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Another math challenge

Rob
Graphisoft
Graphisoft
I just do not know how to tackle this problem and/or if it is solvable with given variables. Basically I would like to calculate the coordinates of the tangential fillet center of two segments.
if someone could help I would appreciate
ta

X1,Y1,X2,Y2,X3,Y3 - known
Xa,Ya,Xb,Yb - known
Xc,Yc - ???

mathquestion-fillet.jpg
::rk
8 REPLIES 8
Frank Beister
Moderator
You should be able to calculate it by pythagoras and TAN. I have added some length (L,r), which are known out of your settings
bim author since 1994 | bim manager since 2018 | author of selfGDL.de | openGDL | skewed archicad user hall of fame | author of bim-all-doors.gsm
Rob
Graphisoft
Graphisoft
I appologise, I have forgotten to mention that alpha is not known, it's there just to demonstrate the centerline of the angle.
::rk
Frank Beister
Moderator
Are you shure that the both distances L are identical? Elsewhere the center line and alpha is not the only problem. 😉
bim author since 1994 | bim manager since 2018 | author of selfGDL.de | openGDL | skewed archicad user hall of fame | author of bim-all-doors.gsm
Rob
Graphisoft
Graphisoft
yep, the L distances are identical
::rk
Laszlo Nagy
Community Admin
Community Admin
I think if you know the coordinates of all three corners of the triangle, you should be able to find out any of the angles of the triangle. Alpha will be half of the angle at point 2.
Loving Archicad since 1995 - Find Archicad Tips at x.com/laszlonagy
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Rob
Graphisoft
Graphisoft
I think if you know the coordinates of all three corners of the triangle, you should be able to find out any of the angles of the triangle. Alpha will be half of the angle at point 2.
thanks laszlo for your suggeston, it did not solve the problem, however it gave me the starting idea and finally solution after couple hours of sketching and calculating.
::rk
LiHigh
Newcomer
You can also avoid Alpha:

!-------------------------------------------------------
! (X0, Y0) is the center of the arc
! D = distance betwen center and point (X2,Y2)
!--------------------------------------------------------

R^2 = (Xa-X0)^2 + (Ya-Y0)^2
R^2 = (Xb-X0)^2 + (Yb-Y0)^2

L^2 = (Xa-X2)^2 + (Ya-Y2)^2
L^2 = (Xb-X2)^2 + (Yb-Y2)^2

D^2 = R^2 + L^2


5 unkowns & 5 Equations, solvable.......
Howard Phua

Win 10, Archicad 19 INT
Anonymous
Not applicable
I know the post I'm adding to is old, but it's related.

I was working on a much simplified detail symbol which shows the clip polygon with filleted corners. I managed to figure out how to draw the clip polygon, . Simple enough, you just put the clip coordinates into memory and draw a polygon but I could not figure out how to add the nodes to create filleted coners.


I attached the library part, but following is the"for next" loop that extracts the clip polygon.


add2 AC_RefCoord[1][1], AC_RefCoord[1][2]
for i=1 to AC_ClippNodes
if i<>AC_ClippNodes then
put AC_ClippCoord[1],AC_ClippCoord[2],1

next i

poly2_ AC_ClippNodes, 1,
get(nsp)