Jerry Avins <jya@ieee.org> wrote: (snip of lossless capacitor question)> I won't give it away, but instead pose a mechanical version. Two > identical flywheels, moment of inertia I, can be coupled by a dog > clutch. Initially, one rotates at angular velocity w radians/second and > the other is stationary. Without exerting any forces outside of the > system, the flywheels are coupled by the dog clutch. Momentum being > conserved, they then rotate together at w/2. Before they were coupled, > the total energy was 0.5Iw^2, all in the rotating part. After the > coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or > 0.25Iw^2. What happened to the rest of the energy?I know about the capacitor, but I am not so sure about this one. (Also, I don't know the dog clutch.) There are many ways to lose energy, and it will find one. If the clutch is even a tiny bit slow, there will be friction before it completely grabs. Otherwise, there might be sound producing vibrations. Or, due to the infinite torque the shaft could break or the flywheel fly (maybe that is where the name comes from) apart. If the shaft isn't infinitely stiff, it could absorb some of the energy, either as heat (friction) or stored in the bonds holding it together. If there is no loss, then it will be an oscillating system. -- glen
Filter Output Energy
Started by ●October 18, 2011
Reply by ●October 18, 20112011-10-18
Reply by ●October 18, 20112011-10-18
On 10/18/2011 3:35 PM, glen herrmannsfeldt wrote:> Jerry Avins<jya@ieee.org> wrote: > > (snip of lossless capacitor question) > >> I won't give it away, but instead pose a mechanical version. Two >> identical flywheels, moment of inertia I, can be coupled by a dog >> clutch. Initially, one rotates at angular velocity w radians/second and >> the other is stationary. Without exerting any forces outside of the >> system, the flywheels are coupled by the dog clutch. Momentum being >> conserved, they then rotate together at w/2. Before they were coupled, >> the total energy was 0.5Iw^2, all in the rotating part. After the >> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or >> 0.25Iw^2. What happened to the rest of the energy? > > I know about the capacitor, but I am not so sure about this one. > (Also, I don't know the dog clutch.) There are many ways to > lose energy, and it will find one. If the clutch is even a tiny > bit slow, there will be friction before it completely grabs. > > Otherwise, there might be sound producing vibrations. > > Or, due to the infinite torque the shaft could break or the > flywheel fly (maybe that is where the name comes from) apart. > > If the shaft isn't infinitely stiff, it could absorb some > of the energy, either as heat (friction) or stored in the > bonds holding it together. If there is no loss, then it will > be an oscillating system.Go to the head of the class. :-) A dog clutch can't slip. It consists of meshing teeth ("dogs") that can be disengaged by being moved axially apart and engaged by moving axially together. (A car's manual transmission uses dog clutches to lock appropriate gears to their shafts. The gears themselves remain always in mesh.) http://www.youtube.com/watch?v=C0aiUBf1yGo&noredirect=1 http://www.youtube.com/watch?NR=1&v=vq11CusULlk Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●October 18, 20112011-10-18
On Oct 18, 5:09�pm, Jerry Avins <j...@ieee.org> wrote:> On 10/18/2011 3:35 PM, glen herrmannsfeldt wrote: > > > > > Jerry Avins<j...@ieee.org> �wrote: > > > (snip of lossless capacitor question) > > >> I won't give it away, but instead pose a mechanical version. Two > >> identical flywheels, moment of inertia I, can be coupled by a dog > >> clutch. Initially, one rotates at angular velocity w radians/second and > >> the other is stationary. Without exerting any forces outside of the > >> system, the flywheels are coupled by the dog clutch. Momentum being > >> conserved, they then rotate together at w/2. Before they were coupled, > >> the total energy was 0.5Iw^2, all in the rotating part. After the > >> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or > >> 0.25Iw^2. What happened to the rest of the energy? > > > I know about the capacitor, but I am not so sure about this one. > > (Also, I don't know the dog clutch.) �There are many ways to > > lose energy, and it will find one. �If the clutch is even a tiny > > bit slow, there will be friction before it completely grabs. > > > Otherwise, there might be sound producing vibrations. > > > Or, due to the infinite torque the shaft could break or the > > flywheel fly (maybe that is where the name comes from) apart. > > > If the shaft isn't infinitely stiff, it could absorb some > > of the energy, either as heat (friction) or stored in the > > bonds holding it together. �If there is no loss, then it will > > be an oscillating system. > > Go to the head of the class. :-) > > A dog clutch can't slip. It consists of meshing teeth ("dogs") that can > be disengaged by being moved axially apart and engaged by moving axially > together. (A car's manual transmission uses dog clutches to lock > appropriate gears to their shafts. The gears themselves remain always in > mesh.)http://www.youtube.com/watch?v=C0aiUBf1yGo&noredirect=1http://www.youtube.com/watch?NR=1&v=vq11CusULlkso, then, what's the deal with the two capacitors? are the wires that connect them the imperfectly lossy dog clutch? what if they were superconductors? does the charge just slosh back and forth between the two capacitors forever? r b-j
