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June 4

Htrae

Resolved

Confirmation please. I would think that walking from the center of a face to the corner, even if the surface is perfectly flat, would result in me going from level ground to a 45 degree incline by the time I made it to the edge. Is this accurate? 70.177.189.205 (talk) 00:32, 4 June 2011 (UTC)

A 45 degree incline from what? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:46, 4 June 2011 (UTC)

45 degrees relative to the hypothetical tangent of the sphere - yes. It won't gravitationally flat, gravity will be strongest in the centers of each face. This would give you the perception of always walking uphill or downhill. Plasmic Physics (talk) 00:54, 4 June 2011 (UTC)

No, that's wrong. At the center of an edge it would be 45 degrees. At a corner it would be 60 degrees. Looie496 (talk) 00:56, 4 June 2011 (UTC)

But gravitationally, most of the attraction is to the material closer to you. So, this means that someone standing on a corner wouldn't feel "down" as towards the center of the cube, but rather somewhere between the center and their current location. This means the angle wouldn't seem quite as steep. I'd be interested to see the actual results, if somebody wants to "run the numbers". StuRat (talk) 01:02, 4 June 2011 (UTC)

No need; the symmetry alone dictates that the attraction be to the center, Taruts. Clarityfiend (talk) 01:11, 4 June 2011 (UTC)

Both 45°and 60° are incorrect.

Use a coordinate system such that the origin is at the center of the cube, and the surfaces are at x=±1, y=±1 and z=±1. At the point (1,1,1), gravitational "up" must be in the direction (1,1,1) from symmetry; consider rotations of the cube by 360/3° around a line that passes through the origin and the point (1,1,1). A vector of unit magnitude that points in the gravitationally "up" direction at (1,1,1) is thus $A={\frac {1}{\sqrt {3}}}(1,1,1)$. The unit normals to the adjacent surfaces of the cube, i.e., the unit normals to the planes x=1, y=1 and z=1, are (1,0,0), (0,1,0), and (0,0,1), respectively. We thus have $A\cdot B=1/{\sqrt {3}}$, where B is any of the three unit normals to the adjacent surfaces of the cube. But then the identity $A\cdot B=\left\|A\right\|\,\left\|B\right\|\cos \theta$, along with $\left\|A\right\|=1$ and $\left\|B\right\|=1$, leads to $\theta =\arccos(1/{\sqrt {3}})\approx 54.7356103^{\circ }$. Red Act (talk) 03:51, 4 June 2011 (UTC)

What about the edge? As far as I know every triangle has a total internal angle of 180 degrees, taking away 90 degrees made by two adjoining faces, leaves another 90 degrees. Due to the symmetry of the resulting isosceles triangle, the remaindingg 90 degrees is divided into two equal sections of 45 degrees. Plasmic Physics (talk) 09:03, 4 June 2011 (UTC)

A formula for the incline at the mid-edge would be 45° cos (θ). Theta represents the surface vector, where zero degrees means pointing away from the mid-face, and 90 degrees means pointing towards a corner. Plasmic Physics (talk) 09:21, 4 June 2011 (UTC)

What ever the incline is at the corners, it is less than 45 degrees, not 54. Red Act is basing his calculations on the tetrahedron. Plasmic Physics (talk) 10:00, 4 June 2011 (UTC)

No, I most certainly am not basing my calculations on a tetrahedron. You're correct about the 45° for mid-edge, but you're confused and mistaken about the corners. Red Act (talk) 13:22, 4 June 2011 (UTC)

Confirmation from my "common sense": If I made it to the corner, using the above concept of symmetry, I should be at the peak of a 3 sided mountain, and in each direction the slope would be 45 degrees down (I assume I would be standing feet on corner, head directly away from center of mass/cube) no? If I continue down this slope in a path equidistant to the 2 ridges (edges) I should halfway to the far corner find myself on level ground no? And if I continue on this perfectly flat face of the world in any direction, continuing to any edge/corner find myself again taking my last step on a 45 degree slope? In essence, walking on perfectly flat terrain, I find my journey from a mountain, taking me into the center of a valley and back up to a very steep peak? 70.177.189.205 (talk) 12:37, 4 June 2011 (UTC)

Red Act: why is it then that your calculated value equals the angle between a face and an edge for a tetrahedron as given in the table on that page?

70.177.189.205: No, it wouldn't be 45 degrees down, to have your head directly away from the centre of mass, you'd have to look straight up. Yes, your final premise is correct. Plasmic Physics (talk) 13:58, 4 June 2011 (UTC)

It's the same number because sometimes two different problems happen to have the same answer. Red Act (talk) 14:50, 4 June 2011 (UTC)

What do you mean by "each direction"? If you're at the peak of a mountain that descends at a slope of 45° in every direction, then the peak must be conical. But if three ridges lead to the peak, then you can descend the mountain faster by heading down in between the ridges, rather than heading down a ridge. In this case you're descending at an angle of about 54.7356° if you head down between a pair of ridges toward the corner at the opposite end of a face, and descending at an angle of $\arccos(2/{\sqrt {6}})\approx 35.2644^{\circ }$ if you head down one of the ridges. I calculated that latter angle as being the angle between the vector (1,1,1) and any of the vectors (1,1,0), (0,1,1) or (1,0,1).

I think your common sense here amounts to you being able to tell that the angle you're looking for is in the ballpark of 45°, so it's intuitively feeling to you like it must be 45°. It really works better in this case to use a little analytic geometry on the problem, so that you won't be led astray by your intuitive guess. Red Act (talk) 14:48, 4 June 2011 (UTC)

Just to be extra clear: the slope calculated as 54.73 is the amount the surface drops down from horizontal toward the centers of the cube faces. The corresponding slope at the edges is calculated the same way, but the edges are not 1 unit but the square root of 2 units away from the center for the same calculation. Thus the slope is gentler, only 35.26 degrees down, going down these ridges. Now it so happens that 54.73 + 35.26 = 89.99 and change - indicating that there is a right angle between the edge and the face when you look at a diagonal slice through the corner of a cube. Wnt (talk) 14:51, 4 June 2011 (UTC)

This question and its answers don't make sense to me. Let's suppose I'm standing on a cube that's maybe 10 feet on each edge, and is positioned horizontally. So I walk the few feet from the center to a corner. How am I suddenly on a "45 degree incline"? ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:08, 4 June 2011 (UTC)

Well, people are debating the direction of the gravity vector at the corner of a gravitating cube relative to its direction at the center of one of the faces. Obviously you needn't be at that angle yourself (you're entitled to lean or lie down) but that's the way you'd stand naturally. 129.234.53.36 (talk) 16:14, 4 June 2011 (UTC)

Do you mean the gravity (such as it is) from the cube itself? ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:36, 4 June 2011 (UTC)

Yes. The question is about Htrae, a cubical planet. Red Act (talk) 16:44, 4 June 2011 (UTC)

"I don't think you understand the gravity of the situation", might be the correct response for Mr. Bugs ;) The Header for the question was for to describe a planet sized cube and the slopes described were for a traveler, with respect to gravity's effect on his travels on the six faced world. Thanks to all for the help. you confirmed my theory and thank you for helping to resolve my oversimplification of the 45 degree vs the actual angles involved. 70.177.189.205 (talk) 16:54, 4 June 2011 (UTC)

Actually, we never did figure out the interesting part of the question: the direction of gravity as you start walking down the mountainside. We know gravity points to the center at the vertex, at the middle of the edge, and at the center of a face. But we don't know exactly where it points when you're at some other arbitrary position. Like the lower gut grumbling, I feel an integral coming on...

Now the problem should be basic enough to state. Defining the cube as +- 1 unit in three dimensions, the magnitude of the gravitational force should be +1 -1 ʃ +1 -1 ʃ +1 -1 ʃ K/r^2 U dx dy dz) where r^2 for the inverse square law = ((x-x0)^2 + (y-y0)^2 + (z-z0)^2), K is proportional to gravity and U is the unit vector toward the point x y z being examined, i.e. ((x-x0),(y-y0),(z-z0))/sqrt ((x-x0)^2 + (y-y0)^2 + (z-z0)^2)) . To convert this into gravitational components in three directions I think we just consider x-x0 or y-y0 or z-z0 depending on which, thus for example +1 -1 ʃ +1 -1 ʃ +1 -1 ʃ K (x-x0)/((x-x0)^2 + (y-y0)^2 + (z-z0)^2))^1.5 U dx dy dz). Ach, but now I have to remember what is done to try to solve a beast like that. Ah, yeah, I remember. Google. Which gets me page 14 of this. http://www.congrex.nl/09m01/papers/11_TU_Delft.pdf You know you were sunk when you don't really understand the answer when you find it. "Adapted from Nagy, 1966, Geophysics, 31, 362-371": g_x = ϱG |||x tan−1 (yz/xr) − y ln(z + r) − z ln(y + r) | ^x2 _x1 | ^y2 _y1 | ^z2 _z1. I suppose if you work through the derivatives that has to come out to the integral above, but I wouldn't have guessed it.. Well, anyway, being out of my expertise I'll see if this gets further comment. Wnt (talk) 21:34, 4 June 2011 (UTC)

I calculated the inclines for the corners: ascending the corner along an edge gives arctan (1/√3) = 30°, ascending along a face gives arctan (2/3) = 33.690°. These values were calculated by taking the diagonal cross section of a cube with a side length 2. A cross section should give a hexagon with side length 1, then three sides are extended to form a triangle with side length 3. The height of the resultant pyramid is half the distance between two opposite corners of the cube, calculated using Pythagoras' theorem twice, equating to √3. Using trigonometric theory twice, the inclines were calculated for a triangular pyramid with base length 3 and height √3.

