So rare is a content that goes down to the nitty gritty details in order to add solid arguments to a wandering debate. And if only a fraction of the audience can validate or disprove the argument, well that’s science going to business. I will you’d educate more explaining why you would integrate or differentiate or why you drive the calculation the way you do but you cannot suit every need in one video in a limited time. That’s excellent. I’ve watched it through and my only concern is the coefficients alpha and beta that seem to prove something when they come a bit ad-hoc it seems. Thank you for the damn calculation and your time recording it.
Wow. I didn't know any of the calculation for this, had a little trouble trying to understand all the formulaes (not because of your explanation, you were very clear), I'm just not that well educated in maths and physics. Based on observation, I though the force needed to make the beads change direction, would translate back into the chain, in form of tension, divided to both sides, like half the orbit of the moon, symplistic speaking. The shorter side (the chain leaving the pot), has less mass, and therefore achieves a higher acceleration, bringing the chain arch up. For a given shape, what would change this force is only the velocity of the chain: Higher velocity means a bigger force needed to change the direction, which would be translated in tension on the chain. I tought, if so, the lateral movement shouldn't matter, and that I was just wrong for some reason i couldn't understand.. I just simply couldn't prove I was right. Or wrong! I really wish I studied more.. This was just a lucky guess, that barely hit near the answer. To think Galileo did all the calculations of our solar system even before he could observe.. That is science!
At a "meta-level" the chain fountain/Mould effect surely is something that engineers from the nineteenth early twentieth century would be well acquainted with - belt-drives were used to distribute power in factories to drive machinery all over the place. When the speed of the belts become too high compared to the tension, the belts start to "overshoot" around the wheels (there are a couple of videos on here illustrating this phenomena). Since that cannot be good for anything (power-transmission, stability of operation, workplace safety...) this surely must have been a point of emphasis in engineering education? Thankfully we got more efficient and safer power-distribution systems, but then this quirky mechanics got left outside the curriculum and generally behind and with time forgotten...
I appreciate the math, but skimmed through it only. You claim to have gotten similar results as Biggins and Warner, but they claim a fountain can only rise from rigid chains such and these, and that it needs a hard support surface. Do you assert this as well? I'm curious as to what your analysis says. And what of different chain configurations, can your analysis predict whether a chain configuration will/can yield a fountain? If so, can easily tell which fountain will be higher? I'm curious as to the extent of your calculations. What all can it predict? Thanks in advance. I'd like to know because my findings says Biggins and Warner are wrong and that no force from the surface is needed. Does it exist? Yes, but it doesn't give the fountain its rise is my findings. They claim it's tension and this additional force from the surface. Do you take the same position? Just curious. Thanks in advance!
There's also an increasing fountain height to calculate as demonstrated by Steve Mould. I suggested aerodynamic lift is developed by the top arc shape & bead chain speed. Can you calculate the Bernoulli or Euler formula for resultant increasing lift?
No, as vel of chain increase as the end reaches down, if pulls harder the stationary chain. So height increase. But notice it doesn't decrease, bcz the vel never decreases of any link of chain
One may choose formulas based on understanding of the problem. If the fundamental physics / issues of the problem are misunderstood, incorrect formulas may be chosen, and thus the calculation becomes meaningless. Especially with several complex systems coming together to form an even more complex system. Torque, Inertia, chain-whip-motion mechanics...
I'm guessing everyone here is either a professor or a professional physicist who has had some kind of physics background , can anyon explain me the science (or atleast their hypothesis) behind the propulsion system of the UFO's told by Bob lazar?
I like this approach and this video, I really do. My comments come from a place of respect. So at one point, to find the 'stationary' solution of the equations, you set the first derivative equal to zero. On a potential energy graph like y=x^2, this will be a point of stable equilibrium. But it seems to me that in your model, this catenary is an unstable equilibrium. Since, empirically, the physical chain fountain reaches (/approaches) a somewhat stable state, there must be some negative feedback in the physical system, not a part of this model, which counters the inevitable wobbles away from the unstable equilibrium, and indeed, brings the system towards that point from a state that's really quite far from it in state space, the state of the end of the chain dangling from the edge of the container. If you could show that this stationary solution is a stable equilibrium, an attractor, that'd be massive. Otherwise, you're not showing any reasoning, or any mechanism, by which the system goes from the state 'chain-draping-over-the-edge' to the state 'chain-fountains-up'. Best of luck.
