Great Circle Problem: Derive Equation for Route from A to B on Sphere

[usaplasticman]

Homework Statement

derive/create an equation for a "great circle" route r(t) from a given point A to a given point B along the surface of the sphere with center (0,0,0) and radius = 15

test point 1: A=(2,10,11) to B(14,5,2)
test point 2: A=(10,5,10) to B(0,-12,9)

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

The "great circle route" from point A to point B on a sphere is the intersection of the sphere with the plane determined by A, B, and the center of the sphere.

The sphere with center (0,0,0) and radius = 15 has equation $x^2+ y^2+ z^2= 225$.

The plane determined by points (2,10,11), (14,5,2), and (0, 0, 0) has equation 35x- 150y+ 130z= 0. With two equations in three variables, you can solve for two in terms of the third, using that last variable as parameter.

 

[LCKurtz]

Halls has given you one method; here's another. Parameterize the straight line between A and B (very easy) and call that parameterization $\vec r(t)$. Then use$$
\vec R(t) = \frac {15} {|\vec r(t)|}\vec r(t)$$for the arc of the great circle between the two points.

 

[nixed]

A general expression for the great circle of radius R, passing through points A= R ( a_x, a_y, a_z) and B= R (b_x, b_y, b_z) (where (a_x, a_y, a_z) is the unit vector a along OA and (b_x, b_y, b_z) is the unit vector b along OB) is given by

x/R = [Cosψ]. a_x + [Sinψ].{(a_z²+a_y²)b_x - (a_zb_z+a_yb_y)a_x}/sinδ

y/R= [Cosψ].a_y + [Sinψ]{(a_z²+a_x²)b_y - (a_zb_z+a_xb_x)a_y}/sinδ

z/R= [Cosψ].a_z + [Sinψ]{(a_y²+a_x²)b_z - (a_yb_y+a_xb_x)a_z}/sinδ

Cos δ = a_xb_x + a_yb_y + a_zb_z (δ is the angle between OA and OB vectors)

These are derived by setting up a new, right handed, orthonormal set of basis vectors e₁, e₂, e₃ oriented such that e₁ is along OA, and e₃ is perpendicular to the great circle. In this basis the equation for any point Q on the gt circle is simply:

Q/R = Cosψ e₁ + Sinψ e₂

as ψ increases from 0 to δ we sweep along the great circle from A to B. For the complete circle
0≤ψ≤2π radians. To get the equations for the x, y, z components of the circle we just relate the new basis e₁ etc to the old basis i,j,k. Explicitly:

e₁= a = a_xi + a_yj + a_zk

e₂= e₃ X e₁

e₃= a X b /Sinδ =


(1/sinδ).
\begin{vmatrix}
i&j&k\\
a_x&a_y&a_z\\
b_x&b_y&b_z\end{vmatrix}

(nb remember the sinδ factor: since the unit vectorsa and b are not in general orthogonal a X b= sinδ ≠ 1 so to get a unit vector e₃ we must divide by sinδ)

then x/R = i.Q/R etc

 

[nixed]

So the answer to the original problem:

A(2,10,11) B(14,5,2) both are points on sphere radius 15. So a= (2/15,10/15, 11/15) and b=(14/15,5/15,2/15) are unit vectors which give us the components a_x, a_y etc.

∴Cosδ = (28+ 50+22)/15² ∴ δ = 1.1102 rad = 63.61 deg

the shortest arc length along surface of sphere from A to B = R.δ =15 x 1.1102 units. this arc is part of the great circle passing through A and B whose equation is:

x/R= [cosψ] 0.13 + [sinψ] 0.97574 ∴ x = [cosψ] 2 + [sinψ] . 14.6361

y/R = [cosψ] 0.6666 + [sinψ] 0.041345 ∴ y = [cosψ] 10 + [sinψ] . 0.620174

z/R =[cosψ] 0.73333 + [sinψ] 0.97574 ∴ z = [cosψ] 11 + [sinψ] . (-3.2249)

ψ=0 corresponds to point A, ψ=δ to point B.

case 2:

A= (10,5,10) B=(0,-12,9) so length of arc Cos δ= (0 - 60 + 90)/15²

∴δ= 1.43706 rad i.e 82.34 deg. The arc length A to B is R.δ= 21.55 units

the great circle which passes through these two points is

x= 10[cosψ] - 1.34535[sinψ]

y= 5[cosψ] -12.7808[sinψ]

z = 10[cosψ] + 7.7357[sinψ]

0≤ψ≤2∏

for both sets of equations for all values of ψ; x²+y²+z² = 15²

as they must for points on the surface of the sphere.

 

[LCKurtz]

Have you noticed that you responded to a thread more than a year old?

 

[nixed]

Yes, I noticed the age of the thread, but I posted because I have been working on the 'ant and honey' problem on a spherical bowl and needed to work out the great circles going through given points on a sphere and couldn't find the information easily on-line so had to derive it. Having done so I thought i'd put it here for reference in case others are looking for the info!