Sunrise Direction

This note is about the actual direction of Sun rise relative to the geographic East. The Sun rises from the East. This would have been true if the Earth axis of rotation was not tilted with respect to its orbital plane around the Sun.

But the fact is, it is tilted and makes an angle of θ=23.44 \theta = 23.44^{\circ} with the normal to the orbital plane that passes through the Earth's center.
Another fact about the Earth axis of rotation, is that it always points in a fixed inertial direction with respect to our galaxy. Therefore, when it rotates about the Sun, the axis projection on the orbital plane rotates, with respect to the Sun's direction, clockwise at a constant rate when viewed from the North.

On June 21st. the projection of Earth's axis on the orbital plane, points directly towards the Sun. After June 21st, the angle this projection makes with the direction of the Sun ( ϕ \phi ), increases linearly with time, completing 360 360^{\circ} in one year.

On a given day, we can determine the angle ϕ \phi by the number of days N N that have passed since June 21st. The projection of the Earth's axis on the orbital plane will make an angle of ϕ=(N/365.25)360 \phi = -(N / 365.25) * 360^{\circ} with respect the line connecting the Earth's center and the Sun's center. Consider a point on the surface of Earth at a latitude of L L degrees North of the Equator. (L L is negative for a location that is South of the Equator). Considering these variables, what will be the direction of Sunrise on a certain given day, with respect to the East direction. We have three coordinate reference frames to relate, the first one (Frame 1) 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 (Frame 2) is attached to Earth, centered at the Earth center, with its z-axis along the axis of Earth rotation. The x-axis orientation, results naturally when we look at this frame as being the result of two consecutive geometric rotations, 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.

Finally, the third frame (Frame 3), is the frame attached to a moving point on the surface of Earth, with its three axes pointing towards the standard directions of local East, local North, and local Vertical. Thus, let r1=[x1,y1,z1]T r_1 = [x_1, y_1, z_1 ]^T be the coordinate vector in the orbital plane, and r=[x,y,z]T r' = [x', y', z' ]^T be the corresponding coordinate vector in the frame resulting from the first rotation, and let r2=[x2,y2,z2]T r_2 = [x_2, y_2, z_2 ]^T be the corresponding coordinate vector in Frame 2. We can relate these three vectors as follows

r1=Rr r_1 = R' r'

where R R' is the rotation matrix about the z-axis by an angle ϕ \phi , and is given by

R=[cosϕsinϕ0sinϕcosϕ0001] R' = \begin{bmatrix} \cos \phi && -\sin \phi && 0 \\ \sin \phi && \cos \phi && 0 \\ 0 && 0 && 1 \end{bmatrix}

In addition,

r=Ryr2 r' = R_y r_2

where Ry R_y is the rotation matrix about the y-axis by an angle θ \theta , and is given by

Ry=[cosθ0sinθ010sinθ0cosθ] R_y = \begin{bmatrix} \cos \theta && 0 && \sin \theta \\ 0 && 1 && 0 \\ -\sin \theta && 0 && \cos \theta \end{bmatrix}

Hence,

r1=RRyr2=R1r2 r_1 = R' R_y r_2 = R_1 r_2

where

R1=RRy=[cosϕcosθsinϕcosϕsinθsinϕcosθcosϕsinϕsinθsinθ0cosθ] R_1 = R' R_y = \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}

Next, we consider Frame 3. We note that the unit vectors pointing East and North and Vertically Up, at a given point that has a latitude of L L , is given by

vEast=[sinϕt,cosϕt,0]T v_{East} = [ - \sin \phi_t , \cos \phi_t , 0 ]^T

And

vNorth=[cosθLcosϕt,cosθLsinϕt,sinθL]T v_{North} = [ - \cos \theta_L \cos \phi_t , - \cos \theta_L \sin \phi_t, \sin \theta_L ]^T

And

vVertical=[sinθLcosϕt,sinθLsinϕt,cosθL]T v_{Vertical} = [ \sin \theta_L \cos \phi_t, \sin \theta_L \sin \phi_t, \cos \theta_L]^T

Where ϕt \phi_t is the rotation angle counterclockwise from the x-axis of the Earth reference frame, and θL=90L \theta_L = 90^{\circ} - L .
Together, these three vectors form an orthornormal basis for the reference frame we name Frame 3, and can written with respect to frame 2 as

R2=[vEast,vNorth,vVertical] R_2 = [ v_{East}, v_{North}, v_{Vertical} ]

i.e.

R2=[sinϕtcosθLcosϕtsinθLcosϕtcosϕtcosθLsinϕtsinθLsinϕt0sinθLcosθL] R_2 = \begin{bmatrix} -\sin \phi_t && -\cos \theta_L \cos \phi_t && \sin \theta_L \cos \phi_t \\ \cos \phi_t && - \cos \theta_L \sin \phi_t && \sin \theta_L \sin \phi_t \\ 0 && \sin \theta_L && \cos \theta_L \end{bmatrix}

