cliff@jarthur.Claremont.EDU (Clifford Stein) (04/23/91)
What's the purpose of polarized wall outlets with an AC signal? I don't understand. Is it really bad to force something to plug in the wrong way? I am serious. With a real AC signal (sinusoidal waves) I can see no reason at all for it. How can the electrical equipment tell? Curious --Cliff -- cliff@jarthur.claremont.edu | "Ted Striker? Never heard of him. Wait! cliff@jarthur.uucp | That's not exactly true. We were like ...uunet!jarthur!cliff | brothers." cstein@hmcvax.bitnet | --Buck Murdoch
jws@cica4.mlb.semi.harris.com (James W. Swonger) (04/23/91)
Why: Some appliances (like TVs, for instance) have a "hot chassis", i.e. the chassis is not isolated from the power line. Given this, it's better that the chassis be attached to the neutral wire than the hot wire. Polarized plugs decide this for you. A flat file and a minute or two will allow you to exercise your God-given right to plug things in backwards.
kludge@grissom.larc.nasa.gov ( Scott Dorsey) (04/23/91)
In article <11864@jarthur.Claremont.EDU> cliff@jarthur.Claremont.EDU (Clifford Stein) writes: >What's the purpose of polarized wall outlets with an AC >signal? I don't understand. Is it really bad to force >something to plug in the wrong way? You are thinking that the reference ground in your house is at zero volts "the Ground they tell you about in Physics class" and that the AC power alternates at + and - 84 volts around that ground (so that your peak voltage would be 168 volts from one side of the plug to the other, or 120VRMS). Just like they show you in class. If this were the case, then you'd be able to get 60 volts (rms) between each side of the plug and ground. But this isn't the case. Try it with a meter. You get 120V on one side, and zero on the other (hopefully) if you measure from the two prongs to the building ground. You can think of it as having a DC component of 84V and an AC component of 84V. This means that if you stick a fork into one prong, you won't feel anything and if you stick a fork into the other you are likely to be killed. The polarized plug is usually a safety thing to make sure that the switch turns off the hot side of the line instead of the cold side. In Europe, they often turn both off (which is the real solution). Yeah, it's a stupid question. But nobody ever told me any of this stuff when I was an EE student either, so don't feel bad. --scott
kucharsk@Solbourne.COM (William Kucharski) (04/24/91)
In article <1991Apr23.125408.17745@mlb.semi.harris.com> jws@cica4.mlb.semi.harris.com (James W. Swonger) writes: >Why: > >Some appliances (like TVs, for instance) have a "hot chassis", i.e. the >chassis is not isolated from the power line. Given this, it's better that >the chassis be attached to the neutral wire than the hot wire. Polarized >plugs decide this for you. And on a related note, my winner for the prize of all-time poorly designed products: the Carver C-9 Sonic Hologram Generator. Why? It had a non-polarized AC plug and a polarized convenience outlet on the back. And yes, flipping the AC plug in the wall would indeed cause the large slot of the polarized outlet to become hot! When I discovered this I quickly painted one side of the AC plug's sheath white to avoid surprises later... -- | William Kucharski, Solbourne Computer, Inc. | Opinions expressed above | Internet: kucharsk@Solbourne.COM | are MINE alone, not those | uucp: ...!{boulder,sun,uunet}!stan!kucharsk | of Solbourne... | Snail Mail: 1900 Pike Road, Longmont, CO 80501 | "It's Night 9 With D2 Dave!"
grege@gold.gvg.tek.com (Greg Ebert) (04/24/91)
In article <11864@jarthur.Claremont.EDU> cliff@jarthur.Claremont.EDU (Clifford Stein) writes: >What's the purpose of polarized wall outlets with an AC >signal? I don't understand. Is it really bad to force >something to plug in the wrong way? > >I am serious. With a real AC signal (sinusoidal waves) I can see no reason >at all for it. How can the electrical equipment tell? > Hot-chassis sets that I have fixed/examined/destroyed always had the (presumably) neutral lead connected to the chassis. It makes it a bit safer for homes where curious children like poking metal objects into vent holes, etc. I say presumably because you shouldn't ASSUME that the outlet is properly wired. My friend had such a stereo in his garage (minus the case). I used to reverse the plug on him, until one day he reversed it again and I got zapped !
