Zoom and magnify relation

Discussion in 'Digital Photography' started by juky, Apr 5, 2008.

  1. juky

    juky Guest

    Dear all,

    I have a mathematical question regarding relation between zoom factor
    and magnification. My video camera shows me the current zoom factor
    selected. Focal is 3,5-91 mm. I'm looking for mathematic formulas
    which give me the zoom factor to be selected so that an object in the
    shot will appear halved, double, etc. I premise the object distance is
    unknown.
    Is there any way?

    Thank you.
    Juky
    juky, Apr 5, 2008
    #1
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  2. juky <> wrote:
    >I have a mathematical question regarding relation between zoom factor
    >and magnification.


    Those two have nothing to do with each other.
    The zoom factor is simply the longest focal length divided by the
    shortest focal lenght of a zoom lens. For your example below the zoom
    factor would be 26x which seems to be rather large.

    The (apparent) magnification on the other hand depends on the focal
    length and the size of the sensor. Commonly 50mm focal length is
    considered a 1x magnification or 'normal' lens on a 36mm sensor
    (correspondingly ~33mm focal length on DX-size sensors).

    > My video camera shows me the current zoom factor
    >selected. Focal is 3,5-91 mm.


    That doesn't make much sense because the zoom factor is a constant for a
    given lens and does not change. Do you mean it shows you the current
    focal length?

    >I'm looking for mathematic formulas
    >which give me the zoom factor to be selected so that an object in the
    >shot will appear halved, double, etc. I premise the object distance is
    >unknown.


    Assuming you are talking about the focal length rather than the zoom
    factor, then the magnification and the focal length are 1:1. A doubling
    in focal length also doubles the apparent magnification.
    If you know the size of the sensor or the 'normal' focal length (like in
    50mm for 36mm sensor) for your camera then you can also directly convert
    a given focal length to an apparent magnification factor, e.g. a 400mm
    lens corresponds to a 8x magnification on a 36mm sensor.

    jue
    Jürgen Exner, Apr 5, 2008
    #2
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  3. juky

    juky Guest

    On Apr 6, 12:40 am, Jürgen Exner <> wrote:
    > juky<> wrote:
    > >I have a mathematical question regarding relation between zoom factor
    > >and magnification.

    >
    > Those two have nothing to do with each other.
    > The zoom factor is simply the longest focal length divided by the
    > shortest focal lenght of a zoom lens. For your example below the zoom
    > factor would be 26x which seems to be rather large.
    >
    > The (apparent) magnification on the other hand depends on the focal
    > length and the size of the sensor. Commonly 50mm focal length is
    > considered a 1x magnification or 'normal' lens on a 36mm sensor
    > (correspondingly ~33mm focal length on DX-size sensors).
    >
    > > My video camera shows me the current zoom factor
    > >selected. Focal is 3,5-91 mm.

    >
    > That doesn't make much sense because the zoom factor is a constant for a
    > given lens and does not change. Do you mean it shows you the current
    > focal length?
    >
    > >I'm looking for mathematic formulas
    > >which give me the zoom factor to be selected so that an object in the
    > >shot will appear halved, double, etc. I premise the object distance is
    > >unknown.

    >
    > Assuming you are talking about the focal length rather than the zoom
    > factor, then the magnification and the focal length are 1:1. A doubling
    > in focal length also doubles the apparent magnification.
    > If you know the size of the sensor or the 'normal' focal length (like in
    > 50mm for 36mm sensor) for your camera then you can also directly convert
    > a given focal length to an apparent magnification factor, e.g. a 400mm
    > lens corresponds to a 8x magnification on a 36mm sensor.
    >
    > jue


    The additional information I can get right now about video camera is
    the image sensor=1/4" CCD
    juky, Apr 6, 2008
    #3
  4. juky

    Paul Furman Guest

    David Ruether wrote:
    > "juky" <> wrote in message news:...
    >
    >> I have a mathematical question regarding relation between zoom factor
    >> and magnification. My video camera shows me the current zoom factor
    >> selected. Focal is 3,5-91 mm. I'm looking for mathematic formulas
    >> which give me the zoom factor to be selected so that an object in the
    >> shot will appear halved, double, etc. I premise the object distance is
    >> unknown.
    >> Is there any way?
    >> Thank you.
    >> Juky

    >
    > ???
    > Well, there is the basic (and most important, for practical purposes...)
    > visual effect of zooming as seen in the viewfinder. I can't see much
    > use for anything else, especially since any "zoom factor" indicator
    > needed to set the "ZF" on the camcorder would be only an
    > approximation, if present at all. Further, applying a zoom factor (if
    > you could do it accurately) to a FL gives a good idea of lens coverage
    > and changed magnification only at long focal lengths. At short FLs, it
    > doesn't (I.E., going from a 35mm equivalent FL of 400mm to 800mm
    > would almost exactly double the size of the image, and reduce to very
    > nearly one forth the area covered at a given distance - but going from
    > a 35mm equivalent FL of 20mm to one of 40mm would not do the
    > same thing, due to simple geometry).


