The Rocket Equation: all maths, no distracting astronaut
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- Опубликовано: 3 сен 2020
- Pick up the main Rocket Equation video exactly where this one warps-up: • The Rocket Equation: M...
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We have a complete guide to the mathematics in this video as well as some student worksheets and activities.
www.think-maths.co.uk/rocket
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Huge thanks to Chris Hadfield and their amazing team for making these videos possible. chrishadfield.ca/
With thanks to NASA and the team at the Kennedy Space Center Visitor Complex via agreement with Trunkman Productions Ltd and The Cosmic Shambles Network.
To discover more about the KSC please visit www.kennedyspacecenter.com/
Cosmic Shambles: cosmicshambles.com/
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Maybe you shouldn't have gone to the space museum. It's totally astronaut infested and they regularly bomb photos and ruin math equations.
Ruined it with the "just kidding" part.
@@Exachad lol he remove d it
hahahahahahaha!
@@black_jack_meghav I'm always open to good suggestoins :-)
If Kerbal Space Program based math has taught me anything, it's that the natural log version is actually more useful than the exponential version in most cases. You usually want to know the amount of delta-v that a rocket can produce (final velocity here, since you assumed vi = 0), and that's based on payload mass and engine parameters (exhaust velocity here, though it's usually altered somewhat into a quantity called specific impulse).
Who needs "real life" or "actual experience" when you can just do maths and assume things to be constant! :P
I love how the title says no distracting astronaut.
Every time you say "dmv" I picture the long lines at the Department of Motor Vehicles.
Oh, you made that joke while I was typing...
A lot of people would like to subtract the D.M.V. 😂
wait.. I didn't get it ? : p
@@cheesywiz9443 Department of Motor Vehicles
@@jellomochas haha yeah I googled it but thanks!
Lol
1:50 Matt: "I know I've been told I integrate like a physicist". Mathematicians: does something every mathematician knows can be proven, and will therefore not do it themselves. "yeah yeah, why are you mentioning this? go on".
2:29 Matt (nonchalantly, without any explanation): "Uuhh, that's a constant so I can integrate there". Mathematicians (without firm grasp of mechanics, like me): "did you just arbitrarily assume the exhaust velocity to be consant? Without proof?! That's dangerous man!! People will die!!!"
It's been a while since I've done multivariable calc, but wouldn't accounting for both variables come out with the same result?
For launching, this doesnt work since you need to account for drag and aerodynamic pressure at your exhaust. But once you are in outer space, the exhaust velocity can be assumed constant and the equation is valid
@@louiswouters71 unless you have flow separation in your nozzle I believe it will expand the same in an atmosphere as it will in space, which would be fully dictated by the chamber pressure, nozzle throat diameter and nozzle exit diameter. The vehicle will indeed experience drag, but this doesn't affect Ve. The vehicle also experiences gravity, but since the exhaust and nozzle are practically at the same altitude, they are practically feeling the same gravity and equally affected and Ve, which is defined relatively to the coordinate system of the rocket, is unaffected. It does get funky when you change the chamber pressure or gas composition by throttling the engine
@@petroleus I'm not saying that drag influences the exhaust velocity but it will make your calculated delta V useless.
@@louiswouters71 Add 1000 m/s and most of the time you will be fine.
Sort of like the opposite of continuously compounded interest: Instead of adding interest, you are subtracting exhaust
Reminds me of Full Frontal Nerdity when Matt talks about vortexes..."let's just stare at the equations"
Phenomenal - it's phenomenal how something seemingly so complex becomes something so simple, yet powerful. No pun intended.
This was perfectly done. Give us his fantastic description of going into space, but also the math that we are all here for.
yeaaaaaaah
invite famous person and then make a video where he didn't distract you :D
Finally, I can go back to the equally important astronaut who was keeping me from my daily fix of math
"a lot of people like to subtract the dmv" hahahaha
As a 11th grade physics student, the calculations are almost understandable. So close yet so far
You'll get there soon! After Calc 3 and Diff Eq integrations get a lot easier? Or more doable since you know what you are doing (at least a bit more).
@@sketchyAnalogies What does this have to do with differential equations? There's no derivative to be found in here.
Louis Wouters even when integrating its still called a differential equation.