That proves that in all aproach vector at the peak has inclines less than 45°. No complicated calculus necessary. Plasmic Physics (talk) 23:28, 5 June 2011 (UTC)

Your error starts in your third sentence here. For example, with the coordinate system as above, you can form a regular hexagon from the points on the cube (-1,1,0), (-1,0,1), (0,-1,1), (1,-1,0),(1,0,-1) and (0,1,-1), but the length of the sides of that hexagon are √2, not 1. Red Act (talk) 02:55, 6 June 2011 (UTC)

Red Act: The angle between a face and an edge cannot be the same for a tetrahedron and this pyramid, as you are suggesting. If it was, then it would imply that there is more than one solution for the Opposite in a triangle with a known angle between Hypoteneuse and Adjacent. It just makes sense that a shorter pyramid has a lesser incline than a taller one. Plasmic Physics (talk) 23:51, 5 June 2011 (UTC)

For the cube to be symmetrically truncated, the six edges involved are bisected, a bisected edge is 1 long not √2.

I see what you mean, I didn't use the hypoteneuse. Plasmic Physics (talk) 07:21, 6 June 2011 (UTC)

I never claimed that the angle at the corner of a cube between an edge and the face it intersects at that corner is the arcos(1/√3). That angle is 90°. That's one quick easy way to tell that your answers of 30° and 33.690° above must be wrong, because30 + 90 + 33.690 ≠ 180. Red Act (talk) 02:58, 6 June 2011 (UTC)

I don't understand your first statement. Why are you adding 30 to 33.690? They aren't on on the same plane i.e. they belong to different triangles. Plasmic Physics (talk) 03:06, 6 June 2011 (UTC)

I recalculated the inclines based one the new base length of the hypothetical triangle, and found an edge approach incliine of 35.26439° and a face approach incline of 54.73561°, this confirms both your values. This still leaves me with the question: how can a tetrahedron be equally steep to this pyramid? Plasmic Physics (talk) 07:54, 6 June 2011 (UTC)

A tetrahedron can be formed from (0,0,0), (1,1,0), (0,1,1), (1,0,1). Each of these are sqrt(2) distant from one another. Note that the vertices of a cube are thus produced by two inverse tetrahedrons - the other being 0,0,1 etc. - like a three-dimensional Star of David.

• The center of the tetrahedron is at the average position (0.5,0.5,0.5) and is sqrt (3/2) from any vertex.
• The midpoint of any edge is at an average of two of the four points, e.g. (0.5,0.5,0) and is 0.5 away from the center, and sqrt(2)/2 away from the vertex.
• The midpoint of any face is the average of three of the four points, e.g. (1/3,2/3,1/3) and therefore is sqrt (3*(1/6)^2) = sqrt(1/12) away from the center, and sqrt (2*(1/3)^2+(2/3)^2)) = sqrt (2/3) away from the vertex, and sqrt ((1/6)^2+(1/6)^2+(1/3)^2) = sqrt (1/6) away from the midpoint of the edge.

Therefore:

• The angle at which an edge ascends from the vertical (I used tangents instead of cotangents here without thinking about it) is tan-1(0.5/(sqrt(2)/2)) = cos-1(1/sqrt(2)) = 35.26438 degrees.
• The angle at which a face descends is tan-1(sqrt(1/12)/(sqrt(2/3)) = tan-1(sqrt(1/8)) = 19.47122 degrees.
• This makes the opposing edge and face angle an odd 35.26438+19.47122=54.73560 degrees; but the third angle in that triangle is the angle between two faces at the midpoint of the edge.
• At the midpoint of the edge each face drops down tan-1(sqrt(1/12)/(sqrt(1/6)) = tan-1(sqrt(1/2)) = 35.26438 degrees. The angle between them is 2*35.26438 = 70.52876 degrees. Now 54.73560+54.73560+70.52876 = 179.99996. Note: actually reading tetrahedron finds the edge face angle is 54.7356 degrees and the angle between faces is 70.5288 degrees ... and was absolutely necessary in order to finally get the bugs out of my math. :----< Wnt (talk) 14:52, 7 June 2011 (UTC)

Oh yes, and to come back to coincidences with the cube, there are two main numbers that are the same for both, the 55 degrees and the 35 degrees, which are the arctangents of 2 and 1/2 and add up to 90 degrees. The 55 degree figure is the amount that the cube faces slope down from the horizontal or the tetrahedron edges slope down - note that the tetrahedron's edges pass through three faces of a cube that meet at one vertex. The same number is also the (remaining) angle between the edge and the opposing face of the tetrahedron at the vertex. It is also the amount that the cube edges are bent up from the vertical. It is also half of (90 + the amount a tetrahedron face slopes down). Wnt (talk) 15:28, 7 June 2011 (UTC)

Water

I am writing an article about the world water content.

"how much water does a zoo use per day."

How much water would an wild elephant take during one day.

regards lesley freeman — Preceding unsigned comment added by Waterlessdams (talk • contribs) 01:38, 4 June 2011 (UTC)

Mammals' daily requirements for water for hydration varies significantly according to physical activities undertaken, air temperature during that activity, and relative humidity. It is likely that the daily consumption by elephants in captivity is well known, particularly those that are kept in zoos. But for elephants in the wild I would expect little to be known because of the difficulty of measuring how much an elephant in the wild actually drinks during its daily wanderings. I suggest you go looking for some information about how much an elephant in captivity drinks in a day. Dolphin (t) 04:34, 4 June 2011 (UTC)

What species is this fungi?

I found it in a jungle.--Inspector (talk) 03:44, 4 June 2011 (UTC)

A jungle in China? Looie496 (talk) 04:28, 4 June 2011 (UTC)

In any case it looks like some type of cup fungus, but there are a lot of them. Looie496 (talk) 04:35, 4 June 2011 (UTC)

anatomy

which is the most constricted part of gastrointestinal tract? — Preceding unsigned comment added by Akash541 (talk • contribs) 06:49, 4 June 2011 (UTC)

There are three 'pinch points' (sphincter muscles) in the GI tract all evolved for their particular purpose. The highest is the pyloric sphincter cardiac sphincter where the oesophagus enters the stomach. The next is the pyloric sphincter where the stomach empties into the duodenum and the last is the endpoint, the anus. My opinion is that the anus is the strongest of those three and (thank goodness) the most constricted when not defaecating. Richard Avery (talk) 07:03, 4 June 2011 (UTC) I have removed your duplicated question

Is not defecation a "normal circumstance", Richard? I wouldn't be too happy if my anus was constricted at that time. (I can't believe I'm referring to "my anus" in a place where the entire world online community can read it, but there you go.) -- Jack of Oz [your turn] 08:08, 4 June 2011 (UTC)

Perhaps the significant thing is that, unlike the earlier pinch points, you (hopefully) have conscious control over releasing that final constriction. (I'm happy to keep talking about it, since you started it. HiLo48 (talk) 08:22, 4 June 2011 (UTC)

Right I take your point Jack, I've reworded my reply. (Hmm, we're talking about 'down under' in down under!)

@ HiLo, of course there are some people who can release their oesophageal sphincter at will to belch, vomit or ?swallow swords. Right I'm off to have breakfast. Richard Avery (talk) 08:28, 4 June 2011 (UTC)

I think that releasing the pyloric sphincter is not that difficult either - not particularly more difficult than gaining control of the anus was earlier in life, I would say. It rather appalls me that so many people have become dependent on taking special pills for gastric reflux, when a simple motion can alleviate the pressure. (Though IMHO the pylorus has certain tastes of its own which are hard to override - likes sour, hates scratchy stuff - thus milk and cereal are not good to eat in the evening...) Wnt (talk) 14:58, 4 June 2011 (UTC)

Err, that's a hard-working pyloric sphincter that has to work both sides of the stomach :). The connection between the esophagus and the stomach is the lower esophageal sphincter (cardia). Also note that there's a higher pinch-point at the upper esophageal sphincter which is in your throat. That's technically not part of the GI tract, but I mention it as some people use that term to encompass the entire digestive tract from mouth to anus. Matt Deres (talk) 11:55, 4 June 2011 (UTC)

DC vs AC

I am currently reading a book made by an electrician expert and he says this

"Low-frequency (50- to 60-Hz) AC is used in US (60 Hz) and European (50 Hz) households; it can be more dangerous than high-frequency AC and is 3 to 5 times more dangerous than DC of the same voltage and amperage.Low-frequency AC produces extended muscle contraction (tetany), which may freeze the hand to the current’s source, prolonging exposure. DC is most likely to cause a single convulsive contraction, which often forces the victim away from the current’s source."

