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It occurs to me that you could run a 230V 100Watt bulb at 115V (be equiv 50W) and it could possibly last several years.. Any of you UK guys want to sell your bulbs to us USA guys? .. Group buy??:)
What type fixtures you-all got over there? An "A" type we call it here...
..just a thought...
It's a quarter of 100W. (Physics) So 25W:(
and our standard 100W have a bayonet type fitting ( 2 small pins stick out of the side, twist and lock ) and im sure the USA is eddison screw fitting
all the best ....., Karl
Why not just go to ebay.co.uk ?
Jim
why not just buy the 137 volt LONG LIFE INDUSTRIAL that they sell to the NYC subway and schools ? Yes folks, those annoying big tungsten lamps that lit the dim hallways in your elementary school leaving plenty of spaces for the bullies to hide were probably good for 10-20 years at minimum.Quote:
Originally Posted by JimBo
cheaper then the UK post, which I must admit, seems faster then teh US snail.
Steve
Yeah, I remember them.. bah--
Still , might be kinda large to cram into the kitchen fixture;)
I dont see how 1/4 x watts fits into Ohms Law even regarding hot tungsten's increased resistance . Please explain, Dr laser..
The color from them would be terrible dimmed down that much.
The color temperature from a tungsten source at full power is 2800k if you dim it down it will get warmer and warmer and not put out much light and just be a really crappy light source. If you want long life look at led or fluorescent. Although the fluorescent has a crapy color also on the other end of the spectrum.
Chad
I've put a diode in series with a 120V bulb which effectively reduced it to half cycle (power) and it wasnt an un-pleasant color ; it lasted quite a long time.. the flicker at 30 Hz was what I didnt like, therefore the question..
I like a warm color..
I tried to find 230V 100Watt bulbs and only found 1 in the US .. I may order it just for shits and grins..:D ..and see how it does..
I just dont see how he says it will emit 1/4 power..should be 50%.. his 'physics' are in error.. unless it has something to do with quantum photon distribution:cool: I doubt it..
Interesting that you would mention this... I did a lot of volunteer work for a theater when I first started with sound and lights. In the house we used 230v lamps so they would last much longer.
I also don't understand why it would be 25watts. The relationship is fairly linear. And this wouldn't apply to an inverse square law...:confused:
Also putting a diode in line with the light would make it flicker at 60hz, since you are only taking away 1/2 of a cycle, you still have 60hz.
-Max
If youre running at 60 Hz and you cut off the top (or bottom) of the sine wave then you get 30 Hz to the load.
But 1hz is one cycle which includes 1 peak and 1 troff. If you have 60hz and cut off all the peaks you still have 60hz.
I may be wrong here, but I am almost positive this is right...
-Max
I'm tired, I'm going to bed, I dont have time for electricity 101.. so I'll get back to this tomorrow...bye;)
Power = volts x amps. (Watts law)
Amps = volts / resistance. (Ohms law)
So with simple substitution we get the following:
Power = volts x volts / resistance, or simply volts squared / resistance.
Now, resistance is a constant for your light bulb. So the only thing that changes is the voltage you run it at. But since power varies as the square of voltage, if the voltage changes by a factor of 2 the power will change by a factor of 2 x 2, or 4.
So if you decrease the voltage by 1/2, the power will drop by 1/4.
Wasn't that easy?:D
Adam
Lessee P=E x I: 50=115 x 0.435, 100=230 x 0.435. The current doesnt change on these bulbs, the voltage and filament resistance does.
You have one 115V 50watt bulb and one 230v 100 watt bulb.
R= E/I so the filament resistances are respectively 264.3 Ohms and 528.7 Ohms.
________________________
If you have 60 Hz (cycles per second) thats 8.3 mS positive voltage cycle and 8.3 mS negative voltage cycle. Removing the negative voltage cycle will leave a zero volt condition for 8.3 mS during the complete cycle resulting in only positive voltage cycles at .. n/m--
I you have a bridge rectifier connected to 60 Hz AC you get 120 Hz ouput.
If you put a single diode in series with that circuit, you only get 30 Hz out. Clear as mud?:)
I think AC vs pulsating DC is where the confusion is occuring.
