How to use with FAI scoring rules

[hyperknot]

I'd like to use this with FAI scoring rules, to evaluate record flights. With FAI rules triangle closing needs to be within 400 meter radius fixed. How can I modify it?

So far I find no solution other than copying cli.js and hard coding a different value in my version of cli.js

[mmomtchev]

You need to edit scoring-rules.config.js and add a scoring rule with a closingDistanceFixed: 0.4

[hyperknot]

I added the change to the scoring rules here:
https://github.com/hyperknot/igc-xc-score/blob/6bca0fe83ed608a09261b2eb8eab982f4ce5e167/scoring-rules.config.js#L124

My problem is that it's not doing the optimal calculation. The best point would not be on the track, but between the "head" and the "tail" of the tracklog, as illustrated here. Do you think you can modify the app to make such calculations?

Are FFVL and XContest calculating the closing differently? Also, what is the difference between code: fai and tri?

[mmomtchev]

An fai triangle must be almost equilateral with none of its sides being smaller than 28% of the total perimeter.
Do you have a link to a document explaining those rules? Normally all points must lie on the track.

[hyperknot]

Sure, it's here: https://www.fai.org/sites/default/files/civl/documents/sporting_code_s7_d_-_records_and_badges_2022.pdf

It's not an easy read, but I believe it's like on my drawing:

[mmomtchev]

Ok, so in fact all turnpoints and the start/finish points are 400m cylinders?

[hyperknot]

Yes. But for the turnpoints it makes no sense to make them cylinders, it's easier to calculate them as points. The trick is the start/end cylinder.

[mmomtchev]

In fact this rule can be interpreted as a 800m closing rule? It is a bijection:

• All points that a separated by 800m or less can be inside a 400m radius cylinder
• No points separated by more than 800m can be inside a 400m radius cylinder

[hyperknot]

I think a straight 800 meter line is not optimal. The optimal solution would be a "buffer" operation around both polylines and then finding the furthest point on this polygon.

[mmomtchev]

I think that it is. You think that a 800m rule will leave out points that will be included in the buffers? I don't see any case where this is possible?

[hyperknot]

It's very close to optimal but not optimal. I don't know how much we want to optimise those few meters though. I made a Geogebra sketch. The left point is the 800 meter midpoint, the right point is the optimal one.

A very simple algorithm I could imagine is:

1. Calculate the start/endpoint as it does currently.
2. In a loop: Move the start/end 1 meter in a random direction, check if still FAI and still within 400 meter of both. If yes and score is bigger than the previous one, move it.

[mmomtchev]

2r is a better closing then the angled r+r - it keeps more of the track

[hyperknot]

It's true but FAI calculates by the triangle sides, which in our case is determined by the center of the circle.

[mmomtchev]

Do you mean the turnpoints or the closing? The placement of the closing center does not influence the score, only the part of the track that is included or not influences the score. But this is not true for the turnpoints, this was my first question?

[hyperknot]

The FAI document is really quite low in details, but here is how I understand it:

• You put 3 circles, such that they are 400 meter radius and they include 1-1-2 points of the track.
• You select the closest point of each circle to make up the triangle.
• For the 2 mid-points, this basically makes no sense, you select a point, put a circle and then select the original point again. Instead it makes sense to select the track point itself.
• For the start/end point, now I believe it'd be point K on this updated sketch:

[mmomtchev]

All FAI rules use 400m cylinders. This means that if you are following a predefined task, you must cross into the 400m cylinder for the turnpoint to be considered attained.
When scoring a free triangle - which means that you define the best possible task given your existing flight track - it is in your best interest to define the turnpoint 400m away from the actual triangle vertex. This rule will require some adapting of igc-xc-score.
The closing is defined as a single point that must be not further than 400m from each side of the track. This rule should be perfectly satisfied by using 800m as closing distance.

I wonder if there is some clever way to calculate the size of the biggest triangle that has vertices lying on equal circles around the given triangle - there probably is - as this will be much easier to implement.