New poll regarding ARCA stabilization method

The movie is pure art, likely built using AGI, a specialized software for this kind of things.
(You can also use AGI to display real mechanical motion governed by Newtonian laws but this is not the case of the above mentioned animation.)

Arrows without feathers work, but arrows with feathers work better. They fly straighter, and are more accurate. I'm not a good enough archer for the feathers to make a difference, but I still buy arrows with feathers because they look cool.

Make an experiment.

Take an arrow preferably without feathers, put a few (heavy) iron nails around its rear side and wrap adhesive tape around nails and the arrow in order to make the arrow tail heavy. Finally shoot the bow (not inside the room, outdoors). Tell me how the arrow behaves.
Do not draw the string back too much (do not use too much force) at least at the first try.

In this video http://www.youtube.com/watch?v=dqnSNjLW0rE Stabilo is pulling a tiny mass behind it (notice that the diamond thing below it collapses in the water, at 3:52)

Yeah, arrows are balanced pretty carefully. I don't want to get a bunch of nails in my hand. Also, I left my
bow and arrows back in the states.

I heard something: "A launch attempt over the Back Sea will follow in less than 48 hours!"
Can anybody confirm this information?

Yes, it's pure art, but it gave me an idea of what they are trying to do.

I don't have any diagrams for the Helen rocket, only guessing (I'm not actually involved in any space project, I just found this discussion thread interesting; my background is anti-aircraft missiles but these are aerodynamically stabilised). I also don't have much time available to write the equations involved.


The formula is overly simplified and even so it would apply only for a tiny fraction of time. You could use differential equations for the movements, forces etc., integrate them over time and get some complicated equation. As you said, I don't think they have an analytical solution, so you are better off with discrete computer simulation + trial and error until finding something that looks stable in simulation.

My assumptions were not necessarily about mass but mainly about the time this would be valid (very tiny) and the length of the cable (really long so that small horizontal rocket movement doesn't affect the cable verticality, or it make it negligible).

The point I wanted to make is that if the structure is vertical initially, a destabilisation torque affecting the angle of the rocket relative to the cable would be countered by a stabilisation torque given by the tension in the cable and this angle. If such rocket-cable angle would not exist, it is a pendulum fallacy. But my view is that with a heavy enough payload and depending on the engine force, aerodynamic forces and the G force of the payload it could lead to a tension in the cable at a slight angle from the structure's acceleration. Now it needs to be determined what the angle between the structure's acceleration and the engine's force (rocket vector) would be and whether this angle creates big enough stabilising torque.

Of course, even if you have such stabilising torque, you have to sort out possible oscillations, spinning etc.


Is your simulation for a pendulum rocket with rigid connections? As I said above, flexible connections, IMHO, don't necessarily lead to pendulum fallacy.

Interesting, how can you prove that flexible connections do not lead to pendullum rocket fallacies. The mathematical model is much more complicated than that corresponding to rigid connections so if you do not run a numerical simulation is practically impossible to tell how your flexible system behaves.

What I can tell you is that if you split the initial rigid connection in 2, 3 ,4, 5, ... articulated rigid segments then things do not change at all, you obtain the same behavior. The rocket together with its counterweight is turning upside down.

Today at Constanta ARCA Team and Romanian Navy discussed the detailed procedures for the Helen launch.

I will try one more time to show that the ARCA method is unstable in a vacuum. If the thrust vector is not in line with the Cg there will be a torque component to the thrust that will cause the rocket to rotate. For the thrust to be in line with the Cg one of the following two things must be true:

1. The rocket would have to be pefectly straight and the rocket exhaust would have to be perfectly straight, or
2. The end of the rocket motor would have to be at the Cg of the rocket.

Option #1 is not possible to acheive in practice. Therefore, the ARCA method has to acheive option #2. Now let's consider a rigid rocket consisting of two masses -- m1 and m2, which are seperated from the Cg by a distance of d1 and d2. It would look something like the following:

(m1)=======Cg==================(m2)
<-------d1------->|<--------------d2------------------>

By definition, the values of m1, m2, d1 and d2 would be related by the following equation:

m1*d1 = m2*d2

The only place that the motor can be without producing rotational torque would be at the Cg. This rocket is similar to a single-stage ARCA rocket where m1 is the mass of the rocket engine, Cg is the location of the end of the rocket nozzle, m2 is the mass of the tethered object, and d2 is the length of the tether.

In 2 dimentsions, the tether consists of two lines that are attached on either side of the rocket body and converge at a point below the rocket. This distance is d2 in my diagram. It doesn't matter if there is a single line after d2 that connects to the mass m2, or if the two lines connect directly to m2. The important fact is that the two tether lines form a triangle with a base defined by the width of the rocket and a length of d2.

There are several factors that cause this method to be impractical and fail. They are as follows:

1. As the rocket consume fuel the value of m1 will be reduced. This would move the Cg away from m1 and cause the rocket to rotate uncontrollably.

2. Even if the rocket were to maintain the thrust vector at the Cg there is nothing to guide it in the direction of travel. If the rocket started out with a rotation there would be no forces to stop the rotation. Even if the rocket was not rotating, there is no control to guide it in the desired direction.

3. The rocket exhaust will strike the tethered mass, which will cause drag and random rotational forces. In a vacuum, the rocket exhaust will strike the tethered mass at the same velocity that it exited the rocket nozzle. There are no air molecules in space that would slow down or scatter the rocket exhaust molecules.

Dave

According to THIS ARTICLE:

The pendulum belief is a fallacy because it stems from the implicit (and false) assumption that simply because the weights and "hanging" devices are arranged in roughly the same way in both the rocket and the pendulum, they will behave in the same fashion. However, the forces exerted are different. While gravity does act similarly in both physical systems, the supporting force exerted onto the pendulum by its hanging point is constrained to remaining aligned with said fixed point; this is unlike the force exerted onto the rocket by its engine, whose direction instead depends on the rocket's overall orientation or attitude.,

I have predicted success in the form of a dampening sinusoidal motion for the Helen rocket system.

Hi,

This is a static analysis of the forces for the active stage(AS), where:
CG = center of gravity of the AS (in void AS will rotate around this point)
G1 = force of gravity of AS (in void does not affect the AS rotation)
F = motor trust of AS
d1 = distance from the CG and F application point (somewhere in the combustion chamber)
G2 = force of gravity for the next stages
d2 = distance from the CG and next stages anchor point

It results the longitudinal(y) and the transversal(x) components of the forces which affect the AS rotation.
Fy - G2y is on the CG direction(in void no rotation from that)
Fx it's due to not a perfect alignment of trust to CG

So the rotation of the AS is affected by Fx and G2x using d1 and d2 lever.
At equilibrium:
Fx*d1 = G2x*d2

Fx is constant and G2x will increase with the angle dragging back the AS to the vertical.

For the dynamic analysis I can say that the inertia of the next stages and payload will drag the cord more close to the longitudinal axis of AS. So the (G2x + inertia x component) will decrease but it always be there.

My prediction for CG of AS trajectory it's an helicoidal ascending move around one axe parallel with the gravitational field direction. The radius of the elicoid will have an amortized oscillation.

So they have my vote!

Ever tie a string to a cats tail with a piece of paper tied to it?

Monroe