The first step is to establish a coordinate system to take measurements. As a matter of convenience we shall let the principal point be the origin of this coordinate system, and make all measurements in "pixels". We can convert from pixels to meters, or feet, in the final analysis. The cartesian Coordinates, in pixels, of the corners relative to the focus are:

Table 1.

Spherical coordinates relative to the center are given in Table 2. Radii are in pixels, and angles are in degrees measured from the "east" (mathematical convention).

Table 2.

The equation of an ellipse in polar coordinates is given as

a(1 - e²)

R(θ)  =  ———————

(1 - ecos(θ-θ₀))

where “a” is the semi-major axis, “e” is the eccentricity of the ellipse, and θ is the angle measured counter-clockwise from the east to some point on the perimeter. The angle θ0 indicates the orientation of the semi-major axis relative to horizontal axis. The minus sign in the denominator between terms indicates the origin of measurement is from the left (or west) focus.

The equation of the ellipse contains three unknowns. The data measured from the graphics, however, will allow us to establish a set of equations that will help us eliminate two of the unknowns, and solve for the third. Then we can solve for the other two.

Let us first consider the ratios

R₃      (1 - ecos(θ₁-θ₀))

— = ———————

R₁      (1 - ecos(θ₃-θ₀))

and

R₄      (1 - ecos(θ₂-θ₀))

— = ———————

R₂      (1 - ecos(θ₄-θ₀))

If we rearrange these equations to solve for the eccentricity we get

R₃

(1 - ecos(θ₃-θ₀)) — = (1 - ecos(θ₁-θ₀))

R₁

R₃           R₃

—  - e — cos(θ₃-θ₀)) = 1 - ecos(θ₁-θ₀)

R₁            R₁

R₃              R₃

—  - 1 = e — cos(θ₃-θ₀)) - ecos(θ₁-θ₀)

R₁              R₁

or

R₃ - R₁ = e(R₃cos(θ₃-θ₀) - R₁cos(θ₁-θ₀))

and finally:

R₃ - R₁

e = ————————————

R₃cos(θ₃-θ₀) - R₁cos(θ₁-θ₀)

Likewise, we have

R₄ - R₂

e = ————————————

R₄cos(θ₄-θ₀) - R₂cos(θ₂-θ₀)

As a matter of convenience we chose the corners diametrically opposed to each other; i.e., North/South and East/West. This leads to a convenient simplification given below.

If we equate these two equations, we only then have to solve for the angle, θ₀. Before doing so, let us consider one more simplification. The angles from the east axis to each of the points maintain a constant relationship. Primarily, The Northern corner is always 180° away from the southern corner, as is the angle between the east and west corners. Thus, θ₃ - θ₁ = 180° and θ₄ - θ₂ = 180°. Using the property

cos(α - 180°) = -cos(α)

in Equations (4) and (5) they reduce to

R₁ - R₃

e = ————————

(R₁ + R₃)cos(θ₁-θ₀)

and

R₂ - R₄

e = ————————

(R₂ + R₄)cos(θ₂-θ₀)

Now we have

(R₁ + R₃)cos(θ₁-θ₀)      (R₂ + R₄)cos(θ₂-θ₀)

———————— = ————————

R₁ - R₃                      R₂ - R₄

Just as opposite angles maintain a 180° phase relationship, adjacent angles also maintain an constant phase relationship. In the case above θ₂ = θ₁ + 112.7°. We also take advantage of the trigonometric identity

cos(α+β) = cos(α)cos(β) - sin(α)sin(β)

Making the subsitution θ₂ = θ₁ + 112.7° in the above equation and applying the identity, we have

R₁ + R₃                      R₂ + R₄

———cos(θ₁-θ₀) = ————(cos(θ₁-θ₀)cos(67.3°) - sin(θ₁-θ₀)sin(67.3°))

R₁ - R₃                       R₂ - R₄

Dividing by the cosine term on the left, and by the ratio of radii on the right, we have

(R₁ + R₃)(R₂ - R₄)

——————— = (cos(67.3°) - tan(θ₁-θ₀)sin(67.3°))

(R₁ - R₃)(R₂ + R₄)

Solving for tan(θ₁-θ₀) leads to the result

(R₁ + R₃)(R₂ - R₄)

tan(θ₁-θ₀) = cot(67.3°) - ————————————

(R₁ - R₃)(R₂ + R₄)sin(67.3°)

and the solution

{θ₀ = 48.34°}

This means that the axis of the ellipsoid is inclined at an angle of 48.34° from east, or 41.66° true (relative to north). Returning to the solutions for eccentricity we find

{e = 0.215}

As stated above, the incentricity is the sine of the bank angle. The inverse sine of the eccentricity then gives us a bank angle of

{φ = 12.4°}

It immediately follows from Equation (1), then, that the semi-major axis is

{a = 1,643 pixels}

and the semi-latus rectum is

{l = 1,567 pixels}

Given that the aircraft is now in a bank of φ = 12.4°, the field of view is also tilted at an angle of 12.4°. The projection of the film plane onto the ground plane intersects along the latus rectum, with the focus at the center. Recall that for congruent triangles (f/d) = (F/D). For the film of a K-21 camera (f/d) = (6.25/8.6). The latus rectum is twice the size of of the semi-latus rectum, or 3134 pixels, and is equal to the diameter of the circular field of view of the projected film plane. Thus, the distance from the aircraft to the principal point (the focus of the ellipse) is found to be

F = (f/d)D = 2,277 pixels

The aerial photograph has a scale of 1,169 pixels per 4,000 meters. The above distance is then equivelant to 7,791 meters, or 25,561 feet. The altitude and range to the principal point are related by the cosine of the tilt angle, A = Fcos(φ), from which we arrive at an altitude of 24,964 feet/7609 meters.

The aircraft is flying in a direction perpendicular to the ellipse's major axis, which is aligned along 48.3°. The aircraft's course is then 90° toward the south, or 138.3° true. As for the aircraft's attitude, we have to examine the orientation of the ellipse's major axis, and the polygon of the imaged scene. The major axis of the imaged scene bisects the angles of the smaller sides.

When the photo was taken the aircraft's geographic location was 5493 feet/1674 meters to the southwest of the principal point.

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