Also he said DC just stops the heart while AC is making it fibrilate which makes the heart harder to get it back to work. yet I know exactly the opposite :). DC is deadlier in terms of same voltage/current because it is constant and not like AC alternating. --Leonardo Da Vinci (talk) 07:08, 4 June 2011 (UTC)

You know how? And what do you mean 'DC is deadlier in terms of same voltage/current because it is constant and not like AC alternating'? Anyway there is some discussion at Electric shock. It doesn't go in to detail in the AC vs DC but I would trust what it does say where sourced more then what 'you know'. Also did I miss something? What exactly is the question? Nil Einne (talk) 09:39, 4 June 2011 (UTC)

There is much difference in opinion over this, and the argument has been going on for a very long time. The comparison between the dangers of AC and DC at similar voltages is complicated by the fact that the effect depends mainly on exactly how the shock occurs. Two people can receive apparently the same shock, but one can walk away unharmed whilst the other doesn't walk again. Does anyone know of any published scientific research on this? I tried using myself as a guinea-pig many years ago, and I can confirm (OR warning) that AC feels more dangerous, but I stopped short of passing dangerous currents in the region of my heart. Dbfirs 11:50, 4 June 2011 (UTC)

See War of the currents for an account of late 19th century demonstrations (paid for by some inventor/industrialist whose name does not come to mind), which showed AC to be more lethal than DC at a range of voltages. Edison (talk) 20:39, 4 June 2011 (UTC)

The same inventor/industrialist who made a Snuff film of poor Topsy? Cuddlyable3 (talk) 23:29, 4 June 2011 (UTC)

How to check for signs of life in a non-pulsatile person

Before this gets unfairly written off as a request for medical assistance, I'd like to clarify that if I was indeed looking for help for myself or someone else, I wouldn't come here, type a question, then wait 20 minutes for an answer while mine or someone else's life quickly dwindles. Now my question is how do you check for signs of life in a person who is non-pulsatile should they become unconscious? How would an average person with no medical knowledge who is not aware of the non-pulsatile person's condition be able to know? 173.2.165.251 (talk) 13:46, 4 June 2011 (UTC)

First of all, I find it hard to imagine a person who has no pulse being concious. From the first-aid viewpoint if you find someone who you believe may have recently lost conciousness and has no pulse and is not breathing, or you are with someone who collapses under the same conditions then do not waste time trying to determine whether the person is 'alive' or not. Call for emergency help, ensure your own safety and commence first-aid cardiopulmonary rescusitation, CPR. If you are not familiar or practised in this procedure then you should get in touch with a local first-aid organisation, enrol in a class and become proficient. You might someday save a life (correctly, postpone a death) but in any case you will not have niggling doubts about what to do if you are ever in the position you originally imagined. Caesar's Daddy (talk) 14:06, 4 June 2011 (UTC)

I should've clarified that the particular type of people I'm talking about are those who are on a ventricular assist device, which if I read that article correctly, some are non-pulsitle. And I believe Dick Cheney is one of those people who use such a device, because the last I heard he really has no pulse. 173.2.165.251 (talk) 14:35, 4 June 2011 (UTC)

Ah ha, well, that changes things slightly. Firstly address the person with a shake of the arm or shoulder to elicit some response, if that is negative then lightly pinch the ear lobe or similar place to elicit a pain response, if there is no reaction then check whether they are breathing by putting your cheek close to their mouth or nose and looking down their body for any respiration movements for about 10 seconds. If both these are negative then call for emergency aid and commence CPR etc. The need for further training still applies. Caesar's Daddy (talk) 15:16, 4 June 2011 (UTC)

Current (2010) AHA standards scrapped the whole "look, listen, and feel" stage after "check for responsiveness"...now one starts compressions much earlier--even before establishing airway or giving first breaths. NB, this is not medical advice, just a statement about their published standards and overhauled training materials. DMacks (talk) 20:06, 4 June 2011 (UTC)

I am sure there is an article about it, but the question reminded me of stage magicians who persuade the audience their pulse has stopped. I don't know how it is done but I doubt it is magic. Kittybrewster ☎ 15:24, 4 June 2011 (UTC)

Advice used to be to hold a mirror over the person's nose and mouth for 10 - 15 seconds. If it frosts up you know they are breathing. I agree with the pinching of the ear lobe - it wakens people who are in a deep sleep. --TammyMoet (talk) 16:35, 4 June 2011 (UTC)

In Charade, George Kennedy stuck a needle in his victim. This is not medically advised. Kittybrewster ☎ 17:03, 4 June 2011 (UTC)

In US CPR courses a few years ago they emphasized making noise, and not being stealthy, when kneeling by a downed person. One was to say loudly "Are you OK?" and shake the person, so that a bypasser would not assume you were mugging the person. Edison (talk) 20:36, 4 June 2011 (UTC)

In the current curriculum for first aid in Germany (well, I was re-certified in February, IIRC), people are taught not to look for a pulse anymore - just to check breathing (by bringing your face close to the nose and mouth, and observing the chest). Apparently, people had a hard time reliably finding a pulse under stress situations. If a patient does not breathe, he won't have a pulse for long, anyways (and, rare anomalies excluded, vice versa). So if there is no breath, you start CPR. BTW, does anybody know if AEDs are programmed to recognise ventricular assist devices? I suspect a shock would not be advised... --Stephan Schulz (talk) 21:17, 4 June 2011 (UTC)

Patients with ventricular assist devices tend to be prone to heart problems, and compatibility of VADs with external defibrillation is an important design consideration. (Here's an interesting case report. The patient was in ventricular fibrillation for seven hours, and his LVAD kept him alive until he could be electroverted. They shocked him three times with external paddles before normal rhythm was restored.) In general, most of the electronics are located a reasonable distance below the heart; the only bits directly in the path of the current are going to be plastic plumbing. (Similarly, implantable pacemakers generally sit well above the level of the heart.)

On the other hand, conventional CPR can be very dangerous for these patients, as the chest compressions can dislodge the tubing that connects the ventricular assist device and cause massive internal bleeding: [1]. TenOfAllTrades(talk) 22:01, 4 June 2011 (UTC)

So are these devices obvious or detectable by exterior examination, if they are then that's one more thing to check and if they are not you would have to play the percentages and get on with CPR. Caesar's Daddy (talk) 22:18, 4 June 2011 (UTC)

If they aren't obvious, then the patient should wear a Medical identification tag. That article does list "Pacemaker or other implantable medical devices" as one of the reasons for wearing one. --Tango (talk) 01:30, 5 June 2011 (UTC)

How about some references, here on the Reference Desk? A quick test is the sternum rub, but some unconscious patients do not respond to it. More formally, here is a handy PDF file discussing how to elicit the gag and cough reflexes, corneal reflexes, and so forth. Comet Tuttle (talk) 03:49, 5 June 2011 (UTC)

Trained medic here. Taking a pulse isn't always easy. We are trained to look for the other signs of life that indicate there is a pulse. You can waste time on a casualty looking for a pulse, and there are other things you can look for. Get low down to the patient, put your ear near his mouth and look for chest movement and listen for breath. Judge the color of the skin of the lips. They change color pretty quickly when there is a problem. Hope this helps. Zzubnik (talk) 08:33, 6 June 2011 (UTC)

It's fairly complicated, establishing if someone is alive or not, pulse and breathing are not very accurate as signs of life. It seems that signs of life now refer more to brain activity, here's a description of techniques used to establishbrain death or in an how stuff works article.

However some tools (other than pulse/breathing) which are used to establish signs of life:

• Glasgow Coma Scale (level of consciousness) wiki web
• Core Temperature
• Pupils will be not move and go wide in death (and visa versa).

JamesGrimshaw (talk) 03:05, 8 June 2011 (UTC)

Calibrating length markers for micrographs?

In scientific articles, micrographs often include a line which is equal to a stated distance (e.g. one micrometre). How do the authors determine the correct length for this line? How is it calibrated? --129.215.4.89 (talk) 16:10, 4 June 2011 (UTC)

The details depend on the type of microscope and imaging system used, but it boils down to geometry. The magnification factor associated with each set of optical elements in a microscope will be known (specified and calibrated by the manufacturer), which means that one can figure out the size of the image formed by those optics compared to the size of the object that you're looking at. (If the field of view with a low-ish power objective is one millimeter across, and the camera sensor is 1000 pixels across, then you know that each pixel in the resulting image is 1 micrometer – 1 micron – wide.) In practice many modern microscopy systems hide all the math behind the scenes, and automagically spit out a calculated size for each pixel in the image. Some software allows the user to draw scale bars on directly.

