Wikipedia:Reference desk/Archives/Science/2014 May 5

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May 5

Aviation rumble

In an aviation context, what is "rumble"? The pilot of Northwest Orient Airlines Flight 710 reported rumble soon before ground impact, but there's no link. I didn't think Rumble (noise) was very likely (why would you report hearing white noise?), but nothing else at rumble was relevant. The ground was snow-covered at the time, so rumbling thunder wouldn't be as likely. Nyttend (talk) 02:06, 5 May 2014 (UTC)

The crash was due to flutter, which is often manifested as a rumbling or buzzing noise coming from the wings or tail. 24.5.122.13 (talk) 02:14, 5 May 2014 (UTC)

I don't know why it said on the Rumble page that a rumble is white noise - White noise is a hiss, whereas a rumble is low frequency noise. I've now changed it. Richerman (talk) 09:14, 6 May 2014 (UTC)

Here's a guide to colours of noise, for comparison. You don't even have to be a synesthesiac. What seems clear for all is that you probably don't hear your plane say "bloop". InedibleHulk (talk) 02:49, 7 May 2014 (UTC)

Concentrated Sun-blackbody equilibrium temperature

If I had a device that was perfectly insulated and lost heat only via blackbody radiation and I put it in the sun and let it reach equilibrium, what temperature would it reach? My intuition tells me it would reach the same temperature as the surface of the sun, but I'm not sure.

What if I used a solar concentrator that doubled the sun (in terms of Solar flux)? I think the temperature of the device would not change (it would simply reach equilibrium faster), but I'm not certain. Ariel. (talk) 02:13, 5 May 2014 (UTC)

The more sunlight the hotter it would get. However, it would never reach the temperature of the Sun because only a portion of it is pointed at the Sun, and the rest radiates heat off into space without getting any sunlight (on the far side) or as much sunlight (on the Sun side, but not pointed right at the Sun). Note that your description is pretty close to an asteroid. StuRat (talk) 02:35, 5 May 2014 (UTC)

I didn't consider the back side of it. For this thought experiment can we just assume it can't radiate from the back? Would angle really make a difference? That's basically the solar concentrator part of the question. Ariel. (talk) 03:16, 5 May 2014 (UTC)

No! Do not assume the device radiates in only one direction, because that dangerous road leads you directly into un-physical conclusions! Heat flows from hot to cold: and if you hypothesize a "one-way radiator" then you are forcing heat to flow from cold to hot! That violates the second law of thermodynamics and is, statistically, very improbable! Even if your surface is very reflective, and well insulated from the other parts of the device, heat will still flow and radiate from all sides. Nimur (talk) 14:58, 5 May 2014 (UTC)

Where's the problem Nimur? Talking about thermodynamics is already a violation of the First law of thermodynamics. Or is it? - ¡Ouch! (^{hurt me} / _{more pain}) 09:28, 6 May 2014 (UTC)

Yes, angle makes a big difference. This is why the tropics are so much hotter, because they get sunlight straight on, versus at a shallow angle near the poles, although both are in sunlight about half the time. In fact, right at the poles, in summer, they get 24 hour sunlight, yet it's still cold there, due to the shallow angle. StuRat (talk) 04:28, 5 May 2014 (UTC)

I know that for Earth, but I'm not so sure it applies to my question. The angle changes the flux of sunlight (so it provides more power), but I'm unsure if it changes the temperature of an item that does not get cooled (so power is irrelevant). Ariel. (talk) 05:35, 5 May 2014 (UTC)

But it does get cooled, by radiation. And unlike the sunlight coming in from only one direction, it will radiate equally in all directions. StuRat (talk) 15:09, 6 May 2014 (UTC)

Clarification: By "put it in the sun" I assumed you meant "placed it in the sunlight". However, it now occurs to me you might have meant "placed it inside the Sun". Which is it ? StuRat (talk) 02:39, 5 May 2014 (UTC)

I meant in sunlight :) On Earth, if it matters. Ariel. (talk) 03:16, 5 May 2014 (UTC)

