Finding the Great Circle between Two Points on a Sphere – INVESTIGATION
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Is this Autograph activity we treat the Earth as a sphere or radius 1, enterned as r = 1 with the option
“Spherical r = Polar” selected.
Angles are measure in degrees
The two points that we wish to show the Great Circe between have been created as two vectors drawn
from the origin (0, 0, 0): [1, b, a] and [1, d, c], and small spheres have been drawn at the ends to signify
the two points under consideration.
The Great Circle is a very involved parametric equation, shown in red.
The planes x=0, y=0 and z=0 have been shown (at 50% transparency) so that the meaning of the angles
can be investigated.
1.
Open the Constant controller and set each of ‘a’, ‘b’, ‘c’ and ‘d’ to zero.
2.
Vary each away from zero so can see what the four angles represent
3.
Which represent Longitude and which represent Latitude?
4.
Set ‘a’ = 45° and vary ‘b’ from 0° to 360° and observe.
5.
What do you notice about the Great Circle, whatever the values of ‘a’, ‘b’, ‘c’ and ‘d’?
6.
Find out roughly the Latitude and Longitude of Beunos Aires and London, and hence set up the
Great Circle route that planes will fly when going to the Summer Olympics in 2016.
DOWNLOAD THE AUTOGRAPH FILE
NOTE
if you want to manipulate the Autograph graph page, because the planes have been
drawn, you should use the HAND mode so that objects are not selected.
If you hold CTRL and sweep the mouse UP and DOWN you can zoom inside the sphere!
If you hold SHIFT and sweep the mouse around the POSITION of the ‘camera’ is moved.
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Finding the Great Circle between Two Points on a Sphere – THEORY
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Links
http://www.physicsforums.com/showthread.php?s=c1151a98e1e0256612ca5043dabb777d&t=571535
&page=2
http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecturenotes/MIT16_07F09_Lec03.pdf see pages 9-10 on transforming coordinate systems.
Setup
Sphere can have radius for simplicity. Two points are given in polar coordinates on the sphere (1, b, a)
and (1, d, c).
To find the great circle which goes through these two points you could find the plane defined by the two
points and the origin (on which the sphere is centered), then find the intersection of the plane and the
sphere. This turns out to be quite messy but should work.
The method described in the first link above is a nicer solution although there are still many trig
functions when you get into the details.
Solution
Let p and q be the position vectors for the two points on the sphere's surface. By definition p and q are
unit vectors.
Let u = p, now we want to find vector v which is a unit vector perpendicular to u in the plane p and q are
in. Cross-product is our friend here. First find the normal:
w = (p x q)/|p x q|
The divider is required to ensure that w is a unit vector because although p and q are unit vectors they
are not perpendicular to one another.
Then v = u x w.
Now in the uv plane the intersection with the sphere is u = cost, v = sint (w = 0). All we need to do now is
transform the coordinate system back to xyz.
This is fairly simple but I needed to remind myself of the logic, that can be found in the second link
above on pages 9 and 10.
This works because uvw are orthogonal unit vectors defining a coordinate system just like ijk do.
𝑢1
𝑥
(𝑦) = (𝑢2
𝑢3
𝑧
𝑣1
𝑣2
𝑣3
𝑤1 cos 𝑡
𝑤2 ) ( sin 𝑡 )
𝑤3
0
𝑅 = 𝑢 cos 𝑡 + 𝑣 sin 𝑡
Components
x = cos(a)cos(b)cos(t) + (cos(a)sin(b)(cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d)) sin(a)(sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c)))sin(t)/k,
y = cos(a)sin(b)cos(t) + (sin(a)(cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d)) - cos(a)cos(b)(cos(a)cos(b)cos(c)sin(d)
- cos(a)sin(b)cos(c)cos(d)))sin(t)/k,
z = sin(a)cos(t) + (cos(a)cos(b)(sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c)) - cos(a)sin(b)(cos(a)sin(b)sin(c) sin(a)cos(c)sin(d)))sin(t)/k
k = sqrt((cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d))^2 + (sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c))^2 +
(cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d))^2)
Equation for Autograph
x = cos(a)cos(b)cos(t) + (cos(a)sin(b)(cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d)) sin(a)(sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c)))sin(t)/sqrt((cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d))^2 +
(sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c))^2 + (cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d))^2), y =
cos(a)sin(b)cos(t) + (sin(a)(cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d)) - cos(a)cos(b)(cos(a)cos(b)cos(c)sin(d) cos(a)sin(b)cos(c)cos(d)))sin(t)/sqrt((cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d))^2 + (sin(a)cos(c)cos(d) cos(a)cos(b)sin(c))^2 + (cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d))^2), z = sin(a)cos(t) +
(cos(a)cos(b)(sin(a)cos(c)cos(d) - cos(a)cos(b)sin(c)) - cos(a)sin(b)(cos(a)sin(b)sin(c) sin(a)cos(c)sin(d)))sin(t)/sqrt((cos(a)sin(b)sin(c) - sin(a)cos(c)sin(d))^2 + (sin(a)cos(c)cos(d) cos(a)cos(b)sin(c))^2 + (cos(a)cos(b)cos(c)sin(d) - cos(a)sin(b)cos(c)cos(d))^2)
Simon Woodhead
September 2013