ssave@ole.UUCP (Shailendra Save) (04/24/91)
In article <1991Apr23.125408.17745@mlb.semi.harris.com>, jws@cica4.mlb.semi.harris.com (James W. Swonger) writes: > Some appliances (like TVs, for instance) have a "hot chassis", i.e. the > the chassis be attached to the neutral wire than the hot wire. Polarized One of the other important reasons is that the fuse/circuit breaker is always on the neutral side, and so when it blows, the appliance is completely isolated and no part of it is live. The other way around, you could still get a shock. Shailendra
rando@skipper.dfrf.nasa.gov (Randy Brumbaugh) (04/24/91)
In article <11864@jarthur.Claremont.EDU>, cliff@jarthur.Claremont.EDU (Clifford Stein) writes: > What's the purpose of polarized wall outlets with an AC > signal? I don't understand. Is it really bad to force > something to plug in the wrong way? There is only AC potential at one blade of an ordinary 120v AC outlet (the hot side). Inside the outlet it is wired with black wire to make it easy to tell from the neutral, which should be near ground potential. For safety, outlets are designed so that it is possible to plug in a connector only one way. This allows all switches, fuses, etc. to be placed in the hot, high potential side. Also, a light socket is much safer if the small button contact at the bottom is hot and the outside shell is neutral -- much less shock hazard. If a plug is somehow forced in wrong, the equipment will still work, but an electrical hazard is created. Randy Brumbaugh rando@skipper.dfrf.nasa.gov
jeh@dcs.simpact.com (04/25/91)
In article <1991Apr23.152315.22825@news.larc.nasa.gov>, kludge@grissom.larc.nasa.gov ( Scott Dorsey) writes: > In article <11864@jarthur.Claremont.EDU> cliff@jarthur.Claremont.EDU (Clifford Stein) writes: >>What's the purpose of polarized wall outlets with an AC >>signal? I don't understand. Is it really bad to force >>something to plug in the wrong way? > You are thinking that the reference ground in your house is at zero volts > "the Ground they tell you about in Physics class" and that the AC power > alternates at + and - 84 volts around that ground (so that your peak voltage > would be 168 volts from one side of the plug to the other, or 120VRMS). Just > like they show you in class. > If this were the case, then you'd be able to get 60 volts (rms) between > each side of the plug and ground. But this isn't the case. Try it with a > meter. You get 120V on one side, and zero on the other (hopefully) if you > measure from the two prongs to the building ground. fine so far. > You can think of it as having a DC component of 84V and an AC component > of 84V. you lost me here. The hot side goes from +84 V to -84 V and back, referenced to neutral, for a peak-to-peak swing of 168. There is no "DC component" anywhere. Couldn't very well be, since your power is coming from a transformer winding. (Check it with a scope, set to DC coupling, making sure that you don't overload the front end; the 10-meg resistor in the normal 10:1 probe will suffice.) (Meanwhile, the other leg of your 240V residential service is going from -84 V to +84 V, so you can get 240V RMS for your electric dryer or air conditioner. In such an outlet you typically have two hot leads and a ground, not neutral. Some of the 120V outlets and builtins in your house are connected to one of the legs, and some to the other. The neutral wire coming into your house is really just a center tap on a 240V secondary winding, which in turn is usually excited by one side of a three-phase delta transformer on the power pole.) > This means that if you stick a fork into one prong, you won't feel anything > and if you stick a fork into the other you are likely to be killed. The > polarized plug is usually a safety thing to make sure that the switch turns > off the hot side of the line instead of the cold side. In Europe, they often > turn both off (which is the real solution). total agreement here. Also, light sockets (and other things with exposed contacts) are supposed to be wired so that the narrow blade of the plug (the hot side) goes to the switch and then to the harder-to-reach of the contacts. Suppose you're removing a dead light bulb and forgot to turn the switch off first. It's possible to touch the metal threads of the base of the bulb while they're still in contact with the socket, but since that's supposed to be the neutral side of the socket, it won't matter. --- Jamie Hanrahan (x1116), Simpact Associates, San Diego CA Chair, VMS Internals Working Group, U.S. DECUS VAX Systems SIG Internet: jeh@dcs.simpact.com, or if that fails, jeh@crash.cts.com Uucp: ...{crash,scubed,decwrl}!simpact!jeh
grege@gold.gvg.tek.com (Greg Ebert) (04/25/91)
In article <1991Apr24.111307.2296@dcs.simpact.com> jeh@dcs.simpact.com writes: > [...] >you lost me here. The hot side goes from +84 V to -84 V and back, referenced >to neutral, for a peak-to-peak swing of 168. There is no "DC component" >anywhere. [...] Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. You are absolutely correct about there NOT being a DC component. If there *was* a DC component in your AC service, it would quickly saturate every inductive load. Hmmmm, sounds like a nasty prank ;-) .