The Sun is the same brightness per unit area no matter how far away it is. So if you set up a cunning shell of mirrors (not even with lenses, though the net assembly is a lens or something like one) that shoots sunlight at an object from 180 degrees, then you could make it so that an object is lit up the same as if it were in very low orbit around the sun, and if you insulate the back, it should be the same temperature. Wnt (talk) 04:32, 5 May 2014 (UTC)

Now, to continue your thought experiment, to get a temperature equal to that of the Sun, you would need the object to be surrounded by Suns on all sides, so that no space was visible (distance to the Suns wouldn't matter). Note that I didn't say the temperature of "the surface of the Sun", since many of the photons emitted from the Sun come from below the surface, where it's much hotter. The corona is also hotter, but it doesn't emit much light, at least not in the visible range. So, the Sun really can't be modeled as if it were a solid sphere of constant temperature. StuRat (talk) 04:40, 5 May 2014 (UTC)

(Replying to both of you) This is kinda what I suspected, but I was trying to confirm that the intensity (distance, angle, or concentrator) of the sunlight has no affect on the equilibrium temperature of the object. So basically while photons from the sun don't have enough power (in the scientific meaning) to burn something (in the ordinary light on earth), their temperature is actually quite high. I assume there has to be an "equivalent effective temperature" for the Sun, including all the layers - I don't suppose you know what it is? Ariel. (talk) 05:35, 5 May 2014 (UTC)

Color temperature#The Sun discusses the effective temperature of the Sun. StuRat (talk) 13:56, 5 May 2014 (UTC)

The effective temperature of photons that come from the Sun will be smaller. To see this intuitively, consider operating a reversible heat engine that takes in heat from the Sun, converts part of to work and dumps waste heat in the Earth's environment. In the limit of maximum efficiency, the heat engine will be reversible. Suppose that we attempt to realize such a heat engine by bringing photons from the Sun to the Earth and using these photons to extract the work. If we can do that with the same maximum efficiency, the photons will have the same temperature as that of the Sun.

Imagine putting a box in the Sun and filling that with photons. When you extract that box, the volume that the box occupied will be filled with the photons in the Sun and that process is irrevesible. You could have extracted the box in a slow versible way by letting the photons perform work as they fill in the gap left by the box. That work that you are not extracting is not available when use only the photons from the Sun, therefore the maximum efficiency will be less than given by ther Carnot formula.

So, what is lost is the work done by radiation pressure. To see this more formally, consider a box filled with photons at temperature T. The internal energy is proportional to T^4, so we can write E = a T^4. Then according to the fundamental thermodynamic identity we have dE = T dS - P dV. If we keep the volume of the box constant, we have dS = dE/T = 4 a T^2 dT. Integrating gives S = 4/3 a T^3 = 4/3 E/T.

When you use an amount of energy E from photons from the Sun, the entropy thus decreases by E/(3/4 T_sun) which has to be balanced by waste heat that is dumped into the environment. This means that the effective temperature is 3/4 T_sun, the difference between this and T_sun can be traced back to the radiation pressure of 1/3 E/V which is not available in practice to perform work. You could argue that you do have an entropy of E/T if you capture phtons emitted by the Sun on Earth, when keeping the momenta of the photons which point away from the Sun. It is only when you fully thermalize the photons that the entropy jumps to 4/3 E/T. In the former case you can extract the work done by radiation pressure, in the latter case you can't. Count Iblis (talk) 14:39, 5 May 2014 (UTC)

This question can be answered using a standard equation: Planetary equilibrium temperature. That equation is derived from the Stefan–Boltzmann law for radiated power, using a method similar to what Count Iblis outlined above; it accounts for geometry and radiant efficiency (albedo).

This type of question - the temperature of an object floating in space - is often asked as a homework question in introductory classes on astronomy and planetary science. In an advanced class, you might also incorporate terms related to the blackbody temperature of the cosmic background (which is not absolute zero); so the efficiency of radiation is lower than ideal; but this only makes a very tiny change to the result. Nimur (talk) 14:42, 5 May 2014 (UTC)

Vacuum tube filament conditions

I must have made a mistake somewhere, but I can't see where.

