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December 18

Organ pipe

If the air temperature rises with height due to poor church HVAC then would there be only one main pitch equal to putting the average speed of sound of the standing wave in the pitch formula or will there be two main pitches one for the speed of sound at each end or will the fundamental be smeared into an infinite number of pitches more or less equally loud all lying between a pitch based on the speed of sound of the most heated part and the speed of sound of the least? 166.199.7.54 (talk) 00:51, 18 December 2023 (UTC)[reply]

Are you asking whether a temperature gradient across a column of air causes its resonant standing wave to vary in frequency accordingly? Remsense 02:11, 18 December 2023 (UTC)[reply]
Correct, if a pipe is x Hertz when everything's 20C what sound will come out when it's 20 on average but the air has a temperature gradient? 166.199.7.54 (talk) 02:40, 18 December 2023 (UTC)[reply]
I've thought about this for about 15 minutes, and I think your "two frequencies at each end" guess is going to be my guess also. Ideally, there's only meaningfully a "pitch" in the direction that changes in air pressure are propagating (i.e. down the length of the tube) Remsense 04:38, 18 December 2023 (UTC)[reply]
As well as the standing wave producing the sound, there is also movement of air due to the action of the bellows. I suspect that quite quickly the air in the pipe will have a uniform temperature closely related to the intake temperature of the blower. Martin of Sheffield (talk) 09:20, 18 December 2023 (UTC)[reply]
Even if the temperature varies considerably in the pipe, there is only one pitch. The pressure waves will travel through the pipe with a varying velocity, the speed of sound, which increases with the temperature. But the pitch is determined by the standing waves, whose frequency only depends on the total travel time of the wave through the pipe. Compare this to a pendulum clock. The speed of the pendulum varies over its trajectory, but the clock ticks with only one rhythm.  --Lambiam 10:15, 18 December 2023 (UTC)[reply]
Right! I think my mistake was thinking about the standing wave as if it were a traveling wave, not the sum of traveling waves. Remsense 15:29, 18 December 2023 (UTC)[reply]
When the air temperature rises, the speed of sound in air increases proportionally to the square root of the absolute temperature and the resonant length of the metal Organ pipe expands. The sound pitch does not change as much as either effect alone would cause because they partially compensate one another. Philvoids (talk) 18:45, 18 December 2023 (UTC)[reply]
This is not the case: the expansion/contraction of the aerophone body caused by change in temperature is negligible proportional to the corresponding expansion/contraction of the column of air inside. Remsense 18:51, 18 December 2023 (UTC)[reply]
The length of the column of air inside the pipe is essentially the length of the pipe containing the air.  --Lambiam 19:47, 18 December 2023 (UTC)[reply]
Right, but the point I was making is that change in the length of the pipe due to the expansion or contraction of the aerophone itself is negligible. I suppose I misspoke when I said the air was expanding/contracting, I should've said it was becoming more or less dense. Remsense 19:49, 18 December 2023 (UTC)[reply]
That is correct. A pipe made of an alloy of tin and lead, as many church organ pipes are, will expand by about 0.05% when the temperature rises by 20 degrees. The speed of sound will increase by about 3.4%.  --Lambiam 20:29, 18 December 2023 (UTC)[reply]
I assume that the Fourier analysis in terms of wavelength of the waveform inside the pipe depends only on the shape of the pipe but not on the conditions of the air inside it. The lowest eigenmode (I guess that's what's meant by "standing wave" above) has wavelength twice the length of the pipe (if the later is closed at both ends and essentially one-dimensional). However, the conversion to frequency (the dispersion relation) depends on temperature, as demonstrated in the video linked above. With a temperature gradient, the air molecules should oscillate with varying frequency along the length of the pipe. The next question is how the oscillation inside the pipe is coupled to the air outside. If the coupling is homogeneous along the pipe then we should here a continuous superposition of frequencies over a range determined by the temperature range. If the coupling occurs at a single point (the labium or an open end (are organ pipes open?) come to mind) then the pitch will be determined by the temperature at that point. The actual situation will be somewhere inbetween these extremes, and we won't discuss the effect on timbre. --Wrongfilter (talk) 20:31, 18 December 2023 (UTC)[reply]
It feels like we should link to Q factor, a parameter that describes (among other things) the "perfectness" of a resonator - in other words, how ideally the resonating energy is exactly concentrated at one single harmonic frequency. Real physical resonators have a bit of a spread - a "bandwidth" - over which the energy is spread. Things like imperfections or nonuniform conditions along the extent of the resonant cavity (including temperature) will cause the resonator to have a lower "Q" factor - and consequently a wider bandwidth. Audibly, this means a tone that is less "clear" and it will sound different - perhaps even sounding "out of tune" in severe situations. Nimur (talk) 20:43, 18 December 2023 (UTC)[reply]
Assuming a constant frequency , the wavelength is given by in which is the speed by which the waves propagate – in this case the speed of sound. If that speed varies along the length of the pipe, so does the wavelength, which makes it difficult to assign a meaning to "the Fourier analysis in terms of wavelength of the waveform inside the pipe".  --Lambiam 08:36, 19 December 2023 (UTC)[reply]
What do you mean by "constant frequency"? The air molecules obviously do not travel the entire length of the pipe but oscillate by a fairly small amount around a mean position. The frequency of these oscillations can vary along the pipe, and it is these oscillations that eventually couple to the air outside. You can of course always analyse any 1D wave form in terms of Fourier waves (sines and cosines); what's the worst that could happen? Maybe the coefficients become time dependent, but the wavelengths of the modes remain the same. Compare to a guitar string that is plucked in the middle so that predominantly the lowest mode with wavelength twice the scale length of the guitar is excited, this determines the pitch. Now increase the tension of the string; the wavelength remains the same but the pitch increases. Wavelength is determined by geometry, frequency by the physical properties of the oscillating medium. --Wrongfilter (talk) 08:58, 19 December 2023 (UTC)[reply]
Sorry, that doesn't sound right. A standing wave is a product of a position-dependent oscillation and a time-dependent oscillation, or a sum of two travelling waves. It can't have a position-dependent frequency, or waves would pile-up somewhere. What happens is that the wavelength varies along the length of the tube. Assuming there's an antinode at both ends of a simple, one-dimensional tube, the node in the middle gets displaced towards the end with the lower sound speed. The period of the fundamental of the produced sound is still twice the sound travel time along the length of the tube.
As for your guitar comparison: consider a guitar where the thickness of the string changes along its length. To excite primarily the fundamental, you have to pluck it near the thick end. PiusImpavidus (talk) 13:15, 19 December 2023 (UTC)[reply]
I don't think there's a contradiction here. A standing wave with a displaced node can still be Fourier analysed, it will just have a different overtone spectrum compared to a symmetric wave. And I guess you can rearrange the Fourier sum to arrive at an expression with a wavelength that is formally position-dependent. But that just gives the spatial waveform inside the tube. The main question still remains: what does it sound like? Heating the pipe uniformly does not change the spatial form of the standing wave, but it does change the pitch. Applying a temperature gradient doesn't necessarily change the pitch if applied in the right way, but it will change the spectrum of the wave inside the pipe, as well as that of the sound wave outside the tube. I'm not sure we can get any further than that without a quantitative analysis. --Wrongfilter (talk) 16:26, 19 December 2023 (UTC)[reply]
A few thoughts: Organ pipes can be open-ended or stopped, in which case they sound an octave lower. According to a document from Harrison's website, Peterborough Cathedral is heated continously during the winter, and the difference from floor to ceiling is only 2° C. In other cases a difference was noted of 15° C between the player at floor level and the Swell division 15ft above; another tuner noticed a difference of 11° C between Great and Swell divisions. The relative humidity matters almost as much as the temperature. Get both wrong, and this happens: "The organ is in an organ chamber some 25 feet above the only outlet from the forced-air heating. Six months after the heating was installed, every soundboard split, slides stuck, runnings everywhere, and tuning almost impossible. The difference between floor level and the organ was at least 20° F (11° C). Remedial work to the organ cost more than £55,000." Church Heating and the Pipe Organ. MinorProphet (talk) 22:52, 18 December 2023 (UTC)[reply]
And pipes can be flue or reed. Flue pipes have an antinode at the window whereas reeds have a node at the reed. And after you've considered flue/reed stopped/open, then the shape of the resonator is critical to the tone: parallel, conical (getting narrower) or conical (getting wider). To get back to the OP's original question though, very shortly after speaking any pipe will be filled by the air from the blower/bellows and will be consistent in temperature and humidity along the length. The pipe, however, is quite a different matter! Martin of Sheffield (talk) 23:05, 18 December 2023 (UTC)[reply]
In many English churches, in winter, the temperature of the pipes or the air in the pipes is not an issue. Far more important is the need to warm up the organist to operating temperature. -- Verbarson  talkedits 19:00, 19 December 2023 (UTC)[reply]

