L3 internal frame construction method

Imagine holding the rocket in place with your hands while the motor fires.
Think about the forces acting at the point where the thrust of the motor is transferred to the rocket.
Then imagine letting go and how that changes the forces on that part of the rocket.

The conclusion is that the shear force applied to the centering ring of the rocket is only equal to the motor thrust at the point where the motor is still thrusting and the velocity of the rocket is not changing - e.g. when drag + gravity = thrust.

FredA said:

So without mass, and velocity, you cannot calculate any resistive forces on the rocket.

You can go though the math dance...but in the end you will see that all the force of the motor is "resisted" by drag and gravity.
Less mass, you accelerate faster and build more drag....
It must zero out.

The motor puts out a force over time (thrustcurve) and the rocket resists it....all of it...

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I went through some of the math dance in post #10, showing this is not correct - even if for no other reason than the motor is not massless. The thrust goes into accelerating the motor as well as the airframe.

Trivial counterexample: Ignition of the rocket; it starts to accelerate. Drag is essentially zero. Gravity provides 1G of deceleration. Thrust exceeds weight or it wouldn't be moving or accelerating. Acceleration in Gs is the thrust to weight ratio minus one. IIRC, that's required understanding for a L2 written test.

If all the thrust of the motor were resisted by drag and gravity, the rocket would still be stationary.

Please go back and read post #10 to look at a little bit of the physics of the situation.

As pointed out by another person, once one gets into things like vibration and accoustic modes, it gets a whole lot more complex than what I showed. It's been too many years since I've been in the heavy swing of physics for me to want to tackle that problem without cracking open a few books.

Gerald

Zebedee said:

Imagine holding the rocket in place with your hands while the motor fires.
Think about the forces acting at the point where the thrust of the motor is transferred to the rocket.
Then imagine letting go and how that changes the forces on that part of the rocket.

Click to expand...

yes, this is why the thrust curve, and acceleration graph from the rocket flights are different from each other.
The change in mass, is not present for a horizontally tested motors. If they are vertical, sometimes they filter the results with a mass correction.

There is a guy who I've seen that takes the altimeter data, baloon data from national weather service, and filter out the mass and drag forces, taking the angle of flight into account to get an accurate (or more accurate) thrust curve, filtering out mass and drag. Since gravity is a constant, the angle of flight is needed so you can adjust the affect of gravity on the sensor, since it reduces the seen force from gravity. Most of the time you see 26 - 28 ft. per s/s/ rather than 31. Getting an inflight thrust curve for my research M motors is hard without actually measuring the chamber pressure.

bill_s said:

Almost. Built a complete wood frame (balsa) and covered it with flat pieces of posterboard. Rocket was powered by a D motor. This kept it from warping and allowed building the motor mount structure inside a frame.



You're both wrong. The motor must lift and accelerate itself. However, if the remaining force is 100 lbs., it will press with 100 lbs. into the rocket, no matter what the breakdown of where that goes: drag, gravity, and accelerative.

I don't get the impression over-building is seen as a bad thing in HP rocketry. The heavier it is, the bigger motor it can take and be seen again. If you don't over-build, you're liable to end up having to carry your L1 rocket on the roof of your car, it's so big.

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I was concerned with the force applied to the centering ring, and how to "figure out, what it would be". So it was from a force applied scenario, not a thrust / forces distribution scenario. I am cool with being wrong when I learn from it.

Zebedee said:

The conclusion is that the shear force applied to the centering ring of the rocket is only equal to the motor thrust at the point where the motor is still thrusting and the velocity of the rocket is not changing - e.g. when drag + gravity = thrust.

Click to expand...

there are times when (drag + gravity > thrust). You can encounter this in dual thrust motors, or long burn motors.

You can take the acceleration data from an altimeter.
Determine Cd over a span of V by looking at the deceleration due to drag post burn out.
Subtract out the effect of gravity accounting for the mass change during the burn.
And you get the thrust curve.

A couple of altimeter vendors provide this post-flight analysis.

Of course, non-vertical flights make this a little more difficult to separate Drag from Gravity.

FredA and the 6 PE guys are correct. So is GT. If you assume a massless motor, the CR feels 100% of the motor thrust. For a real system, the centering ring feels T(1-mt/mr) where T is the thrust of the motor, mt is the mass of the motor and mr is the total mas of the rocket.

Force does not get converted into acceleration, force causes acceleration.

As the rocket gains drag, this changes the stress or force distribution along the rocket. Follow this thought experiment:

You get a 1000# force placed at the bottom of your feet upward. You are held in place with a plate over your head. Your are going to feel alot of stress and the weakest bone from your head to feet may break.

If this plate is removed you will be accelerated upward. The very top of your head will feel no stress. Your neck will feel more, your back will feel even more, your lower legs will feel alot and your feet will feel the same stress as in the held position.