If you want to check your math, you can also calibrate the scale using a test object with features of a known size. For high-precision work, one can purchase a stage micrometer; essentially a test slide marked with a very tiny ruler: [2]. If you're in a biology lab, someone probably has a hemocytometer, which incorporates a grid of regularly spaced lines. TenOfAllTrades(talk) 19:28, 4 June 2011 (UTC)

There are also grids available, but I suppose that's too simple, and you were asking about instances in which no grids were used. DRosenbach ^{(Talk | Contribs)} 20:31, 5 June 2011 (UTC)

EHEC again, receptor

Shiga-like toxin says the receptor the toxin binds to is called Gb3. (I can't find an article on it.) What is that receptor good for if no toxin is around? I guess it does something useful and is not only loitering around waiting for The Wrath Of The Cucumbers to come along. 77.3.146.181 (talk) 20:29, 4 June 2011 (UTC)

Googling around, Gb3 appears to be globotriaosylceramide, a member of the glycosphingolipids. DMacks (talk) 20:39, 4 June 2011 (UTC)

Hmmm, it seems to interfere with HIV infection.^[3] The α-galactosyltransferase that synthesizes it is one of the markers for HIV resistance. Wnt (talk) 23:48, 4 June 2011 (UTC)

Sleep in older people

I've read that older people need as much sleep as younger adults, but they just wake up earlier so they end up sleeping less. Is it true that the amount of sleep needed by humans is not diminished even though they generally sleep less as they age? Is it known, biologically, why older people tend to sleep less? --173.49.79.135 (talk) 20:54, 4 June 2011 (UTC)

Here is an article in PubMed that speaks directly to the issue. The abstract says that the need for sleep is the same, regardless of age, but that the ability to sleep is damaged. Bielle (talk) 21:04, 4 June 2011 (UTC)

Well, as for my original research on getting old myself, I can't say that I need less sleep, nor do I actually sleep less. Might be other people accumulated too much of a bad conscience that won't let them have a sound sleep. 77.3.146.181 (talk) 21:06, 4 June 2011 (UTC)

if you look at the article, you will find no mention of "conscience", but only of physical disorders that lead to less than the ideal amount of sleep. Bielle (talk) 21:13, 4 June 2011 (UTC)

Now I see clearly why we should mark comments with an "(ec)". I didn't have a look at the article or your comment before I wrote mine. 77.3.146.181 (talk) 21:28, 4 June 2011 (UTC)

It is very clear that older people don't sleep as long, on average, as younger people, but whether they need less sleep is much less clear. There are research studies pointing in both directions. My take on the overall data is that in the elderly, a sleep duration of around 7 hours leads to the best performance on fatigue-sensitive tasks, and solid sleep is better than broken sleep. In biological terms, some of the causes of altered sleep have been identified (altered circadian rhythms, for example), but I don't think anybody has identified a functional reason why the elderly would sleep less. Looie496 (talk) 21:28, 4 June 2011 (UTC)

It is very clear? The old people I know seem to sleep about the same amount as other adults... --Tango (talk) 23:18, 4 June 2011 (UTC)

I have read some time ago that young adults of up to 25 years of age need far more sleep than they usually get. They need 9 to 10 hours of sleep, while they typically get 7 hours sleep. Also, I've read that before the 20th century, people did sleep a lot longer than we do today. Count Iblis (talk) 23:38, 4 June 2011 (UTC)

As an "older" person (in my 60s) I simply must add to this discussion that a reason I sleep for shorter periods is the basic need to urinate more frequently. I'm pretty confident that this is not a rare condition. (NOTE: I am NOT seeking medical advice) OR from other "older" folks would be welcomed. HiLo48 (talk) 23:42, 4 June 2011 (UTC)

I'm not that old, but I also have to urinate quite frequently. But I still manage to sleep 8.5 hours per day, I just have to go to bed about 9.5 hours before I have to wake up. So, I think this is the fundamental problem: Most people go to bed way too late so that any disruption of sleep is going to lead to less sleep than they ideally need. I think another factor is that people don't get enough exercise. If you don't do a few hours per week hard exercise like fast running, your sleep cycles may not kick in with full force. Count Iblis (talk) 00:06, 5 June 2011 (UTC)

3 questions about Endocrinology

1.Why do we separate between Genitals and Gonads?. 2.What comes first in the Fetus? - The Genitals or the Gonads?. 3.In what word shall we use to describe both Genitals and Gonads as "1 reproductive package"?, whether it be Male, Female, or of an Inter-sex...?

sorry for the ignorance.

1000 thanks. 109.67.42.106 (talk) 21:01, 4 June 2011 (UTC)

According to our article Gonads: The gonad is the organ that makes gametes. The gonads in males are the testes and the gonads in females are the ovaries. Gonads in females are internal and gonads in men are external. They are a part of the genitals; the word "genitals" or "genitalia" covers all the parts. Bielle (talk) 21:10, 4 June 2011 (UTC)

I don't think very many people use the word genitals in such a way as to include the ovaries. The testicles, yes. I think the usage notes at [[4]] (WARNING: if you're at work and you think your IT department might snoop around in browser caches, don't click) are a bit dubious, frankly. --Trovatore (talk) 21:18, 4 June 2011 (UTC)

Isn't there a Greek\Latin word for BOTH organs? (both Vagina\Ovaries | Penis\Testicles)

Thanks.—Preceding unsigned comment added by 109.67.42.106 (talk) 03:00, 5 June 2011

Sex organs or genitalia most certainly include gonads, birth canal parts, and penis. Sex organs are, naturally, any organs that an individual does or does not have depending on the individual's sex (gender). --PeeKoo (talk) 11:20, 5 June 2011 (UTC)

Note that carpel and stamen are also sex organs. --PeeKoo (talk) 11:23, 5 June 2011 (UTC)

June 5

Foods S.PROVEN to include Somatotropin?

can somebody list some?,

Thanks?. — Preceding unsigned comment added by 109.67.42.106 (talk) 02:57, 5 June 2011 (UTC)

This is addressed in Growth hormone, which says that most of the supplements actually advertise that somehow they "release" HGH. The only food I can think of that is sure to contain somatotropin is pituitary... but note that even primate GH is inactive in humans. So if you want GH that is active in humans you have to cannibalize human pituitaries Dawn of the Dead style. No, wait, that doesn't work, because the stuff has to be injected to avoid being broken down by the digestive system. In general - human proteins do many fascinating things, but they are not readily accessible as herbal medicine. Wnt (talk) 06:21, 5 June 2011 (UTC)

Please help me identify unknown bacteria

I initially thought I had Serratia fonticola, but my instructor basically hinted that this is wrong since we didn't see this baterium during lab. I'm thinking I should have a very common bacterium (perhaps an enterobacter). Also, because our reagents are getting old, my Indole red test may be negative or positive; I don't know.

Here's my results: Gram negative rod. Methyl Red: POS. Vogues-Proskauer: NEG. hydrogen sulfide production: NEG. Motility: POS. Citrate: POS. O/F glucose: positive (turned yellow). O/F glucose + 1/2" oil on surface: positive (turned yellow). facultive respiration. Oxidase: NEG. Catalase: POS. Nitrate reduction: POS. Nitrite to Ammonia: POS. Urea hydrolysis: NEG. Casein protien hydrolysis: NEG. Starch hydrolysis (amalase production): NEG. Tryptophan degradation (using tryptone broth): NEG. Phenylalanine deamination: NEG.

All of the following tested positive for acid and gas production: glucose, glycerol, lactose, sucrose, mannitol, maltose. Thank you for the help161.165.196.84 (talk) 04:20, 5 June 2011 (UTC)

There's a policy here about doing homework problems - just out and out giving you the answer is frowned upon. More to the point, I don't know what to do with this information anyway. So to start with let's see if we can figure out the logic to this problem.

• The IMViC test series is meant to identify coliform bacteria. The initials stand for:

• Indole test "Indole red may be positive or negative, I don't know"
• Methyl Red test "Methyl Red: POS"
• Voges–Proskauer test "Vogues-Proskauer: NEG."
• Citrate test "Citrate: POS"

Now even at this point I should point out, there are already unwarranted assumptions. By chance I was just reading about Richard Lenski's famous ten-year E. coli selection experiment, in which he found that in one flask, the bacteria suddenly started to use citrate as a food source. Before long they became quite competent at it, even "speciating" into a large Cit+ population and a small specialist population that could only use glucose. But according to the tests above, Cit+ means your bacterium is not E. coli! Since surprises like this also exist in nature, it's hard to be confident about such tests ... and yet, generations of microbiologists have somehow managed to do extraordinarily good work with them. It boggles the mind. Wnt (talk) 06:10, 5 June 2011 (UTC)

Thank you for the response. To clarify, I do understand Wikipedia's hw policy; I just need some direction. My lab manual says I should look at Citrobacter, but this guy produces hydrogen sulfide (so that's not it). I've gone through the Bergie's manual a few times now, to no avail. I'm thinking I have an enterobacter, but I need help figuring out which one. I KNOW I don't have Bacillis subtillis, Providencia, Morganella, Proteus, Serratia, and probably not Klebsiella or Shigella — Preceding unsigned comment added by 161.165.196.84 (talk) 06:23, 5 June 2011 (UTC)

I don't think it really boggles the mind that much. The fact that such things can and do exist in nature doesn't meant they are common. It's obvious that Cit+ E. coli are rare enough in nature (which was after all one of the reasons the discovery was significant), that the test works the vast majority of the time. The reason the trait is so rare would likely be because in most environments E. coli live in (particularly those likely to lead to E. coli ending up in food or water) the trait isn't beneficial enough compared to the cost. That's hardly surprising or difficult to understand, except perhaps to the founder of conservapedia. Nil Einne (talk) 06:52, 5 June 2011 (UTC)

Well, Citrobacter does not necessarily produce H₂S ... here's an article about 25 strains which don't.^[5] In a pinch, it's a case you could argue. Wnt (talk) 04:50, 6 June 2011 (UTC)

On the other hand, nitrate reduction:POS would seem to me to indicate this is among denitrifying bacteria, and nitrate to ammonia:POS I think means that ccNIR nitrate reductase must be present, though I didn't research alternatives. I don't see anything about ccNIR in Citrobacter. Wnt (talk) 05:00, 6 June 2011 (UTC)

Half-life of Proton-AntiNeutron bound state

The class of exotic atoms consisting of Protons, Electrons, and anti-neutrons would seem potentially interesting, given they might provide a way to store anti-neutrons in a normal matter world. But only if the half-life is long enough. I presume that the anti-neutrons wouldn't get close enough to neutrons in surrounding matter to annihilate. Is that correct? How long would bound proton-antineutron nuclei last?

