6
\$\begingroup\$

How do I calculate the aperture size and area, considering an 35mm equivalent focal length value and non-equivalent aperture f-number (in terms of exposure, no multiplication of the crop factor of the sensor).

For example, I have a 1.5x crop sensor. The lens is equivalent to 50mm in 35mm terms and the camera has an aperture of 2.8. What is the aperture size and area?

I know that we probably shouldn't use the crop factor in the equation because the focal length given is already in 35mm terms. The aperture is as give by the company specifications, and I don't know if it needs multiplication or not.

\$\endgroup\$
0

7 Answers 7

13
\$\begingroup\$

The aperture size is a property of the lens only and does not depend on the crop factor. It does depend, however, on the actual focal length of the lens (not the "equivalent" focal length).

So you need to obtain the actual focal length by dividing by the crop factor.

actual-focal-length = equiv-focal-length / crop-factor

You can then calculate the size (diameter) of the aperture (strictly speaking the size of the entrance pupil, which is the image of the aperture) by dividing the focal length by the f-number (which does not need to be divided by the crop factor).

aperture size = actual-focal-length / F-value
              = (equiv-focal-length / crop-factor) / F-value

So in your case the calculation is:

aperture size = (50mm / 1.5) / 2.8
              = 11.9mm
\$\endgroup\$
5
  • \$\begingroup\$ So you need to obtain the actual focal length by dividing by the crop factor ...Sorry, I don't follow... The size of the aperture of that lens (50mm ƒ/2.8) on APS-C, Medium Format, 4/3's is... 17.857mm. Why are you suggesting to multiply (or divide) by the "crop factor"? \$\endgroup\$
    – BBking
    Oct 14, 2014 at 12:07
  • 2
    \$\begingroup\$ @BBking read the question again, the lens has a 50mm FF equivalent field of view on a 1.5x crop sensor, hence it's really a 33mm lens, hence the entrance pupil diameter (on all formats) is 11.9mm \$\endgroup\$
    – Matt Grum
    Oct 14, 2014 at 12:10
  • 4
    \$\begingroup\$ Well, I'd suggest to use a 35mm lens in the answer to avoid confusion. Because it's not right. For any prime lens, you don't obtain the focal length by dividing by the crop factor. I read the question, but would have answered it differently. \$\endgroup\$
    – BBking
    Oct 14, 2014 at 12:14
  • 1
    \$\begingroup\$ @BBking it is right, you do obtain the real focal length by dividing the 35mm equivalent length by the crop factor compared to 35mm (e.g. Nikon 1 kit lens is a 27-80 equivalent, divide by the crop factor (2.7) and you get 10-30mm which is what's printed on the lens barrel). I could change the answer but the question specifically asked how to calculate the aperture size when you have an equivalent focal length for real f-number. \$\endgroup\$
    – Matt Grum
    Oct 14, 2014 at 21:12
  • 1
    \$\begingroup\$ This answer makes no sense. You first say apeture is not dependant on the crop factor. Then you say focual length is dependant on crop factor and apeture is dependant on focal lengtth. SO APETURE IS DEPENDANT ON CROP FACTOR. \$\endgroup\$
    – Denis
    Feb 4, 2020 at 12:41
12
\$\begingroup\$

How do I calculate the aperture size and area

You divide the focal length by the aperture/F-stop value.

Infact, that's what the F-stop/aperture value is. It's a divider.

Sometimes written as ƒ2.8 (as an example) but a lot of people leave out the vinculum and should be written as ƒ/2.8.

Replace the ƒ with the focal length and that's the diameter of the pupil entrance. Examples below:

35mm, ƒ/4   = 8.75mm
50mm, ƒ/5.6 = 8.93mm
100mm, ƒ/11 = 9.09mm

It doesn't matter what sensor size the lens is on. It's not like the pupil entrance changes size if you put a lens on a different format.

The "crop factor" is only for field of view equivalence. Each format is different and none are superior.

And yes, the area is standard mathematics πr² (Pi, r squared) and does not change. So if the F-stop gives you the diameter, halve it to get the radius.

Do I use the crop factor in calculating aperture size and area?

No. The reason for this is answered here.

\$\endgroup\$
2
\$\begingroup\$

Your lens is a 35/2.8, but that doesn't mean the aperture is necessarily f/2.8. Just that f/2.8 is the widest it can be opened.

The f-number is the focal length divided by the aperture diameter:

f_number = focal_length / diameter

So, solving for the diameter:

diameter = focal_length/f_number.

So, in this case your diameter is 35mm/2.8 = 12.5mm. If the lens is set to f/2.8.

If the lens is set to f/8, however, then the diameter becomes 35/8 = 4.375mm.

Figuring out the area of the aperture does depend on the geometry of the aperture opening (number of blades, etc.) but most of us would probably just simplify it to using the area of a circle and πr2. So, wide open, the aperture area is:

π(12.5/2)2 ≈ 123mm2

\$\endgroup\$
2
\$\begingroup\$

Generally, we don't really care about the actual aperture size and area. We care a little bit more about the apparent size of the aperture as viewed through the front of the lens, as this affects depth of field.

For exposure, all we care about is the f number, like f/2.8, because the same f/number and shutter speed result in the same exposure regardless of sensor size — because exposure is per area.

For depth of field, it is approximately true that multiplying f-number by crop factor gives comparable equivalence. But this depends on a number of assumptions. (Details in the linked question and answer.)

And, in fact, more on all of this at Why don't comparisons of aperture take sensor size into account?

