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Horizon & Ellipse Measurements

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The projection of a sphere onto a flat screen is an ellipse or related conic curve. This shape depends on the position of the sphere, in units of its radius and relative to the line of sight, and the extent of the angular field of view. If the camera points directly towards the centre of the sphere then the projected outline is exactly a circle. As the camera is tilted further away from the centre of the sphere, the ellipticity and the pixel area of the projected shape both increase.

In some restricted circumstances the elliptical shape of disk image can be measured, and thus the camera's field of view and displacement of the sphere can be inferred. In other cases, when the field of view is narrow or the sphere lies close to the line of sight, the projected disk is approximately circular. These circumstances are only distinguishable when the outline of the globe is measurable around most of its circumference. If only a small arc is visible, or if the horizon looms large and flat, then the global measurements are indecisive and practically uninterpretable as either lower and upper scaling limits.

Suppose that an ellipse is centred at the origin and its major axis is in the x direction. The semimajor axis length is a and the semiminor axis is b. Points (x,y) on this standard, centred ellipse obey the relation

x² / a² + y² / b² = 1.

Then suppose that we transform the coordinates by rotating the ellipse through an anticlockwise angle φ, and shift the centre of the ellipse to a point (X*,Y*). The coordinates of points on this transformed ellipse (X,Y), are related to the old coordinates via

x = (X−X*) cos φ + (Y−Y*) sin φ,
y = -(X−X*) sin φ + (Y−Y*) cos φ.

After substituting these expressions into the equation of the ellipse in centred coordinates, it can be shown that the equation of the displaced and rotated ellipse has the following polynomial form,

C_xx X² + C_yy Y² + C_xy XY + C_x X + C_y Y + C₀ = 0.

The constant coefficients depend on the parameters of the ellipse, its rotation and displacement as follows,

C_xx = b² cos² φ + a² sin² φ,
C_yy = b² sin² φ + a² cos² φ,
C_xy = 2 cos φ sin φ ( b² − a² ),
C_x = -(2C_xx X* +C_xy Y* ),
C_y = -(2C_yy Y* +C_xy X* ),
C₀ = C_xx X*² +C_yy Y*² +C_xy X* Y* − a² b².

Thus, if we know the axes of the ellipse (a, b), its orientation (φ) and location of its centre then we can determine the coefficients of the equation of this ellipse. Conversely, if we measure enough points on the outline of a picture of an ellipse then we can infer the C coefficients, and thereby solve for the parameters (a, b, φ, X*, Y*). We obtain the solution as follows.

k_xx ≡ C_xx / C₀ ,
k_yy ≡ C_yy / C₀ ,
k_xy ≡ C_xy / C₀ ,
k_x ≡ C_x / C₀ and
k_y ≡ C_y / C₀.

We measure the screen coordinates of representative points on the ellipse, and substitute them into the polynomial equation of the ellipse. The k and C coefficients are initially unknown, but must be the same for all measurements. If a measurement labelled i has screen coordinates (X_i,Y_i), then we can obtain a linear relationship between the k coefficients,

k_xx X_i² + k_yy Y_i² + k_xy X_iY_i + k_x X_i + k_y Y_i = -1.

Measuring five unique points on the ellipse provides five independent linear equations (i=1,2,3,4,5), which we can solve for the all the k values. The parameters of the ellipse and its transformation are, explicitly,

X* = ( k_xy k_y − 2k_x k_yy ) / ( 4 k_xx k_yy - k_xy² ) ,
Y* = ( k_xy k_x − 2k_y k_xx ) / ( 4 k_xx k_yy - k_xy² ) ,
φ = ½ atan( k_xy , k_yy − k_xx ) ,
a² = ½ C₀ { k_xx + k_yy + [ (k_yy − k_xx)² +k_xy² ]^½ } ,
b² = ½ C₀ { k_xx + k_yy − [ (k_yy − k_xx)² +k_xy² ]^½ } ,
C₀ = 4 ( k_xx X*² +k_yy Y*² +k_xy X* Y* − 1 ) / ( 4 k_xx k_yy − k_xy² ) .

Suppose that we have a sphere centred at (x₀,y₀,z₀) relative to the camera, and points measured on the surface of the sphere are labelled as (x_i,y_i,z_i). A ray drawn from this surface point back to the centre of the camera will intersect the image plane at the pixel location (X_i,Y_i), where (0,0) represents the centre of the screen. These rays obey the equations

x_i / z_i = X_i ⁄ F ,
y_i / z_i = Y_i / F.