The youtube auto-generated transscript is good. It is unbelievably good. To acces the transscript: click the group-of-three-dots icon underneath the video. That opens a menu with the choices 'report' and 'open transscript'. The transscript column has a three-dots icon of its own, clicking that opens the option 'Toggle timestamps'.
@@dangraff8467 i saw a simulation of it. it was in favor of mehdi. i mean mehdi came as close and precise as he can to explain it but he wasn't correct i guess. There are some unanswered questions left.
NO WAY!! Too many considerations, so may simplification. That can't translate reality! My take is: 1. Roughly, the driving force pulling the chain down is equal to the sum of masses of the falling beads times gravitational acceleration. Hence, the pulling force will increase until the first bead in the chain reaches the floor, after that the pulling force will be constant. 2. A chain in tension is an efficient system to transfer mechanical forces, therefore each bead resting in the pile will be yanked out in succession by this pulling force. 3. If losses are low, like friction forces, beads will need to be yanked out in very rapid succession, with huge upwards momentum, and that will create the fountain effect. 4. In other words, the bead's initial kinetic energy will have to be transformed into potential energy and again into kinetic energy downwards, if energy losses are small in the system. The math to prove this should be much simpler.
So, after all that, no definitive answer ? The reason the chain rises above the containers lip, is due to centrifugal forces-nothing more. I can't believe all these !so called! academics getting this wrong. Every force has an equal and opposite force. Therefore, the force the chain is applying to the base, the base applies an equal and oppsoite force to the chain. Therefore, the total force is equal to 0. So there is 0 upward force being applied to the chain from the base, as the bases upward force is equal to the chains downward force. Regardless of these fancy calculations, and what the !academics! have to say, you cannot lift yourself off the ground by pulling on your own boot laces.
48:10 ... impose the condition that y'(0)=1. There is no point on the catenary where y'=1. So it seems, by imposing this condition, you're introducing a logical contradiction into your model... EDIT: should have had the patience of another minute, yes, it's the 'degenerate' catenary of straight up-and-down. Sorry.
At 0:50 u are wrong. The inertia explanation is correct. The Cambridge explanation is rubbish. There is never a bonus-kick at the container/jar/beaker. There is indeed a bonus-kick when a falling chain-link hits (collides with) the floor of the laboratory, & this adds to the downward pull of a chain, & increases the size of the fountain etc (by say 1%). But that extra (bonus) force is initiated by the falling/colliding link itself. Meanwhile, back in the jar, a rising/yanked link will indeed get a kick from the floor of the jar (or from the links supporting that link), but, that kick is not initiated by the rising link, it is initiated by the preceding link. The preceding link has a limited amount of impulse to give. The kick from the jar results in a similar (but opposite) kick being given to the preceding link. But the kick from the jar is not a bonus-kick. The initiating impulse from the preceding link is not added-to by the kick from the jar. The impulse of the kick experienced by the rising/yanked link has been borrowed from the preceding link. In reality, borrowed from the full/whole chain. Hence, at the jar, links or beads, it makes no difference. In reality, there is no rising/yanked link. Duznt happen. What i mean is, we mostly have a rising/yanked arc, made of many links. In slow-mo u can see that there is mostly a long arc of moving links (talking bout metal beads here). And the movings usually involve being dragged gradually horizontally for a time, & gradually upish. There is no sudden kick -- it is a gradual kkkiiiiiiiiiiiiiiiiiicccccccccckkkkkkkkkkkkkkk. Look in slow-mo at a bead chain fountain. There are say 4 adjacent beads acting as pseudo-links being yanked/jerked up. (a) But, these pseudo-links are not rigid, they have a lot of give. (b) And, these & the trailing say 5 or even 10 beads are moving horizontally & have daylite under them. (c) The pseudo-links never have a solid floor to kick (down) against. (d) And, even if the last bead in the pseudo-link does enjoy an upwards (very very weak) kick, this kick is off a slippery & lose bead or two, & the kick moves thems beads sideways (or forwards)(or backwards)(they are after all lose in every way). So, beads do not provide anything like a solid surface to kick offa. Sheeeesh! Stone the crows! Milo give me strength! Are we blind! The fountain (ie the arch)(the half