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June 2

Black swallowtail species

Can anyone say how big are the following species: Papilio indra, Papilio joanae and Papilio brevicauda? Because the respective articles don't say anything about size. (Pictures are OK, as long as they are not any bigger than the ones in the articles.) 2601:646:8A00:A0B3:1C3:741A:FF1C:6A88 (talk) 01:30, 2 June 2019 (UTC)[reply]

Well, that last one is pictured on or just above a dandelion flower. It appears to be about 3 times wider than said flower. --Khajidha (talk) 04:58, 2 June 2019 (UTC)[reply]

~~So, about 2 inches in wingspan? That's on the small side for a swallowtail! :-)~~ Scratch that -- depending on the flower's size, this could make it anywhere between 2 inches (which is nice and small) and 6 inches (which is huge)! A better measure of the size is needed! And how about the other two? 2601:646:8A00:A0B3:E143:3EF:8DB:B5B7 (talk) 05:58, 2 June 2019 (UTC)[reply]

P. Indra wingspan ranges 58–72 mm, P. brevicauda wingspan 57–73 mm (both taken from the "Butterflies of Canada" links in the articles). Can't find anything yet on P. joanae. Mikenorton (talk) 07:58, 2 June 2019 (UTC)[reply]

P. joanae wingspan 82–102 mm from here. Mikenorton (talk) 08:05, 2 June 2019 (UTC)[reply]

Thanks! So all of them are fairly small by swallowtail standards. :-) 2601:646:8A00:A0B3:E143:3EF:8DB:B5B7 (talk) 10:24, 2 June 2019 (UTC)[reply]

About the same size as the swallowtails that we have in England (P. machaon brittanicus) - I was lucky enough to see one of these rare butterflies last weekend. Mikenorton (talk) 11:37, 2 June 2019 (UTC)[reply]

Lucky you -- here in San Jose it's mostly P. multicaudata (which is twice the size of P. machaon -- BRRR!) and P. rutulus (smaller, but still too big for my liking -- although P. joanae is about the same size but I like it!) 2601:646:8A00:A0B3:E143:3EF:8DB:B5B7 (talk) 02:17, 3 June 2019 (UTC)[reply]

List_of_Proton_launches_(2010–present)

Hi,

The last launch hasn't been updated yet. It was a success: https://www.nextspaceflight.com/launches/details/977 2001:EE0:40E1:6B5B:54EF:628E:26F2:6B84 (talk) 05:48, 2 June 2019 (UTC)[reply]

good. You obviously have some relevant information, and you know how (and are wiling) to edit since you just did here, so feel free to edit the article. That is the way wikipedia works. Gem fr (talk) 05:57, 2 June 2019 (UTC)[reply]

Longer lasting incandescent light bulbs

Wouldn't using a thicker filament, made of a material with more resistance, so it achieves the same temperature and hence color, allow it to last longer ? Application might be for appliance bulbs, like in an oven, where more modern bulbs won't work well. Is the issue that more heat would conduct back down the filament into the base ? SinisterLefty (talk) 13:02, 2 June 2019 (UTC)[reply]

3 watt bulbs can last a century 24/7 and are about as bright as a candle. Sagittarian Milky Way (talk) 13:25, 2 June 2019 (UTC)[reply]

See Incandescent_light_bulb#Light_output_and_lifetime. For long lamp life the filament is made of tungsten which is the metal with the highest melting point, 3,695 K (6,191 °F). Choice of operating voltage allows a tradeoff between light output (proportional to V^3.4) and lifetime (proportional to V^-16). The wearout mechanism is evaporation of the filament, especially at irregularities where hot spots arise; this effect can be reduced by filling the bulb with inert gas e.g. argon or alloying the tungsten with rhenium. Filament sagging between supports may be reduced by doping the tungsten with potassium. See Incandescent light bulb#Filament. Edison's investigation of lamp filament evaporation led to the discovery of the Edison effect, thermionic emission and invention of the vacuum tube. DroneB (talk) 14:32, 2 June 2019 (UTC)[reply]

