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From: Graeme Zimmer on 7 Jul 2010 00:40 Thanks Fred, I would have thought that as the Capacitor closes (eg C increases ) the Voltage decreases, but net Charge remains the same.. Sort of like going from High Impedance to Low Impedance. It's trading Voltage for Current, which is what we want. I agree the attraction between the plates will drive the Cap like a motor. Is this good or bad? ...................... Zim
From: Jim Thompson on 7 Jul 2010 10:48 On Tue, 06 Jul 2010 20:57:21 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Wed, 07 Jul 2010 08:18:12 +1000, Adrian Jansen <adrian(a)qq.vv.net> >wrote: > >> >>Jim Thompson wrote: >>> On Tue, 06 Jul 2010 12:59:35 -0700, John Larkin >>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>> >>>> On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> >>>> wrote: >>>> >>>>> On Jul 6, 6:53 am, John Larkin >>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >>>>>> wrote: >>>>>> >>>>>>> On Jul 5, 9:41 pm, John Larkin >>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>>>>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>>>>>> between them, back and forth, forever. Just don't use resistors. >>>>>>> It has to be identical size capacitors, otherwise 'all the charge' >>>>>>> can't be transferred without adding/losing energy... >>>>>> Not so. >>>>> Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >>>>> a 2 uF capacitor. The first state of the system holds twice the >>>>> energy of the second. >>>> >>>> Well, depends on words now. I can transfer "all the charge that's in >>>> C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but >>>> the numerical amount of coulombs must change if the cap values are >>>> different, to conserve energy. I can move the charge back into C1, and >>>> return the system to its original state. >>>> >>>> My point was that you can move charge between caps, without losing >>>> energy, but not by using resistors. >>>> >>>> John >>>> >>> >>> Depends on the definition of "depends" :-) >>> >>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>> >>> C1*V1 == C2*V2 >>> >>> ...Jim Thompson >>If you conserve energy, then you must have >> >>C1*V1^2 = C2*V2^2 > >By analogy, consider a billiard ball striking another (at rest) ball 1/4 >diameter away from the line of centroid motion (glancing blow / cut). >Does energy and momentum get conserved? Show the math. I CAN do the math! Can others here? Why do you think I was dumping on Larkin? First Law of Thermodynamics: You always lose :-) ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Obama: A reincarnation of Nixon, narcissistically posing in politically-correct black-face, but with fewer scruples.
From: Chieftain of the Carpet Crawlers on 7 Jul 2010 11:46 On Wed, 07 Jul 2010 07:48:35 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >On Tue, 06 Jul 2010 20:57:21 -0700, >"JosephKK"<quiettechblue(a)yahoo.com> wrote: > >>On Wed, 07 Jul 2010 08:18:12 +1000, Adrian Jansen <adrian(a)qq.vv.net> >>wrote: >> >>> >>>Jim Thompson wrote: >>>> On Tue, 06 Jul 2010 12:59:35 -0700, John Larkin >>>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>>> On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> >>>>> wrote: >>>>> >>>>>> On Jul 6, 6:53 am, John Larkin >>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >>>>>>> wrote: >>>>>>> >>>>>>>> On Jul 5, 9:41 pm, John Larkin >>>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>>>>>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>>>>>>> between them, back and forth, forever. Just don't use resistors. >>>>>>>> It has to be identical size capacitors, otherwise 'all the charge' >>>>>>>> can't be transferred without adding/losing energy... >>>>>>> Not so. >>>>>> Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >>>>>> a 2 uF capacitor. The first state of the system holds twice the >>>>>> energy of the second. >>>>> >>>>> Well, depends on words now. I can transfer "all the charge that's in >>>>> C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but >>>>> the numerical amount of coulombs must change if the cap values are >>>>> different, to conserve energy. I can move the charge back into C1, and >>>>> return the system to its original state. >>>>> >>>>> My point was that you can move charge between caps, without losing >>>>> energy, but not by using resistors. >>>>> >>>>> John >>>>> >>>> >>>> Depends on the definition of "depends" :-) >>>> >>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>>> >>>> C1*V1 == C2*V2 >>>> >>>> ...Jim Thompson >>>If you conserve energy, then you must have >>> >>>C1*V1^2 = C2*V2^2 >> >>By analogy, consider a billiard ball striking another (at rest) ball 1/4 >>diameter away from the line of centroid motion (glancing blow / cut). >>Does energy and momentum get conserved? Show the math. > >I CAN do the math! Can others here? Why do you think I was dumping >on Larkin? > >First Law of Thermodynamics: You always lose :-) > > ...Jim Thompson A pair of billiard balls colliding, even at perfect tangency, does not do so without losses. In the glancing blow, the coefficient of friction of the ball media comes into play as does the coefficient of friction of the ball-to-cloth interface. These two factors mean that there will not be a 100% transfer of energy from one ball to the other without some loss. Otherwise, 'English' would not work. Nor would 'Throw'. Two very important weapons in the billiard artist's arsenal. Bank shots would also have several issues. Also, using a 'term' like "a quarter ball" can be ambiguously interpreted. Just give the number of degrees in the future. Question/poser for you all: Can a 'cut shot' be made that is less than 90� with these friction effects in place? The answer is yes, but do you know how or why to get there?