\$\endgroup\$
1
  • \$\begingroup\$ The only answer which mentions the DOF and equivalence, because often we equate Fstop with the bokeh effect, which indeed is affected by crop factor \$\endgroup\$
    – hbogert
    Aug 17, 2020 at 10:47
1
\$\begingroup\$

Original question was how to calculate 35mm equivalent for the lens. It is simple math and physics.

If you change one side of formula you have to change other too. The only thing that is constant is lens iris diameter. For example, a full frame 35mm f/1.4 lens has an aperture diameter of 25mm. This same lens on crop sensors, let's say on 1.5x crop, will act as a 52.5mm lens. Because the diameter of the iris did not change 52.5mm / 25mm = 2.1. So if you want to truly represent 35mm equivalent, then a 35mm f/1.4 full frame lens will be 52.5mm f/2.1 on a 1.5x crop sensor camera.

If you want to achieve exactly the same photo on a crop sensor you also need to calculate ISO equivalent. It is all about how much light the sensor has available. A full frame sensor has way more available light with same lens because the sensor is bigger. So for example if I take shot on full frame sensor with ISO 800 and f/2.8 you would need to set your crop sensor camera to ISO 360 and f/1.8 - this will give you exactly same image regarding to depth of field and noise ratio.

Manufacturers are lying to consumer when they sell for example Micro Four Third lenses and give you 35mm equivalent only for focal length. They never include f-stop 35mm equivalence. The f-stop doesn't magically change. Unfortunately we are bound by math and physic rules. So buying lenses for APS-C or Micro Four Thirds cameras, people have to be careful because they think they are buying longer lenses. Take, for example, the Panasonic 12-35mm f/2.8 — Panasonic will advertise it as 35 mm equivalent 24-70 f/2.8; that is totally wrong because in fact the lens is 24-70 f/5.6. So you will never get same image as 24-70 f/2.8.

So again when you calculate 35 mm equivalent you also need to calculate f-stop 35mm equivalent to be accurate because math will never work if you change only one side of the equation.

\$\endgroup\$
1
  • \$\begingroup\$ With regard to DoF the 35/1.4 will act as a 53/2.1 on a 1.5X crop. With regard to exposure it will still be an f/1.4 lens. That's because exposure (field density strength) doesn't change as we crop the FoV. \$\endgroup\$
    – Michael C
    Feb 8, 2017 at 20:18
1
\$\begingroup\$

The crop factor does not affect the aperture.

The aperture is given by the physical construction of the lens. It is a function of the focal length and the pupil.

f = focal length
D = Diameter of the pupil

Fnumber(N) = f/D

There are no other variables involved.

enter image description here

If you physically change the focal length, for example by using a teleconverter, then you do need to make adjustments to the Fstop.

\$\endgroup\$
-5
\$\begingroup\$

Right, the focal length remains the same no matter what size the sensor is.

The effective focal length however is based on sensor size, it's called crop factor, but that was implied in the OP's question initially, so I didn't find any reason to repeat it.

F-stop is based on the effective focal length, as I said, the aperture diameter is fixed, however effective focal length is not, even though the physical focal length is always fixed.

But crop factor is an abstraction, it's a way to calculate how to get a certain result. So yes, crop factor includes changes to F-stop, for-instance:

Sigma 18-35mm F/1.8 is the physical dimensions and obviously can't be changed.

However it's an APS-C sensor only lens, so the 35mm equivalence is approximately: Sigma 27-50mm F/2.8

The DoF is determined by the range to subject. When increasing the distance, because of the crop factor on any given focal length, then the DoF is increased accordingly. The total amount of light gathered by the sensor for the same exposure time and aperture, is also less by the same crop factor.

F-stop (not aperture) needs crop factor applied as well. Hope I have express myself more clearly now. And the downvotes, only shows, some are into alternative facts. Those seeking the truth, will however try it out and perhaps like those links. Considering all in all, I'll consider the downvotes medals of honour, the definitive proof I hit a nerve.

Videos and links to support the fact, can be seen here:

https://www.youtube.com/watch?v=lte9pa3RtUk

https://www.youtube.com/watch?v=f5zN6NVx-hY

https://www.youtube.com/watch?v=DtDotqLx6nA

https://www.dpreview.com/articles/2666934640/what-is-equivalence-and-why-should-i-care

https://www.youtube.com/watch?v=At_HegUQ3yQ

http://www.josephjamesphotography.com/equivalence/index.htm

\$\endgroup\$
6
  • \$\begingroup\$ Please edit your original answer instead of writing a second one. \$\endgroup\$
    – Caleb
    Mar 3, 2016 at 20:21
  • \$\begingroup\$ Also, this answer is flawed in the same way as your original answer. Please cite a reputable source that agrees with you. Or, try using the same lens on both a crop sensor camera and a full frame camera. You'll find that both report the same maximum aperture. Or, please explain how crop factor is an abstraction implies crop factor includes changes to F-stop. \$\endgroup\$
    – Caleb
    Mar 3, 2016 at 20:25
  • \$\begingroup\$ the effective focal length or effective aperture is an abstraction to describe the 35mm equivalence. You can see Tony Northrup demonstration on youtube if you like. \$\endgroup\$
    – Goat
    Apr 1, 2016 at 12:42
  • \$\begingroup\$ Added some more links to "reputable sources". \$\endgroup\$
    – Goat
    Feb 6, 2017 at 21:45
  • \$\begingroup\$ There's no such thing as exact equivalence. In terms of DoF the f-number is affected by the change in shooting distance about equally as the f-number multiplied by the crop factor at the original subject distance with the FF camera. But in terms of exposure the f-number is NOT affected by the sensor size. exposure is a measure of field density - that is light per unit area. \$\endgroup\$
    – Michael C
    Feb 8, 2017 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.