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January 22

gravitons

how do they whizz here there asnd everywhere at infinite speed to convey the force of gravity between all objects in the universe?--178.105.166.117 (talk) 01:25, 22 January 2016 (UTC)[reply]

The graviton is a hypothetical particle (not discovered yet), which is supposed to travel at the speed of light. Tgeorgescu (talk) 01:41, 22 January 2016 (UTC)[reply]

One thought experiment which I still find fascinating is that given light takes about 8 minutes to travel from the sun to the earth; IF somehow it was possible for the sun to just instantly wink out of existence, we would not have any way of knowing this had happened for 8 minutes. The sun would still be visible in the sky and the earth would keep orbiting around the now non existent sun for another 8 minutes. After that 8 minutes, we would "see" the sun disappear, and instead of following it's previously curved orbit, the earth would start to travel in a straight line the direction it was last traveling as the sun vanished. That is to say, Gravitons do not have infinite speed. Vespine (talk) 02:37, 22 January 2016 (UTC)[reply]

The thing is, no two points in spacetime can interact faster than c, so to an observer on Earth the Sun only stops existing when they see it go bye-bye. There's no way the observer could "know" the Sun stopped existing while the Sun's light is still getting to them. --71.119.131.184 (talk) 21:51, 22 January 2016 (UTC)[reply]

We actually have an article on the Speed of gravity. There is no direct measurement of the speed of gravity, though this may be possible in the future with measurement of gravitational waves. The indirect measurements that have been attempted show a speed consistent with the theory that changes in gravitational fields (and perhaps gravitons themselves) propagate at exactly the speed of light. Someguy1221 (talk) 04:06, 22 January 2016 (UTC)[reply]

If things worked any other way, there would be severe causality issues that I strongly suspect would be obvious at the galactic level. SteveBaker (talk) 04:33, 22 January 2016 (UTC)[reply]

(Strictly speaking, I don't think anything would be egregiously broken if gravity propagated slower than the speed of light—not that there's any reason to suspect such a thing.) TenOfAllTrades(talk) 19:28, 23 January 2016 (UTC)[reply]

If space is literally bent near massive objects, then the explosion of one of those objects should alter space in some kind of "ripple effect" in its gravity - extending, as hypothesized, at the speed of light. Unfortunately for us, our ability to witness the sun going nova would be very short-lived. And the earth wouldn't head off in a straight line, as the massive amount of debris from the explosion would, at the very least, start pushing us away from the sun's former location. Or more likely it would incinerate and disintegrate the earth, which would become part of the debris from the sun. On the plus side, this is not likely to happen any time soon. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:22, 22 January 2016 (UTC)[reply]

Just a note, in my hypothetical situation, it's not enough for the sun to go nova for the gravity to disappear. In a nova only a small fraction of the mass is "used up" and the mass, whether whole or "exploded" would have the same gravity. Of course the energy of the explosion would blast everything "outwards", but disregarding that the earth would still orbit the nova, it wouldn't start traveling in a straight line. Vespine (talk) 22:41, 24 January 2016 (UTC)[reply]

Parachute jump

Suppose you are dropping paratroopers into a fairly small DZ on a windy day (but without wind shear); if you know the drop altitude AGL and the wind speed, can you calculate how far upwind of the DZ should you drop the paratroopers so they will land near the center of the DZ with a minimum of steering? 2601:646:8E01:9089:F4FA:EFCF:C9DF:C509 (talk) 02:45, 22 January 2016 (UTC)[reply]

I'm certain you can. The military would no doubt take that into consideration when deploying paratroops. Pretty much all modern parachutes are steerable, so it might not be much more than if it's light wind, deploy a bit up wind from the DZ, if there's more wind, deploy a bit further. Do that a few times and you'll quickly get the hang of how far away you have to deploy for a particular wind speed for a particular parachute. Vespine (talk) 03:05, 22 January 2016 (UTC)[reply]

Here is how the U.S. Army does it. Shock Brigade Harvester Boris (talk) 03:06, 22 January 2016 (UTC)[reply]

Thanks! So if the wind is blowing at 15 knots and you jump from 1000 feet, you'll drift about 450 meters, right? 2601:646:8E01:9089:F4FA:EFCF:C9DF:C509 (talk) 08:17, 22 January 2016 (UTC)[reply]

It's not quite that simple:

1) Wind speed, and even direction, varies with altitude. So, you'd need to account for the drift at each level.

