That Poly would have been a great truck engine, and no problem fitting it.
Stop in for a cup of coffee
- Thread starter toolmanmike
- Start date
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Yep. They were and used them in big cars through 66.
Poly Super Pac reported it.
In decent light look at the lens. The codes will tell you what they are. There are fakes - sometimes they can be figured from the codes not making sense. LOL.
US market lamps will have HR for Halogen Replaceable. They are designed for the HB2 bulb, almost the same as H4.
ECE market lamps will have HCR and take H4 bulb.
The higher the voltage they get, the more current they will draw. On battery power they're getting 12.8 V at best.
If you go to that Candlepower thread with the H4 bulb comparison, you can see the difference higher voltages make in current and light output.
FISHYPETE
Well-Known Member
The 318 poly is a dependable engine. Have it in my 59 Fury and 59 Sweptside in stock form. And one from a 66 Charger in shed
Had one in my 61 Fury. Ran OK in that tank....
You said the higher the voltage the higher the current? Hmm, Ohms law says otherwise...
That's how I used to think about it too. You know, like going to a 220 V AC motor from a 110.
The voltage in this case is the supply voltage.
P= IV and V=IR
Nominal 60 Watt lamp at 12 Volts draws 5 amps,
and so at 14 volts it should only draw 4.3 amps.
But that's wrong.
With a lamp and many other devices, resistance stays roughly the same, not the power.
Lets say that nominal 60 Watt lamp was rated at 12 Volts.
12 volts = 5 amps x R, R = 2.4 Ohms
then
14 volts = I x 2.4 Ohms, I = 5.8 amps
solving for the new power at 14 Volts
P= 14 volts x 5.8 amps
P = 81 Watts.
All assuming I didn't make a mistake.
zkx14
Duster De-ruster
HmmmmThat's how I used to think about it too. You know, like going to a 220 V AC motor from a 110.
The voltage in this case is the supply voltage.
P= IV and V=IR
Nominal 60 Watt lamp at 12 Volts draws 5 amps,
and so at 14 volts it should only draw 4.3 amps.
But that's wrong.
With a lamp and many other devices, resistance stays roughly the same, not the power.
Lets say that nominal 60 Watt lamp was rated at 12 Volts.
12 volts = 5 amps x R, R = 2.4 Ohms
then
14 volts = I x 2.4 Ohms, I = 5.8 amps
solving for the new power at 14 Volts
P= 14 volts x 5.8 amps
P = 81 Watts.
All assuming I didn't make a mistake.
Hmmmm, OK. Never thought of this with bulbs...That's how I used to think about it too. You know, like going to a 220 V AC motor from a 110.
The voltage in this case is the supply voltage.
P= IV and V=IR
Nominal 60 Watt lamp at 12 Volts draws 5 amps,
and so at 14 volts it should only draw 4.3 amps.
But that's wrong.
With a lamp and many other devices, resistance stays roughly the same, not the power.
Lets say that nominal 60 Watt lamp was rated at 12 Volts.
12 volts = 5 amps x R, R = 2.4 Ohms
then
14 volts = I x 2.4 Ohms, I = 5.8 amps
solving for the new power at 14 Volts
P= 14 volts x 5.8 amps
P = 81 Watts.
All assuming I didn't make a mistake.
Here's the post about the H4 bulb tests.
Interesting headlight bulb test results
He tested at 13.5 Volts and then converts to some of the other common reference voltages.
He mentions that
US standard is for wattage to be stated at 12.8 Volts.
UN/ECE standard for wattage for same type of bulb is at 13.2 Volts.
So based on that, we clearly can't compare 'wattage' of an 9003 or HB2/H4 given in US Standard to an H4 in ECE standard without doing some math.
Fun 'eh?
