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January 9[edit]

Alcohol-based hand sanitizer[edit]

Does alcohol-based hand sanitizer lose alcohol content to a significantly degree over time? If so, is it known how fast alcohol content is lost? In a half-used bottle of hand sanitizer, does the existence of a significant volume of air-filled space in the bottle increase the rate of alcohol content loss? --100.34.204.4 (talk) 04:44, 9 January 2017 (UTC)[reply]

If left open, then yes, the alcohol, being more volatile, will evaporate first. But the amount of air in any portion of a sealed bottle won't hold much of the total alcohol volume as partial pressure. I suppose continuously opening the bottle top would allow the alcohol to escape at each occurrence, cumulatively reducing the percentage by a significant amount, but hand sanitizer bottles are typically designed so they don't require that. StuRat (talk) 05:10, 9 January 2017 (UTC)[reply]

Steam locomotive speed[edit]

What maximum speed a train hauled by a steam locomotive could attain? --IEditEncyclopedia (talk) 06:09, 9 January 2017 (UTC)[reply]

About 200 km/h (125 mph). See Land_speed_record_for_rail_vehicles#Steam. Notably, the world record holder was going downhill at the time and the engine broke in the process. The level grade record holder is not far behind. Someguy1221 (talk) 06:16, 9 January 2017 (UTC)[reply]

If the "could" is a question about possibility, not historical reality, then I suspect the answer would a lot higher for a purpose-build race machine - I see no particular reason why a train driven by e.g. steam turbine shouldn't be able to beat that. --Stephan Schulz (talk) 08:59, 9 January 2017 (UTC)[reply]

Here is one reason. See Hunting_oscillation. So it could be done but you'd need special rail tracks as well. 196.213.35.146 (talk) 10:18, 9 January 2017 (UTC)[reply]

Yes, but that is not caused by the particular motive force - as the other records show, a conventional train on conventional tracks can go over 300 km/h. --Stephan Schulz (talk) 10:28, 9 January 2017 (UTC)[reply]

The Mallard set a record for steam of 202.58 km/hr (125.88 mph) in July 1938. That record was never broken. Akld guy (talk) 09:19, 9 January 2017 (UTC)[reply]

False precision. The dynamometer-car record was only read in whole miles per hour: that's 126 mph. — Preceding unsigned comment added by 69.159.60.210 (talk) 08:03, 10 January 2017 (UTC) ^{[citation needed]}[reply]

No doubt you have a reference for that claim. Akld guy (talk) 03:02, 12 January 2017 (UTC)[reply]

Classical steam locomotives (reciprocating engine, direct drive) have a very large reciprocating mass. This makes it very hard to have their wheels turn at more than about 10 revolutions per second. To go fast, they need big wheels; Mallard's driving wheels were 6 ft 8 in (2.032 m) in diameter and making them much bigger would be hard and give poor acceleration.

The speed could be increased by, for example, using a steam turbine and electrical transmission and some of these locomotives were indeed build, but the technology only became mature in the 1930s. By then, some countries, like the UK, still tried to push the speed of steam (building locomotives like the Mallard), while others like Germany focussed on diesel-electric propulsion (Fliegender Hamburger) or, like Italy, electric propulsion (ETR 200). Soon it became obvious that the future of fast trains would be electric (although diesel would play a transitional role) and development of fast steam locomotives stopped. PiusImpavidus (talk) 10:24, 9 January 2017 (UTC)[reply]

Also, while steam trains could reach 126 mph, they could only maintain such speeds for a few minutes before things would begin to fall apart -- in fact, during that record run, the Mallard actually suffered a broken big-end bearing and a hotbox (both due to frictional heating of the parts from overspeeding). Diesel and electric trains, on the other hand, could maintain their maximum speed indefinitely as long as the track was clear -- which meant a higher average speed even if the maximum speed was the same. Add to this better low-end torque with the electric transmission (hence better acceleration) and a lower center of gravity (hence higher speeds around curves) and you can see why fast steam trains didn't see any further development. 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 12:50, 9 January 2017 (UTC)[reply]

