Light output of dimmed lamps

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(Sometime before I mentioned a personal estimate of requiring 700 Kelvin to see with dark-adapted eyes something visibly incandescing.)

I have heard of people seeing single photons. But if you dark-adapt yourself and see individual flashes, I suspect some of those will come from thermal agitation as well as from photons. But I'll make a guesstimate of how much light it takes to see...
I have heard, I think I remember, that the human retina utilizes (at favorable wavelengths) 20% of photons hitting it. In extreme darkness, I am guesstimating persistence ov vision to be effectively .2 second (varies slightly inversely with lighting level). And I feel like arbitrarily saying that to see things you want to see 5 times a second a photon from every square 1/2 degree of what you are looking at. (I picked that because the angular diameter of the sun and moon seen from Earth are slightly over half a degree).
A very highly dilated human pupil is, as seen when magnified by the cornea, about 10 millimeters in diameter. (I know that 21 CFR 1040.1, regulating laser exposure, says to use 7 mm when pupil diameter affects exposure. I am guessing that people who write regulations are assuming you won't be working with lasers in such extreme darkness as to have your pupils to be as big as possible.)
A surface 1 meter away would have a piece 8.73 mm wide appear half a degree wide. And 1 photon out of every 40,000 emitted would pass through an effectively 10 mm diameter pupil from that distance. 200,000 photons per second from a square 8.73 mm wide is about 26.2 million photons per second emitted per square meter.
The wavelength that scotopic vision is most sensitive to is 508 nm, and one photon of that wavelength has an energy of about 2.44 electron volt. Divide that by Faraday's number (6.3 E18) to get joules - that is 3.87E-19. Multiply by the above 26.2 million photons per second per square meter, and I get a rough "calculation" of 10.1 picowatts per square meter.
Human vision is about 30% as sensitive to "white light in general" (counting only visible wavelengths) as to the most vuisible wavelengths - so now my rough "first calculation" becomes 33.5 picowatts per square centimeter.
================= I try again: I have heard of Venus being able to illuminate ground brightly enough to cast a shadow. Venus has a mgnitude mostly around -4 and the Sun has one around -27 (I forget exactly). 5 magnitudes (the astronomic measurement of brightness of a celestial object) is a factor of 100. That makes Venus having a brightness around that of the Sun times 6.3E-9. The Sun typically delivers about 1 kilowatt per square meter to a perpendicular surface at Earth's surface, with about half of that being visible. 500 times 6.3E-9 is 3.15E-6. Since Venus is almost always within 30 degrees of the horizon when it's really dark, I would say more like 1.5 microwatts of visible light from Venus per square meter of ground on a good evening or pre-dawn.
================= Another attempt: Illumination to match the "surface brightness" of a hypothetical 6th magnitude hypothetical celestial object that is square and half a degree wide. Astronomy books like to say 6th magnitude is the dimmest brightness at which celestial objects are visible to the naked eye, but I have heard reports of 8th magnitude in favorable circumstances.
With my above calculation of Venus delivering 3.15E-6 watt of visible light per square meter of a perpendicular surface, a 6th magnitude object delivers 1/10,000 of that or somewhere around 3.15E-10 watt per square meter. An 8.73 mm wide square (half a degree wide at 1 meter away) doing this would radiate that times pi, divided by .00873 squared, or *roughly* 13 microwatts per square meter. 8th magnitude would make that roughly 2 microwatts per square meter.
============================= Temperature that a blackbody needs have brightness equivalent to that of radiating 1 microwatt per square meter in the 400-700 nm (visible spectrum) range: 1 microwatt per square meter of "white light" (having an efficacy of 243 lumens per watt for equal-power-per-unit-wavelength band from 400-700 nm) illuminates to an extent of 2.43E-4 lux, and such illuminated surface has a surface brightness (in candela per square centimeter) of that divided by 10,000 and by pi - which is 7.7E-9. A blackbody does this at about 687 Kelvin. 702 Kelvin gets double the brightness of 687 Kelvin (same surface brightness as 2 microwatts of "white light" per square meter).
If that 33.5 picowatt per square centimeter figure I got at first has any truth to it, then surface brightness would be 2.7E-13 candela per square centimeter. A blackbody achieves this at about 521 Kelvin.
These temperatures are for calculating for photopic vision. Since scotopic vision requires shorter wavelengths than photopic vision does, the temperature required to make a blackbody visibly glow would be a little higher than these temperature figures.
- Don Klipstein ( snipped-for-privacy@misty.com)
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(A first and "very low side" "calculation" of minimum illumination to see at all, in part:)

