x and min values for x,y,z and calculate energy for each optionand collect these results in excel sheet ...
option No. x y z Annual coiling demand(by DIva)
1 10 10 10
2 10 15 20
3 11 10 19
4 12 14 17
5 15 16 15
6 16 11 14
7 18 12 12
.
.
etc
Regards ...
hossam
Hossam.wefki@gmail.com…
EC
1. Between hours 1:00 to 24:002. Current document units is in Meters3. Conversion to Meters will be applied = 1.0004. [1 of 7] Writing simulation parameters...5. [2 of 6] No context surfaces...6. [3 of 6] Writing geometry...7. [4 of 6] Writing materials and constructions...8. [5 of 7] Writing schedules...9. [6 of 7] Writing loads and ideal air system...10. [7 of 7] Writing outputs...11. ...... idf file is successfully written to : c:\ladybug\Freeformtower_IDF\EnergyPlus\Freeformtower_IDF.idf12. 13. Analysis is running!...14. ......
Done! Read below for errors and warnings:
15. 16. Program Version,EnergyPlus-Windows-64 8.1.0.009, YMD=2015.04.04 23:39,IDD_Version 8.1.0.00917. 18. ************* IDF Context for following error/warning message:19. 20. ************* Note -- lines truncated at 300 characters, if necessary...21. 22. ************* 577 Zone,23. 24. ************* Only last 1 lines before error line shown.....25. 26. ************* 578 Freeformbuilding27. 28. ** Warning ** IP: IDF line~578 Comma being inserted after:" Freeformbuilding" in Object=ZONE29. 30. ** Severe ** Out of range value Numeric Field#5 (Type), value=0.00000, range={>=1 and <=1}, in ZONE=FREEFORMBUILDING31. 32. ************* IDF Context for following error/warning message:33. 34. ************* Note -- lines truncated at 300 characters, if necessary...35. 36. ************* 586 BuildingSurface:Detailed,7341.
…
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.…
indexes later and the k value is just 5. The goal is to output all combinations of length 5 from the list of integers 0 to 19.
I placed the list in brackets in order to make it iterable (I don't really know what this means.) Somehow I need to get the combinations out of the itertools.combination output.…
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
Integer = 0 To 9
val *= 2
lst.Add(val)
Next
Since val is a ValueType, when we assign it to the list we actually put a copy of val into the list. Thus, the list contains the following memory layout:
[0] = 2
[1] = 4
[2] = 8
[3] = 16
[4] = 32
[5] = 64
[6] = 128
[7] = 256
[8] = 512
[9] = 1024
Now let's assume we do the same, but with OnLines:
Dim ln As New OnLine(A, B)
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
ln.Transform(xform)
lst.Add(ln)
Next
When we declare ln on line 1, it is assigned an address in memory, say "24 Bell Ave." Then we modify that one line over and over, and keep on adding the same address to lst. Thus, the memory layout of lst is now:
[0] = "24 Bell Ave."
[1] = "24 Bell Ave."
[2] = "24 Bell Ave."
[3] = "24 Bell Ave."
[4] = "24 Bell Ave."
[5] = "24 Bell Ave."
[6] = "24 Bell Ave."
[7] = "24 Bell Ave."
[8] = "24 Bell Ave."
[9] = "24 Bell Ave."
To do this properly, we need to create a unique line for every element in lst:
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
Dim ln As New OnLine(A, B)
ln.Transform(xform)
lst.Add(ln)
Next
Now, ln is constructed not just once, but whenever the loop runs. And every time it is constructed, a new piece of memory is reserved for it and a new address is created. So now the list memory layout is:
[0] = "24 Bell Ave."
[1] = "12 Pike St."
[2] = "377 The Pines"
[3] = "3670 Woodland Park Ave."
[4] = "99 Zoo Ln."
[5] = "13a District Rd."
[6] = "2 Penny Lane"
[7] = "10 Broadway"
[8] = "225 Franklin Ave."
[9] = "420 Paper St."
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 6:26am on September 9, 2010
ll these 12500 points.
Group 1 would represent the point located at 0, 5, 10, 15, 20 etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc.
I can create the pattern but the selection of points are all the points in row 0 and then all the points in row 5 and so on.
I would like the selection of points to start at the bottom left, and sequentially continue to the right and then continue on the 2nd row (left to right & bottom to top). i am hoping the pattern i am trying to achieve is more understood with the quick screen capture I uploaded.
the end goal is to be able to select all the points in the grid that are in each pattern.
Thanks in advance for any guidance with this. …
Added by Alyne Rankin at 6:53am on October 11, 2017