nklr-dakar 6th stage - old riders rejoice

[Fred Williams]

You Gotta be kidding... Last time I checked, the bulb is a constant load, they don't change resistance. If it did, it would likely go UP when the bulb got hot, so you might draw more current when the VOLTAGE is LOWER!!! This is called a cascade effect, if you lose voltage on the wires, you actually waste more current because the voltage across the bulb is lower, thus increasing the "I" in the P (watts)= I (current) X E (obviously Voltge) equation. remember..... E/R=I, E/I=R RxI=E etc..etc... It's called OHMS LAW. Fred Williams (has more electrons in his head than most of the list) A15 Kawtchasaki

--- In DSN_klr650@y..., Claes Borovac wrote: > I'd say you are wrong here. You do consume more current. Simply, the voltage > over the bulb increases, which gives a higher current through the bulb. The > same current that flows through the bulb flows through the wires, that goes > without saying. > > The question here is, is a bulb a constant current drain? Or not? I would > say, more voltage over the bulb, more current. > > Claes, a swede in Dublin. > > -----Original Message----- > From: Ron Hipkiss [mailto:rhipkiss@f...] > Sent: 23 May 2001 15:52 > To: DSN_klr650@y... > Subject: RE: [DSN_klr650] extra lights > > > Actually, you aren't consuming any more current than with the stock > cables/wires. Current is constant in a series circuit like that, regardless > of the size of the wire. By increasing the gauge of the wire, you are > reducing the resistance, which increases the voltage reaching the bulb. > That gives you the brighter light. > > Cpt. Ron > > -----Original Message----- > From: Claes Borovac [mailto:claes@M...] > Sent: Wednesday, May 23, 2001 3:32 AM > To: Claes Borovac; 'Tom Simpson'; DSN_klr650@y... > Subject: RE: [DSN_klr650] extra lights > > > So yes, you are "consuming" more current that way but you have reduced a > loss in the cable so you are getting more use of your current. Hmm, I'll > stop now. He he. > > Claes, again. > > > Visit the KLR650 archives at > http://www.listquest.com/lq/search.html?ln=klr650 > > Post message: DSN_klr650@y... > Subscribe: DSN_klr650-subscribe@y... > Unsubscribe: DSN_klr650-unsubscribe@y... > List owner: DSN_klr650-owner@y... > > Support Dual Sport News by subscribing at: > http://www.dualsportnews.com > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > > > > [Non-text portions of this message have been removed]

[Claes Borovac]

Yes, as I wrote the bulb does increase in resistance. What you all seem to be missing is that the filament is not a constant load. It does vary with temperature so you have an R(t). Ok? I know this subject fairly well. I do know what ohms law is. In this case we can probably ignore the changes in the resistance of the bulb, ohms law applies, and hence, it will draw more current due to the slightly higher voltage, all easy to figure out through the law that you all know so well. Just realise it is a model of the real world, with constraints for when it is valid. That is what I have been saying all the time. Everyone happy? And no, the bulb will not draw more current when you drop the voltage. Yes, the resistance will drop as you lower the voltage, and the temperature decreases, but seriously unlikely that much to compensate for the voltage drop. Just as well as the resistance will not increase enough to compensate when the voltage increases, so the current will increase. Claes. -----Original Message----- From: Fred Williams [mailto:mtnbikerfred@...] Sent: 25 May 2001 03:39 To: DSN_klr650@yahoogroups.com Subject: [DSN_klr650] Re: extra lights You Gotta be kidding... Last time I checked, the bulb is a constant load, they don't change resistance. If it did, it would likely go UP when the bulb got hot, so you might draw more current when the VOLTAGE is LOWER!!! This is called a cascade effect, if you lose voltage on the wires, you actually waste more current because the voltage across the bulb is lower, thus increasing the "I" in the P (watts)= I (current) X E (obviously Voltge) equation. remember..... E/R=I, E/I=R RxI=E etc..etc... It's called OHMS LAW. Fred Williams (has more electrons in his head than most of the list) A15 Kawtchasaki

