simple spherical algebra question
Rick Coates
Here is a straightforward question that I have not been able to find
in my reference books:
Given two points on a sphere, what is the equation of the
line (great circle) between them? What is the midpoint?
Many thanks in advance,
Rick Coates
tektronix!sequent!islabs!ateq!rick
OR
tektronix!reed!percival!agora!rickc
Todd M. Lewis
I had (have) a similar problem which is probably related to this one:
Given a point on a globe and a direction (which defines a Great
Circle, right?), go X units (degrees, nautical miles, whatever) in
that direction. Now, what is your new location (latitude and
longitude), and what direction are you facing (in, say, degrees
east or west from north)?
This shouldn't be that hard, but darned if I've come up with a
general solution. The same problem comes up in a flight simulator
type of program (i.e., GAME (shhhh! its a simulation!)). Given
a heading, pitch, and bank, add some additional heading, pitch and
bank. What are the combined H, P, and B? I know you can build
simple rotation matricies and concatenate them, but that won't give
you the new H, P, and B values.
If anyone would care to comment by email, I would welcome any
tips, pointers, or even VERY CLEAR, COMPLETE SOLUTIONS! (mmm, Nah!:-)
--Todd M. Lewis
Vallury Prabhakar
#
# Here is a straightforward question that I have not been able to find
# in my reference books:
#
# Given two points on a sphere, what is the equation of the
# line (great circle) between them? What is the midpoint?
#
Wouldn't this line between the two points on the surface be the
geodesic curve? Derivations of the relevant equations may be found
in most books covering differential geometry of surfaces.
The theoretical analysis for the mid-point might be a bit more involved.
Perhaps a good approximation might be obtained by using a parametric spline
representation for the geodesic. Given the tangents to the surface (and
hence the curve) at the two points and the locations of the points themselves,
one could write the parametric equation for say a cubic approximation. Setting
the parameter to 0.5 will then yield the mid-point.
In fact, if the curve is strictly circular, then it is possible to
derive an a priori parametric equation so that the boundary conditions AS
WELL as the mid-point may be matched up.
Hope this helps.
-- Vallury Prabhakar
Richard Bumby
In article <11...@agora.UUCP> ri...@agora.UUCP (Rick Coates) writes:
> Here is a straightforward question that I have not been able to find
> in my reference books:
>
> Given two points on a sphere, what is the equation of the
> line (great circle) between them? What is the midpoint?
>
There is ONLY ONE way to do all problems involving the global geometry
of the sphere -- formulate them in the Euclidean space in which the
sphere is embedded -- and there is ONLY ONE way to solve problems in
Euclidean space -- use vector methods to reduce them to planar
problems. Following these principles, the two give points and the
center of the sphere give 3 points in space, and the equation of the
plane through those points can be found easily from their coordinates.
This plane meets the sphere in the required great circle. The rest of
the problem is done in this plane. The midpoint of the segment
joining the given points is found by a standard method. If you have
followed up to this point you have a picture showing that the point
you want is a point at the right distance from the center on the line
joining the center to this point -- another standard problem.
--
--R. T. Bumby ** Math ** Rutgers ** New Brunswick **
(in one form or another for all kinds of mail)
[bu...@math.rutgers.edu]
Dan Christensen
>
>
>in my reference books:
>
> Given two points on a sphere, what is the equation of the
> line (great circle) between them? What is the midpoint?
The equation for the great circle can be found using the technique
described in an earlier posting.
To find the midpoint, use the following method.
Assume that the sphere is centered at the origin, and has radius r.
(You can translate your sphere to this state and then
undo the translation the resulting point.)
Let's say the two points are (x1, y1, z1) and (x2, y2, z2).
The midpoint of the straight line joining these points is
p = ( (x1+x2)/2, (y1+y2)/2, (z1+z2)/2 ).
Then just scale this point to the surface of the sphere.
The midpoint of the great circle is then
r
--- * p, where |p| is the length of p, sqrt( x^2 + y^2 + z^2 ).
|p|
Hope this helps.
Dan Christensen
jdch...@watcgl.uwaterloo.ca
Eric Smith
> in my reference books:
>
> Given two points on a sphere, what is the equation of the
> line (great circle) between them? What is the midpoint?
>
> Many thanks in advance,
>
> Rick Coates
Suppose you take two points on the sphere. That gives you a circle; call
it the equator. Draw a meridian through one of the points to give you
a reference for lines of longitude. Then that point has coordinates (0,0),
and the other one has coordinates (a, 0), where a=longitude of second pt.
The line between them is { (x,0) : 0 <= x <= a }.
If you want to translate this back into another latitude/longitude coordinate
system, you just have to rotate; in general the equation will be
(x,y) = t(u,v), where t is a parameter, and (u,v) is one of the points.
So the lines on a sphere are like lines through the origin on a plane.
--
Eric R. Smith email:
Dept. of Mathematics 7103...@uwovax.uwo.ca
University of Western Ontario 7103...@uwovax.bitnet
London, Ont. Canada N6A 5B7 (a shared mailbox: put my name on
ph: (519) 661-3638 the Subj: line, please!)
Rick Coates
Many thanks to all those who sent me suggestions - especially the fellow
at Tektronix who mailed me a copy of a couple of pages from an encyclopaedia
of mathematics.
I went ahead and got off the dime and solved the problem myself - as it
turned out, I did have what I needed: my 1968 edition of the CRC Standard Math
Tables, which includes a summary of spherical trigonometry (do the new
editions still have pages and pages of numerical results for trig functions?).
I had thought that this would be a standard problem in cartography and
someone in Netland would have the solution at her/his fingertips.
For those of you who asked why I wanted to do this - the most obvious
application is to plot great circle routes on a map. If I want to
show the course from Sidney to Seattle, I have to incrementally draw the
great circle route between the two, since the map projection is not
linear (the 'straight line' great circle becomes a curved line on the
projection).
