Aperture f-stop multiplier?

[Nigels]

Not really sure where to ask this question. And it may be a silly question but here goes; If the multiplier for one stop of aperture is 1.414 (the square root of 2) then what is the multiplier for a third of a stop?

[Greg Lockrey]

1.1545561

[Joanna Carter]

.4714045

Hmmm, anything less than 1 is going to reduce the result, not increase it

[Greg Lockrey]

Hmmm, anything less than 1 is going to reduce the result, not increase it

yeah, I retunck it....see above. Take sqrt of 1.3333.

[jb7]

ask me one on sport...

[Bill Kumpf]

It is my understanding the multiplier is the “square root of one plus the fraction”

If you want ½ stop increase, the multiplier is the square root of (1+ ½).

Exposure Increase
f/stop Time multiplier Aperture Increase

1/12 1.059 1.040833
1/6 1.122 1.080123
1/4 1.189 1.118034
1/3 1.260 1.154701
1/2 1.414 1.224745
1 2.000 1.414214


Of course it has been a loooong time since I studied math.

[Greg Lockrey]

It is my understanding the multiplier is the “square root of one plus the fraction”

If you want ½ stop increase, the multiplier is the square root of (1+ ½).

Exposure Increase
f/stop Time multiplier Aperture Increase

1/12 1.059 1.040833
1/6 1.122 1.080123
1/4 1.189 1.118034
1/3 1.260 1.154701
1/2 1.414 1.224745
1 2.000 1.414214


Of course it has been a loooong time since I studied math.

My caculator says sqrt of 1.333 (1 + .333) is 1.1545561. Should I get a new one.

[Struan Gray]

1.12246 or so.

You want a number that when multiplied by itself three times gives the correct multiple for a stop's worth of exposure change. Thus you take the cube root of 1.414, which is the sixth root of two.

[Greg Lockrey]

1.12246 or so.

You want a number that when multiplied by itself three times gives the correct multiple for a stop's worth of exposure change. Thus you take the cube root of 1.414, which is the sixth root of two.

That's some heavy siffering. Aren't we dealing with an area of a circle and how it multiplies itself by 2 from one diameter to the next? You're going to have to explain cube's and sixth roots to me when you are just dealing with square areas.

[Struan Gray]

That's some heavy siffering. Aren't we dealing with an area of a circle and how it multiplies itself by 2 from one diameter to the next? You're going to have to explain cube's and sixth roots to me when you are just dealing with square areas.

Others have given the formalism, but since you asked here's my take on things.

The real key is to ask yourself what is meant by "A third of a stop"? A whole stop is a doubling of the light reaching the film, and when we say a third of that most people intuitively think that we are looking for an increase that when made three times will double the light exposure.

Let's say we have an exposure time of 2 seconds. One stop less exposure is simply 1 second. Easy. It is tempting to think that one third of a stop less exposure must therefore be 1.666 seconds, i.e. that subtracting a third of a second gives the right answer because doing that three times gets us to 1 s, which is one stop less.

The problem is that the amount to subtract depends on your starting exposure. If your starting exposure is 4 s, you need to subtract two-thirds of a second. If your starting exposure is 8 s, you need to subtract 1.333 s.

The reason is that a stop is defined as a ratio, and that fractions of a stop need also to be defined as ratios. If you do that, and define a portion of a stop as an amount to multiply your base exposure by, you automatically adjust for the size of the base exposure.

In this case, you need to find a number that when multiplied by itself three times gives you a total factor of one half. That is (1/2)^(1/3), the cube root of one half, or 0.7937. Exposure times need to be multiplied by this number if you want to reduce the exposure by one third of a stop (or by its reciprocal, 1.260, if you want to increase exposure by one third of a stop). For two thirds of a stop, you just do the multiplication twice, and for three thirds, or a full stop, three times.

In the case of apertures the same thing applies. You need to reduce the area of the aperture by a factor 0.7937 if you want to reduce the exposure by one third of a stop. Thus the diameter of the aperture must be reduced by the square root of this ratio, which is 0.8909. This number is a square root of a cube root, which is a sixth root - in this case, the sixth root of 0.5. The f-number is proportional to the reciprocal of the aperture diameter, so the multiplier for the f-number comes out as the sixth root of two, or 1.225.