--- In DSN_klr650@y..., Claes Borovac wrote: > I'd say you are wrong here. You do consume more current. Simply, the voltage > over the bulb increases, which gives a higher current through the bulb. The > same current that flows through the bulb flows through the wires, that goes > without saying. > > The question here is, is a bulb a constant current drain? Or not? I would > say, more voltage over the bulb, more current. > > Claes, a swede in Dublin. > > -----Original Message----- > From: Ron Hipkiss [mailto:rhipkiss@f...] > Sent: 23 May 2001 15:52 > To: DSN_klr650@y... > Subject: RE: [DSN_klr650] extra lights > > > Actually, you aren't consuming any more current than with the stock > cables/wires. Current is constant in a series circuit like that, regardless > of the size of the wire. By increasing the gauge of the wire, you are > reducing the resistance, which increases the voltage reaching the bulb. > That gives you the brighter light. > > Cpt. Ron > > -----Original Message----- > From: Claes Borovac [mailto:claes@M...] > Sent: Wednesday, May 23, 2001 3:32 AM > To: Claes Borovac; 'Tom Simpson'; DSN_klr650@y... > Subject: RE: [DSN_klr650] extra lights > > > So yes, you are "consuming" more current that way but you have reduced a > loss in the cable so you are getting more use of your current. Hmm, I'll > stop now. He he. > > Claes, again. > > > Visit the KLR650 archives at > http://www.listquest.com/lq/search.html?ln=klr650 > > Post message: DSN_klr650@y... > Subscribe: DSN_klr650-subscribe@y... > Unsubscribe: DSN_klr650-unsubscribe@y... > List owner: DSN_klr650-owner@y... > > Support Dual Sport News by subscribing at: > http://www.dualsportnews.com > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > > > > [Non-text portions of this message have been removed] Visit the KLR650 archives at http://www.listquest.com/lq/search.html?ln=klr650 Post message: DSN_klr650@yahoogroups.com Subscribe: DSN_klr650-subscribe@yahoogroups.com Unsubscribe: DSN_klr650-unsubscribe@yahoogroups.com List owner: DSN_klr650-owner@yahoogroups.com Support Dual Sport News by subscribing at: http://www.dualsportnews.com Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ [Non-text portions of this message have been removed]

[Williams Fred]

You are right. I did not make it clear, that by "wasting" I meant you would need to increase the voltage across the whole circuit. I didn't mean the current would magically increase..... I litterlly meant in a cascade situation, that more heat means more R, which requires More E, to maintain I, which causes more Heat.... Etc... Etc.... Fred A15 Kawtchasaki --- dinero_0@... wrote:

> Fred, > > One point, the I(current) cannot increase > unless the voltage > increases from the source(i.e. battery). I'm pretty > sure that bulbs > are rated at a nominal value (lets say 60watts) for > a nominal voltage > (12vdc at 5amps). If you supply them 6vdc they > don't change their > resistance to draw 10 amps and give 60 watts of > illumination/heat, it > will only draw 2.5amps for 30watts (p=IxE). > > I do agree with you that the more resistive the > wires are, the > smaller the voltage that will be available to the > bulb. the smaller > the voltage to the bulb, the less current it can > draw and the weaker > the intensity. If a wire is too small for the > current draw, it will > heat up and become more resistive (those little > electrons moving do > create heat). The more resistive the wires become > the more heat they > create (?cascade effect). The stop, drop and roll > aspect comes in > when the wire becomes soo hot it melts the > insulation on the wire. > > Just my $0.01 > Mike Yuhas > --- In DSN_klr650@y..., "Fred Williams" > wrote: > > You Gotta be kidding... > > > > Last time I checked, the bulb is a constant load, > they don't change > > resistance. If it did, it would likely go UP when > the bulb got hot, > > so you might draw more current when the VOLTAGE is > LOWER!!! > > > > This is called a cascade effect, if you lose > voltage on the wires, > > you actually waste more current because the > voltage across the bulb > > is lower, thus increasing the "I" in the P > (watts)= I (current) X E > > (obviously Voltge) equation. remember..... E/R=I, > E/I=R RxI=E > > etc..etc... > > > > It's called OHMS LAW. > > > > Fred Williams (has more electrons in his head than > most of the list) > > A15 Kawtchasaki > > > > > > > > > > > > > > > > --- In DSN_klr650@y..., Claes Borovac > wrote: > > > I'd say you are wrong here. You do consume more > current. Simply, > > the voltage > > > over the bulb increases, which gives a higher > current through the > > bulb. The > > > same current that flows through the bulb flows > through the wires, > > that goes > > > without saying. > > > > > > The question here is, is a bulb a constant > current drain? Or not? > I > > would > > > say, more voltage over the bulb, more current. > > > > > > Claes, a swede in Dublin. > > > > > > -----Original Message----- > > > From: Ron Hipkiss [mailto:rhipkiss@f...] > > > Sent: 23 May 2001 15:52 > > > To: DSN_klr650@y... > > > Subject: RE: [DSN_klr650] extra lights > > > > > > > > > Actually, you aren't consuming any more current > than with the > stock > > > cables/wires. Current is constant in a series > circuit like that, > > regardless > > > of the size of the wire. By increasing the > gauge of the wire, you > > are > > > reducing the resistance, which increases the > voltage reaching the > > bulb. > > > That gives you the brighter light. > > > > > > Cpt. Ron > > > > > > -----Original Message----- > > > From: Claes Borovac [mailto:claes@M...] > > > Sent: Wednesday, May 23, 2001 3:32 AM > > > To: Claes Borovac; 'Tom Simpson'; > DSN_klr650@y... > > > Subject: RE: [DSN_klr650] extra lights > > > > > > > > > So yes, you are "consuming" more current that > way but you have > > reduced a > > > loss in the cable so you are getting more use of > your current. > Hmm, > > I'll > > > stop now. He he. > > > > > > Claes, again. > > > > > > > > > Visit the KLR650 archives at > > > > http://www.listquest.com/lq/search.html?ln=klr650 > > > > > > Post message: DSN_klr650@y... > > > Subscribe: DSN_klr650-subscribe@y... > > > Unsubscribe: DSN_klr650-unsubscribe@y... > > > List owner: DSN_klr650-owner@y... > > > > > > Support Dual Sport News by subscribing at: > > > http://www.dualsportnews.com > > > > > > Your use of Yahoo! Groups is subject to > > http://docs.yahoo.com/info/terms/ > > > > > > > > > > > > [Non-text portions of this message have been > removed] >

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[richardm@gowinnt.com]

--- In DSN_klr650@y..., "Fred Williams" wrote:

>Last time I checked, the bulb is a constant load, they don't change >resistance. If it did, it would likely go UP when the bulb got hot, >so you might draw more current when the VOLTAGE is LOWER!!!

