Static load equal 29.865

[tecpilot]

Thanks Genghis,

for your words. You are on the right way. I've splitted the system up into component parts some months before, as you advised.

To get the whole system breaking load Fb=[Cw+Hl] x Lf x Sf x Tf

Fb=Breaking Load

Cw=Components Weight
Hl=Human Load
together = 400kg / 3,92 kN

Lf = Load Factor = 3,5
Sf = Safety Factor = 1,5
Tf = Textile Factor = 2 (Aging and worn out factor of included textile lines and belts)

Fb =[3,92]x3,5x1,5 x 2 = 41,16kN

Ok up to this point only mathematics. All included metall parts must have breaking loads greater than 20,58kN, textile parts needs 41,16kN.

I splitted the parts and the system himself is safe, also in the view of our CAA. The problem is, the main system is hooked into the a/c air frame load hook, approved by STC and FM to 1000kg means 9,8 kN. Not enough to the CAA. Unfortunately no other papers to the hook available. The redundancy systems is attached on the airframe and ok to the CAA.
The problem is the hook. But as i wrote before, in my opinion, if the hook is approved to 1000kg on a FAR/JAR 29 helicopter, he must be certified with a load factor of at least 2,5 and could therefore count as a part with 24,5kN breaking load, greater than the needed 20,58kN. With the certification of the hook on a 29' helicopter the CAA should have accepted the hook with a breaking load of at least 24,5 kN and should therefore also in our case the hook count as described.

The system is designed to carry 3 to 4 persons. It makes no sense to cut it to only 1-2 persons because a 1000kg hook, usually used to transport 800-900 kg nonhuman load isn't safe enough.