Signpost

Revision as of 13:58, 21 May 2008

Huh?

We have all seen those signposts - the ones stuck in the middle of nowhere with a multitude of signs pointing to famous places (London, New York, Yoko, Rowville, etc).

• http://images.google.com/images?q=signpost

They all point across the surface of the earth - presumably to follow a great circle to the desired locale.

I want one which points in a true straight line. As the laser shoots. THROUGH THE EARTH!

Especially from Australia, this would be pretty neat I think, with North America and Europe being the most 'down' you can get from an australian perspective. More interesting stuff would be in the directions of africa, Asia, south america, antarctica, NZ...

Tell me more, tell me more

For bonus points, instead of a signpost, make it out of actual lasers! (and sealed in a cage with appropriate gas so the line of the laser is visible. Say, ALL lasers always visible in red, then you select a city and it goes GREEN!).

Another method for a permenant exhibit might be to start with a point central to the room, (normal signpost), but then ALSO place markers on the walls and floor of the room to indicate the marker - so as to make clearer the path and descent taken.

Some math

Bearing and Arc Distance

Using this site: http://gc.kls2.com/, I calculate the "raw data". That is, the distance on the great circle, and the degrees - ie, the direction I travel in. The required final data required a conversion: the great circle distance into both a straight line (distance, and angle down (0 being tangent to here, 90degrees being straight down). The degrees should remain the same.

Declination and Direct distance

So knowing the arc distance, we can calculate the declination (angle down from tangent to source location), and direct distance via the following:

• We assume the earth is a sphere, with radius 6371km (the avg radius of the earth, whcih varies in fact between 6356.8 and 6378.1km)
• The declination (angle down) is calculatable from knowing the center angle and distances of the isosceles trangle with points of the two locations, and the center of the earth. In this instance, we know the arc length between two locations, so the center angle can be found with: θ = s/r, where θ is the subtended angle in radians, s is arc length, and r is radius
• From the internal angle, the outer angles are discoverable (π radians in a triangle, thus (π-θ)/2 is the internal angle "up" from drop-down angle. We take that from the tangent angle and we get:
  • π/2-((π-θ)/2)
• In degrees: One radian is 180/π, so the inner angle θ=s/r becomes a=180s/πr
  • Because we know r=6371km, we can solve to: a=s*0.00899321606...
• Our final formula becomes: 90-((180-a)/2)
  • With everything plugged in: 90-((180-(s*0.00899321606))/2)

Results

Data table

The raw data from Canberra (CBR) is: