Thank you, Sir, it's wonderful explanation and you should be a brilliant star lecture! The flow of the lecture is; minimising Potential Energy to Lagrange Multiplier to Euler-Lagrange(never boring :)) to Beltramy Identity and finally two ends boundary conditions and the length of the line = l. Only one mistake is eq.23 where the correct is Lambda/(mg) instead of Lambda/C1, but it wouldn't alter the solution :) But the lecture was great without doubts!
Thanks for your words and your feedback. I have been asked this question a few times, so I'll pin this comment to the top. Eqn 23 is correct as written. What I did in a single step (without adequate explanation) is: 1. subtracted λ/c1 from both sides, 2. divided both sides by mg, and 3. absorbed the mg into the λ (which is okay because λ is simply a constant - I could have called it a new constant λ' or λ2 or something else, but I just kept it as λ).
@@Freeball99 this is incorrect because when after you subtract the lambda/c1 you later have to multiply both sides by c1 anyway in order to isolate the y, which cancels out the c1, so absorbing that into a new arbitrary constant is not valid (because it’s not really arbitrary). Then dividing both sides by mg yields lambda/mg, as the original commenter correctly stated
12:44 ... years ago I was tasked with working the length of a suspension bridge, I did it with repeated use of substitution and the answer was full of square roots etc. The engineer tasking me took a piece of string and measured his drawing, It took him 3 seconds! and he said yep That'll do! Someone after asked why I hadn't used hyperbolic functions .... I was straight out of high school and had never heard of them! I appreciate this video :-)
I study mathematics at university and wanted to say that even though your videos are more aimed at applied maths, you have a beautiful style which pure matgs professors could benefit from. The historical aspects were wonderful and so rarely found. I hope you keep posting videos and cover pure maths too.
Your explanation on the variational calculus especially in the first video unlocked the mystery of understanding the meaning of the various paths graphs since my teenager age! After more than 20 years, then now I am fully understand each of the knowledge based on your clear lecture. Thank you Prof
Your lectures are priceless. Very clear and easy to follow. I intend to gobble up more of them. But as I was considering 'going down the rabbit hole' toward a numerical solution, the issue of units came up. And I believe that a parenthesis got dropped on the way from Eq 27 to Eq 28. I think the arguments of the sinh function in Eq 28 must be the same as they are for the cosh function in the earlier solution for y. Thanks again. All for now.
@15:19 i feel " -lambda/mg ". Thank you freeball for solving this problem. i was not able to solve it without beltrami identity. Your presentation is clean and beautiful.
I absorbed the mg into the λ since λ is just a constant (I should have mentioned that). I have pinned a comment to the top of the discussion section which explains this in more detail.
This topic is so interesting and you have made it so more interesting. Thanks a lot for that. I have found that the shape function for the hanging chain problem does not depend on the mass per unit length. However, the same is not evident in the mathematical expression.
Interestingly, the shape of the catenary minimizes the tensile load in the chain. As a result, arches that are built in the shape of a catenary are the strongest as they minimize the compressive loads.
@Amit Correct. The parameter mg/c_1 sets a common scale for horizontal and vertical distances. In units of mg/c_1, the shape will follow the cosh function; see Equation 23.
Incrediby well explained. I wonder if minimising potential energy of chain is equivalent to nature's way of evenly distributing the force/tensions in the rope.
Thank you very much for your video lectures, dear Andy. I have a question in this one: when we determined the eqn of catenary in the end, we had three unknown constants there, including the lambda. For that, you said we determine the third condition using the constraint equation. But then we wrote an eqn with l and c1 and c2. What about the lambda? And what do we do with this 3rd condition if we leave the lambda aside?
You have 3 equations and 3 unknowns so this is solvable though it will likely require a numerical solver in all but perhaps the simplest cases. I have shown an example in a different video of how to treat this for a pendulum problem where I substituted the constraint equation into the other two equations of motion. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-wPk16G7mkZY.html
Since lambda is just a constant, I have absorbed the mg into lambda since a constant divided by a constant is still a constant. What I have done in a single step is to define a new constant, say, lambda2 = lambda / mg and then recognized that lambda2 is a dummy variable so I could just call it lambda.
