Inverted Archs
So I'm working on generating positions in this circular fashion and I've the outer circle and linear parts working. I'm not sure how to get the inverted archs to work appropriately. I was hoping to get some help on that. If you look at the picture I got the green and yellow parts working but not sure how to calculate the red? Script is below.
http:\\www.jokermartini.com/myStuff/circle.jpg
Feel free to play around and change up the columns variable to any number higher than 3.
steps = 6 height = 0.0 width = 30.0 len = 30 radiusX = 15 radiusY = 15 fourCorners = #() rounded = .5 delete objects /* Create Circle */ columns = 4 --don't got lower than 3 columnAngle = 360.0 / columns subCounter = 0 clearlistener() /* Create Corner Points */ for c = 1 to columns do ( calcX = radiusX * cos((c-1) * columnAngle) calcY = radiusY * sin((c-1) * columnAngle) cornerPos = [calcX,calcY,height] point pos:cornerPos size:4 wirecolor:blue append fourCorners cornerPos subCounter += 1 /* Create Sub Points */ subColumns = (steps*columns+columns) subAngle = 360.0 / subColumns for s = 1 to steps do ( /* Circular Calculations */ subCounter += 1 calcXRadial = radiusX * cos((subCounter-1) * subAngle) calcYRadial = radiusY * sin((subCounter-1) * subAngle) circularPos = [calcXRadial,calcYRadial,height] point pos:circularPos size:2 wirecolor:green /* Linear Calculations */ calcXLinear = radiusX * cos((c) * columnAngle) calcYLinear = radiusY * sin((c) * columnAngle) nextCorner = [calcXLinear,calcYLinear,height] linearPos = (cornerPos + ((nextCorner-cornerPos)*(((s as float)/(steps+1))))) point pos:linearPos size:2 wirecolor:yellow append fourCorners circularPos /* Inverted Circular Calculations */ ) )
Comments
Thanks bud.
Thanks bud.
John Martini
Digital Artist
http://www.JokerMartini.com (new site)
:)
A simple solution how to get position of "MIRROR-RED" point.
bga