pierson@cimcad.enet.dec.com (Dave Pierson) (04/26/91)
In article <1991Apr24.111307.2296@dcs.simpact.com>, jeh@dcs.simpact.com writes... >In article <1991Apr23.152315.22825@news.larc.nasa.gov>, > kludge@grissom.larc.nasa.gov ( Scott Dorsey) writes: >> In article <11864@jarthur.Claremont.EDU> cliff@jarthur.Claremont.EDU (Clifford Stein) writes: >>>What's the purpose of polarized wall outlets with an AC >>>signal? I don't understand. Is it really bad to force >>>something to plug in the wrong way? >> You are thinking that the reference ground in your house is at zero volts >> "the Ground they tell you about in Physics class" and that the AC power >> alternates at + and - 84 volts around that ground (so that your peak voltage >> would be 168 volts from one side of the plug to the other, or 120VRMS). Just >> like they show you in class. >> If this were the case, then you'd be able to get 60 volts (rms) between >> each side of the plug and ground. But this isn't the case. Try it with a >> meter. You get 120V on one side, and zero on the other (hopefully) if you >> measure from the two prongs to the building ground. > >fine so far. > >> You can think of it as having a DC component of 84V and an AC component >> of 84V. > >you lost me here. He said think of it as. He didnt say it was... > The hot side goes from +84 V to -84 V and back, referenced >to neutral, for a peak-to-peak swing of 168. There is no "DC component" >anywhere. Couldn't very well be, since your power is coming from a transformer >winding. urmm, well, sorta... The neutral is REQUIRED by the National Electric Code to be tied to ground (as defined). This forces the neutral to "stay put". this can be considered DC, if one likes. The hot will swing plus 168/minus 168 around the neutral. [exception: some "220/240" circuits are actually two phases of a grounded neutral three phase circuit. The voltages will decrease moderately, and phasing will show up....] I can think of two advantages of "forcing" (via wide blade) the connection "polarity": reduced capacitive leakage and reduced 60 Hz "noise", where this is relavant. ======================================================= Someone [Shailendra?] whose posting i lost referred to the breaker/fuse being in the neutral, for safety. I suspect that was a typo, as the explanation was correct. The fuse/breaker MUST be in the HOT line, else the load is hot-to-ground even with the fuse/breaker open. [this commentary applies to grounded neutral circuits only.... Marine, RR and others are sometimes full floating...] thanks dave pierson |the facts, as accurately as i can manage, Digital Equipment Corporation |the opinions, my own. 600 Nickerson Rd Marlboro, Mass 01752 pierson@cimnet.enet.dec.com "He has read everything, and, to his credit, written nothing." A J Raffles
dana@locus.com (Dana H. Myers) (04/26/91)
In article <2226@gold.gvg.tek.com> grege@gold.gvg.tek.com (Greg Ebert) writes: >Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. I know I'm the one who claimed that power is dissipated in capacitors a while back (ugh!), but I do not believe 120VRMS is 340V p-p. It is 170V p-p. -- * Dana H. Myers KK6JQ | Views expressed here are * * (213) 337-5136 | mine and do not necessarily * * dana@locus.com | reflect those of my employer *
sidney@coed.coastal.ufl.edu (04/26/91)
> > I know I'm the one who claimed that power is dissipated in capacitors a >while back (ugh!), but I do not believe 120VRMS is 340V p-p. It is 170V p-p. > >-- The 120vrms power is 340v p-p. The voltage equation for our 120 volt system is 120*sin(377t). The rms value is the square root of the integral of the voltage squared over a full period. For sine wave voltages this works out to .707 * the peak voltage. For 120vrms power this is .707 * 170 = 120 volts. The 170 volt peak voltage occurs on both the positive and negative portions of the sin wave. Thusly the peak to peak voltage is 340V not 170.
