ements between mid axes of elements ( being perpendicular to both). On the attached image, the links are the small elements connectiong nodes 4 to 5, 6 to 7 and 8 to 9. All nodes (1 to 9 including 1', 2', 3') are defined in Karamba as fixed supports but nodes 1,1', 2, 2', 3, and 3' have hinges added with the beam joint component. The freed rotations are shown on the figure.
I wondered if that was the correct way of defining such a structure in Karamba bearing in mind that nodes 1', 2' and3' are free nodes in the reality.
Thanks again for your help !
Yousef…
2 & 3 are showing the base curves for the inside and outside starting point if the form. Both the inside and outside outline starting curves are made up of a set of 8 tangential arcs.
Added by David Hines at 1:20am on October 23, 2015
ems in the same way. Lofting was particularly difficult, you had to have a separate loft component for every lofted surface that you wanted to generate because the component would/could only see one large list of inputs. Then came along the data structures in GH v0.6 which allowed for the segregation of multiple input sets.
If you go to Section 8: The Garden of Forking Paths of the Grasshopper Primer 2nd Edition you will find the image above describing the storing of data.
Here you will notice a similarity between the path {0;0;0;0}(N=6) and the pathmapper Mask {A;B;C;D}(i). A is a placeholder for all of the first Branch structures (in this case just 0). B is a place holder for all the second branch structures possibly either 0, 1 or 2 in this case. And so forth.
(i) is a place holder for the index of N. If you think of it like a for loop the i plays the same role. For the example {A;B;C;D}(i) --> {i\3}
{0;0;0;0}(0) --> {0\3} = {0}
{0;0;0;0}(1) --> {1\3} = {0}
{0;0;0;0}(2) --> {2\3} = {0}
{0;0;0;0}(3) --> {3\3} = {1}
{0;0;0;0}(4) --> {4\3} = {1}
{0;0;0;0}(5) --> {5\3} = {1}
{0;0;0;1}(0) --> {0\3} = {0}
{0;0;0;1}(1) --> {1\3} = {0}
{0;0;0;1}(2) --> {2\3} = {0}
{0;0;0;1}(3) --> {3\3} = {1}
{0;0;0;1}(4) --> {4\3} = {1}
{0;0;0;1}(5) --> {5\3} = {1}
{0;0;0;1}(6) --> {6\3} = {2}
{0;0;0;1}(7) --> {7\3} = {2}
{0;0;0;1}(8) --> {8\3} = {2}
...
{0;2;1;1}(8) --> {8\3} = {2}
I'm not entirely sure why you want to do this particular exercise but it goes some way towards describing the process.
The reason for the tidy up: every time the data stream passes through a component that influences the path structure it adds a branch. This can get very unwieldy if you let it go to far. some times I've ended up with structures like {0;0;1;0;0;0;3;0;0;0;14}(N=1) and by remapping the structure to {A;B;C} you get {0;0;1}(N=15) and is much neater to deal with.
If you ever need to see what the structure is there is a component called Param Viewer on the first Tab Param>Special Icon is a tree. It has two modes text and visual double click to switch between the two.
Have a look at this example of three scenarios in three situations to see how the data structure changes depending on what components are doing.
…
{8} become {1) etc
(see image attached)
I've played around with series components and feeding the data into a tree branch in stages (0-4 5-9 etc), however the amount of branches i need to merge (5 in this case) is a variable that plan on changing in the future so I want a more sophisticated solution than manually copying groups of series components and feeding them into separate tree branches.Thanks in advance for any help…