cted curvesso the subcurve 0;0 to 0;1 should be a subc. from curve 0the subc. from 1;1 to 1;2 should be a subc. form curve 1
and so on....thanks in advance…
cy of design communication and the control of information-flow are as important as the creativity of ideas. In response to the concurrent digital evolution emerging in the architectural industry world-wide, the Faculty of Architecture at The University of Hong Kong will host a two week intensive summer program named Digital Practice.Led by professors from The University of Hong Kong, as well as invited practitioners with expertise in practice of cutting edge digital techniques, the program offers participants opportunities to experience applications of computational tools during different stages of an architectural project, i.e. concept design, form finding and optimization, delivery, management and communication of design information under the team-based working environment. By learning advanced computational techniques through case studies in the context of Hong Kong, participants are expected to go beyond the conventional perception of technology, considering users and tools as a feedback-based entity instead of a dichotomy. The program, which is taught in English, includes a series of evening lectures related delivered by teaching staff and invited local architects.對於高品質的建築專案,創意之外,專案過程中高效的設計資訊管理和交流成為項目設計深化和實施必不可少的環節。今天,數字化技術不但改變了建築師的繪圖工具,影響了設計的過程,而且提供了工程建造和管理實施的更有效、更高效的手段。針對建築的數位化演進,香港大學建築學院將於2011年暑假期間,在香港大學建築學院舉辦“數位化實踐”國際研習班。在香港大學建築學院教授及有著相關豐富經驗的外聘實踐建築師的指導下,學員將有機會體驗在專案的不同階段(如概念設計、設計形式的生成、優化,設計資訊的管理和交流),如何有效地應用各種運算智慧化技術(從設計的數位化生成和建築資訊類比到物理模型),提升設計實施的品質,增加設計團隊對於方案的控制。我們將挑戰對於“技術”的傳統認知,即相對於使用者它不僅是工具,更是與使用者互動的媒介,二者形成一個有機的合體。研習班期間會安排系列講座,展現數位化技術在實踐工程中的廣泛應用。…
lues of the Radiation result?
I gathered the radiation results in ranges of five Groups.
1. 67 - 271 -> list of points
2. 271 - 475 -> list of points
3. 475 - 679-> list of points
4. 679 - 883-> list of points
5. 883 - 1085-> list of points
Now, I want the points which are located f.e. in group 1.
So at the end I want sort points by radiation values in lists.
Could anyone helpe me?
Thank you !
Leonie
attached my file
…
n. Of course I have not figured out how to create a typical composite wall section that will always be perpendicular to this line yet.
I am planning starting some kind of Safdie Habitat 67 study with Grasshopper. Parametric modeling usually focus on surfaces and blobs, but can be used to create other things as well.
Basically what I need to do is:
1) A system of rectangles that defines the bottom outline of a habitable unit. [easy]
2) How to convert this bottom outline into a box with a composite wall, and slab system and with windows and stuff. [should i try to create a boolean system from this outline? or a system with extrusions along the rectangle?]
3) A system that generates the staircases and ramps. [basically extrusion along the lines that defines the corridors and staircases]
hmmmmm...
Martin…
Added by Martin Hedin at 11:44am on October 21, 2011
estion:
1. I would add these steps to the installation instructions
2. I would double quote the path, I got confused by whether I should include the "_" at the end of the path (%appdata%\Ladybug\OpenStudio_).
Anyway, I downloaded the libraries, and pasted all the dll to the OpenStudio_ folder needed.
But it throws an exception on line 67:
Runtime error (IOException): file does not exist: C:\Users\...\AppData\Roaming\Ladybug\OpenStudio_\openStudio.dll Traceback: line 67, in script
One thing is that the box link for the libraries give you a zip that does include a dll that looks very similar... except it's cased differently: it is Upper Camel Case (OpenStudio.dll) and not Lower Camel Case (openStudio.dll)
Another thing is that even once I have renamed the above dll to match the casing, I still get this exception, even though os.path.isfile(os.path.join(openStudioLibFolder, "openStudio.dll")) returns True.
If I type the path on the run command it does land at the appropriate file...
I'm confused, any ideas?
…
ay how many valid permutations exist.
But allow me to guesstimate a number for 20 components (no more, no less). Here are my starting assumptions:
Let's say the average input and output parameter count of any component is 2. So we have 20 components, each with 2 inputs and 2 outputs.
There are roughly 35 types of parameter, so the odds of connecting two parameters at random that have the same type are roughly 3%. However there are many conversions defined and often you want a parameter of type A to seed a parameter of type B. So let's say that 10% of random connections are in fact valid. (This assumption ignores the obvious fact that certain parameters (number, point, vector) are far more common than others, so the odds of connecting identical types are actually much higher than 3%)
Now even when data can be shared between two parameters, that doesn't mean that hooking them up will result in a valid operation (let's ignore for the time being that the far majority of combinations that are valid are also bullshit). So let's say that even when we manage to pick two parameters that can communicate, the odds of us ending up with a valid component combo are still only 1 in 2.
We will limit ourselves to only single connections between parameters. At no point will a single parameter seed more than one recipient and at no point will any parameter have more than one source. We do allow for parameters which do not share or receive data.
So let's start by creating the total number of permutations that are possible simply by positioning all 20 components from left to right. This is important because we're not allowed to make wires go from right to left. The left most component can be any one of 20. So we have 20 possible permutations for the first one. Then for each of those we have 19 options to fill the second-left-most slot. 20×19×18×17×...×3×2×1 = 20! ~2.5×1018.
We can now start drawing wires from the output of component #1 to the inputs of any of the other components. We can choose to share no outputs, output #1, output #2 or both with any of the downstream components (19 of them, with two inputs each). That's 2×(19×2) + (19×2)×(19×2-1) ~ 1500 possible connections we can make for the outputs of the first component. The second component is very similar, but it only has 18 possible targets and some of the inputs will already have been used. So now we have 2×(18×2-1) + (18×2-1)×(18×2-1) ~1300. If we very roughly (not to mention very incorrectly, but I'm too tired to do the math properly) extrapolate to the other 18 components where the number of possible connections decreases in a similar fashion thoughout, we end up with a total number of 1500×1300×1140×1007×891×789×697×...×83×51×24×1 which is roughly 6.5×1050. However note that only 10% of these wires connect compatible parameters and only 50% of those will connect compatible components. So the number of valid connections we can make is roughly 3×1049.
All we have to do now is multiply the total number of valid connection per permutation with the total number of possible permutations; 20! × 3×1049 which comes to 7×1067 or 72 unvigintillion as Wolfram|Alpha tells me.
Impressive as these numbers sound, remember that by far the most of these permutations result in utter nonsense. Nonsense that produces a result, but not a meaningful one.
EDIT: This computation is way off, see this response for an improved estimate.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 12:06pm on March 15, 2013