An inverted pendulum puzzle
Вставка
- Опубліковано 15 вер 2024
- Sign up for Backblaze and get unlimited storage for Mac or PC for $6/month (plus a 15 day free trial): thld.co/backbl...
►Items from this video (support the channel)
Double Pendulum: stemerch.com/c...
Floating Globe: stemerch.com/c...
Don't be a jerk shirt: stemerch.com/c...
Sierpinski Triangle Poster: stemerch.com/c...
"What is mathematics?" (affiliate link): amzn.to/2KYdcXg
More book recommendations: stemerch.com/c...
Support the Channel: / zachstar
PayPal(one time donation): www.paypal.me/...
Join this channel to get access to perks:
/ @zachstar
►Follow me
Instagram: / zachstar
Twitter: / imzachstar
2D Graphing Software: www.desmos.com...
Animations: Brainup Studios ( brainup.in/ )
Check out my Spanish channel here: / zach star en español
►My Setup:
Space Pictures: amzn.to/2CC4Kqj
Magnetic Floating Globe: amzn.to/2VgPdn0
Camera: amzn.to/2RivYu5
Mic: amzn.to/35bKiri
Tripod: amzn.to/2RgMTNL
Equilibrium Tube: amzn.to/2SowDrh
►Check out my Amazon Store: www.amazon.com...
I put a list of all the items from this video in the description (like the double pendulum, floating globe, and "What is mathematics" book) but as always you can find it all at STEMerch.com :)
most of your videos aren't paced and filled for three year olds. maybe delete.
are you selling the triforce fractal on the background
The pendulum must be arbitrarily long as well. To the point of triviality by simply saying no matter the path I can have a long enough vertical pendulum that will not fall in any given finite time.
@@user-cj4fu8qq9b yep! That's the Sierpinski triangle poster
thank u Zach ur a g
When the graph came up, I figured "that's assuming it's continuous" and even though I couldn't have given a proof, I had a feeling there was a discontinuous point on the graph. Cool solution!
Yeah, the "subtle issue" felt bloody obvious
It's not discontinuous though unless you simplify gravity out of the model
@@errorerror6918 That means you got good mathematical intuition :)
@@iwersonsch5131 I think it is continuous, even without gravity. Without gravity, and with constant cart acceleration, the whole thing is a normal pendulum rotated 90. If the pendulum starts infinitesimally close to flat, then any vertical force on it is infinitesimal. To go very near the axis, then turn away from it, at some point you must be stationary. The distance from the X axis can be upper bounded by an exponential growth curve, starting when you were stationary. So continuous. A sufficiently minute distance from the x axis will stay small.
Same, I was really confused for a second
last time I was this early pendulums obeyed the laws of gravity
There's nothing that can effect the pendulum beside the car thing
@@beaclaster gravity
ua-cam.com/video/u1LdsYhTgPI/v-deo.html Delta function
Wait, so the problem assumed gravity didn't affect the pendulum (i.e. no gravity, i.e. the cart is in space)?
*Alejandro Nava* 00:07 "Assume there's no friction or air resistance affecting the pendulums rotation" i.e. gravity isn't mentioned and I would assume it is affected by gravity.
Always when i have some unknown problem with math, the first try is "let's look if the function is continuous", it's always the pitfall when your dealing with controlled systems or signal processing.
you're*
@@RishabhSharma10225 imagine this mistake is so common I see AI doing it.
I love that problem! :) The lagrangian approach is great! :3
Make a video on the Lagrangian approach then Papa Flammy
Lagrangian video pls. U the man
So does the Lagrangian approach show that it is possible or not? I think the verdict is still out on that one. I'm on team rod-can-always-stay-up.
What about my name?
thought of doing the same!
This doesn't prove that we can't always keep the pendulum in the air, it only shows that it can't be proved using the intermediate value theorem. So is there an example of some motion where we can show it's impossible to keep the pendulum from falling?
Seems easy enough,
(EDIT: not easy enough, in fact wrong)
this is a common unstable system discussed in control theory, unstable meaning that if you stop moving the pendulum and it isn't exactly at the equilibrium point (90 degrees), then it will fall. Therefore any slow trajectory not starting at ninety degrees is pretty much guaranteed to fail, here is my proposal:
For starters, let's define the equilibrium as zero degrees, the problem remains the same, but now the pendulum is allowed to move from -90 to +90 degrees, notice how the system can be considered symmetrical, this allows me to use the absoute angle for simplicity.
1 - Stand still, the linearized differential equation for the inverted pendulum indicates that the accelaration away from the unstable equilibrium point is directly proportional to the angular distance from it, in other words, arbitrarily small deviations take arbitrarily larger times (albeit finite) to fall. Therefore, after wating for some finite time the larger (absolute) angles are guaranteed to fall, and we can pick our cutoff point from there. Let's pick 10 degrees, so after waiting time = T1 any abs( theta(0) ) >= 10 degrees causes the pendulum to fall to abs( theta(T1) ) = 90.
Now we have a range of initial angles ranging from -10 to 10 (exclusive), so if a pendulum starting at theta = 10 falls to theta2 over the course of time T1 we know the only possible states for the pendulum are - theta2 < theta(T1) < theta2.
2 - Move backwards, we know we have plenty of room backwards and we will accelerate fast enough and for long enough that a pendulum at -theta2 will fall to +90 ninety degrees. Since our states are bounded we know the minum angle possible is theta(T1) > - theta2, so regardless of initial condition the pendulum is guaranteed to fall.
3 - Move fowards all the way to point B.
Solved in finite time and space, this problem would be more interesting if bounded by some time constraints.
Actually intermediate value is enough once you account for physics of the problem. In particular that the cart can only go left or right, not up or down. It's easy to find practical counterexample if you're allowed to pull the cart down to counteract gravity, that's where the discontinuity occurs. But that's not allowed in original problem.
@@TheMrCarnification I don't get your step 1. Indeed we can wait long enough until only begin angles with abs(theta)
@@jeroenodb ah, you are right.
If a pendulum at 10 degrees falls to 90 then a pendulum infinitesimally close to 10 will also fall somewhere infintesimaly close to 90.
Although waiting narrows the range of starting positions, those starting positions will be spread over the same range after waiting.
Now that I think about it moving does the same thing, moving to the left reduces the range of starting positions, but spreads whatever existing safe positions there are on the left to the right.
Seems like there will always be a viable starting position. I can't think of any trajectory that would solve it (other than waiting infinite time and then moving, but that's not a reasonable solution).
Thank you, now I like the problem even more!
@@kasuha So are you saying that the graph of initial angle and final angle is continuous once you take into account the physics of the problem (in which case the assumption in the book wasn't wrong afterall)?
Zach took a course in Real Analysis, expect more continuity videos
"It seems like a physics problem, but we're going to treat it like a maths problem."
There's a difference?
In this case treating it like a physics problem would mean you would have to calculate the forces on the pendulum. In particular you'd have to calculate the momentum of the pendulum and the force gravity will exert on the pendulum (which depends on the angle so involves trigonometry) and finally the force the pivot point on the moving cart exerts on the pendulum.
All these things are messy and complicated, but by working more abstractly we've figured out that in the general case we can't solve this problem and we've done it very quickly and easily.
@@kosie1991 Well that's why you have the lagrangian double derivative equation. You have to take into account all those variables, but at the end of the day, it might be simpler than trying to analyze an abstract graph that you aren't even sure is continuous. Sometimes mechanical solutions are easier than mathematical ones.
Physics is just applied math, chemistry is just applied physics, biology is just applied chemistry.
I don't know whether that's true. I just wanted to say it because it sounds neat.
@@kosie1991 No, Im studying Physics and this is a totally normal approach to prove properties of physical systems. Technically every physical proof is a methematical proof. The difference is only that in some physical proofs (mainly in quantum mechanics, electro dynamics, ...) physicist make wild assumptions and just pretend its a mathematical proof.
