I was wondering if anybody could tell me the size of these bulbs in inches, mm, or or the standard light bulb size classification.
Also, what size socket works with them and if anyone could point me in the direction of finding sockets for these bulbs.
Thanks.
Dear Jay,
I'm pretty sure you are talking about a #53 lamp, (Radio Shack 272-1117) CM figure "F" below.
http://www.mouser.com/catalog/catalogUSD/641/114.pdf
It's possible this would work for you. (Mini-bayonet base, holder or socket?)
http://www.radioshack.com/product/index.jsp?productId=2062369
Hope this helps.
Sincerely,
Joe Satnik
The Radio Shack light bulbs come in either 14.4 volt or 18.4 volt in the bayonette style for that size.
Don't use automotive bulbs as they are rated for 12 volt DC.
You can use less voltage to make a bulb dimmer but don't go over the voltage rating as that will shorten the life of the bulb.
Lee F.
#53 lamp is standard for Williams and WBB.
Thanks for the info guys. If WBB uses the #53 bulb, why do they list it as 12v whereas the Chicago Miniature catalogue lists them as being 14.4v? Not a big deal, just curious.
Dear JayOcks,
It has to do with automobile lead-acid batteries which are nominally 12V.
The charging system voltage regulator takes the battery up to 13.9V, perhaps even higher in colder weather.
Also, to charge a battery up in a reasonable amount of time, your regulator has to be set a bit above the target full voltage.
If your battery is disconnected while the engine is operating, the charging system can spike the voltage up even higher.
In the end, the 14.4 V spec is a safety margin to keep from burning out the lamps in your "12V" automobile.
Hope this helps.
Sincerely,
Joe Satnik
Joe,
For your info an auto's regulator will normally charge in the 13.4 volt to 14.9 volt range and be considered perfectly safe for automotive work, sometimes due to an extreme load on the alternator it could be as low as 13.2 volts. New car batteries should be 12.6 volts when fully charged.
Most 12 volt bulbs can handle 18 volts for short periods of time.
A lower volt bulb will heat up faster than a higher volt bulb and can melt some softer plastics.
Lee F.
I think th eShack also has 24 V bulbs. But they would be dull.
Dear All,
Well, I'd guess that 18 Volt lamps would be dull, as the traditional power Williams/WBB locos fly around the track at pretty low voltages (hence the 14.4 Volt lamps, not the 20 Volts max of a Lionel traditional transformer).
You'd probably have a hard time seeing the glow of a 24 Volt bulb at normal speed.
One way to brighten the lamps up would be to wire the loco's dual motors in series instead of parallel.
Two more advantages of series wired motors:
1.) Better control at low speeds,
2.) More power, less distortion for the True-Blast II sound system.
Hope this helps.
Sincerely,
Joe Satnik
You've all been very helpful.
One last question:
Anyone know how or why WBB is able to charge around $15 for 50 of the bulbs when the Chicago miniature bulbs are .67 when you're buying 100 or more?
Not complaining, but this is a great price, I haven't been able to find miniature bulbs anywhere else at this price.
Old stock, haven't upped the price yet?
Overstock, want to reduce inventory?
Another thing about using 24 volt bulbs is that it may slow down your engine's top speed, as it uses more power to light up.
An 18 volt bulb should be the highest volt bulb to use with any O gauge train.
J O, the prices for light bulbs could be old stock left over from Williams. That's a very good price, but who needs 50 light bulbs?
Lee F.
Lee,
Power = Volts x Amps
Look at the specs for each lamp, compare power (VxA) and candlepower. They should be proportional.
The 24 volt lamp draws half the amps for the same power 12V lamp.
Hope this helps.
Sincerely,
Joe Satnik
Joe,
I think that you are mistaken on the 24 volt bulbs drawing half the power.
A 24 volt bulb will actually draw more power(watts) because the resistance is higher.
Lee F.
Dear Lee,
Ohms law: (V is voltage in "Volts", I is current in "Amperes" or "Amps", R is resistance in "Ohms", P is electrical power in "Watts".)
V/R = I
I x R = V
V/I = R
P = V x I
P = V x V/R (or V squared/R)
P = I x I x R or I squared x R
Same power with twice the voltage needs half the current.
Your homework assignment: What happens to the resistance (of the lamp filament) in this case?
Hope this helps.
Sincerely,
Joe Satnik
The 24 V might have more resistance, but the difference would be so small I don't think it is going to a big difference.
I used a 28 V bulb on an O22 switch, and it seemed to work. A 14 V bulb runs so hot I was worried it would melt the lantern.
Dom,
How did you power your O22 switches? If constant power, not track power, what was the voltage that you used?
Lee,
Perhaps I should rephrase my previous question on lamp resistance.
Start with an original lamp.
If you wanted to design a new lamp that would use the same amount of electrical power, yet be supplied with twice the voltage, how would the resistances of the original ( = Ro ) and new ( = Rn ) lamp filaments compare?
Hint: Use equation P = V x V/R,
Hope this helps.
Sincerely,
Joe Satnik
I use about 14 V on the side plug on the switch.
Using algebra,
Po = Pn Original power same as new power.
Po = Vo x Vo/Ro power is V squared over R
Vn = 2 x Vo New voltage is twice original
Pn = 2Vo x 2Vo/Rn = Po = Vo x Vo/Ro substitute
4 x Vo x Vo/Rn = Vo x Vo/Ro combine twos
4/Rn = 1/Ro divide out (cancel) Vo squared from both sides
1 x Rn = 4 x Ro Cross multiply
Rn = 4 x Ro Drop 1 x (redundant)
So, same electrical power, @ twice the voltage and half the current, the resistance is quadrupled. (X 4)
Hope this helps.
Sincerely,
Joe Satnik
Dom,
What are the O22 lamp part numbers, 28 and 14 volt?
Joe Satnik
Right now i am using a Shack bulb. Only problem is the bulb is taller than the usual O22 bulb.
But Lionel did sell a line of replacement bulbs for 18 volts because of TMCC/L. That might be a source.
Quote from: Joe Satnik on May 14, 2010, 12:00:39 PM
Using algebra,
Po = Pn Original power same as new power.
Po = Vo x Vo/Ro power is V squared over R
Vn = 2 x Vo New voltage is twice original
Pn = 2Vo x 2Vo/Rn = Po = Vo x Vo/Ro substitute
4 x Vo x Vo/Rn = Vo x Vo/Ro combine twos
4/Rn = 1/Ro divide out (cancel) Vo squared from both sides
1 x Rn = 4 x Ro Cross multiply
Rn = 4 x Ro Drop 1 x (redundant)
So, same electrical power, @ twice the voltage and half the current, the resistance is quadrupled. (X 4)
Hope this helps.
Sincerely,
Joe Satnik
Joe,
I am totally lost when it comes to algebra! I may know some geometry but that is about it, mainly basic math. Also I think that you are trying to describe an electronic circuit and how it works rather than a basic electric circuit.
Basically when you increase resistance in a circuit you increase the power consumed. So the point that I was trying to get across is that a lower volt light bulb should let you have more power to the motor, this will be noticed with a lower watt transformer.
Lee F.
Lee,
You said:
"Basically when you increase resistance in a circuit you increase the power consumed."
No, just the opposite.
Power = V x V / R
If the Voltage on a load stays the same, and its resistance increases, the power decreases.
/R means "divide by R".
Hope this helps.
Joe Satnik