# relationship between a great circle arc and a latitude circle arc at a given latitude

My spherical geometry is a bit rusty but looking at the figure below:

... my intuition tells me that angles $$\phi$$ and $$\theta$$ (measured in radians) are connected with the following equation:

$$\phi = \theta*cos(\delta)$$

... where $$\delta$$ is the angle corresponding to the green arc. At least the above holds true for the special cases $$\delta=0$$ and $$\delta=\pi/2$$. Does the above equation hold true in general and what is the terminology that describes the various angles and circles I've drawn?

# update on terminology

It is now clear that the angles can be more properly described as follows:
• angle $$\theta$$ is the arc between points $$A'$$ and $$B'$$ on a latitude circle at latitude $$\delta$$ (see the drawing on this answer which makes this point clear)
• angle $$\phi$$ is the arc between points $$A'$$ and $$B'$$ on a great circle
• It seems about right. Depends on whether the angle changes linearly or sinusoidally. I would guess sinusoidally but I'd have to do geometry to check – Don Thousand Oct 9 '18 at 20:06
• Careful........ You've labelled $\phi$ as if it were the angle between two central rays, but then corresponding to that central angle you have connected the two points where those rays hit the sphere by a latitude circle. Although it's true that the length of a great circle arc is equal to the corresponding central angle (in radians, assuming radius $=1$), it is NOT TRUE that the length of a latitude circle is equal to the corresponding central angle. – Lee Mosher Oct 9 '18 at 20:23
• @LeeMosher $\phi$ is the angle between the two central rays; that's the only thing that interests me. I am not sure I fully comprehend the point you are making. – Marcus Junius Brutus Oct 9 '18 at 20:29
• If the length along the latitude circle is not of concern, in particular if there is no risk of confusing $\phi$ with the length of that latitude segment, then fine. – Lee Mosher Oct 10 '18 at 0:53
• @LeeMosher thanks; I now see your point. I am denoting with $\phi$ the angle between the rays OA' and OB' that reside on the plane defined by points O, A' and B'. I.e. $\phi$ is an ordinary two-dimensional Euclidean geometry angle. – Marcus Junius Brutus Oct 10 '18 at 2:08

## 3 Answers

The distance between the endpoints is scaled down like the circle radius by a factor of $$\cos\delta$$. However, that rather makes $$\sin \frac\phi2=\cos\delta\sin \frac\theta2.$$

• I am trying to derive $\theta$ given $\phi$ and $\delta$. How do I do that using your formula? My feeling is that the formula I give is accurate, at least for small values of $\phi$ and $\theta$ – Marcus Junius Brutus Oct 9 '18 at 20:31
• I just figured why this holds; will try to upload a diagram – Marcus Junius Brutus Oct 10 '18 at 19:37

My attempt of a proof:

In any circle, with center O, radius r, points A and B on the circumference and angle AOB = $$\alpha$$, the straight segment from A to B is : $$AB_{straight}^2 = 2 r^2 \times (1-cos(\alpha))$$ (1)

This is a straight application of the cosine theorem.

We also know for any angle $$\alpha$$ that $$cos(\alpha) = 1-2 sin^2 (\frac{\alpha}{2})$$ (2)

By combining (1), (2) we get : $$AB_{straight}^2 = 4 r^2 \times sin^2(\frac{\alpha}{2})$$ (3)

Apply (3) twice on the segment $$A'B'$$ from the original diagram by Marcus Junius Brutus :

• once for the great circle with $$r=R$$ and $$\alpha = \phi$$
• once for the latitude circle with $$r=R \times cos(\delta)$$ and $$\alpha=\theta$$

That gives $$4 R^2 \times sin^2(\frac{\phi}{2})$$ = $$4 (R \times cos(\delta))^2 \times sin^2(\frac{\theta}{2})$$

which simplifies to $$sin(\frac{\phi}{2}) = cos(\delta) \times sin(\frac{\theta}{2})$$

The accepted answer by Hagen von Eitzen is correct. This is just to supply a proof. The below proof uses only the basic trigonometric definitions of $$cos$$ and $$sin$$ as well as a very elementary theorem about isosceles triangles.

In the figure below points $$A'$$ and $$B'$$ have been renamed to $$A$$ and $$B$$ and the points $$A$$ and $$B$$ in the original drawing have been omitted as they are not necessary for the proof. In particular the angle $$\theta$$ of the original drawing (that was defined using points on the great circle) is the same as the angle $$\theta$$ of the new drawing that is defined using points on the latitude circle.

Refer to the figure below and observe the following elements:

• a great circle with center $$O$$ (therefore $$O$$ is also the center of the sphere)
• a latitude circle at latitude $$\delta$$ with center $$O'$$
• the straight line segment $$AB$$ (a line segment, not an arc) with point $$\Gamma$$ as its midpoint
• three right angles shaded grey and outlined red
• the right triangle $$AO'O$$ with the right angle being the one at point $$O'$$
• the right triangle $$O'\Gamma{}A$$ (which resides on the plane defined by the latitude circle) with the right angle being the one at point $$\Gamma$$
• the right triangle $$A\Gamma{}O$$ with the right angle being the one at point $$\Gamma$$
• angle $$\phi$$ is the angle $$AOB$$, the angle $$AO\Gamma{}$$ being exactly $$\phi/2$$
• angle $$\theta$$ is the angle $$AO'B$$, the angle $$AO'\Gamma{}$$ being exactly $$\theta/2$$

Observe that the angle $$OAO'$$ is identical to the angle $$\delta$$ and that $$OA$$ is a ray of the sphere, so we can write $$OA=R$$

We have the following equations:

1. $$O'A = R\cdot{}cos(\delta)$$ ; since $$OA$$ is the hypotenuse and equal to $$R$$, and since $$OAO'=\delta$$ as already noted
2. $$A\Gamma{}=O'A\cdot{}sin(\theta/2)$$

From $$1$$ and $$2$$ we obtain:

1. $$A\Gamma{}=R\cdot{}cos(\delta)\cdot{}sin(\theta/2)$$

We also have (from the right triangle $$A\Gamma{}O$$):

1. $$A\Gamma{}=R\cdot{}sin(\phi/2)$$ ; since $$OA$$ is the hypotenuse and equal to $$R$$

From $$3$$ and $$4$$ we have:

$$sin(\frac\phi2)=cos(\delta)\cdot{}sin(\frac\theta2)$$

$$\blacksquare$$