Got the following from Stern's website:
Headlamp bulb light output is severely compromised with decreased voltage. The drop in light output is not linear, it is exponential with the power 3.4. For example, let's consider a 9006 low beam bulb rated 1000 lumens at 12.8 Volts and plug in different voltages:
10.5V : 510 lumens
11.0V : 597 lumens
11.5V : 695 lumens
12.0V : 803 lumens
12.5V : 923 lumens
12.8V : 1000 lumens ←Rated output voltage
13.0V : 1054 lumens
13.5V : 1198 lumens
14.0V : 1356 lumens ←Rated life voltage
14.5V : 1528 lumens
from here: Daniel Stern Lighting Consultancy and Supply
Interesting headlight bulb test results
He tested at 13.5 Volts and then converts to some of the other common reference voltages.
He mentions that
US standard is for wattage to be stated at 12.8 Volts.
UN/ECE standard for wattage for same type of bulb is at 13.2 Volts.
So based on that, we clearly can't compare 'wattage' of an 9003 or HB2/H4 given in US Standard to an H4 in ECE standard without doing some math.
Fun 'eh?
Got the following from Stern's website:
Headlamp bulb light output is severely compromised with decreased voltage. The drop in light output is not linear, it is exponential with the power 3.4. For example, let's consider a 9006 low beam bulb rated 1000 lumens at 12.8 Volts and plug in different voltages:
10.5V : 510 lumens
11.0V : 597 lumens
11.5V : 695 lumens
12.0V : 803 lumens
12.5V : 923 lumens
12.8V : 1000 lumens ←Rated output voltage
13.0V : 1054 lumens
13.5V : 1198 lumens
14.0V : 1356 lumens ←Rated life voltage
14.5V : 1528 lumens
from here: Daniel Stern Lighting Consultancy and Supply
I thought about it 'cause I was confused about it for a long while. Now I can explain it and my head doesn't hurt.Hmmmm, OK. Never thought of this with bulbs...
Now the other question is.
who's still awake?
Speaking of class.
Did anyone else hear the newly released live tape of Tom Lehrer at a show?
It includes another version of his song of all the elements in the periodic table.
It's a classic version.
One Aristotle would recognize.
All four elements.
who's still awake?
Speaking of class.
Did anyone else hear the newly released live tape of Tom Lehrer at a show?
It includes another version of his song of all the elements in the periodic table.
It's a classic version.
One Aristotle would recognize.
All four elements.
So you are saying because a bulbs resistance stays the same no matter the voltage?
Unconventional
Well-Known Member
I thought it was very amusing, the little shorty 3 day cruise from San Pedro to Ensenada and back. It was the New Years Day edition. All the Mai Tais you could drink at the Captain's cocktail party, so I did. First night was pretty much straight to Ensenada and dock. I wake up the next morning, take a short walk to the upper deck, there was a heavily armed Federale Soldier staring at me from a gun tower on the dock
. Surprise!
Unconventional
Well-Known Member
Yes.
I suppose the heat must change the resistance somewhat, but not enough to worry about.
With enough voltage from a good power supply on it, it will draw enough current to burn out.
That makes sense?
Yes....
dukeboy440
Well-Known Member
Same shop posted another, 514 hpAlmost 500 horse
I was going to ask about the heat......
Yowza....
Above my pay grade.
We'ld have to ask someone with expertise in light bulbs. I mean real ones, not physics book ones!
Ddaddy
Just doing what I do
I always though about light and voltage this way...That's how I used to think about it too. You know, like going to a 220 V AC motor from a 110.
The voltage in this case is the supply voltage.
P= IV and V=IR
Nominal 60 Watt lamp at 12 Volts draws 5 amps,
and so at 14 volts it should only draw 4.3 amps.
But that's wrong.
With a lamp and many other devices, resistance stays roughly the same, not the power.
Lets say that nominal 60 Watt lamp was rated at 12 Volts.
12 volts = 5 amps x R, R = 2.4 Ohms
then
14 volts = I x 2.4 Ohms, I = 5.8 amps
solving for the new power at 14 Volts
P= 14 volts x 5.8 amps
P = 81 Watts.
All assuming I didn't make a mistake.
Higher voltage means more electrons per unit time that can be converted to photons.
Amps determine how fast the supply of electrons can be replenished (or sustained at steady draw rate) to maintain the photon emission rate (luminosity).
In other words, higher voltage means more photons (light output) as long as it is not restricted by resistance to amperage flow that chokes off the electrons being converted to photons.
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