Thanks so far, but I'm not finding any discussion on filament thickness versus longevity. A thicker filament should take longer to evaporate (sublimate, technically), but would also have less resistance, causing it to run cooler and give off less and redder light. However, this could be compensated for by using a filament material with more electrical resistance. So, why isn't this method used in practice ? SinisterLefty (talk) 14:58, 2 June 2019 (UTC)[reply]

Because no-one cares. Many incandescent bulbs failed by gross physical damage, rather than filament burnout. Even more failed from slight physical shock when lit. The pressure for their design was to first make them brighter, then make them adequately longer lasting, then make them more robust. Making them incredibly long lasting is certainly possible, but as it compromises their efficiency, there's just not a demand for it. There are some lamps which are intended to be very long-lasting. Some of those for warning lights on antenna masts have twin filaments and auto-switchover, just to avoid the cost of climbing (and mostly to delay the need to climb until Summer's good weather).

With compact fluorescents there's also a trade-off of longevity vs other factors, which began as warm-up time and also mercury content. Some of the first of these were the Philips SL range, the glass "jam jars" which were noted for their longevity (no electronics). I had one fail recently after 30+ years of nightly use (outdoors, timeswitched) – so I replaced it with another old stock one of similar age, but unused. OTOH, it takes a long time to come to full brightness, and they can't be made in Europe any more as they contain too much mercury. Andy Dingley (talk) 16:26, 2 June 2019 (UTC)[reply]

So wouldn't thicker filaments makes sense on those antenna tower warning lights and other cases where frequent replacement is extremely inconvenient ? They could combine this with the switch-over system of dual filaments. SinisterLefty (talk) 17:35, 2 June 2019 (UTC)[reply]

Filament thickness (that the OP has named 3 times) is not a free design variable. Given a required voltage, power and filament material, a thicker filament wire means a longer wire to maintain the same resistance. A longer wire means more surface area which generally means more, not less, evaporation to shorten lamp life. The claim "having less resistance causes it to run cooler" is true for a constant-current supply but untrue for a constant-voltage supply, which is the usual case. (Joule's first law of resistive heating follows Ohm's law.) The reference already given describes the coiled-coil filament construction that does extend lamp life by reducing filament evaporation. I suggest that LED lighting is a better upgrade to tower warning lights than hoping for some new material for filaments. DroneB (talk) 18:17, 2 June 2019 (UTC)[reply]

Re: "Given a required voltage, power and filament material, a thicker filament wire means a longer wire to maintain the same resistance". Yes, but I don't suggest using the same material, but rather a material with more resistance. That doesn't seem like an impossible task. How about a carbon rod ? They were used in carbon arc lamps, but that is in normal air, and with an air gap.

Also, there seems to be a logic error when you said that a longer filament will increase the evaporation rate and thus shorten the life of the filament. If the sublimation rate is approximately constant over the length of the filament, then a 10x as long filament will have 10x the evaporation rate, but the same evaporation rate per unit length, which is what actually matters in determining how long the filament lasts. And if that filament is thicker, it will last longer. However, there is another issue, that the evaporated material redeposits on the inside of the bulb, lowering efficiency. This is why I didn't suggest a longer filament, but rather one made with a material with more electrical resistance. SinisterLefty (talk) 02:40, 3 June 2019 (UTC)[reply]

indeed. Actually all variables are connected. Since voltage is a given, once you set the power, you have no choice of resistance: it is set. Because you want light, there is an optimal temperature you want (let say, 2700K); but you set the power, so, the surface through which the power is dissipated is also set: smaller surface<->higher temperature and viceversa. Longer filament lineary increase both surface and resistance, thicker filament increase surface lineary but lower resistance by the square, and this give an equation with single solution for thickness and length to achieve a given resistance for a given material (and we saw that resistance is set)(*). This leave only one free varaible: what the filament is made of. Except you want something that will not melt, nor evaporate too quickly, nor break too easily, at the given temperature; something with decent availability and as cheap as possible; so tungsten it is... and you are done. (*)well, actually things are more complicated, the geometry of the filament is another variable to consider, and there are others I dont even know about; but you get the ideaGem fr (talk) 20:31, 2 June 2019 (UTC)[reply]

