Draw an arrow on a sphere.

If you are not to close to a pole, and your arrow is not too long, you might try scaling wx by 1/cos(latitude)

Thanks ozo,

That is an improvement in the right direction but not quite enough I think. Can anything more be done?

Paul

You may also need to scale x by cos(lat), so

w = 0.1 (Width of the shaft of the arrow (for example))
n = 2 (Number of w's the point of the arrow should extend (for example))
x =  cos(lat)*(x2 - x1) (Extent of line in x direction)
y = y2 - y1 (Extent of line in y direction)
d = sqrt((x*x)+(y*y)) (Length of line)
wx = w * y / d / cos(lat) (Extent of shoulder in x direction)
wy = w * x / d (Extent of shoulder in y direction)
nwx = n * wx (Extent of point in x direction)
nwy = n * wy (Extent of point in y direction)

it may be better to have a different cos(lat) for each of
(x1-wx,y1-wy)
(x1+wx,y1+wy)
(x2-nwx+wx,y2+nwy+wy)
(x2-nwx+wx+wx,y2+nwy+wy+wy)
(x2,y2)
(x2-nwx-wx-wx,y2+nwy+wy+wy)
(x2-nwx-wx,y2+nwy+wy)
(x1-wx,y1-wy)
but if the arrow is long enough or close enough to a pole for that difference to be significant, then it may be better to use a different approach than tweaking a planar solution

Hi ozo,

Strangely, that made it worse, but just to 'see what happens' I tried 'x = (x2 - x1)/cos(lat)' and it looks significantly better. The shoulders of the arrow at least now look square to the shaft. The head looks a bit strange though.

What would you suggest as an alternative to 'tweaking a planar solution'? My schoolboy maths reminds me that Pythagoras and some trig doesn't strictly apply on a sphere. I guess a simple mapping of the original endpoints to cylindrical coordinates may be a good start but can we work purely in spherical? How does one calculate a perpendicular vector in spherical coordinates?

Paul

I would probably do it in 3-D coordinates then project it on to a sphere

Which direction are you calling x and which direction are you calling y?

Hi ozo,

If the choice is arbitrary I would choose the same as Google Earth which is

x (Longitude) e 180 and <= 180
y (Latitude) e 90 and d 90

x=0 at Greenwich meridian y=0 at equator.

Given a solution I should be able to work on the coordinate system if it is not what I want. I have done some maths since school but not a huge amount. :-)

So you would translate the endpoints into true 3d (x,y,z) cartesians, draw the arrow between them and map the results back onto the sphere? How would you project it back to the sphere?

I was considering rotating the sphere so the arrow is north-south, draw the arrow, rotate back to original position. The 1/cos(lat) adjustment can still be applied if required but at least the true shape should be preserved which is my primary aim. This 'should' only require conversion from r/theta/phi to lat/long as the first rotation is purely notional.

Paul

To project a point P = (x, y, z) onto a sphere, radius r, centred at the origin O:

1. Find the length of OP = R = sqrt(x^2 + y^2 + z^2).

2. Scale the co-ordinates of P to get a point, P', such that P' lies on OP and |OP'| = r - i.e. P' = P * r / R.

3. Translate the co-ordinates of P' = (x', y', z') into a latitude and longitude on the sphere.

Obviously you could combine all three steps into one calculation, and even apply it to the equation of the line joining your two endpoints, but that might start to look ugly.