Can Air Resist a Piston? A Serious Discussion

[Sailor Al]

Anderson is fine, and if you have Fundamentals handy (as you seem to based on your link) then it has some good chapters on elementary compressible flow.

The discrepancy to which you allude absolutely can be attributed to a typographical error: I have sitting on my desk right now the 5th edition of said book and the arrow points down in both figures you reference. Some publishing staff member likely just got the direction of the arrow wrong when they prettied up the figure for the new edition. All the equations that follow that figure have the signs correct.

Anderson is by no means perfect, but there's no reason to doubt the credibility of the man's textbooks. They're the de facto standards for teaching introductory aerodynamics in the English-speaking world for a reason.

WOW! can you sen me a scan of that page please? email in my Origins doc. How does he explain how the resultant force is away from the surface when the pressure is towards the surface. There's no explanation in Sixth edition.

 

[Sailor Al]

WOW! can you sen me a scan of that page please? email in my Origins doc. How does he explain how the resultant force is away from the surface when the pressure is towards the surface. There's no explanation in Sixth edition.

I have just downloaded the Fifth edition of Fundamentals It contains exactly the same error:
Figure 1.14 on p. 20 has the pressure arrow TOWARDS the surface,

Figure 1.18 on p. 22 had the pressure arrow AWAY from the surface.

It is not a typo introduced in Sixth edition. I still doubt Anderson's credibility on this subject.

 

[Sailor Al]

This doc.pub is an amazing site - I don't know how it exists.
I have just downloaded the Third edition, and the discrepancy is there as well:

(I'm also very impressed with this forum interface, it's brilliant.)

 

[A.T.]

For a finite tube, of course the shock will eventually exit the tube and allow some pressure equalization

That's what I was originally considering as steady state, including re-circulation outside of the tube.

Assuming an infinitely long tube, this force will be maintained forever, even at steady state.

Yes I agree. But it will be a finite force that is approached in steady state.

 

[Sailor Al]

vdoc.pub is an amazing site - I don't know how it exists.
I have just downloaded the Third edition, and the discrepancy is there as well:
View attachment 325906
(I'm also very impressed with this forum interface, it's brilliant.)

OK, on closer examination, I concede: you may be correct. In the previous editions the direction of the arrowhead was ambiguous as it was partly obscured by the hatching of the background. It does appear that in tidying up the graphics, the ambiguity of the arrow's direction has been incorrectly resolved in the Sixth edition.
But the fundamental question remains unresolved: How does this pressure on the surface resolve to the resultant aerodynamic force R, shown in Figure 1.17? There is no explanation.

 

[A.T.]

But the fundamental question remains unresolved: How does this pressure on the surface resolve to the resultant aerodynamic force R, shown in Figure 1.17? There is no explanation.
View attachment 325911

The absolute pressure is positive everywhere, so the pressure force always points towards the surface. Air molecules cannot pull at the surface perpendicularly to it.

The resultant force R is the vector sum of all pressure forces, from both sides of the airfoil. If they are greater at the bottom than at the top, R points up.

 

[Sailor Al]

The absolute pressure is positive everywhere, so the pressure force always points towards the surface. Air molecules cannot pull at the surface perpendicularly to it.

The resultant force R is the vector sum of all pressure forces, from both sides of the airfoil. If they are greater at the bottom than at the top, R points up.

My point is that Anderson makes no attempt to explain why the pressure, that in all his figures is represented as arrows of equal length pointing towards the surface, is greater at the bottom and lower at the top.
He promises on p. 284:
" Keep in mind that the true physical sources of aerodynamic force on a body are the pressure and shear stress distributions exerted on the surface of the body, as explained in Section 1.5."
But here, in section 1.5, he makes no attempt to explain how the pressure differences arise.
And that's not a slip in the graphical editing of a blurry figure.
Hence my concern about his credibility.

 

[A.T.]

But here, in section 1.5, he makes no attempt to explain how the pressure differences arise.

Explaining why the pressures are different is rather complicated, depending on what "explaining" means. There are endless discussions on what the most fundamental "explanation" of lift is.

On a very general level, if a body is not symmetrical to the relative flow, it just very unlikely that the pressure forces perpendicular to the flow balance exactly.