I forgot to multiply that 40,000 by 5 twice - to get 5 photons through the pupil per .2 second. Multiply everything by 5 to correct.

Now it's .16 nanowatt per square meter.
This still sounds awfully low and not just a little lower than the second and third "calculations" that ended up close together in "determination" and I suspect something could still be wrong.

Changes to 542 Kelvin, and this "low side" could still be too low. The second figure of 687 Kelvin and the third figure of 702 Kelvin stay the same.

- Don Klipstein ( snipped-for-privacy@misty.com)
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Even then, some people never see the art.

Wow. I'm surprised nobody's answered this question more precisely yet. Maybe it's time for experiments with large groups of people in very dark rooms answering questions like "Do you see a large dim object?" with good statistical analysis, as in modern parapsychology.
Nick
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snipped-for-privacy@ece.villanova.edu says...

More likely, this work was already done long ago, in works so old that they aren't available on-line.
I know similar experiments were done in the early days of photography, but my ancient photography books are all in boxes while we remodel.
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("calculations" half out of my hat as to how little illumination or surface brightness can achieve threshold of human vision)

I found a link:
http://webvision.med.utah.edu/imageswv/table3.jpeg
within,:
http://webvision.med.utah.edu/Psych1.html
This is a chart indicating at least roughly that "absolute threshold" (lower) for human scotopic vision is 1E-7 candela per square meter. Assuming a lambertain perfectly white reflector or lambertian source, this means pi times 1E-7 lumens per square meter. And how many lumens did I say were in a watt of "white light" by one measure - 243? For now I get 243 for equal-power-per-unit-bandwidth from 400-700 nm. Divide 3.14159E-7 by 243 and that means about 1.3 nanowatts per square meter. This is with making assumptions! For one thing, different versions of "white light" have different ratios of ability to stimulate scotopic vs. photopic vision, and photometric units are defined in terms of photopic vision of a specifically defined (and apparently to me agreed upon at a CIE convention) "standard oberver".
But 1E-7 candela per square meter is 1E-11 candela per square centimeter. 570 Kelvin achieves this from a blackbody according to a homebrew BASIC program of mine with the blackbody equation and the photopic function. Since I also have CIE Xbar, Ybar and Zbar functions in that program, I get from that a 1931 CIE chromaticity of .713, y=.2869, which is a shade of red similar to that of monochromatic light at 634 nm to "the standard observer", or close to the color of a usual red HeNe laser. This stimulates scotopic vision a lot less than any kind of white or "white" light does. And when I tried dimming a clear high wattage lightbulb in a dark room while dark adapted while viewing very closely and with a magnifying glass, I found the color of the filament to change from orange-red to grayish as the lightbulb was dimmed more extremely - this means to me that scotopic vision is the component of human vision that sees the coolest possible incandescent blackbody. So, I expect a blackbody to require a temperature higher than 570 Kelvin to be visibly incandescent.
I plan to add the scotopic function to the homebrew software that I mentioned above. Watch this space for an answer as to what temperature achieves scotopic equivalence of a blackbody to "white light" at 1E-7 candela per square meter.
============================== And I think that such a "rock bottom" is not the bottom of the usable range of a light dimmer. I think the bottom would be more like:
a) Achieving such a level of illumination 1 meter from a 150 watt floodlight, 1 foot from a 100 watt A19, or something like that, as opposed to having the filament surface achieving that sort of intensity. Maybe even as low as having a 100 watt "A19" (regular shape and size lightbulb) "Soft White" achieving that kind of intensity - requiring ballpark 100 times more filament brightness, achieved by 650 Kelvin uncorrected for orangish red light having lower scotopic/photopic ratio than "white light".
