## Austin's Math

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### Re: Austin's Math

Polly's flywheel on a train is an excellent analogy! This might be the easiest approach to tackle this.

What we have? Groundtrack and altitude information. Let's use this.

In a vertical cylindrical coordinate system there is a linear axis vertically, the "train", and radial system horizontally, the "flywheel".
For the moment constant altitude change is assumed, so 1g in this direction.
On the flywheel it is classic centripetal ar=Vt²/r. Note, that Vt is meant to be tangential velocity, so the groundspeed.
Then it a simple addition of vectors, into which changes of vertical or tangential speeds can easily be incorporated.
Once again a drawing would have been easier than words.

Excellent video and a classic of its own. Now I know how to impress in the gym! Ysop

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### Re: Austin's Math

So can I assume you don't agree with my maths ?

And that as you consider it an excellent video you agree with it ?
"Stay away from negative people.They have a problem for every solution." - Albert Einstein
S&J

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### Re: Austin's Math

If the kinematic description is off, the result will be off.

No, I was referring to Alan's post. Ysop

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### Re: Austin's Math

I don't get the impression people are very interested in actually understanding this as the relevant argument keeps being ignored, but alas - other than the enthusiastic people interested in math would suggest, this actually is the key to understanding the issue In a vertical cylindrical coordinate system there is a linear axis vertically, the "train", and radial system horizontally, the "flywheel".
For the moment constant altitude change is assumed, so 1g in this direction.
On the flywheel it is classic centripetal ar=Vt²/r. Note, that Vt is meant to be tangential velocity, so the groundspeed.
Then it a simple addition of vectors, into which changes of vertical or tangential speeds can easily be incorporated.

The 'simple addition of vectors' you're invoking here is known as Galilean Invariance - physics is the same in two inertial systems, and so you can compute in the easy inertial system and then apply a x2 -> x1 + v*t transformation to go in an inertial system that moves with a relative velocity. At schools we train people to attack problems like this, but...

For aerodynamics Galilean invariance cannot be invoked like this.

A system in which you're at rest with the air isn't just as force-free as one in which you're moving with a constant velocity because air exerts forces

Consider a balloon floating in still air - its buoyancy balances gravity, that's all. Now transform to a system in which the balloon moves downward at Mach 2 through the air- far from being balanced, it will be ripped to shreds because the second system isn't just the same good old force free inertial system.

So we cannot do the simple addition of vectors because there will be additional forces.

How does that work?

Consider a plane that circles at constant altitude, with the wings an unrealistic 90 degree bank angle. There's some alpha angle, but no beta angle. Superimpose a downward motion at constant speed - suddenly there's a beta angle that lets a side-force appear, and if the plane is sufficiently well-shaped, there's a force driving beta to zero. So you can't simply fly in any attitude that happens because you changed coordinate system - if you change the relative motion through the air, additional forces and moments come up.

So there's no two valid coordinate systems in which the problem can be analyzed, there's only the one in which air is at rest, and the situations 'plane flies a circle at constant altitude' and 'plane flies a descending circle' are different because there are additional aerodynamic forces in the second situation that do not appear in the first.

Now we imagine a situation in which the plane flies straight at 120 kt, or it flies descending at 200 kt (in each case alpha and beta = 0 because of the forces which drive these to zero). Which of the trajectories can be bent into a circle with less g-force (whatever course change you make, you need to add a velocity component 90 deg to your current velocity)?

Standing in a train, facing the loco, you measure the force on the spring balance needed to spin the weight at 100 rpm. By your reasoning the value would be different when the train is moving at 120mph compared to when it was standing at the platform.

Ah, but the train moves the air inside it - which restores Galilean invariance. Try standing on the roof of the moving train in the wind and whirling something around and compare with someone standing at the station...
Thorsten

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### Re: Austin's Math

Actually it only a kinematic problem. There is a velocity vector from the trajectory and acceleration-vector is time derivative of velocity.
I would shift all considerations about air and forces to the very end, when trying to answer, if the wing can produce such an acceleration. Ysop

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### Re: Austin's Math

Actually it only a kinematic problem.

And that is why exactly?

What you're saying is that a balloon can move at Mach 2 vertical velocity without feeling any force because it's only a kinematical problem. No, it's not, as simple as that. You can of course derive a solution for space - but that's not the original problem. The problem 'as posed' and the problem 'in space' are fundementally different.
Thorsten

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### Re: Austin's Math

If you had a balloon moving at speed, Newtons laws say that it would continue at that speed until a force was applied.
At Mach2 in the atmosphere there would be an enormous drag force, so it would slow down rapidly. (and probably burst under the forces)
Mach2 in space is meaningless, as the speed of sound is zero,
Alant

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### Re: Austin's Math

Ysop wrote in Sun Mar 12, 2023 9:19 am:Actually it only a kinematic problem. There is a velocity vector from the trajectory and acceleration-vector is time derivative of velocity.
I would shift all considerations about air and forces to the very end, when trying to answer, if the wing can produce such an acceleration.

