ggle A
7. Toggle A
8. Toggle 2
9. Toggle 3
10. Toggle A
11. Toggle A
12. Toggle 3
I was thinking to use somehow slider and animate option....but without luck
Any idea would be appreciated…
etting when I merge the three trees, but what I would like to get is:
essentially a tree with 27 branches, each with a single list of either 11 or 21 points.
{0} (N=11)
{1} (N=11)
...
{10} (N=21)
{11} (N=21)
...
{17} (N=11)
{18) (N=11)
{27} (N=11)
Any help would be greatly appreciated.
All the best,
Matt
…
Added by Matt Schmid at 3:06pm on December 4, 2010
First use a series component with start=1, step=42, count=3
Use the output to create a new series component with start=existingseries, step=1, count=11
this, you'll have no horizontal force at the roller, but you will have it at the pinned support. If you wouldn't, then the structure will be displaced.
Usually, in 2 dimensional structures, if you want to know if an articulated structure is isostatic (as opposed to hyperstatic, which is what you have right now) is to use the following formula:
b+c-2·n=0;
b being the number of bars, c the number of constraints you have and n the number of nodes. In your case: b=19, c=3 (displacements constrained in X, Z at your pinned support and only constrained in Z at your roller support) and n=11, so: 19+3-2·11=0.
I recommend you to download the app SW Truss, as it's very useful to check your results instantly.…
your projects in class this week. Please use the comments to sign up for a spot number. You will be required to work in class until at least 9 PM and you must meet with me before you go to get credit for the class.…
s
(distance from Spine to Profile)
4) Create Circles at each floor
5) Rotate Each Floor by twisted amount
(according to height of floor)
6) Divide each floor by number of Flutes
7) Flip Matrix of Flute Points (version 0.7)
(if using v0.6 then search flip matrix on site for method)
(Rows to Columns)
8) Interpolate Curve through Flute Points
9) Mirror Flute Curves
10) Create display grid
11) Make a vector 2Pt from the floor centre
to the corresponding display point…
nch, xno items in one list)2 divide the list lenght value by the numer of items per branch needed3A generate a list with the series component: the step equal to the target numer of items per branch; the no of items equals the number of target branches
3B generate a list with the series component: the first number of the series equals to the number of items needed (-1 to account for the 0 index); the step size again equal to the target number of itmes per branch as 3A4 feed 3A & 3B to a domain component thus identifying the start -3A- and end -3B- of the domains by which the list will be subdivided5 use a subset component with the domains above thus creating 19 branches with lists having 5 items eachfor lists which are subdivided into branches when the target number of branches is not a multiple of the number of items contained in the list:6 identify if the target number of branches is a multiple of the list by using the modulus component fed by the list lenght -1- and the target number of branches7 identify last index in the 3B series with the item component (reversed to take the last value fed)8 add 6+7 above which dill define the start of the domain that will pick up the remanent items not accommodated in 59 add (+1) to 7 above to define the end of the domain that will pick up the the remanent items not accommodated in 510 feed 8 & 9 to a domain component11 include 10 as part of the subset in 5I'm now trying to understand the components mentioned by Michael...
sn
…