December 19

Pi

Hi all, further to my earlier question about engine capacity:

A manufacturer's vehicle manual states its 12-cylinder engine has bore and stroke of 95 * 100 mm, giving approximately 8.520 litres.[1] (needs Ctrl + Scroll)

Given the capacity of an engine in litres = (bore2 [mm] * stroke [mm] * (π/4) * no. of cylinders) ÷ 1,000,000, what value of π (say to 10 decimal places or more, or a fraction) was used to arrive at a capacity of 8.520 l? If someone could kindly re-arrange the formula in terms of pi that would be great. Cheers, MinorProphet (talk) 11:24, 19 December 2023 (UTC)[reply]

π = capacity * 4,000,000 / (bore2 * stroke * cylinders)
I think, said Ouro (blah blah) 11:35, 19 December 2023 (UTC)[reply]
Hang on a bit, you've dropped a factor of 103. When you square the bore you also need to square the 10-3 giving 10-6. Then allow for the stroke being in mm and your constant is 10-9. Martin of Sheffield (talk) 11:43, 19 December 2023 (UTC)[reply]
Hence the I think bit. Just have to move the decimal point around. --Ouro (blah blah) 11:59, 19 December 2023 (UTC)[reply]
Bore and stroke are in millimetres, volume in cubic decimetres, so that's 1003. The factor 106 is correct. PiusImpavidus (talk) 13:43, 19 December 2023 (UTC)[reply]
Doh! Aren't litres banned by the SI these days? Bl**dy metric system, doesn't do what it claims is its strong point. (Good catch though). Martin of Sheffield (talk) 14:52, 19 December 2023 (UTC)[reply]

Thanks, that's great. Going with Ouro's original version, which apparently works...

8.520 * 4,000,000 (= 34,080,000) / (952 (= 9025) * 100 * 12 = 10,830,000) =
3.14681440443213296398891966759 (Windows Calc), which seems hugely wrong, considerably worse than...
3.142857142857... which is 22/7
3.1415926535897932384626433... is in the infobox for Pi.
3.1415929203539823008849557522124 is 355/113, which I somehow thought is what the manufacturer used. Using 355/113, my own calculation of the capacity is approx. 8.5058628309 litres.

So, how did they arrive at π = 3.14681440443213296398891966759, the first answer? And if my calculation of 8.50586 litres is right, why state the capacity as 8.520 litres? Maybe it's just a rough estimate for mechanics and not intended to correspond directly to actuality. But I've just noticed that the manual (in German) says Hubvolumen 8 520 ebem. I misread that as eben, i.e. 'exactly', 'precisely', although German has tens of meanings for 'eben. But, assuming it isn't a misprint, what does ebem mean? (I've had no luck, maybe a job for the Language Desk.) MinorProphet (talk) 13:03, 19 December 2023 (UTC)[reply]

Thinking about an answer to my earlier question, the manufacturer may well have used pistons with concave heads to lower the compression ratio to cope with lower octane fuel, which would have increased the capacity compared with flat-topped pistons.
If so, 8.520 - 8.506 = 0.014 l = approx. 14 cc extra with concave heads, so we need to deduct that figure from the stated capacity to arrive at a better approximation of π: (there will be errors, I hate doing maths in public)
π = 8.520 - 0.014 (=8.506) * 4,000,000 (=34,024,000) / (952 (= 9025) * 100 * 12 = 10,830,000 =
3.1416435826408125577100646352724 which really isn't too far off our infobox value of
3.1415926535897932384626433.
Maybe that could be one answer. MinorProphet (talk) 13:54, 19 December 2023 (UTC)[reply]
Don't use excessive precision. Even if bore and stroke are exact, the volume only has three or four significant digits (more likely three, with the trailing 0 the result of first rounding to tens of cubic centimetres, then converting to litres, forgetting to drop that 0). That suggests 3.145<π<3.149. But you have to consider that the bore and stroke may be rounded too. I wouldn't be surprised if the actual bore is systematically slightly larger than the nominal value. Or it could be some fudge factor for dead volume or to take wear into account. I expect that for π they simply used whatever value their calculator or software has built-in, which would at worst be about 7 decimal digits. PiusImpavidus (talk) 14:04, 19 December 2023 (UTC)[reply]
Probably 12" slide rules, this is the Maybach HL85 designed in 1938 (Faber-Castell made a million slide rules every year in the late 1930's).[2] Otherwise thanks for your comments. While checking some engine capacity figures, I've been trying to account for the relatively big differences found in various sources, both RS and not. It could be that the ebem / eben in the manual means 'actual volume, despite theoretical calculations'. MinorProphet (talk) 14:45, 19 December 2023 (UTC)[reply]
No, capacity is on swept volume, not affected by the combustion chamber's shape. Greglocock (talk) 17:33, 19 December 2023 (UTC)[reply]
The observation that the shape is irrelevant was also made in the earlier discussion at Wikipedia:Reference desk/Archives/Science/2023 December 8 § Cylinder capacity (search for "Cavalieri's principle").  --Lambiam 22:43, 19 December 2023 (UTC)[reply]
(edit conflict) Cylinder capacities are swept volume, not total enclosed volume? You can ignore the headspace. For taxation reasons the swept volume is often a shade under the commonly stated volume: 1,490 cc for a 1.5 l car, 49.5 cc for a 50cc moped. Martin of Sheffield (talk) 14:52, 19 December 2023 (UTC)[reply]
@MinorProphet I would suggest an alteration from cbcm: "cubic centimeter" see the "Hansa-Cabrio" adv. here p.6, from some aberration on the use of "ebem" (root sometimes flat, horizontal, smooth ([3]), sometimes probably used meaning "the dimension referred to" as in f. d. ebem here --Askedonty (talk) 15:06, 19 December 2023 (UTC) (Obviously this would be better placed on the Language Desk, anyway at least all ebem occurrences related to cylinders/volume I've found are from the era of the phasing out of the Fraktur typesetting.) --Askedonty (talk) 16:36, 19 December 2023 (UTC)[reply]
ebem is not a word. Both examples clearly read cbcm, even if the OCR doesn't agree, and there's no need to look for any other meaning than cubic centimetre. In the document originally linked by MinorProphet, the c does look like e due to jpg artifacts suggesting the presence of horizontal bars. But if you look at an actual e in Fraktur you'll see that the horizontal bar is actually tilted. --Wrongfilter (talk) 16:49, 19 December 2023 (UTC)[reply]
That's if Stahl und Eisen 1913 doesn't know it's not a word, or they are making sense of "worin r den spezifischen Widerstand f. d. ebem (read: cubic centimeter), L die Länge und S den Querschnit", etc. "Widerstand" being "r", electrical resistance. --Askedonty (talk) 17:23, 19 December 2023 (UTC)[reply]
I interpret this as "für den Kubikzentimeter", i.e. per cubic centimeter. The formula makes sense physically, I don't immediately see what the dimensions of R are supposed to be. --Wrongfilter (talk) 17:54, 19 December 2023 (UTC)[reply]
Yes, regarding a steel foundry it will physically make sense. I only remain dissatisfied with my eyesight after having the search function focused on "worin". Without bothering about the OCR it's still not a cbcm to see. --Askedonty (talk) 18:10, 19 December 2023 (UTC)[reply]
You're right, that's it. I find a "6,5 em" (6,5 cm) below and also a "Bezeiehnung", so the typeface is not one I'm fully familiarized with. Besides of it, indeed, the resistance here is not "electric", rather something akin to: thermal inertia. --Askedonty (talk) 18:49, 19 December 2023 (UTC)[reply]

