sloan@uw-tanga.arpa (09/16/85)
From: Kenneth Sloan <sloan@uw-tanga.arpa> Here's a new subject to think about, start aguments, etc. Let's say I have a robot that pushes a box. I put a certain amount of energy into it, and get most of that energy out as work performed on the box (the rest being lost to maintain the robot's life support systems). Now I set up another one of these, and place it alongside the first robot. I have them push in the same direction so that the forces add. Now the output of this system is a moving box with the same direction and twice the speed. I'm putting twice as much energy in and getting twice as much energy out. Here's the question... If I place them on opposite sides of the box, the pushes will cancel. Now I appear to be getting no energy out of this system, at least not in the form of a moving box. I am still putting as much energy into the system. All I did was move one of the devices. What is happening to the energy? Is there an output in another form of energy? Is it building up in one of the devices somewhere? -Ken Sloan
FAILOR%LLL@LLL-MFE.ARPA (09/17/85)
From: Bruce Failor <FAILOR@LLL.MFENET>
To: Physics@SRI-UNIX.ARPA
In-Reply-To: Message from "Physics@SRI-Unix" of Tue 17 Sep 85 00:47:42-PDT
I think that it is pretty clear that some slippage is occuring somewhere
or the motor in the robot is going to break. For example, the robots "feet"
slip and thus heat up the floor or a belt inside of him slips and begins to
heat up. If there is no slippage--the motor will burnout or die some other
way. I don't think that there is any mystery here.
Bruce Failor
-------rdp@teddy.UUCP (09/20/85)
In article <546@sri-arpa.ARPA> sloan@uw-tanga.arpa writes: >From: Kenneth Sloan <sloan@uw-tanga.arpa> > >Let's say I have a robot that pushes a box. I put a certain amount of >energy into it, and get most of that energy out as work performed on >the box (the rest being lost to maintain the robot's life support systems). > >Now I set up another one of these, and place it alongside the first >robot. I have them push in the same direction so that the forces add. >Now the output of this system is a moving box with the same direction >and twice the speed. I'm putting twice as much energy in and getting >twice as much energy out. > >Here's the question... If I place them on opposite sides of the box, >the pushes will cancel. Now I appear to be getting no energy out of >this system, at least not in the form of a moving box. I am still >putting as much energy into the system. Wrong, you are putting no "energy" into the system, : 1 2 Energy = - * Mass * Velocity 2 Since the net velocity is 0, then the total energy is also 0. Note also that no work is done (save the robots grinding clutches and the like), since work is a measure of the rate of enrgy expenditure. NOw, this is not to say that no power will be drawn from the robots power sources. As alluded to above, these poor little suckers are going to be weeping, wailing and gnashing gears trying to go nowhere, generating all sorts of heat, etc. Let's re-do your experiment a bit to help illustrate. Put our box on a frictionless table. now attach a line and let it pass over the edge of the table, where it is attached to a weight. Sure enough, the box will move (more accurately accelerate) under the influence of the force exerted by pulling on the weight. What has happened here is the potential energy of the weight is being converted into kinetic energy of motion. NOw, on the opposite side of the box, attach another line with a weight of identical mass, hanging over the edge of the table. Voila! the box does not move. The potential energy of the ssystem does not change because it is not being converted into kinetic energy of motion. To argue that both weights are exerting "energy" violates the conservation law of energy, since this energy is would be irretrievably lost. What is happening, very simply, is that there is a perfect balance of forces, resulting in no net force, causing no net motion, therefore no net energy use. Dick Pierce
tmb@talcott.UUCP (Thomas M. Breuel) (09/20/85)
In article <546@sri-arpa.ARPA>, sloan@uw-tanga.arpa writes: > From: Kenneth Sloan <sloan@uw-tanga.arpa> > > > Here's a new subject to think about, start aguments, etc. > > Let's say I have a robot that pushes a box. I put a certain amount of > energy into it, and get most of that energy out as work performed on > the box (the rest being lost to maintain the robot's life support systems). > > Now I set up another one of these, and place it alongside the first > robot. I have them push in the same direction so that the forces add. > Now the output of this system is a moving box with the same direction > and twice the speed. I'm putting twice as much energy in and getting > twice as much energy out. > > Here's the question... If I place them on opposite sides of the box, > the pushes will cancel. Now I appear to be getting no energy out of > this system, at least not in the form of a moving box. I am still > putting as much energy into the system. All I did was move one of the > devices. What is happening to the energy? Is there an output in > another form of energy? Is it building up in one of the devices > somewhere? > > -Ken Sloan There are several types of energy in the system: -- internal energy (the robot's internal battery) -- kinetic energy (the energy stored in the motion of the boxes and the robots) -- thermal energy (energy that is essentially lost) -- stress energy (stored in the 'shape' of materials) If one robot pushes one box, it converts internal energy into kinetic energy. During the motion, part of that kinetic energy is converted into thermal energy by friction. At the end of the motion, the robot can recover what is left of the kinetic energy and convert it back into internal energy, or can just convert it into thermal energy with a simple mechanical brake. (Note that by conservation of momentum, the kinetic energy is stored in the relative motion of robot-box to the planet that they are on). If there is no friction and no internal losses, the whole process does not require any net energy. If two robots push two boxes into one another, they convert internal energy into stress energy. As in the case of kinetic energy, in the real world, some of that stress energy is necessarily lost and converted into heat, both during the deformation (friction) and while the force on the bodies is maintained (deformation). When the robots stop pushing, they again have the choice of converting the stress energy back into internal energy or of converting the stress energy into heat. Altogether, the 'paradox' exists only because many people still have pre-Newtonian ideas about 'energy', 'momentum', 'force' &c. Thomas.