BTW, what are these sort of atoms called? I can see no reference to them in Wikipedia.88.104.247.65 (talk) 10:08, 5 June 2011 (UTC)

Do you have a reason to think such a combination would be bound? (I'm no QCD expert myself, so if it's blindingly obvious, no big.) --Trovatore (talk) 10:28, 5 June 2011 (UTC)

I don't know about any kind of bound state either, but I would expect that the proton's down quark and one of the anti-neutron's down anti-quarks as well as one of the proton's annihilate quickly, leaving a meson made up of an up quark and a down anti-quark (e.g. a pion or a rho meson). Icek (talk) 13:37, 5 June 2011 (UTC)

That's correct. The proton and the anti-neutron would annihilate each other into a set of mesons almost instantaneously. Dauto (talk) 14:42, 5 June 2011 (UTC)

A neutral pion isn't dissimilar, in that it contains a quark and the corresponding anti-quark, and is stable enough to exist as a known particle. The mean lifetime is only about 10^-16s, which you may consider "almost instantaneous", but it's still far from zero. I know very little about the subject, but it seems plausible to me that a proton-antineutron particle could have a comparable lifetime (which is, of course, far too short for the purpose proposed by the OP). --Tango (talk) 16:45, 5 June 2011 (UTC)

The neutral pion decay is governed by the electromagnetic interaction while the proton-neutron decay is governed by the strong interaction. That makes the latter decay much faster than the former. My back of envelop guesstimate gives about 10^-23s which is pretty much instantaneous by any standard. Dauto (talk) 17:55, 5 June 2011 (UTC)

I guess the thrust of the question was the extent to which the quarks and anti-quark could "see" each-other and annihilate rather than being constrained inside the proton/anti-neutron. If a quark-antiquark pair itself has a measurable half-life, I think that's quite encouraging that a proton/anti-neutron might have have a longer one, though maybe a second is pushing it! Dauto, can you share the back of your envelope?--88.104.247.65 (talk) 20:33, 7 June 2011 (UTC)

Hmmm, how about a He-4 nucleus, an electron, and an antiproton? The presence of the electron means this em 'atom' can never have spin 0, so the nucleus and antiproton are always spinning around each other at some rate and never touch, right...? (somehow I don't think this works that way, but I had to ask) B) Wnt (talk) 05:05, 6 June 2011 (UTC)

No, not right. The anti-proton will be in a s shell state around the nucleus and will have a non-vanishing probability to be found at the origin where it will annihilate one of the nucleons in short time. Dauto (talk) 05:59, 6 June 2011 (UTC)

In your usual atom, electrons don't fall into the nucleus because they're small and bouncy and won't stay put. An antiproton is much heavier, so it's pulled to the nucleus much more easily, and there I guess it's annihilated with a proton or neutron. – b_jonas 09:20, 6 June 2011 (UTC)

EHEC, and now it's biogas

Newest reports now blame biogas. Experts think it is possible that bacteria could mix in the tanks and thus generate new strains. Biogas lobby instantly denies and states that the substrate would be heated for at least one hour to 70C. Now I wonder if that heating takes place at the begining of the process, at the end or two times. I wonder further if that heating would not consume a lot of energy and void the benefits of biogas. Does anyone know some facts and details about that? 77.3.180.185 (talk) 11:16, 5 June 2011 (UTC)

I take it you are talking about the 2011 E. coli O104:H4 outbreak? I cannot find any reports connecting this with biogas - could you provide a link? 81.98.38.48 (talk) 11:44, 5 June 2011 (UTC)

So it probably only available in German language. AFP via Google and the press release from the biogas lobby 77.3.180.185 (talk) 13:10, 5 June 2011 (UTC)

The story you provide only appears to quote one person going by a machine translation. It's also not clear how much expertise they have in the matter (they work for a medical laboratory although in what role is unclear, even what the medical laboratory does is unclear). There's something on the Agricultural and Veterinary Academy which I'm not clear about but I think it's either saying it's unlikely or it's something to look in to (which they probably say with all possible leads) Nil Einne (talk) 15:46, 5 June 2011 (UTC)

The medical laboratory is doing those kinds of test as they are proposing. So if they are right, they would have dug up a really big business for themselves and other laboratories like theirs. Also, they, as well as those from the veterinary academy are not saying the actual EHEC strain does provably come from a biogas facility, only that it was possible that it did. Anyway, my question is about the heating of the substrate for biogas generation and its costs in terms of energy. 77.3.180.185 (talk) 16:19, 5 June 2011 (UTC)

Would seem to be a very big COI then (it's questionable if they even have to be right or just spread enough FUD to get business). And this still doesn't show whether or not they are likely to be experts on the evolution & spread of new E.coli strains who have any idea what they are talking about (your first statement was 'Experts think it is possible') I wouldn't normally expect those who are primarily involved in the testing side of things to be. (As I said about the Agricultural and Veterinary Academy, I can't really understand that part but it sounds to me like they are mostly just saying 'it may be possible, we will look in to it' which from a political POV is better than saying 'the idea is nonsense and the people suggesting it have no idea what they are talking about' even if they really think that.) Also, many things are possible, it doesn't mean it's likely. Nil Einne (talk) 03:16, 6 June 2011 (UTC)

Ha, so Germany may have more deaths due to an industrial accident involving green technology than with nuclear technology. And I think they are also against irradiation of food. :) . Count Iblis (talk) 14:52, 5 June 2011 (UTC)

Whatever you think of German "ecological" politics, believe me, it's worse. 77.3.180.185 (talk) 16:25, 5 June 2011 (UTC)

That hypothesis is newly a week old. It is more likely a fault with the water recycling treatment plant used in bean sprouting. [6]--Aspro (talk) 16:53, 5 June 2011 (UTC)

Since no one else has answered. About the second question, this is not something I have much experience with but while the heating process obviously won't help the equation in favour of biogas I don't see any reason to think it's going to make it completely untenable. Any energy extraction and production system has lots of costs that would be nice to do away to improve the energy efficiency equation with but are ultimately part of the cost. This would include biogas and I doubt the heating cycle is the biggest cost. Since from the sound of it this isn't a new requirement, if biogas is worth it then this would presumably be with the heating cycle requirements. Nil Einne (talk) 03:25, 6 June 2011 (UTC)

About the first question, a simple search for 'biogas heating 70' finds plenty of discussions. [7] for example suggests pre and after heating is done in some circumstances (and also looks at the effictiveness of the sanitisation). [8] shows the heating cycle and suggest it is primarily part of the production/fermentation cycle rather than simply intentional heating for sanitisation purposes. [9] appears to also have two 70 °C preheating cycles although there's little discussion of the biogas production so it's not clear if 70°C is reached there. [10] appears to use a post-heating pasteurisation but doesn't mention much of the initial process. [11] uses a CHP system to achieve 70-90°C apparently to increase production efficiency. [12]and [13] appears to only use pre-heating. From these, I think it's clear it depends a lot on the precise process. Generally pre-heating is the norm, which makes sense from a biological/sanitisation POV. (Remember other outputs would often be fertiliser or compost.) Post-heating may also be done. And in some cases the heating may be part of the fermination process. Nil Einne (talk) 04:08, 6 June 2011 (UTC)

It sounds like one person spouting off with no evidence (except, perhaps, a check from Chevron in his pocket). Yes, biogas uses cow manure or other yucky stuff that could contain E. coli. And 70 C is not really that impressive a method of sterilization, especially when we consider how often accidents happen in waste disposal. (Just work downwind of a city incinerator late at night. It's amazing how many "accidents" happen...) But it could probably much more easily be somebody trying to fertilize the vegetables directly with cow manure or some improperly processed derivative. Or a cow that got loose and wandered into the field. Or... well, just about anything. The investigators have a long, thankless job ahead of them. Wnt (talk) 05:18, 6 June 2011 (UTC)

What is the name of alkane that has 1000 and 10000 carbon atoms?