grege@gold.gvg.tek.com (Greg Ebert) (04/26/91)
I (Greg Ebert) wrote: Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. dana@locus.com (Dana H. Myers) writes: > I know I'm the one who claimed that power is dissipated in capacitors a >while back (ugh!), but I do not believe 120VRMS is 340V p-p. It is 170V p-p. > For sinusoids, Vpeak = Vrms * sqrt(2), or Vpeak = 120*1.414 ~ 170 Volts. Vpeak-peak = 2 * Vpeak (for symmetrical waveforms), or Vp-p= 340 Volts. Stick a scope on the hot lead (but please be careful !). Perhaps some confusion arises because on a standard 240V circuit (I mean single-phase), *both* leads are hot, and *both* are 340 V p-p with respect to ground. One might infer that a 120 V circuit is half of this.
jeh@dcs.simpact.com (04/27/91)
In article <1477@rust.zso.dec.com>, pierson@cimcad.enet.dec.com (Dave Pierson) writes: > urmm, well, sorta... The neutral is REQUIRED by the National Electric > Code to be tied to ground (as defined). A bit of clarification is in order here: This connection MUST ONLY BE AT THE SERVICE ENTRY POINT TO THE HOUSE!!! (ie the main breaker box that's associated with the meter) Everywhere else, ground and neutral are to be kept separate. --- Jamie Hanrahan (x1116), Simpact Associates, San Diego CA Chair, VMS Internals Working Group, U.S. DECUS VAX Systems SIG Internet: jeh@dcs.simpact.com, or if that fails, jeh@crash.cts.com Uucp: ...{crash,scubed,decwrl}!simpact!jeh
sukenick@sci.ccny.cuny.edu (SYG) (04/28/91)
>> What's the purpose of polarized wall outlets with an AC >> signal? >(the hot side). Inside the outlet it is wired with black wire to "white is for weddings, black is for funerals" >make it easy to tell from the neutral, which should be near ground >potential. >If a plug is somehow forced in wrong, the equipment will still work, >but an electrical hazard is created. At the fuse box, ground (bare, green and/or cable shield ) and neutral (white wire) are tied together. Ground is usually tied to a pole in the ground and/or maybe the metal plumbing. The difference between ground and neutral is that neutral carries current, and since wires have some resistance, may be a few volts above ground. Some devices have powered parts which may be exposed in use or if damaged. It'd be better if these parts were at or near ground rather than 120 volts AC..
chap@art-sy.detroit.mi.us (j chapman flack) (04/28/91)
In article <11864@jarthur.Claremont.EDU> cliff@jarthur.Claremont.EDU (Clifford Stein) writes: >What's the purpose of polarized wall outlets with an AC >signal? I don't understand. Is it really bad to force >something to plug in the wrong way? >I am serious. With a real AC signal (sinusoidal waves) I can see no reason >at all for it. How can the electrical equipment tell? The equipment can't tell, but under certain abnormal conditions the *USER* can tell, and would much prefer having the equipment wired properly. One of those two conductors (the one connected to the WIDER slot in a U.S. 120VAC receptacle, provided the installer was competent) is very solidly bonded to ground potential. The other isn't. Equipment is designed so the grounded conductor ("neutral") is the one you're more likely to come into accidental contact with (e.g., the screw threads of a lamp socket), or the one that's more likely to come in contact with the equipment case in the event of some internal failure, etc. There may be additional reasons as well, but those are good enough for me. I was able once to tell my former employer that he had hot & neutral reversed, simply by leaning on the panel he'd built. (Ouch!) That episode earned me the title of "Official Ground-Fault Detector." -- Chap Flack Their tanks will rust. Our songs will last. chap@art-sy.detroit.mi.us -Mikos Theodorakis Nothing I say represents Appropriate Roles for Technology unless I say it does.
