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September 8

Ball valve

Right now the construction of my pilot plant is going more-or-less OK, but there's one slight problem: the shutoff valves to the main pump are frozen solid in the fully open position after almost a whole year of disuse. The one on the discharge side (which is the more critical) I was able to wiggle free, but the one on the suction side (being a cheaper brand) remains frozen despite my utmost efforts to the contrary, as does the only spare for it. So can anyone advise me on how to free a jammed ball valve (if possible, without introducing any chemical contaminants into the piping)? 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 10:59, 8 September 2017 (UTC)[reply]

What is your pilot plant for? Blooteuth (talk) 12:35, 8 September 2017 (UTC)[reply]

Synthetic turquoise. 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 06:41, 9 September 2017 (UTC)[reply]

An impact driver might loosen it - but it might also break the valve handle and stem (and/or the pipe joints). Moving it every so often so it would not seize up would been preferable, but a little too late for that now. WegianWarrior (talk) 12:38, 8 September 2017 (UTC)[reply]

For the same drawbacks but less expense (if you don't own an impact wrench), you might try increased the lever arm length to get more torque. Putting a piece of pipe over the handle to create a cheater bar might work, but they are dangerous, so consider the alternatives listed there, too (and kicking the end of the cheater bar instead of putting your body weight into it is a bit safer). Or just use a large pipe wrench to grab hold of the handle (other types of wrenches typically aren't as large). There's also changing the temps of the inner and outer materials, in opposite directions, to get one to expand and one to contract, but that's tricky to do. StuRat (talk) 15:25, 8 September 2017 (UTC)[reply]

You really only have three choices. [1] Drop your unneeded "without introducing any chemical contaminants" requirement, remove the valve from the system, soak it in a penetrating oil for a few days, then clean it before reinstalling. [2] Hit it with a blowtorch and see whether it comes unstuck or whether you just ruined your valve with a blowtorch. [3] Apply more and more force until it either breaks loose or simply breaks, causing you to buy a new one.

There is another lesson to be learned here. From now on, whenever you build a system, make a maintenance plan and follow the plan. In your case, one of the items should be moving every valve once every two weeks. Make a checklist with things like "inspect pipe 23A for external corrosion" and "clean air filters on Unit 17B". It's part of the difference between being a professional and being a garage shop. --Guy Macon (talk) 15:55, 8 September 2017 (UTC)[reply]

Thanks! I think I'll go with #3 -- I happen to have a big pipe wrench, and I'll use it. (#1 is kind of hard to do, given that the valve is a one-piece design, and #2 is dangerous because the plant is housed in a wooden building -- you don't want to use a blowtorch around wood if you don't absolutely have to!) 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 06:41, 9 September 2017 (UTC)[reply]

I'm reminded of the time I had a car with a flat tire, tried to change it, and one bolt just would not come loose. I used the breaker bar method, and it still wouldn't budge with me pushing on the end of the rod. So then I jumped up and down on the end of it. That sheared the head off the bolt. At least I was able to change the tire then, and the spare stayed on just fine with 4 of the 5 bolts. StuRat (talk) 17:31, 10 September 2017 (UTC)[reply]

Vespula germanica

"V. germanica disperses at a rate of no more than 1000 m per year" — What exactly does that mean? And where does the name come from? Unfortunately, this is not really explained in the article. Best regards--Tuchiel (talk) 13:16, 8 September 2017 (UTC)[reply]

Vespula comes from the same root as the word wasp, being the Latin word vespa. Germanica means "of Germany". --Jayron 32 13:19, 8 September 2017 (UTC)[reply]

@Jayron32: Thanks. Yes, I already knew the meaning, but I wanted to know why they chose the term "germanica" as a reference to Germany.--Tuchiel (talk) 14:37, 8 September 2017 (UTC)[reply]

The species was named by Johan Christian Fabricius, a Danish biologist, in 1793, undoubtedly because he observed them in Germany. As for the meaning of the sentence, it says that if you find these wasps in a given set of places at a given time, and then look around one year later, you probably won't find them more than 1000 meters away from the original set of places. Looie496 (talk) 14:43, 8 September 2017 (UTC)[reply]

@Looie496: Wouldn't that be actually an interesting / relevant complement to the article?--Tuchiel (talk) 14:57, 8 September 2017 (UTC)[reply]

I made an edit to that effect, linking to the biological dispersal article (though I am sure it could be better-formulated). Tigraan^{Click here to contact me} 17:22, 8 September 2017 (UTC)[reply]

