headlight bulbs

just had a headlight bulb burn out in my 97 ram 4x4 this weekend. i went
ahead and replaced both at the same time as usual. as i was logging the
service into my service journal i looked back to see when the last time the
bulbs were replaced on this truck. at 275,350 miles and 10+ years of rough
service this is the first time i have ever put new headlight bulbs in the
truck. i was amazed, the factory bulbs lasted a decade.

i decided to look at the service records on my other dodge trucks. the 2001
3/4 ton 4x4 diesel at 160k is still on the factory bulbs. the 1991 dakota
that has 177k known miles (had a broken odometer for a while) is on its 2nd
set of bulbs. the 1993 4x4 dakota with 220k is on its 5th set of bulbs. i
am now curious as to why the 93 goes through so many more bulbs than the
others. any theories?
michael

Most likely, higher system voltage. This will be a product of the voltage
regulator on the alternator. You are probably over 14 volts when running.
Steve

That could be caused by many things. The two primary ones are a loose
headlight assembly and / or a voltage problem. If the headlight assembly or
the bulb socket is loose, that will cause the bulb to vibrate excessivly
which can cause a significant reduction in bulb life. If the regulator is
not controlling the voltage properly and either allowing an excessivly high
voltage or voltage spikes, that to can cause a significant reduction in bulb
life.

--
If at first you don't succeed, you're not cut out for skydiving

Since Power = Volts x Amperes; an increase in the applied voltage will
result in a proportionate increase in current, and an increase in the power
that is a *product* of the two. An increase in consumed power will cause
the filament(s) to burn hotter, and reduce the life of the lamp.

Another factor is the number of on/off cycles. A lamp filament undergoes an
extreme change in temperature from off to on, causing a mechanical stress in
the filament. These mechanical stresses in the filament decrease the life
of the filament.

Bryan

"Since Volts = Amperes x Resistance" would make more sense. What you
wrote has volts and current on the same side of the equation.

Punkin, a.a.d.t's resident electrical engineer, taught us that blue
light is brighter than yellow light! Therefore, we can turn down the
juice and just tint the glass blue.

You are both correct, kinda sorta. For the purpose of this discussion
however, Bryan is uisng the "correct" interpretation of Mr. Ohm's law.
Since the resistance of the bulb's filament is a "constant" in this case,
the question (variable) is the voltage which ultimately determines the
current and thus the filament temperature which if too high will result in an early failure.

That's a discussion for another day.

Mike

Once a cock sucking whore, always a cock sucking whore.
But as usual you can do nothing but stir shit and prove yourself to be
the ass that you truly are.
Additionally, I never once indicated or implied any such thing fuck
nuts.

No, come on Mike, lets hear it.

Both of Georg Ohm's expressions are true. I was solving for E (volts) using
P (watts) and I (current/amperes). Ohm's pie and a nifty calculator is
here: http://www.the12volt.com/ohm/ohmslaw.asp

Another way of tracking the consumed power (assuming the resistance of the
load doesn't change appreciably) would be to use:
P = I^2 x R
or
P = E^2 / R

The only trouble is, AFIK, no lamp manufacturer specifies lifetime vs power
consumption. However, suffice to say it will be less if the specified
voltage (and current) is exceeded, and longer if the voltage (and current)
is less than specified. There is some mention of lifetime vs applied
voltage here: http://tinyurl.com/2opm7l. If concerned, the OP should check
the applied voltage with engine running (above idle speed) to verify it is
not significantly higher than it should be.

Bryan

Sorry, he had a 50/50 chance, he picked the wrong equation.

"FYI white or blue light is brighter then yellow." http://groups.google.co.in/group/alt...a2a718fd5687de

That's NOT what you wrote. You were (weren't) showing that an increase
in applied voltage will result in a proportionate increase in current.

Some here don't know that the voltage should be checked across the bulb,
while the bulb is in the socket and turned on. They'll take out the
bulb, measure 12V at the empty socket, and think it's all bitchen.

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode test 33Switch to Hybrid Mode Switch to Threaded Mode

2 2nd June 06:09 steve lusardi External User Posts: 1	headlight bulbs Most likely, higher system voltage. This will be a product of the voltage regulator on the alternator. You are probably over 14 volts when running. Steve

5 2nd June 06:10 beryl External User Posts: 1	headlight bulbs "Since Volts = Amperes x Resistance" would make more sense. What you wrote has volts and current on the same side of the equation. Punkin, a.a.d.t's resident electrical engineer, taught us that blue light is brighter than yellow light! Therefore, we can turn down the juice and just tint the glass blue.

7 2nd June 06:10 azwiley1 External User Posts: 1	headlight bulbs Once a cock sucking whore, always a cock sucking whore. But as usual you can do nothing but stir shit and prove yourself to be the ass that you truly are. Additionally, I never once indicated or implied any such thing fuck nuts.

8 2nd June 06:10 azwiley1 External User Posts: 1	headlight bulbs No, come on Mike, lets hear it.