Hi Adam,

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Adam Wulkiewicz wrote On 28-11-2014 3:26:

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Adam Wulkiewicz wrote On 28-11-2014 3:26:

I'm not sure if there is something like "great elliptic arc" or "great ellipse". AFAIU shortest paths on spheroid aren't fragments of spheres or ellipses.

http://en.wikipedia.org/wiki/Great_ellipse

According to wikipedia (http://en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid), geodesics which supposed to be the shortest paths between points, in general aren't closed.

That is right.

(From wikipedia)

If the Earth is treated as a sphere, the geodesics are great circles (all of which are closed) and the problems reduce to ones in spherical trigonometry. However, Newton (1687) showed that the effect of the rotation of the Earth results in its resembling a slightly oblate ellipsoid and, in this case, the equator and the meridians are the only closed geodesics. Furthermore, the shortest path between two points on the equator does not necessarily run along the equator.

Though I'm not really sure if the side of this geodesic can be checked comparing azimuths calculated using Vincenty formula. And how precise it is.

In other words, it would be true if all intermediate points of a segment was going through the same coordinates. Are you sure that this is the case? I'm not. My test seems to prove that this is not true or at least that a method (Vincenty) which is known for giving precise results for an ellipsoid gives different results than SSF. Of course this is more visible for greater flattening.

OK. Vincenty is precise but very slow (maybe you have to compare performance too). At least distance calculations and comparisons I did in the past, comparing with Andoyer, Vincenty was much slower.

If we are going to use side-information on ellipsoids we have to have a reasonably fast algorithm, the side calculations are used very very often. The SSF is already slower of course than the cartesian calculation.

If the difference is very small, and the results of SSF-only are mutually consistent, and we don't use it together with distances (I don't think we do - but we use fractions which have to be comparable) it is not impossible that we can still use it (needs more research). It is just a different method to calculate the side, and if we can use that for turn-calculations and point-in-polygon calculations, it might give the correct results. Other libraries or packages first project to a Cartesian coordinate system, do the operation and convert back - you will also loose some information there too.

I don't state that it is possible, we just can research the possibilities.

Furthermore the formulas given (on that blog) and probably also what is implemented for SSF are about spheres. For the ellipsoid x,y,z have to be calculated differently, taking axis lengths into account. That might (most probably will) result in different results. It has to be implemented. That is the phase where we are.

Hmm, do you mean by this that we could place the plane with more precision on a spheroid, as perpendicular to the surface in 2 endpoints? Which maybe could result in a plane oriented differently? But still use the SSF for calculation, so assuming that the segment is a fragment of a sphere?

That segment is a fragment of an ellipse, the cut of a plane with the spheroid is an ellipse.

But for the rest, sure, I mean that. We should do that.

So the difference would be the way how the orientation of the plane was calculated. It might as well give the same result because he definition of geographical coordinates is the same as for spherical - latitude is an arc between XY plane and a line perpendicular to the surface. So it's possible that this plane would have the same orientation. But it's definietly something worth checking.

But the xyz coordinates are different so the plane will be different.

Do you have an idea how we could verify this?

No... or basically it is probably already clear now. The methods are different.

I came out with somethign like this. Please say if I'm missing something or if it's completly wrong.

A hypothesis:

If we gathered points of geodesic (coordinates where left side changes to right side between the segment's endpoints), form a path, smooth it e.g. by running bg::simplify(), and then calculate the length by suming all distances between the path's points, we should be able to check which curve is the shortest when measured on spheroid and sphere. This way we could check which one better represents the shortest path on both models.

It is known that the path over a great ellipse is not the shortest path. See in "Geometrical Geodesy: Using Information and Computer Technology" (google gives a preview, search for great elliptic arc", quote: "The great Elliptic arc is not the shortest possible connection between two points".