circle) is simply due to the inertia of a chain. If a stationary chain lays over the top of a wall, it forms a sharp V-bend. If the chain is pulled down on one end, then it will rattle up & over the wall, with some speed. If the speed is sufficient, the sharp V-bend will become a U-bend, due to centrifugal/centripetal forces opening up the V. If speed increases, the U-bend opens further. If speed increases a lot, the rising chain leaps up clear of the wall, & stays clear, the zenith/crest depending on gravity, & the U-bend opens some more. And we have our fountain. The height of the crest/fountain depends on speed (& is probly higher if chain heavier). The radius of the U-bend/fountain depends on speed (& is probly larger if chain heavier). Steve Mould & others say that the radius of curvature cancels out in the equations, & doesn't play any roll in these dynamics. Steve is wrong. If he were correct, then that would mean that the radius can vary greatly, for any given setup. It might mean that if u initiated the fountain with a given radius then it would affect the final (maximum)(steady state) radius. But i reckon that the radius (for a given setup) tends to one number. And i reckon that the radius (for a given setup) varies with speed (ie varies with drop). Just koz the radius cancels out in some dynamical equations duznt mean that radius duznt play any roll in the dynamics. Steve Mould shows how a horizontal tight pattern of rows of chain (beads) gradually moves away to the west (due to the kick-effect) when the end of the chain is pulled east. Yes, there is some kick effect, due to the pseudo-links. But, the greater part of the effect is due to a chain's inclination to retain a pattern. Here, we see that at each end of each row the chain forms a loop, looping around & back throo say 220 deg, such that the rows are hard up to each other. As the end of the chain is drawn east, the end loops move across & back & across & back etc. Now, Steve knows that a chain has a memory, ie it tends to retain a pattern. In this case, that pattern is a 220 deg loop. And, as the loop or loops traverse across & back etc, the loops slowly push the rows west. That westwards slow pattern-push is a different animal to the quick jerk of a pseudo-link kick. The slow pattern-push duznt suffer from the hi losses suffered by a pseudo-link kick. Pseudo-links are not rigid, they have give. Give aint a problem for a slow westward pattern-push. I said that yes there is some kick-effect. But, i have already said that these kicks are not bonus-kicks. They borrow from the overall power of the chain system. They don’t add to the fountain. It is virtually impossible to load/store a chain in a jar without having loops. And loops result in jumps. If u have a close look (in slow-mo) u can see that chains have a lot of horizontal movement before exiting vertically, & the chain sometimes suffers little jumps in the jar, as each leg of each loop jumps (horizontally) over its mate. Jumps magnify the fountain, kicks dont. OOPS. No, i am wrong. These little jumps are not bonus-jumps, just like the kicks are not bonus-kicks. Any vertical impulse gained by a jump must have been borrowed from preceding links. Also, the horizontal movements seen in the container/jar/beaker are a waste of impulse. But, in any case, we don’t need bonus-jumps nor bonus-kicks to achieve a growing fountain, & to achieve a very high fountain. All we need is lots of speed. The chain will crest at any height, depending on speed. There is no limit. The idea that something special (like a bonus-kick) is needed (if the chain is to rise above the edge of the jar) is silly. One thing that everyone has missed, in every youtube re the chain fountain (that i have seen), is that vertical jump effects & vertical kick effects (etc) are cumulative, and lasting (tautology alert). Or at least partially cumulative & lasting. Air friction, & link friction see to that. Anyhow, cumulative/lasting effects are the reason for some of the slightly weird gyrations/waves. For example, it (lasting accumulation) is why we eventually get a persevering, vertical component of arc (a mini fountain) when the test seems to be strictly in the horizontal.
So rare is a content that goes down to the nitty gritty details in order to add solid arguments to a wandering debate. And if only a fraction of the audience can validate or disprove the argument, well that’s science going to business. I will you’d educate more explaining why you would integrate or differentiate or why you drive the calculation the way you do but you cannot suit every need in one video in a limited time. That’s excellent. I’ve watched it through and my only concern is the coefficients alpha and beta that seem to prove something when they come a bit ad-hoc it seems. Thank you for the damn calculation and your time recording it.
