Reply To: Huben Hammerless K1

Forums PCP Airguns Huben Hammerless K1 Reply To: Huben Hammerless K1

Link

RiverDevil
Participant
Member

wlbryce:
First I want to thank you to do such a job,let all of us know about it clearly.

U r much more honest than Chinese People,Chinese Society is full of cheating now,I won’t believe any business man immediately of what they said ,just like HD MASTER said that he shot much better than you did,I won’t believe that.They are less of honest and trustworthy. 

Back to what I want say is that what I told you here,some of them is what HD MASTER said(I have pictures,but I don’t know how to upload it) just like the grooves was made in the magazine,he told me that and I just imagine its function and how it works.If I was wrong,I apologied for the misleading. 

I want to tell u all about how the valve works,its diameter is about 8mm(HD told us,about 8mm,maybe 8.2mm,I forgot it),and when it is released,it moves about 9.5mm(Also HD said,I forgot what he said,only remember that it is less than 10mm),and what is the energy it gets from the air pressure,I can calculate for you:

F=PS
P=15MPa=15*(10^6)    
if the Constant Pressure Valve was set at 17MPa and when it was released the pressure maybe will go down to 13MPa
I use the average (17+13)/2=15MPa
Of coz the pressure decline is not “linear”,I can only use the approximate method to get the average one,I think it won’t be a big deviation.
S=pi*r*r=3.14*4*4*(10^-6)
then the F=15*3.14*4*4=753.6N
W=FS  
then W=753.6*0.0095=7.16j  
If we ignore the Friction Force of the Valve

And a traditional PCP that is working within a spring and a hammer,what is the energy the hammer got from the spring???
The spring max force is about 50N and we suggest that at last the force is zero.
So the average force is about 25N because of the spring force is “linear”——–Fspringforce=kx
And how far dose the hammer move?about 20mm,yes it is nearly double of HUBEN’s 9.5mm.
But what is the energy the hammer got from and at last the energy was used to open the one-way valve?
Then,W=FS=25*0.02=0.5j
  
When the HUBEN released the valve,it got 7.16j(the internal force is much more powerful than the traditional one just like FX cyclone) and at last it will hit something and stop,obviously the structure inside HUBEN needs a buffer part,I mentioned this problem and HD MASTER answer me  that there is a polyurethane part to absorb the energy,but I m wondering again,what will it become to when it was hit thousands times because it can only reduce a part of the hitting force.I have ever seen that HUBEN said that it needs no service,how could a company say such confident words???Dose Apple ever said like that?Is it arrogant?Who else can say that my product will never go wrong???Or I misunderstand about “never need servicing”?

Forgive my poor English,I haven’t used it for a long time since my graduation from the school…