The 'usual' equation for the shape of a catenary is, I believe, y = acosh(x/a) which is much simpler than the one that you finish up with. Are they effectively the same? I like the maths though.
The shape of the two are the same (ie the cosh function). In our case, we need to fix the cosh function curve to the starting and ending points PLUS we need to generalize it for chains of different length. This is why we end up with the constants c1, c2 and λ.
Another gem of a video. I do have one question, though. In the introduction to the Lagrangian multiplier, you say that G, the constraint, must be taken to be zero, so that it effectively changes nothing when added, but here J doesn't seem to satisfy this condition. I'd appreciate some clarificafion, if you've got the time. Thanks anyway.
This really requires a mathematical derivation of the method to explain it. The basic idea is that by adding λG in this form (where G=0) does not affect the necessary condition for finding an extremal of the modified functional. Unfortunately I just don't have the ability to explain it using text only, so will have to save it for a video.
Hey, nice FreeBally. Nice to hear a Cape Town accent. I have a more important optimisation problem to solve, which is that of an elastic catenary with fixed weight at some point in the catenary. The problem space is maximum highball free-lining, where we deliberately span open spaces with elastic climbing ropes and the walk them. The ultimate boundary condition of course is elastic strength failure of the freeline. However that failure condition is not the solution to the fundamental problem. What is the catenary solution for an elastic rope in the presence of one or more discrete local masses?
Originally a Durban accent, now with a California influence... Interesting question. I'll need to give it some more thought, but fundamentally, the discrete mass produces a discontinuity and would thus produce a boundary condition. Thus we would have to treat the rope on each side of the mass separately (if it is symmetric, we could just solve for one half). Also, it is not immediately clear to me whether this problem would have an analytical solution, but certainly we could set up the governing equations. It's a difficult problem to explain in these comments; really requires a video. I had planned to make a video demonstrating the case of a beam with a discrete mass. This will be fairly similar to the problem you have posed in that it was demonstrate how to treat discrete masses - after all, a rope is just a beam without any flexural rigidity (EI = 0). Perhaps I'll have to promote to the top of my To Do List.
"elastic catenary" might be thought of a chain where each link is a spring with a finite spring rate. A realistic problem in practical engineering, e.g., suspension bridge.
Hey, brilliant video mate, I thought of this problem a few weeks ago but couldn't solve it, now I have a name and a solution! I have a question/comment though, at 15:30 in equation 23 shouldn't lambda's coefficient be 1/mg?
Since lambda is just a constant, I absorbed the mg (which is also a constant) into the lambda. It's a different lambda from the line before (I didn't want to call it something different) but it's just a constant all the same and will come from solving the given conditions.
@@Freeball99 Ah, fair enough. you have lambda/C1 though, wouldn't it be more efficient to absorb that too? or is this a sort of canonical form and these specific constants have meaning behind this equation? And wouldn't it be better to specify C1/mg as another constant, then it takes the form y=a*cosh((x+b)/a)-c, which imo looks much cleaner?
Thank you for the insightful lecture! 🙏 Quick question: At 5:00, when discussing the variation of (I + lambda*J) = 0, shouldn't we express J as an integral equation - L = 0, as discussed in your Lagrange multiplier video? Just looking for clarification! Thanks again! 🌟
This is a very good question and really requires a video deriving each method to get a proper understanding of why this is. I haven't made the video yet because it will be very mathematical, dry and likely not very popular. In short, however, these two problems you mentioned are not the same type of problem due to the different nature of the constraints. This problem (the hanging chain problem) is known as a CONSTRAINED OPTIMIZATION problem because the constraint is written as an integral and serves as an overall constraint of the system. On the other hand, the pendulum problem is known as a CONSTRAINED COORDINATES problem since the constraints involve the coordinates themselves (could also include derivatives) and apply to every point within the domain, rather than being satisfied in an integral sense like for the hanging chain. While both problems are solved using the Lagrange multiplier method, the reason that we do not subtract L from the constraint in this video is due to the difference in the types of constraints. So the methods used in each of these two videos are correct, i.e. in the case of an integral constraints, we DO NOT write G in the form G=0 while in the pendulum problem, we DO write G in the form G=0. Hope this makes sense.
