Tangent points to SXI in GSE

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The aiming point of the satellite is always on (x, 0 , 0) in the GSE system
A relation can be derived to correlate the image plane coordinates with GSE polar coordinates

Idea

We will try to convert the shue model to a 3D surface in spherical coordinates by rotating over the symmetry axis (x) with the φ angle. We will then convert this to cartesian coordinates, thus parametrizing the surface. We then compute the normal vector of the surface for each point (θ,φ) and find the tangent direction in the FOV of the satellite by requesting that the normal vector and the vector that connects the point to the satellite are perpendicular (dot product is zero). This gives us a set of (θ,φ) - a curve in 3D space. This is the tangent curve. We can find the projection of this curve to the image plane by calculating the angle between the aim point and each point of the curve, separately for the x and y direction.

Analytical implementation

Shue model in spherical (phi rotation is symmetrical when defined on the yz plane - Geocentric Solar Ecliptic System):

r (θ, ϕ) = r_{0} {(\frac{2}{1 + \cos θ})}^{α}

Cartesian coordinate Shue model:

x = r c o s θ = r_{0} {(\frac{2}{1 + \cos θ})}^{α} \cos θ

y = r s i n θ c o s ϕ = r_{0} {(\frac{2}{1 + \cos θ})}^{α} \sin θ \cos ϕ

z = r s i n θ s i n ϕ = r_{0} {(\frac{2}{1 + \cos θ})}^{α} \sin θ \sin ϕ

Parameterized vector:

R = (r_{0} {(\frac{2}{1 + \cos θ})}^{α} \cos θ) \hat{i} + (r_{0} {(\frac{2}{1 + \cos θ})}^{α} \sin θ \cos ϕ) \hat{j} + (r_{0} {(\frac{2}{1 + \cos θ})}^{α} \sin θ \sin ϕ) \hat{k}

Partial differentials:

R_{θ} = \frac{\partial R}{\partial θ} = r_{0} {(\frac{2}{1 + \cos θ})}^{α} [\sin θ \frac{(a - 1) \cos θ - 1}{\cos θ + 1} \hat{i} + \cos ϕ (\cos θ + \frac{\sin^{2} θ a}{1 + \cos θ}) \hat{j} + \sin ϕ (\cos θ + \frac{\sin^{2} θ a}{1 + \cos θ}) \hat{k}]

R_{ϕ} = \frac{\partial R}{\partial ϕ} = r_{0} {(\frac{2}{1 + \cos θ})}^{α} \sin θ (- \sin ϕ \hat{j} + \cos ϕ \hat{k})

Normal vector:

η = R_{θ} \times R_{ϕ} = r_{s h u e}^{2} \sin θ [(\cos θ + \frac{\sin^{2} θ a}{1 + \cos θ}) (\cos^{2} ϕ + \sin^{2} ϕ) \hat{i}] + \dots \hat{j} + \dots \hat{k}

Tangent points for satellite position r_sat:

\hat{η} \cdot \vec{r_{s a t}} = 0 ⟹ (x_{s a t} - x) r_{0}^{2} {(\frac{2}{1 + \cos θ})}^{2 α} \sin θ (\cos θ + \frac{\sin^{2} θ a}{1 + \cos θ}) + \dots = 0

Attention

This is unsolvable analytically, or with series expansion.
==> Numerical solutions and discrete comparison of fit ==> Numerical Solution & Performance

Conversion to image plane - angular difference from boresight:

must have information of camera orientation
Assumption: orientation of el-az of camera same as X-Y plane of GSE:

\cos θ_{F O V} = \frac{\vec{r_{a i m}} \cdot (x, 0, z)_{p o i n t}}{| | \dots | |}

\cos ϕ_{F O V} = \frac{\vec{r_{a i m}} \cdot (0, y, z)_{p o i n t}}{| | \dots | |}

Correct way : Use rotation matrix from the GSE to the camera frame when available - Coordinate transformation from GSE to SXI