Thanks all for your various intelligent and helpful contributions. However, going with cbcm=cm3, I am still at a loss to explain how anyone, let alone the manufacturer, works out a bore and stroke of 95 * 100 mm to displace 8.520 litres? Furthermore, the late Stefan Zima (or his editor) on p. 375 states the volume of a single cylinder of these dimensions to have 0.903 litres. Multiplying by 12 cylinders gives 8.496 litres. But this appears to be as equally 'wrong' as 8.520, since
π = 8.496 * 4,000,000 (=33,984,000) / (952 (= 9025) * 100 * 12 = 10,830,000 = 3.137950... which is far too small a value, even less than the back of an envelope approximation of π = 3.14. I fear that I'm going to have to re-calculate all his values to see if he is consistently using the same value of π.

  • Zima, Stefan (2021) [1992]. "Hochleistungsmotoren 1933 bis 1950". In Eckermann, Erik; Treue, Wilhelm; Zima, Stefan (eds.). Technikpionier Karl Maybach - Antriebssysteme, Autos, Unternehmen. [Originally published as 'Hochleistungsmotoren - Karl Maybach und sein Werk', ed. Zima & Treue] (in German) (3rd ed.). Wiesbaden, Germany: Springer. ISBN 978-3-658-25118-5. + url

Also, if anyone has access to Zima's earlier book (I think a reprint of his Dr.-Ing. thesis), I would be grateful to learn about it.

  • Zima S. (1987). Entwicklung schnellaufender Hochleistungsmotoren in Friedrichshafen. (Reihe: Technikgeschichte in Einzeldarstellungen, Band 44/1987) ["Development of high-speed high-performance engines in Friedrichshafen". (Series: History of technology in individual representations Volume 44/1987) ] (in German). VDI-Verlag. ISBN: 9783181500446.

— Preceding unsigned comment added by MinorProphet (talkcontribs) 07:38, 20 December 2023 (UTC)[reply]

If the person performing the calculations used a slide rule, as was common practice among engineers until well into the 1970s, one can expect this level of inaccuracy.  --Lambiam 10:27, 20 December 2023 (UTC)[reply]
Have you considered the possibility that the volume (which, as far as I understand, is the relevant number to characterise the engine) was measured rather than calculated? The bore and stroke values may have been target values that were not met precisely in manufacturing — in fact, the bore only needs to deviate by 0.08 mm (0.08 per cent) to account for the difference in volume. Blaming π among all the things that could go wrong seems a bit off the mark. Also, when you say that 3.137950... is "far too small", do you realize that the deviation from the actual value is still only 0.12 per cent? --Wrongfilter (talk) 10:43, 20 December 2023 (UTC)[reply]
Thanks again for everyone's helpful comments. Throughout this and my previous query I have only been trying to work out exactly how anyone has come up with any of these figures. I have been waylaid by wondering what they were using for π. I'm sorry for any confusion I have caused. Here's another example from Zima 2021, p. 375: he says that one cylinder of the 6-cylinder NL 42, bore & stroke 90 * 110, has a volume of 0.699 litres. Multiplying this by 6 cyls. = 4.194 litres. The HL 42 manual states 4 198 cm3. But using infobox π/4,
1 cylinder = 0.78539816 * 90 * 90 * 110 / 1,000,000 = 0.699 789 7605 litres. Multiplying that by 6 gives you 4.198 738 563 litres, and that's essentially Maybach's own value. So, using relatively high precision throughout the calculation and taking only the first three decimal places of the answer, there is consistency. Which is all I've been looking for.
However, returning to my OP, Zima states for the V-12 HL 85 a bore & stroke of 95 * 100, giving 1 cylinder = 0.708 litres.
  • My calc of 1 cyl. taking infobox π/4 = 0.78539816 * 95 * 95 * 100 / 1,000,000 = 0.708 821 8394 litres.
  • * 12 cylinders = 8.505 862 0728 litres: but as I said in my OP, the HL85 manual says 8.520 litres. However, if you use 1 cylinder = 0.710 litres (which is 0.7088 rounded up), you get exactly 8.52 litres, which appears again to be somehow consistent, or at least a figure I can reproduce. I'm vaguely happier, anyway. I realise the exact capacites aren't hugely important, but how they are reached is: now I can apply the same procedures to the rest of the figures. Cheers all for the moment, MinorProphet (talk) 11:49, 20 December 2023 (UTC)[reply]

Why is argininosuccinic acid a basic amino acid?