What Zach probably meant was, that it involves a more 'mathematical like' approach. But its still treating it like a physical problem.
Im not even sure if one can try to prove the statement via classical mechanics, since the 'driving conditions' (accelerations, etc.) are unknown and can be completely wild.
@@kosie1991 So, you'd have to do a bunch of maths. Sounds like that's a maths problem.
Mathematician: There must be a solution for every f(x)
Mathematician2: Oh no for some f(x) you might hit a discontinouity
Engineer: Ah guys a inverted pendulum is inherently unstable if you drive slowly it will fall down anyway!
Drop the x. The function itself is f not f(x). f(x) is a specific value of this function not the function itself.
Mathematician 2 is wrong. f *is* continuous for all object paths. This is easily proven by realizing that all complicated paths can be broken down into simpler ones, and all simple paths are most assuredly continuous.
"The function is continuous"
Unit step function: "Allow me to introduce myself"
6:22 bold assertion.
I love how when he was giving the first explanation I literally said to myself, "But how do we know it's continuous?"
Yes, but the pendulum problem can still be solved because the function in this case is still continuous. The observation is that the only way to force acceleration is to move the cart, but we can make the pendulum arbitrarily insensitive to acceleration by fine-tuning the stopping point to be close to 0 or 180 degrees. So, the function indeed has middle value.
A really nice question! In this case, though, we don't just have a free rolling marble on a flat semicircle with sticky ends. We have gravity applying a force downwards and movement applying a force left or right, and that changes the whole situation.
The closer the pendulum gets to 180° or 0°, the less it will be affected by the cart movement as opposed to gravity. Spinning this further, we do actually get to a situation where given any cart movement with finite acceleration, there will be an angle at which the pendulum will not fall down on one side, but won't either swing over to the other side until the acceleration changes signs.
The key to getting this answer of "yes" lies in understanding that the restriction not only applies _at_ 0° and 180°, but also affects how the curves can behave _near_ those values. The areas near those edge values apply a pull, which, divided by the force applied by the cart, tends towards infinity and as such can counteract any cart movement at some in-between value.
I thought I was being an idiot when I tried to prove that the final angle was a continuous function of the initial one, but I guess I was right to be careful this time lmao
I believe it actually does stay continuous! I've written a post describing how this video is wrong. If you could take a look at it, I would appreciate it!
@@EmersonPeters Any place where we could read that post?
@@EmersonPeters Starting angle of 0 degrees upright may produce a 30 degree end angle while a starting angle of 1 degree upright may produce a 35 degree end angle. If this was a simple pendulum and not one attached to a movable base, I would agree. (giving it enough degrees of freedom to behave chaotically as the energy has different ways it can disperse, and if the base has a motor...)
Personally, not like I've analytically solved it, I didn't even watch the video so I'm not sure what all the constraints even are, I'm just commenting here...I don't know, but it seems reasonable that they aren't continous in all cases. (With a double pendulum, it is surely obvious. here I'm mostly imagining the base as supplying the extra degree to make the analogy) I'm sure that some are, like fixing the wheel base, or constant accelerations, etc...edge cases.
Since you mentioned you wrote a post, I'd like to see your analytical work on the problem, since I'm too lazy to do it. Please link when you have the time.
@@EmersonPeters It's obviously not continuous in some case. Indeed, we can easily take a motion (x(t)) where it s not continuous. Take a long distance between A and B.
Phase 1:
Take x(t) = 0 for a long time (for example: 10m) the pendulum will fall if theta is not close to 90°.
Phase 2:
After that, you put a big constant acceleration that make move the pendulum. Because theta is close to 90°, there is a maximum time where however the initial angle, the pendulum will touch the ground.
So case 1: initial theta is far from 90° and he fall during the phase 1
Case 2: initial theta is near to 90° and he fall during the phase 2.
If I m not clear, I can make calcul if you want and show up the proof.
(sorrt for my english, i m french)
EDIT: my comment is wrong, see comment below
@@st_s3lios860 But then, wouldn't there be some position really close to 90 degrees such that the pendulum falls at the very end of phase 1, and phase 2 perfectly catches it before it actually hits 0 or 180 (depending on the direction of motion in phase 2)? Surely, for any length of phase 1, you can just get arbitrarily close to 90 degrees to make it take that amount of time to actually diverge.
Really interesting problem, something to really think about and change your perspective!
This was one of my most memorable control systems labs at uni.
Yeah, I had thought that Zack (with his engineering background) would discuss it. But it looks like his recent (self-)studying of real analysis has rubbed off on him which is why he is preferring continuity arguments
Him: explaining the question in the book
Me: looking at the globe floating glowing globe
Here's an idea:
Let's think about what happens to the angle space over time. This will help us consider all the starting angles at once.
If the cart stays in place, then 90° remains fixed in place, and all the other angles flow outwards to the edges, eventually hitting them. If it accelerates forward at a certain rate, there's some angle, maybe 120°, that remains fixed, and everything left of it flows left towards 0°, and everything right of ot flows right towards 180°.
If you want to find a starting angle that results in a final angle of say 37° after a given period of constant acceleration, just follow the flow backwards.
I think this shows if x''(t) is piecewise constant, then the final angle as a function of the initial angle is continuous and surjective, and even injective everywhere except 0° and 180°.
Now maybe we can use the fact that we can approach as close as we want continuous function in closed interval.
So if we have a given function x''(t), we have a seq of piecewise constant functions xn''(t) and associated solution angles an . Now if the application that associate solution to piecewise constant functions is continuous then the limit of the an may do the job?
I think this idea is very much on the right track
You have to be careful of the possibility that the flow "tears the space apart" - you can define flows for which the initial 0-momentum conditions tear apart the position-space - but this problem is smooth enough to prevent that I think. You also have to make sure that whatever curve the 0-momentum states deform into under one constant acceleration will not be entirely sent out of the good region of phase space once the acceleration changes (since you don't start with the 0-momentum states every time - your argument presently assumes this I think). Given constant acceleration, the curve of 0-momentum conditions deforms into one that looks like a sinusoid, but this sinusoid is a bit differently positioned and scaled for each constant acceleration. I think the important thing is that all curves that you can deform the initial 0-momentum curves into will "get stuck" on the new unstable equilibrium when the acceleration changes, and this applies inductively over any finite number of acceleration changes.
My first thought was, picture this motion: jerk the car back several inches, then have it stay still for 1000 hours, then jerk it forward and hold still for another 1000 hours, then back again, repeat a few times. In order for it to "survive" in the air for the wait times, it would have to be straight up, which you can't guarantee for every wait time... probably.
i dont think this argument pans out quite as youve described it. suppose for the sake of argument that a jerk gives 2 degrees worth of motion and that it takes 1000 hours to fall from 89.9 degrees to 88.1 degrees. then you could start at 91.9, jerk back to 89.9, fall to 88.1, then jerk back up to 90.1, then fall to 91.9, etc.
so since the initial angle is chosen /after/ the wait time, you can just pick the angle that would be most inconvenient for whatever wait time you would choose. maybe you could construct a counterexample in this way, but as it stands it's not complete
Yes this would work, but the question was, if the premise works with any movement of the car.
@@gorgit The question is, precisely, "will you always be able to find an initial position of the pendulum to keep it up, no matter what movement of the car someone proposes"
or equivalently, "is there some way to move the car that will guarantee the pendulum falls down no matter what position it started at".
Damn, the "Get floating globe" infobox popped up a second after I noticed it in the background and thought to myself "what is this magic, I want one of those".
Welcome to Neuralink.
If we try to solve by engineering approach There will be a specific formula changing by
INPUTS
acceleration of the system,
mass in the rod,
initial angle of the rod,
distance.
OUTPUT
Final angle of the rod
Discontinuity does not "sneak in" to anything, you always have to look at the requirements for your theorems
The correct way to prove this is through induction. First take any simple element of movement. It could be a pause in movement, a constant speed, or an acceleration. Now, it is trivial to keep the pendulum up through that movement. And the final angle as a function of the initial angle IS continuous through such a simple movement. This is plain to imagine.
