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majid_sabet
Sep 19, 2008, 4:05 PM
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FOR IMMEDIATE RELEASE
September 18, 2008
Release #08-403
Firm's Recall Hotline: (877) 740-3826
CPSC Recall Hotline: (800) 638-2772
CPSC Media Contact: (301) 504-7908
Safety Alert: Petzl America Warns of Burn Hazard from Headlamps; Product
Should Only Be Used with Non-Rechargeable Batteries
WASHINGTON, D.C. - The U.S. Consumer Product Safety Commission, in
cooperation with the firm named below, today announced a safety alert for
the following consumer products. Consumers should immediately stop using
rechargable batteries with any of the products listed below.
Name of Product: "MYO" and "MYO Belt" Headlamps
Units: About 322,000
Distributor: Petzl America, of Clearfield, Utah
Manufacturer: Petzl S.A., of Crolles, France
Hazard: If the headlamp is used with rechargeable batteries, the cable
connecting the battery pack to the lamp can spark, melt, or catch fire. This
poses a burn hazard to consumers.
Incidents/Injuries: Petzl has received 13 reports of sparking and/or
melting, with 2 reports of flames. One consumer received a minor burn to the
hand, and another experienced singed hair.
Description: This safety alert involves the "MYO" and "MYO Belt" headlamps
with name and model numbers listed below. The "MYO" headlamps have a battery
pack attached to the headband. The "MYO Belt" headlamps have a remote
battery pack attached to the headlamp via a long electric cable.
Name Model Number
MYO E26P
MYO-black E26PN
MYO 3 E27P
MYO 3-black E27PN
MYO 5 E28P
MYO 5-black E28PN
MYO Belt 3 E29 P
MYO Belt 5 E30 P
MYOLITE E31 P
MYOLITE 3 E32 P
MYOBELT SB5 E33 P
MYO XP, blue E83 P
MYO XP, gray E83 P2
MYOBELT XP, blue E84 P
MYO XP BELT, gray E84 P2
The name of the product can be found on the side of the headlamp and on the
packaging. The model number can be found on the packaging.
Sold at: Specialty retailers nationwide from February 2003 through August
2008 for between $40 and $80.
Manufactured in: France
Remedy: Consumers should immediately stop using rechargeable batteries with
the headlamps and contact Petzl to obtain a new warning label. Consumers can
continue to use the headlamps with non-rechargeable batteries.
Consumer Contact: For additional information, contact Petzl America
toll-free at (877) 740-3826 between 8 a.m. and 5 p.m. MT Monday through
Friday, or visit the firm's Web site at http://www.petzl.com
To see this recall on CPSC's web site, including pictures of the recalled
products, please go to:
http://www.cpsc.gov/cpscpub/prerel/prhtml08/08403.html
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reg
Sep 19, 2008, 4:18 PM
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hummm..... why does this happen? only thing i can think of is a fully charged AA battery is at 1.6 vdc.
these batteries are commonly called 1.5 vdc. all AAA AA C D are all "1.5 vdc". but re-cahrgeable's are 1.2 vdc. ohm's law tells us that as voltage goes down, amperage goes up. so a re-chargeable being drained will draw increasingly more amperage causing heat and deterioration of the wire which is probably at a minimum anyway.
majid, does that make sence?
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eliclimbs
Sep 19, 2008, 4:24 PM
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Reg,
I think you may be mistaken. Ohm's law is V=IR, so if the voltage goes down, the current will also go down. I don't know why the rechargable battieries would be a problem.
Eli
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Valarc
Sep 19, 2008, 4:32 PM
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The current from rechargeable batteries is indeed higher than that from the standard throw-away kind. I've seen lots of consumer electronics manuals that specifically warn you not to use rechargeable batteries. I even had it happen once - fried a cheap digital camera from overheating.
Kinda scary to think about getting burned by your headlamp, though...
