct terms due to these conflicting systems ...

Ah well :rolleyes:
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#6 Jonathan Benny

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Posted 25 May 2007 - 11:41 PM

So Chuck is correct but just got the term mixed up ?


No, Chuck was incorrect by stating a filter factor of 1.2 equals 4 stops (even though it may have been a result of mistakenly using a different scale to express the effect of a filter on light). I do understand the confusion, though.

He was referring to your calculation which yielded a filter factor of 1.2 which has no relation to 4 stops absorption.

You were correct when you stated that a filter factor of 16 equals 4 stops.

I have a question, though, about your calculation:

From measurement I know the prism soaks up %17 of the light coming through it to the film ...
ie. I need to increase exposure (time or aperture) by:

1 / 0.83 = 1.2 (0.83 = 1 - 0.17)

So thats a %120 increase or an equivalent filter factor of 1.2


How did you come up with a 120% increase? How can a 17% absorption equal a 120% necessary increase (which by the way, is not a filter factor of 1.2)? You must have meant 20%, or 120% of the original value. But I still fail to see how a 17% absorption equals 20% compensation?

In any case: if the prism is taking 17% of your light, that is how much you need to compensate for: if you're not doing scientific work, just round it up to 20% and call it 2/10ths of a stop compensation.

Great topic, btw.

AJB
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#7 Nick Mulder

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Posted 26 May 2007 - 02:10 AM

How did you come up with a 120% increase? How can a 17% absorption equal a 120% necessary increase (which by the way, is not a filter factor of 1.2)? You must have meant 20%, or 120% of the original value. But I still fail to see how a 17% absorption equals 20% compensation?

yer right - %20 increase - but the maths requires this to be expressed as 1.2 (120/10)... Just another communications breakdown due to the nomenclature of mathematics

%17 absorption = 100-17 transmission ie. %83

1/0.83 = 1.2048192771

So I mean to say a %20 increase in exposure is required to offset a filter with a factor of 1.2

check that wiki page I've linked - you'll see a filter factor of 1.3 refers to %25 absorption

Its no mistake that if you divide the % increase by 10 and add 1 you'll end up with the filter factor - and the inverse logically also follows...

eg. say you had some whacky 'raccoon-rainbow-moonshine' filter with a filter factor of 3.46 - take away 1 and multiply it by 10 to find that the filter requires a %246 increase in exposure to account for it's raccoon light sucking qualities

just round it up to 20% and call it 2/10ths of a stop compensation

This is another misconception about stops - they work in base2 log's - pretty sure you cant think in percentages like this ...

Anyways, you are right though - I'm not doing scientific work and my ultra $$ 'cine' lightmeter cant even cope with the resolution of the calculations I'm making here - a bit disappointed to tell you the truth !
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#8 Dominic Case

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Posted 26 May 2007 - 02:37 AM

It's obviously easy to become confused if you switch from percentage factors to log values - and then again to stops.

The "17% loss equals a 20% gain" is an old trick. If you start with $100 and pay 17% tax, you are left with $83. To recover the $17, when you only have $83 left to work with, you need to earn 20% of $83.

(I find a lot of people can do this stuff more easily if it relates to something more countable than light!).

ND3 is just a label that the filter manufacturers use, rather than an actual measurement of density (0.30). It's not really a mistake.

One of the benefits of using logs for filters is the intuitive convenience of being able to add the values to get a total result. If you have a 0.30 ND and a 0.90 ND, then together they have the effect of a 1.20ND. Couldn't be easier. Also, since a stop is 0.30 ND, then one third of a stop is 0.10ND, and (if you need to be that precise) that is the filter correction you need to compensate.

One other thing - it's log base 10 that you are working in, not log base 2. In base 10, log 2 = 0.3010 (which is where the 0.30ND =1 stop comes from). log 10 = 1 , so a 1.0ND filter reduces the light to 1/10th. (compare with 0.9 being 3 stops or 1/2 x 1/2 x 1/2 or one eighth.)
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#9 Nick Mulder

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Posted 26 May 2007 - 02:54 AM

One other thing - it's log base 10 that you are working in, not log base 2. In base 10, log 2 = 0.3010 (which is where the 0.30ND =1 stop comes from). log 10 = 1 , so a 1.0ND filter reduces the light to 1/10th. (compare with 0.9 being 3 stops or 1/2 x 1/2 x 1/2 or one eighth.)