    I drew a little diagram in CAD for an AP-S sensor doubling the focal
    length & got these numbers assuming the angle of view is a simple
    x/hourglass geometry:
    10mm 135deg 1.4x
    20mm 100deg 1.6x
    40mm 62deg 1.6x
    80mm 33deg 1.8x
    160mm 17deg 1.9x
    320mm 9deg 2.3x <-maybe rounding errors?
    640mm 4deg 2.0x
    1280mm 2deg

    > Additionally, the focus distance
    > must be kept constant since some lenses change their FLs with focus.
    > So, fergit it...;-)
    Paul Furman, Apr 6, 2008
    #4
  5. On Apr 5, 3:21 pm, juky <> wrote:
    > Dear all,
    >
    > I have a mathematical question regarding relation between zoom factor
    > and magnification. My video camera shows me the current zoom factor
    > selected. Focal is 3,5-91 mm. I'm looking for mathematic formulas
    > which give me the zoom factor to be selected so that an object in the
    > shot will appear halved, double, etc. I premise the object distance is
    > unknown.
    > Is there any way?
    >
    > Thank you.
    > Juky


    There are several meanings for the term "magnification" in cameras.
    In pure optics it means the ratio of the image size to the object
    size, and unless you are into macro and microphotography, the number
    is less than (usually MUCH less) than one.

    I think the meaning you are questioning is the ratio of the current
    focal length to the shortest (lens zoomed all the way out). Just
    divide the current focal length by the shortest possible focal length.
    Don Stauffer in Minnesota, Apr 6, 2008
    #5
  6. David Ruether <> wrote:

    > So, if we assume a 24mm x 36mm FF film
    > frame, the diagonal (for finding the angle of coverage on the
    > format diagonal for a particular focal length lens) is 24mm
    > squared (576) plus 36mm squared (1296) added together (to
    > get 1872) - and the square root of this is 43.27mm. Let's call
    > it 44mm...;-) To find the angle of coverage of a lens with this
    > format, a right triangle can be made by drawing a vertical line
    > perpendicular to the frame diagonal line, and intersecting it in
    > its center (as for a non-shifted lens of a given focal length, with
    > the focal length being the length of that new line). That gives us,
    > for an 18mm lens, a vertical line side of the triangle of 18mm,
    > and a horizontal line side of the triangle of 1/2 of 44mm, or
    > 22mm. We now have everything needed except a calculator...;-).
    > In a right triangle, the opposite side divided by the adjacent
    > side (not the diagonal hypotenuse) gives the "tangent" for the
    > angle in degrees between the vertical line and the hypotenuse,
    > or 1.22. The tangent angle for that value is 50.66 degrees,
    > which is one half of the full angle (remember that the frame
    > diagonal was split in two to solve for the angle in a right
    > triangle?), so the full angle is 101.32 degrees...
    > Continuing --


    > 9mm = 135.50 degrees, or 1.34x more than an 18mm lens.
    > 18mm = 101.32 degrees, or 1.62x more than a 36mm lens.
    > 36mm = 62.86 degrees, or 1.85x more than a 72mm lens.
    > 72mm = 33.98 degrees, or 1.96x more than a 144mm lens.
    > 144mm = 17.37 degrees, or 1.99x more than a 288mm lens.
    > 288mm = 8.74 degrees, or 2x more than a 576mm lens.
    > 576mm = 4.37 degrees, or 2x more than a 1152mm lens.
    > 1152mm = 2.19 degrees, or 2x more than a 2304mm lens.


    > So you can see that as focal lengths are halved or doubled, the
    > angle of view and image magnification does not follow those ratios
    > toward the wide angle end of the possible lens focal length range,
    > though it does for long focal lengths for the format...


    Tangents and sines approximate to linearity and each other as angles
    become small (you could have done your calculation using sines instead
    of tangents).