The guanidyl group present can only absorb one proton AFAIK, as all the nitrogens in this functional group are sp2. There are 3 acidic groups and 2 basic groups. Surely arginosuccinate should be classified as an acidic amino acid? But when I look up databases online, it is classified as a "basic amino acid based on pKa" but I am not sure if this is a robotic classification or a human one. EDIT: I also checked out of the NIH ref originally used by the article on wiki, which did not support the initial claim of "basic amino acid." Nevertheless, I see other databases continue to consider it a basic amino acid. Yanping Nora Soong (talk) 23:44, 19 December 2023 (UTC)[reply]

Quick correction, two of the nitrogens (the -NH2 primary amines) are sp3, not sp2 hybridized. One has to take into account their lone pairs when determining hybridization. --OuroborosCobra (talk) 02:40, 20 December 2023 (UTC)[reply]
The guanidinyl NH2 is sp2-hybridized because of resonance. This is the nature of the guanidinium group, which I was taught was the basic equivalent of a carboxylic acid, if a bit stronger (pKb of arginine is 0.2 to 1.2 depending on how you calculate). You would not say that the -OH group in a carboxylic acid group is sp3, would you? My question is, am I missing something here, or is the NIH database entry wrong? Yanping Nora Soong (talk) 18:21, 20 December 2023 (UTC)[reply]
Quick note about provenance not science: that part of the NIH database entry is pulled from the T3DB entry. DMacks (talk) 20:13, 20 December 2023 (UTC)[reply]
That database states its physiological charge is -1, so why would it be classified as a basic amino acid? Was this amino acid classified by unsupervised AI? Yanping Nora Soong (talk) 20:32, 20 December 2023 (UTC)[reply]

December 20

Control of mitochondrial gene expression

The mitochondrial genome is circular plasmid (prokaryotic in origin) while the eukaryotic genome is linear. Are there well-known mechanisms by which the eukaryotic nucleus turn individual mitochondrial genes on and off and exert transcriptional control on the circular mitochondrial genome? When cancer cells inhibit mitochondrial activity, would they for example, just repress the activity of POLRMT instead (on chromosome 19)? Yanping Nora Soong (talk) 02:09, 20 December 2023 (UTC)[reply]

Here is an article reviewing what is known. Summary: it is complicated.  --Lambiam 20:19, 20 December 2023 (UTC)[reply]
Thank you so much! Saving this for further reading (after my exam). Yanping Nora Soong (talk) 20:39, 20 December 2023 (UTC)[reply]

Odd-indexed chemical element stability

Why does the chemical element with odd atomic number each have at most two stable isotopes? Because of the Mattauch isobar rule? Here the the term "stable isotopes" includes also the "observationally stable" ones, such as the only natural isotope of gold (197Au, theoretically it should be able to decay through alpha emission). The case of potassium (19K) is quite stunning as it's the only near-miss to this rule: despite that potassium-40 does decay (unlike two other natural isotopes of potassium), its half-life is very long (comparable to the Age of Earth, so long that any amount of primordial potassium-40 should have survived until now). 2402:800:63AD:E8CE:481:F5E6:57F9:5FDA (talk) 11:28, 20 December 2023 (UTC)[reply]

See Even_and_odd_atomic_nuclei. There are only 48 odd-even and 5 odd-odd isotopes overall for 41 chemical elements. So, as a rule there is only one stable isotopes for each odd numbered element. Ruslik_Zero 20:37, 20 December 2023 (UTC)[reply]
Only 40 chemical elements with odd atomic number have stable or near-stable isotopes (1 to 83, except 43 and 61), also, there may be one or two stable isotopes for each odd number of protons, the same holds for each odd number of neutrons. 2402:7500:916:185A:756A:B526:20F:B8D6 (talk) 03:24, 24 December 2023 (UTC)[reply]
It's pretty much the Mattauch isobar rule, I think. In almost all cases, odd-odd isotopes will not be anywhere near stable: the exceptions are either very light nuclides (where beta decay leads to a too lopsided proton-neutron ratio) or cases with bad spin mismatches. So if nearly all the odd-odd cases are ruled out immediately, that doesn't give you a whole lot of odd-even options; going beyond two takes you too far away from the valley of stability again. Potassium basically lucked out by having two stable odd-even isotopes 39K and 41K at the middle of the valley, plus 40K whose decay is inhibited by a spin mismatch.
When the Solar System was only a few million years old, though, Cl, Tc, La, and Bi would also have been examples, thanks to the quite stable (but not stable enough to be primordial) 36Cl (note 35Cl and 37Cl are stable), 97–99Tc, 137La (139La is stable and 138La almost is), and 208,210mBi (209Bi is almost stable). Double sharp (talk) 05:02, 24 December 2023 (UTC)[reply]

Fractions

Why fractions in unit values on Wikipedia articles cannot be used with metric units, unlike imperial units? Is a value e.g. 5 12 km unacceptable, and why? --40bus (talk) 20:19, 20 December 2023 (UTC)[reply]

That sounds like a Manual of Style issue, not a science issue? But I can easily see that because metric is intrinsically base-10, it is more self-consistent to use decimal values. DMacks (talk) 20:26, 20 December 2023 (UTC)[reply]
Because 5+12 km actually means 5+12 × 103 m, which is a messy mixture of vulgar fraction and decimal exponential. It might work on road signs, but not in an encyclopaedia, please. -- Verbarson  talkedits 21:08, 20 December 2023 (UTC)[reply]
I've never heard fractions described as "vulgar" until now. ←Baseball Bugs What's up, Doc? carrots→ 01:12, 21 December 2023 (UTC)[reply]
Fuck yeah. DMacks (talk) 01:37, 21 December 2023 (UTC)[reply]
See Fraction § Simple, common, or vulgar fractions. -- Verbarson  talkedits 09:01, 21 December 2023 (UTC)[reply]
Vulgar fractions may be, but are not always, improper fractions.
It is more common to mix decimal exponents and fractions like this: 1012, meaning the square root of ten. -- Verbarson  talkedits 09:12, 21 December 2023 (UTC)[reply]
MOS:FRAC has "Metric (SI) measurements generally use decimals, not fractions (5.25 mm, not 5 1⁄4 mm)." Guidance from NIST such as [4] uses examples with decimal points and provides guidelines for how to format decimal numbers with SI units. It does not have any examples or recommendations for using fractions with SI units (although it doesn't explicitly say anything against it either). It does have an item 7 which aims to avoid ambiguity between mathematical operations and other symbols, which could apply here. In general, the metric system is designed to be a consistently decimal-based system, where things are always grouped by tens or divided into tens. (Our article metric system begins "The metric system is a system of measurement that is a decimal system.") Mixing it with other common fractions somewhat defeats the purpose. --Amble (talk) 22:46, 20 December 2023 (UTC)[reply]

December 21

Why the gravitational constant is the physical constant which is the hardest to measure?

