}
0,3
{7}
1,2
{8}
1,3
{9}
2,3
{10}
0,1,2
{11}
0,1,3
{12}
0,2,3
{13}
1,2,3
{14}
0,1,2,3
I want it to be parametric so I can have it for any set of indexes
I have tried to solve it with points and then removing duplicate points but it's not helping much
any help is truly appreciated
…
ed according to list C.
I tried replace members, but it consists of geometry, so it doesn't worked.
In list C I filtered all the indices that need to be replaced.
All items in {0;0} from C need to be replaced by List B {0;0}
All items in {0;1} from C need to be replaced by List B {0;1}
And so on...
In the end everything needs to be fed into a orient component.
G is the geometry behind list B
A is the new list (but I don't know how to create)
B is the list A of all 60 panels
To wrap up the question:
I need to have entry 1 from list B in a new list on position: 0 1 2 3 5 6 7 10 11 12 15 16 21
I need to have entry 2 from list B in the same new list on position: 4 8 9 13 14 etc. etc. etc.
I wonder how I can do this :) Thanks in advance!
…
lane that looks like the outline of a gear wheel
2. Scale a bunch of copies of the curve to different sizes (I use 11 different sized curves.)
3. Move each curve vertically to a different Z-height
4. Rotate each curve to get the desired wavy/wiggly effect
5. Create a Loft surface using all the curves.
A critical step when creating the Loft is to add the curves in order - either top to bottom or bottom to top.
Step 4 can be omitted if you want a constant curvature throughout the final part. In this case all you have to do is Twist by the desired amount the Loft surface made from un-rotated curves. …
Added by Birk Binnard at 2:18pm on October 15, 2016
segments (ie. polylines)
2 = conic section (ie. arcs, circles, ellipses, parabolas, hyperbolas)
3 = standard freeform curve
5 = smoother freeform curve
The higher the degree, the less effect a single control-point has on the curve, but the further that weak effect reaches. Degree=5 curves are smoother, but it's also harder to add local details to it without adding a lot of control points. Rhino supports curves up to degree=11, but you almost never need more than 5.…
rcle A 0---1---2Circle B 0---1---2---3---4---5---6---7---8---9---10---11I was actually trying to connect through separated lines, point (A,0) to (B,11) and (B,1),point (A,1) to point (B,3) and (B,5), and (A,2) to (B,7) and (B,9).I was able to do it using a list item selecting the points and creating a line between them, but i would be to long to do it if i would like to divide in many more points, and try to connect them using list item... i went looking around for a bit, i think its possible to do it with a path mapper or a flip matrix, althought i have no clue how to make it.
You will find enclosed, some screenshots...If anyone could give me any tip, i thank you all in advance.Have a nice day.+Joan…
Added by Joan tarragon at 7:38am on December 5, 2011
7 -18.2
8 5.02
9 12.4
10 18.1
11 7.01
12 5.11
13 2.35
this data is waveform data.
i want to pick out only climax(for example in upper data, number 1 and 10) from this data.
i have this problem...that is, i do not know to do what.
i think my English is strange...so you are difficult to understanding my question.…
hat position you would like in order:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
You still must show up on time and you still must stay and work on your project during the duration of class.
See you all next week, I hope I won't still have this cough!
-Joseph Iwaskiw…
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R Light Studio - Georg Sochurek Schnelle Beleuchtungsoptimierung mit HDR Light Studio: HDRI-Environment-Maps in Echtzeit ganz einfach erstellen und bearbeiten 10:45 - 11:45: C4D und Adobe Photoshop - Christoph Schindelar Prototyping für einen Wintersportartikel-Hersteller mit Cinema 4D 12:00 - 13:00: Autodesk 3ds max - Daniel Daghofer Software für 3D-Modellierung und Animation 13:30 - 13:30 Pause 13:30 - 14:30: MODO - Mathias Zellerhof Produktdesign leicht gemacht - mit modo 701 von TheFoundry 14:45 - 15:45: Rhino 5 - Dipl. Ing. Fabio Palvelli Advanced NURBS-Modelling mit Rhino 5 für Designer und Architekten 16:00 - 17:00: Autodesk Maya - David Wuchte Umfassende 3D-Animationssoftware 17:15 - 18:15: 3D Coat - Christoph Schindelar Feinmodellierung (Sculpting) & Design von Handy-Designelementen mit 3D Coat Zusätzlich haben Sie die Möglichkeit in den Pausen / im Anschluß div. Hardware zu testen / anzusehen: Monitore: CG Serie von EIZO (inkl. Swing Sensor/Colorimeter) EIZO Foris mit EIZO Spyder EIZO Touchscreen NEC Spectraview Reference inkl. X-Rite Colorimeter Stiftdisplays: WACOM Cintiqs 3D Mäuse: SpaceMouse Pro SpacePilot Pro…
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.…