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April 6

Can someone, please, correct this messed up navbox?

There is an obvious error in navbox in article Stainless_steel headed: "Iron-carbon alloy phases", where it lists "Graphite (allotrope of carbon)", as a phase but that is erroneous:

That navbox element is incorrect because graphite is formula C, and is not a phase of steel alloy. In fact graphite is a form of pure carbon, and is of no more relevance to stainless steel phases than diamonds would be... Garshepp (talk) 04:04, 6 April 2013 (UTC) — Preceding unsigned comment added by Garshepp (talk • contribs) 04:01, 6 April 2013 (UTC) Garshepp (talk) 04:04, 6 April 2013 (UTC)[reply]

It IS a phase in ~~pig iron~~ cast iron, and the infobox is about ALL phases in ALL iron-carbon alloys, not just steel. I thought pig iron and cast iron were the same thing -- apparently not. Live and learn!24.23.196.85 (talk) 04:17, 6 April 2013 (UTC)[reply]

Tire pressure sensors

Last night after I put air in my tires because the light on my dashboard told me to, I started wondering how the car knew. I haven't found a clear explanation of the mechanics of it. So, could someone educate me? Where are the sensors and how do they test the pressure? Thanks, Dismas|^(talk) 11:57, 6 April 2013 (UTC)[reply]

See tire-pressure monitoring system. The sensor is inside the tire, around the valve stem (there's an image here: www.exam iner.com/article/tire-pressure-monitoring-it-s-not-a-whole-bunch-of-hot-air-it-s-the-law spam filtered, so manually fix that link yourself). That monitors the pressure (the pressure differential across the valve) and communicates by radio back to a module inside the car. -- Finlay McWalterჷTalk 12:14, 6 April 2013 (UTC)[reply]

Actually, there are two common methods. One is the direct measurement technique that Finlay describes - but the older (and much cheaper) method is to have the computer count the number of times each wheel rotates and if (on average) one is rotating faster than the others, then its circumference must be a little less - which is most likely due to the pressure being lower. SteveBaker (talk) 19:39, 6 April 2013 (UTC)[reply]

Stars

If a red dwarf crosses paths with a white dwarf, which one is more likely to capture the other? 24.23.196.85 (talk) 19:40, 6 April 2013 (UTC)[reply]

A red dwarf usually has less mass, so it would be deflected more. However, they would each capture each other, meaning they would orbit a barycenter between the two, but closer to the white dwarf. Note, however, that there is some overlap of solar masses, ranging from 0.075 to 0.500 for a red dwarf and from 0.140 to 1.330 for a white dwarf, so, while I described the average pairing, you could have pairs of equal or even the reverse mass ratios. StuRat (talk) 19:47, 6 April 2013 (UTC)[reply]

Thanks! 24.23.196.85 (talk) 20:24, 6 April 2013 (UTC)[reply]

You're quite welcome. Can we mark this Q resolved ? Also, you might want to check out Pluto and Charon (moon), which have a similar mass ratio and thus orbit a common barycenter. StuRat (talk) 21:08, 6 April 2013 (UTC)[reply]

I think we can call this bit closed, before things escalate. — The Hand That Feeds You:^Bite 20:51, 7 April 2013 (UTC)[reply]

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

StuRat, can you please refrain from answering questions on subjects you know little about, which apparently includes fundamental physics? Your answer is almost completely wrong.

If a red dwarf and white dwarf cross paths, unless there's a third body, they will not capture each other. This is because the two bodies would have enough kinetic energy to escape each other's gravitational pull. In Newtonian gravity with 3 bodies, there are 2 situations that can happen:

1. The two bodies have too much energy. In this case, they don't orbit each other, but escape each other's gravity and keep on going. Their trajectories describe a hyperbola.

2. The two bodies have too little energy. In this case, the bodies orbit each other in an ellipse whose shape and size never changes.

Because energy is conserved, you can't go from 1 to 2 or vice versa unless some third body robs the system of energy. This is what happens during gravitational capture. However, the 3-body problem has no analytic solution and is quite hard to analyze, so there's no simple description of what happens during gravitational capture.

StuRat is correct, however, in implying that it's meaningless to say whether one body captures the other. The Moon doesn't orbit Earth; it orbits a common barycenter in between the Earth and Moon, but much closer to Earth. --140.180.248.141 (talk) 23:02, 6 April 2013 (UTC)[reply]

With stars, you have more complex motions than just 2 or 3 bodies. They are also orbiting the galactic center, and there are gravitational waves affecting their motion within the arms. And, there also appears to be a dark matter effect on them, as their motion can't completely be explained without it. So, the actual capture process would be quite complex. I chose not to describe possible capture processes, since the question didn't seem to be about that. As I read it, they were only asking what the eventual orbits would be, IF a capture did occur. Now, please desist with the insults. They do not belong on this page (or anywhere, really). StuRat (talk) 23:15, 6 April 2013 (UTC)[reply]

So, if a red dwarf were to collide with a white dwarf, would the result be a pink dwarf? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:36, 6 April 2013 (UTC)[reply]

No, the result would be a larger red dwarf. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 00:07, 7 April 2013 (UTC)[reply]

The effect of the galactic center, gravitational waves, and dark matter are utterly negligible when considering an encounter between two stars. LIGO failed to detect any gravitational waves, and it can measure a 4 km length to within 10^-18 m (about 1000 times smaller than a proton). Dark matter and the galaxy accelerate both bodies by almost exactly the same amount. Only the tidal force can have any effect, and it's negligible for galactic distances.

