a furniture with diagonal joints.
Let me express myself graphically.
http://www.plataformaarquitectura.cl/2011/01/28/munich-l%e2%80%99illa-diagonal-dear-design/05-78/
take a look at the picture. how could I make panel wood with a diagonal cut in each end, so when I place them together they fit perfectly?
a diagonal cut with a CNC machine?!
…
سلام آقای جان بیلی کاوه اشکوه هستم اپراتور سی ان سی از ایران وطراح دکوراسیون داخلی اگر ممکن هستش فایل اصلی میزcosinosa baram mail konid adressmail man hast
(kavehoshkooh2290@gmail.com)بسیار متشکرم
Added by kavehoshkooh at 3:25pm on September 11, 2015
m b1 As Brep = Nothing
If (heights(b) > heights(b+1)) Then
b0 = breps(b+1)
b1 = breps(b)
Else
b0 = breps(b)
b1 = breps(b+1)
End If
Dim bDiff As Brep() = Brep.CreateBooleanDifference(b0, b1, 0.1)
If (bDiff IsNot Nothing) AndAlso (bDiff.Length > 0) Then
breps(b) = bDiff(0)
End If
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 7:32am on October 16, 2012
pe and its surface.
However, I don't have that much knowledge about both grasshopper and Mathematica.. I mean I can only make assumptions and think about relations of certain functions but that's all.
If you can help me on this, I would appreciate it so much.
You can see a screenshot of the code and model of the demonstration from mathematica in attachment.
And here is the mathematica code;
Manipulate[ Module[{\[CurlyEpsilon] = 10^-6, c1 = Tan[a1], c2 = Tan[a2], c3 = Tan[a3], c4 = Tan[a4], c5 = Tan[a5], c6 = Tan[a6]}, ContourPlot3D[ Evaluate[ c6 Sin[3 x] Sin[2 y] Sin[z] + c4 Sin[2 x] Sin[3 y] Sin[z] + c5 Sin[3 x] Sin[y] Sin[2 z] + c2 Sin[x] Sin[3 y] Sin[2 z] + c3 Sin[2 x] Sin[y] Sin[3 z] + c1 Sin[x] Sin[2 y] Sin[3 z] == 0], {x, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {y, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {z, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, Mesh -> False, ImageSize -> {400, 400}, Boxed -> False, Axes -> False, NormalsFunction -> "Average", PlotPoints -> ControlActive[10, 30], PerformanceGoal -> "Speed"]], {{a1, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(1\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a2, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(2\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a3, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(3\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a4, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(4\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a5, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(5\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a6, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(6\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, AutorunSequencing -> {1, 3, 5}, ControlPlacement -> Left]…