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August 22

How Often Do Hurricanes Hit Puerto Rico Redux

How often do hurricanes hit Puerto Rico? Redux Tropical Storm Irene barrels toward Puerto Rico DANICA COTO, Associated Press Updated 10:24 p.m., Sunday, August 21, 2011

http://www.chron.com/business/article/Tropical-Storm-Irene-barrels-toward-Puerto-Rico-2134551.php

μηδείς (talk) 03:35, 22 August 2011 (UTC)[reply]

According to our last discussion, I think several people brought up some convincing data that it was once every three years or so. --Jayron32 11:47, 22 August 2011 (UTC)[reply]

Three was my subjective guess, I think it was less often. The editor who posted the question was going on Sept 1, so this is good news, since it's highly unlikely a secon storm will so quickly follow the same path. μηδείς (talk) 16:17, 22 August 2011 (UTC)[reply]

I bet somewhere there's a map of the hurricane frequency for the whole hemisphere or more - if someone could get or make such a thing for Hurricane it would be absolutely awesome. We have some maps there of all the hurricane tracks, but it's hard to convert that mentally to frequency; yet that data could be used directly to infer frequency if we make the somewhat unreasonable approximation that it is "frequency of being within X miles of a hurricane track". Obviously a map based on the actual extent of each hurricane would be better if we could. Wnt (talk) 16:51, 22 August 2011 (UTC)[reply]

Jeff Masters at http://wunderground.com has the historical data set and (almost all?) the code to make such a map if he doesn't already have such a map, and I'm sure he'll be willing to share it, but you probably want to catch him next week. 208.54.5.213 (talk) 04:48, 23 August 2011 (UTC)[reply]

It looks like Beeblebrox has missed the boat, although he could still go to Florida later this week. Count Iblis (talk) 23:08, 22 August 2011 (UTC) ''Italic text[reply]

Weak interaction

This is based on a question I had asked around a year ago, and have been thinking about since. From the standard formulation of QM, ${\langle \chi |\psi \rangle }^{*}={\langle \psi |\chi \rangle }$ . So $|\langle \chi |\psi \rangle |^{2}=|\langle \psi |\chi \rangle |^{2}$ . This is another way of expressing time-reversal symmetry: if a particle is in a certain state ψ, it has a certain probability of being found in the state χ at a later time. If we reversed the situation, the probability that the particle goes from χ to ψ is the same. But weak interactions aren't symmetric in time. I was wondering how this is accommodated in the mathematics of QM. 74.15.137.168 (talk) 05:26, 22 August 2011 (UTC)[reply]

{\langle \chi |\psi \rangle }\neq {\langle \chi |\psi \rangle }^{*}={\langle \psi |\chi \rangle }

. which means that QM requires time reversal but doesn't require time symmetry. Dauto (talk) 14:23, 22 August 2011 (UTC)[reply]

Life Expectancy After Surviving Childhood?

Is there any data showing the average life expectancy of someone in the United States after surviving to adulthood? My understanding is that childhood deaths really bring down the average. Historical data going back to the 70s would be great. --CGPGrey (talk) 09:07, 22 August 2011 (UTC)[reply]

WHO have a mass of data by age range by country by year etc., but their site seems to be having problems right now. [1]--Aspro (talk) 11:47, 22 August 2011 (UTC)[reply]

Here (page 7) is a life table for the total US population for the period 1999-2001. It says the life expectancy at birth was 76.83 years, which the life expectancy at age 18 was 59.70 (which corresponds to an age of death of 77.70). So, people dying in childhood drag down the average by about a year. The probability of dying before your 18th birthday, according to that table, is 1.178% (the chance of dying before your first birthday is 0.695%). That's very low, which is why the effect on the average is also very low. It would have been much greater 100 years ago, say. --Tango (talk) 19:51, 22 August 2011 (UTC)[reply]

You want to take a look at the article life table. Quest09 (talk) 19:58, 22 August 2011 (UTC)[reply]

I'm an actuary, so no, I don't really want to take a look at that article. The OP might though. Please see WP:INDENT! --Tango (talk) 21:03, 22 August 2011 (UTC)[reply]

Killing the brain while keeping the body alive

Is this possible? Can you give a person brain death but then keep their body alive, functioning, and growing? Kind of like an empty shell so to speak? ScienceApe (talk) 16:11, 22 August 2011 (UTC)[reply]

Brain death covers some of this. --Jayron32 16:18, 22 August 2011 (UTC)[reply]

Also organ harvesting. 76.254.20.205 (talk) 16:32, 22 August 2011 (UTC)[reply]

A) If the brain was totally dead, the rest of the body couldn't be kept alive long. This is because the brain stem regulates breathing, heartbeats, etc. These functions can be temporarily replaced by machines, but not over the long term.