Among some physical constants:

  1. speed of light in vacuum
  2. Planck constant
  3. gravitational constant
  4. vacuum permittivity
  5. vacuum permeability
  6. impedance of free space
  7. elementary charge
  8. electron mass
  9. proton mass
  10. Boltzmann constant
  11. Avogadro constant

Why gravitational constant is the hardest physical constant to measure? 61.224.159.150 (talk) 03:19, 21 December 2023 (UTC)[reply]

That question is discussed in detail here:
--136.54.106.120 (talk) 04:23, 21 December 2023 (UTC)[reply]
Almost half of these (speed of light, Planck constant, elementary charge, Boltzmann constant, Avogadro constant) get a free pass. They are not measured; their values were defined by the 26th meeting of the CGPM.  --Lambiam 06:00, 21 December 2023 (UTC)[reply]
Apart from vacuum permeability and impedance of free space, they had to measured first to sufficient accuracy, or there would be an unacceptable uncertainty in our units of measurement.
The gravitational constant is hardest to measure because gravity is so weak. To measure the gravitational constant, you have to measure the gravitational force between two objects and determine their masses in a way independent of that gravitational force. The gravity between Earth and a lead ball is pretty strong, we can easily measure that, but we cannot determine the mass of Earth in any other way than by measuring the gravitational force. Measuring the gravitational force between two lead balls is difficult, as that force is so small and easily disturbed by the electromagnetic force. Which is unavoidable, as the lead ball must be movable by that force, or we couldn't determine its mass. PiusImpavidus (talk) 10:37, 21 December 2023 (UTC)[reply]
This is only for SI units, by your logic, you can say that in Planck units all of (speed of light, Planck constant, gravitational constant, vacuum permittivity, vacuum permeability, impedance of free space, Boltzmann constant, Avogadro constant) get a free pass and not measured. 2402:7500:900:5961:A03B:547A:DD54:8F35 (talk) 02:31, 22 December 2023 (UTC)[reply]
Yes, it should be obvious that meetings of international standards bodies cannot actually change fundamental facts about physics. --Trovatore (talk) 03:38, 22 December 2023 (UTC)[reply]
The question itself implies the use of another system than Planck units. We can eliminate the dependency on any specific system by rephrasing the question and asking why the the ratio of the electrostatic and the gravitational forces between two protons, one of Martin Rees's "Six Numbers", is so hard to pin down. This number differs by at least 36 orders of magnitude from the other five.  --Lambiam 09:40, 22 December 2023 (UTC)[reply]
The dimensionless physical constants includes the fine-structure constant and the gravitational coupling constant, which are the (-2)nd powers of: the elementary charge in Planck charge and the electron rest mass in Planck mass, respectively. (We can compare the Planck units and the Hartree atomic units) 2402:7500:916:185A:756A:B526:20F:B8D6 (talk) 03:29, 24 December 2023 (UTC)[reply]
This gives a value for the dimensionless gravitational coupling constant of 1.751810(39)×10−45.  --Lambiam 10:44, 24 December 2023 (UTC)[reply]
This video Watch gravity pull two metal balls together on YouTube was uploaded an hour ago. DB1729talk 17:39, 22 December 2023 (UTC)[reply]

December 22

Bamboo that blossoms once every 130 years

I read the article Bamboo_blossom. There is no reference to how that synchronous interval counting works. Ant ideas? Zarnivop (talk) 00:02, 22 December 2023 (UTC)[reply]

A bamboo grove is often one genetic individual that spread out by underground rhizomes. Even if there are several individuals, they are from the same cohort. Now, just how they know how many years have passed is a different question. You can read the unsatisfying synchronous flowering article. Abductive (reasoning) 09:17, 22 December 2023 (UTC)[reply]
We are not talking a single grove here. All that genus flowers simultaneously once every 120 year. Zarnivop (talk) 05:34, 23 December 2023 (UTC)[reply]
I heard that populations of bamboo distantly removed from each other might be out of sync. Don't know about this genus. But the "cohort" rule still applies. Abductive (reasoning) 14:48, 23 December 2023 (UTC)[reply]
Quoting from Bamboo blossom § Mechanism [my underlining for emphasis]:
The lack of environmental impact on the time of flowering indicates the presence of some sort of "alarm clock" in each cell of the plant which signals the diversion of all energy to flower production and the cessation of vegetative growth.[1] This mechanism, as well as the evolutionary cause behind it, is still largely a mystery.
 --Lambiam 09:23, 22 December 2023 (UTC)[reply]
The evolutionary pressure is explained in one of the links to one of the associated articles. Basically bamboos with double the intemittency of their competitors will produce more seeds when they bloom, this pressure is one way, so it doubles and doubles and doubles, to 128. https://web.archive.org/web/20150517210059/http://phenomena.nationalgeographic.com/2015/05/15/bamboo-mathematicians/ Greglocock (talk) 22:58, 22 December 2023 (UTC)[reply]
It is an essential part of the article, yes, but it's not relevant to "mechanism" header, which is sorely lacking. Zarnivop (talk) 05:36, 23 December 2023 (UTC)[reply]
The problem is, how do you study it? A grad student has five years to collect all their data and write it up, and you would need variability. Back in August I created a stub on a species, Strobilanthes flexicaulis, (not a grass); it has one variety that does synchronous flowering every six years, and one variety that just flowers yearly. Something like it would be more amenable to study. Abductive (reasoning) 14:52, 23 December 2023 (UTC)[reply]

Only a mere 40 or so years ago, there appeared in The Times a letter from Ronald Neame wondering about just this phenomenon, the cause of which "the local soothsayers and village sybils" of Selling were at a loss to explain.[citation needed] The previous day I had coincidentally found myself on a chalk escarpment overlooking a wide vista of "The Garden of England", and was unfortunately moved to versify my reply; and although the letters editor of that august paper decided not to print my scribblings, I reproduce my effort here just because there isn't enough poetry on the RefDesks. And anyway, as quoted above, "This mechanism...is still largely a mystery."

 
                Selling

High on a hill in Kent
                       I stood
And watched the world go by,
Wondering what all my questions meant
And whether the answers I found were good
If they all led back to Why?

Marvelling at the seasons'
                              change
I laid my questions down —
Somehow all the need for reasons
Fled as I watched the Autumn arrange
His bridal Earth's gold gown.

But the tyrant horizon pushed
                                  my gaze
Past clouds where sunlight cowered:
These loves must fail, desires be hushed,
Our memories will fade of Summer's haze
Because the bamboo has flowered.

MinorProphet (talk) 21:49, 23 December 2023 (UTC)[reply]

(cuogh) Thanks. Irrelevant, but the effort is appreciated. Zarnivop (talk) 09:48, 25 December 2023 (UTC)[reply]

December 23

Is producing tetraploid human embryo possible?

I refer to current technology, and to an embryo that will thrive.

Zarnivop (talk) 05:53, 23 December 2023 (UTC)[reply]

Our Tetraploid complementation assay says: "Such a tetraploid embryo can develop normally to the blastocyst stage and will implant in the wall of the uterus. The tetraploid cells can form the extra-embryonic tissue (placenta, etc.); however, a proper fetus will rarely develop." (links omitted). Note that is only using one kind of tetraploid (fusing the two cells of a two-cell embryo), so it's a doubling of all normal chromosomes rather than a "four parents" model. The most advanced research mentioned there is back from 2009, leading to full-term mice. Presumably more has been done in the last 14 years, so someone who has more-recent info please update our article. DMacks (talk) 06:07, 23 December 2023 (UTC)[reply]

Current dhole population?