Here we define the constant F = the effective separation between the image plane and the centre of the camera, measured in pixels. The angular field of view is defined by half-angles tan θ_x = ΔX/2F and tan θ_y = ΔY/2F, where ΔX and ΔY are the pixel width and height of the screen.

We define units of measurement of the sphere (not its projected image) in terms of the spherical radius, r=1. It follows that

( x_i - x₀ )² + ( y_i - y₀ )² + ( z_i - z₀ )² = 1.

The horizon of the sphere consists of points where the line-of-sight ray meets the sphere at right angles. In terms of a vector product, the LOS ray and the radius line are perpendicular,

(x_i, y_i, z_i) • (x_i - x₀, y_i - y₀, z_i - z₀) = 0.

Furthermore, for a particular image the distance from the camera to a horizon point is the same for all horizon points. Call this distance l, and then we can recast the equations of the ray passing to horizon point i,

x_i = X_i l / ( X_i² + Y_i² + F² )^½ ,
y_i = Y_i l / ( X_i² + Y_i² + F² )^½ ,
z_i = F l / ( X_i² + Y_i² + F² )^½ .

Substituting these expressions into the perpendicular ray equation provides the value of

l = ( x₀ X_i + y₀ Y_i + z₀ F² ) / ( X_i² + Y_i² + F² )^½.

The camera, the centre of the sphere and each horizon point form the vertices of a right angled triangle, with l₀² ≡ x₀² + y₀² + z₀² = l² + 1. After substitution of the ray equations and some rearrangement of the radius equation, we obtain the equation describing the projection of the horizon in pixel terms, (X_i,Y_i). This turns out to be a general description of a conic section,

D_xx X² + D_yy Y² + D_xy XY + D_x X + D_y Y + D₀ = 0 ,

where the coefficients are

D_xx = (x₀² − l²) F² ,
D_yy = (y₀² − l²) F² ,
D_xy = 2 x₀ y₀ F² ,
D_x = 2 x₀ z₀ F³ ,
D_y = 2 y₀ z₀ F³ ,
D₀ = ( z₀² − l² ) F⁴ .

If this horizon curve is closed then it is an ellipse or circle. If the horizon curve is open then it is a parabola or hyperbola, in which case it extends infinitely off the edges of the screen and its parameters may be difficult to measure precisely. For the sake of simplicity we will assume that it is an ellipse, but the same formal mathematical derivations apply to the parabolic or hyperbolic cases.

Top view illustration of the parameters defined for the globe and the screen where its image is projected. Rays from the camera to the outline of the sphere (horizon) are shown in yellow; they meet the 3D surface of the sphere at right angles. The 3D position of the globe is (x₀,y₀,z₀), measured in units where the radius is 1. Each point on its horizon can be described with 3D coordinates (x_i,y_i,z_i), and the corresponding 2D point on the image outline is (X_i,Y_i). The ideal distance between the screen and the observer, measured in pixels, is F. The half angle of the horizontal field of view is θ_x. The distance from the observer to every horizon point on the sphere is l.

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Let us assume the model of a sphere projected onto the movie screen as described above, and then take the normalised coefficients to be those of a projected ellipse. Explicitly,

k_xx=C_xx/C₀ =D_xx/D₀ ,
k_yy=C_yy/C₀ =D_yy/D₀ ,
k_xy=C_xy/C₀ =D_xy/D₀ ,
k_x=C_x/C₀ =D_x/D₀ ,
k_y=C_y/C₀ =D_y/D₀ .

By a process of algebraic substitutions and eliminations, it can then be shown that

X* = F x₀ z₀ / ( z₀² − 1 ) ,
Y* = F y₀ z₀ / ( z₀² − 1 ) ,
C₀ = F⁴ ( l² − z₀ ) / ( z₀² − 1 )⁴ ,
a² = F² l² / ( z₀² − 1 )⁴ ,
b² = F² / ( z₀² − 1 )³ ,

From these results we can calculate the area of the image ellipse on the movie screen. For an ellipse in general, A=πab. For the image projected from a sphere,

A = π F² l / ( z₀² − 1 )^7/2 .

The eccentricity, e, of an ellipse is generally given by the implicit equation b² = a² ( 1 − e² ). Thus the eccentricity of the elliptical image of a sphere is given by

e² = ( l₀² − z₀² ) / ( l₀² − 1 ) .

