e vertical load V = q * l/2 and the sum-vector of H and V points in the tangent direction of the cable (as zilic mentioned before).
For example:
The wight of the wet fabric q = 4 kg/m == 0.04 kN/m, l = 2 m and you choose f ~ 0.20 m, then
H = 0.04 * 2^2/(8*0.2)= 0.10 kN and V = 0.04 kN
If you reduce f to f ~ 0.05 m (5 cm) then H will become 4 times more > H = 0.40 kN, V = 0.04 kN
The formular is a very good approximation, but it does`nt include the elongation of the cable itself. If you want to be exact, you have to add df (due to the elongation of the cable under load) to f (the sag of the cable without load) and, of course, you have to add the wight of the cable itself to the wet fabric load too.
pbau
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u mean they are just randomly?
For Q 3, I have checked my model and the base curve is located on the ground. Could you help me check again? (see attached)
Thanks a lot!!
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r(int i = 0; i < dataList.Count; i++){ dataTree.Add(dataList[i], new GH_Path(i)); } return dataTree; }
Other that that:
https://www.google.gr/url?sa=t&rct=j&q=&esrc=s&sour...
https://www.google.gr/url?sa=t&rct=j&q=&esrc=s&sour...(v%3Dvs.110).aspx&ei=NbLkVdrSHoHasgHv2qjYCw&usg=AFQjCNEviU8qD6_VSaf5-6DWYL9vvVHExQ
http://www.codeproject.com/Questions/325178/Difference-between-ilis...
http://www.codeproject.com/Articles/832189/List-vs-IEnumerable-vs-IQueryable-vs-ICollection-v
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