Retreating blade stall analogy

[Ripley]

Today I was thinking at an orbiting vessel around a planet, in a near equatorial orbit, and (don't ask me why) I suddenly thought about the [ame=http://en.wikipedia.org/wiki/Retreating_blade_stall]retreating blade stall[/ame] problem.
At least, I saw some analogies...from my space-mechanics-ignorance, that is.

The vessel flies along the same path as the orbited body for 1/2 orbit, then, during the other half, it goes towards the "opposite" direction than the orbited body...just like a retreating blade...
Could it experience some perturbations or anything else?

...Does it make any sense?

...And don't :rofl: at me! You're warned!

 

[Scruce]

:rofl: ... Im joking . Isnt that what the tail rotor on the helicopter is for?

 

[Hlynkacg]

No the tail-rotor is to counteract the torque effect that comes from getting the transmission/rotor system up to speed. When the head is in an unloaded state the tail-rotor is effectively idle. On a dual rotor helicopter the two heads spin in opposition thus negating the need for a tail rotor.

Think of each rotor blade as a wing where airspeed is a function of RPM. In a hover, the airspeed of each blade is equal, but as the helicopter begins to move forward the relative airspeed for the leading blade to increases, and the trailing blade decreases.

The faster the helicopter goes the greater the disparity, eventually the disparity will become so great that the leading blade will be effectivly super-sonic and the trailing blade will be in a stall.

Needless to say, this makes maintaining control of the aircraft problematic.

 

[Eli13]

:rofl: ... Im joking . Isnt that what the tail rotor on the helicopter is for?

No but its a main factor in helicopters airspeed. I'm sure it says something about that on the Wikipedia article.

 

[Mister Mxyzptlk]

Cyclic: The cyclic control is usually located between the pilot's legs and is commonly called the cyclic stick or just cyclic. On most helicopters, the cyclic is similar in appearance to a joystick in a conventional aircraft. By contrast, the Robinson R22 and Robinson R44 have a unique teetering bar cyclic control system and a few early helicopters have had a cyclic control that descended into the cockpit from overhead, one example being the HC-2 "Heli Baby", HC-102. [3] The control is called the cyclic because it changes the pitch of the rotor blades cyclically. That is, the pitch or feathering angle of the rotor blades changes depending upon their position as they rotate around the hub so that all blades will change their angle the same amount at the same point in the cycle. The change in cyclic pitch has the effect of changing the angle of attack and thus the lift generated by a single blade as it moves around the rotor disk. This in turn causes the blades to fly up or down in sequence, depending on the changes in lift affecting each individual blade.

This is the mechanism that increases AoA on the retreating blade to maintain equal lift during the rotation of the blades. The "swash plate" is the doughnut like assembly that is critical to this function.

 

[tblaxland]

I think I know what you mean. For example, take an object orbiting the Earth in an prograde equatorial orbit. On the side of the Earth closest to the Sun, the object's heliocentric velocity is lower than on the opposite side of the Earth. What you are asking is, are the gravitational perturbations greater on that side than on the other? I think they would be. Whether or not they are large enough to be significant is another question. For a typical low to mid earth orbit they would not be, especially since the difference in perturbations averages out over the course of a year. For very high orbits, the perturbations would be significant and they the basis for defining the [ame=http://en.wikipedia.org/wiki/Hill_sphere]Hill sphere[/ame].

 

[JEL]

I believe this illustration agrees with you:
http://upload.wikimedia.org/wikipedia/en/6/61/Lunar_perturbation.jpg
If you look at what it says about Line of apsides, X1.

 

[Ripley]

Ok, thanks, so I was not totally wrong! :thumbup:

I was not thinking at "gravitational perturbations" by other bodies, but more simply about the relative "speed" of a prograde equatorial orbit around both "sides" (Sun-related sides) of a planet. Tblaxland's reference to "orbit height" makes me think somehow of heli blades' lenght...if I explain myself right...

Wikipedia quote:
Retreating blade stall starts at the tip region and develops inboard

 

[JEL]

There's a difference though; while the single rotor-blade's tip-velocity is greater than it's root-velocity, it's the other way around for an orbiting object. The farther away you orbit, the slower you orbit. That's why the ISS orbits the earth much faster than the moon, for example (if you consider the moon to be the tip of the rotor-blade and the iss to be the root)