umdhep@eneevax.UUCP (Todd Aven) (09/21/85)
I hope that you don't really believe that crap about E= -0.5M*v**2! Tell me about the energy in the electric field driving the poor little electrons in your CRT to their demise. I don't recall ever seeing a velocity in that sense associated with a field. ============================================================ |Todd Aven MANAGER@UMDHEP.BITNET | |Softwear Sweatshop AVEN@UMCINCOM (arpanet, bitnet)| |High Energy Physics UMDHEP@ENEEVAX.UUCP | |University of Maryland | |College Park, MD 20742 (301)454-3508 | ============================================================
ewiles@netex.UUCP (Ed Wiles) (09/21/85)
In article <546@sri-arpa.ARPA>, sloan@uw-tanga.arpa writes: > From: Kenneth Sloan <sloan@uw-tanga.arpa> > > [omitted] > > Here's the question... If I place them on opposite sides of the box, > the pushes will cancel. Now I appear to be getting no energy out of > this system, at least not in the form of a moving box. I am still > putting as much energy into the system. All I did was move one of the > devices. What is happening to the energy? Is there an output in > another form of energy? Is it building up in one of the devices > somewhere? > > -Ken Sloan You bet! The energy is being translated into heat, inside the drive devices of the robots. (Carefull, you may melt down!) Eventually, the heat would have to be radiated, in some fashion, to the universe at large. (Entrophy wins again!) I suggest you look up the laws of Thermodynamics. E. L. Wiles
gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>) (09/22/85)
> Altogether, the 'paradox' exists only because many people still > have pre-Newtonian ideas about 'energy', 'momentum', 'force' &c. This is not being helped any by our public schools. Feynman presents a truly damning indictment of public school textbooks in his semi-autobiography, "Surely You Are Joking, Mr. Feynman". In it, he tells of one science book he reviewed that gave several examples of different types of systems, e.g. bicycle, clock, muscles, etc., asking in each case "What makes it go?" As I read the examples, I had the same general response that Feynman did, "Oh, goody, they're going to discuss the innards of these things!", and the same disgust at the textbook's answer "Energy makes them go". I also recall, when I was a Physics graduate student, my Department was trying to write a research proposal concerning magnetic critical phenomena and was attempting to work the magic word "energy" (buzzword of that year) into the proposal; we finally settled on suggesting that the research would "lead to a better understanding of energy levels" or some such silliness, inserted just to tickle the right response out of the NSF. I think, on balance, government has not been a force for positive good in either education or science research; to the contrary, it has crippled minds, wasted resources, and entrenched mediocrity. There are exceptions to this general trend, but they are just that.
kim@mips.UUCP (Kim DeVaughn) (09/23/85)
[ ... go ahead, eat my bits ... ] > From: Kenneth Sloan <sloan@uw-tanga.arpa> > Here's a new subject to think about, start aguments, etc. > > Let's say I have a robot that pushes a box. I put a certain amount of > > Now I set up another one of these, and place it alongside the first > robot. I have them push in the same direction so that the forces add. > > the pushes will cancel. Now I appear to be getting no energy out of > this system, at least not in the form of a moving box. I am still > putting as much energy into the system. All I did was move one of the > devices. What is happening to the energy? Is there an output in > another form of energy? Is it building up in one of the devices Heat. If your robot's motors have sufficient torque, their treads (or wheels, or whatever) will "slip" against the surface they're "pushing against" (i.e., friction). Less torque, and the motors will heat up due to "electrical friction" (i.e., hysteresis, I-square*R losses, etc.) The surface is part of the system (and if it were "removed," you'd have your robots in space, and there would be no motion irrespective of which side the 'bots were on). /kim > -Ken Sloan *** REPLACE THIS LINE WITH YOUR MESSAGE *** -- UUCP: {decvax,ucbvax,ihnp4}!decwrl!mips!kim DDD: 415-960-1200 USPS: MIPS Computer Systems Inc, 1330 Charleston Rd, Mt View, CA 94043
jheimann@BBNCC5.ARPA (09/23/85)
This message is empty.