I think that ALL conductive materials have more resistance the hotter they get.

>This is called a cascade effect, if you lose voltage on the wires, >you actually waste more current because the voltage across the bulb >is lower, thus increasing the "I" in the P (watts)= I (current) X E >(obviously Voltge) equation. remember..... E/R=I, E/I=R RxI=E >etc..etc...

A voltage drop in the wires will decrease the voltage across the bulb. The lower voltage across the bulb means a lower current flowing through the bulb. Lower voltage * lower current = less power. The temperature-to-resistance relationship is not enough to totally compensate. ---- Consider this: We'll assume that a mfr. has rated a bulb 60W @ 13.8 volts. 13.8 to 14.2 volts is a healthy charging system. 60W/13.8v = 4.35 Amps From this, we can calc the bulb's resistance to be 3.17 Ohms since 13.8v/4.35A = 3.17 Ohms Now... Let's add a lossy wire to the equation. .6v is a good number. We will now have only 13.2v across the 3.17 Ohm bulb since we pissed away .6v in the wire. This means that the circuit's current is now 4.16 amps since 13.2/3.17 = 4.16. 4.16A * 13.2v = 54.9 Watts. This is why lossy wiring dims the bulb. But wait - It gets worse... The cooler running bulb now has a lower color temperature (reddish- brown) thanks to the undervolting. A greater percentage of an undervolted bulb's total light output is in the infrared spectrum when compared to a properly volted bulb. IR light does us no good. Double-whammy. But wait - It gets worse... As the operating temp of the bulb gets lower, the halogen gasses lose their ability to do the job of "collecting" the tungsten atoms that condense on the inside of the glass and redepositing them on the filament. The end result is that the glass gets brown with age, just like a non-halogen bulb. Dirty brown glass blocks light. Triple-whammy. RM

[Russell Dyck]

I'm thinking about adding some extra lights to the KLR. What's the max wattage I could go with, and whats a good cheap set (if there is such a thing. The bike is a 02 and all stock. Russ

[guymanbro]

I got a pair of "off-road" driving lights from Walmart for $20. 55W each and had them wired (with much help from Devon) so they ran without the stock headlight. Night time visibility compared to stock? Still had the same throw (distance down the road) but could easily see both sides of road and a lot of the brush (much wider light). Lots of options available though, dat brooklyn bum --- In DSN_klr650@yahoogroups.com, "Russell Dyck " wrote:

> I'm thinking about adding some extra lights to the KLR. > What's the max wattage I could go with, and whats a good cheap set > (if there is such a thing. > The bike is a 02 and all stock. > > Russ

[ephilride@aol.com]

By the numbers: Birthday boy, 40 year old, Alfie Cox (40) wins today's 142 miles special with an average speed of 57 miles per hour. Today's outstanding ride brought Alfie back into contention in 4th overall, 3'33" behind leader Sainct (32). Roma (30) and Meoni (45) are tied for 2nd overall 17 seconds behind. At 28 years of age, youngster Despres is in 5th, 4'23" behind. Big gap to Coma (26) in 6th, 21 minutes back (who no doubt lost time today helping a dune stuck Sala) closely followed by De Azevedo (28), Sala (39), Lundmark (41) and in 10th place Bucy (40), 29 minutes off the pace. 6th through 10th place is anybody's race and positions will no doubt be swapped as the event progresses. First non-KTM is Honda rider Mitsuhashi in 23rd, 55 minutes back. The top five places are likely to be as fluid as the sand dunes they race across as the days unfold. The v-2 boys will have to manage their tires carefully. The one lung riders will have to resist the temptation to run at full song all day long just to stay in contact with the most potent Dakar machines ever built -- the mighty LC-8! Will the LC-8 with its decided horsepower advantage be the undoing of the LC-4 as a viable Dakar rig? Will endless dunes and power sapping sand allow horsepower to win over rider skill? Will Meoni, the crafty one, the wise one, the most resent master of this event, win tomorrow? Could it be Roma's turn, riding with his heart instead of his head as he tries to win his first Dakar, who presses tomorrow and establishes himself as the best of the best. Or will he roll out of the throttle a little, conserve his tires, conserve his machine, conserve his body and wait for the others to make a mistake? Tomorrow's performance on the longest special (363 miles) of the world's greatest off-road race will go a long way in defining who owns Dakar this year. -knot