Chandra Education - Distance Between the Two Offset Points

pi = 3.1416 (Sourced from calculator, rounded to one more specific digit than the piece of information than the radius of the earth.)
r = radius of arc on the surface the earth = rearth =6378 km
(# degrees in the arc/360): As previously discussed, the crux of determining the distance between the two points along the arc lies in determining this angle. Let's begin our investigation by looking at the information that we have and figuring out a way to get at the information that we need. Here, the easiest way to start is by drawing as detailed a pictures as possible so as to simplify our problem.

Here we can see that our movement between the two points can be broken down into two perpendicular components that are parallel to either the equator eastern/westward movement) or the prime meridian (northern/southern movement). The blue dot indicates where the components meet. We now have what appears to be a right triangle smoothed over the face of a sphere. Our problem would be simple if the Pythagorean Theorem worked for curves!

However, note that if we want to arrive at the same point by first moving north, and then moving west, the length of our westward displacement is different! (The extreme case being the destination point being very close to the pole.) This arises from the decreasing length of the lines of lattitude as we approach to poles. So, if we were to connect the four points of the two sets of displacements that one can take to get from A to B with straight lines, they would not form a square. Instead they form a trapezoid.

This picture shows the starting, ending and meeting points connected by straight lines. As a result of the straight line connections, the trapezoid shown in blue is located within the sphere.

Here, we connect the points on the surface of the sphere with the center of the sphere, forming a pyramid with a trapezoidal base. Now, let's redraw our pyramid so as to be able to identify the angles and leg lengths that comprise it.

The picture to the left displays the pyramid as seen with the center of the earth between the observer and the part of the earth with the curved triangle on the surface. The pyramid has also been rotated somewhat so that it sits on the triangle formed by the points AOC. We'll call the Green Dot (MAD) point A, the Gray Dot (NY) point B, the Blue Dot (crossing point of components when moving longitudinally first) point C, the Pink Dot (crossing point of movement components when moving lattitudinally first) the Black Dot, the center of the earth, O, and the dot along the line AC that we will call P, that forms the right triangle ABP. Let's start by determining exactly which attributes of the trapezoid we will have know for certain to solve for the diagonal length. The endpoints of the diagonal will make an angle with the origin that can be used to determine what portion of a great circle it is equivelent to on the surface of the sphere. Let's quickly note that if two of the legs of a triangle are equal in length, that triangle is said to be eqilateral. This special kind of triangle has the unique property of having angles of equal size opposite the legs of equal length. Also notice that we are working with is an isosceles trapezoid; the shorter base is "centered" with respect to the longer base as a result of the equal decrease in length of the lines of longitude with increasing lattitude, so the sides AD and CB are equal in length, as are the angles that those lines make with AC.

The length of AB = ((AP)²+(BP)²)^1/2
AP	BP
AP = AC - ((AC-DB)/2)	BP = 2rsin(BOP/2)
AC = 2rsin(AOC/2)	BOP = latB-latA
DB = 2rsin(DOB/2)
AOC = longB-longA * cos (latA)
BOC = longB-longA * cos(latB)
AP = 1.065 * 10⁴ km	BP = 6.184 * 10³ km
(AP)²+(BP)²)^1/2 = 1.232 x 10⁴ km

Then the Angle AOB = 2(sin^-1((length AB/2)/r)).

= 2(sin^-1((1.232 x 10⁴ km/2)/6378 km) = 149.95 deg