Let us consider how the shape and size of the image ellipse vary as the camera pans. The range to the sphere, l₀, remains constant. The displacement of the sphere in the camera direction depends on the angle (τ) at which the camera is tilted away from the centre of the sphere,

z₀ ≡ l₀ cos τ .

It is then found that

A = π F² l / ( l₀² cos² τ − 1 )^7/2 ,
e² = l₀² sin² τ / ( l₀² − 1 ) .

We can distinguish a few qualitative cases depending on the angle of tilt:

1. When the camera points directly towards the sphere, τ=0, the image area is a minimum (A=πF²/l³) and the shape is perfectly circular (e=0).
2. As the tilt angle τ increases from zero up to the half-angle of the sphere (γ), both the area and eccentricity of the image ellipse increase. One axis of the ellipse is parallel to the tilt direction (X*,Y*).
3. At a maximum angle of τ = γ the eccentricity reaches e=1, and thus the image has become a parabola, not an ellipse. A parabola is an open shape, and the horizon curve invariably exceeds the borders of the movie frame, regardless of the field of view. In the parabolic condition, z₀=1, which means that some part of the sphere (with radius = 1) extends behind the position of the camera. The expressions above show that X*, Y* and A become undefined in this case.
4. Beyond this limit, z₀>1, the horizon is hyperbolic. Qualitatively this is an open curve like in the parabolic condition, but flatter as more of the globe is behind the camera. Measuring the curvature on any local part of the hyperbola doesn't directly indicate the equivalent apparent radius of the sphere if it were projected at the centre of the frame.

If we measure the projected ellipse from an image, as described in the sections above, then the equations of the C coefficients can be solved to obtain the relative position of the sphere and the angular field of view (expressed in terms of F).

For convenience, define the ratio λ = l / z₀, which is a constant for each image. It can be related to the measurable parameters of the onscreen ellipse,

λ² =  1 − 4 (k_xx − k_yy) / ( k_x² − k_y² ) .

If the field of view of the image is unknown, then it can be calculated from

F² =  [ λ² / (1 − λ² ) ] [ ( k_x² − k_y² ) / ( k_xxk_y² − k_yyk_x² ) ]

The positive solution of F can be substituted to obtain expressions for the coordinates of the sphere's centre:

x₀ / l = ½k_xF (1−λ²) / λ,
y₀ / l = ½k_xF (1−λ²) / λ,
z₀ / l = 1 / λ     as defined above.

From these three expressions, we can relate the range to the centre of the sphere, the range to the horizon, and the angle γ, which is half the total angle that the sphere subtends on the sky.

l₀² / l² = [¼F²(1−λ²)² (k_x²−k_y²) + 1 ] / λ² ≡ 1 / cos² γ .

As we have defined the measurement units in terms of the radius of the sphere, r=1, the ranges to the centre and to the horizon are thus given by

l₀ = 1 / sin γ ,
l = 1 / tan γ .

A spherical object appears as an elliptical, parabolic or hyperbolic outline on a movie frame. In theory it is straightforward to measure five different points on that outline, calculate the coefficients of the conic section, and infer the field of view and the 3D position of the sphere in units of its radius.

In practice there are several possible sources of difficulty, imprecision and ambiguity in the measurements or in the inference.

Firstly, the image does not contain spatial information finer than a pixel. Therefore it is impossible to pick sample points more precisely than within one pixel inside or outside the ideal horizon curve. Uncertainties will propagate into the inferred conic coefficients and thence to the derived parameters of the sphere.

Secondly, the horizon may not be completely measurable. In the best possible scenario, we can pick measurement points that are roughly evenly spaced on all sides of an ellipse. Both the semimajor and semiminor axes are well constrained from both the left and right (top and bottom) sides. However if one side of the globe falls off the edge of the image, or is invisible because of obstruction or shadow, then that side is poorly constrained. The inferred ellipse has a lot of scope for variation in the direction of the unmeasured side, for very tiny deviations on the choices of reference points in the measurable sides.

