G’day and welcome to my channel. Here I will be discussing aerospace topics which I really struggled to understand during my studies. I often need physically intuitive explanations to truly understand a topic, and these aren't always available in textbooks or lectures. Here I will share some of the explanations that I developed or collected in the hope that they will also be useful for others. A warning - a physically intuitive understanding is only one piece of the puzzle, you still need to understand the maths to fully master a topic, but I will leave that to others who can explain it better.
Future topics (please suggest others!); - the Tisserand plane for (aero)gravity assists - hypersonic waveriders - thermal protection systems - J2 pertubation - GEO (in)stability points - Oberth effect and chemical vs electric propulsion - mechanical impedance - space telecomms for dummies (mechanical engineers) - why rockets crackle
Well done mate. I had naively thought the Lagrange points were points where the gravity was equal, although if I ever thought about it, that won't make sense.
Someone has explained it to me, but I still only have a rough grasp of: why spacecraft at a L point dong just go there and stay there. It's seemed to me that this would minimize station-keeping fuel use. But, apparently, orbiting the L points instead is actually more efficient, station-keeping wise. I suppose it's to do with the, in real life, changeable nature of the L points. I wonder if you would perhaps include a visualization of the changing location a L points over long timespans? This video came very close to doing that, but it was mostly about what happens if you perturb a spacecraft at the L point, not so much about the dynamic nature of that point itself over long time frames. Obviously, aside from station-keeping concerns there are other good reasons for orbiting that point instead of going right there, like there might be several spacecraft who want to be there, and they can't all inhabit the same point in space! Good video, enjoyed this.
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!
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!
Congratulations Master for your beautiful presentation and interesting manter! I learnt so much! Thank's a lot and my best regards! Jacareí-Sao Paulo- Brazil.
I have a question, u mentioned that pressure at the exit nozzle must match the atmospheric pressure. So how do you reduce the atmospheric pressure. Also the part abt shock was a bit confusing Looking forward to your reply
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
It is criminal that you are so under-subbed! I did my part in trying to rectify that grievous error by subscribing. Lol. Great videos, very informative and easy to follow. Keep up the great work, don't give up, and I'm sure you'll go far on this platform. I can't wait to see how far you go!
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
The biggest question I have is just “how do you reduce the pressure at nozzle exit?” Does the air coming out of the nozzle play any part in this? I can certainly understand a pressure controlled room or environment, but I don’t think that’s really what’s meant or else it’d be kind of useless.
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. >.<
@@TheLordGojira 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.
I'm out of mind! Accidentally watched this video, now I'm forced to subscribe this channel. ❤❤❤ More info expected on DSN communication. Best wishes, keep educating.
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.
A couple questions: 1. How much gravity do LaGrange Points have? Is there a direct relationship to the strength of that points gravity when compared to the celestial bodies that created it? 2. Are LaGrange Points taken into account when trying to predict the orbits of the planets? 3. Do Lagrange Points create gravitational lensing? Efforts are currently underway to use the suns gravity as a telescope, thanks to the gravitational lensing the sun creates. If these LaGrange Points have the necessary gravity, could they be used the same way as, "Gravity Telescopes", so to speak?
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.
@@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..
Hi! I'm currently designing a Nozzle for a CubeSat, I used MOC to obtain the design of the wall. I then passed it to ANSYS and i have an underexpansion effect which was expected if my ambient pressure is 0. So, I want to know how much I can expand the exit area to get the pressures to balance. I want to model this just like your video where I can have two charts and see the change in Mach and Pressure changing the velocity and exit area, what program do you use or what can you recommend for this. Thank you!
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!
Great explanation with very explanatory graphics. Many thanks. A question though: what are design parameters for a nozzle to create certain effects? E.G., if I wanted te create maximum shockwaves, regardless wat happens to the airspeed at the exit, what would that mean for the design of the nozzle?
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.