As a 9th grade Geometry student trying to at least be able to understand these equations, I still have a long ways to go lmfao
@@louiswouters71 I know it’s been two years but a differential equation is one where you have a relationship between derivatives and you want to find the function(s) that make it true.
MATT! You obviously dropped a negative sign somewhere. The logarithm term is ln(M0/Mf), you have it upside down. The logarithm of any number greater than zero but less than one is negative. The way you have the equation, your rocket is going backwards! 😆
I am going crazy over here trying to find the mistake... in the other video he ends up putting a minus in front of ve, and he says that's because it goes the other way. But that's not it, because in the diagram (and momentum equation) he has it right as dm*(v-ve)...
Where is the problem then? The integration limits also look correct...
it's way more sneaky than I thought... Matt is taking a shortcut in his derivation, but basically his dm is positive as exhaust mass, but it results in a negative variation of the rocket mass... so the limits of integration would have to be swapped, or dm become something like -dM
Interesting way to say "m nought" 😂
oh thank god! I was worried I was gonna have to ask Chris to politely quiet down because I couldn't hear the maths
Have you got a version with more band, and less alarm klaxon?
lmao
Doing similar in uni just now for my maths module. Separable differential equations.
The math of the staged nature of rocket flight/the rocket equation is a core mechanic of the board game “Leaving Earth”. I have no association with the publishers/creators other than buying the game, but feel the audience for this vid (and Matt, you too) would enjoy it.
The force of friction getting out of the atmosphere is a bit problem. So for how long it takes to get out, it is not good to try to go too fast, because unless you are extremely aerodynamic, the force will just increase and you would get less delta v for fuel. The rocket equation mentioned is probably for non-atmosphere considerations. Then yes generally rockets use a lot of fuel and generally chip off dead weight in stages to up their acceleration. There really needs to be a better way to do so, but travel is generally recognized as observing conversation and requiring fuel.
I couldn't figure out when you taped the new page onto the board, so I went back and checked. And it was the wind that did that! (2:57)
Now do it with changing gravity and changing air resistance. That's when it gets real fun.
Anakin, if he were a mathematician: Here's where the real fun begins!
I've done this calculation, and I believe it is not explicitly solvable. I had to use computational methods.
Yeah it’s much spicier this way although the approach is still essentially the same, although I agree with dreggory
totally worth it watching it here! hahaha
Roughly 1/8th decided to watch maths as well - Thanks for brief intro into [ln] - I'll probably use it soon in maths highers :D
More -band- astronaut!
So I've been told that what happened at 1:48 is actually not supposed to happen, because the velocity of the exhaust is v-ve+dv so if you multiply it out there should be a cancellation with one of the terms in front.
I wish I was there :( I really enjoyed the video though, stuff like this is really fun!
You can easily derive it from the conservation of impulse (p).
So d/dt p = 0, p = m*v
Therefore (m*v)' = 0 = m'*v + m*v'
Well you normally say m'=0, and you get m*v' = m*a = 0, which is first law of Newton.
But if m' is not 0, then you get
m'*ve = - m*v'
dm/m = - dv/ve
If ve is constant, then we get
v(m) = ve * ln(m(t)/m0)
100% More Integration!
50% Less Astronaut!
Now we just need 90% more Marching Band!
Damn, I'm surprised I actually understood all of this
So this is the famous "rocket science" everyone is talking about?
Well r/whoosh me if you want, but: every "Ve" in that equation should be a function, not a constant and that is equation only for 1 stage rockets, so it would be much more complex...
Also this doesn't count drag and other aerospace thingies.
The rocket science is trying to design an engine and working out Ve (or specific impulse in some cases) for that engine at different altitudes. The last bit is fairly straight forward. The first bit is difficult (yes I have tried it).
Parker Science.
The left channel was way too low. Thanks for reuploading.
Just wondering, why is the mass and velocity of three fuel dm and dv, respectively.
Pretty sure there's a mistake here. The original diagram should have written the new rocket mass as m + dm, not m - dm, and the mass of the ejected fuel as -dm instead of dm. If you are using dm as a differential of the rocket's mass (which is the assumption you make later when you're integrating), then the change in the rocket's mass has to be labelled as dm and not -dm. This gives an extra minus sign in the exponential at the end.
You are right . The sign of log term on the right would come out to be negative.
