Poss. Symmetry Break?
Moderator: scott
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Happy to admit that i've no sorry clue what the flip i'm currently on about. Probably goes nowhere. Ain't been getting much sleep this last week..
The system i described above can be further simplified with just a rack and pinion - the rack being a linearly-accelerated mass, via the rotating 'pinion', ie. a gear wheel.
So the effective moment of inertia felt on the wheel is a function of the mass of the rack, and the gearing ratio - how much it's accelerated for a given angle of input torque.
Bringing it all back into a rotating frame again, suppose we have a large inner ring of mass. Then, much further out, we have a much thinner, lighter ring of mass.
Despite their different weights, each ring can have the exact same value of MoI - the heavier ring, closer in, is undergoing less angular acceleration, while the lighter ring, further out, is undergoing a much greater angular acceleration for the same given RPM.
Suppose we now add a pair of clutches, so that either ring can be engaged to the main shaft, seperately from each other...
With either ring engaged, the inertia you measure on the axle is the same. And if you calculate your input energy as a function of that angular inertia times RPM, you'll get a consistent answer regardless of which ring is being torqued.
But internally, the two rings have completely different energies! Half the mass at twice the radius still has twice the energy for a given RPM!
Instead of clutches, we could just have two wheels side-by-side; both having identical MoI's, and input energies, yet different internal energies - one a function of more mass at less acceleration, the other, of less mass under greater acceleration.
There can be no paradoxes, yet i'm quite comprehensively metagrobolized - the former wheel has 1 J for every J given.. but the latter has 2 for 1.. Give it ten and it actually has 20 - each Joule input or output at the axle translating to 2 J of work out at the rim.
Unfurling it back into linear-only, the handles of a scissorjack experience inertia as a mass at the other end is accelerated. But if you couldn't see the mass being accelerated, and were asked to calculate how much energy you'd input to the handles - how much work you've performed - purely as a function of the horizontal resistance force felt at the handles, and the angle through which they've turned, you'd have no way of knowing whether you'd accelerated a large mass to a low speed, or a lighter mass to a higher speed. In other words, what you've actually input is ambiguous momentum.. its actual energy value depends upon the specific mass / velocity distribution... which cannot be determined merely via the horizontal force component acting on the handles alone...
Twilight zone or Twiglet zone?
The system i described above can be further simplified with just a rack and pinion - the rack being a linearly-accelerated mass, via the rotating 'pinion', ie. a gear wheel.
So the effective moment of inertia felt on the wheel is a function of the mass of the rack, and the gearing ratio - how much it's accelerated for a given angle of input torque.
Bringing it all back into a rotating frame again, suppose we have a large inner ring of mass. Then, much further out, we have a much thinner, lighter ring of mass.
Despite their different weights, each ring can have the exact same value of MoI - the heavier ring, closer in, is undergoing less angular acceleration, while the lighter ring, further out, is undergoing a much greater angular acceleration for the same given RPM.
Suppose we now add a pair of clutches, so that either ring can be engaged to the main shaft, seperately from each other...
With either ring engaged, the inertia you measure on the axle is the same. And if you calculate your input energy as a function of that angular inertia times RPM, you'll get a consistent answer regardless of which ring is being torqued.
But internally, the two rings have completely different energies! Half the mass at twice the radius still has twice the energy for a given RPM!
Instead of clutches, we could just have two wheels side-by-side; both having identical MoI's, and input energies, yet different internal energies - one a function of more mass at less acceleration, the other, of less mass under greater acceleration.
There can be no paradoxes, yet i'm quite comprehensively metagrobolized - the former wheel has 1 J for every J given.. but the latter has 2 for 1.. Give it ten and it actually has 20 - each Joule input or output at the axle translating to 2 J of work out at the rim.
Unfurling it back into linear-only, the handles of a scissorjack experience inertia as a mass at the other end is accelerated. But if you couldn't see the mass being accelerated, and were asked to calculate how much energy you'd input to the handles - how much work you've performed - purely as a function of the horizontal resistance force felt at the handles, and the angle through which they've turned, you'd have no way of knowing whether you'd accelerated a large mass to a low speed, or a lighter mass to a higher speed. In other words, what you've actually input is ambiguous momentum.. its actual energy value depends upon the specific mass / velocity distribution... which cannot be determined merely via the horizontal force component acting on the handles alone...
Twilight zone or Twiglet zone?
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
The point in all this is simply that gearing amplifies effective inertia, along with velocity and acceleration.