Reply by ●October 18, 20112011-10-18
On 10/18/2011 4:55 PM, robert bristow-johnson wrote:> On Oct 18, 5:09 pm, Jerry Avins<j...@ieee.org> wrote: >> On 10/18/2011 3:35 PM, glen herrmannsfeldt wrote: >> >> >> >>> Jerry Avins<j...@ieee.org> wrote: >> >>> (snip of lossless capacitor question) >> >>>> I won't give it away, but instead pose a mechanical version. Two >>>> identical flywheels, moment of inertia I, can be coupled by a dog >>>> clutch. Initially, one rotates at angular velocity w radians/second and >>>> the other is stationary. Without exerting any forces outside of the >>>> system, the flywheels are coupled by the dog clutch. Momentum being >>>> conserved, they then rotate together at w/2. Before they were coupled, >>>> the total energy was 0.5Iw^2, all in the rotating part. After the >>>> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or >>>> 0.25Iw^2. What happened to the rest of the energy? >> >>> I know about the capacitor, but I am not so sure about this one. >>> (Also, I don't know the dog clutch.) There are many ways to >>> lose energy, and it will find one. If the clutch is even a tiny >>> bit slow, there will be friction before it completely grabs. >> >>> Otherwise, there might be sound producing vibrations. >> >>> Or, due to the infinite torque the shaft could break or the >>> flywheel fly (maybe that is where the name comes from) apart. >> >>> If the shaft isn't infinitely stiff, it could absorb some >>> of the energy, either as heat (friction) or stored in the >>> bonds holding it together. If there is no loss, then it will >>> be an oscillating system. >> >> Go to the head of the class. :-) >> >> A dog clutch can't slip. It consists of meshing teeth ("dogs") that can >> be disengaged by being moved axially apart and engaged by moving axially >> together. (A car's manual transmission uses dog clutches to lock >> appropriate gears to their shafts. The gears themselves remain always in >> mesh.)http://www.youtube.com/watch?v=C0aiUBf1yGo&noredirect=1http://www.youtube.com/watch?NR=1&v=vq11CusULlk > > so, then, what's the deal with the two capacitors? are the wires that > connect them the imperfectly lossy dog clutch? > > what if they were superconductors? does the charge just slosh back > and forth between the two capacitors forever? > > r b-jIf the wires connecting them are perfect superconducting wires then you're subjecting them to enormous dV/dt and dI/dt. As you back those wires away from that approximation, slowing both those transients down, you're incurring ohmic losses. -- Rob Gaddi, Highland Technology -- www.highlandtechnology.com Email address domain is currently out of order. See above to fix.
Reply by ●October 18, 20112011-10-18
On 10/18/2011 8:07 PM, Rob Gaddi wrote:> On 10/18/2011 4:55 PM, robert bristow-johnson wrote: >> On Oct 18, 5:09 pm, Jerry Avins<j...@ieee.org> wrote: >>> On 10/18/2011 3:35 PM, glen herrmannsfeldt wrote: >>> >>> >>> >>>> Jerry Avins<j...@ieee.org> wrote: >>> >>>> (snip of lossless capacitor question) >>> >>>>> I won't give it away, but instead pose a mechanical version. Two >>>>> identical flywheels, moment of inertia I, can be coupled by a dog >>>>> clutch. Initially, one rotates at angular velocity w radians/second >>>>> and >>>>> the other is stationary. Without exerting any forces outside of the >>>>> system, the flywheels are coupled by the dog clutch. Momentum being >>>>> conserved, they then rotate together at w/2. Before they were coupled, >>>>> the total energy was 0.5Iw^2, all in the rotating part. After the >>>>> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or >>>>> 0.25Iw^2. What happened to the rest of the energy? >>> >>>> I know about the capacitor, but I am not so sure about this one. >>>> (Also, I don't know the dog clutch.) There are many ways to >>>> lose energy, and it will find one. If the clutch is even a tiny >>>> bit slow, there will be friction before it completely grabs. >>> >>>> Otherwise, there might be sound producing vibrations. >>> >>>> Or, due to the infinite torque the shaft could break or the >>>> flywheel fly (maybe that is where the name comes