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Thanks for watching and the kind words. I am Australian though 🐨

An air molecule is very different from an entire automobile.

Thanks for the kind words and thanks for watching. The critical variable is the pressure ratio, rather than the exact value of the inlet or outlet pressure. So yes, similar transitions / changes could be observed by varying either the inlet or outlet pressure while holding the other constant.

Thanks for the kind words and thanks for watching!

Thanks for the kind words and thanks for watching!

Thank you and thanks for watching

Glad the video was helpful! Thanks for watching

Glad it was helpful!

Thanks for watching and the kind words!

yes because of the mesh that was shown at the start. there it cancels out, but only because the fake/imaginary value from the rotation is added sun = earth + rotation

Thanks for watching, and apologies for the very slow reply. This one slipped through the cracks. Orbiting of un-stable Lagrange points is yet another thing on my list of things to cover when I have time to make more videos. I think a useful (if not rigorously correct) way to think about it is like the difference between an orbiting and non-orbiting spacecraft around Earth. A sub-orbital vehicle that wants to stay at a fixed point in space will have to constantly fire its thrusters to stop from falling back down and to counteract any perturbing forces. Look up some videos of multiple kill vehicles for a visualisation of this. On the other hand, an orbiting spacecraft is expending no energy to remain in orbit. It just has a velocity and Earth's gravity is constantly bending its trajectory in a way that it just keeps rotating around Earth. All it has to do is make small station keeping / correction burns from time to time to balance out disturbances. Likewise at an unstable Lagrange point. If you aim to 'hover' at exactly the point, you will frequently have to thrust against disturbances before they become run-away instabilities (like shown in this video). These kind of thrusting maneuvers directly against the direction of motion take more energy than glancing re-direction burns. Kind of like in cricket or baseball how hitting the ball directly back where it came from is much harder than lightly redirecting it. An orbit around a Lagrange point is kind of like an orbit around Earth, but with different forces (in the rotating reference frame). Instead of gravity deflecting the velocity, it's the Coriolis and centrifugal effects, and so the orbit shape looks more like a bean than a circle. Again, like gravity at Earth, we're letting these other forces do most of the work, and just making small corrections from time to time. Let me know if that helps!

@@TheGravityAssistant Yes, that was helpful. Thanks for taking the trouble to reply.

Thank you! It's on my to-make wish list, but unlikely to happen any time soon unfortunately. My understanding of NRHO orbits at the moment is not deep enough for me to concisely explain them in a coherent way. The things higher up the list are topics I'm in a better position to explain!

Thanks very much for the kind words. All the best to you too

Hi, atmospheric pressure at the exit can be reduced by increasing altitude (e.g. a rocket or plane flying higher). Alternatively by conducting a rocket engine test in a vacuum chamber, or with a pressure reducing device such as an inducer. You could instead imagine that we increase the pressure of the fluids going into the engine, as it is the ratio between fluid total pressure and atmospheric pressure that matter. What was confusing about shocks, could you elaborate a bit more please?

12:58 since u said shock is an inefficient compression process, and since shock travels in opposite direction to the flow in the nozzle, how does it's increase help match the nozzle exit pressure to the atmospheric pressure p.s- what is supersonic expansion

Apologies for the slow reply. Can you please restate your questions, I don't really understand what you're asking.

@@TheGravityAssistant how does increase in shock help exit pressure to match atmospheric pressure

Thanks for watching, glad it was helpful

Thanks for the kind words! I have plans to make more videos, when life permits, unfortunately can't say when..

Thanks for watching!

Thanks for watching, very glad to hear that it was helpful!

Very glad to hear that :). Thanks for watching!

Thanks for the kind words and for watching!

Thanks for watching and the kind words. I am in the process of making more videos, but unsure when I will be able to complete them unfortunately. Work and life is keeping me very busy these days

What is important is the ratio between the pressure in the combustion chamber (high pressure) and the ambient (low) pressure at the exit. If you want to achieve supersonic flow at the exit, you can increase the high pressure or decrease the ambient pressure (or both). You're right that if we want a rocket nozzle to work at sea level, it doesn't make sense to try and lower the environment pressure - so first stage rocket engines usually have very high combustion chamber pressures to achieve a large pressure ratio. Conversely, 2nd and 3rd stage engines which operate in near-zero pressure environments can have much lower chamber pressures and still achieve supersonic flow / a high pressure ratio. Anything divided by almost zero is very big.