 

[Chestermiller]

Which bit offended you. Do you think his PDE's are soluble? (There's a $1M prize on offer if you can show that a solution even exists!)
Muddying the waters? Yes, his response was not an answer to the question. That's muddying the waters in my book.
Are you able to answer the questions I raised?

Who says the equation I derived is unsolvable, you? I'll show the derivation of this PDE later.

The solution to this equation for the present situation of a suddenly started piston of velocity v is

$u=v\left(t-\frac{x}{c}\right)$ for x < ct

and

$u=0$ for x > ct

where u(x,t) is the axial displacement of the material cross section that was at axial location x at time t = 0 (so, x = 0 always corresponds tothe piston face, and the actual axial location of the piston face at any time is u(0,t)=vt).

In this solution, the speed of sound c satisfies $$c=\frac{c_0}{(1-\frac{v}{c})^{\frac{\gamma+1}{2}}}$$ with $$c_0=\sqrt{\frac{\gamma RT}{M}}$$

This solution satisfies all the features of behavior that you were expecting, and, from this solution, we can also easily determine the pressure on the piston face.

 

[Sailor Al]

Explaining why the pressures are different is rather complicated, depending on what "explaining" means.

Explaining why the pressures are different is the subject of my paper "[Link to unpublished paper redacted by the Mentors]" . It really quite simple and nothing to do with Bernoulli, Newton, streamlines, flow fields and PDEs.

 

[Sailor Al]

Who says the equation I derived is unsolvable, you? ....

This solution satisfies all the features of behavior that you were expecting, and, from this solution, we can also easily determine the pressure on the piston face.

Please, show me a worked solution to the piston in a cylinder problem posed at the start. Use any cylinder area, say 1 sq m, piston velocity Mach0.5, others as specified.

 

[Sailor Al]

Back to the question of force...
If you know the shock Mach number, $M_s$, that uniquely determines the pressure ratio across the shock (for a given gas), $p_2/p_1 = f(M_s,\gamma)$. If you have pressure and area, then you have force.

Some good sources
[1] Liepman, H. W., Roshko, A.1957. "Elements of gasdynamics." Dover.
[2] Anderson, J. D. 2003. "Modern Compressible Flow with Historical Perspective." McGraw-Hill.
[3] https://www.grc.nasa.gov/www/k-12/airplane/normal.html

I have spent an hour speed reading Liepman and trying to get a sense of the subject. Having skimmed over pistons in closed cylinders and pipes with varying cross-sections, I arrived at Chapter 3 "One-Dimensional Wave Motion" and thought, "That sounds like my problem" but my brain went pop! when I got to para 4 :

"The fluid flows through the shock with speed u_a
What does that even mean?

 

[Chestermiller]

Please, show me a worked solution to the piston in a cylinder problem posed at the start. Use any cylinder area, say 1 sq m, piston velocity Mach0.5, others as specified.

Whoa pardner. Hold yer horses. Your 60 years of experience (I too have 60 years under my belt) should tell you that good modeling practice requires proceeding in small bite-sized chunks, gradually adding complexity. We haven't even reached agreement on the solution I presented, and you haven't even seen the derivation of my PDE yet. So, what do you want to do first (a) test the solution I presented to make sure you are comfortable that it satisfies the differential equation and boundary conditions or (b) have me present the derivation of the PDE for your consideration?

 

[snorkack]

Consider the case of an infinite (and non-expandible) tube in which the gas travels at a steady velocity, with negligible head loss. And the velocity of the gas is small relative to speed of sound.
Now suddenly put the piston in the way of the tube by way of closing a valve.
The air right upwind of the piston is instantly stopped.
The air hammer will travel upwind along the tube from the piston at the speed of sound - but since the initial velocity of the airflow was small, the pressure jump at air hammer will also be small. It will travel upwind at speed of sound, not faster - and it will not be a true shockwave. At the air hammer front, the kinetic energy of air moving at a slow speed is converted to potential energy of the stationary compressed air. But the force from the air hammer will be steady.
Since the piston/valve is in the middle of the tube, a second air hammer will travel downwind with equal speed and exert equal force, just in suction.