A familiar example would be those toy cars with so-called 'friction motors' - inside is a flywheel driven by a high gear ratio with the rear wheels; you swipe it across the carpet a few times to rev it up, then off it goes..
So when the car is pushed along, we're accelerating much of its mass by a greater acceleration and displacement than the car itself. The car thus has more momentum and energy than its rest mass times its apparent velocity. The novelty fun factor in the toy is precisely my point of interest - that if we can't see and thus don't know a system's gearing / power ratio, then we can only predict our input momentum - not its distribution of mass and velocity, or thus how much energy we're actually inputting..
Still not sure where this is going yet, just freefalling thru the rabbit hole of 'what constitutes inertia'..
But at the bottom of the hole is a black box, with a lever sticking out. The lever has a constant inertia across its given range of travel. The proposition here is simply that inside, we could be accelerating a high mass to a low velocity, or vice versa... or that we could perform each in turn, without being able to determine either from our input energy alone. Conceivably, the internal distribution of mass and velocity could change each time the lever is worked, hence for a constant input energy, our actual output energy, inside the box, is indeterminable.. it could contain a little bit of mass going really fast, or a slow lumbering mass, but only its net momentum can be precisely determined.
But even that measure would be based on the assumption that the sign of the momentum wasn't changing internally - alternate pulls of the lever might be decelerating mass, for all we know. Pull the lever one time, a flywheel's accelerated, then, next pull, another's spun up in the opposite direction, then they clutch together and halt.. and we'd be none the wiser. No matter how carefully we measure our input work, the momentum and energy inside the box could be anything, within a progressively-climbing range, or nothing at all.
So i guess the angle here is simply what Fletcher pointed out earlier - that inertia, linear or angular, is as much a function of gearing (power ratios), as mass, or its radial distribution.
That gives us something else to play with. Likewise, maybe gravity's inertial equivalence could come into play here..
A familiar example would be those toy cars with so-called 'friction motors' - inside is a flywheel driven by a high gear ratio with the rear wheels; you swipe it across the carpet a few times to rev it up, then off it goes..
So when the car is pushed along, we're accelerating much of its mass by a greater acceleration and displacement than the car itself. The car thus has more momentum and energy than its rest mass times its apparent velocity. The novelty fun factor in the toy is precisely my point of interest - that if we can't see and thus don't know a system's gearing / power ratio, then we can only predict our input momentum - not its distribution of mass and velocity, or thus how much energy we're actually inputting..
Still not sure where this is going yet, just freefalling thru the rabbit hole of 'what constitutes inertia'..
But at the bottom of the hole is a black box, with a lever sticking out. The lever has a constant inertia across its given range of travel. The proposition here is simply that inside, we could be accelerating a high mass to a low velocity, or vice versa... or that we could perform each in turn, without being able to determine either from our input energy alone. Conceivably, the internal distribution of mass and velocity could change each time the lever is worked, hence for a constant input energy, our actual output energy, inside the box, is indeterminable.. it could contain a little bit of mass going really fast, or a slow lumbering mass, but only its net momentum can be precisely determined.
But even that measure would be based on the assumption that the sign of the momentum wasn't changing internally - alternate pulls of the lever might be decelerating mass, for all we know. Pull the lever one time, a flywheel's accelerated, then, next pull, another's spun up in the opposite direction, then they clutch together and halt.. and we'd be none the wiser. No matter how carefully we measure our input work, the momentum and energy inside the box could be anything, within a progressively-climbing range, or nothing at all.
So i guess the angle here is simply what Fletcher pointed out earlier - that inertia, linear or angular, is as much a function of gearing (power ratios), as mass, or its radial distribution.
That gives us something else to play with. Likewise, maybe gravity's inertial equivalence could come into play here..
Last edited by MrVibrating on Sat Apr 23, 2016 2:30 pm, edited 2 times in total.
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Going back to the wheel and scissorjack - it's basically just a linear jack, but operated by a rotary handle.
Obviously cranks and conrods have their varying angle of torque, so just consider a constant-angle transmission instead (rack & pinion or chain & sprocket, etc.).
The question is, are we performing a linear or angular acceleration? Think carefully, cos the implications of this are quite unintuitive.... the balance of RKE or linear KE we calculate depends upon the force / angle power ratio; how much mass we accelerate how far for a given angle of rotation.
My contention here is that if we cannot see the linear mass being accelrated - ie. we're oblivious to whatever's actually manifesting the inertia, beyond the axle's resistance to accelerations - then based on the available information, all we can say for sure is that we're definitely inputting RKE as a function of half the angular inertia times the angular velocity.