@@TheGravityAssistant I see, that makes intuitive sense in retrospect. If the starting total pressure is higher, to the point that the reduced static pressure is close to atmospheric conditions while holding onto a lot of dynamic pressure, it would just keep moving through similar conditions as though not much had changed. If the static pressure is similar but the temperature is much higher, does that mean the compressed gas is actually released in the exhaust at densities lower than STP? I’ll have to try to wrap my head around that… I’ve been trying to look into a means of numerically estimating the motion of airflow based on pressure differentials and figuring out how far an open air supersonic flow will continue traveling at supersonic speeds before reaching equilibrium with normal air, and I’m not quite sure what-in open air away from the turbine-stops a flow from forming normal shock outside of just powering through it with high pressure…? And how the duration of that air burst impacts this. Do you have any suggestions on where I should look to that? I’m not really sure how to mathematically define the behavior of a compressible thermodynamic gas to move from a region of high pressure to low pressure in the first place. >.<

@@Goji-Moji To answer your first question, the ideal gas law PV = nRT can be rearranged to P = rho * R * T, where rho is density and R = R_universal / M_molecule. From that we can see that yes, gasses at higher temperature for a given pressure will have lower density. It also makes sense when you think about a candle or a camp-fire, their combustion is happening at 1 atmosphere of pressure, and the heated gasses float away because of their reduced density (and the effect of buoyancy). I'm not sure I totally understand your second question. I'm going answer the question 'how does a supersonic jet stream break down and dissipate', and hope that helps :). When the supersonic stream exits a nozzle, it is at the same static pressure as the atmosphere, but it is has a much high velocity than the ambient air, so when it moves past there will be a large shear force. This shear force acts to slow down the jet, (and accelerate the ambient air a little bit). Viscous shear and turbulence dissipates the supersonic jet's high dynamic pressure. Have a look at figure 12 and 13 of this paper: iopscience.iop.org/article/10.1088/1742-6596/1240/1/012019/pdf And look at the bottom of the exhaust stream from this photo: space.stackexchange.com/questions/29758/temperature-and-pressure-of-rocket-exhaust Similarly, the temperature will eventually equalise once the hot exhaust gases have radiated or convected their excess heat away to the rest of the atmosphere.

That's for the kind words and thanks for watching!

Thanks for watching! I have more videos planned and some draft scripts written. The problem is finding the time to make them :(

Read the other comments and youll have your answer....

Unfortunately it doesn't work in KSP (at least it hasn't work when I have tried). I suspect it is because of the basic aerodynamics model in KSP. AGA really needs a good supersonic / hypersonic model in order to work.

@@TheGravityAssistant what about with Ferram Aerospace Research installed?

@@nikelinq2899 I'm not familiar with how FAR models hypersonic flight. If you feel like giving it a go, I'd be very interested to hear the outcome!

Hi, thanks for watching! To answer your questions: 1 - Lagrange points don't 'have' any gravity of their own, as they are just points in space with no mass. The two bodies (e.g. the Sun and the Earth) have mass and so generate gravitational fields. The Lagrange points are just empty points in space where the two gravity fields of the bodies interact in a way that allows a third body with a small mass to orbit around the main body in a way that would not 'normally' be possible if there was no second body. 2 - No - similar to point 1, the Lagrange points do not attract / repel / interact with anything as they are just empty points in space that are a result of two interacting gravity fields. It is the behaviour of the gravitational fields around a Lagrange point that causes the peculiar motion of the third body. The gravity of each of the planets however, must be taken into account when predicting the orbits of the other planets, as the gravitational fields of each planet propagate infinitely and interact and change the motion of each of the planets. 3 - No, because Lagrange points don't have any mass (and therefore gravity) of their own, they will not cause lensing.

