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    Re: Camera distortion of sky images. was Re: NG's "Midnight Fun
    From: George Huxtable
    Date: 2010 Jun 16, 15:14 +0100

    Marcel works fast.
    His first fit of the calibration function of Greg's camera, with only one
    adjustable parameter, is-
    Altitude in moa = 120 arc-tan ( pixels / 18161.80)
    This would model the case where all was explained by the simple tangent
    projection of sphere on to plane, with no additional distortion from the
    lens itself.
    His alternative fit, allowing adjustment of the "constant" multiplier to
    get the best result, produces-
    Altitude in moa = 133.94 arc-tan ( pixels / 20308.33)
    If there's a significant difference in the goodness-of-fit between those
    two, then the alteration of the multiplier term shows that there is a bit
    of distortion from the lens which has been added in as well, contributing a
    small addition, of about one-tenth, to the overall distortion. Which is
    perfectly plausible.
    But I suspect that it's hardly possible to make such a fine distinction
    between these two fits, given that the distortion is so small, in the
    present case.
    I presume that before the analysis was made, predicted altitudes of the
    Sun's centre had already been corrected for dip and refraction,.if we're
    looking into such fine distinctions.
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ----- Original Message -----
    From: "Marcel Tschudin" 
    Sent: Wednesday, June 16, 2010 9:01 AM
    Subject: [NavList] Re: Camera distortion of sky images. was Re: NG's
    "Midnight Fun
    | George, you wrote:
    | > Two alternative fitted equations are shown, one linear, one
    | > with little to choose between them, but it looks to me as though only
    | > line-of-fit has been drawn, the second-order fit.
    | My old version of Excel can only show the equation for the fit if the
    | fitted line is also drawn. This figure shows therefore both lines but
    | they are so close to each other that they can't be distinguished.
    | further on you tried the following:
    | > I have used that graph to check out my suggested model, in which radius
    | > the array varies as the tan of the subtended angle from the centre
    | > In this case, because each measurement has been equally disposed about
    | > centre point (a sensible practice which minimises distortion) the
    | > angle is half the altitude.
    | >
    | > So I have tried a function as follows-
    | >
    | > H(pixels) = 18140 Tan (altitude / 2) (in degrees)
    | >
    | > or, for those that prefer it the other way,
    | >
    | > Altitude in degrees = 2 arc-tan ( pixels / 18140)
    | >
    | > And I can report that that function fits exactly on top of the data and
    | > fit-line that's already drawn in, so that it's just not possible to see
    | > difference.
    | Yes, the tangent function correlates in this case better than
    | polynomials. I fitted your suggested function the following way to the
    | original data:
    | Altitude in degrees = 2 C arc-tan ( pixels / D)
    | Setting first C=1 (constant) results in a best fit for
    | D = 18161.80 (compared to what you estimated from the figure 18140).
    | The resulting standard deviation is only marginally greater than the
    | one obtained for the linear fit having two parameters (0.56 moa vs
    | 0.52 moa).
    | Fitting then the parameters C and D results in
    | C = 1.116192
    | D = 20308.33
    | The standard deviation is about the same than the one obtained for the
    | fitted second order polynomial having three parameters (0.22 moa).
    | For his 50mm lens Greg could therefore also use instead of the two
    | equations already provided the following ones:
    | either
    | Altitude in moa = 120 arc-tan ( pixels / 18161.80)
    | or, more precise
    | Altitude in moa = 133.94 arc-tan ( pixels / 20308.33)
    | Marcel

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