    --
    Chris Malcolm DoD #205
    IPAB, Informatics, JCMB, King's Buildings, Edinburgh, EH9 3JZ, UK
    [http://www.dai.ed.ac.uk/homes/cam/]
    Chris Malcolm, Apr 11, 2008
    #6
  7. juky

    Paul Furman Guest

    David Ruether wrote:

    >
    > 9mm = 135.50 degrees, or 1.34x more than an 18mm lens.
    > 18mm = 101.32 degrees, or 1.62x more than a 36mm lens.
    > 36mm = 62.86 degrees, or 1.85x more than a 72mm lens.
    > 72mm = 33.98 degrees, or 1.96x more than a 144mm lens.
    > 144mm = 17.37 degrees, or 1.99x more than a 288mm lens.
    > 288mm = 8.74 degrees, or 2x more than a 576mm lens.
    > 576mm = 4.37 degrees, or 2x more than a 1152mm lens.
    > 1152mm = 2.19 degrees, or 2x more than a 2304mm lens.
    >
    > So you can see that as focal lengths are halved or doubled, the
    > angle of view and image magnification does not follow those ratios
    > toward the wide angle end of the possible lens focal length range,
    > though it does for long focal lengths for the format...


    OK so field of view doesn't change as quick for the wide angles.
    Starting with a super long lens, halving the focal length pretty much
    doubles the field of view down to 150mm, then it doesn't get as much
    wider as one would think.
    Paul Furman, Apr 11, 2008
    #7
  8. Paul Furman <> writes:

    >> So you can see that as focal lengths are halved or doubled, the
    >> angle of view and image magnification does not follow those ratios
    >> toward the wide angle end of the possible lens focal length range,
    >> though it does for long focal lengths for the format...


    >OK so field of view doesn't change as quick for the wide angles.
    >Starting with a super long lens, halving the focal length pretty much
    >doubles the field of view down to 150mm, then it doesn't get as much
    >wider as one would think.


    However, the sizes of objects in the image *is* accurately predicted by
    the focal length ratio, all the way to the widest wide angle, as long as
    you're still talking about rectilinear lenses (not fisheyes).

    Dave
    Dave Martindale, Apr 11, 2008
    #8
  9. Dave Martindale <> wrote:
    > Paul Furman <> writes:


    >>> So you can see that as focal lengths are halved or doubled, the
    >>> angle of view and image magnification does not follow those ratios
    >>> toward the wide angle end of the possible lens focal length range,
    >>> though it does for long focal lengths for the format...


    >>OK so field of view doesn't change as quick for the wide angles.
    >>Starting with a super long lens, halving the focal length pretty much
    >>doubles the field of view down to 150mm, then it doesn't get as much
    >>wider as one would think.


    > However, the sizes of objects in the image *is* accurately predicted by
    > the focal length ratio, all the way to the widest wide angle, as long as
    > you're still talking about rectilinear lenses (not fisheyes).


    Except for the widening towards the edges in wide angle rectilinear
    perspectives, which has nothing to do with the lens, being a simple
    consequence of pinhole projection geometry when the image is viewed
    from a narrower perspective than it was taken from (as it usually is
    with wida angles).

    --
    Chris Malcolm DoD #205
    IPAB, Informatics, JCMB, King's Buildings, Edinburgh, EH9 3JZ, UK
    [http://www.dai.ed.ac.uk/homes/cam/]
    Chris Malcolm, Apr 11, 2008
    #9
  10. A long time ago I wrote:
    >>> However, the sizes of objects in the image *is* accurately predicted by
    >>> the focal length ratio, all the way to the widest wide angle, as long as
    >>> you're still talking about rectilinear lenses (not fisheyes).


    "David Ruether" <> writes:
    >> How can this be? If you shoot a very large grid of small squares from a
    >> constant distance (assuming constant sensor size and distance - and with
    >> the subject grid parallel with the sensor), and keep halving the lens focal
    >> lengths, the angles of view will progressively not follow the FL reduction
    >> ratio (as I showed earlier), resulting, I would think, in less than the
    >> number of squares visible in any direction in the image than a straight
    >> following of the ratio would predict. In other words, the size of subjects
    >> as rendered in the image is also not proportional to the FL ratio when it
    >> comes to short FL lenses (they are larger...). Am I missing something?;-)


    >Ah, I just finished a MUCH more detailed coverage of this, which
    >is going up now on my web page, listed in the "Articles" page as,
    >"On Lens Angles Of View, Magnification, And Perspective. The
    >direct URL is --
    >www.donferrario.com/ruether/lens-angle-of-view-and-perspective.htm


    Sorry I didn't comment sooner - I never saw this reply. But you are
    wrong when you say "resulting ... in less than the number of squares
    visible ... than a straight following of ratio would predict".

