To: heedm
“I still have the feeling that he wasn't familiar with the head and assumed it to be like that on the 206 (link-horn connection in-line with teetering hinge). Some of your descriptions make it seem like the swashplate input links are phased wrong, and that this is the sole cause of all the problems”.
Regarding Ray Proutys’ understanding of the Robinson head he wrote an article on it in the AHS technical forum. I also believe he had some involvement in the early design.
Regarding the swashplate input links I have no problem as they cause the swashplate to operate just like on a two blade Bell. Push the cyclic forward and the front end goes down over the nose and the back end goes up over the tail. The similarities between the Robinson and the Bell stop at that point.
On the Bell when the blade is disposed over the lateral axis and cyclic is pushed forward the pitch horn which is 90-degrees ahead of the blade causes the blade angle to be at its’ lowest pitch. Conversely the other blade is at the highest pitch. Precession causes the blades to flap down over the nose and up over the tail.
On the Robinson with the blades disposed in the same manner and forward cyclic is pushed the front of the swashplate will move down just like the Bell. The Robinson blade will not be at its’ lowest pitch like the Bell. The blade will have to travel another 18-degrees until it is at its’ lowest pitch. The Bell is rigged for forward cyclic with the blades disposed as above over the lateral axis. The Robinson blade is advance 18-degrees from that position in order to establish forward cyclic settings but I’m sure you were well aware of that fact. I’m also sure that Ray Prouty is also aware of it as well.
On the infamous Robinson (Cantrell) web site they show a Robinson head and address gyroscopic precession. When you get to the explanation they have an illustration addressing precession on the Robinson but the illustration is for a Bell swashplate. Now if you look above at the positioning of the blades on both helicopters the inputs are wrong for the Robinson because it implies that the Robinson has a 90-degree phase angle. If it does the blade will dip down and to the left with forward cyclic. However Nick sez that pitch flap coupling will cause the blade to respond in 72-degrees as opposed to 90-degrees.
Now I think that if I am correct about the offset the test will prove it, contrary to what you might think.
You have access to a 206 and a Robinson. Perform this test. On both helicopters place the blade over the longitudinal axis. Move the cyclic forward from the rigged neutral position. The Bell blades will not move but the Robinson blades will. Now advance the Robinson blades until the pitch horns are directly above the lateral axis. Move the stick forward and backward from the rigged neutral position and the blades will not move.
Now I don’t have your educational background but I have always been told that with maximum change in cyclic pitch the blades will respond 90-degrees later. If what Nick said and Frank Robinson alluded to by saying it was too technical to explain to non engineers and pilots then I am wrong and all of my instructors and engineers that I have worked with are also wrong.
Based on the weight of the blade I figured that the centripetal force would be between 50,000 and 72,000 pounds, which would cover a helicopter somewhat bigger than an S-58 and up to a CH-37, which has a centripetal force of 72,000 pounds (I use your term not mine). The first figure is quoted in the Blue Book and the Sikorsky service-training department provided the second number.
Lets’ assume the blade is at rest. Due to the mechanical linkage of the blade to the head one point becomes a fulcrum and the weight of the blade places a small amount of stress on the rotorhead. If we placed a strain gage at the point of stress it can be measured. This is the same strain gage that measured the centripetal force yet there is no movement of the blade. If the blade(s) are not rotating can you have centripetal force? Now we start to rotate the blade(s) and the force starts to build until it reaches maximum. The original force was not centripetal and with the blades at speed it is now centripetal. You say tomahto and I say tomato.
It is easy to accept your premise but in order to do that I would have to totally alter my belief system.
Regarding the detached blade, when it happens the blade will fly off. If centrifugal force is zero on your calculation where does the energy come from to propel the rotor off of the rotorhead?
Is it possible that under this condition the centripetal force will drop to zero and an energy transfer takes place with all of the centripetal force being transferred into the blade?