I obtained some 1S5 directly heated vacumm tubes. According to RCA publications, the filament in these is a simple high purity tungsten wire (presumably round) with a microscopically thin coating of alkaline earth oxides. The construction of this vacuum tube is such that the filament can be seen (just, as it's diameter is tiny) at the ends, about 10% of the way in from the ends, and about 25% in from one end. Upon energisation of the filament only with the rated filament voltage (1.5 V), in a darkened room you can just see it glowing a dull red. This means its' temperature is then about 700 K, and loss of heat by conduction into the supports must be negligible (as otherwise it would not glow near the ends), and virtually all heat is lost by radiation.

I measured the filament cold resistance (300 K, ie a room temperature of 27 C) carefully with a digital multimeter, compensating for lead resistance. It was 9.75 ohms. I measured the current drawn at 1.50 V - it was 52.23 mA, indicating a hot resistance of 28.73 ohms. Using Worthing's formula, R = ρ_1K.T^Pw, where ρ_1K, = 5.97 x 10^-7, Pw = 1.205, I calculated that the filament resistivity at 300 K is 5.77x10^-8 ohm.meters. The filament length by inspection is 25 mm. I then calculated the filament cross-section area (1.48x10^-10 m²), its' diameter (0.014 mm), and its surface area (1.078x10^-6 m2).

Using Worthing's formula again, from the hot resistance, I calculated the temperature as 736 K, close enough to that estimated by visual inspection.

From RCA and textbooks, emissivity of oxide coated filaments is about 0.3. But due to radiation being proportion to the 4th power of temperature, the emissivity value doesn't matter much.

Using the formula (stefan boltzman law), P = A σ ε (T⁴ - T_A⁴), where P is the electrical power disipation, A is surface area, σ is the Stefan-Bolzman constant, 5.67x10^-8 W/m²K⁴, I found interatively the temperature at which, with the surface area it has, the electrical power equals the thermal radiated energy. It worked out as 1204 K (should be 736 K), at which I = 28.05 mA (should be 52.2 mA), and P = 42.0 mW.

Where have I gone wrong? Have I overlooked something? Why is the temperature rise by Stefan-Boltzman about twice what it should be?

Strictly speaking, I should have inserted the anode temperature as it intercepts the filament radiation, and not assumed radiation to ambient. But allowing for anode temperature rise would make the error worse. I left the anode and grids open circuit. So there is no net electron flow, so there is no evaporative cooling of the filament. 144.138.223.88 (talk) 04:42, 5 May 2014 (UTC)

Hmmm, I'm thinking the power emission should be 1.5 V x 52.23 mA ~= 75 mW. Your calculation of the surface area 1.078x10^-6 m2 looks right, but I don't know your figures for this filament really are right. Dividing I get 70,000 W/m2 of emitted power (some of which is IR). Dividing this by σ ≈ 5.67 × 10−8 W/(m2K4) yields 12345 x 10E13 K⁴ and taking the square root of this twice yields 1054. So I didn't find a flaw on the first run-through myself. Still, even a small error in the estimation of the filament diameter could explain it all. Wnt (talk) 05:40, 5 May 2014 (UTC)

Yes, the electrical power in, which should equal power emitted, is 78.3 mW. I notice you did not allow for the ambient temperature. As the ambient is 300 K, it will radiate back to the filament, so the net emission is a little less than 78.3 mW. But, no, a small error in diameter doesn't change the picture, as the surface area changes in proportion. Increasing the diameter by 10% changes the filament temperature by only 6 K. You need to increase the filament diameter, or its length, by almost a factor of 10x to get rid of the error. 121.221.131.231 (talk) 10:30, 5 May 2014 (UTC)

Our table for incandescence doesn't even go down below about 820 K. Are you sure the very low 700K figure accurately represents the cold temperature of the filament, and you're not just seeing that it's small and dim? Wnt (talk) 12:55, 5 May 2014 (UTC)

You mean the hot temperature. Obviously, the 700 K figure was a rough one. I could only just see a faint red line in a darkened room. The light sensitivity of eyes varies from person to person in any case. My wife can detect an oven's hotplate glow several seconds before I can. The Wikipedia article on thermal radiation has a table Subjective color to the eye of a black body radiator lists 750 K as "a faint red glow." but does not indicate the data is for a darkened room. The point, is the visble effect, the measured current, and the manufacturer's design description for the tube type all point to a temperature 700 to 750 K or so. Yet my calculation based on the Stefan-Boltzman law indicates 1200 K and also indicates the filament should draw only about half the current that it does. Why? What have I done wrong or forgotten? 121.221.131.231 (talk) 14:19, 5 May 2014 (UTC)

