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Let $\theta$ be latitude, $\phi$ be longitude. I need to find the formula for the great circle passing ($\theta_0$, 0) and (0, $\phi_0$). This seems a easy and common problem, but I can not find any reference for it to check my answer. The closest problem may be the great-circle navigation problem. Can anyone help?

I guess the answer to be $\frac{\phi \cos \theta}{\phi_0 \cos \theta_0} + \frac{\theta \cos \phi}{\theta_0 \cos \phi_0} = 1$. But I have difficulty to prove it

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3 Answers

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Using the Spherical Law of Cosines, the equation for a great circle with pole $(\theta_p,\phi_p)$ is$$ \tan(\theta)\tan(\theta_p)+\cos(\phi-\phi_p)=0\tag1 $$Since $(0,\phi_0)$ satisfies $(1)$, we have$$ \cos(\phi_0-\phi_p)=0\tag2 $$We can choose $\phi_p=\phi_0+\frac\pi2$ ($\phi_0-\frac\pi2$ simply leads to the opposite pole). Since $(\theta_0,0)$ satisfies $(1)$, we have$$ \begin{align} 0 &=\tan(\theta_0)\tan(\theta_p)+\cos(\phi_p)\\ &=\tan(\theta_0)\tan(\theta_p)-\sin(\phi_0)\tag3 \end{align} $$therefore, $\tan(\theta_p)=\frac{\sin(\phi_0)}{\tan(\theta_0)}$. Plugging back into $(1)$, we have$$ \begin{align} 0 &=\tan(\theta)\,\frac{\sin(\phi_0)}{\tan(\theta_0)}+\cos(\phi-\phi_p)\\ &=\tan(\theta)\,\frac{\sin(\phi_0)}{\tan(\theta_0)}+\sin(\phi-\phi_0)\tag4 \end{align} $$Thus, we get the equation for the great circle to be$$ \tan(\theta)\sin(\phi_0)+\tan(\theta_0)\sin(\phi-\phi_0)=0\tag5 $$

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You may solve this problem using a central cylindrical projection.

• Wrap a cylinder around the equator ($\theta=0$).
• Map points radially from the globe to the cylinder. Assuming unit radius the point with latitude $\theta$ and longitude $\phi$ is mapped onto $(1,\phi,z=\tan\theta)$ in cylindrical coordinates.
• In general the great circle is mapped to an ellipse on the cylinder, then when the cylinder is cut and unrolled the ellipse becomes a sine wave with the equator as its mean value.

Now the sine wave, having the equator as its midline, will take the form

$z=A\sin(\phi_0-\phi)$

and we just plug in the second point $\phi=0,z=z_0=\tan\theta_0$ to determine $A$. Thereby

$z=(\tan\theta_0)(\sin(\phi_0-\phi))/(\sin\phi_0)$

All that remains is to put back $z=\tan\theta$ and take the arctangent to get an explicit equation for latitude in terms of longitude:

$\color{blue}{\theta=\tan^{-1}\left(\dfrac{\tan\theta_0\sin(\phi_0-\phi)}{\sin\phi_0}\right)}$

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There are two vectors here, for the point $(\theta_0, 0)$:

$v_1 = (\cos \theta_0 \cos 0 , \cos \theta_0 \sin 0, \sin \theta_0) = (\cos \theta_0, 0, \sin \theta_0)$

and for the point $(0, \phi_0)$:

$v_2 = (\cos 0 \cos \phi_0, \cos 0 \sin \phi_0, \sin 0) = (\cos \phi_0, \sin \phi_0, 0)$

The normal to the great circle passing through these two points is:

$n = v_1 \times v_2 = (-\sin \theta_0 \sin \phi_0, \sin \theta_0 \cos \phi_0, \cos \theta_0 \sin \phi_0)$

A point on the sphere is $P = (\cos \theta \cos \phi, \cos \theta \sin \phi, \sin \theta)$

Hence the equation of the great circle is $P \cdot n = 0$, i.e.,

$ \cos \theta \sin \theta_0 (- \sin \phi_0 \cos \phi + \cos \phi_0 \sin \phi ) + \sin \theta \cos \theta_0 \sin \phi_0 = 0 $

This simplifies to,

$ \cos \theta \sin \theta_0 \sin( \phi - \phi_0) + \sin \theta \cos \theta_0 \sin \phi_0 = 0 $

And from this last equation, we can express $\theta$ in terms of $\phi$ and have:

$ \theta = - \tan^{-1} \dfrac{ \tan \theta_0 \sin( \phi - \phi_0) }{\sin \phi_0}$

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