Exactly this!

The original problem was:
he calculates the G-force on a PC-12 doing a spiral dive from the radius of the ground track but he uses the airspeed, from both vertcal and horizontal components, to figure the G-force. Is he correct ? I think only the groundspeed should have been used.

We have a given trajectory x(t), y(t), z(t) and want to calculate the G-force (= net acceleration (force)) that acted on the aircraft. Thats just the second derivative: x_dot_dot(t), y_dot_dot(t), z_dot_dot(t). Since in this case the vertical speed is assumed constant (z_dot(t) = const.) it yields z_dot_dot(t) = 0. That means the constant vertical speed has no influence on the G-force experienced in the aircraft. It only depends on x_dot(t) and y_dot(t) (= groundspeed).
Austin is wrong here.

If there is a balloon that could move at Mach 2 vertical velocity (and keep that velocity) you would measure the same G-force as in a stationary balloon.
amue

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### Re: Austin's Math

Also consider this in support of my argument that the centrifugal forces exerted on the airframe at point of failure is not 14g but in fact 9.1g

And that being that this is NOT the only force being exerted on the airframe. The video seems to conveniently ignore the aerodynamic forces of lift and drag that would be existing.

Now we don't know the AoA so can't work out these values however in simple terms

Total force = lift+ drag+ centrifugal

If we know that the airframe fails at 14g and lift and drag exists then centrifugal must be less than 14g..... Around 9.12 I'd say.
"Stay away from negative people.They have a problem for every solution." - Albert Einstein
S&J

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### Re: Austin's Math

Also consider the 'vomit comet' it enters into a parabolic downwards curve.

The people inside experience zero g..... But that doesn't mean the airframe experiences no stress forces now does it ?

Also

My understanding is that g-forces occur with any changes to inertial travel, be that direction or speed increase\decrease.

It's why g-forces are felt in space, when applying thrust, braking or changes to direction.

Turning in a circle is a constant change of direction and it's why g-forces are felt.

If you go over top on a 'big wheel \ Ferris wheel ' you feel lighter, heavier when you come around the bottom.

So for a downward spiral the vertical component would act as a reduction to the horizontal rotational force developed, resulting in less g-force being felt.

even to me writing it, this sounds bizarre
"Stay away from negative people.They have a problem for every solution." - Albert Einstein
S&J

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### Re: Austin's Math

At Mach2 in the atmosphere there would be an enormous drag force, so it would slow down rapidly. (and probably burst under the forces)

That's the whole point - in air you can't simply make a change to a coordinate system offset at constant speed and not see forces.

If there is a balloon that could move at Mach 2 vertical velocity (and keep that velocity) you would measure the same G-force as in a stationary balloon.

Yet it would be ripped apart. Wasn't the original problem about why the plane was ripped apart (at least that's what I understood...)?

Anyway - if you want to apply Galilean invariance where it doesn't apply, look who doesn't care... Thorsten

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### Re: Austin's Math

Thorsten wrote in Sun Mar 12, 2023 6:30 pm:Wasn't the original problem about why the plane was ripped apart (at least that's what I understood...)?

No. The problem was how to calculate the G-force from the given trajectory and why a uniform (vertical) velocity component in the trajectory has no influence on the experienced G-force.
amue

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### Re: Austin's Math

IF I have clicked into the right spots (those 10min vids for one important sentence are tl,ds), it is about the PC-12 accident caused by either icing or partial disorientation. Austin's conclusion is, that the aircraft experienced 14 Gs, which an iced wing would never haven been possible to generate, so it must have been disorientation. I just leave this conclusion as it is.

edit: Posting crossed with amue. Wording unchanged, as it is complementary. Ysop

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### Re: Austin's Math

he problem was how to calculate the G-force from the given trajectory

As we've established a couple of times and as some of you have even acknowledged, inertial force (deviation from a geodesic trajectory) aren't the only forces acting that can destroy a plane. In the case of the balloon at Mach 2, we have zero inertial force, yet plenty of drag force. S&J seems to say the same thing here

Also consider the 'vomit comet' it enters into a parabolic downwards curve.

The people inside experience zero g..... But that doesn't mean the airframe experiences no stress forces now does it ?

So - what you're saying is that you're not interested in assessing the total forces acting on the plane at all (and none of the calculations has hence any real relevance for why the plane broke apart)?

Okay...

Thing is, things don't break because they are subject to a certain inertial force - they break because of internal stresses that distribute the forces - wing bending moment for instance. And if you try to calculate such stresses, you'll find that to first order they all go like rho v^2 with v being total speed relative to the air. So the seemingly stupid Pythagorean addition of the vertical component is actually what you might be interested in when trying to answer 'why did this plane break up?' The stress on the wing is actually higher when you have a downward velocity than when you just circle.

So there's that...
Thorsten

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