19:00 after you've substituted the identity and simplified the integrand, shouldn't the parentheses still be around the x+C_2? I have little experience with hyperbolic functions but that's something which shouldn't have changed? A surprisingly welcome video nontheless!
What I did in a single step here was: 1. Multiply each of the terms by mg/c1 2. Recognize that the 2nd term is mg/c1 * c2, which is just a constant so I absorbed it into c2 - I probably should have made this more clear in the video. I could have instead named it c3, but I didn’t want to start adding new unknowns so it was easiest just to include it as part of c2. This is a commonly used technique when multiplying some unknown constant by additional constants and I use it in some of my other videos (I might have explained it in one of those).
Masterfully explained! The method of Lagrange multipliers really shine in problems like this where the medium is a continuum instead of a set of discrete point masses. Just curiously though, if I were to solve this numerically, would there be alternatives to Newton's / Broyden's method?
Brilliant explanation, I agree. However, I think it’s important to note that the solution is independent of mass and energy. This can be easily implemented by replacing lambda with m*g*lambda_2 in Formula (4) at minute 6:10. Then, you can divide by m*g and it disappears, since the other side equals zero. Am I right? 🙂
@@breitbandfunker4332 I'm pretty sure that works out the same. You can see the final solution in 21:53, if you replace λ with mg λ_2, then c_1 will always be divided by mg. Substituting c_3 = c_1/mg, you will not see mg in the final solution. Although, you are right, explicit independence relations are always good in physics.
@@Zxymr well, I think if you absorb mg in the beginning, it vanishes completely in λ_2. Then you can determine the constants only with starting and endpoint. It should be completely independent from mass and g-factor, which means, any chain has the same shape - even on alien planets!
Using the Lagrange multipliers is it possible to add symmetries to a Lagrangian by finding the conserved quantities by Noether's theorem and setting the variation to 0?
Hello, thanks for the video. , may I ask what the problem is called when one end of a chain is fixed somewhere and the other is free ? So that it is like a pendulum though flexible and bendable at any point instead of stiff. Thanks
Not sure what this is called. If the chain is massless, then perhaps a mass on a string. If the chain has mass, I'd probably call it a compound pendulum
I'm still struggling but it's still a great video. Thanks. 14:19 Does anyone know why Eq21 doesn't have a C_3 on the right hand side after he solved the integral? Nevermind. It's possible to combina C_2 and C_3 so there is only a constant. I saw it later. I'll leave the post in case anyone is confused by the same thing.
Hi, I think you made a mistake @Freeball99 at 2:05. Your formulation for I integral is wrong. So I is the minimal potential energy across the chain from A to B so we could write: I = ∫ dEp (dEp is potential energy for a infinite decimal point/segment across the chain) I = ∫ dm * g * y I = ∫ ρ * dV * g * y but ρ = m/V = m / (A*l) dV = A*ds (constant density and area across the chain) So replacing we get: I = ∫ m / (A*l) * (A * ds) * g * y I = ∫ mgy / l * ds Obviously it will not influence further y calculations, but it’s something which I noticed in university: mathematicians tend to forget that integrating across any dimension is like multiplying by it 😉😅
Hi sir, i have a problem of rope with concentrated load. The problem is,I have a rope with 6 meters length, at the first end of rope (Point A pinned at x=0, y=0) and the second end of rope (Point B pinned at x=4,y= -1) then divided the 6 meters rope to put a 2 concentrated load with the same mass( .5kg ) so there has 3 segments and each segment is 2 meters. The question is. what is the x and y position of 2 concentrated load after i hang it into the rope? please help.. thank you in advanced.
Catenaries appear in many different applications. Clearly in the design of suspension bridges and in designing power lines. They are used in architecture when designing arches - arches designed in the shape of an (inverted) catenary are the strongest arches you can build (since the catenary shape minimizes the compressive load). Catenaries also used in ship building for optimizing the design of anchor chains.