The actual energy of the linearly-accelerated mass depends upon its mass, and the leverage ratios of the jacks and rotary transmission. A given angular increment of rotation could accelerate 1 kg by 2 m/s, in which case we've input 2 J... but all we know is that we've input 2 kg/m/s of momentum. Maybe it actually divided into 2 kg accelerated by 1 m/s... so the same 2 kg/m/s of momentum has been input... only now we've actually only input 1 Joule, not 2...
We just can't know - our internal energy is inherently indeterminable! All we can do is calculate the external RKE as a function of half its angular inertia times angular velocity, but parity between this measure and the actual internal energy is a variable fuinction of the power ratio.
Craziness eh? You just can't trust MoI. It can lie. Lie right to your face. It has its own allegiances.
Obviously cranks and conrods have their varying angle of torque, so just consider a constant-angle transmission instead (rack & pinion or chain & sprocket, etc.).
The question is, are we performing a linear or angular acceleration? Think carefully, cos the implications of this are quite unintuitive.... the balance of RKE or linear KE we calculate depends upon the force / angle power ratio; how much mass we accelerate how far for a given angle of rotation.
My contention here is that if we cannot see the linear mass being accelrated - ie. we're oblivious to whatever's actually manifesting the inertia, beyond the axle's resistance to accelerations - then based on the available information, all we can say for sure is that we're definitely inputting RKE as a function of half the angular inertia times the angular velocity.
The actual energy of the linearly-accelerated mass depends upon its mass, and the leverage ratios of the jacks and rotary transmission. A given angular increment of rotation could accelerate 1 kg by 2 m/s, in which case we've input 2 J... but all we know is that we've input 2 kg/m/s of momentum. Maybe it actually divided into 2 kg accelerated by 1 m/s... so the same 2 kg/m/s of momentum has been input... only now we've actually only input 1 Joule, not 2...
We just can't know - our internal energy is inherently indeterminable! All we can do is calculate the external RKE as a function of half its angular inertia times angular velocity, but parity between this measure and the actual internal energy is a variable fuinction of the power ratio.
Craziness eh? You just can't trust MoI. It can lie. Lie right to your face. It has its own allegiances.
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
So i've been thinking this last week, it seems we can draw up different 'classes' of MoI, as regards what's actually causing the resistance to angular accelerations...
Class 1:
---------
Common MoI as a function of MR^2, and all its geometric derivatives.
Class 2:
---------
Leveraged accelerations trade force for displacement; inertia for velocity. A high angular inertia at low acceleration could be caused by a lower inertia under greater acceleration.
Class 3:
---------
While the former two classes describe real inertia / momentum, any conservative resistance to acceleration may be considered as a pseudo-inertia - springs, gravity etc. - (so a dissipative load such as a brake wouldn't qualify).
Not sure how useful that is, and i'm not about to start describing everything in these terms... but maybe it helps open up the discussion a little..
Class 1:
---------
Common MoI as a function of MR^2, and all its geometric derivatives.
Class 2:
---------
Leveraged accelerations trade force for displacement; inertia for velocity. A high angular inertia at low acceleration could be caused by a lower inertia under greater acceleration.
Class 3:
---------
While the former two classes describe real inertia / momentum, any conservative resistance to acceleration may be considered as a pseudo-inertia - springs, gravity etc. - (so a dissipative load such as a brake wouldn't qualify).
Not sure how useful that is, and i'm not about to start describing everything in these terms... but maybe it helps open up the discussion a little..
Perhaps this is useful?
* When a body is not in translatory equilibrium then the torque equation differs between points in the plane of motion.
* Precission (or a change in MoI?) may result in either a net torque or a change in spin-rate.
Basically what you've been exploring, but maybe it sparks something new.
What happens when a wheel wants -or is about to- precess but can't because of a fixed axis?
Perhaps we can turn this around and say:wiki wrote: The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
τ = dL/dt
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
τ[1] + ... τ[n] = τ[net] = dL/dt
For rotation about a fixed axis,
L = I·ω,
where I is the moment of inertia and ω is the angular velocity. It follows that
τ[net] = dL/dt = d(Iω)/dt = I·dω/dt = I·α
where α is the angular acceleration of the body, measured in rad/s². This equation has the limitation that the torque equation describes the instantaneous axis of rotation or center of mass for any type of motion – whether pure translation, pure rotation, or mixed motion. I = Moment of inertia about the point which the torque is written (either instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then the torque equation is the same about all points in the plane of motion.
