Centre of three positions on earth

[compressor stall]

Out with the spheical trig textbooks for this one.....:eek:

How do you calculate the Centroid (the point from which each vertex is equidistant) of a triangle formed by three points on the earth's surface that are connected by great circles?

Euclidian approximations won't cut it at these latitudes with the distances being covered. I want to establish a new central point that is equidistant from three existing locations.

Once that is done, what about the centroid of a polygon and finally a pentagon?:ooh:

Thanks in advance, and apologies in advance if it keeps you up half the night (like is has done me already :{ )

[bookworm]

How do you calculate the Centroid (the point from which each vertex is equidistant) of a triangle

The centroid (http://mathworld.wolfram.com/TriangleCentroid.html) of a triangle is not "the point from which each vertex is equidistant". That would be the circumcentre (http://mathworld.wolfram.com/Circumcenter.html). Which do you want the spherical equivalent of?

[enicalyth]

G'Day Stallie!

I don't have my books to hand and would rather not derive equations from first principles but...

I can recommend this applet if used with a bit of gumption. The bit you want is called "dads" if I recall correctly and is to better than metre accuracy.

http://www.naco.faa.gov/index.asp?xml=naco/online/compsys

First derive a convenient point on each spherical side using the routine <segment-distance> or take the mid-point. Then using these three calculated points and the three apexes (apices??) find where they intersect using routine <segment-segment>.

I did write an Excel spreadsheet using the Andoyer-Lambert method to do exactly what you ask and I stole the formulae from the Admiralty Manual of Navigation but I'm as far from my computer as you are from a reference library.

[compressor stall]

Thanks for the clarification above - my terminology might not be up to scratch.

Of three points (A,B and C) on the earth's surface, forming a triangle (sides 999nm, 349nm, 745nm) I want to find the point that is equidistant from A, B, and C.

This point would naturally need to be on the earth's surface as I want to leave fuel there! Accuracy is only required down to a few nm which is just as well as there are issues regarding the 14000' height difference between B and A, and the assumption that the earth is spherical.

Enicalyth, thanks for the pointers - will check when I get back in tonight. :ok:

[enicalyth]

Stallie!

Here is a worked example to find the circumcentre Alice Springs, Adelaide, Brisbane. I have left a case of Bundaberg there and a Toohey's or two. I used compsys21 dads and routines <inverse> and <bearing-bearing>. the former does distances and azimuths; the latter does intersections; check the answer with <inverse>.

To find the circumcentre of the spherical triangle YBAS-YPAD-YBBN

B = YBAS = S23* 48’ 24” E133* 54’ 08”
A = YPAD = S34* 56’ 42” E138* 31’ 50”
C = YBBN = S27* 23’ 03” E153* 07’ 03”

Mc = S29* 23’ 53.457” E136* 05’ 23.767” Az McA = 160.123*
Az McO = 070.123*

Mb = S31* 22’ 22.400” E146* 06’ 56.332” Az MbC= 058.718*
Az MbO = 328.718*

Ma = S25* 54’ 50.523” E143* 21’ 55.978” Az MaB = 281.692*
Az MaO = 191.692*

Intersection McO-MbO = S26* 55’ 55.076” E143* 07’ 50.089”
Intersection MbO-MaO = S26*55’ 54.796” E143* 07’ 49.909”
Intersection MaO-McO = S26* 55’ 55.184” E143* 07’ 49.189”

Assumed circumcentre is S26* 55’ 55.019” E143* 07’ 49.729”

Distance BO = 535.024nm
Distance AO = 535.025nm
Distance CO = 535.028nm

Therefore the circumcentre has been found.

http://i28.photobucket.com/albums/c220/enicalyth/circumcentre.jpg

[bookworm]

Nice work enicalyth!

Do you know if the compsys 21 source (or at least its algorithms) are available for those of us that need it for other than Windows?

[OzExpat]

Well done enicalyth! I can now see how you filled in all those boring long-haul hours! ;)

[enicalyth]

aw shucks ozzy

I told everybody I slept.