--Inspector (talk) 14:01, 5 June 2011 (UTC)

1000 is kiliane, but the table of length prefixes for generic long-chain molecules in the IUPAC_Blue_Book does not have 10000. DMacks (talk) 15:45, 5 June 2011 (UTC)

As DMacks mentions, there doesn't appear to be an official term for a 10,000 carbon atom alkane, but note that the factor of tens are based around the SI prefixes. There isn't currently an SI prefix for 10000, but at one time there was myria-, so were there ever a need to have a name for a 10,000 carbon atom alkane, something like "myriane" might be proposed (possibly with "diriane", "tririane", ... for 20000, 30000, etc.). Note that you'll only need those terms if it's a linear alkane of 10,000 carbon atoms. A branched hydrocarbon would be something like "260,375,1503,2673,2654,4536,4675,5430,7650,7895-decahectylnonaliane". As final note, at that size you're really not in the standard hydrocarbon regime, but rather in the land of polymers. Most chemists would likely view a saturated 10000 carbon atom molecule (linear or branched) not as an alkane, but as a polyethylene molecule. - 174.31.219.218 (talk) 17:50, 5 June 2011 (UTC)

Its also kinda moot; the ability to do reasonable organic (non-biological) synthesis on molecules is so small that most organic chemists don't deal with discrete pure substances composed of identical 10,000 carbon-atom chains. As noted by 174.31, very long carbon chains generally fall in the realm of polymer chemistry and polymers are composed of a range of molecular sizes, one sample of polyethylene may have an average chain length of, say, 10,000 monomer units, +/- 1000 units with a certain level of branching, while a different sample may feature a chain length of 20,000 +/- 500 units with less branching, or something like that. You just don't deal with bulk substances composed of pure 10,000 carbon strait-chain hyrdrocarbons. The only other molecules that get that large are biomolecules like nucleic acids and proteins and stuff like that, and that sort of stuff is dealt with by biochemistry, and biochemical molecules do not follow the IUPAC naming standard for obvious reasons. What would be the IUPAC-standard name for hemoglobin? Does it matter? --Jayron32 18:57, 5 June 2011 (UTC)

Hair colour

I was wondering why natural human hair colour has such a narrow spectrum - no greens or blues, no proper red (whatever we may call it). And then I wondered if any of our close relative primates have hair/fur outside of these boundaries? Cheers --Dweller (talk) 18:58, 5 June 2011 (UTC)

There are only a three pigments availible for human hair color. The relative amounts of each determines your hair color. There is brown eumelanin, black eumelanin and pheomelanin, which is redish in color. Whether you have blond, brown, brunette, mousy brown, black, strawberry blond, gray, red, auburn, carrot orange, etc. is determined by how these three pigments are combined in your hair. There is a discussion of this at Human_hair_color#Genetics_and_biochemistry_of_hair_color and in links from there. --Jayron32 19:19, 5 June 2011 (UTC)

Sorry I wasn't clear in my question... I wasn't asking what the mechanism was for producing those colours, I was asking why we evolved with such a narrow spread... --Dweller (talk) 19:30, 5 June 2011 (UTC)

All three pigments are different types of melanin. They are very similar and the genes for them vary by only very small changes. Evolving completely new pigments would require much bigger genetic changes, which makes it much less likely to happen. --Tango (talk) 19:36, 5 June 2011 (UTC)

The technicolor specimens of hominidae got picked off by even the legally blind hawks. DRosenbach ^{(Talk | Contribs)} 19:49, 5 June 2011 (UTC)

What's more remarkable is that this same limited palette is used by so many other mammals - see equine coat color, for example. Even Russian Blue cats are actually just using diffuse black pigment, though looking at one I'd think there must be a bit of structural iridescence at work. Other animals like birds and butterflies have carotenoids and flavonoids; one group of birds invented turacoverdin from scratch. But I just can't think of a mammal right now that goes beyond the black-brown-orange-yellow range of melanins in hair. But I'm probably forgetting something obvious. Wnt (talk) 05:32, 6 June 2011 (UTC)

Mandrills are colorful according to our article, though I have no idea about the pigments they use. – b_jonas 09:13, 6 June 2011 (UTC)

A mandrill's skin is indeed colorful, with blues, reds, pinks, and purpes, but its pelage is dark grey. -- 110.49.251.120 (talk) 15:19, 6 June 2011 (UTC)

Blue in mandrills (and in the vervet monkey's magnificent nutsack) is structural. --Sean 17:01, 6 June 2011 (UTC)

Nice explanation about mandrills' color, both of you. Thank you. – b_jonas 18:28, 6 June 2011 (UTC)

I was going to research this Illyrian poodle^[14] a bit further, but stumbled quickly onto a fairly convincing-looking literature review in, of all places, a cycling forum: [15] which quotes from (in case that link disappears) "P. Sumner and J.D. Mollon Colors of Primate Pelage and Skin: Objective Assessment of Conspicuousness - American Journal of Primatology 59:67-91 (2003) http://vision.psychol.cam.ac.uk/jdmollon/papers/sumner_mollon2003.pdf" which in turn cites Nassau 1983 and Slifka et al, 1999. Wnt (talk) 17:47, 7 June 2011 (UTC)

East and west reversed when looking at the Moon?

The article Near side of the Moon says that although while standing on the Moon, east and west are where you'd expect them to be, when looking at the Moon on the sky, they are reversed. I couldn't understand why, and looking at Talk:Near side of the Moon#Orientation of the Moon, I see I'm not the first one to wonder at this. I thought it could be because of a mirror effect - after all, when standing on the Moon, the lunar surface is below your head, but when looking at it on the sky, it's above your head - but then I looked at the "Blue Marble" photograph on the article Earth, which is a genuine photograph of Earth from space, and it showed east on the right and west on the left, just as you'd expect them to be. Could anyone explain why they are supposed to be the other way around on the Moon? JIP | Talk 19:49, 5 June 2011 (UTC)

The article is not very well written, but maybe what it's getting at is that if you think of the Earth and Moon at a time when the Greenwich meridian is pointing directly at the Moon, then the Western edge of the Earth is "opposite" the Eastern edge of the Moon, and vice versa. It's a similar idea to the (supposed) left-right reversal of the image in a mirror. AndrewWTaylor (talk) 20:07, 5 June 2011 (UTC)

(ec)The statement makes no sense. The moon's orientation in the sky depends on where you are standing on the earth, since the earth is round. Dauto (talk) 20:08, 5 June 2011 (UTC)

I think the question is about the idea of east and west on the Moon itself, which is not obviously a well-defined notion as it stands; some convention must be chosen. You could say that the Moon's "north pole" is the pole that's in the same direction from the Moon's center that the Earth's north pole is from its center. Or, you could say that the Moon's north pole is the pole where the Moon, turning under your feet, turns the same direction that the Earth turns under your feet at the Earth's north pole (that is, counterclockwise, if I've done that right in my head). I think the two definitions would give the opposite answer. --Trovatore (talk) 20:12, 5 June 2011 (UTC) No, I take it back, I think they give the same answer. Still, it's a point that needs further clarification. --Trovatore (talk) 20:14, 5 June 2011 (UTC)

I'm assuming the North Pole on the Moon is the pole that points the same way from the Moon's center as the North Pole on Earth points from the Earth's center. I have no experience on being on the Moon, so the only way I can visualise this is imagining looking at another Earth on the sky instead of the Moon. However, I cannot help but imagine that if I am standing on the Earth, facing north, then east is on my right and west is on my left, both on the Earth under my feet and on the Moon up in the sky. Would the Moon look different if I was facing south? "North is up, south is down" is, to the best of my knowledge, an arbitrary decision and not some irrevocable basic law of physics. Surely it's possible to view the Earth from space as upside down from the "Blue Marble" image? JIP | Talk 20:32, 5 June 2011 (UTC)

When you look at a mirror, why is left and right exchanged, but not up and down? (I know the answer, but I hope the question will lead you to think into the right direction.) 77.3.180.185 (talk) 20:19, 5 June 2011 (UTC)

Please don't forget to end-small next time to prevent everything afterwards from being registered as small. DRosenbach ^{(Talk | Contribs)} 20:29, 5 June 2011 (UTC)

When I look at myself in the mirror, what I am seeing is light that is coming towards the same direction as me, but bounces off the mirror, reversing direction. So it's only natural that left and right are reversed. But when I look at the Moon, what I am seeing is light that is coming towards the same direction as the Moon, just as if I were looking at some other object above my head. There is no mirror up in the sky where the Moon would be reflected. JIP | Talk 21:20, 5 June 2011 (UTC)

When I stand in front of my mirror, Glasgow is miles away to my left. The handsome person in the mirror agrees that Glasgow lies in that direction. In other words, lateral inversion is perceived only in my own reference frame, not in the real outer world. Cuddlyable3 (talk) 07:58, 6 June 2011 (UTC)

The North pole of the moon is defined as that pole pointing in the same direction as the Earth's North Pole. If you are in the Northern hemisphere and the word Moon were written across the face of the moon at its equator as you looked at it, and there were a man walking from the emm toward the enn, he would be walking to the east on the moon but appear to be moving toward the west horizon of the Earth. Simple. μηδείς (talk) 20:52, 5 June 2011 (UTC)