webber@utcs.uucp (R. D. Webber) (09/24/85)
In article <546@sri-arpa.ARPA> sloan@uw-tanga.arpa writes: >From: Kenneth Sloan <sloan@uw-tanga.arpa> > > >Let's say I have a robot that pushes a box. I put a certain amount of >energy into it, and get most of that energy out as work performed on >the box (the rest being lost to maintain the robot's life support systems). > >Here's the question... If I place [two robots] on opposite sides of the box, >the pushes will cancel. Now I appear to be getting no energy out of >this system, at least not in the form of a moving box. I am still >putting as much energy into the system. All I did was move one of the >devices. What is happening to the energy? Is there an output in >another form of energy? Is it building up in one of the devices >somewhere? > >-Ken Sloan Well, for one thing, unless you're assuming no slip at the robots' 'feet' and 'hands', there's some friction. There's also likely to be energy loss due to slip and chatter in internal actuators. You may, however, wish to consider these negligible. In that case, my thought is that, since work is force through displacement, there is no work done and therefore no energy loss apart from 'life' support. Oh, yeah, also some work done in deforming the structures of the box, the robots, and whatever the robots react against, but that's negligible unless you've got some REAL MOTHERF***ING big robots with a lot of power. Bob
usenet@ucbvax.ARPA (USENET News Administration) (09/30/85)
>> Here's the question... If I place them on opposite sides of the box, >> the pushes will cancel. Now I appear to be getting no energy out of >> this system, at least not in the form of a moving box. I am still >> putting as much energy into the system. ... >> >> -Ken Sloan No, you aren't putting as much energy into the system. If you measured it, you would find that the power consumption of an electric motor drops when you prevent it from turning. The reason this is not intuitive is that our experience is mostly with inefficient machines. In particular, 1. Electric motors lose power to electrical resistance, regardless of whether they are doing work. I think this effect is smaller than you would think. Does anyone know how efficient common electric motors are? 2. Muscles are peculiar in that how tired they get is dependent mostly on how much force they exert. I think this is because they are composed of fibers, each of which repeatedly contracts momentarily. To see that force is not generally related to energy consumption, think of stretching a rubber band between two immobile points. I don't see any obvious difficulties in building an electric motor with superconducting magnets. It also seems to me that such a motor would be essentially 100% efficient. It would use power when turning against a load, but not when it is prevented by force from turning. Now, I have a complaint. This was a previous answer to the same question: >You bet! The energy is being translated into heat, inside the drive >devices of the robots. (Carefull, you may melt down!) Eventually, the >heat would have to be radiated, in some fashion, to the universe at >large. (Entrophy wins again!) I suggest you look up the laws of >Thermodynamics. This is at best misleading. I'm sure the tone is unintentional, but I must say it is bullyish. "I suggest you look up ..." is too commonly heard in this newsgroup. Ken Rimey rimey@dali.berkeley.EDU
dgary@ecsvax.UUCP (D Gary Grady) (10/01/85)
> > . . . "I suggest you look up ..." is too commonly > heard in this newsgroup. > > Ken Rimey > rimey@dali.berkeley.EDU Well said! Many otherwise intelligent postings are marred by a sort of smarty-pants tone. I grit my teeth every time I see the sarcasting opening, "Sorry, but..." -- D Gary Grady Duke U Comp Center, Durham, NC 27706 (919) 684-3695 USENET: {seismo,decvax,ihnp4,akgua,etc.}!mcnc!ecsvax!dgary
mikes@AMES-NAS.ARPA (10/03/85)
From: mikes@AMES-NAS.ARPA (Peter Mikes) Subject: cancelling forces ---------------------------------------------------------------- Commenting on the whole series : <10492@ucbvax.ARPA> >> Here's the question... If I place them on opposite sides of the box, >> the pushes will cancel. Now I appear to be getting no energy out of >> this system, at least not in the form of a moving box. I am still >> putting as much energy into the system. ... >> >> -Ken Sloan No, you aren't putting as much energy into the system. If you measured it, you would find that the power consumption of an electric motor drops when you prevent it from turning. Actually, the useful concept, which applies equally well to two oposing robots or two antenas, is called IMPEDANCE. When the position of two antenas is changed so that they interfere rather then cooperate, then the impedance of the system goes up - therefore - if rest of the system was unchanged, less (or none) energygets radiated. However, the original formulation said: " I am putting same energy in.." and that can be done ( e.g. the current drive is a contraption with feedback which will increase the voltage of the power supply to overcome an increase in impedance ) - then if it is done the dissipation will increase and the relative proportions of where the energy goes (emg, motion, heat...) depends on the ratio of impedances. Nice illustration of characteristic impedance is a long tube, closed rigidly on far end and covered with a membrane on the near end. Imagine you are knocking rhytmically on the membrane: When you are 'in sync ' ( ressonant condition , low impedance ) the membrane is soft and gives easily and is accelerated by each push. Then somebody will prolong the tube by half wavelenth - so that as you knock, the wave from previous push returns, reflected from the far end and the membrane feels hard, your knuckles ache and you know that impedance is high. Unless you try harder you do not accomplish as much and in any case, your energy gets dissipated and wasted... The moral if this nice problem is : QUANTIFY ! P.S. that 'somebody' is called fate - all it takes is 1/2 of lambda...