I'm sorry, I don't understand your question. Could you please clarify what you mean?

@@TheGravityAssistant I'm guessing that Rat King is referring to the graph at 7:27 looking like a tanh(x) curve, and the graph at 10:38 looking like a cosh(x) curve, and wondering why those two functions describe the situation..

@@carultch Yes, that is what im trying to say, Why does this flow structure mimic these hyperbolic curves?

Hi, thanks for watching. I hope that I am understanding your first question correctly - it is not possible/practical in reality to match the exit static pressure of the nozzle with the vacuum pressure (~0) in space. It is possible to calculate a maximum theoretical exit velocity from a nozzle where all the thermal energy is converted to kinetic energy and the exit temperature is 0 K (and therefore the exit static pressure is 0), but this would require a very long (and heavy) nozzle with a very big expansion ratio. Additionally a real gas would condense into liquid or solid at a temperature >0 K and so the maximum ideal gas expansion could not be achieved. Additionally + 1, viscous losses with the long nozzle wall would remove some amount of energy from the flow, further preventing us reaching this theoretical idea. In practice, all space nozzles are under-expanded, and the designer has to make a trade-off between factors such as performance (Isp), mass, packaging (length and diameter) etc. etc. Regarding the plots of Mach / pressure / temperature - I made them in matlab using the geometry of a simple converging-diverging nozzle, and the isentropic flow relations (please see the link below). If you know the cross section area at each position along the nozzle, you can use equation 9 (the Mach-area relation) to calculate the Mach number at each position. Once you know the Mach number at each position along the nozzle, you can use equations 6 and 7 to calculate the temperature and pressure at each location. www.grc.nasa.gov/www/k-12/airplane/isentrop.html Hope that helps!

Thank you! Thanks for watching

my friend gold is visible only for the ones who search for it

Not many people want to learn rocket science i guess..

Thanks for watching and the kind words. That's an interesting question and I had to take some time to think about it. On a fundamental level, the most important variable that we can control is the pressure ratio between the total pressure in the chamber and the pressure of the environment at the exit. Please see the image I linked at the end of this message. This following paragraph assumes that we have a constant nozzle geometry, like shown in the linked image. At point 7 we have perfectly expanded flow at the exit and no shocks. As we move upwards to point 6, the environment pressure is a little bit higher than the final nozzle pressure, and so some 'weak' oblique shocks are needed to match the exit pressure to the environment pressure. As we increase the environment pressure more, the weak oblique shocks eventually turn into a 'strong' normal shock at point 5. This shock occurring at point 5 is the 'strongest' shock that a converging-diverging nozzle (of a given geometry) can create. If we increase the environment pressure more, the normal shock has to move into the nozzle and it gets weaker because the subsonic expansion now contributes to some of the pressure recovery. If we want to create a stronger normal shock at the nozzle exit, we need to increase the mach number of the flow at the nozzle exit, and so we need a bigger nozzle exit. However, with the same chamber-to-environment pressure ratio as before, this nozzle will now be over-expanded and the normal shock will actually occur inside the nozzle (situation 4). And so we will need to increase the chamber pressure to move the shock back to the nozzle exit and return to situation 5. So to summarise, if our goal is to make the strongest shock possible, we want a huge expansion ratio nozzle and a (quite high) pressure ratio between the chamber pressure and environment pressure that allows us to achieve situation 5. aerospaceengineeringblog.com/wp-content/uploads/2016/06/NozzleConDiv.jpg This was a longer message than I expected! I hope it is helpful.

Thank you so much for your elaborated answer. Is there I way I can email you directly?

@@ariebos7872 No worries. I can be reached at lyle.tac@gmail.com, and I will reply when I can

There will be!

@@TheGravityAssistant So there is part 1, part 3, but no part 2?

@@24pavlo Correct, there will be a part 2 on oblique shock waves, but it is still in progress.

@@TheGravityAssistant take your time! :)

Thanks for the kind words! I'll do my best to put out some new videos in the near future, work has been hectic the last few months

Oops! Yes you are correct. Good catch!

Thanks for watching. The flow field is solved and animated in MATLAB.