From: John Larkin on 7 Jul 2010 11:50 On Wed, 07 Jul 2010 08:18:12 +1000, Adrian Jansen <adrian(a)qq.vv.net> wrote: > >Jim Thompson wrote: >> On Tue, 06 Jul 2010 12:59:35 -0700, John Larkin >> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>> On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> >>> wrote: >>> >>>> On Jul 6, 6:53 am, John Larkin >>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >>>>> wrote: >>>>> >>>>>> On Jul 5, 9:41 pm, John Larkin >>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>>>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>>>>> between them, back and forth, forever. Just don't use resistors. >>>>>> It has to be identical size capacitors, otherwise 'all the charge' >>>>>> can't be transferred without adding/losing energy... >>>>> Not so. >>>> Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >>>> a 2 uF capacitor. The first state of the system holds twice the >>>> energy of the second. >>> >>> Well, depends on words now. I can transfer "all the charge that's in >>> C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but >>> the numerical amount of coulombs must change if the cap values are >>> different, to conserve energy. I can move the charge back into C1, and >>> return the system to its original state. >>> >>> My point was that you can move charge between caps, without losing >>> energy, but not by using resistors. >>> >>> John >>> >> >> Depends on the definition of "depends" :-) >> >> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >> >> C1*V1 == C2*V2 >> >> ...Jim Thompson >If you conserve energy, then you must have > >C1*V1^2 = C2*V2^2 Right. If you dump all the energy from one charged cap into another, discharged, cap of a different value, and do it efficiently, charge is not conserved. John
From: John Larkin on 7 Jul 2010 11:59
On Wed, 07 Jul 2010 08:46:17 -0700, Chieftain of the Carpet Crawlers <theslipperman(a)thebarattheendoftheuniverse.org> wrote: >On Wed, 07 Jul 2010 07:48:35 -0700, Jim Thompson ><To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: > >>On Tue, 06 Jul 2010 20:57:21 -0700, >>"JosephKK"<quiettechblue(a)yahoo.com> wrote: >> >>>On Wed, 07 Jul 2010 08:18:12 +1000, Adrian Jansen <adrian(a)qq.vv.net> >>>wrote: >>> >>>> >>>>Jim Thompson wrote: >>>>> On Tue, 06 Jul 2010 12:59:35 -0700, John Larkin >>>>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>> >>>>>> On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> >>>>>> wrote: >>>>>> >>>>>>> On Jul 6, 6:53 am, John Larkin >>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >>>>>>>> wrote: >>>>>>>> >>>>>>>>> On Jul 5, 9:41 pm, John Larkin >>>>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>>>>>>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>>>>>>>> between them, back and forth, forever. Just don't use resistors. >>>>>>>>> It has to be identical size capacitors, otherwise 'all the charge' >>>>>>>>> can't be transferred without adding/losing energy... >>>>>>>> Not so. >>>>>>> Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >>>>>>> a 2 uF capacitor. The first state of the system holds twice the >>>>>>> energy of the second. >>>>>> >>>>>> Well, depends on words now. I can transfer "all the charge that's in >>>>>> C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but >>>>>> the numerical amount of coulombs must change if the cap values are >>>>>> different, to conserve energy. I can move the charge back into C1, and >>>>>> return the system to its original state. >>>>>> >>>>>> My point was that you can move charge between caps, without losing >>>>>> energy, but not by using resistors. >>>>>> >>>>>> John >>>>>> >>>>> >>>>> Depends on the definition of "depends" :-) >>>>> >>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>>>> >>>>> C1*V1 == C2*V2 >>>>> >>>>> ...Jim Thompson >>>>If you conserve energy, then you must have >>>> >>>>C1*V1^2 = C2*V2^2 >>> >>>By analogy, consider a billiard ball striking another (at rest) ball 1/4 >>>diameter away from the line of centroid motion (glancing blow / cut). >>>Does energy and momentum get conserved? Show the math. >> >>I CAN do the math! Can others here? Why do you think I was dumping >>on Larkin? >> >>First Law of Thermodynamics: You always lose :-) >> >> ...Jim Thompson > > A pair of billiard balls colliding, even at perfect tangency, does not >do so without losses. > > In the glancing blow, the coefficient of friction of the ball media >comes into play as does the coefficient of friction of the ball-to-cloth >interface. > > These two factors mean that there will not be a 100% transfer of energy >from one ball to the other without some loss. > > Otherwise, 'English' would not work. Nor would 'Throw'. Two very >important weapons in the billiard artist's arsenal. Bank shots would >also have several issues. > > Also, using a 'term' like "a quarter ball" can be ambiguously >interpreted. > > Just give the number of degrees in the future. > > Question/poser for you all: > > Can a 'cut shot' be made that is less than 90� with these friction >effects in place? > > The answer is yes, but do you know how or why to get there? Billiard balls roll, which means that at the instant of contact there is a transverse scraping, like a clutch engaging, which wastes more energy than a classic elastic sphere conservation-of-momentum physics problem. The angular momentum transfer is a lot like transferring charge between two capacitors by connecting them with a resistor, but worse because of the felt. Air hockey is closer to classic elastic collision. John |