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Oct 12, 1988, 6:01:51 PM10/12/88

to

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

Oct 14, 1988, 10:27:44 AM10/14/88

to

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

Oct 14, 1988, 4:58:32 PM10/14/88

to

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?

#

#

# 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

Oct 14, 1988, 6:41:36 PM10/14/88

to

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]

Oct 15, 1988, 4:26:44 PM10/15/88

to

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?

>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

Oct 15, 1988, 9:03:08 PM10/15/88

to

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?

>

> Many thanks in advance,

>

> Rick Coates

> 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!)

Oct 16, 1988, 10:45:16 AM10/16/88

to

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

Oct 16, 1988, 12:26:05 PM10/16/88

to

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.

Oct 17, 1988, 7:43:51 PM10/17/88

to

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?

>

> Many thanks in advance,

>

> 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

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.

Oct 18, 1988, 10:51:27 AM10/18/88

to

In article <11...@agora.UUCP> ri...@agora.UUCP (Rick Coates) writes:

>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.

Oct 20, 1988, 3:19:35 AM10/20/88

to

Re: Great Circle calculations, a while back I had need to assist a HAM

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.

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)

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