Help with direct tunnel distance between two lat /long coordinates. My brother wants to take a sign to the Sign Post Forest in Canada's Yukon. He wants it to show the distance to London, but directly through a tunnel that only exists in his head.  There's not a lot in his head.  So it's the mathematically shortest possible distance through the Earth between the Sign Post Forest and the flag pole on Buckingham Palace.

The two locations are here, obtained from Google Maps:-


There's a whole load of "Similar Questions" coming up  on my screen, but none seem to calculate the direct distance via a hypothetical tunnel.  They're all  on the surface. It seems that the size of the Earth is needed, and I found some parameters for it from this GPS /WGS84 document:-
equatorial radius (WGS84)- 6378137 m
polar radius (derived) - 6356752.3 m
 A: For future reference, the most accurate method to perform this type of calculation is by using an ellipsoidal Earth model, starting with the geographic coordinates (latitudes, longitudes and ellipsoidal heights $\phi, \lambda, h$) of your points, transforming them to cartesian (X, Y, Z) coordinates, then calculating the 3D distance with the Pythagorean theorem. Spherical trigonometry is an approximation that, although good, can give errors up to about 0.5% for Earth coordinates. 
The formulas to convert geographic coordinates to cartesian are:
$$X = \left(N\left(\phi\right)+h\right)\text{cos }\phi\text{ cos }\lambda$$
$$Y = \left(N\left(\phi\right)+h\right)\text{cos }\phi\text{ sin }\lambda$$
$$Z = \left(\frac{b^2}{a^2}N\left(\phi\right)+h\right)\text{ sin }\phi$$
Where
$$N\left(\phi\right)=\frac{a}{\sqrt{1-e^2\text{ sin}^2\phi}}$$
$$e^2=1-\frac{b^2}{a^2}$$
$$a=6378137\text{ m (in WGS84)}$$
$$b=6356752.314\text{ m (in WGS84)}$$
Then we can find the distance in 3D between the 2 points with the Pythagorean theorem:
$$d = \sqrt{(X_2-X_1)^2 + (Y_2-Y_1)^2 + (Z_2-Z_1)^2}$$
We can also find the orientation (zenith angle and geographical azimuth $\theta, \alpha$) of the 2nd point in the 1st point's local reference frame with the following rotation matrix and spherical coordinate conversion:
$$\left[\begin{matrix} x \\ y \\ z \end{matrix}\right]=\left[\begin{matrix} -\text{ sin }\lambda_1 & \text{ cos }\lambda_1 & 0 \\ -\text{ sin }\phi_1 \text{ cos }\lambda_1 & -\text{ sin }\phi_1 \text{ sin }\lambda_1 & \text{ cos }\phi_1 \\ \text{ cos }\phi_1 \text{ cos }\lambda_1 & \text{ cos }\phi_1 \text{ sin }\lambda_1 & \text{ sin }\phi_1 \end{matrix}\right]\left[\begin{matrix} X_2-X_1 \\ Y_2-Y_1 \\ Z_2-Z_1 \end{matrix}\right]$$
$$\theta = \text{acos }\frac{z}{\sqrt{x^2+y^2+z^2}}$$
$$\alpha = \text{atan2 }(x,y)$$
Note that the x and y axes are reversed relative to the mathematical couterclockwise polar coordinate system in the atan2 function, since geographical azimuths are calculated clockwise from the North.
For your specific points with geographic coordinates 1(60.063229°, -128.713104°, 699 m) and 2(51.501326°, -0.141968°, 74 m), we find the following answers :
$$d=6491.075 \text{ km}$$
$$\theta=120.5066°$$
$$\alpha=33.7938°$$
A: Your answer comes in two steps.
First step is to use Spherical Trigonometry to find the distance on the surface (measured in degrees or radians) between the two given locations, also given in degrees. This is a simple problem, involving a single use of the Law of Cosines for spherical triangles.
Second step is much simpler, since it’s regular high-school plane geometry. You want to use that angular distance, the length of an arc, to find the length of the corresponding chord. For this, of course, you need the radius of the Earth.
I would go into more detail, but I just got back from dinner, and am still spinning from a giant martini that I consumed. If this answer was too sketchy, get back to me. And of course, others may give a complete detailed answer.
A: First compute the latitude and longitude in radians:


*

*Watson Lake: $\phi_1=1.048301$, and $\lambda_1=-2.246467$; 

*Buckinham Palace: $ \phi_2 = 0.898867$, and $\lambda_2=-0.002477$.


Then, use the formula here and the radius of the Earth ($r=6371$ km) to get surface distance:
$$\Delta\sigma = \bigl(\sin\phi_1\cdot\sin\phi_2+\cos\phi_1\cdot\cos\phi_2\cdot\cos(|\lambda_2-\lambda_1|)\bigr) = 1.0650,$$
and 
$$ \text{Surface Distance}= r\Delta\sigma=6785.0\text{ km}.$$
Finally, using the law of cosines we can calculate the shortest possible distance through the Earth
$$ \text{Direct Tunnel Distance}= r\sqrt {2-2\cos (\Delta\sigma) } = 6468.9 \text{ km}.$$
A: Let's assume that Google polar coordinates have center in Earth center of gravity, that it's close to its geometric center, and that Earth is revolution ellipsoid with equatorial radius $\approx$ 63781400 m and polar $\approx$ 63567500 m. From a bit of calculation for ellipse diameters and knowing that both points are close to sea level, we'll get what we want with good accuracy.
Knowing distance from center of polar coordinates to both your destination points makes finding distance very easy by law of cosines —  angle is $\approx 0.3389 \pi$ ,  and we get that this distance is something around $6463.75$ km.