Now what happens if you start with stationary air in a long tube and set the piston in motion?
Same. Air hammers will propagate upwind and downwind with the same speed and force as if the air had been moving and piston stationary. The forces on piston will only change if and when either of the air hammers will have reflected from somewhere and travelled back to the piston.

 

[Chestermiller]

The Answer you got at PhysicsStackExchange is the simplified version of the result I presented. Notice their term $t-\frac{x}{c}$.

 

[A.T.]

Explaining why the pressures are different is the subject of my paper "[Link to unpublished paper redacted by the Mentors]" . It really quite simple and nothing to do with Bernoulli, Newton, streamlines, flow fields and PDEs.

From your paper: "..., no attempt has been made to ascribe values to the size of the pressure variations....."

While your simple idea is not completely wrong, it's not the full picture, and doesn't by itself provide the correct quantitative results in general.

 

[boneh3ad]

Consider the case of an infinite (and non-expandible) tube in which the gas travels at a steady velocity, with negligible head loss. And the velocity of the gas is small relative to speed of sound.
Now suddenly put the piston in the way of the tube by way of closing a valve.
The air right upwind of the piston is instantly stopped.
The air hammer will travel upwind along the tube from the piston at the speed of sound - but since the initial velocity of the airflow was small, the pressure jump at air hammer will also be small. It will travel upwind at speed of sound, not faster - and it will not be a true shockwave. At the air hammer front, the kinetic energy of air moving at a slow speed is converted to potential energy of the stationary compressed air. But the force from the air hammer will be steady.
Since the piston/valve is in the middle of the tube, a second air hammer will travel downwind with equal speed and exert equal force, just in suction.

Now what happens if you start with stationary air in a long tube and set the piston in motion?
Same. Air hammers will propagate upwind and downwind with the same speed and force as if the air had been moving and piston stationary. The forces on piston will only change if and when either of the air hammers will have reflected from somewhere and travelled back to the piston.

What you've described is the inverse of the piston problem I described and only varies from it by a change of reference frame. The "air hammer" you describe is still a shock and is still supersonic, per the equations I derived.

On the other side, an expansion wave is generated, lowering pressure. That one does travel sonically.

Also, there's no such thing as a suction force. All pressure forces act toward a surface.

 

[boneh3ad]

My point is that Anderson makes no attempt to explain why the pressure, that in all his figures is represented as arrows of equal length pointing towards the surface, is greater at the bottom and lower at the top.
He promises on p. 284:
" Keep in mind that the true physical sources of aerodynamic force on a body are the pressure and shear stress distributions exerted on the surface of the body, as explained in Section 1.5."
But here, in section 1.5, he makes no attempt to explain how the pressure differences arise.
And that's not a slip in the graphical editing of a blurry figure.
Hence my concern about his credibility.

That's because the explanation is not simple. Conceptually, you can make claims about compression and rarefaction, but that's not the full story. If you want to know how those pressure changes occur and what they are quantitatively, the answer isn't simple anymore. Various concepts throughout that book need to be used together to do the actual aerodynamic analysis of a wing (or sail).

 

[boneh3ad]

I have spent an hour speed reading Liepman and trying to get a sense of the subject. Having skimmed over pistons in closed cylinders and pipes with varying cross-sections, I arrived at Chapter 3 "One-Dimensional Wave Motion" and thought, "That sounds like my problem" but my brain went pop! when I got to para 4 :
View attachment 325929
"The fluid flows through the shock with speed u_a
What does that even mean?

That's a simple change in reference fame. In one, the shock propagates through stationary air with velocity $u_1$. In the other, you subtract that velocity so the shock is stationary and the flow appears to flow through it with velocity $u_1$. The later is the most straightforward problem mathematically since you've made it steady.

 

[snorkack]

What you've described is the inverse of the piston problem I described and only varies from it by a change of reference frame. The "air hammer" you describe is still a shock and is still supersonic, per the equations I derived.

On the other side, an expansion wave is generated, lowering pressure. That one does travel sonically.

Also, there's no such thing as a suction force. All pressure forces act toward a surface.

Certainly in gas. So dropping the approximation of constant ambient pressure...

I am going to use $v$ to represent velocities in the lab reference frame and $u$ to represent velocities in the frame of the compression wave ahead of the piston.