    Say you go from 36 to 18 mm lens focal length. The FOV goes from 62.86
    to 101.32 degrees, less than a factor of 2 increase. But, as long as
    you are shooting a flat wall with a camera aimed perpendicular to that
    wall, you will still get twice as many squares visible. The true field
    width, on the plane containing the grid, *is* doubled.

    If you don't believe me, draw a scale diagram and measure. Or do the
    math using similar triangles. You've decreased the lens-image distance
    by a factor of two, while keeping the image width the same, so the
    tangent of half the FOV has *exactly doubled*, even though the angle
    itself has not doubled. On the subject side, the triangle is similar
    to the image-side triangle, so the subject-side tangent of half the FOV
    is *also* doubled. But the lens-subject distance has not changed, so
    the amount of the subject visible has exactly doubled.

    Or the handwaving argument: you haven't doubled the FOV angle, but the
    additional squares you can see are being increasingly foreshortened by
    the very wide angle they are off-axis, so you get more of them in each
    degree of extra visual angle. The two effects cancel, and you get
    exactly twice as many squares in not twice as many degrees.

    By the way, your argument would be correct if the squares were drawn on
    a sphere centered on the lens, since the squares would always appear a
    certain number of degrees wide everywhere in the field. But we're
    assuming a flat subject, not a spherical one.

    As it happens, I write camera software for video games, so I can create
    any wideangle lens I want, no matter how extreme, as long as it is a
    distortion-free rectilinear lens. When I want, say, "50% more field of
    view", I do *not* multiply the camera FOV by 1.5. I divide the
    effective lens focal length by 1.5 instead. (This is actually
    implemented by multiplying the tangent of half of the FOV angle by
    1.5). This gives the desired effect: making all objects 2/3 the size
    they were, and showing 1.5 times more width of any plane perpendicular
    to the line of sight - even though it does not double the FOV angle.
    I just recognize that FOV is nonlinearly related to what I really want
    to control, which is magnification.

    Dave
    Dave Martindale, Apr 15, 2008
    #10
  11. juky

    Paul Furman Guest

    Dave Martindale wrote:

    > Or the handwaving argument: you haven't doubled the FOV angle, but the
    > additional squares you can see are being increasingly foreshortened by
    > the very wide angle they are off-axis, so you get more of them in each
    > degree of extra visual angle. The two effects cancel, and you get
    > exactly twice as many squares in not twice as many degrees.
    >
    > By the way, your argument would be correct if the squares were drawn on
    > a sphere centered on the lens, since the squares would always appear a
    > certain number of degrees wide everywhere in the field. But we're
    > assuming a flat subject, not a spherical one.


    This is sort of like what is explained below... about the egghead
    effect. There is no distortion of a flat field subject in a super wide
    rectilinear view.
    Paul Furman, Apr 15, 2008
    #11
  12. Paul Furman <> wrote:
    > Dave Martindale wrote:


    >> Or the handwaving argument: you haven't doubled the FOV angle, but the
    >> additional squares you can see are being increasingly foreshortened by
    >> the very wide angle they are off-axis, so you get more of them in each
    >> degree of extra visual angle. The two effects cancel, and you get
    >> exactly twice as many squares in not twice as many degrees.
    >>
    >> By the way, your argument would be correct if the squares were drawn on
    >> a sphere centered on the lens, since the squares would always appear a
    >> certain number of degrees wide everywhere in the field. But we're
    >> assuming a flat subject, not a spherical one.


    > This is sort of like what is explained below... about the egghead
    > effect. There is no distortion of a flat field subject in a super wide
    > rectilinear view.


    Except for the inevitable distortion that happens if you view a very
    wide angle shot from a position where the image is encompassed by our
    eye from a smaller angle (which it usually is), and which is a
    widening of things at the edges. It's just a natural feature of
    changing viewing geometry, nothing to do with lenses, happens with
    pinhole wide angle views :)

    --
    Chris Malcolm DoD #205
    IPAB, Informatics, JCMB, King's Buildings, Edinburgh, EH9 3JZ, UK
    [http://www.dai.ed.ac.uk/homes/cam/]
    Chris Malcolm, Apr 17, 2008
    #12
  13. On Apr 17, 8:03 am, "David Ruether" <> wrote:
    > "Chris Malcolm" <> wrote in messagenews:...
    > > Paul Furman <> wrote:
    > >> Dave Martindale wrote:
    > >>> Or the handwaving argument: you haven't doubled the FOV angle, but the
    > >>> additional squares you can see are being increasingly foreshortened by
    > >>> the very wide angle they are off-axis, so you get more of them in each
    > >>> degree of extra visual angle. The two effects cancel, and you get
    > >>> exactly twice as many squares in not twice as many degrees.
    > >> This is sort of like what is explained below... about the egghead
    > >> effect. There is no distortion of a flat field subject in a super wide
    > >> rectilinear view.

