掌握编程过程中遇到的思路方面和技术方面的问题. 内容包括以下几个方面:
反向逻辑思维能力的培养;
建立清晰的编程逻辑思维能力;
GH 的程序设计理念;
并行数据结构深入理解和控制.
Grasshopper course of McNeel Asia focus on the cultivation of students flexible use of programming techniques, the ability to solve practical problems. Our course deep into the whole process of programming, from programming thinking model, the components principle to usage details do detailed explanation, help students complete mastery programming encountered in the process of thinking and technical aspects, include the following content:
Ability of reverse logical thinking;
Establishment of clear programming logical thinking ability;
The program design concept of Grasshopper;
Understanding parallel data tree structure and how to control it.
更多详细内容... More details…
授课讲师 Instructor 课程由Grasshopper原厂McNeel公司在中国地区的两位 Rhino 原厂技术推广工程师 – Dixon、Jessesn联合授课。课程结束后对达到授课预定目标的学员颁发唯一由Grasshopper原厂认证的结业证书.
Dixon & Jessesn, McNeel Asia Support engineer, by the end of course student who achieve the intended target will get the authentication certificate from McNeel Asia.
课程报名 Register this course 课程即日开始报名, 开课一周前停止报名, 名额满提前报名结束. This course begin to sign up, stop sign up a week ago, with the quota ahead over.
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课程日期 Schedule 7/15-7/20 Beijing 北京 7/26-7/31 Shanghai 上海 7/07-7/12 Shenzhen 深圳
课程范例演示 Samples of Grasshopper course demo
Note: pls follow below comments by Jessesn to see the samples…
f parameters.
For all other combinations, only one row is supported and only one column has horizontal loads.
The second problem is that the deflection of beams is not consistent. It seems that only the outer beams are being deflected and the inner beams remain straight.
I think the problem lies in the way that I constructed my geometry. Instead of filtering out the points that I needed of the made geometry, I made extra arrays of points to define the points of support and the points of wind loading. These points have the exact coordinates, but apparently karamba does not link these two but sees them as separate.
What modifications should I make to get this frame working?
[1]: https://i.stack.imgur.com/M6S2R.png[2]: https://i.stack.imgur.com/ZAri9.png[3]: https://i.stack.imgur.com/CGZF8.png[4]: https://i.stack.imgur.com/nTHX6.png…
jogs between the curves. My Grasshopper program uses the curves to add orientation curves. These curves are then broken down into 150 segments with a "Divide Equal" command", then merged and fed to my Kuka|prc core. 24 commands in total.
The program works well on the Kuka in T1 setting so I can adjust the speed. Albeit my hand is exhausted holding the buttons... and this will not work when I am not in T1 mode.
When trying to program I found that if I did not have each curve broken into an equal number of segments (i.e. 150) the commands would not merge. I found that each LIN command on the kuka is completed in the same time interval. Unfortunately this means that some of my shorter jog curves can take longer to run than my cutting curves. And my velocities are not constant throughout the program because the curves are different lengths. I need the cutter when in the material to have a consistent cutter speed to optimize my machine time (as well to speed up the jogs).
Is there a better way to program this (or component to use) so I can manipulate the velocity with Kuka|prc? i.e. should I be using Divide Equal and Merge?
Thanks
Joanne…
compute and display).
To check which is your case, disable all components' preview first, run task manager and then perform mesh - > surface convesion. Now constatly look how much memory rhino consumes. If you'll see that it finished computing, try to turn on preview - your pc should now boil.
*checked - 10 000 surfaces from mesh = ~150 mb of ram, so this is probably not a problem.…
square units. Then you have an integral number of fragments on each side. This means that if all fragments need to have the same surface area, you can only have the following possibilities for side A:
1 fragment = 100 square units
2 fragments = 50 square units each
3 fragments = 33⅓ square units each
4 fragments = 25 square units each
5 fragments = 20 " "
6 fragments = 16⅔ " "
etc.
For side B, the numbers are mostly different
1 fragment = 300 unit²
2 fragments = 150 unit²
3 fragments = 100 unit²
4 fragments = 75 unit²
For side C they are different still. Unless you join fragments across on both sides of the edges of the box, I very much doubt you'll be able to pull this off.
The solution I attached will create fragments as identical as possible, but it's a very boring outcome...
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David Rutten
david@mcneel.com
Poprad, Slovakia…