Case 1: Impulsively-started piston
Consider first the impulsively started piston. It is initially at rest, and, at some time $t=0$, it is impulsively accelerated to a velocity of $v_p$ (moving left to right, per the diagram in the original post). No penetration requires that no fluid crosses the face of the piston, so locally at the piston face, the flow is moving with $v=v_p$, but in the far field it is still moving at $u = 0$, creating a velocity discontinuity. It is a classical result that such a change in a 1D duct occurs discontinuously, i.e., as a result of a shock. This shock propagates with a speed $v_s$.

Consider a change in frame of reference such that the coordinate system follows along with the shock, i.e., subtract $v_s$ from everything. Then the velocity upstream of the shock is $u_1=0-v_s=-v_s$ and downstream of the shock is $u_2 = v_p- v_s$. One of the standard methods for determining the Mach number of a shock is to look at the velocity ratio across it (in its reference frame), which follows the relation
$$\dfrac{u_2}{u_1} = \dfrac{2 + (\gamma-1)M_1^2}{(\gamma + 1)M_1^2}$$
or
$$\dfrac{v_p- v_s}{-v_s} = \dfrac{v_s-v_p}{v_s} = 1 - \dfrac{v_p}{v_s} = \dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2}.$$
From there,
$$\dfrac{v_p}{v_s} = 1 -\dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{(\gamma + 1)M_s^2 - 2 - (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s^2} \right)$$
Note that $M_s = v_s/a_1$ where $a_1$ is the speed of sound in the stagnant air in the tube, so you can eliminate $v_s$ above and end up with
$$\boxed{\dfrac{v_p}{a_1} = M_p = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s} \right).}$$

Here, $M=u/a$ is the Mach number, $\gamma$ is the ratio of specific heats, and $a$ is the local speed of sound. You can uniquely determine the Mach number of the shock that propagates away from the piston based on the velocity of the piston and the conditions in the tube at the beginning. Indeed, the term $M_p = v_p/a_1$ represents a Mach number for the piston relative to the calm air in the tube (i.e., the lab frame).

You can easily solve the boxed equation for $M_s$ for a given piston Mach number.
$$M_s^2 - \left(\dfrac{\gamma + 1}{2}\right)M_p M_s - 1 = 0.$$
That's a straightforward quadratic equation. If you pick a very low piston Mach number, say, $M_p=0.1$, you still get a supersonic shock Mach number of $M_s=1.06$.

A fundamental truth about shocks is that they increase the temperature, pressure, and density of the air into which the propagate. Thus, the piston face experiences a pressure greater than ambient even if it moves at constant velocity, and a constant force is required to maintain that constant velocity (assuming the tube is sufficiently long).

If it seems odd that the shock is supersonic, that's likely because the shock Mach number is measured based on its speed relative to the medium into which it is propagating (here, the stagnant air). But the shock is propagating through the heated air behind it, which has a higher temperature and, thus, a higher speed of sound. It is still subsonic relative to that medium.

Some good sources
[1] Liepmann, H. W., Roshko, A.1957. "Elements of gasdynamics." Dover.
[2] Anderson, J. D. 2003. "Modern Compressible Flow with Historical Perspective." McGraw-Hill.
[3] https://www.grc.nasa.gov/www/k-12/airplane/normal.html

But the total force on the piston, as pointed out, is the difference of the forces exerted by the compression wave on upwind side and the expansion wave on downwind side. So does the mathematics of the downwind expansion wave differ? An expansion wave travels into warmer air and makes it colder (thus diminishing the sound speed in it). Does it mean that an expansion wave travel exactly at the speed of sound downwind of it - neither faster nor slower?

 

[boneh3ad]

Certainly in gas. So dropping the approximation of constant ambient pressure...

But the total force on the piston, as pointed out, is the difference of the forces exerted by the compression wave on upwind side and the expansion wave on downwind side. So does the mathematics of the downwind expansion wave differ? An expansion wave travels into warmer air and makes it colder (thus diminishing the sound speed in it). Does it mean that an expansion wave travel exactly at the speed of sound downwind of it - neither faster nor slower?

Yes, the mathematics of the expansion wave are different. They'd be related to the Prandtl-Meyer expansion.