Furthermore if a hypothesis about a different coordinates (shape) of geodesic on spheroid and sphere was true, when calculating the length of geographical geodesic on spheroid we should get smaller result than when calculating the length of spherical geodesic. And the other way around spherical geodesic should be shorter on sphere.

And finally if we also took a naiive cartesian-like segment between endpoints in the coordinates space, e.g. the result of a cartesian side_by_triangle. The length of this segment "projected" on the surface of a sphere or spheroid should be different than the "true" distance between endpoints calculated using haversine or vincenty. What is more, it should be longer than the spherical segment on a sphere and geographical segment on a spheroid. It's because the shortest path e.g. on a sphere is defined by a great circle so any other path should be longer.

I've implemented this in test-geoside and this is what I've got:

All results are for a segment (-31 -31, 31 31).

distance is the result of a bg::distance(s1, s2, strategy)

length is the sum of distances between points of a geodesic/curve of left/right edge points.

length geo is the length of a curve of side change according to azimuths calculated with Vincenty formula

length sph is the length of a curve of side change according to SSF

length car is the length of a curve of side change according to side_by_triangle (also verified algebraically)

A letter in brackets is a strategy used to calculate distances

(V) - vincenty, (A) - andoyer : so assuming that points are on spheroid

(H) - haversine mean radius = (2a+b)/3, (M) - haversine max radius = a : assuming that points are on sphere

1. a=1 b=1 - unit sphere

distance (V) = 1.4910384559056726

distance (A) = 1.4910384559056726

distance (H) = 1.4910384559056729

distance (M) = 1.4910384559056729

length geo (V) = 1.4904379394565843

length sph (V) = 1.4904379394565843

length car (V) = 1.4897397736057592// the shortest?length geo (A) = 1.4904379394565841

length sph (A) = 1.4904379394565841

length car (A) = 1.4897397736057592 // ?

length geo (H) = 1.4904379394565841

length sph (H) = 1.4904379394565841

length car (H) = 1.4897397736057592 // ?

length geo (M) = 1.4904379394565841

length sph (M) = 1.4904379394565841

length car (M) = 1.4897397736057592 // ?

All geo and sph distances are the same, ok.

But the length of cartesian segment is the smallest. It's shorter than the distance calculated using directly. It's quite late so it's possible that I'm missing something. But it seems that either my assumptions are bad or the method is wrong or not precise enough.

But let's go forward...

2. a=1 b=1-1/300 - unit Earth

distance (V) = 1.4867431728537253

distance (A) = 1.4867328035365481

distance (H) = 1.4893817465102219

distance (M) = 1.4910384559056729

length geo (V) = 1.4856765746677782// smaller than sph, oklength sph (V) = 1.4861482692673844

length car (V) = 1.4854692836326653 // the shortest again?

length geo (A) = 1.485666275142842 // also smaller than sph

length sph (A) = 1.4861379754374746

length car (A) = 1.485459072084804 // ?

length geo (H) = 1.4883117296448467// but this shouldn't be smaller than sphlength sph (H) = 1.4887818973016325

length car (H) = 1.4880813951276908 // ?

length geo (M) = 1.4899672488101912// also this shouldn't be smallerlength sph (M) = 1.4904379394565841

length car (M) = 1.4897366580811147 // ?

So the length of geographical segment seems to be smaller no matter how distances are calculated. This is rather impossible since the shortest path on a sphere is a great arc fragment. Ok, maybe the method's error is too big in this case. Maybe the flattening is too small to get meaningful results.

Cartesian the shortest again.