I understand what you mean, but I still cannot visualise it when imagining another Earth orbits the Earth instead of the Moon. (The Earth is the only planet where I know intrisincally where "east" and "west" are supposed to be when on the planet's surface, because I have not been on the surface of any other planet.) JIP | Talk 21:04, 5 June 2011 (UTC)

A standard map of the Earth is oriented in the way you would see it from the outside: East is right, west is left (always taking north to be up). With the night sky things are reversed because we're looking at the celestial sphere from the inside. That's why maps of the night sky are drawn such that east is left when north is up; that's also the standard orientation of astronomical images. Now, with the moon you have two options: Either you consider the moon as hanging in the night sky and you define east and west to be the same directions you assign to the sky, i.e. east is left - I think this is referred to as the "astronomical orientation". Alternatively, you consider the moon to be a physical body, like Earth, and you use the same orientation you would for a globe: east is right - that's called the "astronautical" orientation. The latter is the one that is officially in use today (defined by the IAU in 1961 [16]). Observers of the moon actually prefer the terms "preceding" and "following", referring to the daily motion of the moon (the way the moon would move through the field of view of a fixed telescope). --Wrongfilter (talk) 21:40, 5 June 2011 (UTC)

Where does this notion of "east is left" come from? Does the sky somehow have a different coordinate system as the bodies in it? The only possible way I could see all this as making any sense is viewing the visible sky as a thin, transparent plastic band, with "east is that way" and "west is that way" written on it on the outside. When we look at it from the inside, we see those directions the opposite way. But the Moon is not a thin, transparent plastic band. It is a spheroid, and so it surface wraps around it, and would be the right way around on the near side, which is what we see from the Earth. JIP | Talk 19:21, 6 June 2011 (UTC)

What Wrongfilter wrote makes sense to me. It's natural for the celestial sphere's coordinate system to have the opposite handedness from Earth's on a map, since they can be treated as concentric and we look at them from opposite sides. If north is up on both maps, then east will be left on one and right on the other. For the Moon/Sun/Mars/etc. there are multiple conflicting ways of assigning the names "north" and "east", all with some claim to naturalness, and the only solution is to pick one and standardize it. The maps are never mirrored from what we see; this is purely about word choice. -- BenRG (talk) 22:17, 6 June 2011 (UTC)

Does this mean that when north is up, the same direction is right and the same direction is left both under our feet and up in the sky, we're just labelling them "east" and "west" differently? JIP | Talk 19:38, 8 June 2011 (UTC)

Force = mass x ____

Perhaps I'm just understanding this incorrectly, but suppose I have a post submerged into the ground and I run into it with a horizontal beam fitted to a car that is traveling at a constant speed -- won't the speed of the car have an effect on the force applied to the submerged post. My confusion lies in the idea that force is equivalent to mass x acceleration, and in my example, the car is traveling at a constant speed (and is experiencing no acceleration) -- so why isn't it mass x velocity or something like that? DRosenbach ^{(Talk | Contribs)} 20:27, 5 June 2011 (UTC)

The car is slowed (changes velocity) by the impact of the post. That change in velocity times the mass involved gives the force. μηδείς (talk) 20:55, 5 June 2011 (UTC)

But is that the absolute value? How can it be that a car accelerating at that (positive) absolute value provides the same force as one deccelerating at the (negative) absolute value? And that's sort of using the impact (i.e. the "transfer of force," so to speak) to retroactively provide the acceleration. DRosenbach ^{(Talk | Contribs)} 21:07, 5 June 2011 (UTC)

The force provides the acceleration, rather than the acceleration providing the force. So here, the post decelerates the beam, whilst the beam accelerates the post. - Jarry1250 ^{[Weasel? Discuss.]} 21:18, 5 June 2011 (UTC)

Since the OP has the post embedded in the ground, the force of decelerating the beam and car accelerates planet Earth a little. But fear not Earthpeople, the effect merely reverses the acceleration your planet received when the car started. Cuddlyable3 (talk) 07:51, 6 June 2011 (UTC)

Mass x velocity is something called momentum. This is not force. Is that, perhaps, your confusion? The force you hit the post with is dependent on how quickly you change speed. So, if your initial speed was 50 miles per hour, and you drop to 0 miles per hour in 1 second, you would impart the exact same force on the pole as if you were traveling 100 miles per hour and stopped in 2 seconds. In reality, your stopping time would not double as your speed doubles, indeed it will probably only go up a small amount, so lets say that at 50 miles per hour, your stopping time is 1 second, but at 100 miles per hour your stopping time is 1.1 seconds. You clearly impart more force in the second example because your acceleration (deceleration, whetever, same math with a negative sign) is significantly higher, 100/1.1 is a bigger number than 50/1. --Jayron32 00:28, 6 June 2011 (UTC)

WKD colour loss

Hey all. I kept some leftover bottles of WKD Original Vodka for a couple of months. They were originally the typical blue colour [17], but when I got round to drinking them yesterday, they were colourless. Tasted exactly the same, but colourless. What chemical could be responsible for this? What's the story? Regards, - Jarry1250 ^{[Weasel? Discuss.]} 21:24, 5 June 2011 (UTC)

They weren't by any chance exposed to direct sunlight ? This can cause many colors to fade. Another possibility is that the blue coloring came out of suspension or solution. In that case, I'd expect to see some blue residue, perhaps at the bottom of the bottle. StuRat (talk) 01:03, 6 June 2011 (UTC)

I tried to look up the WKD food coloring, but didn't find it easily. You don't suppose that bottle has an ingredients list? I was thinking that using methylene blue might be rather clever as it can reduce liver necrosis, but methylene blue makes for blue urine, but WKD apparently (from the web search) makes for bright yellow urine. Twas a nice idea while it lasted. Wnt (talk) 05:41, 6 June 2011 (UTC)

:) Um, I don't have the box, only the bottles themselves, and they have only a very vague statement about ingredients. Darn. - Jarry1250 ^{[Weasel? Discuss.]} 16:33, 6 June 2011 (UTC)

They were left outside in a box to keep cool, but the ones hemmed in on all sides are no different from the ones in a little more space. There was no residue in any of the 8 bottles. - Jarry1250 ^{[Weasel? Discuss.]} 16:33, 6 June 2011 (UTC)

Well, then, we're down to a chemical reaction, it sounds. Perhaps the dye reacts slowly with some other component to form clear products. StuRat (talk) 22:47, 7 June 2011 (UTC)

Sound-guided missiles

It seems most new military aircraft are "stealth" and harder to target by heat-seeking or radar guided missiles. Has anybody ever developed a missile that flies towards the source of a loud sound? Would this be possible and how could it work?--92.251.222.88 (talk) 21:41, 5 June 2011 (UTC)

This sounds like a good idea, but it has some problems. For one thing, most anti-aircraft missiles fly over the speed of sound. Even when they don't, the sound of the missile flying at hundreds of miles an hour would overwhelm the microphone. Also, new stealth aircraft are also designed to minimize sound as well as radar. And an aircraft could fly at supersonic speeds and the missile would be useless. I think the next anti-stealth weapon will be low-frequency radar. --T H F S W (T · C · E) 22:29, 5 June 2011 (UTC)

Stealth aircraft are only really stealthy when flying over countres that use pre-1970s air defense systems. Modern air defense systems, like the MIM-104 Patriot or the S-400 (SAM) don't have problems shooting down the most advanced stealth aircraft from more than 100 km distance. Count Iblis (talk) 01:23, 6 June 2011 (UTC)

Could you provide a reference for that? I don;t think stealth fighters are designed with 1970s technology being the primary opposition. Googlemeister (talk) 16:47, 6 June 2011 (UTC)

The Lockheed F-117 Nighthawk design was started in 1975 and first flight of a demonstrator was in 1977, so he's probably right. Alansplodge (talk) 12:59, 8 June 2011 (UTC)

Well F-117s are all retired. I was referring to more modern designs, like the F-35 or the F-22 (even though the F-22 is something like 15 years old now). Googlemeister (talk) 21:22, 9 June 2011 (UTC)

Forget aircraft. I want a noise-seeking missile to take out motorcycles and overly-loud cars. --Carnildo (talk) 23:22, 7 June 2011 (UTC)

See Japanese war tuba and acoustic location. I don't know if any torpedoes used ASDIC. 92.24.128.171 (talk) 18:07, 8 June 2011 (UTC)

Safety of BPA-free plastic water bottles

The plastic water bottles sold these days all seem to claim to be BPA-free. However, they all seems to have some "plasticky" smell, suggesting chemical emissions. What material(s) are BPA-free plastic water bottles usually made of? Do we actually know that these materials are safer than the polycarbonate they replace, or are we just trading one known hazard for some other unknown ones? --173.49.79.135 (talk) 23:59, 5 June 2011 (UTC)

It's largely a ruse that water bottles need be free of bisphenol-A because it's so dangerous, because many other things still contain it, such as most of the esthetic white dental fillings (composite, glass ionomer, etc.) Are we looking at a threshold? How much is "too much"? Apparently, it doesn't make too much of a difference. DRosenbach ^{(Talk | Contribs)} 03:39, 6 June 2011 (UTC)

I'd go with your instincts here. If you can smell it and/or taste it, then a significant quantity of chemicals must be given off by the bottle. Whether those chemicals pose a health risk is probably unknown, but it seems reasonable to avoid them, wherever possible, say by using glass bottles. StuRat (talk) 04:02, 6 June 2011 (UTC)

June 6

To charge a phone on a bicycle

Hi, is there any gadget I could get anywhere online (or preferably off) that will charge my phone while I pedal my bike? I remember bike odometers that hooked a little wheel to one of the tires (therefore, called "flywheels?") in order to spin the numbers. Could that same small wheel utilize the spinning of the tires to recharge my phone?