Any answers? 20 upper (talk) 13:35, 23 December 2023 (UTC)[reply]

Our article (dhole) cites the IUCN Red List, which identifies Cuon alpinus as endangered, with a population of fewer than 2500 adults in the wild.
The Red List website more specifically puts the number between 949 and 2215, with notes describing the challenges to establishing more precise numbers. Those figures date to 2015, with the caution that the population is exhibiting a downward trend. TenOfAllTrades(talk) 13:43, 23 December 2023 (UTC)[reply]
Thank you. I wonder why the article didn't mention that. I'm really disappointed that such important and instructive information has been omitted. 20 upper (talk) 14:26, 23 December 2023 (UTC)[reply]
The link to the Red List is right in the footnote for the population number, and easy enough to follow. I'd say it's a judgement call whether or not to include a more-specific eight-year-old estimate with very wide uncertainty in a general encyclopedia article. Honestly, the portion of the Wikipedia article on Distribution and Habitat probably tries to capture too many details already—or at least is in need of a cleanup. The information isn't organized by any clear geographical or chronological scheme. TenOfAllTrades(talk) 14:43, 23 December 2023 (UTC)[reply]
The first sentence is completely out of context and doesn't belong there. This is not a research paper, so the article should not go into this much detail. One sentence even stated that "One pack was sighted in the Qilian Mountains in 2006", I'm sure none of us care about that, and I don't see what it adds to the subsection as a whole. Will see what I can do. Any suggestions? 20 upper (talk) 15:12, 23 December 2023 (UTC)[reply]
What's your basis for "none of us care about that"? ←Baseball Bugs What's up, Doc? carrots→ 17:00, 23 December 2023 (UTC)[reply]
I mean to say that it's not important information. 20 upper (talk) 18:05, 23 December 2023 (UTC)[reply]
I for one respectively respectfully disagree. I concur that the section would benefit from some reorganisation as discussed above, but this particular fact, concerning an increasingly rare species, does not seem to me to be too trivial for mention. {The poster formerly known as 87.81.230.195} 90.205.111.170 (talk) 20:38, 23 December 2023 (UTC)[reply]
I, for another, respectfully suggest you meant "respectfully". :) :) -- Jack of Oz [pleasantries] 22:09, 23 December 2023 (UTC) [reply]
Respect to you, since you're right. Odd how one's brain intends one thing but one's fingers type another. Duly corrected. {The poster formerly known as 87.81.230.195} 90.205.111.170 (talk) 02:28, 24 December 2023 (UTC) [reply]
See also this 2021 journal article which I have added to the "Threats" section. Alansplodge (talk) 14:08, 24 December 2023 (UTC)[reply]
@Alansplodge: Fixed typos you introduced. 20 upper (talk) 18:05, 24 December 2023 (UTC)[reply]

December 24

Which pure metal is more prone to corrosion in air: magnesium or samarium?

Title says it all really. I've seen conflicting reports, so it may depend on humidity and other conditions. Double sharp (talk) 04:54, 24 December 2023 (UTC)[reply]

Pure magnesium exposed to carbon dioxide in the air forms a surface layer of magnesium oxide that protects the metal from further corrosion by air. Under humid conditions pure samarium gets a surface layer of samarium(III) oxide, but, according to our article on this oxide, "this oxide layer spalls off the surface of the metal, exposing more metal to continue the reaction".  --Lambiam 10:09, 24 December 2023 (UTC)[reply]
Damn: that set of samarium teaspoons I bought was probably a bad idea. {The poster formerly known as 87.81.230.195} 90.205.111.170 (talk) 00:31, 25 December 2023 (UTC)[reply]
Interesting. If Sm isn't air-stable, then it provides a nice mnemonic for the 4f row: the first seven (La–Eu) rust away completely in air (though at varying speeds), while the second seven (Gd–Yb) don't because the oxide layer protects them. Double sharp (talk) 03:41, 25 December 2023 (UTC)[reply]
I was just saying to my brother that all I want for Christmas is a nice mnemonic for the 4f row. —Tamfang (talk) 20:54, 26 December 2023 (UTC)[reply]

Etymology of hydrogen

When was the word hydrogen (or hydrogène) coined and by whom? Our article on the element says it was 1783 by Antoine Lavoisier. A discussion on the talk page suggests that the earliest instance of the word was 1787 and that Louis-Bernard Guyton de Morveau might deserve credit as well. This paper supports that conclusion, drawing on another 1947 paper by Harold Hartley which says "The name of hydrogen was not adopted until the whole nomenclature was revised four years later." (in 1787). gobonobo + c 16:24, 24 December 2023 (UTC)[reply]

The French Wikipedia has an article on the treatise Méthode de nomenclature chimique by Morveau and others, published in 1787. The etymology section for hydrogène of the TLFi only mentions the 1787 treatise.[5] While this may have been the first publication using the term hydrogène,[6] Lavoisier praises Morveau in the introduction of this treatise for his major role in designing the new nomenclature. Morveau himself uses the spelling Hidrogène.[7] Since he explicitly credits Lavoisier for the name oxygène, it is unlikely the name hydrogène is due to specifically Lavoisier. Morveau acknowledges the role of unnamed members of the Académie, as does Lavoisier in the introduction. Morveau published a treatise on the nomenclature in 1782;[8] it would be interesting to see which name he proposed there for "the base of inflammable air".  --Lambiam 19:32, 24 December 2023 (UTC)[reply]
Thank you Lambiam. This is very helpful. gobonobo + c 16:13, 25 December 2023 (UTC)[reply]

Does calculating the (upper bound of) neutrino mass, rely (also) on the mass energy equivalence?

HOTmag (talk) 18:15, 24 December 2023 (UTC)[reply]

