So you're think you're good at math?

[Asher]

Do you want to expand by all three directions?

Asher, what Polars do for you is rather than getting into a messy coordinate as cartesians, they let you find a number that you are very interested in, the 'radii' of the object from the com.

Instead of XYZ, you have inside, outside and edge.

Do you want it for n-sided polygons, or do have restricted cases?

Edit, this is probably what you want Asher. They've got the functions here.

http://web2.uwindsor.ca/math/hlynka/Ngon.pdf

They are n-sided polygons. As the users will create them, they can be arbitrarily complex.

That PDF seems to be about creating random points within a polygon, from what I read from skimming.

[Ben Kenobi]

[r1r2sin(theta2-theta1) + r2r3sin(theta3-theta2) + r3r1sin(theta1-theta3)]/2.

For triangles at arbitrary points P1,P2,P3.

[Ben Kenobi]

http://www-math.mit.edu/~gs/papers/PolygonEigenvalues.pdf

This may also be helpful.

[Asher]

Eigenvalues, holy ****. Blast from 2001.

[Ben Kenobi]

http://maven.smith.edu/~orourke/books/compgeom.html

This may also help too.

[Kuciwalker]

In units of the radius of the earth (i.e. dealing with the unit sphere, and converting to radians) I've got

(latC, longC) = center
(latG, longG) = given
d = some_flavor_of_acos(sin(latC)sin(latG)(cos(longC)cos(longG) - sin(longC)sin(longG)) + cos(latC)cos(latC))

by converting to cartesian coordinates and taking the dot product.

We now want to solve for the point (latN,longN) = new such that dist(new,center) = scale*d, and such that dist(new,given) = (scale-1)*d.

Note: I don't understand why in the OP you assuming you'll be able to construct a right triangle using this three points.

edit: also, this is all probably pointless because KH knows the formula for this in his sleep

[Asher]

I can always construct a right triangle.

pointC (centre)
pointG (given)
pointI (imaginary)

the imaginary point would be constructed like so: pointC.lat, pointG.long

It'll create a right triangle every time, with the hypotenuse being the distance between pointC and pointG

[Kuciwalker]

Ooh, I thought you were trying to construct the new scaled point.

[Kuciwalker]

By the way, as it turns out actually solving my system of equations kind of sucks.

[Asher]

Ooh, I thought you were trying to construct the new scaled point.

Nope, I was creating the right triangle 'cause I thought it'd be easiest to leverage trig to determine the new point (and it would be in a true Cartesian plane).

FWIW I found this formula that supposedly determines a lat/long of a point given an origin point, radial, and distance. But when I implemented it in JS, it gave me bizarre and obviously incorrect results:

lat =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))

dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat))

lon= (lon1-dlon +pi) % (2*pi) )-pi

Where lat1/lon1 and lat2/lon2 are the centroid and given point, respectively, d is the distance between the centroid and given point, and "tc" is the radial.

[Kuciwalker]

Always make sure you are using the correct asin/acos/atan (in terms of domain and range).

[Kitschum]

I'm grammar more better at.

[OzzyKP]

I'm grammar more better at.

Finally, something in this thread that makes sense.

[Asher]

It looks like this should work...

Definitions:
Centroid (Lat1, Lon1)
Bearing (theta)
distance (d): angular distance in radians
radius (R): earth's radius

Lat2 = asin(sin(Lat1) * cos(d/R) + cos(Lat1) * sin(d/R) * cos(theta))

Lon2 = Lon1 + atan2(sin(theta) * sin(d/R) * cos(Lat1), cos(d/R) - sin(Lat1) * sin(Lat2))

Will try it out tomorrow.

[KrazyHorse]

Asher, I'm not sure your problem is well-formulated....

An example for why I say this is as follows: Take as your polygon a square "centered" at the north pole with "radius" (distance from north pole to each vertex) of pi/2*R where R is the Earth's radius

First off, the location of the "centroid" is ambiguous; is it the north pole or the south pole?

Secondly, if you "double" this polygon you either end up with 4 overlapping points at the south pole (if you assumed the original polygon was centered at the north pole) or at the north pole (if you assumed a center at the south pole)

Either way, the side lengths have shrunk under a "doubling".

For less extreme cases, it is ALWAYS true that "doubling" the polygon using the method you claim here will scale the side lengths by a factor LESS than 2. This is because a sphere is a surface of constant positive curvature. More worryingly, the side lengths will all change by different factors (sides closer to the centroid will increase in length more than sides far from the centroid)

In other words, if I "double" an equilateral triangle of sides 500 km I will end up, using your method, with a triangle of sides less than 1000 km, and for more complex figures not even the ratio of side lengths is constant.

What is the purpose of this function? We need to know what the appropriate sense of "scale" is; is it side lengths? Is it "radius"? Is it something else?