Wow. I didn't know any of the calculation for this, had a little trouble trying to understand all the formulaes (not because of your explanation, you were very clear), I'm just not that well educated in maths and physics. Based on observation, I though the force needed to make the beads change direction, would translate back into the chain, in form of tension, divided to both sides, like half the orbit of the moon, symplistic speaking. The shorter side (the chain leaving the pot), has less mass, and therefore achieves a higher acceleration, bringing the chain arch up. For a given shape, what would change this force is only the velocity of the chain: Higher velocity means a bigger force needed to change the direction, which would be translated in tension on the chain. I tought, if so, the lateral movement shouldn't matter, and that I was just wrong for some reason i couldn't understand.. I just simply couldn't prove I was right. Or wrong!
I really wish I studied more.. This was just a lucky guess, that barely hit near the answer. To think Galileo did all the calculations of our solar system even before he could observe.. That is science!
At a "meta-level" the chain fountain/Mould effect surely is something that engineers from the nineteenth early twentieth century would be well acquainted with - belt-drives were used to distribute power in factories to drive machinery all over the place. When the speed of the belts become too high compared to the tension, the belts start to "overshoot" around the wheels (there are a couple of videos on here illustrating this phenomena). Since that cannot be good for anything (power-transmission, stability of operation, workplace safety...) this surely must have been a point of emphasis in engineering education? Thankfully we got more efficient and safer power-distribution systems, but then this quirky mechanics got left outside the curriculum and generally behind and with time forgotten...
Hi! I really love this video, is there a pdf version of all the deductions! thx u very much!
I appreciate the math, but skimmed through it only. You claim to have gotten similar results as Biggins and Warner, but they claim a fountain can only rise from rigid chains such and these, and that it needs a hard support surface. Do you assert this as well? I'm curious as to what your analysis says. And what of different chain configurations, can your analysis predict whether a chain configuration will/can yield a fountain? If so, can easily tell which fountain will be higher? I'm curious as to the extent of your calculations. What all can it predict? Thanks in advance. I'd like to know because my findings says Biggins and Warner are wrong and that no force from the surface is needed. Does it exist? Yes, but it doesn't give the fountain its rise is my findings. They claim it's tension and this additional force from the surface. Do you take the same position? Just curious. Thanks in advance!
There's also an increasing fountain height to calculate as demonstrated by Steve Mould. I suggested aerodynamic lift is developed by the top arc shape & bead chain speed. Can you calculate the Bernoulli or Euler formula for resultant increasing lift?
No, as vel of chain increase as the end reaches down, if pulls harder the stationary chain. So height increase. But notice it doesn't decrease, bcz the vel never decreases of any link of chain
One may choose formulas based on understanding of the problem. If the fundamental physics / issues of the problem are misunderstood, incorrect formulas may be chosen, and thus the calculation becomes meaningless. Especially with several complex systems coming together to form an even more complex system. Torque, Inertia, chain-whip-motion mechanics...
I'm guessing everyone here is either a professor or a professional physicist who has had some kind of physics background , can anyon explain me the science (or atleast their hypothesis) behind the propulsion system of the UFO's told by Bob lazar?
I like this approach and this video, I really do. My comments come from a place of respect.
So at one point, to find the 'stationary' solution of the equations, you set the first derivative equal to zero. On a potential energy graph like y=x^2, this will be a point of stable equilibrium. But it seems to me that in your model, this catenary is an unstable equilibrium. Since, empirically, the physical chain fountain reaches (/approaches) a somewhat stable state, there must be some negative feedback in the physical system, not a part of this model, which counters the inevitable wobbles away from the unstable equilibrium, and indeed, brings the system towards that point from a state that's really quite far from it in state space, the state of the end of the chain dangling from the edge of the container.
If you could show that this stationary solution is a stable equilibrium, an attractor, that'd be massive. Otherwise, you're not showing any reasoning, or any mechanism, by which the system goes from the state 'chain-draping-over-the-edge' to the state 'chain-fountains-up'.
Best of luck.
How old are you sir
@@lawliet2263 In the double digits. Why do you ask?
@@hughobyrne2588 what's your background in education?
@@lawliet2263 I've been on both sides, both student and teacher.
Why so nosy?
In all the videos the chain always spirals on the way up. Does the math take this into account?
Wow. 1h video. Is there a pdf for this?