A torque is not necessarily limited to rotation around a fixed axis, however. It may change the magnitude and/or direction of the angular momentum vector, depending on the angle between the velocity vector and the non-radial component of the force vector, as viewed in the pivot's frame of reference. A net torque on a spinning body therefore may result in a precession without necessarily causing a change in spin rate.
Source: https://en.wikipedia.org/wiki/Torque
* When a body is not in translatory equilibrium then the torque equation differs between points in the plane of motion.
* Precission (or a change in MoI?) may result in either a net torque or a change in spin-rate.
Basically what you've been exploring, but maybe it sparks something new.
What happens when a wheel wants -or is about to- precess but can't because of a fixed axis?
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Think i'm prolly just repeating Wubbly's results, but ran thru some sims this afternoon:
MoI = MR^2
At double the radius, MoI rises 4x
If we double the radius but quarter the mass, MoI sould be constant:
Ex. 1:
MoI = 1 kg @ 2 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, then released with 0.496 kg/m/s and 0.123 J @ 2.367 RPM
Ex. 2:
MoI = 250 grams @ 4 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, mass detaches with 0.248 kg/m/s and 0.123 J @ 2.367 RPM
Ex. 3:
MoI = 62.5 grams @ 8 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, mass detaches with 0.124 kg/m/s and 0.123 J @ 2.367 RPM
RKE = 1/2Iw^2 = 0.5 * 4 kg/m^2 * (2.367 RPM * 360°/ minute = 852.12 °/min) = 0.122975 J /N/m.
So we have the right amount of energy for the MoI, despite whatever's causing the MoI. Only MoI times RPM determines RKE.
As we trade mass for radius, the amount of work done remains constant, in terms of energy expenditure, but progressively less momentum is dished out.
So my mistake there was assuming momentum would be constant, and energy variable.. it's actually the other way around..!
I need to digest this result and reconsider whether what i was originally thinking is possible or not.. the idea was to deal fixed units of momentum, for variable energy... so either there's some way to swap suites, or i'm just talking rubbish.. Again.
MoI = MR^2
At double the radius, MoI rises 4x
If we double the radius but quarter the mass, MoI sould be constant:
Ex. 1:
MoI = 1 kg @ 2 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, then released with 0.496 kg/m/s and 0.123 J @ 2.367 RPM
Ex. 2:
MoI = 250 grams @ 4 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, mass detaches with 0.248 kg/m/s and 0.123 J @ 2.367 RPM
Ex. 3:
MoI = 62.5 grams @ 8 m radius = 4 kg/m^2
Torqued with 1 N/m for 1 sec, mass detaches with 0.124 kg/m/s and 0.123 J @ 2.367 RPM
RKE = 1/2Iw^2 = 0.5 * 4 kg/m^2 * (2.367 RPM * 360°/ minute = 852.12 °/min) = 0.122975 J /N/m.
So we have the right amount of energy for the MoI, despite whatever's causing the MoI. Only MoI times RPM determines RKE.
As we trade mass for radius, the amount of work done remains constant, in terms of energy expenditure, but progressively less momentum is dished out.
So my mistake there was assuming momentum would be constant, and energy variable.. it's actually the other way around..!
I need to digest this result and reconsider whether what i was originally thinking is possible or not.. the idea was to deal fixed units of momentum, for variable energy... so either there's some way to swap suites, or i'm just talking rubbish.. Again.
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Isn't this how satellites maintain attitude? By torquing / applying a precessional force to gyros.ME wrote:Perhaps this is useful?Perhaps we can turn this around and say:wiki wrote: The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
τ = dL/dt
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
τ[1] + ... τ[n] = τ[net] = dL/dt
For rotation about a fixed axis,
L = I·ω,
where I is the moment of inertia and ω is the angular velocity. It follows that
τ[net] = dL/dt = d(Iω)/dt = I·dω/dt = I·α
where α is the angular acceleration of the body, measured in rad/s². This equation has the limitation that the torque equation describes the instantaneous axis of rotation or center of mass for any type of motion – whether pure translation, pure rotation, or mixed motion. I = Moment of inertia about the point which the torque is written (either instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then the torque equation is the same about all points in the plane of motion.
A torque is not necessarily limited to rotation around a fixed axis, however. It may change the magnitude and/or direction of the angular momentum vector, depending on the angle between the velocity vector and the non-radial component of the force vector, as viewed in the pivot's frame of reference. A net torque on a spinning body therefore may result in a precession without necessarily causing a change in spin rate.
Source: https://en.wikipedia.org/wiki/Torque
* When a body is not in translatory equilibrium then the torque equation differs between points in the plane of motion.