kludge@grissom.larc.nasa.gov ( Scott Dorsey) (04/29/91)
In article <2226@gold.gvg.tek.com> grege@gold.gvg.tek.com (Greg Ebert) writes: >In article <1991Apr24.111307.2296@dcs.simpact.com> jeh@dcs.simpact.com writes: >>you lost me here. The hot side goes from +84 V to -84 V and back, referenced >>to neutral, for a peak-to-peak swing of 168. There is no "DC component" >>anywhere. >Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. You are >absolutely correct about there NOT being a DC component. If there *was* >a DC component in your AC service, it would quickly saturate every >inductive load. Hmmmm, sounds like a nasty prank ;-) . Hmm... The point that I was making was that the signal was offset. The zero voltage point was at the bottom of the wave instead of the center. I think that imagining this as being a DC offset is a good way to look at it. When you talk about DC, you should ALWAYS talk about what ground level you are referring to, because DC voltage levels are not useful without a reference. If I consider my reference ground to be the +5V terminal of a power supply, then your building ground is probably floating at -5V. (Okay, not a good analogy at all, granted). --scott
kossackj@uncle-bens.rice.edu (Jordan Marc Kossack) (04/30/91)
In article <1991Apr27.174257.21380@sci.ccny.cuny.edu> sukenick@sci.ccny.cuny.edu (SYG) writes: > >At the fuse box, ground (bare, green and/or cable shield ) and neutral >(white wire) are tied together. Ground is usually >tied to a pole in the ground and/or maybe the metal plumbing. >The difference between ground and neutral is that neutral carries current, >and since wires have some resistance, may be a few volts above ground. > So, this makes me wonder if this is a cause of leakage current. When I was working as an electronics tech, we sold some electronic stuff to hospitals and physicians and had to guarantee that the leakage current did not exceed a specified value. To test a unit, we would unground it and measure the leakage current between the metal enclosure and 'actual' ground. At times, a physician or hospital would send one back to us claiming excessive leakage current, but when I measured it on the lab bench, it was OK. Could this have been caused by the neutral side of the circuit at the hospital floating too high above ground? Any ideas? OBTW, I'm not working there any more so I'm not trying to weasel out of meeting the specs. ;-) Thanks. - Jordan -- +-------------------------------------------------------+ | Although robust enough for general use, | | adventures into the esoteric periphery of the | | C shell may reveal unexpected quirks. |
jeh@cmkrnl.uucp (05/01/91)
In article <1991Apr29.132059.9403@news.larc.nasa.gov>, kludge@grissom.larc.nasa.gov ( Scott Dorsey) writes: > In article <2226@gold.gvg.tek.com> grege@gold.gvg.tek.com (Greg Ebert) writes: >>In article <1991Apr24.111307.2296@dcs.simpact.com> jeh@dcs.simpact.com writes: >>>you lost me here. The hot side goes from +84 V to -84 V and back, referenced >>>to neutral, for a peak-to-peak swing of 168. There is no "DC component" >>>anywhere. >>Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. You are >>absolutely correct about there NOT being a DC component. If there *was* >>a DC component in your AC service, it would quickly saturate every >>inductive load. Hmmmm, sounds like a nasty prank ;-) . > > Hmm... The point that I was making was that the signal was offset. The zero > voltage point was at the bottom of the wave instead of the center. I was afraid that was what you were trying to say. Tthe 120 VAC power waveform is NOT offset. (Where on earth did you get the idea that it was???) The zero voltage point (where no current flows, assuming purely resistive loads) is at the center of the wave, not the bottom. To think of it another way, the hot side of a 120V outlet both "pushes" and "pulls" wrt to the neutral side. And the wave looks the same, discounting IR losses in the neutral side, whether measured from hot to neutral or from hot to ground (it had damned well better, since ground and neutral are bonded together at the building service entry point). About one minute's experimentation with a DC-coupled scope will demonstrate these facts. You will see that the hot side goes both positive and negative with respect to neutral or ground. (Make DAMNED sure to use a 10:1 probe and that the outlet in question is wired correctly before doing this. And do not try to test the "inverse waveform" with the scope neutral hooked to the hot side of the power line and the probe to power line neutral or ground! "Thank you for observing all safety precautions." On second thought, I hesitate to recommend this experiment to anyone who so obviously misunderstands the