B) On the other hand, it's quite common for the higher brain functions to cease while the brain stem continues to function. In this condition, the body can be kept alive indefinitely. StuRat (talk) 16:42, 22 August 2011 (UTC)[reply]

In vitro meat may be of interest. Comet Tuttle (talk) 17:37, 22 August 2011 (UTC)[reply]

iceberg

How close to the equator was the most tropical iceberg sighted (drifting south from Greenland or drifting north from Antarctica)? Googlemeister (talk) 16:42, 22 August 2011 (UTC)[reply]

For an answer to the first participial phrase in your parenthesis, see this page, under "What is the extreme range of iceberg locations?" Deor (talk) 17:30, 22 August 2011 (UTC)[reply]

From the supplied map in that link, it seems that one iceberg got as far south as about 28 degN. How about coming from the other way? Googlemeister (talk) 18:48, 22 August 2011 (UTC)[reply]

Mass of all the sunlight...

What's the mass of all the sunlight that strikes Earth in one second? Comet Tuttle (talk) 17:35, 22 August 2011 (UTC)[reply]

Per solar energy, the Earth receives 174 petawatts of incoming solar radiation, so 174 petajoules per second. Per e=mc², that's about 2 kg per second. — Lomn 17:47, 22 August 2011 (UTC)[reply]

Fascinating - does that mean that the Earth's mass is increasing by around a thousand tonnes a year? Quintessential British Gentleman (talk) 18:14, 22 August 2011 (UTC)[reply]

Much of that light is reflected back away from the Earth. Note that the Earth also gains mass each year due to meteors, micrometeors, and the solar wind, and loses mass in the form of hydrogen and helium which bleed off into space. I wonder if there's a chart which compares all these quantities.StuRat (talk) 18:20, 22 August 2011 (UTC)[reply]

If I recall correctly, meteors, micrometeors, and space dust lead to an accretion of about 50,000 metric tons per year. (Which is isn't nearly as much as it sounds like once you spread it out across the entire Earth's surface). Dragons flight (talk) 22:51, 22 August 2011 (UTC)[reply]

(ec) More to the point, the "increased mass" due to sunlight would really only be increased temperature. Photons have zero rest mass (probably; if it's nonzero it's not enough to make a difference in this discussion). So it's not like they can stick to the planet and increase its mass. The only way the Earth's invariant mass would be increasing as a result is as an increase in thermal energy.

But the temperature is not increasing. Well, OK, it's increasing some (climate change) but not the way it would be if the energy from the sunlight just stayed.

What happens is that the Earth radiates the energy back into space, as approximately black body radiation based on its temperature. --Trovatore (talk) 18:28, 22 August 2011 (UTC)[reply]

In principle, it could also be stored as chemical energy. However, the size of the biosphere isn't increasing at any particular speed, and I don't know of any inorganic light-driven chemistry that's particularly active either. --Tardis (talk) 13:01, 23 August 2011 (UTC)[reply]

Thank you! Comet Tuttle (talk) 18:45, 22 August 2011 (UTC)[reply]

Radiation pressure can be used to quantify the momentum transfer due to incident electromagnetic waves. Use caution: the math and physics involved is subtle, and it is very common to apply the formulas incorrectly, resulting in non-physical conclusions. A good example is the Crookes radiometer: there are an abundance of incorrect physics-esque explanations for the apparent momentum-transfer. The same conceptual problems exist with regard to solar radiation incident on the Earth - and the experimental challenges of measuring planet-sized objects are much more difficult. Nimur (talk) 18:52, 22 August 2011 (UTC)[reply]

Looking at photon it looks like E=pc=mc^2, so p=mc as expected. 2 kg/s * 299,792,458 m/s = (5.9736 E+24 kg) (1E-16 m/s^2). Not a huge acceleration, despite the remarkable whack. Wnt (talk) 02:49, 23 August 2011 (UTC)[reply]

You should not use that formula. First of all, it's incorrect, relativistically - because, whether you realize it or not, you have taken a first-order derivative with respect to time (in other words, you are using "dm/dt", the rate of incident "mass" of photons). That's not a relativistically invariant value. You should properly compute the radiation pressure by dividing the Poynting vector by the speed of light, to obtain an expectation-value for the radiation-pressure, in units of pressure. Nimur (talk) 17:01, 23 August 2011 (UTC)[reply]