Math check: from your 0.014 mm I get 1.10×10⁻⁶ m², and then from 21 mW and the Stefan–Boltzmann law, I get 762 K with $\epsilon =1$ and 1029 K with 0.3. --Tardis (talk) 13:06, 5 May 2014 (UTC)

That's with $T_{\text{A}}=0$, which is a good approximation here (even $(300/762)^{4}\approx 2.4\%$). --Tardis (talk) 13:13, 5 May 2014 (UTC)

It's not as simple as that. If emissivity is increased from 0.3 to 1, the consequent lower temperature will decrease the electrical resistance and more current will flow, counteracting the drop in temperature. It's a non-linear system - you cannot simply calculate directly, you must iterate. Calculate the temperature with Stefan-Boltzman law from a gusetimate power input, calculate the new resistance with Worthing's formula for that temperature, from that calculate the power dissipation and then use s-b law again to find the new temperature. Do it all again until you settle on a temperature. For emissivity = 1, I get 979 K (which would glow bright red) and 37.02 mA, so power dissipation has increased to 55.5 mW. That's still a 30% error in current and a 56% error in temperature rise. That degree of error definitely needs explaining. Incidentally, I mistyped the 21 mW figure. It should be 42 mW (1.5 V x 28 mA). 121.221.131.231 (talk) 14:37, 5 May 2014 (UTC)

If you have to iterate, something has already gone wrong. You've already measured the power, and by knowing (at known (room) temperature) the voltage, resistivity, and length of the wire you can deduce its radius and thus surface area. Given also its emissivity, you know everything in the Stefan–Boltzmann law except the temperature. (Being a bit less lazy than I was, you can include $T_{\text{A}}$ and write $T={\sqrt[{4}]{{\frac {P}{A\sigma \epsilon }}+T_{\text{A}}^{4}}}$.) Meanwhile, you also know the hot resistance and can deduce the temperature directly from that.

Of course, you know all that: the resulting discrepancy is the reason for the question. I'm just pointing out that no iteration is required: you are simply calculating the temperature from known data two different ways. You can choose instead to compare the measured current to that which ought to flow in equilibrium, and then you do need to iterate as you say. The distinction is simply whether you apply Worthing's formula to obtain a temperature from the measured current or to obtain a current "from" a calculated temperature. (Meanwhile, you're always using it to derive a temperature from the measured resistance.)

You're missing units on Worthing's formula, which I do not find readily online otherwise; if i assume MKS (i.e., $\rho _{1\,{\text{K}}}=5.97\times 10^{-7}\ \Omega \cdot {\text{m}}$, $\rho =\rho _{1\,{\text{K}}}(T/{\text{K}})^{P_{\text{W}}}$), I get a resistivity (and thus cross sectional area) 10000 times as large, and so a diameter 100 times as large. That 1.4 mm length is of course too large: you would be able to see the wire plainly. However, the units implied by your resistivity are also strange: mΩ-dm or equivalent. So if I suppose that it was in reality Ω-mm, then your resistivity and cross section are 10 times too large and the diameter should be 43.4 μm, which is (just) believably small. Then $A=3.408\times 10^{-6}{\text{ m}}^{2}$ and (using the power for your measured current) I get 1080 K — not much of an improvement, as you hint above.

The next likely culprit appears to be the oxide layer. However, if I instead solve for A from the measured power and Worthing temperature and then compute the diameter it's 206 μm, which is too large even with the larger W diameter I compute above.

Summary: I don't see anything that actually solves the problem. Does any of this analysis inspire anything further from anyone else? --Tardis (talk) 13:15, 6 May 2014 (UTC)

Firstly, an appology over units - I'm a sloppy at typing with the awkward system here. In ohm.meters/K, the constant is 5.97 x 10^-11. The correct constant is easily verified by calculating the resistivity at 300K (I got 5.77 x 10^-8 ohm.meters) and comparing it with published standard values for pure tungsten (Wikipedia gives 5.6 x 10^-8 at 293 K). So I got that right, at any rate.