Now, add any next movement to the object. Because the first movement enabled any output, we can give the second movement any initial angle. And I've already proven that the function is continuous given any simple movement.
Any complicated movement need merely be broken down into simpler movements till the conditions are met.
QED
Interesting approach to the problem. A similar question to this was asked in my classical mechanics test to be solved with lagrangian, which solves it analytically (assuming the integral can be solved).
Why do you assume that the pendulum angle graph is a function/is continuous?
Edit: I should really watch to the end before I comment...
Well it is a function
@@Joffrerap It might not be. It could be that, for a given initial condition, there is mroe than one possible outcome.
@@yonatanbeer3475 there is no randomness in the experiment, how can that be?
@@Joffrerap because the cart can take multiple paths between the start and the end.
@@yonatanbeer3475 yeah but the path is already set. it's not a variable.
Your "proof" is also incomplete.
Looks like you're assuming implicitly that if the initial angle is reduced, the entire curve, at all times, will follow angles that are lesser. But that's not necessarily how it will work out always.
What if as you get closer to just touching 180 instead of crossing it, the curve also deflects from crossing zero altogether? For every motion that you can construct?
You are absolutely correct here and that is exactly what would happen. The tail end of the graph would rise drastically until it crossed 180, then would gradually drop down again, making it a continuous drop down from 180 to 0
Still, going back to the question, it is not that we are to show that for _some_ motions, we can choose an angle that results in the pendulum not touching either side. No, we are supposed to show if we can find an angle for _all_ types of motions.
The counter proof would then be that someone finds _any_ set of movements for the cart that causes the pendulum to fall for all angles.
So the true issue with the proof in the video is not that that particular example might be wrong, but that the proof doesn't provide support that it is actually possible to produce a set of movements for the cart such that any angle always touches either or both 0 and 180 degrees. To simply show how such a set could look like is not proof that it is possible to create it.
My first thought was, what if the cart starts by standing perfectly still for a really long time? Then angles close to straight-up would be the only ones still standing. Then if the cart starts going forward, any upright pendulums would drop back.
yeah! I only thought of that on my second watch-through
Nope- there would be an angle that would be close to the center such that it start falling only after around time t. Around this angle there would be one that would have the pendulum lean towards the direction of acceleration perfectly
@@mishael1339 that's why you start going forwards. for any given angle =/= 90° you can wait a finite amount of time, and in that time the weight will fall. then you just need to accelerate forwards fast enough to make the edge case flip sides, and you have a motion of the cart that gets any angle to fall.
@@arust5147 you got it backwards. Your argument shows that for any given angle there exists a cart motion which topples it. This is not very surprising. However, for a given cart movement, say where you wait 100 years then move to the right, there IS an angle that would work. It's close to, but not equal, 90 degrees.
@@mishael1339 I was trying to construct a cart movement that would topple every possible angle, but now I see that my approach doesn't work; it's however still possible that such a motion exist. I'll come back if I find it.
Pretty sure it is continuous though. If you actually try constructing a cart path that always leads to 0 or 180 you end up with a problem: namely that there's no way to actually collapse the possibilities. Changing the cart's acceleration can move the unstable balancing point but it never removes it. But we can think of time progressing as a sort of 'zooming in' on points near that instability, since over time points always fall away from that unstable point. But no matter how you move that unstable balancing point, or how long you zoom in on it, you'll always have more points since the starting angle was continuous. It makes it somewhat more complicated when you consider position and velocity of the pendulum, but the same argument holds albeit in the 2D state space with a quasistable line instead of a point.
So in effect your picture that breaks continuity doesn't exist in the actual problem space because coming back from near 0 or 180 is actually really hard. Those 'paths' that just barely skim one or the other but don't collapse must spend a lot of time near the balancing point, which means nearby states include all possible future paths of the pendulum, including ones where it doesn't ever fall to 0 or 180
If the cart just waits a few seconds and then moves a bit and waits again, you can make sure no angle works (to survive the first wait, it has to be exactly 90 degrees, but then it moves and waits again so it falls)
It doesn't have to be exactly 90 degrees, just very close, and there exists a spot such that it will start to visably fall, just before the move, with the move righting it to very close to 90° again.
@@safrussalmus9056 If you allow infinitesimals, i'm allowing infinite time.
Being an engineer, I just programmed this into Simulink and used a PID controller
When I was a kid I would try to find the maximum amount I could tilt my chair without it tipping over, and I came to the conclusion that the final angle of the chair is certainly not a continuous function of the initial angle. There are only two possible outcomes after all, or three if you count the mythical middle-ground where the center of mass is exactly above the contact point, with infinite precision.
A very simple illustration of one x(t) which clearly shows that the pendulum ends in either 0° or 180° is a movement which is very very very slow. So no matter the starting angle, it would fall down to either side before it reaches the end.
It would be better if you actually provided such an example (as a function). If the cart is moving at a constant speed, you can just put the rod vertically and have it stay in this position for the whole journey.
9:52 How do the first 3 paths (when unrestricted) accelerate back up (lower angle, to above the cart), when the 5th path is also accelerating upwards (again, to a lower angle)? The only way for the first 3 paths to feel an upwards force would be if the cart is accelerating backwards, which wouldn't accelerate path 4 at all, since its direction is parallel to the acceleration of the cart, and would accelerate path 5 downwards (towards 180 degrees).
I still firmly believe that the final angle is continuous as a function of the initial angle, and this seems to backup my belief.
@@danielyuan9862 no it is not. Imagine a movement, where the angle hits 180 degrees. Now change the starting condition infinitesimally small, where the angle is only infinitesimal below 180 degrees. Therefore, an acceleration exists, which when starting at that point bring the second angle to 0 degrees, whereas the first angle has to stay at 180 degrees. And this is a discontinuous jump, since there always will be an acceleration for which this statement holds true, no matter how close the second angle gets to 180 degrees. Therefore it is discontinuous.
I think it was only a mess up at the starting angles, but you can just push them up to the "same side" and then make the same proof from there.
@@gorgit You gave the clearest correct mathematical explaination
@@RAl2O3 haha, thanks man, im flattered
1:20 there is no air
Who are you ???
@@jakemc4177 wtf?
Who are you ???
Who are You???
@@Gaurav-cy5lu Who are you ???
you dont need a double pendulum, just imagine that, at the beginning the cart goes really slow, if the angle is greater than 90, it will end up as 180, if its less than 90, it will end up as 0. Then, for this case in particular, there is no intermediate values, thus the function isn’t continuous.
He didn't need the fact it was a double pendulum, that was just the only one he had
@@romajimamulo yep, my bad, i wrote the comment before finishing the video
still, you dont even need a pendulum, you just need an special case and you can prove the assumption is wrong
I think that function may still be continuous. For any length of time it takes to get to B, you can find an angle slightly less than or greater than 90 that won't fall all the way to 0 or 180. I'm imagining something like a very steep sigmoid curve.
This comment still provides the kernel of a correct proof that a sufficient angle does NOT always exist.
If the cart remains perfectly still at any point for a sufficient time, then the only angles that will not drop the pendulum are those extremely close to 90°. Follow any such pause with a motion for which all angles within that range are insufficient, and you'll have a case for which no angle suffices.
For example, wait 10s, jerk hard left, then wait 10s again.
@@Mothuzad I think this argument is wrong, since you could find some starting angle such that exactly when the cart jerks, it has fallen exactly far enough to get jerked back up to 90 degrees again.