Now, in the orbital reference frame (Frame 1), the direction to the Sun is pointing in the positive x-direction,

u1=[1,0,0]T u_1 = [1, 0, 0]^T

It follows that the direction of the sun , when expressed with respect to Frame 3, is related to u1 u_1 , by

u1=R1R2u3 u_1 = R_1 R_2 u3 From which

u3=R2TR1Tu1 u_3 = R_2^T R_1^T u_1

Being a rotation matrix, the inverse of Ri R_i is its transpose. Performing the indicated multiplication, results in u3 u_3

u3=[sinϕtcosϕt0cosθLcosϕtcosθLsinϕtsinθLsinθLcosϕtsinθLsinϕtcosθL][cosϕcosθsinϕcosϕsinθ] u_3 = \begin{bmatrix} - \sin \phi_t && \cos \phi_t && 0 \\ - \cos \theta_L \cos \phi_t && -\cos \theta_L \sin \phi_t && \sin \theta_L \\ \sin \theta_L \cos \phi_t && \sin \theta_L \sin \phi_t && \cos \theta_L \end{bmatrix} \begin{bmatrix} \cos \phi \cos \theta \\ -\sin \phi \\ \cos \phi \sin \theta \end{bmatrix}

u3=[cosϕcosθsinϕtsinϕcosϕtcosϕcosθcosθLcosϕt+sinϕcosθLsinϕt+cosϕsinθsinθLcosϕcosθsinθLcosϕtsinϕsinθLsinϕt+cosϕsinθcosθL] u_3 = \begin{bmatrix} - \cos \phi \cos \theta \sin \phi_t - \sin \phi \cos \phi_t \\ - \cos \phi \cos \theta \cos \theta_L \cos \phi_t + \sin \phi \cos \theta_L \sin \phi_t + \cos \phi \sin \theta \sin \theta_L \\ \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 \end{bmatrix}

Now, at sunrise , the z-component of this vector is zero, because the Sun will be coming from the horizon. Setting

u3z=0 u_{3z} = 0

results in,

cosϕt=sinϕsinθLsinϕtcosϕsinθcosθLcosϕcosθsinθL \cos \phi_t = \frac {\sin \phi \sin \theta_L \sin \phi_t - \cos \phi \sin \theta \cos \theta_L} { \cos \phi \cos \theta \sin \theta_L}

Substituting this expression, in u3y u_{3y} ,

u3y=sinϕt((cosϕcosθcosθLsinϕsinθL)/(cosϕcosθsinθL)+sinϕcosθL)+cosϕcosθcosθLcosϕsinθcosθL/(cosϕcosθsinθL)+cosϕsinθsinθL u_{3y} = \sin \phi_t ( (- \cos \phi \cos \theta \cos \theta_L \sin \phi \sin \theta_L) / ( \cos \phi \cos \theta \sin \theta_L) + \sin \phi \cos \theta_L ) + \cos \phi \cos \theta \cos \theta_L \cos \phi \sin \theta \cos \theta_L / (\cos \phi \cos \theta \sin \theta_L) + \cos \phi \sin \theta \sin \theta_L

Simplifying,

u3y=sinϕt(cosθLsinϕ+sinϕcosθL)+cos2θLcosϕsinθ/(sinθL)+cosϕsinθsinθL u_{3y} = \sin \phi_t ( - \cos \theta_L \sin \phi + \sin \phi \cos \theta_L ) + \cos^2 \theta_L \cos \phi \sin \theta / (\sin \theta_L) + \cos \phi \sin \theta \sin \theta_L

Hence,

u3y=cosϕsinθ/sinθL u_{3y} = \cos \phi \sin \theta / \sin \theta_L

Since u3 u_3 is a unit vector, and u3z=0 u_{3z} = 0 , its y-component is the sine of the angle it makes with vEast v_{East} . Therefore, finally, the angle the vector pointing to the Sun at exact sunrise makes with the East is given by,

α=sin1(cosϕsinθsinθL) \alpha = \sin^{-1}( \frac{\cos \phi \sin \theta}{ \sin \theta_L} )

And this angle is the same deviation from the West direction at sunset.

As an example, to determine the sunrise direction in Ottawa, which is at a latitude of 45.42 45.42^{\circ} North, on December 11, we have to compute the number of days between June 21st, and Decemeber 11th, and that is equal to =173 = 173 days.

Therefore, ϕ=(173/365.25)360=170.51 \phi =- (173/365.25)*360^{\circ} =- 170.51^{\circ}

And

θL=90L=9045.42=44.58 \theta_L = 90^{\circ} - L =90^{\circ} - 45.42^{\circ} = 44.58^{\circ}

and,

θ=23.44 \theta = 23.44^{\circ}

Hence, by plugging in these values, we obtain,

α=sin1(cos(170.51)sin(23.44)/sin(44.58))=sin1(0.558971)=33.985 \alpha = \sin^{-1}( \cos(-170.51^{\circ}) \sin(23.44^{\circ})/ \sin(44.58^{\circ})) = \sin^{-1}(-0.558971) = -33.985^{\circ}

The minus sign indicates that the Sun will rise 33.985 33.985^{\circ} south of East.

Note by Hosam Hajjir
4 years, 12 months ago

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This note is really interesting @Hosam Hajjir -- thank you for writing it up!

Silas Hundt Staff - 4 years, 12 months ago

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Follow up questions

  1. Calculate the angle at which the sun rises where you are at today.

  2. At what longitude and latitude will the sun not set in summer? Why?

Calvin Lin Staff - 4 years, 12 months ago

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Is it possible to prove that ϕ\phi varies linearly with time assuming the fact that sun itself is orbitting around the centre of Milky way?

Pranjal Jain - 4 years, 12 months ago

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It is an established fact that the orbit of Earth around the Sun is almost a circle (ellipse with eccentricity of 0.01671123). This implies that the axis projection will rotate at an almost constant rate.

Hosam Hajjir - 4 years, 12 months ago

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With this , i think Calvin that can't we include Astronomy and Astrophysics on brilliant ? I have participated in olympiads on these subjects and feel that these are important parts of maths and physics . In fact it is very intersting science . So , what you think Calvin ??

Utsav Singhal - 4 years, 12 months ago

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