    > > Except for the inevitable distortion that happens if you view a very
    > > wide angle shot from a position where the image is encompassed by
    > > our eye from a smaller angle (which it usually is), and which is a
    > > widening of things at the edges. It's just a natural feature of
    > > changing viewing geometry, nothing to do with lenses, happens with
    > > pinhole wide angle views :)
    > > --
    > > Chris Malcolm

    >
    > I have quite a bit on my web page on lens "distortion" (the articles
    > index is atwww.donferrario.com/ruether/articles.html), but there
    > is really no distortion in super-wide images (which you *almost*
    > get to at the end of the above...;-), just an image with possibly
    > "unexpected/unfamiliar" appearance (I included a couple of samples
    > of these in the article below...;-). Paul Furman made a CAD drawing
    > of a hemisphere as imaged near the corner of a S-W WA image
    > that nicely demonstrates that for sections of "distorted" rounded
    > objects that are taken parallel with the sensor, there are no
    > deformations in the image (as you note above, using a different
    > process). The difference is in semantics - I don't call "distorted",
    > images that are really not...;-) The article that includes the photos
    > and drawing is at --www.donferrario.com/ruether/lens-angle-of-view-and-perspective.htm
    > --
    > David Ruether
    >
    > www.donferrario.com/ruether


    One way to envision a distortion-free lens is to map out the image a
    pinhole camera produces. Any ray entering a pinhole lens MUST
    continue in a straight line, since there is nothing in the lens to
    deviate it. When distortion happens, the chief ray is not following
    the normal "thin lens" mapping that requires the chief ray to transit
    the lens and aperture undeviated.

    As long as the chief ray transits the lens undeviated there is no
    actual distortion in the lens design sense of the word.
    Don Stauffer in Minnesota, Apr 17, 2008
    #13
  14. juky

    Paul Furman Guest

    David Ruether wrote:
    > "Chris Malcolm" <> wrote in message news:...
    >> Paul Furman <> wrote:
    >>> Dave Martindale wrote:

    >
    >>>> Or the handwaving argument: you haven't doubled the FOV angle, but the
    >>>> additional squares you can see are being increasingly foreshortened by
    >>>> the very wide angle they are off-axis, so you get more of them in each
    >>>> degree of extra visual angle. The two effects cancel, and you get
    >>>> exactly twice as many squares in not twice as many degrees.

    >
    >>> This is sort of like what is explained below... about the egghead
    >>> effect. There is no distortion of a flat field subject in a super wide
    >>> rectilinear view.

    >
    >> Except for the inevitable distortion that happens if you view a very
    >> wide angle shot from a position where the image is encompassed by
    >> our eye from a smaller angle (which it usually is), and which is a
    >> widening of things at the edges. It's just a natural feature of
    >> changing viewing geometry, nothing to do with lenses, happens with
    >> pinhole wide angle views :)
    >> --
    >> Chris Malcolm

    >
    > I have quite a bit on my web page on lens "distortion" (the articles
    > index is at www.donferrario.com/ruether/articles.html), but there
    > is really no distortion in super-wide images (which you *almost*
    > get to at the end of the above...;-), just an image with possibly
    > "unexpected/unfamiliar" appearance (I included a couple of samples
    > of these in the article below...;-). Paul Furman made a CAD drawing
    > of a hemisphere as imaged near the corner of a S-W WA image
    > that nicely demonstrates that for sections of "distorted" rounded
    > objects that are taken parallel with the sensor, there are no
    > deformations in the image (as you note above, using a different
    > process). The difference is in semantics - I don't call "distorted",
    > images that are really not...;-) The article that includes the photos
    > and drawing is at --
    > www.donferrario.com/ruether/lens-angle-of-view-and-perspective.htm


    I had to try drawing it to convince myself based on David's description.
    It's the way the wide angles 'chop up' 3D objects that creates
    distortion but flat subjects are not distorted. Long lenses see 3D
    objects more like as if it was an isometric view with close to the same
    angle for nearer & further where wide angles have more difference in
    angle of view.
    Paul Furman, Apr 17, 2008
    #14
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