I remember reading, back in the 1980s when I professionally edited a locomotive-centred publication, that in contrast to Mallard's official record-breaking but somewhat self-destructive run (alluded to by Someguy1221, Akld guy, PiusImpavidus and 2601), contemporary US passenger express locos not infrequently ran at up to 120mph without problems, and could readily have pushed the record beyond that of Mallard had it been desired. I suspect little publicity was sought for these performances because they were probably done unofficially to make up time from late running, and the record was not pursued on either passenger or test runs for fear of frightening off customers. (Sadly, the extensive railway library I built up back then was the property of my employers, not myself, so I have long since lost access to it.) {The poster formerly known as 87.81.230.195} 2.122.62.241 (talk) 23:42, 9 January 2017 (UTC)[reply]

Weeeeellll...it's possible, but since the records don't exist, it's hard to say for certain. The experimental PRR S1 (only one built) was reported in Popular Mechanics to have exceeded 133 mph, and various sources have made claims for speeds as high as 156 mph. The Milwaukee Road class A locos were designed for a cruise in excess of 100 mph and a top speed of at least 120 mph (and probably could do slightly better than that, though again we don't have official records). The Milwaukee Road class F7 locos were even faster and more powerful, expected to exceed 100 mph in daily scheduled service and having been observed at at least 125 mph.

As our article notes, the F7s had to maintain the fastest average speed of any scheduled steam locomotive in history, completing the 78.3 miles between Portage and Sparta (Wisconsin ~~Washington~~) in 58 minutes, for an average stop-to-stop speed of 81 mph. TenOfAllTrades(talk) 22:23, 10 January 2017 (UTC)[reply]

You mean Sparta, Wisconsin -- there's no Sparta in Washington, and the F7s (to my knowledge) were never used on the Olympian Hiawatha (the only Milwaukee train going to Washington). 2601:646:8E01:7E0B:6CD5:FDD3:C2B8:18E8 (talk) 04:00, 12 January 2017 (UTC)[reply]

Yes, definitely. Fixed! I also entirely forgot the PRR T1s, for which there are anecdotal claims of operation at speeds up to 140 mph. TenOfAllTrades(talk) 04:06, 12 January 2017 (UTC)[reply]

There's some mention here [1] of others possibly exceeding the speed. It also mentions how other stuff like DRG Class 05 seems to have come very close. Nil Einne (talk) 15:01, 11 January 2017 (UTC)[reply]

Feynman Lectures. Lecture 6 (6-3, 6-4) [2], Lecture 41 (41-4) [3], Lecture 43 [4]. Probability = 33%[edit]

According to the example of Lect. 6 we can calculate the probability of distance after 30 steps with length S_rms=1. The probability to go farther σ = P(D>σ) = $\int _{-\infty }^{-\sigma }p(x)\,dx+\int _{+\sigma }^{+\infty }p(x)\,dx=2\int _{+\sigma }^{+\infty }p(x)\,dx=2\int _{+\sigma }^{+\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}\,e^{-x^{2}/2\sigma ^{2}}\,dx$ .
For σ=S_rms√30 we have P(D>σ)=0.317.
If the step is fixed as +1 or -1 , then we have the probability = 0.362 JPG xmcd.

According to my previous question [5] if we have e.g. 100 atoms in 100 m³ the probability not to find any atom in 1 m³ = (99/100)¹⁰⁰ = 0.366.

According to Lecture 43 (43-1) the probability that the molecule avoids a collision for a time equal to τ (average time between the collisions) is e⁻¹≈0.37.

Is there any connection between all these probabilities?

Username160611000000 (talk) 11:01, 9 January 2017 (UTC)[reply]

You wrote yourself that one is some value of the error function, another is 0.99^100 and the last is exp(-1). So no.