I think you're assuming the filament has to radiate as a black body, which may be valid in practice, but I don't actually know that. Maybe there is some metamaterial that acts somewhere between an LED and a black wire? Wnt (talk) 23:31, 2 June 2019 (UTC)[reply]

I don't agree that voltage is a given. For applications like warning lights on antenna towers, surely the voltage can be stepped up or down, as needed. SinisterLefty (talk) 02:52, 3 June 2019 (UTC)[reply]

voltage can be stepped up or down for all applications. It is just not done, because, it very rarely make sense to use non standard power supply. Gem fr (talk) 07:59, 3 June 2019 (UTC)[reply]

It seems to me it's often done. For example, a PC's power supply outputs a range of voltages for each component. Or a car cell-phone charger changes from 12V DC to whatever the cell phone needs, while a wall outlet cell phone charger changes from 120V or 240V AC, typically. 13:53, 5 June 2019 (UTC) — Preceding unsigned comment added by SinisterLefty (talk • contribs)

Re [a thicker filament] could be compensated for by using a filament material with more electrical resistance - there are a couple of incorrect assumptions here. First, if supplied by a constant-voltage source (which is a good approximation of the main electrical network in most countries), a lower resistance will actually provide more Joule heating (because the intensity crossing it will be higher), so you are looking for a higher resistance. Furthermore, what material would that be? Having an extremely high melting point is non-negotiable in order to have most of the radiation spectrum in the visible (the typical value with tungsten filaments is only about 5% of radiation in visible wavelengths, and it falls down very quickly at lower temperatures).

The easiest way to go about it (but not necessarily the cheapest) would be to have a transformer change the voltage input imposed to the filament (while keeping the same material). Tigraan^{Click here to contact me} 09:44, 3 June 2019 (UTC)[reply]

I don't read Joule heating to say that lower resistance provides more heating: "Joule's first law ... states that the power of heating generated by an electrical conductor is proportional to the product of its resistance and the square of the current ...". That would seem to indicate that the more the resistance, the more the "power of heating", which I take to mean generating more heat and light. But, I agree with your statement that we "are looking for a higher resistance". I've suggested a carbon rod as a material with more resistance and a high enough melting point. But changing the voltage (and hence the current) also works. Perhaps not practical for each bulb, but it could be done once for the entire string of lights on the antenna. SinisterLefty (talk) 13:44, 5 June 2019 (UTC)[reply]

well, I am not sure of what you meant, but Joule's law is also that the power of heating generated by an electrical conductor is proportional to the square of the voltage and the INVERSE its resistance. And this is the formula to use under a fixed voltage, as accurately stated by Tigraan. And you are NOT looking for a higher resistance. Just for a proper resistance to get the power you want for the light bulb, that is, a resistance equal to the square of the voltage divided by the required power (for instance, a 60 W light bulb under USA's 120 V need a 120x120/60= 240 ohm filament, no more, no less). Gem fr (talk) 16:45, 5 June 2019 (UTC)[reply]

@SinisterLefty: When you change the resistance, you change the current (unless the generator is fixed-current). If the generator is fixed-voltage,

U=RI

is constant, hence

P=RI^{2}

increases with R because I increases.

Such steady-state electricity problems are (IMO)) easier to understand by using a current-voltage characteristic representation. You plot the curve for the resistance (a line going through 0 with slope 1/R) and the curve for the generator (a vertical line at given U); the intersection point ("operating point"? not sure of the translation) gives you the actual voltage and intensity that goes through the circuit; the area of the rectangle between 0 and the operating point gives you the transferred power. You can clearly see that with a higher slope (i.e. lower resistance) the operating point is pushed further up and hence the transferred power is higher. On the other hand, if the generator is fixed-current, its characteristic curve is horizontal (not vertical), so decreasing the resistance causes a switch of operating point letwards and hence the transferred power is lower. Tigraan^{Click here to contact me} 09:39, 6 June 2019 (UTC)[reply]