Wikipedia:Reference desk/Archives/Science/2023 December 19

= 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. (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)
 * π = capacity * 4,000,000 / (bore2 * stroke * cylinders)
 * I think, said Ouro (blah blah) 11:35, 19 December 2023 (UTC)
 * 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)
 * Hence the I think bit. Just have to move the decimal point around. --Ouro (blah blah) 11:59, 19 December 2023 (UTC)
 * 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)
 * 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)

Thanks, that's great. Going with '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)
 * 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)
 * 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)
 * 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). 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)
 * No, capacity is on swept volume, not affected by the combustion chamber's shape. Greglocock (talk) 17:33, 19 December 2023 (UTC)
 * The observation that the shape is irrelevant was also made in the earlier discussion at (search for "Cavalieri's principle").  --Lambiam 22:43, 19 December 2023 (UTC)
 * 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)
 * @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, 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)
 * 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)
 * 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)
 * 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)
 * 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)
 * 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)

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.
 * + 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. — Preceding unsigned comment added by MinorProphet (talk • contribs) 07:38, 20 December 2023 (UTC)
 * 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.
 * 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)
 * 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)
 * 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)

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)


 * 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)


 * 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)
 * 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)
 * 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)