Most textbooks, websites, etc give a variation of Mitchell's formula for resistivity of metals ( k(1 + αT) ) as a function of temperature. That is an acceptable approximation at common engineering temperatures, say zero to 100 C. For larger temperature ranges, Mitchell is hopelessly inaccurate. If you consult textbooks on the design of incandescent lamps and vacuum tubes, you'll find Worthing's formula and constants. If you consult physics textbooks you will find more complex formulae, as Worthing is no good at very low temperatures. I have some textbooks so didn't bother looking on the web.

I have since realised a posible component of my error. Books on vacuum tubes give design filament temperatures as measured by optical pyrometry - i.e., essentially what I did by eyeballing. This measures what is called "brightness temperature." To get the actual temperature of a gray body, you need to apply a correction for emissivity. For e = 0.3 the correction is approx 0.3^-0.25 ie true temperature is about 35% higher, ie the temperature by eyeballing is not about 700 K, its about 900 K. This means either the filament in the tube I tested is not tungsten or the emissivity is close to 1, and the manufacturer quoted the brightness temperature without stating it's the brightness temperature, which is pretty sloppy. Problem: I tried Worthing constants for other possible materials and the error was much the same. I have been unable to verify the 0.3 emissivity value, beyond a single reference in a book on vacuum tubes. And there's still a large descrepancy.

Yes, you can calculate based on the measured power input to get resistance, from that temperature, and from that the radiated power. Then, and only then, no interation is required. But if you do that, the radiated power calculated with Stefan-Boltzman is a lot less than the electrical power input.

I don't think the error can be a thick oxide layer. Every textbook on vacuum tubes says it's a semiconductor, and thus has to be made microscopically thin, otherwise the operation of the tube is seriously compromised.

The plot thickens! 121.221.131.231 (talk) 17:34, 6 May 2014 (UTC)

If the emissivity correction makes it look like it's a lower temperature, then it's not a grey body but a blue body, isn't it? (i.e. reflecting the non-red colors back into itself) I find the idea of an oxide layer much more interesting. After all, the filament is continually being heated and cooled in equal amounts, so there ought to be some gradient within it between the center and edge of the filament, especially if the edge is nonconductive. Wnt (talk) 19:12, 6 May 2014 (UTC)

Why on earth would you suspect a blue body? The temperature involved is far too low to produce anything but insignificant green or blue radiation. A body that has emissivity less than unity is by definition a gray body, and as such radiates less energy than a black body at the same temperature. The percieved colour is the same, but the intensity is less. For that reason a correction for emissivity is required for grey bodies. Since the glow is visibly even throughout the filament length, there is insignificant temperature gradient along its length. This means that the radial gradient is even more insignificant, as the path length radially is vastly shorter (in the ratio of 24 mm to 0.014 mm), even if the thermal resistivity of the thin oxide coating is large compared to the tungsten.124.182.16.27 (talk) 01:55, 7 May 2014 (UTC)

Alright, suppose some of the filament's energy is being reflected internally. It still has to radiate the same in watts, or else it will explode. Holding in some of the power could force its interior to go to a hotter/bluer temperature, but if all frequencies escape equally... that makes the filament's color even more off from what you observe than you already calculated, doesn't it? Wnt (talk) 23:02, 9 May 2014 (UTC)

An RCA guide from 1943 says that oxide coated filaments have a "relatively thick coating" of " alkaline earth compounds on a metallic base." It says they often are operated at about 800 C (about 1070 K) and have a dull red glow as described by the OP. The tube in question was in use in portable dry-cell powered receivers by the early 1940's and in common use until the early 1950's. Edison (talk) 23:05, 6 May 2014 (UTC)

Carefull. Your ref is comparing oxide coatings with thoriated tugsten. And the oxide coating is EXTREMELY thick compared to thoria coatings, which were produced by diffusion and were thought to be only one molecule thick.