"If you have two 100 watt bulbs, one designed for use atQuote:
Originally Posted by mliptack
110V and the other at 220V, the 220V bulb will have four times the
resistance of the 110 watt bulb. This is because the power used by a device
or resistance R when used at a voltage V and current I is given by:
P = IV = V^2/R
So when the 220V bulb is used at 110V, the power used is 25 watts, since the
voltage is halved and power goes like the square of the voltage. Since the
resistance of the filament increases with temperature, the current is
somewhat higher than 1/2 the current at 220V, but you will see a very dim
filament.
So you will use less electricity if you replace the bulbs in your house with
220V bulbs, but your house will be very dark."
This is getting f'cking ridiculous. I was going to replace a 60 watt 120V bulb with a 100 watt 230V bulb. I was using the 50watt example to simplify.
Nevermind- I found one. I'll either do that or wire my dual fixtures in series instead of parallel and use 2 120V 100watt bulbs..:mad: :rolleyes: It may be dark.. we'll see...:)
wow.
You know that it'll also put out a higher percentage of IR to visible light (lower efficiency)? how about installing a compact fluorescent instead?
it's likely the bulb will run a bit over 25% because the filament temperature and thus resistance will be lower than when it's at full power.
Yeah, ok this all makes sense. Thanks everyone. I am running several compact florescents already, I just like a regular tugsten for color balance in some rooms.. and tired of changing light bulbs. I could fork out for the 'long life' bulbs but are they really? I mean, does anybody hook an hour-meter to a 2000-hour bulb vs a std 750-1000? heh--
Wrong. The filament resistance is the same. The filament *IS* a resistor, remember? It's constant. The voltage changes, because you're plugging it into a 115V socket instead of a 230 volt socket. And since the voltage changes, so does the current. (Lower voltage through the same resistor = less current.)Quote:
Originally Posted by steve-o
And if you take the 230V 100 watt bulb and put 115 volts across it, you will get .218 amps flowing through it, giving you 25 watts. (.218 x 115 = 25 watts)Quote:
You have one 115V 50watt bulb and one 230v 100 watt bulb.
R= E/I so the filament resistances are respectively 264.3 Ohms and 528.7 Ohms.
Seriously, where did you get the idea that the resistance would change based on the applied voltage? Resistance is a constant. (Except for something special like a thermistor which is designed to change it's resistance as it's temperature changes, but that's *not* how an incandescent bulb works.)
Adam
OK Adam I'm talkin' about 2 different bulbs. - A 100watt 230 and a 115V 50 watt... ok?
I think your wording is what confused Adam, you state that the voltage and the resistance CHANGE, when really they only differ between the two bulbs.Quote:
Originally Posted by steve-o
-Max
Exactly....................... mebad......................
Quote:
Originally Posted by steve-o
My initial reply (The one that listed Ohms and Watts law) was to explain to Steve why the bulb would only put out 1/4 power. (If you recall, he wanted to buy 230 volt bulbs and run them at 115 volts to make them last longer, but he thought that since the voltage was half, the power output would also be half.) Several people corrected him by saying it would be 1/4 power, but no one explained why, and Steve was still thinking that the power output would be 50 watts. (See his post above)Quote:
Originally Posted by mliptack
His reply to my Ohms law post certainly sounded like a rebuttal to my explanation, but if he was indeed talking about two different bulbs then I retract my pendantic second post. :o
Adam
I should have explained that I was talking about 2 diff bulbs-- sorry Buffo--:)
No prob dude... It's all good. :D
Adam
Shit, Man youre like 2X theposts The verbal D is kicking in again:) :)
Lookin foward to selem.. SR got me fixed and M3 too --gots some ideas....if theyres time... /;
Dude-
Well, you've got quite a few months to assemble your kit yet. SELEM 2008 will be in August. I'm hoping to have a scan head on my HGM 5 by then... Might have a couple other toys to bring too.;)
Yeah - post count is geting ridiculous... :o I'm seriously addicted to this place.
Adam
Yeah me too..