Hmmmm... I'm not very familiar with that concept, but the article calls it an energy flux in W/m^2. Taking it times however m^2 the Earth gets gives us watts, or energy per second. Dividing energy by c gets us mc, which is what I used above. I don't see any relativistic correction. Originally I ignored relativity, being satisfied to use a clock on Earth, but truly if your clock ticks as some different rate you should count a different number of photons and get the same answer, I'd think. Wnt (talk) 16:58, 24 August 2011 (UTC)[reply]

Yeah, that would be applying the lorentz transform to the "dm/dt". You can relativistically correct any formula by applying the appropriate Lorentz transform (if you can remember how to do it correctly, which I can rarely do); but, if you just use a relativistically invariant formula, you never need to do that. S / c is always valid. Nimur (talk) 22:29, 24 August 2011 (UTC)[reply]

Galaxies, solar systems and planets systems

Is the speed of a star around a black hole and the speed of a planet around a star and the speed of a moon around a planet all based on the same formula of mass and the distance apart or are there differences between each system? --DeeperQA (talk) 20:23, 22 August 2011 (UTC)[reply]

They are all due to the same property of the universe, gravitation. In the case of a moon/planet system, we can use the equations defined as Newton's law of universal gravitation to model the interaction. In the other cases you mentioned, more sophisticated mathematical equations are needed in order to accurately describe the interaction, but they are based on the same physics. You can, if you want, apply the identical formulation of gravitation to the planet/moon system, but the math is more difficult, and you obtain the same result as the Newtonian formula, out to several decimal-places. Nimur (talk) 20:43, 22 August 2011 (UTC)[reply]

If other objects are sufficiently small and far away, you can treat any of these three scenarios as a two-body problem and use Kepler's laws of planetary motion to get a very good approximation. Your star around the black hole would also need to be sufficiently far enough away that the BH is not tearing pieces off of the star. Googlemeister (talk) 20:53, 22 August 2011 (UTC)[reply]

Well, there are several relativistic corrections to Newton's laws that may be quite a bit larger for the black hole scenario, because the black hole will produce a much larger curvature of space-time in its vicinity. Looie496 (talk) 21:16, 22 August 2011 (UTC)[reply]

Only particularly near to the black hole. A black hole with a mass equal to the sun, say, will have exactly the same gravity as the sun as long as you are more than one solar radius away from it (ie. as long as you wouldn't be inside the sun if it were actually the sun). --Tango (talk) 22:58, 22 August 2011 (UTC)[reply]

Is the solar equivalent radius the black hole's event horizon? --DeeperQA (talk) 06:59, 23 August 2011 (UTC)[reply]

electricity and light?

do electricity and light travel at the same speed or is light slightly faster? Thornydevil ^Munchies? 21:11, 22 August 2011 (UTC)[reply]

It's potentially important to specify the medium, as both light and electricity travel at different speeds through different substances. That said, if we consider that electrons are particles with rest mass, it's clear that they cannot be accelerated to c (the speed of light in vacuum) -- thus, it's reasonable to make a general statement that electricity does not travel as quickly as light. For practical numbers, the propagation of an electrical signal through a copper line travels at about 2e8 m/s, as contrasted with light's 3e8 m/s. — Lomn 21:17, 22 August 2011 (UTC)[reply]

It's misleading to bring up the speed of the electrons. Nothing you said is actually wrong, but it could lead a naive reader to think that the speed of the electric signal is the same as the speed of the electrons in it.

In fact, the signal travels much much much faster than the so-called drift velocity of the electrons. The usual help to intuition is to imagine a pipe full of tennis balls. If you push a tennis ball in one end, another one pops out the other end, and the delay in that happening is much smaller than the time it would take the tennis ball to get to the other end at the speed the tennis ball itself is moving. --Trovatore (talk) 22:14, 22 August 2011 (UTC)[reply]

Sorry, I should have specifid in the air. Thornydevil ^Munchies? 21:31, 22 August 2011 (UTC)[reply]

We have an article, speed of electricity, that deals with this. The basic answer is that light is faster but sometimes not much faster. Looie496 (talk) 21:18, 22 August 2011 (UTC)[reply]

Thanks for the help! Thornydevil ^Munchies? 02:52, 23 August 2011 (UTC)[reply]

Ball screw

A diagram of a Recirculating Ball mechanism

How does a recirculating ball nut and lead screw achive zero backlash?--92.28.71.6 (talk) 21:58, 22 August 2011 (UTC)[reply]

In a regular "geared" system, the tighter the gears are meshed the more friction there is, the looser they are the more slack there is, so inevitably you'll have some friction and some slack. It seems to me that due to the reduced friction of the Ball screw design, the mechanism can just be made within much tighter tolerance. I'm not sure it's correct to say "zero" backlash, it might be so small as to be insignificant, but even at 80% or more efficiency, it's not technically "zero" backlash. Vespine (talk) 23:48, 22 August 2011 (UTC)[reply]