The closeness of the last two can somewhat be explained by

\lim _{n\to +\infty }(1+a/n)^{n}=\exp(a)

. Tigraan^{Click here to contact me} 16:14, 9 January 2017 (UTC)[reply]

- Thank you very much. Username160611000000 (talk) 18:08, 9 January 2017 (UTC)[reply]

Feynman Lectures. Lecture 43. Ch.43-6 Thermal conductivity [6][edit]

...
The thermal conductivity κ is defined as the ratio of the rate at which thermal energy is carried across a unit surface area, to the temperature gradient:

${\frac {1}{A}}{\frac {dQ}{dt}}=-\kappa {\frac {dT}{dz}}.$ (43.41)

Since the details of the calculations are quite similar to those we have done above in considering molecular diffusion, we shall leave it as an exercise for the reader to show that

$\kappa ={\frac {knlv}{\gamma -1}},$

— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Using arguments of Ch. 43-5 I try next: $J={\tfrac {n_{\text{hot}}\cdot v_{\text{hot}}\Delta T-n_{\text{cold}}\cdot v_{\text{cold}}\Delta T}{\Delta T}}$

If . . $n_{\text{hot}}=n_{\text{cold}}=n$ . . we have $J=n(v_{\text{hot}}-v_{\text{cold}})$

$-(v_{\text{hot}}-v_{\text{cold}})={\tfrac {dv_{a}}{dx}}\Delta x={\tfrac {dv_{a}}{dx}}\cdot 2l$

${\tfrac {d(v_{a})^{2}}{dx}}\sim {\tfrac {dT_{a}}{dx}}$

$m{\tfrac {d(v_{a})^{2}}{dx}}=3k{\tfrac {dT_{a}}{dx}}$

$m{\tfrac {2vdv_{a}}{dx}}=3k{\tfrac {dT}{dx}}\Rightarrow {\tfrac {dv_{a}}{dx}}={\tfrac {dT}{dx}}\cdot {\tfrac {3k}{2mv}}$

$\therefore J=n(v_{\text{hot}}-v_{\text{cold}})=-n{\tfrac {dv_{a}}{dx}}\cdot 2l=-n{\tfrac {dT}{dx}}\cdot {\tfrac {3k\cdot 2l}{2mv}}$
$=-{\tfrac {dT}{dx}}\cdot {\tfrac {3nkl}{mv}}=-{\tfrac {dT}{dx}}\cdot {\tfrac {3nkl{\tfrac {v}{2}}}{mv{\tfrac {v}{2}}}}.$
$J\cdot {\tfrac {mv^{2}}{2}}=-{\tfrac {dT}{dx}}\cdot {\tfrac {3nklv}{2}}$
$J\cdot {\tfrac {mv^{2}}{2}}={\tfrac {dN}{dt}}{\tfrac {1}{A}}\cdot {\tfrac {mv^{2}}{2}}={\tfrac {dQ}{dt}}{\tfrac {1}{A}}$

$\therefore {\tfrac {dQ}{dt}}{\tfrac {1}{A}}=-{\tfrac {dT}{dx}}\cdot {\tfrac {3nklv}{2}}$ .

$(\gamma -1)=2/3$

Is it correct ? I'm not sure that in 1-dimentional case we can write mv²=3kT. For 1 degree of freedom we have 1kT.

Username160611000000 (talk) 18:05, 9 January 2017 (UTC)[reply]

Why are the leaves still red?[edit]

In January, a tiny plant in my yard still has red leaves. They look like maple leaves, specifically the Acer pseudoplatanus photo with that article, and there are maple trees across the street. However, full-size trees have pretty much lost their leaves, or at least the ones left on trees have turned brown. This plant, a couple of inches tall, is even red below the leaves (the "trunk"). I never saw that.— Vchimpanzee • talk • contributions • 22:11, 9 January 2017 (UTC)[reply]

Were the leaves green before? Apparently the color leaves turn depends on what was in them when the chlorophyll dies off, apparently red leaves have left over "food" in them while brown leaves are more depleted. Vespine (talk) 22:43, 9 January 2017 (UTC)[reply]

A further possible factor: being so small, the plant may be in a more benign microclimate than the conditions to which nearby full-sized trees are overall subject. The article Deciduous may point to some clues as to why your yard-sheltered plant is privileged.