basics of AC power. If you get yourself killed, don't come complaining to me about it.) The other 120 VAC leg in your house is just the inverse of the first so you can get 240V between them. As I said before, the two "legs" are just the ends of a 240V secondary winding on a local distribution transformer, and your house neutral lead is this winding's center tap. With this arrangement there is NO WAY the zero voltage point on either leg could be anywhere but the center of the waveform. There is simply nothing in the system that could possibly produce an offset such as you describe -- no matter what you call it. --- Jamie Hanrahan (x1116), Simpact Associates, San Diego CA Internet: jeh@dcs.simpact.com, or if that fails, jeh@crash.cts.com Uucp: ...{crash,scubed,decwrl}!simpact!jeh
jgk@osc.COM (Joe Keane) (05/01/91)
In article <1991Apr23.152315.22825@news.larc.nasa.gov> kludge@grissom.larc.nasa.gov ( Scott Dorsey) writes: > If this were the case, then you'd be able to get 60 volts (rms) between >each side of the plug and ground. But this isn't the case. Try it with a >meter. You get 120V on one side, and zero on the other (hopefully) if you >measure from the two prongs to the building ground. This experimental evidence is correct. > You can think of it as having a DC component of 84V and an AC component >of 84V. But this explanation is wrong. In article <1991Apr29.132059.9403@news.larc.nasa.gov> kludge@grissom.larc.nasa.gov ( Scott Dorsey) writes: >Hmm... The point that I was making was that the signal was offset. The zero >voltage point was at the bottom of the wave instead of the center. I think >that imagining this as being a DC offset is a good way to look at it. Try another experiment. Put a scope on the hot line, assuming it's rated for the voltage. Even with DC coupling, the zero voltage point will still be in the center. I think what you want to say is that there is a _common mode_, not DC, voltage component. That is, the voltages on the lines are not balanced. That's why you get a lot of voltage on one side and none on the other. I think balanced voltages makes more sense, that's generally what you get here with a 220V outlet. The advantage is that you don't lose power in the neutral line, which doesn't serve any useful purpose. -- Joe Keane, amateur mathematician jgk@osc.com (...!uunet!stratus!osc!jgk)
dana@locus.com (Dana H. Myers) (05/03/91)
In article <2237@gold.gvg.tek.com> grege@gold.gvg.tek.com (Greg Ebert) writes: >I (Greg Ebert) wrote: > >Almost. 120V RMS sinusoids are actually 340 v p-p, or 170 v peak. > >dana@locus.com (Dana H. Myers) writes: > >> I know I'm the one who claimed that power is dissipated in capacitors a >>while back (ugh!), but I do not believe 120VRMS is 340V p-p. It is 170V p-p. >> And then later I wrote that I was mistaken. Sorry. Sinusoidal 120VRMS is 170V PEAK, and 340V PEAK TO PEAK. I knew that and I still said the wrong thing. Side comments: I should comment I have received several messages from people intent on correcting me. Some of them were informative and only slightly snide. One of them was completely uninformative and thoroughly inflammatory (Hi Steve!). This is 'sci.electronics'; the 'sci.' means something like 'science'. Flaming isn't very scientific. If everyone acted they way they do on e-mail in real life, we'd all be assholes. In general, I believe that if I had made that comment in a personal conversation, most folks would politely disagree, etc. In the fantasy world of e-mail, however, I was accused of getting my amateur radio license in a ceral box and referred to as Mr. Science. I think this isn't very scientific. I am certain we have all made errors at some time or another, and reacting in such an arrogant manner as I have seen is very hypocritical. -- * Dana H. Myers KK6JQ | Views expressed here are * * (213) 337-5136 | mine and do not necessarily * * dana@locus.com | reflect those of my employer *
chap@art-sy.detroit.mi.us (j chapman flack) (05/04/91)
In article <1991Apr26.035007.2804716@locus.com> dana@locus.com (Dana H. Myers) writes: > > I know I'm the one who claimed that power is dissipated in capacitors a >while back (ugh!), but I do not believe 120VRMS is 340V p-p. It is 170V p-p. According to a peak-reading meter I just stuck in the wall outlet (here in the usa), I have a nice symmetrical -163.4 - 163.4, or 326.8 V p-p. Check it: 120 VRMS * sqrt(2) ==> 170 V peak (assumes a sinusoidal wave) 170 V peak * 2 ==> 340 V peak-to-peak So my line voltage is a mite low in the middle of the afternoon, which isn't too surprising.... -- Chap Flack Their tanks will rust. Our songs will last. chap@art-sy.detroit.mi.us -Mikos Theodorakis Nothing I say represents Appropriate Roles for Technology unless I say it does.