Not that I know of, what gave you hat idea? NadVolum (talk) 18:52, 24 December 2023 (UTC)[reply]
I know that it's not that simple to calculate the neutrino mass, for example it was only possible to calculate the sum of squares of the masses of all three types of neutrinos, and that what the physicists have only achieved is the upper/lower bounds of this sum. For some reason (which is a bit complicated to tell here about) I wanted to know if those calculated values were calculated directly (e.g. by rules related to inertial mass or to gravitational mass and likewise), or indirectly - i.e. by calculating the neutrino energy and then by using the mass energy euivalence.
What did you mean by saying "Not that I know of"? Have you ever dealt (in depth) with the neutrino issue? HOTmag (talk) 22:29, 24 December 2023 (UTC)[reply]
Neutrino masses cannot be calculated, they have to be measured, and the measurements are rather indirect. What is calculated is how some observable quantity depends on the neutrino mass; these predictions are then compared to observations or experimental results to derive a best-fit value for the masses (or upper or lower limit if the measurements are not sufficiently sensitive to distinguish between different mass values). There are two main methods as indicated in the artice you linked to. The first is the observation of neutrino oscillations, which involves the computation of quantum mechanical transition probabilities. I suspect that these computations are not impossible to understand for someone who's done a course or two in quantum mechanics, but I haven't looked at them in detail myself. The other method uses the temperature fluctuation spectrum of the cosmic microwave background. The full computations here are very complex and involve several parameters in addition to the neutrino mass, but according to [9] the constraints on the neutrino masses come essentially from the suppression of small-scale fluctuations compared to large-scale fluctuations. Up to you to decide whether any of this fits in with your directly/indirectly dichotomy. --Wrongfilter (talk) 23:59, 24 December 2023 (UTC)[reply]
Thank you for the new link. The first method you mentioned is used for calculating the lower limit, which I didn't ask about. Before I asked my question, I had already seen the second method indicated in the article I linked to, and I had also read the abstracts of the sources linked to in that article (while the full researches are not available in the web), but unfortunately those abstracts, as well as the new article you've added, don't hint if the way to measure the upper limit (I actually asked about) relies also on the mass energy equivalence. HOTmag (talk) 08:40, 25 December 2023 (UTC)[reply]
You'd do better to engage with people in a cooperative way rather than complaining. Your question might as well have been one of the productions of The Engine from Gulliver's Travels and the article didn't mention anything like that, and yet you just attack when asked how you came by it. Why should anyone bother answering? NadVolum (talk) 22:14, 25 December 2023 (UTC)[reply]
What? Complain? Attack? I'm quite surprised. My first words to the only user who gave me the new link, were "Thank you". I tried to find any words of any attack or of any complaint, but found nothing, so I'm a bit confused now about what you meant.
Maybe you've ascribed - a complaint or an attack - to me, becuase of the (actually) innocent question I asked you, which you didn't interpret well? So let me be clear: I asked this question - about whether you had dealt (in depth) with the neutrino issue, just out of curiosity and hope, after I had read your first response "not that I know of": Your first response made me hope (and I still hope) you were a person who had dealt (in depth) with the neutrino issue: I thought that if you were the person I was looking for, I would ask you (on your talk page) some further questions about the neutrino. That's why I asked you the question mentioned above (which you haven't answered yet BTW). Anyway, if you are not the person I'm looking for (even though I still hope you are), I will be grateful if you could tell me about such a person, because I have some other questions about the neutrino.
HOTmag (talk) 08:04, 26 December 2023 (UTC)[reply]
None of the methods used to derive limits on the neutrino masses uses the mass-energy equivalence explicitely and primarily. However, the cosmological methods do rely on special and general relativity, of which mass-energy equivalence is also part. It becomes most explicit in the fact that particle masses contribute to the energy budget, i.e. have to be taken into account when writing down the energy density, which along with pressure is the source of gravity. That's almost trivial and doesn't describe the actual method, which rather tries to determine when neutrinos (from the cosmic neutrino background) became non-relativistic, i.e. when their average kinetic energy dropped below their masses (my preferred formulation is ). For a current list of upper limits to the sum of the neutrino masses see here (with indication of methods and data sets as well as links to the original papers); for a summary review see here (Sect. 26.2.3 and 26.2.4 in particular). Note that these methods constrain the sum of the masses; neutrino oscillations constrain the differences of squared masses. I'm not aware of any method that would constrain the sum of squared masses (but my knowledge of the topic is far from complete). --Wrongfilter (talk) 09:30, 26 December 2023 (UTC)[reply]
Thanks a lot. I appreciate your full answer. HOTmag (talk) 10:43, 26 December 2023 (UTC)[reply]

December 25

Replacing a dishwasher

Is this a built-in dishwasher? Standard installation from say Home Depot won't cover it, will they? Hopefully I'll get a better photo, but this one will have to do for now. Thank you. 68.174.155.22 (talk) 04:15, 25 December 2023 (UTC)[reply]

That looks built-in to me. Looking at [10], you can provide your zip code and the Home Depot will connect you with a pro that can install the new one. RudolfRed (talk) 04:21, 25 December 2023 (UTC)[reply]
I would interpret this as indicating that Home Depot will install a built-in dishwasher, as long as a water supply with a shut-off is located appropriately, and local codes don't require a licensed plumber to do it. I'd suggest calling Home Depot or using the chat feature on their website to confirm. CodeTalker (talk) 05:50, 25 December 2023 (UTC)[reply]
Is it a drop-in, which HD doesn't do? "Basic hook up service does not include hook up or removal of built-in cooktops, wall ovens, drop-in appliances, range hoods and air conditioners." [11] The thing about "Pro" services is that it seems to be just another expensive marketplace for odd jobs. 68.174.155.22 (talk) 23:47, 25 December 2023 (UTC)[reply]
What is a "drop-in" appliance? I've never seen the term before and can't readily find an explanation online. This page from one major manufacturer refers to dishwashers only as "built-in" on the one hand, "portable" or "freestanding" on the other. --142.112.220.136 (talk) 14:16, 26 December 2023 (UTC)[reply]
Beats me. The term is used with ranges, [12] which may be installed flush with surrounding surface rather than beneath the counter. Another drop-in range claims it may be used in "an island or peninsula." 68.174.155.22 (talk) 22:30, 26 December 2023 (UTC)[reply]
As used here, it is an appliance designed to hang from the countertop, without bottom support.  --Lambiam 09:58, 27 December 2023 (UTC)[reply]

December 26

Why so many unpublished studies?

I was reading this article from Scientific American, How Two Pharmacists Figured Out That Decongestants Don’t Work, and something caught my attention:

"the oral decongestant monograph panel reviewed a few published studies and multiple unpublished studies for phenylephrine".

Why would there be so many unpublished studies? I'm not so concerned about the particular drug in question but science in general. What's the point of conducting a study if it doesn't get published? Do such studies get rejected because of some sort of flaw? Did the authors not find a publication willing to publish their study? Do scientists conduct studies with no intention of publication?

Sorry, I'm asking too many questions and speculating on the possible answers I guess what I am really asking is

  1. How common is it to have unpublished studies?
  2. What are the most common reasons for a study not to be published?

Thanks! Pealarther (talk) 11:04, 26 December 2023 (UTC)[reply]

The vast majority of science done by industrial firms is never published. Suppose you want to find a new decongestant and test thousands of potential drug candidates. Most won't work and these negative results will sit in company databases but are unlikely to be suitable for inclusion in reputable journals. Some aspects of such work may be reported in internal documents but that's not "publication": secrecy is the norm so as not to alert competitors to what has been tried and failed. Mike Turnbull (talk) 11:47, 26 December 2023 (UTC)[reply]
It is not difficult to get results published because there are competing vanity publishers who have no interest in whether a book is suitable for publication or good enough to sell. In professions that are obsessed more with quantity rather than with quality, the Publish or perish phenomenon arises and may contribute to a Replication crisis where reported results of scientific studies are difficult or impossible to reproduce. Some reasons for rejection of results from Wikipedia include unverifiable claims, lack of peer review, outdated material or conflicts of interest that may lead to covert advertising. Scientists who find the peer-review and publication process to be too slow often choose to release results in Preprints. There are research results in mathematics, physics, astronomy, electrical engineering, computer science, quantitative biology, statistics, mathematical finance and economics that have been published only on the online-accessible preprint server arXiv that receives about 16,000 articles per month. Philvoids (talk) 12:06, 26 December 2023 (UTC)[reply]
It's one of the known issues with clinical trials. Historically, an organizations (public or private, academic or commercial) could conduct a clinical trial, and be left to their own devices and schedule on when or whether to publish their results, and how to conduct their analyses of the data.
This led to a number of problems, in various flavors of publication bias and selective reporting. A drug company might choose not to publish (or to slow-walk publication of) a result that failed to show efficacy for their novel drug candidate. An academic lab might lose interest in a study that failed to produce a novel result, and shelve their data in order to preserve and pursue funding for more 'interesting' studies. A clinical trial group might choose to only report a subset of their findings to amplify or suppress certain results. Worse, they might opt to re-analyze their data to identify 'effects' that they hadn't planned for or predicted when setting up the trial: HARKing and other sorts of post-hoc analysis.
It's only fairly recently that clinical trial pre-registration has become the norm (at least in North America and Europe). TenOfAllTrades(talk) 14:45, 26 December 2023 (UTC)[reply]
Trials that fail to show a useful result are particularly liable to never be published. This is a very real problem and probably affected a number of those studies. And some of those test may have been done because they hadn't heard about other failed trials so it was a wast of effort too. Documenting negative results is very important but it is only recently that people have started to recognize this and there are now journals that have been set up specifically to publish them. The bias in favor of only positive results also contributes to the replication crisis mentioned above. Hopefully the trial pre-registration mentioned in th previous response will also contribute towards rectifying this problem. NadVolum (talk) 11:01, 27 December 2023 (UTC)[reply]