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reg
Sep 19, 2008, 4:47 PM
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eliclimbs wrote: Reg, I think you may be mistaken. Ohm's law is V=IR, so if the voltage goes down, the current will also go down. I don't know why the rechargable battieries would be a problem. Eli
actually it's I=V/R. here's an example: a 100 watt light bulb at 120 vac draws .833 amps (100/120=.833). if you lower the voltage to 115 amp draw goes up to .869 (100/115=.869)
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eliclimbs
Sep 19, 2008, 5:01 PM
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reg wrote: eliclimbs wrote: Reg, I think you may be mistaken. Ohm's law is V=IR, so if the voltage goes down, the current will also go down. I don't know why the rechargable battieries would be a problem. Eli actually it's I=V/R. here's an example: a 100 watt light bulb at 120 vac draws .833 amps (100/120=.833). if you lower the voltage to 115 amp draw goes up to .869 (100/115=.869)
Reg,
Not trying to flame, but... I=V/R is equivalent to V=IR. Second, I think your units don't add up. V is volts, and R is Ohms. You're dividing Watts (j/s) by Volts. Please correct me if I'm wrong.
Eli
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sungam
Sep 19, 2008, 5:02 PM
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Am I the only one laughing at the thought of some dude working at petzl opening a complaint email and reading
"dear petzl, I used your headlamp, and after the doused the flames out I realized...
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djlachelt
Sep 19, 2008, 5:06 PM
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reg wrote: eliclimbs wrote: Reg, I think you may be mistaken. Ohm's law is V=IR, so if the voltage goes down, the current will also go down. I don't know why the rechargable battieries would be a problem. Eli actually it's I=V/R. here's an example: a 100 watt light bulb at 120 vac draws .833 amps (100/120=.833). if you lower the voltage to 115 amp draw goes up to .869 (100/115=.869)
Actually, V=IR is the same as I=V/R.
However, your example is all messed up.
V represents Volts (not Watts).
I is the current (in Amps)
R is the resistance in the circuit (not the Voltage). You plugged the voltage into the denominator of your equation (in place of resistance - which we don't know about this circuit)... that is why you got the opposite conclusion from Eli.
Eli is correct. If the Voltage (V) goes down so does the Current (I)... no matter whether you use V=IR or I=V/R
None of which goes to explain why a rechargable battery would cause these lamps to be a hazard.
-Jon
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Valarc
Sep 19, 2008, 5:27 PM
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The voltage has very little to do with it.
In reality, the issue is the internal resistance of the batteries. Rechargeable batteries have a much lower internal resistance, meaning that the total resistance in the circuit will similarly be lower. Lower resistance at a similar voltage = higher current. Usually this difference isn't enough to make a difference, as the resistance in the circuit is much larger than the battery's internal resistance, but sometimes it's not, and that difference can cause a big increase in current, and all of the lovely overheating that goes with it.
Lots of headlamps use LEDs, which are generally low resistance. In an effort to minimize losses due to heat, petzl probably tried to keep the resistance of their circuitry as low as possible, and this put them in the realm where the internal resistance of the batteries becomes important. All a giant guess, naturally.
(This post was edited by Valarc on Sep 19, 2008, 5:28 PM)
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reg
Sep 19, 2008, 5:34 PM
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djlachelt wrote: reg wrote: eliclimbs wrote: Reg, I think you may be mistaken. Ohm's law is V=IR, so if the voltage goes down, the current will also go down. I don't know why the rechargable battieries would be a problem. Eli actually it's I=V/R. here's an example: a 100 watt light bulb at 120 vac draws .833 amps (100/120=.833). if you lower the voltage to 115 amp draw goes up to .869 (100/115=.869) Actually, V=IR is the same as I=V/R. However, your example is all messed up. V represents Volts (not Watts). I is the current (in Amps) R is the resistance in the circuit (not the Voltage). You plugged the voltage into the denominator of your equation (in place of resistance - which we don't know about this circuit)... that is why you got the opposite conclusion from Eli. Eli is correct. If the Voltage (V) goes down so does the Current (I)... no matter whether you use V=IR or I=V/R None of which goes to explain why a rechargable battery would cause these lamps to be a hazard. -Jon
i wasn't thinking correctly at the time - they are the same. but your both wrong. not sure if this discussion is of any interest to anyone but us. if volts times amps = watts and 120v x .833a=99.96watts then if voltage drops to 115, amps have to go up in order to supply 99.96 watts. a 100 watt light bulb will draw 100 watts even if voltage drops. voltage and amprege have an inverse relationship. wattage does not change. that's why power companies charge based on watts.