* Precission (or a change in MoI?) may result in either a net torque or a change in spin-rate.
Basically what you've been exploring, but maybe it sparks something new.
What happens when a wheel wants -or is about to- precess but can't because of a fixed axis?
I would imagine that forcing a gyro to precess against its plane of motion would exert a braking or acceleratory torque - same principle those "Powerball"gyros use..
https://www.youtube.com/watch?v=Fq2YzSwfKKY
(i've actually got one here, they're insane little toys)..
re: Poss. Symmetry Break?
I had absolutely no idea what just I saw there...never heard of such thing.
I've found some other video
(...could be considered somewhat... hmm disturbing?)
I've found some other video
(...could be considered somewhat... hmm disturbing?)
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Poss. Symmetry Break?
Gee , - will have to get one or two of them !
OK grandson , - get this going good and fast and jump off that diving board with your Superman pose and I'll work the video !
See if you can change your angle of attack to the water as you near it and I'll try and get a close up if you turn your head in my direction .
OK ? - I'd let you do the filming of me doing it but you know how complicated this camera is !
Oh , also if you feel like it , try and notice how it reacts under water ! : )
OK grandson , - get this going good and fast and jump off that diving board with your Superman pose and I'll work the video !
See if you can change your angle of attack to the water as you near it and I'll try and get a close up if you turn your head in my direction .
OK ? - I'd let you do the filming of me doing it but you know how complicated this camera is !
Oh , also if you feel like it , try and notice how it reacts under water ! : )
Have had the solution to Bessler's Wheel approximately monthly for over 30 years ! But next month is "The One" !
-
- Devotee
- Posts: 1605
- Joined: Sat Mar 17, 2012 4:50 am
Re: re: Poss. Symmetry Break?
ME wrote:I had absolutely no idea what just I saw there...never heard of such thing.
I've found some other video
(...could be considered somewhat... hmm disturbing?)
Make sure you buy two or at least remember to work both arms. You wouldn't want to end up like this:Forearms like Popeye :-)
I don't believe in conspiracies!
I prefer working alone.
I prefer working alone.
- Wubbly
- Aficionado
- Posts: 727
- Joined: Sat Jun 06, 2009 2:15 am
- Location: A small corner of the Milky Way Galaxy
- Contact:
re: Poss. Symmetry Break?
Reminds me of the Atwoods where we used a 1 kg mass difference, dropped 1 meter, to create variable amounts of momentum.Mr.V wrote:So my mistake there was assuming momentum would be constant, and energy variable.. it's actually the other way around..!
The energy used was the same, but the momentum generated was different.
http://www.besslerwheel.com/forum/downl ... er=user_id
Here's a different example of using 9.81 Joules of energy to create variable amounts of momentum by varying the mass of the system.
http://www.besslerwheel.com/forum/downl ... er=user_id
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Only just noticed something silly - i'm sure this must've come up here before, but would appeciate some help resolving it:
- two identical 1 kg masses on a collision course, each travelling at 10 m/s
- from the external frame, each mass has 10 kg/m/s of momentum, and 50 Joules of KE
- from the rest frame of either mass, the other is approaching with 20 kg/m/s and 200 J
..so momentum's conserved, but where's that extra 100 J come from?
- two identical 1 kg masses on a collision course, each travelling at 10 m/s
- from the external frame, each mass has 10 kg/m/s of momentum, and 50 Joules of KE
- from the rest frame of either mass, the other is approaching with 20 kg/m/s and 200 J
..so momentum's conserved, but where's that extra 100 J come from?
-
- Addict
- Posts: 2879
- Joined: Sat Jul 31, 2010 12:19 am
- Location: W3
Re: re: Poss. Symmetry Break?
Yep prolly wouldn't wanna be using it in public..ME wrote:I had absolutely no idea what just I saw there...never heard of such thing.
I've found some other video
(...could be considered somewhat... hmm disturbing?)
But a cool gift for wives and gf's..
Two identical cars are in the same scenario as above. They crash head on. What is the energy of the collision ? Both cars deform the same extent.MrVibrating wrote:Only just noticed something silly - i'm sure this must've come up here before, but would appeciate some help resolving it:
- two identical 1 kg masses on a collision course, each travelling at 10 m/s
- from the external frame, each mass has 10 kg/m/s of momentum, and 50 Joules of KE
- from the rest frame of either mass, the other is approaching with 20 kg/m/s and 200 J
..so momentum's conserved, but where's that extra 100 J come from?
Is this the same as one car traveling the same speed colliding with a concrete wall ?