Tuchiel's follow up query –". . . why they chose xxx . . ." – illustrates a widespread unconscious false assumption. Scientific names are not chosen by some overarching committee of worthies; they're (usually) chosen by the individual or (in more recent times) perhaps a team of authors who first published a scientific description of the species concerned. Often (though not always) they will have explained their choice, and its meaning in a vernacular language, in that description. Nearly all of the species of animals and plants in Europe (and many common ones elsewhere) will have been named by the 18th-century originator of the binomial system, Carl Linnaeus, or one of his near contemporaries and successors. {the poster formerly known as 87.81.230.195} 2.122.61.201 (talk) 23:51, 8 September 2017 (UTC)[reply]

The Danes call Germnany Tyskland. It's entirely possible The author was using wikt:germane intentionally (as there was a related species with a plainer name). Or he could also be making a pun. μηδείς (talk) 10:02, 11 September 2017 (UTC)[reply]

The Latin for Germany is Germania. You can decide for yourself if this might have been relevant to a guy who called himself Fabricius. Looie496 (talk) 15:20, 11 September 2017 (UTC)[reply]

Caesar named it Germania, that was not a traditinal name. They're not quite sure why he did. Still has absolutely no bearing on the possibility was the "germane" one, or that the good doctor Tyskensis was making a good pun, and was aware of the double meaning. Or maybe the fly kept seizing Schleswig-Holstein? μηδείς (talk) 21:30, 11 September 2017 (UTC)[reply]

After looking for a plant with this species epithet, that didn't mean "to do with Deutschland" and considering that the form meaning "germane" would germana, not german-ic-aI withdraw the hypothesis. μηδείς (talk) 02:08, 12 September 2017 (UTC)[reply]

@Medeis: I'm sorry, but your postings are just so confusing to me! What are you trying to say? And what pun are you exactly referring to? I really don't get it – I'm not a native speaker though. By the way: a wasp is not a fly! Best--Tuchiel (talk) 17:29, 12 September 2017 (UTC)[reply]

Feynman Lectures. Exercises. Exercise 14-3 JPG

. .

...
14-3. The particle of Ex.14-1 moves from the point (0,-1,0) to the point (0,+1,0) on a frictionless track under the action of the force F (plus a certain force of the constraint). Find the work done by the force F if the truck is
a) A straight track along the y-axis
b) A circular track in the z-y plane
Is this a conservative force?

F=1.5yi + 3x²j - 0.2 (x²+y²)k

— R. B. Leighton , Feynman Lectures on Physics. Exercises

In Solutions of 1965 and 1978 there are 2 different answers. I don't understand, why. It seems both are correct. They use another path to show that the force is not conservative : a circle path in x-y plane png.
1965:
$W=\int \limits _{M_{1}}^{M_{2}}F_{x}dx+\int \limits _{M_{1}}^{M_{2}}F_{y}dy=3\int \limits _{-1}^{1}x^{2}dy=3\int \limits _{-1}^{1}(1-y^{2})dy=4{\text{ J}}$

1978:
$W=\int \limits _{M_{1}}^{M_{2}}F_{x}dx+\int \limits _{M_{1}}^{M_{2}}F_{y}dy=1.5\int \limits _{M_{1}}^{M_{2}}ydx+3\int \limits _{M_{1}}^{M_{2}}x^{2}dy=$

$\textstyle 1.5\int \limits _{\pi }^{0}\cos ^{2}(\psi )d\psi -3\int \limits _{\pi }^{0}\sin ^{3}(\psi )d\psi =$

$\textstyle 1.5[{\tfrac {1}{2}}\psi +{\tfrac {1}{4}}\sin 2\psi ]_{\psi =\pi }^{\psi =0}-3[\cos \psi -{\tfrac {1}{3}}\cos ^{3}\psi ]_{\psi =\pi }^{\psi =0}=1.6{\text{ J}}$

$x=\sin \psi {\text{ , }}y=\cos \psi {\text{ , }}dx=\cos \psi d\psi {\text{ , }}dy=-\sin \psi d\psi$ . Username160611000000 (talk) 15:23, 8 September 2017 (UTC)[reply]

I think I understood. In first case the integral must be not $\textstyle 1.5\int \limits _{0}^{0}ydx$ , but $\textstyle 1.5\int \limits _{0}^{1}-{\sqrt {1-x^{2}}}dx+1.5\int \limits _{1}^{0}+{\sqrt {1-x^{2}}}dx$ .Username160611000000 (talk) 05:19, 9 September 2017 (UTC)[reply]

Sorry, it is impossible to help you as you did not formulate the problem that you are trying to solve with these integrals. Ruslik_Zero 18:15, 9 September 2017 (UTC)[reply]

Delay timer mains power

I want to create a circuit that has a delay timer so that it turns on after a certain amount of time, say 5 minutes. I have the timer dial from an old oven which is rated for mains voltage and turns the power on for the duration it is set for, then turns the power off. Can I just connect this to a relay so that when the timer turns off the relay switches the power to the other circuit on? Would this work or have I missed something important? 77.28.149.135 (talk) 16:12, 8 September 2017 (UTC)[reply]

It would work. But then for £1 or so from eBay, I can buy a modern module that does all sorts of timing, and has pushbutton start.