As an aerospace engineer, I think your videos are fantastic! The animation is awesome and the detailed explanation of the theory is pretty clear. 😇 I can't wait to see more videos.(I love hypersonic topics🤩)
Great informative video, but at 5:07, the formula you use is T^2/2pi, where I believe I should have been (T/2pi)^2 or T^2/4pi^2. can anyone confirm this? thank you
In this video you put a lot of emphasis on the rotating coordinate system. I believe you are not doing your audience any favors with that. Please allow me to explain. Before I get to my main point, let me first acknowledge the history of the Lagrange points problem. Today we live in the age of computers that can do high precision numerical analysis, allowing us to create visualizations such as the potential surface representation. Before the age of electronic computers the choice of way to bring the Lagrange point problem to a solution was determined by computational efficiency. Using a rotating coordinate system was more efficient, so that is what physicists used. Choice of computation method and how to physically understand a phenomenon do not necessarily coincide. Very often they do, but not always. Example: there is the branch of electronic engineering that deals with the equipment for generating, or receiving, or processing electric oscillations. These circuit boards have resistors, capacitors and inductors. The electric oscillations in those circuit board can be modeled with sines and cosines, but when using sine-cosine notation it is tedious to keep track of phase. As we know, sines and cosines can be represented with exponential notation, by using complex number notation. To make the calculations more efficient electronics engineers introduce imaginary current and imaginary voltage, allowing them to move the calculation to complex number space. The electronics engineers are aware that the imaginary current and imaginary voltage have no physical counterpart; they are purely computational tools. To *explain* to students what is happening in the circuit board the teacher uses only the physical current and the physical voltage. In Celestial mechanics: The intuitive understanding of celestial mechanics is in term of the quantities momentum, angular momentum, and kinetic energy. Example: Halley's comet: The major axis of Halley's comet's orbit is about four times greater than the minor axis. Let's start from the aphelion of Halley's comet. Moving at its slowest velocity of all its orbit the comet starts falling to the Sun. The comet has only a bit of radial velocity, but it is enough to not hit the Sun. Moving down the gradient of the Sun's gravitational field the comet is constantly being accelerated. As the comet approaches its perihelion the gravitational acceleration becomes ever larger, and that acceleration vector is shifting to perpendicular to the instantaneous velocity. At perihelion there is a local maximum in how much the orbit curves. The ascending journey back to the aphelion is the time inverse of the descend to perihelion. I think we will all agree that the above is the way to understand the orbit of Halley's comet intuitively. Another expression would be 'visceral understanding'. Conversely, any attempt to present the mechanics of Halley's orbit in terms of a rotating coordinate system would be absurd. In my opinion this generalizes to all of celestial mechanics. Why does an asteroid orbiting near L4/L5 of Jupiter never escape? Well: that is due to celestial orbital mechanics, which arises from the inverse square law of gravity and inertia. The inverse square law of gravity gives rise to a Kepler orbit, and a Kepler orbit loops back onto itself. Let's say you are a celestial object, and you are orbiting right at the L4 point of a primary and secondary. Some perturbation (from yet another planet, far away, in that same system) makes you slide away from the L4 point. Well, there is no escape; the orbital mechanics brings you right back to where you started from; a Kepler orbit loops back onto itself. My point is: there should not be a tacit assumption that there is a 1-on-1 relationship between efficient computation and physical understanding. They often coincide, but sometimes they don't. Some years ago I created a physics simulation for orbiting motion along the L4/L5 points. I programmed the calculation for motion with respect to the inertial coordinate system. Using inertial coordinate system or rotating coordinate system, the computer performs the computation in real time either way, so for programming the computer there is no need to use a rotating coordinate system. For programming the numerical analysis: using the inertial coordinate system is simpler. As to the display of that simulation: that is a two-panel display. One panel for the inertial point of view, the other for the rotating coordinate system point of view. The computer performs that coordinate transformation in just a couple of processor cycles, so of course I added that coordinate transformation.