In the worst possible cases, the majority of the horizon is off the edge of the screen and we can only measure a tiny angular arc, which appears almost straight. The axes of the curve become practically unmeasurable, and even the type of conic section may be indeterminate. Attempts to read the local radius of curvature are misleading and provide neither upper nor lower limits, because the field of view is unknown and the aspect ratio of the ellipse/hyperbola is unknowable. As the camera tilts away from a sphere, the area and eccentricity of the onscreen ellipse both increase. An image that is tilted far offscreen is thus magnified and distorted compared to the appearance it would have when centred in the field. When only a small, flat arc is visible and the field of view is unknown then the distortion and magnificiation are inestimable.

Circular approximations are only useful when the disk is entirely within the field of view and close to the centre of frame. If the disk is measurably indistinguishable from a circle then we automatically know that the sphere is close to the line of sight and the field of view is narrow.

Demonstrations:

• Tatooine horizon: The elliptical fits are extremely difficult because of the lack of curvature and the smallness of the arc that this horizon subtends. As an exercise in objective measurement, it is almost impossibly fiddly. This is probably a hyperbolic case, and the few (apparently) elliptical measurements spurious.
  • 1: reference points appear to sit well on the horizon, but the bottom side of the inferred ellipse cuts off part of the visible disk. The inferred semiminor axis is too small. This is not a realistic solution.
  • 2: This is a marginally better outcome, because at least the far horizon is off the bottom of the screen. However a realistic fit could be more elogated in the vertical direction. The inferred semi-axes are (a,b) = (259,1412)
  • 3: More elogated in the vertical direction, but not objectively distinguishable from fit #2. The inferred axes are (a,b) = (956,2708) which is very different from fit #2 but not visibly better as a fit for the onscreen arc.

• Dagobah and X-wing:
  • 1: Fits the visible horizon, but appears prolate in off-screen areas. No objective way to reject this.
  • 2: The first of three presentable fits with a larger off-screen disk area, satisfactory fitting of the onscreen arc, but very different implications for the ellipse centre coordinates, orientation angle and axes:
    (X₀,Y₀)=(182.0,-153.4), φ=63.7º, (a,b)=(740.5,892.1),
  • 3: (X₀,Y₀)=(-19.5,-499.6), φ=-40.4º, (a,b)=(1065.2,1144.7),
  • 4: (X₀,Y₀)=(-3983.3,-5009.4), φ=-41.3º, (a,b)=(2642.7,7148.9).

• Bespin and X-Wing: For such a low resolution frame, this image gives nicely unambiguous results, owing to the measurability of a large arc of the horizon. One half is well lit, and even the night side is dimly visible due to indirect illumination. Fairly consistent ellipses can be fitted. We can discard the ones that visibly contradict the vaguely visible shape of the night side. The example fits listed below were the results of just a few minutes' effort; a little more trial and error would yield great exactitude.
  • 1: (X₀,Y₀)=(121.7,324.9), φ=-0.7º, (a,b)=(73.8,91.1)
  • 2: (X₀,Y₀)=(120.4,315.4), φ=5.0º, (a,b)=(72.8,79.3)
  • 3: (X₀,Y₀)=(121.9,316.0), φ=-8.6º, (a,b)=(71.5,81.1)
  • 4: (X₀,Y₀)=(120.4,323.5), φ=1.6º, (a,b)=(73.2,88.4)
  • 5: (X₀,Y₀)=(120.7,317.7), φ=4.6º, (a,b)=(72.0,83.9)

• Bespin and Shuttle: This is a difficult fit because the visible arc is less than a quarter of a complete disk, and the horizon is blurry due to atmospheric effects.
  • 1: (X₀,Y₀)=(-2446.0,-8187.7), φ=70.8º, (a,b)=(8858.7,2889.6) .... A solution with a large off-screen area, apparently consistent with the measurable arc.
  • 2: (X₀,Y₀)=(500.6,353.2), φ=69.3º, (a,b)=(174.5,454.2) .... A less plausible solution that fits the measured arc OK, but has one short axis that brings the opposite face into the same image, contradicting shadowed portions of the disk in the top-right corner.