 

[Sailor Al]

So, what do you want to do first (a) test the solution I presented to make sure you are comfortable that it satisfies the differential equation and boundary conditions or (b) have me present the derivation of the PDE for your consideration?

Actually I have little experience in modelling practice.
I'm still stuck at understanding why this is a modelling process at all. Is it not a relatively simple one-dimension mechanics problem?

We have all the rules:
F = m * a, a continuum which obeys PV/T = C, we have the speed of the piston v, the speed that the pressure moves, the density of the fluid at NTP, c and the area of the piston.

I don't understand why it's not a relatively simple exercise to build an equation of the pressure in the fluid at time t and position x. Then solve it for x = v * t to get the pressure at the face of the piston and thus the force required to maintain the piston speed at that time.

And I think it can be done without reference to challenging concept of the velocity of the fluid. I've read Brenner's paper* that points out the different interpretations of that term: Lagrangian velocity, and volume velocity. I am of the opinion that the concept of fluid flow velocity is sufficiently challenging that all attempts should be made to avoid any such reference.

Having said my piece, @Chestermiller, if you are confident that following your option a) or b) will lead to the solution I have described above, then please proceed in the sequence you consider most effective.

* I hope this doesn't get me a second warning about referencing an unpublished paper. It's on the Ohio State University website.

 

[Sailor Al]

From your paper: "..., no attempt has been made to ascribe values to the size of the pressure variations....."

While your simple idea is not completely wrong, it's not the full picture, and doesn't by itself provide the correct quantitative results in general.

Absolutely!
As it stands, and as is candidly declared in the paper, it does not provide any quantitative results.
And that is precisely the reason for this thread.
I want to gain an understanding of how the shape and speed of the foil affects the size of the pressure on the sail's surface. I'm starting out small with my infinite cylinder, then I'd like to develop into a more and more realistic environment. I know I don't have the skills. I'm hoping someone brighter and younger can show some interest and come up with some answers.
First of all I have to convince you that the existing literature is incomplete. Maybe that will stir some minds into action.

 

[Sailor Al]

That's a simple change in reference fame. In one, the shock propagates through stationary air with velocity $u_1$. In the other, you subtract that velocity so the shock is stationary and the flow appears to flow through it with velocity $u_1$. The later is the most straightforward problem mathematically since you've made it steady.

I have no problem with the pressure wave moving relative to the cylinder. The cylinder is a solid, nonrotating object. In any (inertial) frame of reference, all parts of the cylinder are moving at the same speed.
The air is a compressible fluid - a continuum. All parts of it may be moving at different speeds. The concept of stationary fluid is not so challenging. None of it is moving, no problem. But the speed of a moving, compressible fluid under the influence of a pressure wave is what makes my brain go pop!
Liepman's statement that the fluid flows through the shockwave with speed u1 cannot be allowed to pass unchallenged.

 

[berkeman]

Liepman's statement that the fluid flows through the shockwave with speed u1 cannot be allowed to pass unchallenged.

Perhaps "unchallenged" is not quite the best term for you to use in this situation...

 

[Sailor Al]

Now what happens if you start with stationary air in a long tube and set the piston in motion?
Same. Air hammers will propagate upwind and downwind with the same speed and force as if the air had been moving and piston stationary. The forces on piston will only change if and when either of the air hammers will have reflected from somewhere and travelled back to the piston.

I have no issue with your upstream “suction” wave. It’s just a way of describing a low pressure wave. The downstream wave is a high pressure wave propagating downstream.
What I’m after is a mathematical function that will solve for the pressure on the piston at time t.

 

[boneh3ad]

I have no problem with the pressure wave moving relative to the cylinder. The cylinder is a solid object. In any (inertial) frame of reference, all parts of the cylinder are moving at the same speed.
The air is a compressible fluid - a continuum. All parts of it may be moving at different speeds. The concept of stationary fluid is not so challenging. None of it is moving, no problem. But the speed of a moving, compressible fluid under the influence of a pressure wave is what makes my brain go pop!
Liepman's statement that the fluid flows through the shockwave with speed u1 cannot be allowed to pass unchallenged.