3. a=1 b=0.9 - unit Saturn

distance (V) = 1.3716008213860109

distance (A) = 1.3618688848319354// now the difference between A and V is greaterdistance (H) = 1.4413371740421503

distance (M) = 1.4910384559056729

length geo (V) = 1.3705358784007811 // smaller than sph, ok

length sph (V) = 1.3711112408432529

length car (V) = 1.3709916809430027 // also this is greater than geo, but the closest to distance

length geo (A) = 1.3608547786450773 // the smallest!

length sph (A) = 1.3614390188832899

length car (A) = 1.3614090781917894

length geo (H) = 1.4403203633519746 // damn, this shouldn't be smaller than sph

length sph (H) = 1.4407566748080314

length car (H) = 1.4400787694784105 // and this is the smallest now

length geo (M) = 1.4899865827779046

length sph (M) = 1.4904379394565841

length car (M) = 1.4897366580811147 // the smallest

So again the geo segment seems to be smaller than sph in all cases, which is AFAIU wrong.

Cartesian still confusing. E.g. the smallest for Vincenty but the length is the closest to directly calculated distance.

3. a=1 b=0.75

distance (V) = 1.2321397094677933

distance (A) = 1.1681145282213292// note the difference between V and Adistance (H) = 1.3667852512468668

distance (M) = 1.4910384559056729

length geo (V) = 1.2294812729423881 // the smallest

length sph (V) = 1.2316859225469647

length car (V) = 1.2319096796713431 // finally, the greatest but the closest to the distance

length geo (A) = 1.165467037015419 // here ok too

length sph (A) = 1.1679406380233488

length car (A) = 1.168887231887985

length geo (H) = 1.3649878832500875 // but this should be greater than sph

length sph (H) = 1.3662347778352022

length car (H) = 1.3655723783015628 // and this too should be greater than sph

length geo (M) = 1.4890776908182768 // same as above

length sph (M) = 1.4904379394565841

length car (M) = 1.4897153217835226

Cartesian confusing again.

4. a=1 b=0.5

distance (V) = 1.1102958385138535

distance (A) = 0.8451906005369858distance (H) = 1.2425320465880607// andoyer can be ignored I guess

distance (M) = 1.4910384559056729

length geo (V) = 1.1064353530360909 // smaller than sph, but the furthest from distance

length sph (V) = 1.1146694840023121

length car (V) = 1.1087899062397581 // this is the closest

length geo (A) = 0.83893092464406038// smaller than sphlength sph (A) = 0.84544333659011361

length car (A) = 0.84807843141588624

length geo (H) = 1.2432889622300805 // greater than sph, still may be a coincidence

length sph (H) = 1.2420316162138203

length car (H) = 1.241438095270917 // the smallest

length geo (M) = 1.4919467546760965 // greater than sph

length sph (M) = 1.4904379394565841

length car (M) = 1.4897257143251006

Similar story, if the length is the smallest it's the furthest from the distance calculated directly.

So I'm guessing that the length and the distance depends mainly on the algorithm used for distance calculation. The variance in the result of calculation using various distance strategies seems to be so big that the possible error based on the method (gathering pixels + simplification) influences the result a lot more than the shape of a path. Hence, we wouldn't check anything like this.

Or do you have some ideas what may be happening?

Not yet.

The assumption about the length of geodesics should at least be good for sphere.

Any help would be welcome.

Vincenty distance is always shorter than haversine which I guess is good.

Should Vincenty give good results for different flattening than Earth's?

Should Andoyer give good results only for Earth?

Might be, it is AFAIK designed for Earth...

It's less precise approximation but it uses flattening and radius but still gives worse results than haversine. Is this a bug or a limitation to flattenings near Earth's?

I also noticed that the greater the flattening, the longer it takes to use Vincenty formula. Maybe we should calculate the threshold based on the flattening?

Furthermore at flattening 0.8 the formula falls into an infinite loop in which lambda and previous_lambda (also sigma) swapps between 2 values all of the time but the difference is greater than the threshold. So we could add a test for this effect. So add some test for local minimum. And maybe some iterations counter just in case. If you're curious, it's for (in radians):

lon1 = -0.54105206811824214

lat1 = -0.54105206811824214

lon2 = 0.89709923552509452

lat2 = -1.0646508437165410

a = 1.0000000000000000

b = 0.20000000000000001

Thanks a lot. This requires more study, will come back later.

Regards, Barend