When you refer to ND as 0.3, 0.6 etc.. that is in base 10

Filter factors work in base 2 - a filter factor of 2.5 is 1 and a 1/3 stops (fact from the net)

log2 (2.5) = 1.3219280948874 (1 and a 1/3rd)

thats why ND0.3 (base10) in one and the same as ND2 (base2)

I understand that the intention may have been so you can add factors up easily but a read of this thread itself shows up the confusion generated by the 'simplification' ...
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#10 Jonathan Benny

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Posted 26 May 2007 - 10:09 PM

This is another misconception about stops - they work in base2 log's - pretty sure you cant think in percentages like this ...


Yes, I got mixed up in thinking 100% as being one stop.

Very silly.

AJB
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#11 Chris Keth

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Posted 27 May 2007 - 03:44 PM

If this stuff is of curiosity to you a lot, you should check out a couple optics books.

transmittance can be expressed as a percentage or as a decimal where 1.0 is all of the light entering is also exiting

opacity is expressed as a decimal aprt of 1.0. It is the reciprocal of transmittance.

Density is the logorithm of opacity, or the logorithm of (1/transmittance)



Since the bolex system passes 83% of light that enters it, it has a transmittance of 83% or of .83

1/.83 gives an opacity of 1.20

The LOG of 1.20 is .081 density or just under a third of a stop (which would be 0.1 density). This checks with my memory from film school of using an adjusted shutter speed of 1/80 where 1/60 was the actual speed.
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#12 chuck colburn

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Posted 27 May 2007 - 04:02 PM

That is not correct.

A filter factor of 16 is 4 stops.

A filter factor of 2 is 1 stop: (you need twice as much light, so to speak, to compensate - or you open up 1 stop, allowing twice as much light in).

4 is 2 stops, 8 is 3 stops, 16 is 4 stops.

I believe you're referring to an ND1.2 (which operates on a different scale related to 1/3rds of a stop) which has a fliter factor of 16: ie 4 stops.

AJB


Ooops
Of course. I was thinking N.D. numbers insted of the reciprocal of the square (half as much/twice as much)
Sorry that I caused confusion.
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#13 Nick Mulder

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Posted 27 May 2007 - 04:27 PM

transmittance can be expressed as a percentage or as a decimal where 1.0 is all of the light entering is also exiting

...

Since the bolex system passes 83% of light that enters it, it has a transmittance of 83% or of .83

1/.83 gives an opacity of 1.20

The LOG of 1.20 is .081 density or just under a third of a stop (which would be 0.1 density). This checks with my memory from film school of using an adjusted shutter speed of 1/80 where 1/60 was the actual speed.


In base2 logs its 0.263 (like you say just under a third of a stop) ... Base2 seems to give you the actual figure in stops

So filter factors are transimittance in decimal - nice to know...
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#14 Chris Keth

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Posted 28 May 2007 - 08:59 AM

In base2 logs its 0.263 (like you say just under a third of a stop) ... Base2 seems to give you the actual figure in stops

So filter factors are transimittance in decimal - nice to know...


Still not right. Think of filter factors as a multiplier for the light to get proper exposure. A filter factor of 2 would require light quantity to be multplied by 2, or increased by one stop. It's opacity that coincides with filter factors.

The transmittance of a filter with a factor of 2 would be 50% or .5 transmittance.

Opacity is found by dividing 1 by the decimal transmittance so 1/.5=2.

Edited by Chris Keth, 28 May 2007 - 09:02 AM.

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#15 Nick Mulder

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Posted 28 May 2007 - 01:39 PM

Still not right. Think of filter factors as a multiplier for the light to get proper exposure. A filter factor of 2 would require light quantity to be multplied by 2, or increased by one stop. It's opacity that coincides with filter factors.

The transmittance of a filter with a factor of 2 would be 50% or .5 transmittance.

Opacity is found by dividing 1 by the decimal transmittance so 1/.5=2.


Crap - really tired at the mo' (48 hour) - getting me reciprocals upside down

cheers for putting me the right way up :rolleyes:
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#16 Chris Keth

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Posted 28 May 2007 - 05:09 PM

Crap - really tired at the mo' (48 hour) - getting me reciprocals upside down

cheers for putting me the right way up :rolleyes:


No problem. I paid attention in "Photographic Materials and Processes" for a reason. :)
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#17 Nick Mulder

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Posted 01 June 2007 - 03:39 AM

No problem. I paid attention in "Photographic Materials and Processes" for a reason. :)


Take some pleasure in that (unless I'm mistaken (again)) you're the only one not to mess up a concept or bit of math in this thread so far :P
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