# Let be a unit-speed curve which lies on a sphere of center p and radius R.

August 31, 2017

Question
1)
Let be a unit-speed curve which lies on a sphere of
show that, if
) + ((

B and

=(

) .

Hints: T, N and B form a basis for = (i.e. 3-space), so

2)

Show that, if (

) + ((

) is a constant,

B and show p is a

3) The point of the problem is to have you explain every line of the
procedure.
Hints: The sphere of radius 1 is parametrized by longitude u and
latitude v:
x(u; v) = (cos(u) cos(v); sin(u) cos(v); sin(v)):
Think about how you can parametrize a circle in 3-space using any
two perpendicular vectors originating at (0; 0; 0). What do you do if
a normal to the
plane they determine? And then how do you get two perpendicular
vectors in the plane?
This procedure draws the great circle arc between two points A and
B on Earth and gives the distance between the points (assuming the
radius of the Earth to be 3970 miles). A and B are given by longitude
and latitude coordinates. For example A=[100,25] means that the
point A has longitude 100 degrees and latitude 25 degrees. We
assume the Earth is a sphere with parametrization
the Earth. Try to understand the inputs and how they are used.

>
> great_circle:=proc(place1,place2,theta,phi)
local point1,point2,angle,A,B,sphere,circ,alpha,unitcp,alp1,alp2,
p1,p2;
A:=map(evalf,[Pi/180*place1,Pi/180*place1]);
B:=map(evalf,[Pi/180*place2,Pi/180*place2]);
point1:=<cos(A)*cos(A)|sin(A)*cos(A)|sin(A)>;
point2:=<cos(B)*cos(B)|sin(B)*cos(B)|sin(B)>;
angle:=evalf(arccos(DotProduct(point1,point2,conjugate=false)));
print(`Distance is`,3970*angle,`miles`);
unitcp:=1/sin(angle)*CrossProduct(point1,point2);
alp1:=ScalarMultiply(point1,cos(t));
alp2:=ScalarMultiply(CrossProduct(unitcp,point1),sin(t));
alpha:=alp1+alp2;
sphere:=plot3d([cos(u)*cos(v),sin(u)*cos(v),sin(v)],u=0..2*Pi,
color=navy):
p1:=pointplot3d(convert(point1,list),color=cyan,symbol=
solidcircle,symbolsize=24):
p2:=pointplot3d(convert(point2,list),color=green,symbol=
solidbox,symbolsize=24):
display({sphere,circ,p1,p2},scaling=constrained,orientation=
[theta,phi]);
end:
Here are other longs and lats (in that order): Ceveland (-81,68,41,
48), New York (-73.94, 40.67), Chicago (-87.68, 41.84), Paris (2.30,
48.83), Moscow (37.62, 55.75),
Atlanta (-84.42, 33.76), Toronto (-79.60, 43.67), Seoul (127.05,
37.52), Beijing (116.35,39.90), Sydney (151.30, -33.90), New Delhi
(77.22, 28.90),
Rio de Janeiro (-42.70, -22.49). How about New York to Moscow?
London to Paris?Cleveland to Sydney?
> great_circle([-73.94,40.67],[37.62,55.75],-21,60);

> great_circle([-81.68,41.48],[151.30,-33.9],-156,75);

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