Almost invariably these sorts of publications, when talking about coated filaments & cathodes, are talking about indirectly heated unipotential cathodes as used in vast numbers of tubes, or they are talking about RF power tubes, or they are talking about wet battery tubes. Normal AC recieving tubes, and wet battery tubes, were designed with cathode areas much larger than needed for operating conditions as a way of improving the finite operating life and keeping cost down. Miniature dry battery tubes were primarily designed to keep current down, and operating life was severely cut back, by providing only just enough emission area, for the anode current in service. Tungsten at operating temperature is fragile, and to prevent too many catastropic failures due to accidental knocks, the operating temperature of miniature dry battery tubes was lower than is optimum. You need to consult publications specifically about these miniature battery tubes. Unfortunately, I found obtaining publications that give actual numerical data rather than near useless words like "thin" and "thick" difficult.

Having said all that, the descrepancy has to be somewhere, and coating thickness is worth following up on. Thanks.

— Preceding unsigned comment added by 124.182.16.27 (talk) 01:17, 7 May 2014 (UTC)

The plot thickens again. I have investigated and on Page 199 of Fundamentals of Engineering Electronics, William G Gow 1937, its says the coating is "a few hundred to a few thousand molecues thick." That would make it in the high nanometer range. In The Oxide-Coated Cathode, G Herrmann, 1951, Vol 1 page 6, its says the coating is at least 100 nm thick but gives no upper limit. On page 39 it says the coating is "40 to 100 μ" thick. Micrometers? Microinches? But in Electron Emission Coating for the Oxide Cathode, C H Meltzer and E G Widell, 1962, page 1, it says the coating is made 0.75 to 3.5 mils thick (ie thousands of an inch, 0.02 to 0.08 mm, or 20,000 to 80,000 nm)! 124.182.16.27 (talk) 06:16, 7 May 2014 (UTC)

Update: I set about measuring the filament diameter. As these tubes are now rare, I wanted to do this in a non-destructive way. So I made a "lash-up" microscope with a couple of magnifying glasses and an old spectacle lens, so I could get some visual sense of diameter through the tube glass. I then snipped off a thin looking white hair from my head. The white colour approximates the cold colour of the filament. I measured the hair with a good vernier and it was about 80 μm. Comparing it under the lash-up microscope with the filament, it is obvious that the filament diameter is greater than the calculated 14 μm, but less than the hair. So the filament is very roughly about 40 or 60 μm, so the coating thickness is about 20 μm. Correcting for this halves the descrepancy in calculating filament temperature. User Edison had something.

Does anybody know a good reference on oxide coating emissivity? 124.182.16.27 (talk) 15:45, 7 May 2014 (UTC)

Vegetable wash and pesticide residue in produce

Is vegetable wash of any kind effective in reducing pesticide residue in produce? (To be effective the amount of pesticide residue remove has to make a significant difference in terms of health risks.) Any data?

Would Vodka (as a food-grade aqueous ethanol solution), applied as a spray before rinsing with water, be effective in removing pesticide from produce?

Thanks. — Preceding unsigned comment added by 173.49.15.150 (talk) 12:05, 5 May 2014 (UTC)

Here are the results of one study. Here is another one. Here is yet another. Here are some general guidelines for cleaning fruits and vegetables. Here is a synopsis of another study. --Jayron32 12:46, 5 May 2014 (UTC)

As an aside, Vodka is always recommended, but usually liberally applied to the cook, rather than the vegetables. --Jayron32 12:48, 5 May 2014 (UTC)

Perhaps after any cooking appliances have been used and turned off. Definitely not before... [1] [2] [3] [4] Nil Einne (talk) 13:19, 5 May 2014 (UTC)

It is a proven fact that alcohol dramatically improves the quality of food. When my family was running a small hotel, we told Cook to stop ordering such high amounts of sherry, about a litre per day, and we monitored the kitchen orders to ensure he complied. Straightaway the food quality deteriorated, so we sacked him, and hired a new cook who we thought had mastered the difficult art of cooking high quality meals without alchohol. But new cook used quite a quantity of sherry too. Beer as well. 121.221.131.231 (talk) 14:54, 5 May 2014 (UTC)

Scientific names

In Bregmotypta Bruce, 1994E - what does the "E" represent? All the best: Rich Farmbrough, 16:51, 5 May 2014 (UTC).