The youtube auto-generated transscript is good. It is unbelievably good. To acces the transscript: click the group-of-three-dots icon underneath the video. That opens a menu with the choices 'report' and 'open transscript'. The transscript column has a three-dots icon of its own, clicking that opens the option 'Toggle timestamps'.
@@cleon_teunissen how to on phone?
so who wins?
@@dangraff8467 i saw a simulation of it. it was in favor of mehdi. i mean mehdi came as close and precise as he can to explain it but he wasn't correct i guess. There are some unanswered questions left.
NO WAY!! Too many considerations, so may simplification. That can't translate reality!
My take is:
1. Roughly, the driving force pulling the chain down is equal to the sum of masses of the falling beads times gravitational acceleration. Hence, the pulling force will increase until the first bead in the chain reaches the floor, after that the pulling force will be constant.
2. A chain in tension is an efficient system to transfer mechanical forces, therefore each bead resting in the pile will be yanked out in succession by this pulling force.
3. If losses are low, like friction forces, beads will need to be yanked out in very rapid succession, with huge upwards momentum, and that will create the fountain effect.
4. In other words, the bead's initial kinetic energy will have to be transformed into potential energy and again into kinetic energy downwards, if energy losses are small in the system.
The math to prove this should be much simpler.
So, after all that, no definitive answer ? The reason the chain rises above the containers lip, is due to centrifugal forces-nothing more. I can't believe all these !so called! academics getting this wrong. Every force has an equal and opposite force. Therefore, the force the chain is applying to the base, the base applies an equal and oppsoite force to the chain. Therefore, the total force is equal to 0. So there is 0 upward force being applied to the chain from the base, as the bases upward force is equal to the chains downward force. Regardless of these fancy calculations, and what the !academics! have to say, you cannot lift yourself off the ground by pulling on your own boot laces.
Sir, please make a video on physics of helicopters!
Good idea. It would take a bit of effort, since I don't know anything about helicopters. (I took a ride in one a few years ago...)
Kumari Nidhi would you ride your bf or a helicopter?
@@lawliet2263 did you get the response
@@kumarinidhi4000 you didnt answer me
48:10 ... impose the condition that y'(0)=1.
There is no point on the catenary where y'=1. So it seems, by imposing this condition, you're introducing a logical contradiction into your model...
EDIT: should have had the patience of another minute, yes, it's the 'degenerate' catenary of straight up-and-down. Sorry.
Gravity A and Gravity B
At 0:50 u are wrong.
The inertia explanation is correct.
The Cambridge explanation is rubbish.
There is never a bonus-kick at the container/jar/beaker.
There is indeed a bonus-kick when a falling chain-link hits (collides with) the floor of the laboratory, & this adds to the downward pull of a chain, & increases the size of the fountain etc (by say 1%).
But that extra (bonus) force is initiated by the falling/colliding link itself.
Meanwhile, back in the jar, a rising/yanked link will indeed get a kick from the floor of the jar (or from the links supporting that link), but, that kick is not initiated by the rising link, it is initiated by the preceding link.
The preceding link has a limited amount of impulse to give.
The kick from the jar results in a similar (but opposite) kick being given to the preceding link.
But the kick from the jar is not a bonus-kick.
The initiating impulse from the preceding link is not added-to by the kick from the jar.
The impulse of the kick experienced by the rising/yanked link has been borrowed from the preceding link.
In reality, borrowed from the full/whole chain.
Hence, at the jar, links or beads, it makes no difference.
In reality, there is no rising/yanked link. Duznt happen.
What i mean is, we mostly have a rising/yanked arc, made of many links.
In slow-mo u can see that there is mostly a long arc of moving links (talking bout metal beads here).
And the movings usually involve being dragged gradually horizontally for a time, & gradually upish.
There is no sudden kick -- it is a gradual kkkiiiiiiiiiiiiiiiiiicccccccccckkkkkkkkkkkkkkk.
Look in slow-mo at a bead chain fountain.
There are say 4 adjacent beads acting as pseudo-links being yanked/jerked up.
(a) But, these pseudo-links are not rigid, they have a lot of give.
(b) And, these & the trailing say 5 or even 10 beads are moving horizontally & have daylite under them.
(c) The pseudo-links never have a solid floor to kick (down) against.