edit: just read this: "With a purely resistive load the relationship between voltage and amperage is directly proportional. If one increases so does the other proportionately. With an inductive load the relationship is inversely proportional increase the voltage and you'll decrease the amperage." we run "high tension" lines (300+KV) across the land from city to city at very low amps but a fixed wattage. guess resistive (electronic) loads act differently. hope we're done now - i'm starting to get indigestion 
(This post was edited by reg on Sep 19, 2008, 5:42 PM)
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sungam
Sep 19, 2008, 5:42 PM
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Valarc wrote: The voltage has very little to do with it. In reality, the issue is the internal resistance of the batteries.
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epoch
Moderator
Sep 20, 2008, 12:53 AM
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Moved this thread to Gear Heads. Also Stickied it for it to remain at the top of the forum.
Also, there's a post by Eric_at_Petzl that has the official release in the General forum.
http://www.rockclimbing.com/...rum.cgi?post=1971840
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wmfork
Sep 25, 2008, 4:33 PM
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Valarc wrote: In reality, the issue is the internal resistance of the batteries.
Well, sort of. Or you can say it has to do with the discharge curve of the batteries. You can draw more current through fresh alkaline batteries than NiMH (for the headlamp application), but it's pretty much unsustainable and drops below the level of NiMH fairly quickly. So the circuit won't generate enough heat over time to burn you. On the other hand, most rechargables provide a higher (and fairly constant) average draw over time, until they are about fully discharged.
The way I interpret the whole thing is: Petzl claims their headlamps are really bright (no doubt tested with fresh alkalines), but it only lasts for a couple minutes on Alkaline (they are unregulated, in contrast to Princeton Tec). And if you wish to have sustained high output, the headlamps do not have adequate heat sink for it, but you should still be OK under cold weather.
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jdefazio
Sep 25, 2008, 5:35 PM
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reg wrote: i wasn't thinking correctly at the time - they are the same. but your both wrong. not sure if this discussion is of any interest to anyone but us. if volts times amps = watts and 120v x .833a=99.96watts then if voltage drops to 115, amps have to go up in order to supply 99.96 watts. a 100 watt light bulb will draw 100 watts even if voltage drops. voltage and amprege have an inverse relationship. wattage does not change. that's why power companies charge based on watts. edit: just read this: "With a purely resistive load the relationship between voltage and amperage is directly proportional. If one increases so does the other proportionately. With an inductive load the relationship is inversely proportional increase the voltage and you'll decrease the amperage." we run "high tension" lines (300+KV) across the land from city to city at very low amps but a fixed wattage. guess resistive (electronic) loads act differently. hope we're done now - i'm starting to get indigestion
These are very incorrect statements. The high tension lines supply constant AC voltage (tension is equivalent to voltage, especially in euro-speak). This is stepped down in intervals (substation, utility pole) to supply your house with a constant voltage as well. A 100W light bulb only consumes 100W for this exact fixed voltage in your house. Bring that same bulb to England (twice the voltage) and rig it to a socket with wires, and it will run brighter and hotter (i.e. consuming more power).
So what it really comes down to is...Magnus's post count will drop off by a factor of 2 when he visits the states since he is used to 240 Vac. ;-)
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sungam
Sep 25, 2008, 6:40 PM
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oh noes!
I guess the fact I won't have my kompooter with me won't help things either, right? 