I don't know what kind of dark sorcery this guy does, but this man made me want to buy every single product from every single sponsor he has ever had
10:40 Yes there is more to the problem than the author originally thought. But you still didn't prove that function is NOT continuous for the restricted pendulum. Maybe the step you took was too large, there are infinite possible starting angle between the blue curve and the white curve, and their output might as well sweep through 0-180. (ignoring quantum physics and plank length, since we are dealing with a mathematical problem)
I don't think it matters how large the step size is between the curves. The white curve (and even green curve) already hits the 0 degree restriction. The point is that you *could* construct an x(t) such that one path of an unobstructed pendulum would pass above 180 degrees and below 0 degrees. Therefore, all paths below it must necessarily hit lower restriction at some point and all paths above it must necessarily hit the upper restriction at some point. That is not to say that if both restrictions are in place that a lower path will rest at the lower restriction (it might hit the upper restriction first), just that if it doesn't hit the upper restriction we know that it will hit the lower.
@@johnfeusi9233 It's been a while. I had to rematch part of the video. But your comment make sense
I'm curious how you would even get a pendulum to go from 170 to 181 and back to 170 smoothly and have a small change in initial position (but no change in motion of the cart) result in ~169 to 179 to 169. Position 180 isn't a natural maximum point, so there would have to be some pretty extreme accelerations going on to get the pendulum to have enough speed to move against gravity. You would have to have a sudden yank of the cart to the left when the pendulum was at 181 to get it to change directions and head back upward, and that same sudden yank to the left would make the 179 accelerate downward, not change directions. So I think it is a faulty assumption that there are smooth adjustments like that. A pendulum with its base (the cart) under extreme x acceleration at 179 experiences an approximately OPPOSITE vertical force compared to one at 181 (only differing by the amount gravity is pulling down on it, which is negligible for extreme enough horizontal cart accelerations), so why would the two curves run parallel to each other? As you get closer and closer to 180 degrees, there would be less and less possibility of actually turning completely around, so the "tail" of those curves would rise drastically.
"You can't just jump to a new position"
Me: Jumps across my room
6:20
Can u make a discord server for us to discuss math and science
Well if the pendulum length isn't set all you would need to do is increase the length of it. Such that it is a little bit longer than the required distance. Start at 90° Then make sure that the cart moves forward at the rate the pendulum falls vertically. (Alternatively you could make a track for the pendulum weight and nudge the cart forward) and the cart would reach point B before the pendulum shaft hits the cart
You should still show that a counterexample exists (i.e. trajectory for the car such that every initial angle ends up with tipped pendulum). I thought I had one, but closer inspection revealed it doesn't work...
Actually, here is an idea: first let the car stand still until all the non-perpendicular angles up to very small error have tipped over. Then set the car into constant acceleration, effectively changing the direction of gravity. This will make the remaining small angles tip over, while remaining large angles have velocities high enough that the small change in gravity would not have significant effect on them, and they will still tip over. Haven't checked whether math works out, but it's the best I could come up with quickly :P
@@tetraedri_1834 you can always find an angle arbitrarily close to 90 degrees that counteracts this I believe. This is a mathematically ideal pendulum so given an x(t) as you described you can find an angle (think 89.999999) so that the cart standing still tips it just enough to where the acceleration will counteract it.
@@hybmnzz2658 That's what I'm also concerned about. But I'm thinking the non-zero velocities could possibly make this impossible. It's like having small initial velocity at the beginning. But as I said, I haven't worked out the math, so I'm not claiming this to be a solution
@@tetraedri_1834 agreed. I am also having difficulty with a concrete counter example. My intuition tells me that if we divide x(t) into intervals of [non zero acceleration] vs [non zero constant velocity] vs [zero velocity] we can always calculate an angle that "responds accordingly" to this sequence of intervals. These "responses" would happen in practice for angles very close to 90 degrees.
An actual formula is annoying because it requires physics and we are talking about mathematical idealism so who cares.
It is refreshing that a simple counter example is so hard to find given that the problem would work if the IVT was satisied like as assumed in the textbook.
I think a counterexample x(t) can only be found by searching for a discontinuous plot like in the video.
There's probably not a counterexample. Ian Stewart, who is the one in the book with the criticism, later conceded that they "were correct to assume continuity in the particular case where the carriage has a flat floor" arxiv.org/pdf/1709.08254.pdf
I think you can also find x(t) so that it's impossible that the pendulum is in the air afterwards no matter the starting angle. For example you could include a longer pause, the pendulum would continue to move during that pause and if it is long enough it has to be at one of the end points
That wouldnt work, as you could just place the pendulum very close to 90 degrees - close enough so it doesnt fall all the way
@@yoni5919 would work with 2 stops, each having a long enough pause in between. Only one specific angle could "survive" the first pause which could then be catched by the second pause!
@@noel.friedrich No it would not, because there is not 1 angle that survives the first stop, but an infinite amount of angles in a very small interval. For example, all algles between 90° plus or minus 3 times 10 to the power of negative 74. That's a very tiny interval, but is still an infinite amount of angles. Some of which (still an infinite amount) would then be brought to be brought again to very close to 90° to survive the second pause.
So what is the answer then? Is there another reason why such angle always exists or is there a path so that no such angle can be found?
The video claims to have a path where no initial angle can keep the pendulum above the cart. At 9:53, all pendulums with initial angle greater than or equal to the initial angle of the white path (about 80 degrees) will end at 180 degrees, whereas it looks like all pendulums with an initial angle below that angle will end at 0 degrees.
However, I do not agree that this graph can correspond to an actual path of a cart, as the first 3 paths (when unrestricted) accelerate back up (lower angle, to above the cart), when the 5th path is also accelerating upwards (again, to a lower angle). The only way for the first 3 paths to feel an upwards force would be if the cart is accelerating backwards, which wouldn't accelerate path 4 at all, since its direction is parallel to the acceleration of the cart, and would accelerate path 5 downwards (towards 180 degrees).
copy paste from a comment I made:
the easy solution is have an acceleration, then stop, making the only valid value one that is in the right and ends up perfectly centered, then you go back and stop, making the middle position as the starting position, and making it end on the right (0).
in this case there is no valid value which gives a non 0 or 180 degrees angle.
since the 180 an 0 degrees angles are "sticky", any function that can force all values to end up on either side during the travel will have no valid solution.
you can chose stopping time and accelerations such that the value that makes the pendulum land in the middle doesn't land in the the small epsilon of valid middle values that won't have the pendulum land on either side.
@@satibel I believe your mistake is assuming that after the first stop the rod has to be upright. This assumption is completely unjustified and is demonstrably false. Indeed, since we have infinite precision (physically it is impossible, but mathematically it is), we can make the rod end up at any desired angle after the first stop. The same goes for the second part of the path
Isn't it quite easy to think of a counter example?:
Lets assume a path, were the car first moves to a spot between a and b and stays there for a very long time. Then it moves further and stops again and then goes to the finish line.
The pendulum will only stay in the air at the first halting point, if its starting angle was choosen in a way, so it will become 90° at the first halting point,, all other angles will fall down eventually whhile the car halts.
Now we choose the second halting point in a way, such that, a starting angle of 90° at the first halting point doesnt lead to a 90° degree orientation at the second halting point.
Therefore the pendulum will fall down for all starting angles at either the first or second halting point
This should also work for very finite halting times, lets assume 2 minutes at each point. This narrows the the possible angle the pendulum can take at the first halting point to a very very small window near 90°. It should still bea easy to choose a path to the second halting point, were no angle close to 90° will lead to another angle close to 90° (for example, fast acceleration und very slow deceleration over a longer period of time)
You are wrong at your first cart stop. You are choosing a path BEFORE you know the angle, therefore, you cannot know in advance how much you have to stay stationary for it to fall down. You can always choose such an angle that after the given time it will fall into a predefined angle to successfully continue the ride, it does not have to be 90 deg. while the cart is stationary, it can still fall slow enough.