[Brian Abraham]

I've used the program for years and love it. Can not see any info for non microsoft users but help desk should answer your questions. Always found them most helpful.
email to : [email protected]

[Brian Abraham]

bookworm,
Just re read the notes and its windows based I'm afraid. Although no harm in sending an email. Don't ask, don't get.

[compressor stall]

Great work enicalyth- I am busy in the field for the next day or two - will have a play with the numbers when I return and let you know how I got on.

Thanks again!

[enicalyth]

Re formulae

Any decent surveying book will have the formulae

eg gps satellite surveying alfred leick isbn 0471059307 and admiralty manual of navigation by hmso

in my own spreadsheets I make use of the gauss method in the book by leick which is a reworked version of the andoyer-lambert technique.

compsys21 is a bit clumsy and modestly accurate i.e. ten metres or thereabouts. [when I say modest it is in comparison with road surveyors who set out curved lines on motorways to millimetric accuracy].

however when it comes to equal time points and circumcentres for aviators it is spot on.

[compressor stall]

Back after an interesting few days away... :oh:

Got the answer I needed. Quickly realised the need for accuracy in decimal angles at these latitudes. It gave me an answer to within 3 miles using only one decimal point, which for my purposes is close enough.

Used the program you gave - a good little unit, but kept getting wildly different co-ords. Finger trouble I think! Switched to Ed Williams avformulary excel spreadsheet and worked a treat.

Unfortunately, given the shape of my long skinny triangle at this lat (between 66* and 81*), the point seems to be outside the triangle :ugh: Rechecking and redoing sums now.

Cheers

CS

[QF5]

I'm sorry if I seem ignorant, but what subject is this calculation for? is it something that you use in the more advanced levels of flying ground school or another course altogether? (I only ask because I've not yet learned anything like this, and with a year left at school before starting training this explanation may be worth bookmarking if needed later on)

Thanks,

[compressor stall]

Hi QF5,

It's not a subject in particular (indeed my recollection of the Australian ATPL syllabus does not even look at spherical trig at a mathematical level).

I happen to have an interest in the concept, and at the latitudes I fly it is extremely relevant. The logistics of flying in my part of the world are huge - and quite often we fly in drums of fuel to midway points to be able to get to our desitination and return. Luckily we can land almost anywhere, hence my thought of trying to find a point central to two destinations.

Good luck with your studies.

CS

[enicalyth]

Stallie's fuel cache being near the pole is where the meridians bunch up close. He needs nav formulae that are accurate over large distances and don't depend too much on tangent of an angle. That gets too large at latitudes near ninety!
In general these days as far as nav is concerned maths is weakly remembered because old-time nav has been supplanted. Given the average pilot's ability to keep track of even his car keys thank the Lord for INS and GPS.
Sometimes it is handy to know a bit about equal time points or where we will cross say a line of latitude or longitude. If we did plane trigonometry at school then fine, you only have a little more to learn. If you didn't then I told you education is going to the dogs, Mother!
Spherical trigonometry... ellipses. hyperbolae, circles and triangles as laid out on the spherical earth is the stuff of the chartmaker. And us if we don't want to faff about with large swatches of paper yet need to know something at the planning stage. Like stallie and his cache.
On a sphere the sides of a triangle become curved and the sum of the interior angles is no longer 180 degrees as it was on the plane blackboard of the schoolroom. But the trig formulae are familiar, usually with only one extra term added to account for curvature.
The real life globe is a sphere somewhat flattened at the poles. So really accurate calculations are a little tougher on a flattened spheroid.
Some work of course you can do off a paper chart. In the RAAF/RAF some charts even have a false equator because some math formulae don't work well near the poles (tangent of an angle being one) but if you are happy to accept the world as a sphere you can put the ruddy equator where you like as far as spatial nav is concerned. Just remember where you put it.
Given how easy a spreadsheet is or an applet, compared to slogging away with paper and pencil many things are solvable in seconds. I'd download compsys21 and keep it handy for future work but don't let stuff get between yourself and the prime objective.
Enjoy! I flew for 16000 hours and never hit anything I wasn't supposed to!!
The "E"