# Area between geodesic and arc of constant lattitude

A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere. Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected constant latitude. Differences between these two lines can be visualized with the use of this Academo program presenting the situation in the context of the map of Earth.

Question:

• How to calculate the area between these two lines?

(Assume for example that the starting point is $(\alpha, \beta_1)=(45^\circ, -120^\circ)$ and the destination $(\alpha, \beta_2)=(45^\circ, 0^\circ))$ - the arc of constant latitude $45^\circ$ has length $120^\circ$.

In wikipedia a formula for an area of a spherical polygon is presented, but the polygon is limited in this case with parts of geodesics. Is it possible somehow transform these formulas of spherical geometry into the case of finding the area between geodesic and arc of constant latitude?

• Here's a strategy: Call your points $A$ and $B$, and let $C$ be the north pole. The area of the "gore" bounded by the longitudes $AC$ and $BC$ and the latitude $AB$ can be found easily by geometry (in terms of the latitudes and longitudes). The area of the geodesic triangle $\triangle ABC$ can also be found by geometry. The area you want is the area of the gore minus the area of the geodesic triangle. if no one beats me to it I'll write up details when time permits. Commented Jan 25, 2017 at 0:41
• @AndrewD.Hwang The plan probably is good, but really is it easy to find area limited by latitude and longtitudes? I suppose if so, from the general formula it would be easy also to see when this area is maximal for $\alpha, \beta$ parametres.. if it is possible to elaborate details I will be very grateful Andrew, there are no many specialists from spherical geometry on MSE , I suppose.. I wonder also whether it is possible also a general solution when $\alpha_1{\neq}\alpha_2$ ( what is the extension of the question) Commented Jan 25, 2017 at 7:32
• If $A = (\alpha, \beta_{1})$, $B = (\alpha, \beta_{2})$, and the sphere has readius $r$, then the "gore" (non-geodesic triangle) has area $r(\beta_{2} - \beta_{1})(1 - \sin\alpha)$ by a theorem of Archimedes on the areas of spherical zones (regions between two latitudes), and the fact that the area of a gore is proportional to the longitude it subtends. Commented Jan 25, 2017 at 11:43
• @AndrewD.Hwang Very interesting theorem, I was unaware of it..thank you Andrew for presenting it. Commented Jan 25, 2017 at 13:48
• The two geodesic sides are $\pi/2 - \alpha$, but the third side is smaller than $\beta_{2} - \beta_{1}$ unless $\alpha = 0$. The exact formula is$$\arccos\bigl(\cos(\beta_{2} - \beta_{2})\cos^{2}\alpha + \sin^{2}\alpha\bigr).$$(To see this, express each point as a unit vector using spherical coordinates, then take the arccos of their dot product.) Commented Jan 26, 2017 at 0:07

Summary: If $A$ and $B$ lie on the latitude line at angle $0 < \alpha < \pi/2$ north of the equator on a sphere of unit radius, and at an angular separation $0 < \theta = \beta_{2} - \beta_{1} < \pi$, then the "digon" bounded by the latitude and the great circle arc $AB$ (in blue) has area \begin{align*} \pi - \theta \sin\alpha - 2\psi &= \text{sum of interior angles} - \theta \sin\alpha \\ &= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right). \end{align*}

If $A$ and $B$ have longitude-latitude coordinates $(0, \alpha)$ and $(\theta, \alpha)$, their Cartesian coordinates (on the unit sphere) are $$A = (\cos\alpha, 0, \sin\alpha),\quad B = (\cos\theta\cos\alpha, \sin\theta \cos\alpha, \sin\alpha).$$ Let $C = (0, 0, 1)$ be the north pole, $G$ the "gore" (shaded) bounded by the spherical arcs $AC$, $BC$, and the latitude through $A$ and $B$, and $T$ the geodesic triangle with vertices $A$, $B$, and $C$.

Lemma 1: The area of $G$ is $\theta(1 - \sin\alpha)$.

Proof: The spherical zone bounded by the latitude through $A$ and $B$ and containing the north pole has height $h = 1 - \sin\alpha$ along the diameter through the north and south poles. By a theorem of Archimedes, this zone has area $2\pi h = 2\pi(1 - \sin\alpha)$. The area of the gore $G$, which subtends an angle $\theta$ at the north pole, is $$(\theta/2\pi)2\pi(1 - \sin\alpha) = \theta(1 - \sin\alpha).$$

Lemma 2: The area of $T$ is $\theta - \pi + 2\arccos\dfrac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}$.

Proof: If $\psi$ denotes the interior angle of $T$ at either $A$ or $B$, the area of $T$ is the angular defect, $\theta + 2\psi - \pi$. To calculate $\psi$, note that the unit vector $n_{1} = \frac{A \times C}{\|A \times C\|} = (0, -1, 0)$ is orthogonal to the great circle $AC$, the unit vector $$n_{2} = \frac{A \times B}{\|A \times B\|} = \frac{(-\sin\theta \sin\alpha, \sin\alpha(\cos\theta - 1), \cos\alpha \sin\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}$$ is orthogonal to the great circle $AB$, and $$\cos\psi = n_{1} \cdot n_{2} = \frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}.$$ This completes the proof of Lemma 2.

The area of the digon is the difference, \begin{align*} A &= \theta(1 - \sin\alpha) - (\theta + 2\psi - \pi) = \pi - \theta \sin\alpha - 2\psi \\ &= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right). \end{align*}

When $\alpha = 0$, the area vanishes for $0 < \theta < \pi$ (because the latitude through $A$ and $B$ coincides with the great circle arc), while if $\alpha$ is small and positive, the area is close to $\pi$ when $\theta = \pi$ (because $A$ and $B$ are nearly antipodal and the great circle arc passes through the north pole).

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jan 29, 2017 at 21:09
• @JyrkiLahtonen What is the method for moving to the chat (as you did)$Is it reserved only for HR users? Commented Jan 30, 2017 at 8:12 • @Widawensen: Only diamond moderators have access to that tool. At a certain point the system will prompt/suggest any user to post the comments in chat after that point, and then a chatroom will be created. Is there something in those comments that you would like to revisit but cannot? Or would rather keep here rather than in chat? The comments still exist here in a deleted state, and undeleting some of them is an option. Commented Jan 30, 2017 at 8:16 • @JyrkiLahtonen If chat room is going to exist the comments may stay in the chat room. Commented Jan 30, 2017 at 8:19 Assume the sphere is of radius$1$, and we have our constant lattitude$\phi$for two coordinates$(\phi, -\lambda_0), (\phi, \lambda_0) \in (-\pi/2,\pi/2)\times [-\pi,\pi]$. Note that the longitudes are positive and negative of a constant value. We can do this because we choose to measure the longitude angle$\lambda$from the coordinates' midpoint. This will be useful for our great circle parametrization. The great circle latitude function,$\gamma$, is tougher to parametrize by$\lambda$. First we should prove a property about this great circle: that the longitudes of its intersections with the equator are$\pi/2$away from the midpoint of the two coordinates' longitudes. We can use Side-Angle-Side Congruence of spherical triangles using the great circle, equator, and midpoint longitude lines. [[Something about latitude lines having equal angle from great circle intersection to the longitude midpoint... but we can't use similar triangle arguments...]] To aid us in our parametrization, we'll keep using that type of spherical triangle. Keep the equator and great circle, and have a great circle (longitude line) for each longitude. Let the arc of the longitude be$a$, the arc of the equator be$b$, and the arc of the great circle be$c$. Corresponding angles are$A$between great circle and equator,$B$between great circle and longitude line, and$C$between equator and longitude line. Note, this means$C = \pi/2$. With a little help from Napier, we get$\tan(a) = \tan(A) \cdot \sin(b)$. We can do some translations into the language of our problem:$A$is the constant angle between the great circle and the equator,$a$is the latitude from the equator, and$b$is the longitude measured from the intersection point of the great circle and the equator. We want to find$a$in terms of our parametrized$\lambda$. We know that$\lambda = b - \pi/2$, because we measure$\lambda$from the longitudinal midpoint. We can also derive the constant$\tan(A)$using known quantities in the great circle, our coordinates: $$\tan(A) = \frac{\tan(\phi)}{\sin(\pi/2 - \lambda_0)} = \frac{\tan(\phi)}{\sin(\pi/2 + \lambda_0)} = \frac{\tan(\phi)}{\cos(\lambda_0)}$$ Therefore, the latitude function$\gamma(\lambda)$of the great circle is: $$\gamma(\lambda) = \tan^{-1}\left[ \tan(A) \cdot \sin(\pi/2 + \lambda) \right] = \tan^{-1}\left[ \frac{\tan(\phi)}{\cos(\lambda_0)} \cdot \cos(\lambda) \right]$$ The constant-latitude function can be parametrized by longitude$\lambda$as: $$\kappa(\lambda) = \phi$$ The solution can gotten by integrating the difference between latitudes, that integral being across longitude. In general: $$\int_{-\lambda_0}^{\lambda_0} \gamma - \kappa d\lambda= \int_{-\lambda_0}^{\lambda_0} \tan^{-1}\left[ \frac{\tan(\phi)}{\cos(\lambda_0)} \cdot \cos(\lambda) \right] d\lambda - 2\phi\lambda_0 = -\cot(A)\int \frac{\tan^{-1}(v)}{\sin\left(\cot(A)\cdot v\right)} dv - 2\phi\lambda_0$$ I've checked around, and I can't seem to find an analytical solution for the integral$\int \tan^{-1}(x)/\sin(k\cdot x) dx$. For your example, the numerical solution is$2.12618$)$-\pi/6 \approx 1.60258$• Hmm, solution without calculation of the integral above probably can be improved to more clarified form, however you have proposed interesting approach to this problem... Commented Jan 25, 2017 at 18:07 • I get a negative number when I plug in$\alpha = \frac{\pi}{4}$and$\theta = \frac{2\pi}{3}$for his solution: $$\pi - \theta \sin \alpha - 2\arccos (\sin \frac{\theta}{2} \sin \alpha) = -0.16$$. Commented Jan 27, 2017 at 19:06 • See Larry my comments under the Andrew's answer.. maybe formula should be improved..it is probably more complicated.. Commented Jan 27, 2017 at 19:28 • However the Andrew's method, I'm sure, leads to the solution, your method, Larry, is also interesting, it would be good to compare results.. Commented Jan 27, 2017 at 19:53 • With improved Andrew's method I've received result$A=0.291$for original data, it is probably better than$1.6$because max. digon$area$for one quadrant of sphere =$\pi$(whole sphere 4$\pi\$ ) but your result Larry covers approximately half of this region what seems to be too much, what do you think about it ? Commented Jan 27, 2017 at 20:31