I think you are correct. As a gedanken experiment you can only have long time if the pendulum at that time is vertical. Once you move off that point particularly in the AB direction the pendulum goes left and cannot get back to vertical without a reverse acceleration (which is not “any” path but a needed path). I think the question is /should be - for a given angle is there a path that works? That answer is yes, simply by an extreme acceleration for pendulum left of midline and for acceleration that keeps pendulum from falling on right or rising to vertical then constant speed thereafter. Also I think there is a path (x(t)) which can “get” the pendulum to any particular position from any given position as well. But not a position for any path. Your contradiction is quite correct given the statement which is not actually what the video header states as the problem.
@@fullfungo
I don't think so.
I choose a path and design it in such a way, that no angle will make it.
I think you greatly underestimate how fast an inversed pendulum will fall down. Try a pencil upside down. It wont stay like this for even 2 seconds unless you have a mathematical-precision-level accuracy close to 90°. Therefore this setup of halting for a long time forces the pendulum into an angle very very very very close to 90°. Same for the second stop. Now i only need a path from the first to second stop, that - for any input between, lets say (89,99°-90,01°) leads to an output thats either 90,01°. And this should be very easy. The cart accelerates first for some time. During this time the pendulum starts falling to the left. Now i only need to make the decceleration small enough such that the inertial-force times vertical deflection is at most equal to the force of gravity * horizontal deflection: m * a_decceleration * sin(angle) * length_pendulum
@@markmajkowski9545
I fully agree with all your statements :)
But i don't quite understand your last sentence. Where is the difference of the statements between the video header and my version? :)
@@Skellborn but we do have a “mathematical-precision-level accuracy” in this problem, because it’s a math problem. In fact you are overestimating how fast it will fall, if you tell me any number, I can give you an angle (not 90) for it to stay this long. And your pencil balancing skill is irrelevant
assuming no maximum acceleration limit, then unless the pengilum starts at 0 or 180 degrees then the cart can travel from A to B no matter the distance with the following method: if the pegilum is in the first quadrant (IE 0
I propose g(theta) could be a function such that the acceleration is dependent on the angle such that gravity is perfectly counteracted and the angle will never change due to gravity or g(theta). h(theta) could be a piecewise function such that if 90
If you carefully rewatch the video, you will find that in this problem you are choosing the starting angle AFTER you have the path, not before.
I knew there was an issue with their being a solution for any x(t) because of this scenario:
Consider if the cart starts by doing nothing for a long time, then it moves to the finish directly.
When the cart is stationary, the pendulum will fall at any angle except 90 degrees. However, if the angle is 90 degrees, it will not stay standing after the cart moves. Therefore, there is no initial angle such that after x(t) the pendulum is still standing.
Wouldn't you be able to influence how the pendulum moves through the motions of the cart?
If it's falling right you can accelerate right to make it fall left
Or if it's falling left you can accelerate left to make it fall left
Then you can pause for a second to let the pendulum fall further in whatever direction it's already falling
the problem with that reasoning is that you can envision a position for the pendulum that's _almost_ balanced, but just when you're about to move, it's fallen enough to be brought into almost-balance again by that movement, which in turn can be used to negate the effects of further movement. So I don't think this reasoning alone proves that the function can be discontinuous.
@@Gamesaucer well, the problem I see with the videos explanation is it assumes that no matter how you move the cart, a certain starting state will always have the same outcome
I don't like the idea of answering "it depends" to problems like this, it's true that depending on what the trajectory of the cart is, there will or will not be an angle where the pendulum stays in the air.
But the question asks if we can always find an angle where the pendulum stays in the air whatever the trajectory of the cart is, so we can deliberately craft a trajectory for which there is a discontinuity in the pendulums final angle in function of the initial angle, and use it as a counterexample, that's enough to answer the problem.
Yooooooooooo! I was watching the part explaining the original proof, an I was thinking, "I think a proof that the function is continuous is in order, IVT only works with continuous functions" and then a minute later you point out the same issue. I feel so smart 😊
I still believe the function is continuous. I'm not sure I can prove that right now, but I can at least counter your argument against it.
Let's look at the graph at 10:00 and actually consider what motion of the cart must be. Around t=3 to t=4 the (unrestricted) dark blue pendulum is just below the horizontal on the left side and is accelerating upwards. This means the cart must be accelerating to the right to pull it back up. But now look at the cyan pendulum at the same time. It's above the horizontal and also accelerating upwards, which means the cart is accelerating _left_ to _push_ it up. The cart cannot cause both of these effects with the same motion.
In fact: the white curve cannot exist at all. At the point where it touches 180° with zero angular momentum, how can it accelerate upwards? The forces acting on it are gravity pulling it down and the force from the cart which can only be parallel to the direction of the pendulum rod (i.e. horizontally), so the resultant force must be downwards.
And that's basically the crux of my belief that the function is continuous: as the pendulum gets closer to the horizontal on either side, it gets arbitrarily hard to pull it back away from that, so no matter how hard one portion of the cart's motion tries to yank the pendulum over to one side, it must be possible to have been close enough to the other side that it will never go all the way.
The problem i see with the corrections is, that in the instance of the white curve f(t) 'barely' touches the 180° mark we have two possible outcomes:
1. The pendulum hits the cart and stays at 180° forever
2. The pendulum does not hit the cart and has an angle theta = 180°-epsilon, but here i could always choose a smaller epsilon which will then result in a second curve g(t) >= f(t). In general i can choose an epsilon so that any final angle i want can be acchieved. This situation will than take place at the next point of the curve were it hits a boundary.
This all assumes however we are not taking into account quantum theory and the therefore discrete values in space. Then you could indeed find a smallest epsilon.
Edit: The green function h(t) requires h(t) >= g(t), which would not allow for any theta_final above 0° as shown in the graph.
But h(t) cant exist in the presented form, when taking into account the boundaries of the problem. Starting from the time of impact t0 we have h(t>=t0)= 180°, which will alsway
I set up a formula where the clockwise torque due to weight was balanced by the counter-clockwise torque due to acceleration, which gives solutions over a range of starting angles. But I would have never guessed that there is a critical angle and acceleration that gives discontinuous solutions. That’s the difference between physics and math.
I think there's an issue with the argument in this video, too: it assumes that the function describing the "unrestricted path of the pendulum" can be arbitrary, but that isn't the case. The easiest way to see this is to imagine a 3-dimensional plot: The x-axis is the initial angle [0,180], the y-axis is the time axis, and the z-axis is the angle of that starting location at that current time [0,180]. Let the function f(x,t) give the position of the point on the z-axis. Because of the specifics of this problem, f has to be continuous (shown below).
Claim: for any time t, there exists an open interval (a,b) such that the image of (a,b) under f is (0,180).
Proof: We wish to show this by showing there cannot be a point where it stops being the case.
At t=0, the interval is (0,180). Also note that at t=0, the angular velocity at any point in the interval is 0.
For any t, the angular acceleration ranges from some negative number (at a) to some positive number (at b). This is because the acceleration only has two contributing factors: gravity and and change from the motion of the cart. Gravity always contributes negative angular acceleration for the low angles and positive angular acceleration for the high angles. The acceleration from the movement of the cart can take three forms, depending on how the cart is accelerating.
If the cart is accelerating forwards (i.e. in the direction of angle 0), then the contribution to the angular acceleration by movement of the cart for angles near 0 is some small, positive value (to see that this approaches 0, split the force vector into components based on the pendulum shaft; as the angle of the shaft approaches 0, the component perpendicular to the pendulum approaches 0, so the acceleration does as well). Thus, the angular acceleration is a nondecreasing continuous function over the whole interval [0,180] that takes value 0 at a.
We can do similar logic for no acceleration and negative acceleration.
Since gravity and the cart's movement are the only two things affecting the pendulum's angular acceleration, the total acceleration ranges from negative to positive, is continuous over the interval, and is monotonic.
This shows that for all t>0, the angular velocity also ranges from negative to positive over the interval (a,b) and is also monotonic and continuous.
Hopefully this is sufficient to convince you that the function f is continuous, and therefore the pendulum must always have a valid value. It isn't something to assume for free, but it is still true.
