): 'Rhino.Geometry.TextEntity' does not contain a definition for 'FontIndex' and no extension method 'FontIndex' accepting a first argument of type 'Rhino.Geometry.TextEntity' could be found (are you missing a using directive or an assembly reference?) (line 92) 2. Error (CS1061): 'Rhino.Geometry.TextEntity' does not contain a definition for 'AnnotativeScalingEnabled' and no extension method 'AnnotativeScalingEnabled' accepting a first argument of type 'Rhino.Geometry.TextEntity' could be found (are you missing a using directive or an assembly reference?) (line 94)…
structures. I think you can get นิสัย and learn more about the business. This architectural dance blurs the line between art and engineering. Because complex algorithms form a space that fosters creativity and collaboration. which is an outstanding platform for future business ideas.…
Added by MichaelD0112 at 1:34am on August 14, 2023
ts will basically be a set of different ellipses:{a1, a2, b1, b2} (with different properties) From that i want to create random Lists of let’s say 15 items (ellipses) Something like that {a1, b2, b1, a1, a2, b1, b1, a2, a1, b1, b2, a1, b2, a1, a1}. But I want to be able to create some constrains. So for example if I have a1 I will be able to have next a2 or b1 but not to b2. I am not sure if this is possible in grasshopper and i was messing around with some logic components but without any luck.
Any help will be greatly appreciated.
…
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
nt to multiply the number of items in the list A, so at the end I will end up with the same number of elements in each lists.
e.g. (for branch 1 in list A I want to have two times the same curve, and the same for the branch 2 and so on )
List A (Data with 88 branches)
{0} N=1
{1} N=1
{2} N=1
{3} N=1...
List B (Data with 88 branches)
{0} N=1
{1} N=2
{2} N=2
{3} N=1...
NEW List A (Data with 88 branches)
{0} N=1
{1} N=2
{2} N=2
{3} N=1...
Any suggestions about how to do this?
Thank you,
Martha
…