real image resolution

Discussion in 'Digital Photography' started by Alan, Dec 22, 2005.

  1. Alan

    Alan Guest

    I want to determine for a given-sized object how many pixels will
    be on what size of the object image at some distance. Does anyone know
    of a good web site that provides an introduction or tutorial to this
    subject?

    I have tried many web searches but cannot find one that
    addresses this in the sea of hits I get.
    Thanks, Alan
     
    Alan, Dec 22, 2005
    #1
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  2. Alan

    Deep Reset Guest

    "Alan" <> wrote in message
    news:...
    > I want to determine for a given-sized object how many pixels will
    > be on what size of the object image at some distance. Does anyone know
    > of a good web site that provides an introduction or tutorial to this
    > subject?
    >
    > I have tried many web searches but cannot find one that
    > addresses this in the sea of hits I get.
    > Thanks, Alan


    Well, you need to know the focal length of the lens, the dimensions and
    resolution of the sensor, the object distance and the size of the object,
    but the rest is simple trigonometry..
    What do you need to know?
     
    Deep Reset, Dec 22, 2005
    #2
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  3. "Alan" <> writes:
    > I want to determine for a given-sized object how many pixels will
    >be on what size of the object image at some distance. Does anyone know
    >of a good web site that provides an introduction or tutorial to this
    >subject?


    It's basic proportional math, if you ignore lens distortion and use a
    pinhole model for the lens.

    To a first approximation, the ratio of the size of the object to the
    distance between lens and object is equal to the ratio of the size of
    the image to the distance between lens and image. In other words:

    size_of_object/object_distance = size_of_image/image_distance

    Rewriting that

    size_of_image = size_of_object * image_distance / object_distance

    Now, you know the size of the object and the object distance. If the
    object is far away (many times the lens focal length), you can just
    assume that the image_distance is equal to the focal length of the lens.
    (Note: you need the *real* focal length, not the "35 mm equivalent"
    focal length).

    For example, suppose you have a P&S digicam whose lens FL is 7 mm on the
    wide angle setting. The subject is a rock 1 m wide at a distance of 10 m.
    The size of the image on the sensor is 1 * 0.007 / 10 = 0.0007 m,
    or 0.7 mm, or 700 um.

    When the object is closer to the lens, the image distance increases.
    You can calculate it by solving the equation

    1/lens_FL = 1/object_distance + 1/image_distance

    For *really* close subjects, you have to worry about measuring distances
    from the lens front and real principal planes. But neither of these
    effects is of any importance for normal-distance subjects.

    Once you've calculated the size of the image on the sensor in distance
    units, you can convert it to pixels by dividing by the pixel pitch.
    Suppose your digicam has a sensor with 2.5 um pixel pitch. Then our
    700-um-wide image from the example above is 700/2.5 = 280 pixels wide.

    You can sometimes find the pixel pitch for particular cameras on review
    sites like dpreview. Or it's often in the EXIF data placed in every
    image by the camera.

    Does this make sense?

    Dave
     
    Dave Martindale, Dec 23, 2005
    #3
  4. Alan

    Marvin Guest

    Dave Martindale wrote:
    > "Alan" <> writes:
    >
    >> I want to determine for a given-sized object how many pixels will
    >>be on what size of the object image at some distance. Does anyone know
    >>of a good web site that provides an introduction or tutorial to this
    >>subject?

    >
    >
    > It's basic proportional math, if you ignore lens distortion and use a
    > pinhole model for the lens.
    >
    > To a first approximation, the ratio of the size of the object to the
    > distance between lens and object is equal to the ratio of the size of
    > the image to the distance between lens and image. In other words:
    >
    > size_of_object/object_distance = size_of_image/image_distance
    >
    > Rewriting that
    >
    > size_of_image = size_of_object * image_distance / object_distance
    >
    > Now, you know the size of the object and the object distance. If the
    > object is far away (many times the lens focal length), you can just
    > assume that the image_distance is equal to the focal length of the lens.
    > (Note: you need the *real* focal length, not the "35 mm equivalent"
    > focal length).
    >
    > For example, suppose you have a P&S digicam whose lens FL is 7 mm on the
    > wide angle setting. The subject is a rock 1 m wide at a distance of 10 m.
    > The size of the image on the sensor is 1 * 0.007 / 10 = 0.0007 m,
    > or 0.7 mm, or 700 um.
    >
    > When the object is closer to the lens, the image distance increases.
    > You can calculate it by solving the equation
    >
    > 1/lens_FL = 1/object_distance + 1/image_distance
    >
    > For *really* close subjects, you have to worry about measuring distances
    > from the lens front and real principal planes. But neither of these
    > effects is of any importance for normal-distance subjects.
    >
    > Once you've calculated the size of the image on the sensor in distance
    > units, you can convert it to pixels by dividing by the pixel pitch.
    > Suppose your digicam has a sensor with 2.5 um pixel pitch. Then our
    > 700-um-wide image from the example above is 700/2.5 = 280 pixels wide.
    >
    > You can sometimes find the pixel pitch for particular cameras on review
    > sites like dpreview. Or it's often in the EXIF data placed in every
    > image by the camera.
    >
    > Does this make sense?
    >
    > Dave


    You can work through the math, if you have the data on the lens and sensor, or you cam
    make a quick estimate from what you see on the viewfinder. If, for example, you judge the
    object's width is 1/5th of the width of the image, and the image is 2000 pixels wide,then
    the image will be about 1/5 X 2000 = 400 pixels wide. It isn't exact, but it may be a
    good enough measure.
     
    Marvin, Dec 23, 2005
    #4
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