New poll regarding ARCA stabilization method

The weight of the entire rocket is 1800 kg

It looks like the real space vehicle is similar to what we saw in the animation ( http://www.youtube.com/watch?v=4qVuVix5kCE )
I heard Dumitru Popescu, the "brain" of ARCA, making a comparison between Helen rocket and a plane that tows gliders. This man does not understand physics or is simply mad, he fails to make logical connections.
The plane and gliders are aerodynamically stable and the plane is under active guidance.
Helen is in a totally different situation. Two unstable stages are aimed to make stable another unstable stage, the first. It is as if with two unstable gliders in the rear of an unstable plane you try to make the plane stable.That is crazy!

Well, we'll see what happens.

I build things. I build rockets that are aerodynamically unstable and are still able to fly. Your rocket is an obfuscated pendulum fallacy, as others have said, and will not fly in a stable manner.

If you didn't want people to analyze your rocket, you probably shouldn't have started a thread asking people to analyze your rocket.

That's a very good point, you are right. What I wanted to say is that I need unbiased feedback from people like you. This feedback is important. If there are others that copy/paste messages on different forums, their feedback is irrelevant.
Your input is important and you will see that it works. If the weather will allow this, we will launch next week.

Actually, I'd say that an unpowered rocket stage is quite stable. It will very reliably fall straight down if you let it, even without active guidance.

No it is not. It will not fall straight down (nose down - tail up). The rocket will have complex rotations (not necessary 360 degree) around its center of mass.
In order to be stable it needs fins, same as a powered rocket.

In vacuum with or without fins a rocket will fall toward the earth spinning around its center of mass.

It will always point with its nose down if, before letting it fall, you spin it quickly around its longitudinal symmetry axis (gyroscopic stabilization).

Why does an arrow without feathers fly straight (they do, trust me, I've shot arrows without feathers without any problem).

So why do any arrows have feathers? If without them they still work OK.

Probably because feather less arrows don't spend enough time in the air to become badly unstable? As it is they are not as accurate.

Perhaps if you fired a featherless arrow straight up as far as you could, on the way down it would be flopping around all over the place. A feathered arrow would come straight down point first.

Arrows have no thrust vector. As long as you center of drag and weight are the right way around it will be stable.

For the same reason a rocket engine with a long stick attached to it flies straightly along a direction.

Any arrow or rocket can be made aerodynamically stable providing you move the CG ahead of CP. So, if you make the rocket or arrow exaggeratedly nose heavy you shift their CGs if front of CPs and they will fly straightly even without fins.

However, a more intelligent approach is to move CG ahead of CP by not adding so much weight in the nose, and this can be achieved using fins or feathers.

In vacuum, as the CP does not exist, the rocket or arrow will be naturally unstable no matter how much weight you add in the nose or how large the attached fins are.

My view on this is that there will be a sufficient torque if the payload hanging down the cables is heavy enough (I don't have the technical details to be able to calculate). However, I don't see what would prevent this structure from oscillating. Assuming that the cable is long enough to consider it vertical (at least initially), any rocket rotation would be countered by a torque, something like below (very simplified model):

t = r * mp * (g + av) * sin(dv)

where:

r = the distance to the from where the cables are anchored to the rocket's centre of mass
mp = the payload mass
g = 9.81 (initially)
av = the structure's acceleration in the vertical plan (given by the rocket's engine force)
dv = the rocket deviation angle from the vertical

and

av = Fr / mt * cos(dv)

where

Fr = the force of the rocket's engine
mt = the total mass of the structure (rocket + payload)

Once deviated from the vertical, the rocket would get back on the vertical with a kinetic energy and it may start oscillating. It's only the air that would slow the oscillation down but as it goes up this is less efficient (and given the design without fins).

In vacuum, I think it will oscillate more and more since g decreases as it goes up and the engine's force may decrease as well. The kinetic-potential energy transformation would force wider oscillation angles. Wider deviation angles have the side-effect of a smaller vertical acceleration which adds to the oscillation angle (catch 22).

A potential for stabilising the oscillations is if the (g + av) actually increases, i.e. av increasing at a faster rate than the decreasing of g. I assume the engine force is decreasing with subsequent rocket stages but the structure also loses mass (fuel, early rocket stages). It could be calculated whether (Fr/mt) increases and at what rate. But at some point the rocket will run out of fuel. Maybe the engine should stop suddenly before (g + av) starts decreasing, otherwise the rocket may destabilise itself.

My conclusion - there are too many variables to simply tell whether this is stable or not. I assume the ARCA guys did the maths (I think it's not too difficult to do a computer simulation of this structure). My vote when for the 2nd option (there is stabilisation torque but I'm not sure it will work in practice).

Anyway, best of luck guys (I'm also Romanian )!

Looking at the graphical simulation, I understand a bit more about how this is supposed to work. I suppose the rocket engine force remains constant during each stage. As it consumes fuel, its mass decreases and the vertical acceleration increases. This would increase the tension in the cables since the payload mass during each stage remains constant. With careful calculation, (g+av) would be increasing and the amplitude of the oscillation reduced (doesn't mean that the energy of the oscillation would also be reduced).

If the cable is long enough, the risk of transferring the oscillation (movement) to the payload is smaller. In the end, this may actually work. Assuming that the ARCA guys calculated the maximum destabilising torque allowed and the minimum/maximum payload mass, I'm changing my vote to "It will work".

Catalin

Without free body diagrams for the rocket and pendullum your calculations do not tell us too much. We need to see the forces you took into account and the second law of dynamics (F=ma) for each body (pendulums and rocket).

From what I can see, looking at the complexity of your formula, you took into account only the case of a pendulum hanging from a body of enormous mass as compared with its own weight (something like m_rocket=1000*m_pendullum).

This is not the case of Helen rocket where pendulums have weights of the same order of magnitude as the first stage.

The real equations which describe the motion of Helen rocket are ugly enough and do not have analytical solutions.



Numerical simulations shows (see the image in 7 steps) that the pendullum rocket is turning upside down.

Even if some perpendicular motion does exist it is not of too much help.

Which graphical simulation? Where can I see it?

Well, the PR exercise at http://www.youtube.com/watch?v=4qVuVix5kCE