b) How about achieving the filament glowing brightly enough to have some significant stimulation of photopic vision - as in glow with a color rather than grayish?
The above links indicate "cone threshold" (threshold of stimulating color-sensing cells in the human eye) of (ballpark) 2E-4 candela per square meter. That is 2E-8 candela per square centimeter. I get 708 Kelvin for that. As for making an A19 "soft white" lightbulb glow with visible color, assuming intensity of the frosted surface being 1/100 of the intensity of a blackbody filament surface: That requires 826 Kelvin. The color is supposedly about that of 614 nm - usually considered a rather reddish orange or notably orangish red.
My personal experience is that in general 120V vacuum-containing incandescents and 75-watt-plus 120V gas-filled incandescents visibly glow, although very dimly, at 6 volts and do not at 3 volts and usually do not to most barely at 4.5 volts. Keep in mind that with severe dimming by a usual light dimmer, most non-true-RMS meters (that sense average voltage and are corrected to RMS assuming a sinewave waveform) will read low when attempting to read the voltage across a dimmed lightbulb. A much smaller number will read high by sensing peak voltage and correcting to RMS assuming sinewave. Only a true-RMS meter with frequency response good enough to catch all significant harmonics of the waveform involved will read accurately the voltage across a lightbulb dimmed by a "usual dimmer". Gas-filled incandescents of lower wattage (more especially lower current) have filament temperature falling more rapidly as voltage is decreased. My experience is that a clear 25 watt 120V gas-filled A19 does not visibly glow at 6 volts, and at 12 volts glows more dimly than most other 120V incandescents of wattage 4 watts or more - especially 25 watt vacuum-containing tubular "showcase"/"refrigerator" (T10) lamps and 15 watt A15 (with C9 filament) as well as higher wattages.
- Don Klipstein ( snipped-for-privacy@misty.com)
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http://webvision.med.utah.edu/imageswv/table3.jpeg
http://webvision.med.utah.edu/Psych1.html
Good stuff. I like this part:
http://webvision.med.utah.edu/imageswv/Kall02.jpeg
and the Parameter Estimation by Sequential Testing (PEST) procedure, which looks easy to automate, like Sequential Probability Ratio Testing in radar.
Nick
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On 24 Dec 2005 10:35:20 -0500, snipped-for-privacy@ece.villanova.edu wrote:

It would emit EM radiation (because of this, an electric heater is NOT 100% efficient). Would it always be in the limited frequency range this is visible to humans?

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Alot of the radiation from incandesants is invisable to the eye, it is in various areas of the spectrum and changes with voltage. Ever notice how incandesants photograph orangeish, cameras have different sensitivities to different wavelents of light. An incandesant outputs only apx 10% of its energy as usefull light, hence it is actualy an electric heater that emits light, where as a flourescent does better. Incandesant output 17-19 LPW, lumen per watt, T8 flourescent up to 110 LPW with 45-60 for CFLs being average.
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There would always be some visible light, but it's very faint at low temps :-)
20 PI=4*ATN(1) 30 LAMBDAU=7.8E-07'upper visible limit (meters) 40 LAMBDAL=3.8E-07'lower visible limit (meters) 50 FOR LT=2 TO 4'log F temp 60 TF^LT'temp (F) 70 TK=(TF+460)/1.8'temp (K) 80 GAM=.0143879/(LAMBDAU*TK)'Pivovonsky and Nagel (1961) polynomial term 90 CFRACU=0'initialize upper cumulative fraction 100 FOR M=1 TO 3 110 TERM/(PI^4)*(((M*GAM+3)*M*GAM+6)*M*GAM+6) 120 CFRACURACU+EXP(-M*GAM)/(M^4)*TERM 130 NEXT M 140 GAM=.0143879/(LAMBDAL*TK) 150 CFRACL=0 160 FOR M=1 TO 3 170 TERM/(PI^4)*(((M*GAM+3)*M*GAM+6)*M*GAM+6) 180 CFRACLRACL+EXP(-M*GAM)/(M^4)*TERM 190 NEXT M 200 P=5.6697E-08*(TK^4)'total emissive power (W/m^2) 210 EFRACRACU-CFRACL'fraction of emissive power in visible spectrum 220 PRINT TF,EFRAC,P,EFRAC*P 230 NEXT
temp visible fraction total power visible power
100 F 6.010933E-22 531.1561 W/m^2 3.192744E-19 W/m^2 1000 2.753038E-07 24540.4 6.756065E-03 10000 .4668266 6.465423E+07 3.018231E+07
Nick
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Joseph Meehan wrote:

