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Apertures, diameters and ratios. What the F?

  • 09-03-2013 12:10am
    #1
    Registered Users, Registered Users 2 Posts: 3,421 ✭✭✭


    My understanding of the f numbers that we use to describe aperture settings is that they are ratios related to focal length. Consider the following example - 50mm lens @ f2.8. The 2.8 bit is the expression of the diameter of the aperture in relation to the focal length of the lens. It is calculated like this

    50mm / 2.8 = 17.86mm This is the aperture diameter. What we want is the area of the aperture (which is a rough circle) so we can use our old friend pi r^2.

    Radius = 17.86mm / 2 = 8.93mm
    pi r^2 => 3.14159 * (8.93^2) = 250mm
    So, a 50m lens at f2.8, has an aperture area of 250mm sq.

    If you now run the same calculation for f4,
    50mm / 4 = 12.5mm (diameter)
    Radius 12.5 / 2 = 6.25mm
    pi r^2 => 3.14159 * (6.25^2) = 123mm
    Therefore the aperture area = 123mm sq

    So you can see the doubling
    f2.8 = 250mm sq
    f4 = 123mm sq
    etc
    etc

    Now here's my question.
    What about a 300mm lens at f2.8. 300mm/2.8 = 107mm (diameter). Can a 300mm lens really have an aperture of 10 cms ? Seems awful big physically to me. Consider that the aperture (actual blades) is normally near the back of the lens. Now think about the diameter of your lens mount. No where near 10cm so what am I missing here?


Comments

  • Registered Users, Registered Users 2 Posts: 719 ✭✭✭calnand


    dnme wrote: »
    My understanding of the f numbers that we use to describe aperture settings is that they are ratios related to focal length. Consider the following example - 50mm lens @ f2.8. The 2.8 bit is the expression of the diameter of the aperture in relation to the focal length of the lens. It is calculated like this

    50mm / 2.8 = 17.86mm This is the aperture diameter. What we want is the area of the aperture (which is a rough circle) so we can use our old friend pi r^2.

    Radius = 17.86mm / 2 = 8.93mm
    pi r^2 => 3.14159 * (8.93^2) = 250mm
    So, a 50m lens at f2.8, has an aperture area of 250mm sq.

    If you now run the same calculation for f4,
    50mm / 4 = 12.5mm (diameter)
    Radius 12.5 / 2 = 6.25mm
    pi r^2 => 3.14159 * (6.25^2) = 123mm
    Therefore the aperture area = 123mm sq

    So you can see the doubling
    f2.8 = 250mm sq
    f4 = 123mm sq
    etc
    etc

    Now here's my question.
    What about a 300mm lens at f2.8. 300mm/2.8 = 107mm (diameter). Can a 300mm lens really have an aperture of 10 cms ? Seems awful big physically to me. Consider that the aperture (actual blades) is normally near the back of the lens. Now think about the diameter of your lens mount. No where near 10cm so what am I missing here?

    It's a bit hard to get your head around it but "theoretically" the opening is that size. But in real life the opening is probably the inner width of the lens. Correct me if I'm wrong.


  • Registered Users, Registered Users 2 Posts: 5,155 ✭✭✭PopeBuckfastXVI


    http://www.matthewlin.com/mynikon/ED300mm/ED300.htm

    Check out the photo of the blades.


  • Registered Users, Registered Users 2 Posts: 3,421 ✭✭✭dnme


    http://www.matthewlin.com/mynikon/ED300mm/ED300.htm

    Check out the photo of the blades.

    Thanks for that Hugh. A good example of an actual 10-11cm diameter (and an interesting article). But I know of a few examples where people have taken lenses apart and found the aperture to be nothing like that size. So there's obviously more than one way to calculate f numbers. Even if you consider Canon's 70-200mm f2.8. AFAICT nowhere along the barrel could they fit a hole 10-11cm's wide. This is just the nerd in me, I like to understand these things :)


  • Registered Users, Registered Users 2 Posts: 6,708 ✭✭✭DaireQuinlan


    The f/stop isn't measured using the physical aperture, but the 'effective aperture', which (broadly speaking) is the size of the aperture taking into account the lens elements in front of it, or the 'entrance pupil'.


  • Registered Users, Registered Users 2 Posts: 3,421 ✭✭✭dnme


    The f/stop isn't measured using the physical aperture, but the 'effective aperture', which (broadly speaking) is the size of the aperture taking into account the lens elements in front of it, or the 'entrance pupil'.

    So to simplify, you could say that the f number represents a percentage of area in relation to adjacent elements ?


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  • Registered Users, Registered Users 2 Posts: 6,708 ✭✭✭DaireQuinlan


    dnme wrote: »
    So to simplify, you could say that the f number represents a percentage of area in relation to adjacent elements ?

    that doesn't really simplify it at all :-D


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