There are probably five sources in the paper from 1994, or one authored by an author identified by the initial E. Without a link to he source it's hard to tell. μηδείς (talk) 17:27, 5 May 2014 (UTC)

Another option is that, in the work where this appeared, four other papers written by Bruce in 1994 had previous been cited. For example "According to Bruce (1994a)..." , and later "This species has three toes (Bruce, 1994b)", etc. The letter labels of the papers written by the same author(s) in the same year are not part of the bibliographic record though. For example, a paper cited as "Bruce, 1994b" by Jones (2000) may be cited as "Bruce, 1994a" by Xu (2001). As Medeis says, if you tell us where you found this quotation, we could probably answer more confidently. SemanticMantis (talk) 19:17, 5 May 2014 (UTC)

• There's an example of this usage here. DuncanHill (talk) 19:21, 5 May 2014 (UTC)

Thanks! Here's that database's bibliography [5]. Sure enough, there are six papers cited there, all by Bruce, N.L., all written in 1994. Since the bibliography doesn't give letter codes, we might assume that it was chronologically the fifth paper published that year. The paper that is the authority for that species is this one [6] (which is listed second at the prior link.) Bibliography and taxonomic authority are tough, and standards don't always exist, and aren't always followed... For example, this species also has (a different) Bruce 1994 as the authority [7] in the same database, but no letter is given, and the full title of the paper is present. One final thought: in botanical naming conventions, there are a few single-letter codes with well-accepted meanings, e.g. 'L.' for Linnaeus. But according to our article, the International_Code_of_Zoological_Nomenclature doesn't seem to use single letters in that way. Perhaps a move towards a PhyloCode would help resolve some of the ambiguitiy. (And despite the link, I remain unconfident as to the true meaning, if any. It could always be an error, the result of poor quality control and quality assurance in the database. I hope someone else might know for sure! ) SemanticMantis (talk) 02:22, 6 May 2014 (UTC)

Yes, I read most of the ICZN's articles on naming, and there was no explanation. I couldn't see how the rather paper-oriented Harvard citation style could pollute the naming conventions, but I guess if someone tries to combine the naming and the cites (they should write Bregmotypta Bruce, 1994 [Bruce 1994E]) then this could happen. I propose to remove the "E" at Sphaeromatidae, and anywehre else I find it. All the best: Rich Farmbrough, 07:50, 6 May 2014 (UTC).

[In this edit respected user Stemonitis removed most of the post-annual letters, so I think we are on safe ground. All the best: Rich Farmbrough, 08:25, 6 May 2014 (UTC).

Medium-Chain Triglycerides preservation

What is the best way to storage products that contain them (such as Coconut oil) for best preservation of these Medium-chain triglycerides? Cooling? Darkness? (If yes, why --- is it only because of oxidation concerns?), thanks. Ben-Natan (talk) 20:05, 5 May 2014 (UTC)

Most sites I found here note a shelf-life for unrefined coconut oil to be about 2 years, and none of them recommend any special procedures (such as refrigeration) beyond what you would use for any other cooking oil. --Jayron32 23:52, 5 May 2014 (UTC)

More, this site: here, in the search above, recommends storage either in the refrigerator or the pantry; so darkness appears to extend the shelf-life, but not cold, per se. The problem is that it has a very high melting point, meaning that coconut oil stored in the fridge can get hard, and need to be thawed at room temperature before use. --Jayron32 23:55, 5 May 2014 (UTC)

Generally the cooler foodstuffs are, the slower any degrading reaction, and most biological decay processes are. Coldness only becomes an issue when it damages the foodstuff (as in strawberries), or when the cold storage is done incorrectly allowing degradation, for example by sublimation. All the best: Rich Farmbrough, 07:57, 6 May 2014 (UTC).

Coconut oil has a very high proportion of saturated fatty acids, so I'd expect it to be more stable than most. Polyunsaturated fat is more active; things like linseed oil and tung oil can even produce substantial amounts of heat when rags soaked in them are left out, to the point of causing fires. Rancidification in any case can be opposed by restricting access to oxygen, such as by sealing containers well or using an inert atmosphere. From the article it also sounds like adding antioxidants like vitamin C is a favored approach. Wnt (talk) 05:52, 7 May 2014 (UTC)