@@Freeball99compressive loads (unless stability related) are not a problem in archs though. The real reason why a catenary shape for load carrying beam structures is desirable, is because we only want the structure to be loaded by membrane forces and not by bending moments. This is because membrane stresses are distributed uniformly over the cross section and therefore make use of "all the material" to carry the load.
I was wondering whether instead of to set the variation of " I + \lambda * J" to zero you would set the variation "I - \lambda*J" to zero. Couldn't that be called the Lagrangian, L= I - \lambda*J ? At the end, since the Lagrange multiplier is a constant number and what is being minimized is linear with respect to \lambda, its value wouldn't be simply the negative of what one gets considering the summing? Thanks for shuch a great video
I have described it in the video above. It is the form that the Euler-Lagrange equation reduces to if the functional, F, is not explicitly dependent on x. The mathematics behind deriving the Euler-Lagrange equation are described in the following video (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-VCHFCXgYdvY.html).
s(b) and s(a) are two different locations/values for the path. The difference between the two is literally the definition of the path length, L. Similarly, x(b) - x(b) would be the distance along the path in the x-direction.
Sorry for the delayed response, but I somehow missed this until now...in this case you have added an extra unknown (the length of the chain) so we need an extra constraint equation. This comes from using the relationship between the length of the chain and the height, y (this comes from the shape of the catenary which you already have an equation for).
@@Freeball99 ok the mg absorption makes sense now thanks, but since there is a c1 multiplying the cosh and nothing dividing y, i really don't see why there still is a c1 dividing lambda
You did something wrong though. You shouldn't have the mass of the chain in the integral. You should have its linear density. Cause potential energy is mgy therefore the integral for the chain would be ρgyds where ρds=dm the infinitesimal mass of the chain and ρ its linear density such that integral ρds=m of the chain. You can even fact check this by looking at 10:17. You have a force being equal to mgy+other stuff, but mgy has units of energy not force. If instead you had ρgy where ρ is the linear density, you would have the correct units :) Other than that the whole Calculus of Variations is solid and thank you for your video!
In this problem, m = ρA is the linear density and not the mass of the chain. I mention this at 2:05 in the video. So, I am integrating ρAgy with respect to x. At the 10:17 mark, the "F" that I am using does not stand for "force", but rather it is the "functional" - that's why the F. In this problem, the functional has units of energy per unit length, but in general could have various other units depending on the quantity that we are trying to minimize.
in catenary which has uniform mass into the length, all forces gravity etc are cancelled out, ive done this to figure out to determined of all the angles in each fixed same length segment as a original chain. without hyperbolic cosine function, we can still to figure out the catenary, here ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YfVsV_jPSrM.html
I'm not familiar with this work, but I'm guessing the weight or mass, gravity, coriolis or spin of planet, and dimensions are taken into consideration for the solution? I only watched 4:25 of the video so I'm just risking errors. I'm no mathematician by any means; but, my opinion is that solutions to problems with force, energy, inertia, movement, stationary objects, etc would have to take into consideration the spin of our planet, solar system, and even galaxial motions.
We cannot grasp anything on It's fullest, we only haver models, and preferably useful ones. So ignoring a lot of negligible influences makes the solution easier - sometimes, it makes them computationally or even analitically viable...
Yes; I understand, the work is done in portions or puzzles. The bigger picture being too complex for the human mind. But, if we never work on the big picture; we will never find the solutions. The ignored portions considered nuisance become missing links.
@@biancabonet You are right, and sometimes the details make all the difference. This is very evident in dynamical systems, for example: the tiniest detail can make a huge difference over time. In the example of the video, however, it could probably be shown that the effects you cited have a very negligible and controlled effect overall, so that the solution is still imensely similar to the one this model already gives.
@@raphaelreichmannrolim25 Thank you. I tend to focus on purity and accuracy and conservation of the big picture. While focusing on retainment of the CMB, I guess you may zoom in...but zoom in to see plank sizes...you would still have to zoom out and see the macro-universe to begin to understand--to see the big picture..
I made a typo here. It should actually just be -λ (not -λ/c1). Since λ is a unknown constant, I can simply absorb the μg (another constant) into this since a constant divided by a constant is still a constant. If you look at the comment pinned to the top of the comment section, it is explained in more detail.