Re the trunk/stem, I've casually noticed myself that some saplings tend to have thinner bark containing a degree of chlorophyll, as contrasted to more mature specimens of the same species, so the sapling's bark can be expected to emulate the leaves' colour changes. {The poster formerly known as 87.81.230.195} 2.122.62.241 (talk) 23:55, 9 January 2017 (UTC)[reply]

There are a number of maple cultivars that have red leaves year round, see for example the later parts of Acer palmatum article. If it has red leaves year round then it may be either one of those, or a natural mutation. Watch that plant for a few years, and see if the trait was transient or persists. Dr Dima (talk) 03:07, 10 January 2017 (UTC)[reply]

Poinsettias are prized for that behavior. StuRat (talk) 04:21, 10 January 2017 (UTC)[reply]

Ah, but the red bits of a poinsettia are bracts rather than leaves. Alansplodge (talk) 13:03, 11 January 2017 (UTC)[reply]

From our article: "a bract is a modified or specialized leaf". StuRat (talk) 16:34, 11 January 2017 (UTC)[reply]

New shoots of many plants are red. As TPFKA says, trees do slightly different things based on their age/height/microclimate/position in the canopy. Generally speaking, understory deciduous temperate trees will have both earlier bud break and keep their leaves longer. This change in phenology is thought to take advantage of the extra light before the canopy_(biology) closes and after it has left in the autumn. Here is a really great freely accessible article on the topic: Differences in leaf phenology between juvenile and adult trees in a temperate deciduous forest Augsperger and Bartlett (2003). All this is to say: it is perfectly reasonable to expect to see seedlings/saplings keep their leaves longer than their full-sized peers. SemanticMantis (talk) 17:24, 10 January 2017 (UTC)[reply]

Update: I forgot about this when it snowed. Now that the snow has melted, the leaves are brown and have fallen off, and the "trunk" is turning brown as well.— Vchimpanzee • talk • contributions • 21:54, 13 January 2017 (UTC)[reply]

Where can you drive through a redwood?[edit]

I didn't find anything on Wikipedia using the normal methods, but Chaz Henry on KKOV says you can't do it any more. A storm has knocked down that tree. I don't know who Chaz Henry is because he never identifies his employer, but sources for this should be easy to find. However, I don't know what Wikipedia article would require updating.— Vchimpanzee • talk • contributions • 22:13, 9 January 2017 (UTC)[reply]

That would probably be the Pioneer Cabin Tree in Calaveras Big Trees State Park. This tree has been all over the local news in Northern California. For example: Remembering California’s storm-toppled historic Pioneer Cabin Tree (from the San Jose Mercury News, January 8, 2017).

I think I mentioned earlier this weekend about how bad the weather was - it's been really wild out there!

Nimur (talk) 22:53, 9 January 2017 (UTC)[reply]

Interested readers, see also: the Wawona Tree and the rest of the trees on the list of largest giant sequoias; and the Chandelier Tree, which is a close relative, a Coast Redwood. Nimur (talk) 22:56, 9 January 2017 (UTC)[reply]

One thing I haven't seen with the pictures that were published: did the Pioneer Cabin Tree crack at the height of the car tunnel? I'm thinking that digging a tunnel through a sequoia may be cool for a hundred years, but in sequoia terms, it's pretty much a killing blow? Wnt (talk) 00:20, 10 January 2017 (UTC)[reply]

Comparing these two pics, taken from near the same vantage point (note the large block of wood on the right side), it looks like it fell backwards from its burn-marked side and the top part of its roots ripped out. One problem with sequoias is that they don't have a particularly sturdy root system. [7][8] ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:24, 10 January 2017 (UTC)[reply]

There are several drive-thru trees in the Redwood Forest in far Northern California. [9] Killiondude (talk) 01:02, 10 January 2017 (UTC)[reply]