The balls sit in the 'V's of the threads of both the ball nut rack & the worm gear. Thus there is no play.--Aspro (talk) 23:56, 22 August 2011 (UTC)[reply]

(EC)Ok, thought about if for one minute and that's incorrect (My 1st post). In a single gear system it's impossible to have zero backlash... With a ball screw it IS very possible to get zero backlash, all you need to do is have SOME of the balls mesh with ONE side of the thread and some of the balls mesh with the OTHER side of the thread. This is easily acheived just by making the ball nut rack pitch very slightly different then the shaft worm gear. That way, the balls on one end of the rack mesh on one side of the worm gear and the balls on the other side meshes the opposite side, there's your "zero backlash". As for Aspro's comment, gears are normally "v" shaped, I don't think that actually has anything to do with reducing the "play". Vespine (talk) 00:02, 23 August 2011 (UTC)[reply]

You seem to be guessing as you go! The balls ensure there is all ways contact on all surfaces (four point contact) – no gaps, no play.--Aspro (talk) 00:09, 23 August 2011 (UTC)[reply]

Hmm, i disagree. I'm not an engineer but I used to fix power tools so have a bit of experience with gears... You're neglecting the fact that this is a three dimensional shape. There is a RING of balls, so the 4 points of contact you talk about has an OPPOSITE 4 points of contact on the other side of the shaft, so there obviously has to be SOME play in the system, or else the thing won't move at all. The "trick" here is to minimize play along the functional axis. The way to do this is to have "mesh" that opposes it self along that axis. If you imagine the ball nut rack has a very slightly narrower thread then the shaft worm, along the top of the worm, the left most ball will be getting pulled towards the centre of the nut toward the right, therefore meshing the LEFT side of the "groove" in the nut and the RIGHT side of the groove in worm. By the time you get to the RIGHT most ball, it too will be getting pulled towards the centre, which is to this ball's left, it will be meshing the opposite sides of the nut and worm cancelling out the "play" along that axis. Vespine (talk) 00:24, 23 August 2011 (UTC)[reply]

Look at the diagram.--Aspro (talk) 00:31, 23 August 2011 (UTC)[reply]

Ok, I am full of it, opposing mesh is used as an "anti-backlash" measure, but it seems not in recilculating ball designs, not that I can find in any of our articles anyway. It makes sense in my head;). The article DOES say however that recilculating ball gears are made to very high precision because of the low friction of the design, up to 90% efficiency. I think this still means that "no gap, no play" is a far too simplistic answer, as is "4 points of contact". You'd be hard pressed to find a "gap" in any geared system with half decent precision, but it doesn't mean there's no play or backlash. A seized gear has NO play. In the real world there's things like wear, lubrication, heat expansion, etc. The very fact that tension in one direction means slack in the OTHER direction on the opposite side of the shaft should be enough to show that the groove or the ball are not the only factors, or even one of the main factors, in the reduction of backlash, otherwise, why not use just ONE rotation around the shaft? I think the main factors in the practical elimination of backlash in this design are the firstly the high precision and secondly the number of turns around the shaft. Vespine (talk) 01:20, 23 August 2011 (UTC)[reply]

I don't know anything about this, but that gear at the bottom of the picture looks much like the gear used as an example in the Backlash article. Shouldn't there be some backlash there? And when looking at the other part, the worm gear part, isn't a worm gear part generally supposed to be effectively irreversible, and thus not able to backlash past that point anyway? Wnt (talk) 00:34, 23 August 2011 (UTC)[reply]

The question was about ball screws, the sector gear is an addition (as found on say a car steering control). A recirculating ball nut and lead screw assembly (on say a large motor car with five and a half turns of the steering wheel between lock and lock) gives a lot of mechanical advantage without the backlash inherent in other gearing systems such a rack and pinion. A little play after this assembly has much less effect on over all directional control. An yes, as you say, because of the mechanical advantage of the worm gear (to the driver) of this configuration, the torque coming the other way has this mechanical advantage working against it but the lack of backlash refers to the lack of play on the side of the steering wheel as well. --Aspro (talk) 01:23, 23 August 2011 (UTC)[reply]

The absence of backlash is indeed due to the inability to circulate the balls by movement of the steering rods due to torque ratio. However, the absence of play is due to the compression of the balls against the walls and one another. --DeeperQA (talk) 09:17, 27 August 2011 (UTC)[reply]