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sungam
Sep 25, 2008, 7:27 PM
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Real haggis... omg, hahaha.
 (not being sarcastic, cracked me up).
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ersatz_radio
Sep 25, 2008, 7:30 PM
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reg wrote: i wasn't thinking correctly at the time - they are the same. but your both wrong. not sure if this discussion is of any interest to anyone but us. if volts times amps = watts and 120v x .833a=99.96watts then if voltage drops to 115, amps have to go up in order to supply 99.96 watts. a 100 watt light bulb will draw 100 watts even if voltage drops. voltage and amprege have an inverse relationship. wattage does not change. that's why power companies charge based on watts edit: just read this: "With a purely resistive load the relationship between voltage and amperage is directly proportional. If one increases so does the other proportionately. With an inductive load the relationship is inversely proportional increase the voltage and you'll decrease the amperage." we run "high tension" lines (300+KV) across the land from city to city at very low amps but a fixed wattage. guess resistive (electronic) loads act differently. hope we're done now - i'm starting to get indigestion
reg, you seem to be consistently confusing resistance with power, and the equation V=IR with the related equation P=IV.
You use the voltage and resistance of a circuit to determine the current with the first equation.
Then you use the second equation to find the power.
You are right that if voltage drops, current will have to increase to maintain a power rating. But there is no magic force that adjusts current levels to cause this increase. In a circuit, current is caused by voltage. Voltage is caused by outside forces, such as chemical reactions in a battery or electromagnetic fields in some other instances. Similarly, power is created by the actions of voltage and current and is strictly a product of those variables.
To reiterate, current is influenced by resistance and voltage in a system. Power is influenced by voltage and current. This does not work the other way around. If you know the power of a circuit, you can use this information to calculate the voltage or current. However, you cannot alter the voltage or current by altering the power
(This post was edited by ersatz_radio on Sep 25, 2008, 7:30 PM)
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ersatz_radio
Sep 25, 2008, 7:37 PM
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reg wrote: -snip- a 100 watt light bulb will draw 100 watts even if voltage drops. -snip-
This is not true. A 100 watt bulb will produce 100 watts under normal circuit conditions. The 100 watt designation simply describes this.
For instance, if you connect two 100 watt bulbs in series, each will only produce 25 watts. This is because you double the resistance of the circuit, halving the current, and halve the voltage across each bulb. Power is current times voltage so it's only 25 watts.
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jdefazio
Sep 25, 2008, 7:54 PM
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ersatz_radio wrote: reg wrote: -snip- a 100 watt light bulb will draw 100 watts even if voltage drops. -snip- This is not true. A 100 watt bulb will produce use 100 watts under normal circuit conditions. The 100 watt designation simply describes this. For instance, if you connect two 100 watt bulbs in series, each will only produce use 25 watts. This is because you double the resistance of the circuit, halving the current, and halve the voltage across each bulb. Power is current times voltage so it's only 25 watts.
^^^^
This iz korrekt 
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tradrenn
Sep 26, 2008, 10:52 PM
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Somebody needs a lappy ( more useful then a cell )
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sungam
Sep 26, 2008, 11:53 PM
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tradrenn wrote: Somebody needs a lappy ( more useful then a cell )
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krosbakken
Sep 27, 2008, 1:16 AM
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sungam wrote: tradrenn wrote: Somebody needs a lappy ( more useful then a cell ) I have a laptop, but I sure as hell aint luggin it around the country with me!
Do it so then it will be easier for us to get a hold of you. DUH. haha 
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sungam
Sep 27, 2008, 10:53 AM
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krosbakken wrote: sungam wrote: tradrenn wrote: Somebody needs a lappy ( more useful then a cell ) I have a laptop, but I sure as hell aint luggin it around the country with me! Do it so then it will be easier for us to get a hold of you. DUH. haha
If I only crash with people who go on rc.com, then I know they have access to rc.com, so all will be well.
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