(d) And, even if the last bead in the pseudo-link does enjoy an upwards (very very weak) kick, this kick is off a slippery & lose bead or two, & the kick moves thems beads sideways (or forwards)(or backwards)(they are after all lose in every way).
So, beads do not provide anything like a solid surface to kick offa.
Sheeeesh! Stone the crows! Milo give me strength! Are we blind!
The fountain (ie the arch)(the half circle) is simply due to the inertia of a chain.
If a stationary chain lays over the top of a wall, it forms a sharp V-bend.
If the chain is pulled down on one end, then it will rattle up & over the wall, with some speed.
If the speed is sufficient, the sharp V-bend will become a U-bend, due to centrifugal/centripetal forces opening up the V.
If speed increases, the U-bend opens further.
If speed increases a lot, the rising chain leaps up clear of the wall, & stays clear, the zenith/crest depending on gravity, & the U-bend opens some more.
And we have our fountain.
The height of the crest/fountain depends on speed (& is probly higher if chain heavier).
The radius of the U-bend/fountain depends on speed (& is probly larger if chain heavier).
Steve Mould & others say that the radius of curvature cancels out in the equations, & doesn't play any roll in these dynamics.
Steve is wrong.
If he were correct, then that would mean that the radius can vary greatly, for any given setup.
It might mean that if u initiated the fountain with a given radius then it would affect the final (maximum)(steady state) radius.
But i reckon that the radius (for a given setup) tends to one number.
And i reckon that the radius (for a given setup) varies with speed (ie varies with drop).
Just koz the radius cancels out in some dynamical equations duznt mean that radius duznt play any roll in the dynamics.
Steve Mould shows how a horizontal tight pattern of rows of chain (beads) gradually moves away to the west (due to the kick-effect) when the end of the chain is pulled east.
Yes, there is some kick effect, due to the pseudo-links.
But, the greater part of the effect is due to a chain's inclination to retain a pattern.
Here, we see that at each end of each row the chain forms a loop, looping around & back throo say 220 deg, such that the rows are hard up to each other.
As the end of the chain is drawn east, the end loops move across & back & across & back etc.
Now, Steve knows that a chain has a memory, ie it tends to retain a pattern.
In this case, that pattern is a 220 deg loop.
And, as the loop or loops traverse across & back etc, the loops slowly push the rows west.
That westwards slow pattern-push is a different animal to the quick jerk of a pseudo-link kick.
The slow pattern-push duznt suffer from the hi losses suffered by a pseudo-link kick.
Pseudo-links are not rigid, they have give. Give aint a problem for a slow westward pattern-push.
I said that yes there is some kick-effect.
But, i have already said that these kicks are not bonus-kicks.
They borrow from the overall power of the chain system.
They don’t add to the fountain.
It is virtually impossible to load/store a chain in a jar without having loops.
And loops result in jumps.
If u have a close look (in slow-mo) u can see that chains have a lot of horizontal movement before exiting vertically, & the chain sometimes suffers little jumps in the jar, as each leg of each loop jumps (horizontally) over its mate.
Jumps magnify the fountain, kicks dont. OOPS. No, i am wrong. These little jumps are not bonus-jumps, just like the kicks are not bonus-kicks. Any vertical impulse gained by a jump must have been borrowed from preceding links.
Also, the horizontal movements seen in the container/jar/beaker are a waste of impulse.
But, in any case, we don’t need bonus-jumps nor bonus-kicks to achieve a growing fountain, & to achieve a very high fountain.
All we need is lots of speed. The chain will crest at any height, depending on speed. There is no limit.
The idea that something special (like a bonus-kick) is needed (if the chain is to rise above the edge of the jar) is silly.
One thing that everyone has missed, in every youtube re the chain fountain (that i have seen), is that vertical jump effects & vertical kick effects (etc) are cumulative, and lasting (tautology alert).
Or at least partially cumulative & lasting. Air friction, & link friction see to that.
Anyhow, cumulative/lasting effects are the reason for some of the slightly weird gyrations/waves.
For example, it (lasting accumulation) is why we eventually get a persevering, vertical component of arc (a mini fountain) when the test seems to be strictly in the horizontal.
I agree
How old are you sir?
Damn calculation done by unsimplified physicist with Texan accent