Thanks for your thoughtful comment, and apologies for my very slow reply. I wanted to take some time to consider your post. Firstly, I agree that in the case of an elliptical orbit such as that of Halley’s comet, it would not make sense to use a rotating reference frame, as the angular velocity of the orbiting body is constantly changing. We could have a variable angular velocity rotating frame, but then that introduces yet another abstract force… However, Lagrange Point analysis assumes a circular orbit of the secondary body (Circular Restricted Three Body Problem). Clearly this assumption will only be valid for planets / moons with a nearly circular orbit about the primary body, and in fact the Lagrange Points become meaningless if the eccentricity of the orbit strays too far from 0. However, this assumption is good enough for many planets in our Solar system. Personally, I don’t find your explanation of L4 stability satisfying. If the stability is simply a result of the inverse-square nature of gravity and a closed Keplerian orbit, why should L4 and L5 be stable while L1, L2, and L3 are not? Why should L4 and L5 become unstable if the secondary body becomes too large? I agree that rotation frame alone is not a total solution, but IMO it can provide an intuitive explanation for something like 80% of the problem. I don’t mean to be rude, but did you watch the entire video? (I know it’s long 😊) At 20:23, 27:45, 32:24, 33:50 I have the side-by-side rotational frame and inertial frame animations that you describe from your previous work. It’s my opinion that humans are not very good an intuiting circular motion in an inertial reference frame, as it’s not something we have day-to-day experience with (additionally, visually keeping track of a quickly rotating object and also analysing its motion at the same time gets tiring quickly). Generally, humans are IN the reference frame that is undergoing rotation. And so, even though the centrifugal force and Coriolis force are not ‘real’, we understand them and their effects in a rotating reference frame more directly. Some daily examples of the centrifugal force being a car or bus going around a corner and ‘throwing’ us outwards, or a roller coaster ‘pushing’ us downwards at the bottom of a loop. In the case of the Coriolis force, kids throwing a ball on a merry-go-round learn that it curves in funny ways, and then in high school we learn about how it causes cyclones and hurricanes due to the Earth spinning. Even things like retrograde motion of the planets are due to us observing them from a rotating reference frame, and the debate around that topic took quite some time to resolve 😉! In any case, the Coriolis and centrifugal forces are basically IF this, THEN that rules which can be visualised quickly. Also, the rotating reference frame allows convenient calculation of the potential surface, which provides a nice graphical illustration of the points and their stability. Feel free to point me towards a reference, but I have never come across such a neat illustration of Lagrange point stability from an inertial point of view. In my mind, the inertial reference frame just provides a useful ‘anchor back to reality’ which is the last 20% of the puzzle in this analogy.
@@TheGravityAssistant I'd like to discuss the mechanics of the motion of a celestial body in a slightly eccentric orbit . To ease the viewer into that I will start with discussing something that happens here on Earth: the physics of using the banking in track cycling. Track cycling team pursuit ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lP7ioj9isw0.html In a team pursuit event all four riders take turns riding in the lead. For each rider: when the turn is done you have to move from the front to the rear as quickly as possible, without losing energy. As can be seen in videos of team pursuit events: this is accomplished by taking advantage of the physics of riding on a banked track. As the four riders enter the turn the rider in front allows himself to climb up the banking. This has two effects: he has to cover a longer distance, and climbing up the banking reduces his velocity. This is intuitive; we all know that when you coast uphill gravity will decelerate you. Once the rider is halfway in allowing himself to be overtaken by the others he starts steering down again. Going downhill gravity is *accelerating* him. At the moment the rider reaches the the other three riders: gravity has assisted in accelerating him back to the (constant) velocity of the other three. In the side-by-side animation starting at 20:22 Slow down the playback speed to 0.25, and in the righthand panel, (inertial point of view) watch the motion of the spacecraft with respect to L4. The spacecraft is periodically pulling ahead of L4 and being overtaken by L4. The spacecraft is oscillating between moving along the inside track (inside the orbit of L4) to moving along the outside track (outside the orbit of L4) When the orbital altitude of the spacecraft increases then gravity is slowing the spacecraft down. So: when moving along the outside track (along highest orbital altitude) the angular velocity of the spacecraft with respect to the Sun is slower than the angular velocity of L4. Conversely, when moving along the inside track (along lowest orbital altitude) the angular velocity of the spacecraft with respect to the Sun is quicker than the angular velocity of L4. In the co-rotating view this oscillation in angular velocity presents itself in the form of small loops in the vicinity of L4. More generally: In physics what we want is to think about phenomena in a way that makes it independent from the particular perspective we happen to be using. We want to look at the quantities that are the same as seen from all perspectives. We should avoid relying on quantities that are perspective-dependent. For orbital motion the following two quantities are what matters: 1. The distance to the center orbital motion (radial distance) 2. The rate of change of angular velocity The radial distance is perpective-independent; it is the same in both the inertial and rotating point of view. The rate of change of the angular velocity is perspective-independent; it is the same in both the inertial and rotating point of view.
amazing video. best one i found on lagrange points. thanks. FYI, adding "0:00" before your other timestamps in your descriptions is how the youtube "chapter feature" gets enabled for easy chapter access from the scroll bar.
Hi! Thank you so much for this wonderful explanation :). I'm a high school student who is really interested in lagrange points and i'm trying to simulate them too. Could I possibly use your simulation of orbits around l4 & l5 in a youtube video im making? it's for a contest - with credit, of course. Either way, your video has really helped me understand the concept better and learn more about it and I really appreciate that sir