2) You are assuming sustained winds, but there are also wind gusts to deal with, which are, by their nature, unpredictable.

3) How far they are blown off course depends on their rate of descent. Therefore, wind will have minimal effect until the parachute is opened.

4) There's also a problem in heavy winds that once they land they could be dragged by the wind. (There is a parachute release mechanism, but using that while being dragged over rough terrain might be difficult. If they could release the instant they hit, then they would be OK, although the released parachutes blowing across the field will give away their position.) StuRat (talk) 08:29, 22 January 2016 (UTC)[reply]

Yes, you did calculate the drift D=KAV correctly using a personnel Load Drift Constant K = 3.0 meters per paragraph 6-32 and figure 6-4. But you also need to take the forward throw into account, as discussed in paragraph 6-36 and table 6-9, which is given as 229 meters for personnel out of a C-5, C-130, or C-17, and then you need to combine them vectorially with regard to wind direction and drop heading as discussed in section 6-89 and shown in Figure 6-8. There seems to be a problem with that figure in the PDF linked above by Shock Brigade Harvester Boris, so you may prefer this copy of the complete US Army Field Manual FM 3-21.38, Pathfinder Operations (with chapter 6, Drop Zones).

Note that there does seem to be an error in that manual in the "Determination of Release Point Location" section (starting with paragraph 6-89). Figure 6-8 shows all five steps, but Step 5 is omitted in the discussion, and the example given under the discussion of Step 4 (paragraph 6-93) belongs under the missing discussion of Step 5 as it is the calculation of Throw Distance, which for rotary-wing and STOL aircraft "equals half the aircraft speed (KIAS), expressed in meters." -- ToE 16:09, 23 January 2016 (UTC)[reply]

So, for a DC-3 flying at 100 knots, the paratroopers would be thrown forward 50 meters? 2601:646:8E01:9089:94DA:2520:D95F:848D (talk) 02:45, 24 January 2016 (UTC)[reply]

It's probably greater than that. Clearly the "equals half the aircraft speed (KIAS), expressed in meters" formula for rotary-wing and STOL aircraft does not extend to the higher drop speeds of the fixed wing aircraft in table 6-9 with forward throws of 229 meters, as there is no way that they have drop speeds anywhere near 458 knots. In fact, table E-1 gives a personnel drop speed of 130 - 135 knots. The formula breaks down somewhere, and I suspect that it is a stretch to apply at the 90 knots used in the example problem. If I had to guess, based purely on intuition, I'd be tempted to scale it with the square of the drop speed, and predict (100/135)^2 * 229 m = 125 m. -- ToE 04:36, 24 January 2016 (UTC)[reply]

It's worth noting that they very often drop supplies and weapons alongside the paratroops - and steerable parachutes don't help those things! So there is a degree of importance to making the drop in the right location. SteveBaker (talk) 17:37, 22 January 2016 (UTC)[reply]

N-A=S what these letters stand for?

In chemistry there is a formula for finding the number of the bonds. N-A=S what these letters stand for? 92.249.70.153 (talk) 14:06, 22 January 2016 (UTC)[reply]

S= N-A, where S is the total number of shared electrons, N is the total number of valence shell electrons needed by all the atoms in the molecule or ion to achieve noble gas configurations and A is the total number of electrons available in the valence shells of all the atoms in the structure. — Preceding unsigned comment added by 81.131.178.47 (talk) 15:32, 22 January 2016 (UTC)[reply]

A link to show the context. Mikenorton (talk) 15:35, 22 January 2016 (UTC)[reply]

Spanish electric connector

Hi, does anyone know whether this old type of connector found in Spain has an official name or not?