from) apart. >>> >>>> If the shaft isn't infinitely stiff, it could absorb some >>>> of the energy, either as heat (friction) or stored in the >>>> bonds holding it together. If there is no loss, then it will >>>> be an oscillating system. >>> >>> Go to the head of the class. :-) >>> >>> A dog clutch can't slip. It consists of meshing teeth ("dogs") that can >>> be disengaged by being moved axially apart and engaged by moving axially >>> together. (A car's manual transmission uses dog clutches to lock >>> appropriate gears to their shafts. The gears themselves remain always in >>> mesh.)http://www.youtube.com/watch?v=C0aiUBf1yGo&noredirect=1http://www.youtube.com/watch?NR=1&v=vq11CusULlk >>> >> >> so, then, what's the deal with the two capacitors? are the wires that >> connect them the imperfectly lossy dog clutch? >> >> what if they were superconductors? does the charge just slosh back >> and forth between the two capacitors forever? >> >> r b-j > > If the wires connecting them are perfect superconducting wires then > you're subjecting them to enormous dV/dt and dI/dt. As you back those > wires away from that approximation, slowing both those transients down, > you're incurring ohmic losses.Even if the wires superconduct, they have inductance. The circuit will then oscillate at a frequency of 4piSqrt(LC) Hz. (Two C's in series.) That frequency is likely to be high enough so that radiation will cause significant damping. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●October 18, 20112011-10-18
On 10/18/2011 12:35 PM, glen herrmannsfeldt wrote:> Jerry Avins<jya@ieee.org> wrote: > > (snip of lossless capacitor question) > >> I won't give it away, but instead pose a mechanical version. Two >> identical flywheels, moment of inertia I, can be coupled by a dog >> clutch. Initially, one rotates at angular velocity w radians/second and >> the other is stationary. Without exerting any forces outside of the >> system, the flywheels are coupled by the dog clutch. Momentum being >> conserved, they then rotate together at w/2. Before they were coupled, >> the total energy was 0.5Iw^2, all in the rotating part. After the >> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or >> 0.25Iw^2. What happened to the rest of the energy? >Jerry, It looks like my post got lost where I responded: "You get hot dogs" Fred
Reply by ●October 18, 20112011-10-18
On 10/18/2011 8:27 PM, Fred Marshall wrote: ...> "You get hot dogs":-) Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●October 19, 20112011-10-19
On Tue, 18 Oct 2011 14:52:57 -0400, Jerry Avins wrote:> On 10/18/2011 2:00 PM, dvsarwate wrote: >> On Oct 18, 1:48 pm, Jerry Avins<j...@ieee.org> wrote: >> >> >>> A real stumper for many beginners is, "If the energy doesn't come out >>> of the filter, where does it go?" There's no problem with digital >>> filters. After all, it's merely numbers. But what about analog filters >>> made of lossless components? >>> >>> >> >> So here is a question about analog lossless components. A capacitor C >> is charged to V volts and thus is storing energy 0.5CV^2. It is >> connected in parallel to an identical capacitor through lossless >> (zero-resistance) wires, and so charge flows from one capacitor to the >> other till each capacitor is holding half the charge and thus has >> voltage V/2 across it. Now the energy being stored is 0.5C(V/2)^2 + >> 0.5C(V/2)^2 = 0.25CV^2. What happened to the rest of the energy? >> >> --Dilip Sarwate >> >> P.S. Answers of E = mc^2 will receive zero credit! > > I won't give it away, but instead pose a mechanical version. Two > identical flywheels, moment of inertia I, can be coupled by a dog > clutch. Initially, one rotates at angular velocity w radians/second and > the other is stationary. Without exerting any forces outside of the > system, the flywheels are coupled by the dog clutch. Momentum being > conserved, they then rotate together at w/2. Before they were coupled, > the total energy was 0.5Iw^2, all in the rotating part. After the > coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or > 0.25Iw^2. What happened to the rest of the energy? > > JerryIMHO the appropriate answer to both Jerry's and Dillip's questions are "you are asking for a reasonable physical explanation of the behavior of a physically unreasonable system". Invoking physics (energy balance) in the same context as physically impossible devices (infinitely strong dog clutches, superconducting arcless switches, etc.) is a good brain teaser, but it's not necessarily fair or in any way realistic. If pressed, my answer to both of them would be to express the "perfect" components as lossy components taken to the limit as the loss goes to zero. Then you'll find that infinite current squared times zero resistance equals infinite power, for just the right sort of zero time that adds up to your lost energy. Or use Jerry's dodge and invoke inductance in the wires (and flex in the shafts attached to the dog clutch) and find that the energy remains stored in the system after all. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by ●October 19, 20112011-10-19