Woooooooo ItZ zAcH sTaR
The answer can be seen easily (and thus the continuity assumption refuted) with a simple counter example. First the cart is motionless for a long time. After this the angle is forced to be either 0, 90, or 180. Only 90 can still succeed the problem, so we assume it is 90. Then the cart moves forward halfway before coming to a stop. First the acceleration pushes the angle off of 90, then the decelleration would push it back to 90 if there was no gravity. But there is gravity and this forces the angle to be larger than 90 degrees. After waiting a long time again, the angle is forced to be 180, and the counterexample is proven. No starting angle can stay up for this motion.
For any given period of time there is a range of angles close to 90, all of which will stay in the air for the given period of time.
I don't think the trajectories you're drawing there as part of the discontinuity proof are physically possible. The pure math approach is fine but if you account for physics of the mechanism, I am pretty sure the continuity re-emerges. You're only allowed to move the cart to the left or to the right, not up or down. So the relative gravity always comes from above the cart, regardless what you do with it.
My take on the problem: the change in angle due to acceleration of the cart is proportional to -a×sin0 and the change due to gravity to g×cos0. This implies that for any constant acceleration there exists a fixed angle where gravity and acceleration cancel each other out. Note that this fixed point is not stable, points close to it accelerate away over time. Thus there should be, for any point on the arc of the pendulum, a starting position near the fixed point that moves to said spot on the arc. This Mapping is continuous. Now, instead of assuming the speed of the cart change continuously, aproximate its acceleration with many constant bits followed by instantanious changes in acceleration (kinda like taking the floor function ). You now choose an angle that the prmendulum is supposed to end up in. Then using the previously established Mapping method, find a point which, during the last constant bit of acceleratio moves to said Final angle. Next find a point that moves to this new point during the previous bit of acceleration And so on. By doing the same for infinitly many bits of acceleration, this should aproximate the real thing better and better. So you can find a starting point ofr any Final possible angle (Sorry for bad grammar/capitalization , im german)
so, what is the actual solution to the question?
The video claims to have a path where no initial angle can keep the pendulum above the cart. At 9:53, all pendulums with initial angle greater than or equal to the initial angle of the white path (about 80 degrees) will end at 180 degrees, whereas it looks like all pendulums with an initial angle below that angle will end at 0 degrees.
However, I do not agree that this graph can correspond to an actual path of a cart, as the first 3 paths (when unrestricted) accelerate back up (lower angle, to above the cart), when the 5th path is also accelerating upwards (again, to a lower angle). The only way for the first 3 paths to feel an upwards force would be if the cart is accelerating backwards, which wouldn't accelerate path 4 at all, since its direction is parallel to the acceleration of the cart, and would accelerate path 5 downwards (towards 180 degrees).
No.
It seems like a matter of working backwards; at some points the movement of the cart will be so severe that the pendulum will have to be pre-emptively counterbalanced such that it is hardly any distance from its terminal values at all. In the world of math, ensuring these extremely sensitive values get achieved at the necessary times seems trivial, as long as "hardly any distance" is anything other than "0 distance".
Love T-shirt. "Don't be a jerk".
Can you explain?
@@honorarymancunian7433 I cane to the comments to find this out too. Given admin admin's answer, I'm assuming that it is a physics notation, describing a "jerk" motion, with the arrow having something to do with velocity, vector math or some such thing, idk, but would be nice to have it explained, I'm interested enough to want to know but too lazy to figure it out with google lol
@@WilliumBobCole It's been four months but jerk is the third derivative of position, which basically means it's a change in acceleration. Jerk is much harder to visually see than velocity and acceleration.
The time angle graph makes absolutely no sense. This drawn changes of angels only can happen with accelaration of the cart. But this would affect the pendulum in a different direction depending if it is above or below 180°. So the lighter blue line must have a completely different shape than the blue/brown/green lines. Also it has maybe a great effect on how close it is to this 180°. Maybe the small difference between wight and lightblue covers the whole range of outcomes without a jump.
Okay, but... You didn't answer the question, you just showed why one possible solution was wrong. Also, the problem as presented in the thumbnail is the inverse of the one you actually talk about.
It said for any x(t). So, even if one is wrong everything is wrong.
It's obviously not possible for every x(t), because one x(t) could involve the cart standing still for a minute to let the pendulum fall, and then moving to B. The only pendulum that would survive the initial wait would be the 90-degree pendulum that's perfectly balanced, and it would fall over as soon as the cart starts to move. Just move the cart slowly enough to let the pendulum finish its fall, and you've got an x(t) that guarantees the pendulum will have fallen before arriving at B no matter the initial position of the pendulum.
@@Ansatz66 actually if the pendulum starts extremely close to the 90 degrees it could take arbitrary amount of time for it to fall, so it's still okay
@@yuanyao5190 : Okay, that's fair, but in that case we can wait a minute to guarantee that the pendulum must be extremely close to 90 degrees, and then make a sharp movement to unbalance the pendulum and wait another minute. Surely there's no way any pendulum could survive that.
@@Ansatz66 wait no, since i already know how you're gonna move , I can position my pendulum so that when you begin the sharp movement its either very close to 0 or 180 (and in opposite direction to the sharp movement) so it will not hit either end after the sharp pmovement
The really sad thing is that there are trivial examples the disprove what the book says.
-Cart stays stationary for some time
-Cart jerks forwards some small distance
-Cart stays still for some time
-Cart finishes journey.
There's the proof by counterexample. I'm surprised there aren't more people pointing this out
EDIT: It's actually not that simple, nevermind.
There are people that have pointed out that does not work because you have theoretically infinite precision.
Now that I think about it, you would theoretically able to set the pendulum just slightly off-center so it doesn't fall when staying still for a long time, but instead sets itself just right to never fall. Whoops.
As others have pointed out, this doesn't work and after thinking about it it actually takes a more complicated set of actions to guarantee the pendulum will fall.
@@burstofsanity
The more complicated set of actions to make it fell doesn't exist. Imagine a cart with a pendulum at every angle. When it is stationary then all pendulums will fall away from 90°. However, there will always be some amount of pendulums that haven't fallen down, and cover the whole range from 0 to 180 degrees as you can be arbitrarily close to that 90 degrees. Then any other acceleration will just move that line that pendulums fall away from, but it doesn't matter as you can still be arbitrarily close to that line. So it's always possible to set the pendulum so it won't fall over.
The solution didn't look right in the first place as it was supposed to be valid regardless of x(t). How would that ever make sense?
Not quite. It is for a given x(t), can you pick an angle.
i.e. if you are given any function x(t), can you use that function to determine an angle.
As x(t) is changed, that angle could change, but the question didn't ask for some angle that works for every x(t). Instead it asked if you are always able to pick and angle for any x(t).
e.g. can you make a function that takes the function x(t) as an input, and determine an angle, such that the pendulum is still in the air.
If it was actually continuous, then it would be the case, that for a given x(t), there would be some angle where you can set it up to be in the air when it reaches point B.
This is because it doesn't matter what x(t) is, if you start at 0, it ends at 0, and if you start at 180 it ends at 180.
If it was a continuous function, then that means there must be some angle that gives a result where the final angle is between 0 and 180, which you could then pick, meeting the requirements of the problem.
The way I thought about it was different.
First, consider the case where it doesn't move at all. Given enough time, only one angle will hold it up, 90° exactly. Likewise, if you're constantly accelerating, again, there's one angle that will stay up, and the rest will eventually fall.
Notice then, that the acceleration changes over time an unknown amount of times with the motion of the cart, and the pendulum would have to move from being near one angle to being near another each time the acceleration changes
But for any path of the cart, it's only staying still or constantly accelerating for a finite amount of time, which means that angles sufficiently close to 90° (or whatever the stable angle is) will remain close to 90° over that time period. It's like all the starting angles around 90° are being stretched out toward 0° and 180°, but there's always a continuum of angles that started closer to 90° and haven't had enough time to fall.
