# Great circle problem

1. May 3, 2013

### usaplasticman

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. The attempt at a solution

2. May 3, 2013

### HallsofIvy

Staff Emeritus
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.

3. May 3, 2013

### 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.

4. Sep 1, 2014

### nixed

A general expression for the great circle of radius R, passing through points A= R ( ax, ay, az) and B= R (bx, by, bz) (where (ax, ay, az) is the unit vector a along OA and (bx, by, bz) is the unit vector b along OB) is given by

x/R = [Cosψ]. ax + [Sinψ].{(az2+ay2)bx - (azbz+ayby)ax}/sinδ

y/R= [Cosψ].ay + [Sinψ]{(az2+ax2)by - (azbz+axbx)ay}/sinδ

z/R= [Cosψ].az + [Sinψ]{(ay2+ax2)bz - (ayby+axbx)az}/sinδ

Cos δ = axbx + ayby + azbz (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 = axi + ayj + azk

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 devide by sinδ)

then x/R = i.Q/R etc

Last edited: Sep 1, 2014
5. Sep 1, 2014

### 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 ax, ay 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.

Last edited: Sep 1, 2014
6. Sep 1, 2014

### LCKurtz

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

7. Sep 2, 2014

### 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!