Request plugin :Trilateration ruby

Bep

Dear Bep,

You can make an equilateral triangle of known side length using the following method:

Use the polygon tool with the number of sides set to 3
Draw a triangle
Double click to select all the entities and make a group
Double click to edit the group, and when in edit mode use the tape measure tool to measure
one side
Type in the desired length
Type yes when asked about resizing group
Explode group

Regards,
Bob

Thank you Bob for responding,

But the ruby i request could help to contruct all types of triangles ,scalene if no sides are equal;isosceles if two sides are equal;equilateral if three sides are equal.

greetings ,

Bep van Malde

Didier Bur

Hi,

@unknownuser said:

Could some of you please make this ruby?

On my to-do-list

Bep

On my to-do-list[/quote]

Thank you Didier, that you are taking up an interest in developing this Ruby.
I am very happy
I can now rest asure knowing that it will be made.

Greetings,

Bep van Malde

Didier Bur

@Bep,
Just a question: Given Side1 from Point A to Point B, and 2 lengthes a and b, the general trilateration case (from Al-Kashi theorem) gives 4 solutions.

How do you would see the script determine which triangle is the "good" one (from the user point of view I mean) ?

Gaieus

What if you allow all four - separately grouped - and let the user delete the unwanted ones.

But also, what would decide which plane they are drawn? In 3D space there are endless possibilities. Or would you then restrict it to the red/green axis? Or maybe not if the initial line is drawn on any face?

Gaieus

Ah, I see, the plane has been solved. Yes, logical. Also the solution for the choice of the triangles sounds quite reasonable!

Didier Bur

What is currently on the drawing board is:

User clicks 2 points to define the first side of the triangle (the red line between the 2 "X"), and click 1 point to define the plane (the third "X"). The plane is computed with a crossproduct between vectors, this is not a problem.
Between the 4 possibilities still remaining, I would suggest that the first length entered would start from the start point of side 1, the second would start from end point of side 1. This leads to 2 solutions. So I also suggest that the point defining the plane will also define the pointing direction of the second side (given by the first length) to get only one solution.
What do you think ?

Bep

Didier,

You are much faster than I am.
I think that we come up with the same sort of solution.

Bep

possibility 1.skp

Didier Bur

Bep,
I'll PM you soon a beta for test. It's almost ready, just several weird angle problems to solve
Regards,

Gaieus

I can hardly wait!