There are other effects that influence gravitational capture. For example, tidal friction induced by one body on the other can make the system lose energy. The red dwarf could lose material to the white dwarf; this process is called accretion. If enough material is accreted, the white dwarf could explode in a Type Ia supernova.

I don't think I insulted you. I said you don't seem to know fundamental physics. I stand by that judgment, and would add that you don't seem to know much about astrophysics either. --140.180.248.141 (talk) 00:01, 7 April 2013 (UTC)[reply]

And do you believe such statements belong here ? Shall I start insulting you ? You've just listed a couple scenarios where a capture could take pace, but, as I said previously, I don't see this Q as being about the capture processes, but only the eventual result of the capture. Where do you see anything in the Q requiring us to describe all possible capture processes ? And, even if it had asked for that, then my answer would be a partial one, not an incorrect one. I just don't see what on Earth you're bitching about. A more constructive way for you to respond would have been: "While StuRat described the eventual result of such a capture, I'd also like to discuss possible processes whereby such a capture might occur...". StuRat (talk) 00:48, 7 April 2013 (UTC)[reply]

You said "however, they would each capture each other". In actuality, they would pass each other without incident unless a third body was involved, because gravitational capture is impossible with only 2 bodies. You said that "the actual capture process would be quite complex" and implied that the galactic center, gravitational waves, and dark matter make it complex. In actuality, all three have an utterly negligible effect. --140.180.248.141 (talk) 01:44, 7 April 2013 (UTC)[reply]

The OP did not specify a universe containing only the two stars. That's your assumption. So, with other stars in the neighborhood, they may have a significant gravitational effect, especially on the scale of billions of years, with orbital distances of multiple light years. (The time frame and distances for the capture was also not specified.) And, over those time scales, the other factors I mentioned, like dark matter and gravitationaly waves, may also have a significant effect on the motion of the stars. StuRat (talk) 02:22, 7 April 2013 (UTC)[reply]

However accurate or otherwise another editor's comments might be, it is not appropriate to comment on or express a sentiment on their level of knowledge, nor on the advisability of their contributing, when such contributions are in good faith, as StuRat's clearly were. Please confine comments to the subject matter; corrections to another editor's contributions are quite sufficient in the context. — Quondum 02:08, 7 April 2013 (UTC)[reply]

Agreed. StuRat (talk) 02:22, 7 April 2013 (UTC)[reply]

Disagreed 100%. In the context of trying to communicate in a forum such a this, pointing out when another editor is beyond their ken is entirely germane, if done in a simply factual matter. I don't see any insult here under any definition. --Mr.98 (talk) 03:02, 7 April 2013 (UTC)[reply]

"Can you please refrain from answering questions on subjects you know little about, which apparently includes fundamental physics?" is not a "factual" statement, it's a gratuitous insult that serves no purpose other than to inflame. If the snippy IP had simply said, "That's not correct", and then explained why, it could have been a different story. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:38, 7 April 2013 (UTC)[reply]

Sturat's first answer was correct, on the obvious assumption the OP wanted cases where the two stars were moving slowly enough to become gravitationally entangled. IP 140 is simply wrong that the barycenter of the earth-moon system lies between the two bodies--it lies ~1000 miles below the earth's surface. μηδείς (talk) 18:39, 7 April 2013 (UTC)[reply]

Barycentric coordinates (astronomy) has the confirmation to Medeis. I wonder where 140.180 studied physics, if at all. His IP geolocated to Princeton, but I hope they don't have such a low level. Not knowing this is not scary if you are not into natural sciences. However, accusing StuRat of knowing little fundamental physics and then making such a mistake is laughable. OsmanRF34 (talk) 20:03, 7 April 2013 (UTC)[reply]

The barycenter is between the centers of the two bodies. The center of a spherical body is usually used for gravitational calculations, and never did I imply that the barycenter is outside Earth's surface. Contrary to what Medeis seems to imply, it is not possible for two bodies to become "entangled" (meaning orbiting a barycenter) in the absence of a third body, no matter how slowly the stars are moving. This is a common misconception about gravitational capture, and since the OP seemed to hold it, people answering this question are responsible for correcting it. Failing to correct it is similar to answering "how many tons of green cheese is the Moon made of?" with the mass of the Moon, without also saying that the Moon is not made out of green cheese. (Formerly 140.180.248.141) --128.112.25.104 (talk) 04:16, 8 April 2013 (UTC)[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Is it possible to capture by aerobraking? With low enough masses, they don't have to cause a type Ia supernova. Well, to tell the truth, in a flat, non-expanding classical Newtonian 2-body universe, even if the stars were atmosphereless spheres that won't touch or change (i.e. subliming into the vacuum), passing does not guarantee that they won't pass again. They can't be infinite distance now can they, there's always a higher real number. No matter the distance a closed orbit is simply a matter of low enough speed for the distance, but not so low that they collide). Sagittarian Milky Way (talk) 18:27, 8 April 2013 (UTC)[reply]