I'm trying to put together scanheads for the event .. check out the ...S thread (for SteveR)/...he sent me them and I just cut and drilled the mounts..now I got amps and a greeny coming.. and the wifey is trying to do a wednesday thing,,,aaaaahhhhh!!!!!!
hehe... Never enough time, eh? I totally know where you're coming from on that score.
I saw the mounts you've been working on. Very nice! Looks like you might want to move the galvos back a bit though. (Either that, or cut the shafts to make them shorter - not sure if cutting the shafts is a good idea though.) The shafts appear to overlap by about an inch or so, which might make mirror alignment a problem. (Then again, maybe it's just the angle you took the picture at - not sure.)
Keep us posted as to your progress. Love to see that greeny running on those scanners...
Adam
It'll be a RGY until I can get enough 'green' for a blue- heh-:)
Ok, just to stir it up some more....
A UK 100W light bulb is less bright than a US one. The lower current means the thinner filament loses heat faster in a UK bulb so the temperature at equilibrium is lower. If you really want BRIGHT incandescent, get a 6V 100W halogen. Run it off a universal input switchmode PSU and it won't matter which country you do it in.
Ok, I'l bite. If the filament is only thinner, and not thinner and longer, I would think the decreased radiant surface area would result in an INcreased temperature. can you explain?
Decreasing the filament size would cause a temperature increase for several reasons.
1) You would be increasing the resistance of the lamp
2) The surface area would also be decreasing (like you said) and therefore the heat dissipation would be greater. Same reason why power density is an issue for laserists. A 40 watt laser could be harmless to our eyes, if the power density is low enough.
Hmmm:cool:..you're still talking.:eek: Hmmm:confused:...ammm :)LOL:D >:)
yes I agree, I was asking The_Doctor to explain his UK dimmer theory.Quote:
Originally Posted by mliptack
It's definitely right, trust me. You can verify it.
Increased resistance alone would drop the power if the voltage stayed the same. What matters is that the consumed power is the same, 100 watts. The 240V filament must be thinner (and longer), because its resistance must be higher. It has more surface area for its mass, so thermal equilibrium happens at a lower temperature because more radiation can occur at that lower temperature than with less surface area in a lower-volt 100W lamp. The same amount of light is emitted, but it looks dimmer because more of it is wasted as IR, and the blue end of the blackbody spectrum is reduced as well. It isn't the output watts that are reduced, but the lumens, which is a measure of luminous efficacy, as judged by eyes, or meters that imitate them.
How would this apply to discharge lamps? I assume it wouldn't but I'm curious if there is something that maybe I am overlooking.
Pass. >:)
Not got much idea on those. As far as I know, the efficiency is based on an optimum pressure for a given voltage and tube length. I guess for a nominal power rating, they have similar efficiencies at 120 or 240, being made to work specifically. The heaters at the ends might make a difference, but I think those will only be powered during startup, to produce free electrons to start the arc.
Discharge light have a big bang at the beginning and then pretty normal voltage to keep the lights going.
HID Xenon gas lasers (miniature) would be very cool. Or Mercury Vapor..
Right the High starting voltage usually in the kV range, is to ionize (I think thats the right term) the gas between the anode and cathode, which causes (in the case of metal halides) the mercury to warm up and vaporize as well. Over the next few seconds the voltage drops down to 70-100v (at least with the Vari-Lites I am familiar with). I just wonder how the resistance works, the way I see it the resistance would be the opposite of an incandescent. I would think the resistance is higher when lamp is cool and lower when it warms.
Any ideas?
Ought to mention I was talking about standard fluorescents, as those are the closest discharge lamp to standard domestic incandescents, as far as purpose goes...
Re arcs, they will take a massive current, if you let them. That's basically the starting point in ion lasers, to get the energy needed to get the population inversion for laser output. In series with most if not all CW discharge lamps there's a ballast, a current limiting inductance, to keep them operating safely. You'd have to use a resistance if it was a DC supply though. Pulse lamps also have an inductance to limit current and control pulse width too, especially in pulsed lasers. Photoflashes rely on the ESR of the capacitor to do most of the limiting, I think.