@@crb233 Right, I meant "near" an angle, so you can hold sufficiently long at stillness, and then sufficently long at constant acceleration, so the ranges that haven't fallen over yet don't overlap.
I'm not a math genius, and didn't even know what continuous functions are. But I did know that the second you showed the graph, my first thought was "why does there have to be points in between? Can't it just jump from 0 to 180?"
He're my first thought: of course in the given example there is such a trajectory. Just let the cart accelerate fast enough to the left until the pendulum is at 90° upright, maybe, a little bit leaning over to the right. Next, accelerate it to the right in a very short time interval, just to stop the angular velocity of the pendulum, and finally with an acceleration just enough to compensate for gravity alone.
I found the graph (and parts of the explanation) very confusing, because the relation between starting and final angle are treated is as if the final angle is a single function of initial angle. In fact, it is highly dependent on the trajectory, so there is really not such a function, as long as the trajectory is not fixed. When connecting the starting angle with the final angle in the graph even enhances this confusion.
I'm a bit puzzled by this one. It seems trivial to find a path of the cart that would ALWAYS make the pendulum fail.
1-wait 5 minute on starting point (only stay up if 90degree perfectly up)
2-move half way. (pendulum at this point is not 90 degree up anymore)
3-wait 5 minute. (definitively fallen now)
4-end the course
So i'm puzzled about how someone could come with the conclusion that it's possible.
I thought about paths like that too but remember there is infinite precision in this cart/pendulum. So I could set the pendulum as arbitrarily close to 90 degree as I want such that it falls to any angle between 0 and 180 in those 5 minutes that we wait. In fact I could pick that angle such that once you start moving and get to the halfway point, it's back to some angle very close to 90 where it will only fall a little bit (or not at all), then you complete the course and the pendulum is still in the air.
@@zachstar If the movement to the middle is instantaneous yes. So i guess you need to move slowly to the middle to make sure the gravity have time to act while the pendulum is displaced by the acceleration.
I found that pretty obvious, you could move really slowly for 5 seconds, then accelerate the cart to the left, then move really slowly for 5 seconds, then accelerate hard to the right. The waiting requires the pendulum to be very near 90 degrees, but the jerky motion would throw it off. I kinda expected that answer to be an easy proof, but graphs are cool too.
Alternatively you could accelerate very fast to the left such that any starting angle would fail, then turn around and roll to the finish.
Easy to come up with a solution. Start with an angle very close to 180. Your hard left movement lifts the pendulum up and over to the other side. Now as you switch to rightward movement, this stop the pendulum from reaching 0.
since the car could accelerate at any rate (even go backwards) there would be several possible final angles for most given initial angles. on top of that, you could start at some angle and then accelerate up and down to balance the pendulum above the 0 and 180 degree threshold and then produce any final angle by timing its descent with the arrival at B. this should mean that the diagram (of the final angle and initial angle relationship) should be an area and not a curve as in 4:20 and 9:40.
If you had to treat it as a physics problem, including all of physics, then the answer would be a very easy "No". Due to Uncertainty, nothing can balance on a point (or an ideal bearing) indefinitely, and the area of the bearing surface influences the maximum duration. Therefore, for any physical system, there is a finite maximum time the inverted pendulum can balance without falling over while the cart remains stationary. Thus, any pattern of movement that includes a stationary period longer than that would result in it falling over. Calculating how long it would take is quite difficult, but simply demonstrating that there is an upper bound for stability is sufficient.
I am a bit puzzled.
He made an argument, that a discontinuity could occur, because there would be a jump when the angle reaches exactly 0° or 180°.
But why is this relevant for the original problem? I don't see how a discontinuity is possible with the original constraints.
If it is, there should be at least one example x(t).
Continuous is only applicable if it’s less chaotic. Makes a lot if sense, only few angles would work for some of them, since they’d skirt the line.
I think you explained it better than I could ever. I didn’t think about the “179 vs 180” issue
Mathematically this is correct. Physically the graph can't be any arbitrary function as any time the pendulum comes close to 180° it tries to converge to it and only a high acceleration cart can keep it from doing so. Thus the graph should be continuous.
I'm not satisfied with the solution to this problem by the end of the video. You have no specific counterexample of x(t) that would make the pendulum have to inevitably hit 0 or 180 degrees, and as for the counter proof, very wacky stuff happens when the angle is close to 0 or 180 degrees, and I still believe the final angle as a function of the initial angle is continuous.
Yes, I agree with this. To be specific, it's impossible to have any function such as the one at 9:41 where the graph "kisses" 0 or 180 degrees, because once it touches either 0 or 180 degrees, it cannot change direction (the acceleration of the cart must be finite).
This seems to be similar to the idea of dead zones in robotics or motors. They are locations in which the motors get stuck because their pressure is tangential to certain positions in the mechanical parts.
I really like your visualisations :) Good work!
To prove at least one final angle between 0 and 180 is always possible, we will first assume all final angles between 0 and 180 are always possible.
My idea was to just find a way for the cart to move that would guarantee the pendulum falls. Just imagine the cart goes really far left before coming back. If the pendulum is right of center it will fall on the initial trip left. If it’s left of center it might stay up going left but then it would fall going back right. And if it’s balanced center it would clearly fall immediately. But I’m not sure that this is really a solid proof.
I wonder how easy it is to construct a x(t) that causes all angles to fall over.
My idea involves making use of the fact that if you stand still the pendulum will fall over except if it is at 90 degrees with no momentum.
So basically the strategy I have involves 3 steps:
1. Stand still for y seconds. If the pendulum fell over, then we're done. If it didn't then we know the initial angle has to be at most some small distance from 90 degrees. Remember you can make the pendulum take an arbitrarily long amount of time to fall over by having the initial angle be close to 90 degrees.
2. Move away from B at a certain velocity until for all initial angles that didn't fall over the pendulum is far enough to the right that waiting y seconds at standstill will make it fall over.
3. Wait y seconds at standstill again
4. Finally move to B. Doesn't matter how as step 3 should've ensure the pendulum fell over.
The only issue I have is trying to make sure that by moving away from B at a certain velocity won't mess things up as we'll give the pendulum a lot more momentum.
Actually, standing still will eliminate a lot of angles, but after step one you still have all angles between 0 and 180 where the rod may end up (after slowly falling from some angle close to 90). Now you still have all possible angles for step 2, and moving at some speed will again eliminate some angles except for some neighborhood of the stable one. And once again you will end up will all angles from 0 to 180 as possible final angles for step 2
I really like the puzzle. There was another puzzle like this in my classical mechanics physics class where a pendulum was tied to a body and the. body would slide. Love the Lagrangian approach.
Didn't watch passed initial problem intrudoction... so I don't know if you're allowed to: Just drive backwards (enough that we are driving into the falling stick) then forwards, since driving will impart a torque counter-wise to the direction of movement...If not allowed to drive backwards....If the pendulum state starts "behind" the perpendicular relative to direction of movement...It will probably fall, but can we brake, then this function space isn't a problem to solve either... Otherwise.. distance, mass, acceleration, drag, pendulum length all will allow for various edge cases which make me think "fall or not" is a red herring... If it starts ahead of the perpendicular, a clever combination of acceleration and no acceleration should be enough, but if we can brake, even easier...
I wish you had showed a possible discontinuity on the graph with the intermediate value theorem where effectively you could have had a vertical line showing that it wasn't a true function.
Great video as always!
I don't know why does UA-cam algorithm recommend me this old video I have already seen but at least I gave it another thought and now I'm 100% sure that the problem not only has a solution but the solution can even be found knowing the motion of the cart.
First observation is that cart motion, actually its acceleration, modifies gravity. Meaning the actual gravity (vertical vector force) combines with the acceleration (horizontal vector force) to form gravity at increased strength and different angle than vertical. But since we combine perpendicular vectors, that gravity never points upwards. We can therefore construct function g(t) -> [a,f] where t is time, a is angle of gravity direction (0,180) and f is gravity force.