Engineers and most physicists prefer to work with everything positive and match what is being given with what is being taken. Some physicists and most mathematicians prefer to count things given as positive, things taken as negative, add the lot up and get nothing.
Either way gets you the right answer; if the first method returns a negative amount, it just means you guessed wrongly whether it was given or taken.
bluerizlagirl or else you might end up with the pointy end down and the flamey end up 🙃
@@bluerizlagirl It's not just a matter of a different sign convention. The way Matt did it was actually incorrect, because in his integration he implicitly assumed that dm was a differential of the rocket's mass m, which means that his labelling in the diagram should have had the rocket's new mass as m + dm with the exhaust mass being -dm. His calculation can be made correct with the diagram as written, as long as when we integrate, it is understood that the quantity we are integrating over is actually the exhaust mass, not the rocket mass, since the diagram labels the small change in the exhaust mass as dm. But in that case, his limits of integration also have to be modified - we have to integrate from exhaust mass = 0 to exhaust mass = (initial rocket mass) - (final rocket mass).
Physicist’s pretty much consider down positive. You don’t really need to consider small things like that if you know the direction that it will be going. Negative just means the direction you guessed was wrong like the other person said.
Imagine if Matt had made this video by taking the beginning of the original video and editing it so that Chris was walking away from the board instead of staying at it, and then had edited that part of the video together with a new recording so that all of that looked like one single continuous recording.
Lmao thanks for posting this.
This is more a critique for Chris Hadfield, but I think Matt would also appreciate it.
If a rocket is a bomb, then any internal combustion vehicle is a bomb. At no point does it explode the propellant: it's burned in a very controlled way. _If it goes wrong,_ a rocket has more stuff inside to go boom, but a car can also go boom.
Using the bomb description, while not entirely unjustified (it _can_ explode), is inadvisable in the same way that "ooo look how weird quantum mechanics is!" is inadvisable. While it may boost enthusiasm, it distorts perception of something we need more people to understand.
derive for gravitational field..!!
I didn't even finish watchint the main video
The velocity sign was right... dm sign was wrong... but I am sure that someone already saw this
Yeah, something doesn't seem right with the final formula, since vf/ve is always positive (since ve is assumed positive in the diagram), the exponential of that will always give a number bigger than 1. Thus mf has to be bigger than m0 which cannot be right
It has been corrected in the original video
Yeah. As soon as the astronaut leaves, the Parker errors creep back in again. I guess it does take a physicist to spot such things.
The time embed in the link doesn't work for me. It just starts at the beginning of the video.
Am I a bad person for having clicked over to listen to the math after only listening to Chris Hadfield for ~1 min...
the real video
Can you imagine hearing this in school 🤯🤦🏾♂️😂😂😂😂
Needs MORE BAND!
Why can we assume the exhaust velocity is constant?
"warps-up"
It's not rocket science...
oh wait a minute....
Priviet, Matt! Good job on the derivation of my equations.
Am I the only one who noticed that this equation says either your final mass has to be greater than your ending mass, or that the ratio of velocities has to be negative?
(I checked the comments, but I might have missed it...)
No, I think you're right -- the ln() term is flipped. When I do the derivation for my classes I change sign on the dm since it's a rate of change of mass, which is negative. ruclips.net/video/VD9wlcV0UDE/видео.html
can we really just say that dm and dv are tiny changes? Like....it's quite a change in velicity from ground to orbit and it takes a lot of fuel. Of course, I know it was done for simplicity's sake
Yes, we can. "dm" and "dv" are not the total change from ground to orbit, but rather the change over some tiny slice of that distance. When you integerate, you're effectively adding up an infinite number of those really tiny slices. That's the point of integration, after all.
@@lostwizard Ahhhh....yes I suppose that is the point of integration. I suppose I just played myself, and interpreted the diagram too literally.
Look it's been a long time since I've had to do real math XD
Math gang!
id Miss Piggy have a role in Math(s) in Space?
Even in the left channel Matt's audio was pretty quiet and the board small, so great there is version without Chris. Why did you even invite him if he didn't want to talk about maths, but cool astronautic things?
1:40 when it comes to discarding -dmdv, the original video goes on to say that approximately 90% of the mass of the launch vehicle was fuel so how can dm be negligible? similarly, when it comes to dv, it is a freaking rocket so the whole point of it is too get a huge dv. so I'm confused as to how we can simply discard this term.