max use temperature in Ar/N2 > max use temperature in vacuum

Hi. This is site[13] lists boron nitride crucibles' maximum use temperature as 900 °C in air and 1800 °C in vacuum. "maximum use temperature in vacuum is higher than maximum use temperature in air" makes perfect sense.

However the max use temperature in Ar/N2 is actually higher than the vacuum temperature. This fact seems counter-intuitive (at least to me).

Q1: What is the reason behind this? (max use temperature in Ar/N2 > max use temperature in vacuum)

Q2: Is there any high temperature crucible materiel for which this isn't true? That is to say, for this hypothetical material: max use temperature in vacuum > max use temperature in Ar/N > max use temperature in air.

Liberté2 (talk) 19:59, 26 December 2023 (UTC)[reply]

  1. 2: no. Inert gases valence being complete, the vicinity with their molecules act like an isolator, a bit as a "punching ball" effect. --Askedonty (talk) 22:15, 26 December 2023 (UTC)[reply]

December 27

Is it possible that europium is more unstable than bismuth?

The only natural occurring isotope of bismuth, Bi209, has a half-life of 2.01*10^19 years, and for europium, Eu151 has a half-life of 5*10^18 years, and the lower bound of the half-life of Eu153 is 5.5*10^17 years, so is it possible that europium is more unstable than bismuth? And hence europium will be the most unstable natural occurring element besides thorium and uranium? 111.253.202.97 (talk) 02:15, 27 December 2023 (UTC)[reply]

Probably not. According to this paper, all theoretical predictions for the alpha half-life of 153Eu are over 10140 years, and no other decay modes (besides spontaneous fission) are possible. Double sharp (talk) 03:26, 27 December 2023 (UTC)[reply]
So what element will be the next "unstable" natural occurring element after uranium, thorium, and bismuth? 2402:7500:944:2AC:45B9:41E0:7299:F0CC (talk) 03:59, 27 December 2023 (UTC)[reply]
Based on the tables in that paper, perhaps rhenium? 187Re is already known to be beta-unstable with a half-life of about 4×1010 years, and 185Re is expected to be alpha-unstable with a half-life on the order of 1024–1025 years. Though I note that the tables don't consider double beta decay (relevant for other elements), so this should be taken as just a guess. Double sharp (talk) 04:07, 27 December 2023 (UTC)[reply]
The longest known half-life of a nuclide is 2.2*10^24 years for Te128, thus maybe the half-life of Re185 is shorter than Te128? (Although their decay modes are different: Re185 is alpha decay while Te128 is double beta decay) 2402:7500:944:2AC:E564:A5C5:C845:76AC (talk) 04:57, 27 December 2023 (UTC)[reply]
Since predictions aren't perfect, it may indeed be that this is the case. (Predictions slightly overestimated the half-life of nearby 184Os.) Double sharp (talk) 05:53, 27 December 2023 (UTC)[reply]
OK, it is for the proton number, the proton numbers with primordial nuclides are 1~83 except 43 and 61, together with 90 and 92, besides, 83, 90, 92 are the proton numbers among them which have no stable nuclides, and perhaps 75 is the next "unstable" proton number after 92, 90, 83. But what is the answer for the neutron number? The neutron numbers with primordial nuclides are 0~126 except 19, 35, 39, 45, 61, 89, 115, 123, together with 142, 143, 146, besides, 21, 142, 143, 146 are the neutron numbers among them which have no stable nuclides, and the next "unstable" neutron number after 143, 21, 146, 142 may be 71 (corresponding to Te123)? If so, what number will be the next "unstable" neutron number after 143, 21, 146, 142, 71? 2402:7500:942:3D81:695F:5302:9A90:FEE6 (talk) 07:47, 27 December 2023 (UTC)[reply]
Probably either 87 (149Sm should be alpha-unstable at 1017–1018 years) or 111 (187Os should be alpha-unstable at 1016–1019 years). Either of these might come before 71, actually. Double sharp (talk) 08:08, 27 December 2023 (UTC)[reply]

Is there a reference I can use to distinguish between electron capture and β+ decay? While writing about isotopes of bromine, I encountered a reference describing the isotope 77Br as primarily undergoing electron capture rather than β+,[1] but NUBASE2020 lists only β+ for most nuclei capable of β+ decay, inluding 77Br.[2]LaundryPizza03 (d) 08:19, 27 December 2023 (UTC)[reply]

References

  1. ^ Kassis, A. I.; Adelstein, S. J.; Haydock, C.; Sastry, K. S. R.; McElvany, K. D.; Welch, M. J. (May 1982). "Lethality of Auger Electrons from the Decay of Bromine-77 in the DNA of Mammalian Cells" (PDF). Radiation Research. 90 (2): 362. doi:10.2307/3575714. ISSN 0033-7587.
  2. ^ Kondev, F. G.; Wang, M.; Huang, W. J.; Naimi, S.; Audi, G. (2021). "The NUBASE2020 evaluation of nuclear properties" (PDF). Chinese Physics C. 45 (3): 030001. doi:10.1088/1674-1137/abddae.

UK metric system (update)

The UK government has announced plans to make it legal to sell wine by the pint. No more messing about with those fiddly little wineglasses, then. -- Verbarson  talkedits 11:58, 27 December 2023 (UTC)[reply]

Last passenger

After pulling the emergency brake in a train (and in particular a British one), is there any way to release the brakes without recharging the brake pipe? I think not (based on what I know of trains), feel free to correct me if I'm wrong! 2601:646:8080:FC40:7042:7EFC:5F93:408A (talk) 12:01, 27 December 2023 (UTC)[reply]

Where was the camera on the Apollo lunar lander?

I'm trying to better understand how NASA filmed the Apollo 11 lunar landing considering obviously there was no one on the Moon to film it. I found this picture which is supposed to show the camera on the lander but I am not seeing it. Nothing that resembles a traditional camera is in the image as far as I can see. Can someone point out where in the image is the camera? A Quest For Knowledge (talk) 12:09, 27 December 2023 (UTC)[reply]