Our article links to Chandelier Tree but mentions it is a "coast redwood not a giant sequoia". (Also mentioned by Nimur above.) Looking at the above list, the species isn't mentioned. A quick search didn't find the species for the other two, so I stopped looking. I get the feeling they're probably coast redwoods though. Nil Einne (talk) 06:08, 10 January 2017 (UTC)[reply]

A simple search finds [10] appears to be the person you're referring to but I'm not sure the relevance of his identity to the question. Nil Einne (talk) 05:11, 10 January 2017 (UTC)[reply]

He wouldn't be the only news source reporting this, so it's not that important. We could assume others reported the same tree.— Vchimpanzee • talk • contributions • 16:15, 10 January 2017 (UTC)[reply]

Not sure why there's need to assume anything. Nil Einne (talk) 19:14, 10 January 2017 (UTC)[reply]

As for giant sequoias tunneled through, see this page. --69.159.60.210 (talk) 08:08, 10 January 2017 (UTC)[reply]

Engineering question: fit, clearance and tolerance[edit]

In reference to something written here (but copied and pasted here), I wonder how to consider this because, in my mind, I've got it backwards:

"When two parts are to be assembled, the relation resulting from the difference between their sizes before assembly is called a fit. A fit may be defined as the degree of tightness and looseness between two mating parts."

The important terms related to the fit are given below:

Clearance

In a fit, this is the difference between the sizes of the hole and the shaft, before assembly, when this difference is positive. The clearance may be maximum clearance and minimum clearance. Minimum clearance in the fit is the difference between the maximum size of the hole and the minimum size of the shaft.

Interference

It is the difference between the sizes of the hole and the shaft before assembly, when the difference is negative. The interference may be maximum or minimum. Maximum interference is arithmetical difference between the minimum size of the hole and the maximum size of the shaft before assembly. Minimum interference is the difference between the maximum size of the hole and the minimum size of the shaft."

I have underlined the part that I don't understand and italicized the corresponding part that I do understand. In terms of what I do understand, interference is when there is what we'd colloquially refer to interference; in other words, item X interferes with item Y. So minimum interference would be a maximum hole and a minimum shaft attempting to pass through that hole. But what I don't understand is why minimum clearance is maximum hole with a minimum shaft attempting to pass -- to be, this seemed to provide the maximum (magnitude of) clearance as 'clearance' is understood colloquially. I suppose it could be a typo, but more likely, I speculate it's merely some mathematical convention of assigning clearance with a negative, and so it's the inverse of what may appear to make sense colloquially -- or perhaps I've just got it more wrong than I think. Thanks to whomever is able to help. DRosenbach ^{(Talk | Contribs)} 23:23, 9 January 2017 (UTC)[reply]

I agree that Minimum clearance in the fit is the difference between the maximum size of the hole and the minimum size of the shaft seems to be a lazy cut and paste. perhaps they just meant maximum. Greglocock (talk) 06:34, 10 January 2017 (UTC)[reply]

I also agree. The source of the error is Indira Gandhi National Open University whose website provides no general e-mail address, only telephone numbers. (One might try contacting directorsoss@ignou.ac.in). Someone near New Delhi could inform IGNOU politely that the mistake in their course material needs correcting. Blooteuth (talk) 12:47, 10 January 2017 (UTC)[reply]

You are right, they meant to write "maximum clearance in the fit". Clearance is positive, there is no weird sign-flip convention. Here is a more clear definition of min/max clearance, with a worked example [11]. Here [12] is another definition of "clearance", which is synonymous with the maximal clearance of the previous ref.

I'm not an engineer, but I'd think relevant definitions of clearance should be added to Engineering_tolerance. Engineering_fit uses the categorical notion of "clearance fit", but does not give any info on clearance as a quantity.

Perhaps this use of min/max clearance is slightly deprecated, but persistent? Because the quality and number of references is surprisingly low... SemanticMantis (talk) 15:47, 10 January 2017 (UTC)[reply]