Wouldn't it be enough to periodically heat them up (i.e. isn't the damage reversible to some extent, as long as the bulb isn't too badly damaged)?
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So far, we seem to have no evidence of any damage.
Nick
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I have heard - sometimes even in literature from halogen lamp manufacturers - that that actually works.
If the bulb is badly damaged from the halogen cycle failing to outrun effects of water vapor or oxygen, then I would say only maybe. And the "repair" may not be good - the tungsten could have come mainly from one region of the filament and it could go mainly to a different region, and uneven filament thickness is bad since it tends to reinforce itself (thinner parts get hotter and evaporate faster). Also, the bulb material gets hotter than normal if the bulb is operated full-blast while blackened. Some halogen bulbs operate under enough stress as it is...
- Don Klipstein ( snipped-for-privacy@misty.com)
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Dimmed bulbs burn out faster? :-)
Dimmed bulbs will burn out SLOWER because the filament does not get as hot and does not degrade as much. Normally the bulb is filled with argon or other inert gas so the fillament lasts as long as possible.
Yes, there is a minimum starting voltage for incandescent bulbs. Get a voltmeter and clamp on ammeter. Wire a dimmer in series with a light bulb. Measure voltage across the bulb and measure the current through the bulb as you turn the dimmer up from the OFF position. See how much voltage and power you need before you can see any visible light (there is infrared output at a much lower voltage). Years ago I did that using an osilliscope instead of a volt meter. You will learn a lot that way. You may not find that in a book, but you can test it for yourself, and the meters you need do not cost all that much.
Stretch
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I wouldn't call that a "starting voltage" in the sense of fluorescents and other types which give no light at all until an arc suddenly starts, altho the dimmer might act like that, passing no current until it reaches some minimum thyristor triggering threshold.
Nick
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Stretch wrote:

The discussion was re halogen bulbs, which have different characteristics from argon-filled bulbs.

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"When shopping for a new dimmer to replace a rheostat model, make sure to specify an electronic dimmer. Electronic dimmers use a circuitry called
TRIAC to control voltage to the light bulbs. TRIACs trim the voltage to the bulb without creating heat, and effectively reduce the energy use while only losing about one percent of the savings to operate the dimmer circuit. A good TRIAC dimmer will reduce the light output over a broad range, and in turn trim the energy use. "
When was this advice written for, 1970? I don't think you could find a rheostat type dimmer to use for home light control if you wanted one. All the ones at any of the home center stores, etc have been electronic for decades.
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snipped-for-privacy@optonline.net wrote:

Hi, Trim the voltage? Or trim the conduction angle? Result is lower voltage on meter but Triac does not tirm the voltage. On theory at least. Rheostat does. Am I wrong? Tony
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I think you're right as long as the dimmer isn't set below 50%. It just reduces the duty cycle (the 50% is because at that point, cutoff would be before the voltage reaches peak).
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You do remember correctly. Running at full on, or more specifically full temp, keeps the tungsten from depositing on the glass, and cycles back to the filament.
In most cases, running for a while at full blast will re-activate the halogen cycle in the bulb, and undo some of the damage done by dimming.
This high temp is also why it is important not to touch a halogen bulb, since the oil from the fingerprint will cause the glass to bubble.
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John Hines wrote:

Ahh ... that's what I suspected.

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