shutter speeds

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mitch brown

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ok stupid questions but i don't know the math so any help greatly app. my 90mm has a sloww shutter speed at 125 how do i figure out how to adjust for it by using 1/3 stops. or other wise what speed is 125 - 1/3 of a stop.
hope i asked this right.
mitch
 

2F/2F

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Same as film speeds: '100-'125-'160, and so on. Double or half each for one full shutter speed difference.
 
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mitch brown

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thanks
let me ask the question another way , a 1/125 shutter speed in mil secs is7.81ms if the shutter timer show that the shutter speed at 1/125 is 11.66 ms how do you figure how to compensate for it.
thanks
mitch
 

RobC

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If the time measured was 11.66ms the speed of that exposure in normal notation is:

1/(11.66/1000) = 1/86 of a second (I think).

The amount to adjust the speed by would be

1/((11.66-7.81)/1000) = 1/260th of a second less exposure.

I would get someone else to confirm my maths because I'm not 100% sure about it.
 

Ray Heath

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are you wanting to know the full and 1/3 shutter speeds or what to adjust a wrong shutter speed to?
 

Arvee

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Mathematically, roughly 8 ms represents 1/125 second shutter speed. 16+ ms represents 1/60 second shutter speed. 11.66 ms is about half way between (12ish to be exactly precise) so I would say your shutter is about 1/2 stop slow, about as precise as one can be with photographic equipment.

HTH,

Fred
 

RobC

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To find a 1/3 stop increase you divide by ³√2 = 1.2599
so 1/86 / 1.2599 = 1/108th of a second which is till to slow.

To find a 1/2 stop increase you divide by ²√2 = 1.4142
so 1/86 / 1.4142 = 1/122th of a second

So in your case with the shutter speed being slow, you need an additional half a stop exposure of the slow shutter speed which should be close enough.
 

RobC

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But given your shutter speed is slow then it could be sticking or it may be out of adjustment. If its sticking then an extra half stop may or may not work.
 

removed account4

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i love threads like this .. :smile:

if you are using an olde betax lens, you would can
set your shutter inbetween your shutter speeds, but seeing
you are probably using a modern shutter,
my suggestion would be to just adjust your fstop a little bit,
and not worry/ stress too much about it. :smile:

have fun!

john
 

RobC

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I knew shouldn't have got into maths. I just realised I got it round the wrong way. You want less exposure not more.
 

Ray Heath

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wow, how soon y'all get techimological

if your using negative film, that you develop, in an old camera, there are lots of other variables

i wouldn't worry too much about 1/2 a stop of over exposure, better too much than not enough

anyway, how do your negs look and print, that's the only test that matters
 

edtbjon

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The first thing that you need to know is that the time is consistent at each and every setting. Or that you do get consistency after a few test exposures. If not, a CLA is really needed.
Second, most old shutters "live their own life" and you need to get to know each and every one closely if you want to gain "total control". This is also a fact that comes with new shutters too. They will differ from one another and the tolerances are pretty wide. I.e. there's a valid idea for e.g. the Sinar shutters, where you use one shutter for all of your lenses.
So, if you shoot b/w or negative color in most cases you could live with it as the film does have the tolerance and it can take the half to full stop overexposure. If you are shooting slides you need to pin down the exposure to 1/3 of a stop. If so, you need to learn how much off each and everyone of your shutters are.

//Björn
 
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mitch brown

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thanks all for all the help. i will be sending it to carol as soon as she is up and running again.
mitch
 

2F/2F

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"thanks
let me ask the question another way , a 1/125 shutter speed in mil secs is7.81ms if the shutter timer show that the shutter speed at 1/125 is 11.66 ms how do you figure how to compensate for it.
thanks
mitch"

'128 is 7.8125 milliseconds (notated as '125 for the sake of convenience), therefore the next slowest full shutter speed ('64 AKA '60) will be twice as long, or to say it another way, it will be 7.8125 msec. longer than 7.8125 msec. So, to find the one-thirds on the way to that next slowest speed, you simply divide the difference between the two speeds (7.8125 msec.) by 3.

7.8125 msec./3 = 2.6 msec. per one-third shutter speed between '125 and '60.

Therefore '125 + 1/3 = 7.8125 + 2.6 = 10.4125 msec.
Therefore '125 + 2/3 = 7.8125 + 5.2 = 13.0125 msec.

Neither of these is 11.66 msec., so just stop down 1/2-stop with your aperture and call it good enough. If, for some reason, you have to use thirds, go only one third with print film, and two thirds with transparency film.
 
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ricksplace

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I have a few shutters that are a little slow at some speeds. Since the shutter is slow, and therefore lets in more light than the indicated speed, I stop down the aperature a half stop or so to compensate. That plus the film latitude usually takes care of the issue. Cheaper than a CLA, as long as the shutter is consistent.
 

RobC

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2F/2F;649387 So said:
That is technically wrong because exposure increments are exponential and not linear. You might think I'm being pedantic about that, but I just don't volunteer to dumb down which some already have.

A third of a stop is ³√2 = 1.2599 used as an adjustment factor and not 1.3333.

The proof is simple.

1.0000 * 1.3333 = 1.3333
1.3333 * 1.3333 = 1.7777
1.7777 * 1.3333 = 2.3702 (this should have been = 2 )

whereas using the correct value:

1.0000 * 1.2599 = 1.2599
1.2599 * 1.2599 = 1.5873
1.5873 * 1.2599 = 1.9998 ( with rounding or working to more decimal places = 2 )

so the constant 1/3 stop factor is 1.2599
the constant 1/2 stop factor is 1.4142
 

2F/2F

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Them thar other-type'n fancy maths done did it ta me a-ginn!

I know that f stops are not linear, but I always figgered that when it came to time, splitting a full shutter speed linearly into thirds made sense, like so: If light is entering for a certain length of time, cutting 1/3 of that time also cuts 1/3 of the amount of light that enters, so 2/3 of the exposure occurs. I viewed it as a division of one full shutter speed, rather than a series of exposure increments, with each based on the last one-third-speed.
 
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RobC

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Them thar other-type'n fancy maths done did it ta me a-ginn!

I know that f stops are not linear, but I always figgered that when it came to time, splitting a full shutter speed linearly into thirds made sense, like so: If light is entering for a certain length of time, cutting 1/3 of that time also cuts 1/3 of the amount of light that enters, so 2/3 of the exposure occurs. I viewed it as a division of one full shutter speed, rather than a series of exposure increments, with each based on the last one-third-speed.

Film response is not linear.
 
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