(The problem was: given two points in spherical coordinates (lambda from
-pi to +pi; phi from -pi/2 to +pi/2) find a function that returns the
phi for a given lambda on the great circle line between the two points;
and what is the midpoint)
Again, thanks for all the replies!
Rick Coates
tektronix!reed!percival!agora!rickc
OR
tektronix!sequent!islabs!ateq!rick
Vidhyanath K. Rao
In article <11...@agora.UUCP> ri...@agora.UUCP (Rick Coates) writes:
>>Here is a straightforward question that I have not been able to find
>>in my reference books:
>>
>> Given two points on a sphere, what is the equation of the
>> line (great circle) between them? What is the midpoint?
have two choices: Translate and scale to get the Unit sphere.
Let the two points be r_1 and r_2. Let s_t = r_1 + t(r_2 - r_1)
and r_t = s_t/|s_t|.
The other method may be quicker if you have some quick and dirty way
of getting sine and cosine of an angle at the same time. [some fpp's
will give you unscaled values]. Find \alpha so that the two
points are u and u \cos\alpha + v \sin\alpha. with u and v
orthonagonal. Let r_t = u \cos\theta + v \sin\theta with \theta
ranging from 0 to \alpha. This is also well suited for theoretica
work.
James E. Prior
> Here is a straightforward question that I have not been able to find
> in my reference books:
>
> Given two points on a sphere, what is the equation of the
> line (great circle) between them? What is the midpoint?
>
> Many thanks in advance,
>
> Rick Coates
There are jillions of spheres that pass through any two given points.
Presuming that you know of a particular sphere, i.e. its center,
The great circle is the intersection of the sphere, and the plane
defined by the sphere's center and the two given points.
You may find it handy to derive the formula for the plane by remembering
that the cross product of the two non-parallel vectors lieing in a plane,
is perpendicular to that plane.
--
Jim Prior osu-cis!n8emr!oink!jep j...@oink.UUCP
Pointers are my friend.
Robert Firth
>For those of you who asked why I wanted to do this - the most obvious
>application is to plot great circle routes on a map. If I want to
>show the course from Sidney to Seattle, I have to incrementally draw the
>great circle route between the two, since the map projection is not
>linear (the 'straight line' great circle becomes a curved line on the
>projection).
Please forgive me if you already know this, but there is a map
projection in which all great circles are straight lines. It is
called the "gnomonic" projection: the hemisphere is projected onto
an infinite tangent plane from a projection point at the centre
of the sphere.
Very few published maps use this projection, since for large areas
the distortion is extreme and for small areas you don't need to
use great-circle trajectories anyway. However, if you have a
digitised world globe somewhere, it's not too hard to write some
software to generate the projection.
Taking the centre of projection as one pole in a system of spherical
surface coordinates (theta, phi), and as the centre of the plane
polar coordinates on the projection plane (r,alpha), then
alpha = phi
r = R tan(theta)
(where R is the scaled radius of the Earth). Clearly, the equator
(theta = pi/2) is mapped onto the circle at infinity, so the projection
cannot quite cover a whole hemisphere.
Hope that helps.
Thad P Floryan
operator with antenna alignment, and developed a great circle distance
calculator program. The development of the solution wasn't immediately
obvious to me, requiring use of haversine formulae. Extracted (below)
is a portion of the tech notes of the program which you might find useful.
The refs to "Bowditch" are to THE AMERICAN PRACTICAL NAVIGATOR.
A second release of the program will occur after the database of all cities
and landmarks on Earth has been entered; included will be graphics routines
to display multiple views of the Earth (e.g. various world projections) with
the chosen great circle(s). If someone uses this for missile targeting data
I'm gonna be P.O.'d! :-)
Thad Floryan [ th...@cup.portal.com (OR) ..!sun!portal!cup.portal.com!thad ]
`` GCDIST TECHNICAL NOTES
The actual calculations use haversine formulae as these avoid the ambiguity
arising through use of trigonometric functions which do not indicate the
quadrant in which the answer lies (ref. Bowditch, p.232).
hav = haversine = 1/2 versine = (1-cosine)/2 = sin^2 angle/2
sec = 1/cos
csc = 1/sin
For reference, the trigonometric formulae are:
cos D = sin A sin B + cos A cos B cos L
sin C = cos B csc D sin L
where:
A = origin's latitude in degrees (+N, -S)
B = destination's latitude in degrees (+N, -S)
L = longitude difference between origin and destination
C = course direction to destination in degrees East or West from North
D = distance to destination in nautical miles/minutes of arc
(1 minute of arc = 1 nautical mile = 1.1507 statute miles)
The haversine formulae used by this program are:
hav D = hav DLon cos L1 cos L2 + hav DLat
hav C = sec L1 csc D ( hav CoL2 - hav D_CoL1 )
where:
DLat = latitude difference between origin and destination
DLon = longitude difference between origin and destination
L1 = origin's latitude in degrees
L2 = destination's latitude in degrees
D = distance to destination in nautical miles/minutes of arc
CoL1 = 90 - L1
CoL2 = if L1 and L2 are same name (both N or S), then 90 - L2
else 90 + L2 if contrary name (one N other S)
D_CoL1 = numerical difference between D and CoL1
C = course direction labelled N or S to agree with L1, and
E or W to agree with DLon. In great circle routing, L2
may be south of L1 yet have a Northerly component
The haversine formulae can be reduced (for computer solution) to:
D = acos ( cos DLat - (1 - cos DLon) * cos L1 * cos L2 )
C = acos ( 1 - (cos D_CoL1 - cos CoL2)/(cos L1 * sin D) )
'' (end of extract)