That is why airframe failures occur close to motor burnout and not at ignition.

jderimig said:

FredA and the 6 PE guys are correct. So is GT. If you assume a massless motor, the CR feels 100% of the motor thrust. For a real system, the centering ring feels T(1-mt/mr) where T is the thrust of the motor, mt is the mass of the motor and mr is the total mas of the rocket.

Force does not get converted into acceleration, force causes acceleration.

As the rocket gains drag, this changes the stress or force distribution along the rocket. Follow this thought experiment:

You get a 1000# force placed at the bottom of your feet upward. You are held in place with a plate over your head. Your are going to feel alot of stress and the weakest bone from your head to feet may break.

If this plate is removed you will be accelerated upward. The very top of your head will feel no stress. Your neck will feel more, your back will feel even more, your lower legs will feel alot and your feet will feel the same stress as in the held position.

That is why airframe failures occur close to motor burnout and not at ignition.

Click to expand...

It might "feel" something, but the CR does not need to resist 100% of the force generated by the motor. It only needs to resist the force of Gravity+mass+drag. when holding the motor from accelerating faster than the rocket.

If this plate is removed you will be accelerated upward. The very top of your head will feel no stress. Your neck will feel more, your back will feel even more, your lower legs will feel alot and your feet will feel the same stress as in the held position.

That is why airframe failures occur close to motor burnout and not at ignition.

Click to expand...

actually, aerodynamic stress would cause it to mostly be felt in the head, and the neck and spine would have to be strong enough... this is what happens under aerodynamic shreds.... There's no reason to shred at ignition, because theres no stress or resistance on the body. ON a 27lb rocket breaking mach, theres +/_ 250lbs of aerodynamic drag assuming a 4"o.d. 3fnc.. that additional resistance is why it shreds, because that's what changes most, not the 10lb drop in mass.

Most of the energy from thrust is seen just above the moment beam... (aka forward most cr.)
this also happens to be near the center of gravity, and near the center of pressure... so.. all things in the right boat and its got to be pretty strong.

ClayD said:

It might "feel" something, but the CR does not need to resist 100% of the force generated by the motor. It only needs to resist the force of Gravity+mass+drag. when holding the motor from accelerating faster than the rocket.

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Clay, it has to resist exactly T(1+mt/mr). If the mass of the rocket is very large compared to the mass of the motor then it has to support almost the entire thrust of the motor. This is basic Newtonian physics. GT showed the equations in his post.

jderimig said:

Clay, it has to resist exactly T(1+mt/mr). If the mass of the rocket is very large compared to the mass of the motor then it has to support almost the entire thrust of the motor. This is basic Newtonian physics. GT showed the equations in his post.

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the centering ring will never never have to resist more than total of aerodynamic drag plus the mass of the rocket. I understand there are shock loads and mass distributions that change things. But, if a motor puts out 100lbs of thrust for 1.5 seconds duration. How would a 6 lb rocket see 100lbs of force on one component?

ClayD said:

the centering ring will never never have to resist more than total of aerodynamic drag plus the mass of the rocket. I understand there are shock loads and mass distributions that change things. But, if a motor puts out 100lbs of thrust for 1.5 seconds duration. How would a 6 lb rocket see 100lbs of force on one component?

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The centering ring must resist drag and the inertial force from the rocket which is m_rocket*(acceleration+1g). That's it. The other inertial force residing the thrust is m_motor*(acceleration+1g). The sum of those three terms totals up to the current thrust of the motor.

In a big rocket like this one, the mass of the rocket is much larger than the motor, so the first two terms dominate (read as: sum up to 90+% of the thrust) and thus we approximate by saying that the centering rings must withstand the thrust of the motor.

G_T said:

Add in one person with a little extremely rusty physics background.

F=MA

Let's say: T instantaneous thrust. D instantaneous aero drag. V instantaneous vehicle mass. M instantaneous motor mass. a is acceleration.

1: T-D = (V+M)a
or rephrased,
2: a = (T-D)/(V+M)

What is the force seen at the thrust transfer point between the motor and the vehicle? That force equals the drag force plus the force due to acceleration of the vehicle mass, but NOT the force due to accelerating the motor mass.

Let's say: f is the force transferred from the motor to the airframe.

3: f = D + Va

Gerald

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Hi Gerald,
Nice explanation. I can follow the math, but I'm pretty rusty on my physics these days...
Could you or someone else explain a little further why the force acting on the transfer point does not include the force due to accelerating the motor mass?
Thanks,
Dave

dvdsnyd said:

Hi Gerald,
Nice explanation. I can follow the math, but I'm pretty rusty on my physics these days...
Could you or someone else explain a little further why the force acting on the transfer point does not include the force due to accelerating the motor mass?
Thanks,
Dave

Click to expand...

Because the motor and transfer point (rest of the rocket) have the same acceleration. Replace the motor with a vector force and that force is Fr = Mr*a.
The net force on the motor is Fm = Mm*a.
Fm+Fr = motor thrust, Mr+Mm = total rocket mass. a = Thrust/total rocket mass.