Aaron R this is because we are only considering a very small bit of fuel so the dm and dv will be tiny, we then add up all the little changes with the integral
@@georgedoran9299 True. I should have thought of that. Thanks
That was surprisingly trivial.
Where's the Diff EQ?
Pretty good, but needs more band
what about the dmdv
Technically they are called second order terms so they don't go into equations with first order term changes
We use the most powerful thing in physics: Ignore higher order terms(or terms which are very small as compared to other terms)
They're both really small, so when they multiply they're practically 0. In the same way that 0.0001 × 0.0001 = 0.0000001
He was doing Parker Integration.
It's the difference between science and engineering.
The word you don't hear in pure math: "Practically, ..."
The word you _do_ hear in engineering: "TLAR"
(That Looks About Right)
also you weren't very loud(in the original video)
It's been 5.5 years since I integrated a formula, I miss this. ~ the jealous med student
It's only been a few weeks since the last time I had to derive a formula and work with integrals and I am soon to be back at it. I wish I could be free of it
~the jealous engineering student
@@petrichor9279 Mind swap
Also, nice username blood of stone (or, the smell of resin)
4:37 But only because it's rocket science ...
who else skipped Hadfield's story?
150K vies there and 18K here. disparities.
No mathematician here but why are we integrating about 2:25 mins in?
Ah good, the annoying astronaut‘s gone
/s
0:30 i am distracted by the fact that the birds in the background dont move..
Glitch in the matrix?
It's not birds, but some dirt on the camera lens.
@ Yeah, I figured.
But it's funny that it was there just as he said "some people found that distracting" (regarding completely different matter)
dmdv is too small to matter. Ahh so we are doing Parker Integration.
As the final calculation shows, dm is 90% of the mass of the rocket and dv is 8,000m/s. So how are those small values?
@@bbgun061 No they are not, dv and dm are meant to be infinitesimals
@@jjtt I guess Matt didn't explain that well enough. Thanks.
I prefer the math.
First I got annoyed when he ignored the dm dv, then I got annoyed when used it to make fun of physicists.
I prefer the version with the astronaut :p
Physics is a cruel mistress.
So, two whole videos about the rocket equation without once mentioning Konstantin Tsiolkovsky?
😦
Uhhh... Wait a minute. Did you lose a sign? Your final equation doesn't work.
m_f/m_0 = exp(v_f/v_e). So m0 = m_fuel+m_rocket, m_f = m_rocket, hence
m_rocket/(m_fuel+m_rocket) = exp(v_f/v_e)
(m_fuel+m_rocket)/m_rocket = exp(-v_f/v_e)
m_fuel/m_rocket + 1 = exp(-v_f/v_e)
m_fuel = m_rocket*(exp(-v_f/v_e)-1)
Now consider the fact that exp(-v_f/v_e) < 1 for any v_f > 0 and quickly decreases with increase of v_f, and you will find that you (a) need negative mass, and (b) that mass of fuel tends to negative unity as v_f tends to infinity.
Sir,
With all the respect
I think u have already considered the dierection of ve and again you are taking the negative sign, that a wrong justification. you did an error and you justified it by taking -3500. At the end . The error: the small maas that came out was you named it dm (well you should have named it sth else say §m , that would have made every thing smooth , coz now you can say that dm = -§m here dm is change mass of rocket with time which is => mf-mi = m-§m - m = -§m . ) so now its bit bizzare to say dm = - dm should be there but thats jus the direct consequence of your choice of notation. So now in the end you should have the correct rocket equation , and would not have to wrongly justify it with that -3500 thing.
I skipped to this video when he stepped forward and started talking. I couldn't hear the working out.
lol i actually didn't watch the other video, all theory no practice.
Like a teenager's sex life.
Sorry.
so nerds are here
My enjoyment was slightly ruined having seen it through (in the other vid) and remembering that he's actually doing it wrong because he's ignoring the different stages and also dropping terms.
But other than that, this was great! Thanks!
I wanna meet the six people who downvoted this and just ask why
Where is the take without the distracting mathematician?
Nice video, but your integration is like the one of a physicist, like a paker square of an integration you could say....