• Endor, the brown orb and destroyers: The horizon of the forest moon is almost too flat to measure. The four fits presented here pass through all the points of measurement, but are implausible because they result in an excessively flattened disk, with the far horizon excluding areas that are actually visible. The moon is also visible in a subsequent reverse-view camera shot from the same location, and this implies that it is a hyperbolic case. The elliptical fits, which were difficult to establish, may be spurious.
  • 1: (X₀,Y₀)=(427.9,34.0), φ=-86.1º, (a,b)=(19.9,598.6)
  • 2: (X₀,Y₀)=(228.5,18.9), φ=-85.6º, (a,b)=(19.7,705.4)
  • 3: (X₀,Y₀)=(401.1,33.8), φ=-86.0º, (a,b)=(17.7,584.0)
  • 4: (X₀,Y₀)=(278.6,13.2), φ=-85.7º, (a,b)=(29.5,779.2)

• Distant view of Endor and DS2: One side of the moon is an easily measurable crescant, while there are stars near the edge of the night side which constrain the fitted ellipse. (We can exclude fits where stars appear inside the curve on the night side.) This disk is very easily fitted, and the circular approximation appears to be applicable.
  • 1: (X₀,Y₀)=(472.4,240.7), φ=-79.8º, (a,b)=(71.8,79.9)
    .... Slightly too wide on the night side, since a few stars appear within the curve.
  • 2: (X₀,Y₀)=(473.3,243.5), φ=-88.3º, (a,b)=(69.6,77.8)
    .... An excellently plausible solution, which fits the sunlit side well and appears consistent with the stars around the night side.
  • 4: (X₀,Y₀)=(479.9,251.8), φ=73.8º, (a,b)=(60.4,73.3)
    .... A possible solution, but perhaps cutting off too much of the night side.
  • 5: (X₀,Y₀)=(479.0,249.8), φ=72.5º, (a,b)=(62.9,73.3)
    .... A possible solution, but perhaps cutting off too much of the night side.
  • 6: (X₀,Y₀)=(475.4,244.0), φ=84.3º, (a,b)=(68.6,77.1)
    .... A possible solution; a bit prolate.
  • 7: (X₀,Y₀)=(475.3,243.4), φ=88.0º, (a,b)=(69.5,76.1)
    .... A more plausible solution; nearly matching the maximum possible outline on the night side.

• Endor and DS2: A small arc of the forest moon's horizon is almost straight and difficult to measure accurately. Extrapolations of the location of the far side of the disk are extremely ill-defined and the horizon parameters practically impossible to interpret.
  • 1: (X₀,Y₀)=(265.8,210.9), φ=-59.4º, (a,b)=(31.3,332.2)
    .... An arc that fits five widely spaced measurements, but it is an extreme underestimate of one axis, and the far side doesn't fit the visible areas of the disk.
  • 2: (X₀,Y₀)=(280.0,188.3), φ=-59.2º, (a,b)=(58.3,414.8)
    .... Essentially the same problem as above: the arc of possible measurements is too narrow and too small a fraction of the overall disk to constrain the far side of the disk.
  • 3: (X₀,Y₀)=(313.9,-1500.3), φ=-36.9º, (a,b)=(1343.6,2332.4)
    .... A more plausible solution, since it doesn't cut off any visible areas of the disk. But could it still be an underestimate, or indeed and overestimate, of the displacement to the far horizon?
  • 4: (X₀,Y₀)=(184.4,-470.6), φ=-49.3º, (a,b)=(537.1,1315.6)
    .... Another plausible solution that fits the visible areas. The implied disk area is much smaller than in the previous solution.
  • 5: (X₀,Y₀)=(261.8,216.0), φ=-59.3º, (a,b)=(24.2,313.2)
    .... An implausible solution, with one axis far too small and the far horizon pulled too short.
  • 6: (X₀,Y₀)=(382.2,-198.4), φ=-56.7º, (a,b)=(440.3,1112.1)
    .... Another plausible solution, with no testable difference from fits #3 and 4.

• Endor and DS2: This is another image where the position and elliptical curvature of the moon are extremely difficult to fit. Measurement may not be feasible at this resolution, with the present restricted field of view.
  • 1: (X₀,Y₀)=(364.7,-99.6), φ=-70.2º, (a,b)=(314.9,1063.4)
    .... An apparently plausible fit for the tiny visible arc of horizon, but it's not certain to be the only possible fit. The inferred ellipse could change dramatically with just a small tweaking of the reference points.
  • 2: (X₀,Y₀)=(252.6,132.8), φ=-69.7º, (a,b)=(57.7,461.7)
    .... An implausible fit with the range to the far horizon underestimated visibly. Such underestimates occur readily, and overestimates must be equally common. But with only a tiny, flat section of horizon visible, we can't tell what is an overestimate and what is a genuinely good fit.

• Wikipedia,
  • Conic Sections
  • Ellipse
  • Parabola
  • Hyperbola