There's also the question of whether you want the relay on before the delay has been started. Your cooker timer would need an additional switch. Andy Dingley (talk) 19:04, 8 September 2017 (UTC)[reply]

It depends on whether the oven timer has normally open contacts and how they are configured to operate when idle and after the time delay expires. If you're having trouble figuring this out, you're strongly advised to not mess with the dangerous voltages involved, and you should follow the advice by Andy Dingley and buy a ready-made timing device that requires you to do no more than plug it in. If you cause a house fire by unapproved wiring, you may find that insurance will not cover losses, and you may put neighbours' lives at risk. Akld guy (talk) 22:17, 8 September 2017 (UTC)[reply]

Hey, I'm from the "burn your house down" school of cookery. Most of the stuff I've learned, I've had my trousers singed, if not quite on fire. Andy Dingley (talk) 00:45, 9 September 2017 (UTC)[reply]

The oven timer has a single-pole switch so that the wiring to your relay will be dangerously live to the mains all the time. I advise including a mains-powered red warning light to show when the countdown is in progress. Blooteuth (talk) 12:19, 9 September 2017 (UTC)[reply]

Did CO2 emissions end the Little Ice Age ?

We can read here:

"The data and simulations pinpointed the early onset of warming to around the 1830s, and found the early warming was attributed to rising greenhouse gas levels. This coincides with the early stages of the Industrial Revolution, in which large volumes of harmful greenhouse gases began to be produced."

"Dr Abram said the earliest signs of greenhouse-induced warming developed during the 1830s in the Arctic and in tropical oceans, followed soon after by Europe, Asia and North America."

This coincides with the end of the Little Ice Age. Count Iblis (talk) 20:02, 8 September 2017 (UTC)[reply]

Well, perhaps, but I'll stay uncommitted, given that such a mechanism clearly can't explain the Medieval Warm Period that preceded the Little Ice Age. I'm skeptical that human-caused greenhouse gas emissions were high enough in 1830 to account for this magnitude of change. Looie496 (talk) 13:54, 9 September 2017 (UTC)[reply]

Fortunately Science does not need simulations anymore since they started drilling ancient ice bodies to get Ice cores which contain an enclose exactly what mixture of air and even dust, ash and the radiation that was in our Atmosphere at a given time in the past. --Kharon (talk) 14:33, 9 September 2017 (UTC)[reply]

Huh? The lead investigator of the work, Nerilie Abram, is apparently well-known for ice core work. Rmhermen (talk) 16:14, 9 September 2017 (UTC)[reply]

I'm not much up to date on this, but there is (was?) a theory that LIA was really the onset of the next glacial, and that human-caused global warming stopped that. 93.139.41.72 (talk) 02:20, 10 September 2017 (UTC)[reply]

Neutron star collision at high speed, follow up question

What if two neutron stars collided head on with center of mass velocity of 99.90 percent of speed of light? Would that cause explosion or fragmentation of the neutron stars? Would a lot of neutrinos be created? (Could neutrinos conceivably carry away a major fraction of mass?)144.35.114.187 (talk) 22:40, 8 September 2017 (UTC)[reply]

Our article Supernova, Section 6.2 Core collapse, states at one point "About 10⁴⁶ joules, approximately 10% of the star's rest mass, is converted into a ten-second burst of neutrinos which is the main output of the event." Clearly, then, neutrinos can carry away significant mass in such stellar events. Without doing the numbers, I'd guess that the scenario you describe would be considerably more energetic than a supernova core collapse, that a very significant proportion of their masses would be converted to neutrinos, and that the two neutron stars would be entirely obliterated in something more spectacular than a mere supernova.

While Stellar collision#Neutron star collisions doesn't go into details, a simple google search finds this paper, but it's addressing a merger of two mutually orbiting neutron stars, rather than the extremely unlikely scenario of two independent neutron stars happening to collide, let alone the vanishingly unlikely one of them moving at such velocities (assuming the collision is natural and not the result of a Super-CERN type experiment by extremely advanced beings).

[Edited to add . . .] Continuing the analogy with CERN, I'd guess that such a collision might well result in some entirely new physics (just as high energy CERN collisions can result in hitherto unknown particles), so your "what if . . ." question may be inherently unanswerable. {The poster formerly known as 87.81.230.195} 2.122.61.201 (talk) 00:07, 9 September 2017 (UTC)[reply]