Tim <tim@seemywebsite.please> wrote:>On Tue, 18 Oct 2011 14:52:57 -0400, Jerry Avins wrote: > >> On 10/18/2011 2:00 PM, dvsarwate wrote: >>> On Oct 18, 1:48 pm, Jerry Avins<j...@ieee.org> wrote: >>> >>> >>>> A real stumper for many beginners is, "If the energy doesn't come out >>>> of the filter, where does it go?" There's no problem with digital >>>> filters. After all, it's merely numbers. But what about analog filters >>>> made of lossless components? >>>> >>>> >>> >>> So here is a question about analog lossless components. A capacitor C >>> is charged to V volts and thus is storing energy 0.5CV^2. It is >>> connected in parallel to an identical capacitor through lossless >>> (zero-resistance) wires, and so charge flows from one capacitor to the >>> other till each capacitor is holding half the charge and thus has >>> voltage V/2 across it. Now the energy being stored is 0.5C(V/2)^2 + >>> 0.5C(V/2)^2 = 0.25CV^2. What happened to the rest of the energy? >>> >>> --Dilip Sarwate >>> >>> P.S. Answers of E = mc^2 will receive zero credit! >> >> I won't give it away, but instead pose a mechanical version. Two >> identical flywheels, moment of inertia I, can be coupled by a dog >> clutch. Initially, one rotates at angular velocity w radians/second and >> the other is stationary. Without exerting any forces outside of the >> system, the flywheels are coupled by the dog clutch. Momentum being >> conserved, they then rotate together at w/2. Before they were coupled, >> the total energy was 0.5Iw^2, all in the rotating part. After the >> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or >> 0.25Iw^2. What happened to the rest of the energy? >> >> Jerry > >IMHO the appropriate answer to both Jerry's and Dillip's questions are >"you are asking for a reasonable physical explanation of the behavior of >a physically unreasonable system". > >Invoking physics (energy balance) in the same context as physically >impossible devices (infinitely strong dog clutches, superconducting >arcless switches, etc.) is a good brain teaser, but it's not necessarily >fair or in any way realistic. > >If pressed, my answer to both of them would be to express the "perfect" >components as lossy components taken to the limit as the loss goes to >zero. Then you'll find that infinite current squared times zero >resistance equals infinite power, for just the right sort of zero time >that adds up to your lost energy. Or use Jerry's dodge and invoke >inductance in the wires (and flex in the shafts attached to the dog >clutch) and find that the energy remains stored in the system after all.Here's another approach: does the mass of the capacitors change when you do this? Given that you believe the answer is yes, then take a list of fundamental particles, figure out if any of them are reasonable candidates for carrying mass away from such a system. There is only one reasonable candidate. Steve
Reply by ●October 19, 20112011-10-19
On Oct 19, 12:17�am, Tim <t...@seemywebsite.please> wrote: <<<<<material snipped>>>>>> If pressed, my answer to both of them would be to express the "perfect" > components as lossy components taken to the limit as the loss goes to > zero. �Then you'll find that infinite current squared times zero > resistance equals infinite power, for just the right sort of zero time > that adds up to your lost energy.Yes indeed. But taking things to the limit is done all the time. Who among the readers of this note has _not_ used 1 as the value of (sin x)/x at x = 0, even though, if pressed, the person might concede that at x = 0 the value of (sin x)/x is undefined and the value 1 that is being used is the limiting value of (sin x)/x as x tends to 0 ? Dilip Sarwate P.S. For the record, here is my take on the solution to the riddle Suppose the connecting wires have resistance R > 0. Then I(t), the current flowing decays away exponentially and the capacitors reach steady state at t = oo. If you calculate the energy expended in the resistance due to I(t), it is _exactly_ the energy that is missing. Curiously, int_0^infty I^2(t) R dt does _not_ depend on the value of R at all (remember that I(t) is of the form exp(-t/RC) etc). So, if we let R -> 0, the initial current is higher, and the decay is more rapid (time constant of circuit is getting smaller) but the energy expended is the same, and using the continuity type arguments that we like, the same amount is expended at R = 0. An instantaneous infinite current delivers finite energy into a zero resistor in zero time! A'i'nt math wonderful?
