shutter speeds

[rob champagne]

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]

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]

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

[wager123]

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]

"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.

[ricksplace]

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.

[rob champagne]

[QUOTE=2F/2F;649387
So, to find the one-thirds on the way to that next slowest speed, you simply the difference between the two speeds (7.8125 msec.) by 3.
[/QUOTE]

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]

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.

[rob champagne]

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.