High Noon

We have two coordinate reference frames to relate, the first one is the reference frame that has its center at the center of Earth, with the z-axis set perpendicular to the orbital plane (the plane in which the Earth orbits the Sun). Also, the x-axis is selected to point towards the center of the Sun. The second reference frame is attached to Earth, centered at the Earth center, with its z-axis along the axis of rotation. The x-axis orientation, results naturally when we look at the second frame as being the result of two consecutive rotations (as the animation depicts), the first is about the z-axis (of the orbital plane). And the second is about the y-axis of the frame resulting from the first rotation.

Thus, let $r_0 = [x_0, y_0, z_0 ]^T$ be the coordinate vector in the orbital plane, and $r_1 = [x_1, y_1, z_1 ]^T$ be the corresponding coordinate vector in the frame resulting from the first rotation, and let $r_2 = [x_2, y_2, z_2 ]^T$ be the corresponding coordinate vector in the final frame (the one attached to Earth, with its z-axis along Earth's axis of rotation).

We can relate these three vectors as follows

$r_0 = R_1 r_1$

where $R_1$ is the rotation matrix about the z-axis by an angle $\phi$ , and is given by

$R_1 = \begin{bmatrix} \cos \phi && -\sin \phi && 0 \\ sin \phi && \cos \phi && 0 \\ 0 && 0 && 1 \end{bmatrix}$

In addition,

$r_1 = R_2 r_2$

where $R_2$ is the rotation matrix about the y-axis by an angle $\theta$ , and is given by

$R_2 = \begin{bmatrix} \cos \theta && 0 && \sin \theta \\ 0 && 1 && 0 \\ -\sin \theta && 0 && \cos \theta \end{bmatrix}$

Hence,

$r_0 = R_1 R_2 r_2 = R_o r_2$

where

$R_o = R_1 R_2 = \begin{bmatrix} \cos \phi \cos \theta && -\sin \phi && \cos \phi \sin \theta \\ \sin \phi \cos \theta && \cos \phi && -\sin \phi \sin \theta \\ -\sin \theta && 0 && \cos \theta \end{bmatrix}$

Now, the unit normal vector to the surface of Earth a at point that has a latitude of $30^{\circ}$ is along the position vector of the point itself and is given by

$v_2 = r_2 = [ \sin \theta_L \cos \phi_t , \sin \theta_L sin \phi_t, \cos \theta_L ]^T$

where $\theta_L = 60^{\circ}$ and $\phi_t$ is the rotation angle counterclockwise from the x-axis of the Earth reference frame. In the orbital reference frame, this same vector has the coordinates

$v_0 = R_o v_2$

And, in the orbital reference frame, the Sun rays are pointing in the negative x-direction, so with respect to our point on Earth, the direction of the Sun is in the positive x-direction, Therefore, the angle between the normal to the surface of Earth and the direction of the Sun is given by

$\theta_N = \cos^{-1} ( v_0 \cdot \hat{i} ) = \cos^{-1} ( v_{0x} )$

where $v_{0x}$ is the x-coordinate of $v_0$ . Using the above expressions for $v_2$ and $R_o$ , this is given by:

$v_0x = \cos \phi \cos \theta \sin \theta_L \cos \phi_t -\sin \phi \sin \theta_L sin \phi_t + \cos \phi \sin \theta \cos \theta_L$

$= a \cos \phi_t + b \sin \phi_t + c$

with

$a = \cos \phi \cos \theta \sin \theta_L$

$b = -\sin \phi \sin \theta_L$

$c = \cos \phi \sin \theta \cos \theta_L$

which is a sinusoid in $\phi_t$ . Since we are seeking the maximum elevation angle above the horizon, then we are seeking the minimum $\theta_N$ . This occurs when $\theta_N$ has the highest cosine value, which is $v_0x$ . The maximum of the expression for $v_0x$ is given by

$v_{0x, MAX} = \sqrt{a^2 + b^2} + c$

Substituting the given values for $\theta$ , $\phi$ , and $\theta_L$ gives the maximum value of the cosine of $\theta_N$ as

$v_{0x, MAX} = 0.9717$

Therefore, the (minimum) value of $\theta_N$ is

$\theta_N = 13.663$

Hence, the maximum angle of elevation of the Sun is

$\theta_E = 90^{\circ} - 13.663^{\circ} = 76.34^{\circ}$