Next we can construct function V(t,b) -> v where t is time, b is angular position of the pendulum (0,180) and v is angular speed of the pendulum at given position and time. V(0,b) = 0 for any angle b. Then notice that due to the fact that the gravity always acts from above, even at most extreme accelerations, for t>0 is V(t,0) always negative (pendulums falling below 0 degrees) and V(t,180) is always positive (pendulums falling beyond 180 degrees). And V(t,b) is continuous function since it is continuous at t=0 and over whole time interval it is only "stretching" the interval (0,180). You can even choose final pendulum position or its final angular velocity (from within the available values) and backtrack it to initial position of the pendulum.
I think there are a few problems with this argument.
Firstly, from the same argument provided in the video, V(t,b) is not a continuous function. As soon as the pendulum hits either ends, the velocity jumps from a finite value to zero discontiniously.
Secondly, the modified force on the pendulum (gravity + acceleration force from the cart) is not always nessasarily on the downwards. Appart from the horizontal acceleration, there is also a vertical component of the normal reaction imparted to the pendulum weight from the connecting rod.
@@ghanshamchandel1854 Pendulums "hitting" ends are "discarded". We start with as many pendulums as there are real numbers between 0 and 180 so we can lose a few, any pendulum hitting the end is not part of the solution anyway. And the "gravity" always points down. Mass or inertia of the pendulum is not part of the "gravity" force. Certainly some pendulums may be accelerated towards vertical in certain configurations but the point is that the "splitting angle", i.e. angle from which all of the pendulums are accelerated away is always between 0 and 180, even taking their inertia into account.
@@ghanshamchandel1854 Actually, the fact that pendulums from within the 0,180 interval are continuously falling over BOTH edges is important part of the proof that the function is smooth and therefore there must be a solution. To lose the solution, you'd have to lift them from one of the ends, then apply strong acceleration towards that direction and make them all fall over the other edge. But you cannot do that without pushing the "almost horizontal" ones upwards and any finite horizontal acceleration does not give you that.
the situation that immediately came to mind is that if the cart ;
1. is still for a while(maybe a second, maybe millennia),
2. moves 50% of the way fast,
3. waits again
4. finally then going to the end, at one point, no matter the angle, the pendulum should fall.
Potentially you could do something like theta=179.99999999 deg so, to stop this you just add
3.b. go back to 25%
3.c. wait again
This should guarantee that the pendulum hits one of the edges
In fact, the assumption that the initial vs. final angle function is continuous excludes all of these problematic x(t)'s. However, this same assumption contradicts the supplementary assumption that x(t) might be arbitrary... instead it is limited to a "conditionally arbitrary" class of functions.
If we can move that cart any possible way, we could manage to prevent the pendulum hit the ground(since the gravity works normally).
There was no air friction means, you can only manipulate pendulums position by using inertia. And I guess its possible to finish this by not touching the ground.
Isn't it true?
In the problem presented in the video the path of the cart is chosen before you place the pendulum, not after.
Zach, this is only one issue with the problem. You gave a very good counterargument to the statement that stipulating the continuity of $f=\theta_\mathit{final}(\theta_\mathit{initial})$ is not enough; even if continuous, it its image must be bounded by the interval (0°,180°). What bothers me is that the continuity of $f$ is accepted at ease, and this assumption is likely incorrect. The continuity assumes that a small change of $\theta_\mathit{initial}$ causes a small change of \theta_\mathit{final}. I have a trouble with this. Inverted pendulum is well-known to be a chaotic system, and thus $f$ must be discontinuous for some choices of $x(t)$.
Except the path is continuous. The unrestricted curves never behave like that.
Suppose they did, then there must be a path that gets infinitesimally close to 180, and then goes back to 0.
But the only force is along the direction of the pendulum. (and the magnitude of this force is limited by the acceleration of the cart) So for a curve that gets really really close to 180, any upwards force is less than the downwards force of gravity. As such, the curve can't change direction.
The answer is clearly NO: There do exist some paths x(t) for which there exists no starting pendulum angle that will end at an angle other than 0 or 180 deg.
Consider an x(t) that remains at point A for several seconds, then moves a small amount towards B, stops again for several seconds, then finishes the journey. During the first rest at point A, any angle other than 90 deg will fall to 0 or 180, whichever is closer. Once it starts moving, if the angle was 90, it will no longer be 90. During the second rest, it will fall back to 0 or 180.
No, there are “semi-stable” angles different from 90 degrees too.
For example, 90.01 will stay upright for some time (the “several seconds” you mentioned)
Therefore, your further steps are not justified.
Thank you for giving such a nuanced solution for a classic problem.
It's easy to create a movement path that cannot be stood up in.
Move the cart at the maximum speed, to make it easier to think about, we can use c the speed of light.
And make AB (c * 1second) + 1 meter.
Now if the pendulum starts at 179.99999999999999... degrees, there is no way for it to avoid ending at 0 degrees, and if it starts at 180, it will end at 180.
I'm sure you can figure out the limit of the distance required if you move at the speed of light, since it is an equation that has a limit, but I'm fairly sure it cannot surpass c+1
If the cart is moving at a constant speed, the rod can simply stay vertical without falling over (since there is no air friction). If it is moving with a constant acceleration, there is a fixed angle, which is rather easy to find, which will prevent the rod from falling too.
@@fullfungo If you're changing velocity as a starting parameter, then breaking the rules of the puzzle.
0:58
But fine, we'll let you start at whaever speed you want. Cool, we'll just have it stay in position for 10 seconds, so any velocity you wanted to start with is now pointless, and you now need to find the equation to get the right angle after waiting 10s.
It's an extra step that you *can* add, but it doesn't affect the puzzle at all.
the real answer is that for any given angle, acceleratioin of the cart * cos(angle) > gravity for the pendulum to not fall, which means that it can be true for any angle except the limits, but for angles
An interesting thought on the intuitive side: it doesn't really seem that out of the ordinary that the function could be discontinuous, considering the problem has a condition whereby the pendulum could be forced to suddenly stop (i.e. potentially go to 0 velocity in a discontinuous manner).
Were we to keep the restriction on the range of angles, but introduce a smooth bouncing condition, I would imagine this kind of thing wouldn't be a problem!
Could you do a video on bode plots? They are really weird and it is hard to tell why they are useful.
I mean, they're certainly complex ...
To whoever is reading this, stay safe and have a great day. Ty for the great videos that make learning actually fun.
Assuming the cart starts perpendicular to the floor, and is perfectly balanced you can actually induce a stable state by first rolling back wards, causing the pendulum to move towards the right, and at the appropriate angle you can then begin accelerating “forward” towards the right so long as you maintain an appropriate acceleration the force of acceleration will be transmitted to the pendulum. This is my answer not watching this video, assuming gravity is perpendicular to the “floor” as well. Obviously this answer may not be accurate to the actual context of the video, but it’s my initial thought. If we assume you cannot start by first reversing then the answer if you can reach the point, is directly related to the distance needed to travel, any acceleration will induce the pendulum to fall, and assuming that greater acceleration causes the pendulums distance to ground to decrease proportionally to the carts distance to destination then this is unlikely. However assuming that it’s a real world sort of problem due to the angle of force applied to the pendulum (assuming it’s the “pull” of the cart at point attached, then this may be feasible)
I learned this intuitively from balancing a broom on my fingertip because a broom irl will act like an inverse pendulum.
I love the fact that this question is literally just a formal proof that someone can pop a wheelie on a bike.
I had my doubts the whole time but I couldn't point out exactly what was wrong. It was obvious to me that the assumption of continuity was wrong but not obvious how. Showing the version without constraints and how adding the constraints breaks continuity made it very clear.
Also intuitively from the start from math experience it felt like the problem was too hard to be solved by invoking the intermediate value theorem.