Fasskreis - Ortskurve gleicher Winkel über Strecke
- Author:
- hawe
φAB+φCD = 180° auf entsprechenden Fasskreisen über AB, bzw CD
construction an intersection of circle arcs of equal angles
(Fasskreis in german does anybody know the translation?)
Rotate AB, (90°- α) x Rotate BA, (-90° + α)
===> MAB=(a1, a2) + (tan²(α / 2) + 1) / (4tan(α / 2)) ((b1 - a1) sa - (b2 - a2) ca, (b1 - a1) ca + (b2 - a2) sa)
FKAB=Circle MAB, r= |A-MAB| (((b1-a1,b2-a2)^2)/4/sin(α)^2)
Rotate CD
===> MCD=(d1, d2) + (-tan²(α / 2) - 1) / (4tan(α / 2)) (-(c1 - d1) sa - (c2 - d2) ca, (c1 - d1) ca - (c2 - d2) sa)
FKCD=Circle MCD, r= |C-MCD| (((c1-d1,c2-d2)^2)/4/sin(pi-α)^2)
If form of trapezoid the angle locus or locus arc of angle (I dont know the correct translation) runs the green bow (horizontal) or the form is parallelogram like the locus curve is shown (vertical curve)
extremely slow up to crash - make a bug report .
Locus of sum of angles over AB + BC