Again, it's a simple change of reference frame. I don't know how else to explain it other than that. A shock can propagate with a velocity past a stationary observer. The frame of reference can also be changed such that the observer moves along with the shock, making the shock appear stationary with the flow passing through it.

Even then, it's actually very easy to create a stationary shock in nature with air flowing through it. Take, for example, any supersonic wind tunnel. Mounting an object in the flow will result in shock waves.

Here is an example of a model aircraft mounted in a supersonic wind tunnel producing oblique and conical shock waves captured via schlieren imaging:

Here is a similar pair of schlieren images of a supersonic inlet in a wind tunnel. This time, the unstarted inlet produces locally normal shocks that are stationary in the lab frame.

 

[boneh3ad]

I have no issue with your suction wave. It’s just a way of describing a low pressure wave. It’s a high pressure wave propagating downstream.
What I’m after is a mathematical function that will solve for the pressure on the piston at time t.

I gave you all the tools you need for that with my post related the piston Mach number and the shock Mach number. You may be able to come up with a closed-form solution by solving that quadratic equation and substituting it into the pressure ratio across a shock of Mach number $M_s$ (available in all of the references I previously gave you).

But aside from all of this, the bigger problem is that this has nothing to do with sail boats or airfoils. Those don't operate in environments with solid walls and no opportunity for air to move around them. And sailboats especially do not generate shock waves.

 

[Sailor Al]

Again, it's a simple change of reference frame. I don't know how else to explain it other than that. A shock can propagate with a velocity past a stationary observer. The frame of reference can also be changed such that the observer moves along with the shock, making the shock appear stationary with the flow passing through it.

A physicist has no problem with the shock wave appearing stationary w.r.t. the cylinder.
But not “the flow passing through it”
As stated, a stationary flow is one thing: all parts of it are stationary. But in the problem context, the flow is a compressible fluid subject to varying pressures. The step from being stationary and moving is not a simple case. When stationary, all parts of the fluid have the same zero velocity. When subject to the pressure of the piston, this homogeneity is destroyed.
I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

 

[berkeman]

I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

Which are you? I'm not finding your educational background in your Profile page. Clarifying that would help us a lot to try to answer your questions...

 

[Sailor Al]

And sailboats especially do not generate shock waves.

I have not propose a “shock wave”, that has serious supersonic connotations.
But a sailboat certainly does create pressure waves.
Go race a sailboat and experience the pressures generated by nearby boats: wind shadow, lee-bow effect, dirty wind astern and leeward etc. These are pressure variations which move propagate through air and move with the boat. That is a pretty good description of pressure waves.

 

[Sailor Al]

Which are you? I'm not finding your educational background in your Profile page. Clarifying that would help us a lot to try to answer your questions...

My BSc. is in Mathematical Physics, Birmingham UK, 1967. I am a retired but still very active software developer, private pilot and enthusiastic keelboat racer.

 

[berkeman]

My BSc. is in Mathematical Physics, Birmingham UK, 1967. I am a retired but still very active software developer, private pilot and enthusiastic keelboat racer.

I'm not familiar with the Mathematical Physics degree in the UK, but why are you saying that such a fundamental concept is a show-stopper for you with your background? Is there something in the mathematics presented so far in this thread that is problematical?

 

[Sailor Al]

I'm not familiar with the Mathematical Physics degree in the UK, but why are you saying that such a fundamental concept is a show-stopper for you with your background? Is there something in the mathematics presented so far in this thread that is problematical?

Yes, I think I have expressed it pretty clearly in my last reply to @boneh3ad .

 

[boneh3ad]

A physicist has no problem with the shock wave appearing stationary w.r.t. the cylinder.
But not “the flow passing through it”
As stated, a stationary flow is one thing: all parts of it are stationary. But in the problem context, the flow is a compressible fluid subject to varying pressures. The step from being stationary and moving is not a simple case. When stationary, all parts of the fluid have the same zero velocity. When subject to the pressure of the piston, this homogeneity is destroyed.
I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

It is a simple case, actually. So simple, that physicists have a name for it: a Galilean transformation.

Not only is this not a contested process (by engineers or physicists) but its accuracy has been confirmed by the successful design of rockets, supersonic airplanes, and engines of all kinds relying on this exact type of analysis with shocks.