Thanks in advance.--Carnby (talk) 19:13, 22 January 2016 (UTC)[reply]

Tough call. This [1] looks like a very WP:RS. It has official names and specification codes for many, many, plugs/sockets. It simply calls this type "old spanish socket". It doesn't have a picture, but the description matches these photos very well, IMO. oops; sorry :( SemanticMantis (talk) 19:56, 22 January 2016 (UTC)[reply]

That's a mirror of an old version of AC power plugs and sockets. -- Finlay McWalterᚠTalk 20:01, 22 January 2016 (UTC)[reply]

D'oh! Thanks, I will look more carefully at the header next time :) SemanticMantis (talk) 20:05, 22 January 2016 (UTC)[reply]

Maybe you can ask our Spanish friends over at Wikipedia:Café/Archivo/Técnica/Actual or elsewhere on la enciclopedia libre. The Quixotic Potato (talk) 00:52, 23 January 2016 (UTC)[reply]

Done--Carnby (talk) 12:33, 23 January 2016 (UTC)[reply]

Top British medical Schools?

Can someone give a simple, easily copied list of the most prestigious British medical schools? This inquiry is for a user who has very limited WP access, and who Wants me to send a list (category), rather than a full-blown prose article. Thanks. ````` — Preceding unsigned comment added by Medeis (talk • contribs) 21:23, 22 January 2016 (UTC)[reply]

Googling "British medical schools" gives me this. Rojomoke (talk) 22:56, 22 January 2016 (UTC)[reply]

Adding "top" to the search gives me this ranked list. Rojomoke (talk) 23:00, 22 January 2016 (UTC)[reply]

List of medical schools in the United Kingdom is the wikipedia article for this. There are only 32 medical schools in the UK. So, there is no problem taking a look at all of them. I'm not sure what part of them are top. The Guardian score linked above could be slightly misleading. All of them seem to get scores that are not far apart.--Llaanngg (talk) 18:13, 23 January 2016 (UTC)[reply]

@Medeis: - (Position: I've done quite a bit of education consultancy work and advice with private tuition and out-of-school education) - All British medical schools essentially teach the same syllabus with the same entry criteria, so there’s a limit to how different they can be. Furthermore, by not attending a London university or Oxbridge you probably get reduced cost of living, so attending a ‘worse’ university may have dramatic compensations.
Honestly, your friend probably wants to worry about getting the grades to be in a position to make that choice. From tuition work I've done I've been startled to see kids in year 10 (age 15) predicted BCC science GCSEs whose parents say they need help deciding whether they should become a doctor or a dentist. 94.119.64.1 (talk) 16:56, 24 January 2016 (UTC)[reply]

Thanks, I'll pass that on. What's actually going on (and the reason I didn't just google it, although I did send the results above) is not an actual choice of schools, but backstory for a story set in Britain. Had I been asked the same question for the US I would have suggested the University of Pennsylvania, which is top-tier, but not a cliche like Harvard or Columbia. But as an American I have no knowledge of British schools other than the obvious cliches. μηδείς (talk) 17:39, 24 January 2016 (UTC)[reply]

"Prestige" doesn't really have the same meaning WRT British academia as the Ivy League etc does in the US, other than a tendency for Oxbridge to consider themselves superior (sometimes with and sometimes without justification). Since most institutions in England and Wales (Scotland and NI are independent for the purposes of education) are charging the legal maximum tuition fee of £9000 per annum, there's a strong levelling effect since if a place starts getting a bad reputation, people will just go elsewhere. If you want a place with all the rowing-and-rugger English cliches, but avoiding Oxbridge, I'd suggest Barts. ‑ Iridescent 18:01, 24 January 2016 (UTC)[reply]

My friend agrees Barts would have been ideal, if not for Sherlock Holmes, and has settled on UCL, and I have sent the text of that article and some images. Thanks, again, for the assistance. μηδείς (talk) 20:07, 24 January 2016 (UTC)[reply]

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