From the constraints I think I can only deduce that we know the length of the sides of the triangles and the fact that three rays come from the origin. The fact that |Va| etc are one don't seem to supply any information since you multiply by OA (so in the end we have the unknown lengths OA OC and OB
If I move point C to lie on line O + Vb*T2 (as an extreme example). I still have the same constraints but point A would be different then I would for instance have point C on line O + Va * T1 (while keeping the shape of the triangle the same).
So there seems to be more than one solution possible (but I might be wrong :-) ).
Locating points on rays given distance bw them
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Quote:Original post by Ron AF Greve
The fact that |Va| etc are one don't seem to supply any information since you multiply by OA (so in the end we have the unknown lengths OA OC and OB
Normalization of ray direction vectors doesnt supply additional information, but it does simplify some equations later on if you're trying this problem as a system of equations.
But I don't think it is solvable to one solution (see my previous post) Besides that if I move the origin along the axe OC I still seem to meet all constraints so as far as I can see there are two reasons why there are multiple solutions (unless you express the solutions in line segments OA, OC and OB (still the fact that I can move C around means there is not one solution).
Quote:Original post by gondarQuote:Original post by ryt
If I have understood correctly its not too hard. Lets take triangle OAB. We know all inner angles and |AB|. We know the angles becase we know vectors, so we just need to take dot product to obtain angles. Than just take sine law and calculate remaining distances |OA| and |OB|. Than make your vectors unit vectors and multiply them with distances |OA| and |OB| to obtain points A and B.
For point C calculate the point between A and B, lets call it D. Since you also know all the angles in triangle ABC and all the distances its easy to find point C. Now just add point C to D and u have real C.
Triangle OAB is not necessarily a right-angled one.
While we clearly can find angle <AOB, since we know normalized OA and OB vectors (the ray direction vectors Va, Vb), it doesnt seem apparent that we know all the inner angles of OAB and thus we can't use the law of sines here.
Please elaborate.
Yea sorry for that.
I think it's solvable:
Consider the 3 rays. If you put the A-B section of line on it, you are able to move it, but only one way, because the position of A determines the position of B.
So you have only one degree of freedom.
If you put C on the middle ray, you lock this freedom, so you have completely locked the triangle (it will be locked, because the movement of C point -before the locking- isn't collinear with the middle ray). Since you can place the C point in two ways (C is inside O-A-B, or outside), you have 2 solutions. This may only be truth in the situation, that the Original Poster has drawn (O-A, O-B, and O-C are positive numbers.)
Why is the iteration a bad idea BTW?
EDIT: maybe there are four solutions, because if we move A, then B can be 2 places (just consider a circle, it intersects a ray two times). But I'm not sure about this one. Maybe it depends on which side C is, so there will be 2 solutions after all.
Consider the 3 rays. If you put the A-B section of line on it, you are able to move it, but only one way, because the position of A determines the position of B.
So you have only one degree of freedom.
If you put C on the middle ray, you lock this freedom, so you have completely locked the triangle (it will be locked, because the movement of C point -before the locking- isn't collinear with the middle ray). Since you can place the C point in two ways (C is inside O-A-B, or outside), you have 2 solutions. This may only be truth in the situation, that the Original Poster has drawn (O-A, O-B, and O-C are positive numbers.)
Why is the iteration a bad idea BTW?
EDIT: maybe there are four solutions, because if we move A, then B can be 2 places (just consider a circle, it intersects a ray two times). But I'm not sure about this one. Maybe it depends on which side C is, so there will be 2 solutions after all.
Why does the A position determine the B position. The only thing I know is that it would move around in a circle around A same for C (since we only know the length A-B and A-C).
Lets assume we know the solution; You got three points forming a triangle you know the three sides (put three things on the floor) now where would the origin be? Seems to me you can't answer that question. The fact that lines intersect at one point on the triangle and the other in the origin doesn't help that is true whereever the origin is.
Lets assume we know the solution; You got three points forming a triangle you know the three sides (put three things on the floor) now where would the origin be? Seems to me you can't answer that question. The fact that lines intersect at one point on the triangle and the other in the origin doesn't help that is true whereever the origin is.
Maybe this is the way to solve it!
You can generate the equation of the curve, that C sweeps, if it's variable parameter is the O-A distance. So the curve is a function of O-A distance. The other parameters (the given things) will be it's constant parameters.
If you have the curve's and the middle ray's equation, intersect the 2 equations, and you will have the O-A distance.
To generate the equation, well that's the hard part. (I won't do it, I have spent way too much time with this problem already)
I hope that helps.
You can generate the equation of the curve, that C sweeps, if it's variable parameter is the O-A distance. So the curve is a function of O-A distance. The other parameters (the given things) will be it's constant parameters.
If you have the curve's and the middle ray's equation, intersect the 2 equations, and you will have the O-A distance.
To generate the equation, well that's the hard part. (I won't do it, I have spent way too much time with this problem already)
I hope that helps.
Quote:Original post by Ron AF GreveBecause the circle intersects the Vb ray (in two points, that's why there is 2 solutions).
Why does the A position determine the B position.
Because the rays are given (the angle between them), the A-B distance is given and if we fix the A point, (So O-A is given too), that's 3 parameters, which define the triangle completely, so B will be determined. If we move A, B will move to, But it can be only 2 places again(on the Vb ray). So the A-B section can only move in one mode/way.
Ok, now I see that in its second post the direction vectors are known (instead of only the absolute values of the direction vectors as in the first post (I missed that), sorry.
Gondar!
If you managed to solve the problem, please, let us know!
I have spent too much time with it, to simply let it go!
If you managed to solve the problem, please, let us know!
I have spent too much time with it, to simply let it go!
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