If so, where is a device that'll do exactly that? I would hope to find one before a long bike-ride. Thanks. --70.179.165.67 (talk) 04:29, 6 June 2011 (UTC)

Though someone smarter than I am on this matter will be able to give you the specifics, I recall that the amount of pedaling on a bike to produce any worthwhile amount of electricity makes the exercise impractical. So even if you were going to ride a marathon's length, there probably isn't a market for selling devices that charge cellphones this way, so you're out of luck unless you want to build your own contraption.--el Aprel (^facta-_facienda) 05:37, 6 June 2011 (UTC)

While the amount of electricity generated by a bicycle is low, and thus insufficient to heat, air condition, or light your home, or run a major appliance, it is sufficient for devices with minimal power requirements, like cell phones. However, the cell phone charger will need to be improved, as current chargers are rather inefficient (you can tell because they get hot). StuRat (talk) 05:59, 6 June 2011 (UTC)

Are you talking about the AC adapter? If so how old is your phone? All phones I've seen in the past 5 or so years, even very cheap ones come with SMPS power adapters (sometimes labelled as travel adapters). You can easily tell because of the weight (and to a lesser extent size and shape). Cheap electronics from China from eBay, DealExtreme etc also always come with (sometimes poorly made) SMPS. Various sources including our article say this has happened world wide, perhaps because of the decreasing cost of the electronics needed for an SMPS and the rising cost of iron (although the iron bit seems to have been removed from our article). Regulation or government pressure on power supply efficiency may have also played a part. Regardless, SMPS power adapter doesn't tend to get very hot. In any case it's a moot point since it's unlikely you're going to output 110 or 220V AC from your dynamo. Nil Einne (talk) 06:27, 6 June 2011 (UTC)

Yes, but inefficiency is possible with any charging system. And, unlike with the wall plug, there isn't much energy to spare in this setup, so efficiency would be key. StuRat (talk) 06:39, 6 June 2011 (UTC)

So? None of that relates to your original (generally false nowadays) claim about how inefficient current chargers are based apparently on some ancient phone with a linear/transformer based AC adapter. I suspect many dynamos output DC in the 4-12V range anyway since modern lighting systems tend to use that and ones with batteries often even use li-ion nowadays. And any slightly decent DC-DC adapter would generally be at least 50% efficient with something like 70-80% more likely and even 80-95% possible (at least from a more constant source, not sure how well they handle a more variable output you may get from a dynamo although again since dynamos for bikes tend to be targeted towards lighting systems it seems likely this is already something handled resonably well). The phone itself and the actually battery also has some loses but there's not much you can do about that.

Really you're approaching this from totally the wrong angle, sure the efficiency of the charging system and power adapter matters, but you don't seem to be considering the dynamo itself. For example [18] shows 60% but only for the best hub dynamos (which was where I found the link). 30% is more likely particularly for bottle dynamos and we're still talking about name brand dynamos. [19] from hub dynamos (where I found all the links) I think shows shows something similar. [20] ditto I think although you have to manually work it out. And this is at 15-20kmh, at high speeds you efficiency generally drops particularly in the crappier dynamos. Unless you plan to be charging your phone all the time, the 'daytime resistance' would likely matter (although on the flipside for a bottle dynamo you can just disengage it).

In other words, unless you're using a good hub dynamo like a Schmidt Original Nabendynamo, power loses in the DC conversion and charging are perhaps not your biggest concern. (Not saying they are not a concern but you obviously should be considering whether you want a better dynamo first.)

P.S. Worth remembering that given the 4.2V most li-ion batteries chargers or older mobile phones use and the 6V that our hub dynamo article and most of the refs above seem to say a typical dynamo for light systems outputs even a simple linear regulator just burning off the excess would achieve 70% efficiency. Of course nowadays phones are moving to microUSB for charging meaning they accept or require 5V so you get 83.3% with a linear regulator (the phone would have to do some conversion internally but it's likely most phones are fairly efficient).

Nil Einne (talk) 10:36, 6 June 2011 (UTC)

By "charging system", I meant to include the efficiency of the dynamo, as well. If this isn't the term you use, how do you collectively refer to it all ? StuRat (talk) 22:45, 7 June 2011 (UTC)

Many bicycle lights are powered by either rim or within-hub dynamos. You would need to build or commission a device that allowed this power source to charge or power a phone. This will increase the friction and thus effort needed to cycle. Fifelfoo (talk) 05:48, 6 June 2011 (UTC)

Another option might be regenerative braking, where you would pull more power from the wheel for charging purposes, but only when braking. This would be a big seller, if it worked. Obviously, such a system would provide more electricity in "stop and go" driving. One simple way to do this might be with the same dynamo, but where it had a user-operated switch to engage it or disengage (perhaps tied to the gear setting). This might work well in hilly areas, where excess speed could be bled off going downhill, but without the unwanted extra drag going uphill. StuRat (talk) 06:03, 6 June 2011 (UTC)

A simple search for 'bike dynamo phone charger' easily finds options both DIY and commercial [21] [22] [23] [24] [25] [26] [27]. So does 'bike phone charger' (mostly finding the Nokia device but also a few others) [28] [29] [30] [31] [32] [33] [34] [35] [36]. Other then specialist or biking companies evidentially Nokia and Motorola make devices too. Nil Einne (talk) 06:15, 6 June 2011 (UTC)

As an alternative, buy one of these traditional mobile phones (as opposed to these fancy new smartphones with large touchscreen), because they run much longer without recharging (despite that they typically have smaller batteries). – b_jonas 09:03, 6 June 2011 (UTC)

This answer is about as good as saying "Install Linux" when someone asks how to do something in Windows! APL (talk) 09:51, 6 June 2011 (UTC)

Another alternative would be simply to get yourself a solar charger..--Shantavira|^{feed me} 09:13, 6 June 2011 (UTC)

Unfortunately, Very few of those are worth the effort. In most cases you'd be better off economically and perhaps even ecologically just buying ones of those AA-to-phone adapters and popping in some AAs whenever you're away from line power. Problem is that solar panels with any sort of useful efficiency are very expensive, so the ones most of those "Solar battery chargers" use are pretty terrible. (Typically these devices are just rechargeable batteries that you charge from line-current, and the solar panel is mostly for show.)

You'll certainly never get enough juice from the sun with one of those devices to justify the cost, or the ecological impact of its own manufacture.

I've heard good things about the Voltaic backpacks, but even at it's theoretical peak output it's not as powerful as my phone charger. And you figure that unless you're pointing it right at the sun you won't get half that. (You don't normally keep your back to the sun while biking. You have to go where the roads go.) So it's still not really worth the effort unless you're honestly going to be away from line current for a while and for some reason can't carry AA batteries. APL (talk) 09:51, 6 June 2011 (UTC)

So, what would happen if you were to wire your bike lights dynamo to a spare car cigarette lighter socket (I think you can buy these from places like Halfords). You could then charge your phone from that using a car adapter/charger suitable for your phone? Astronaut (talk) 13:57, 6 June 2011 (UTC)

This is nitpicking, but the question says "on a bicycle", not "powered by pedalling a bicycle". When I go camping at music festivals I take a small 12V motorcycle battery and use that with the car cigarette socket charger to charge my phone. You could easily fit such a battery on a bike carrier rack - and charge the motorcycle battery up at home every few weeks. -- SGBailey (talk) 15:00, 6 June 2011 (UTC)

Earthquake tilting Earth axis

Originally asked over on the help desk [37] I've copied it here  Chzz  ►  12:14, 6 June 2011 (UTC)

I have been trying to find out more about the Earthquake in Japan. I read that it caused the earth to move 10 inches on it's axis. Is this true and are we tilted more or less on our axis. I have ask our TV stations abiut this and can't get an answer. I'm 71 years old and I took Astronomy Magizine for about 50 years but had to give it up because of my eyesight, it was just too hard to read. So if you can answer my question or tell me who to ask iI would appriciate it. (Redacted) — Preceding unsigned comment added by 75.221.228.71 (talk) 06:01, 5 June 2011 (UTC)

It was actually 10 cms. Please see our article on 2011 Tōhoku earthquake and tsunami.--Shantavira|^{feed me} 12:20, 6 June 2011 (UTC)

Hm, but in 2011 Tōhoku earthquake and tsunami#Geophysical impacts it says 25 cm (9.8 in), with this ref. And this one says 6.5".  Chzz  ►  12:28, 6 June 2011 (UTC)

Indeed this type of thing highlights the flaws with supposedly referenced information. The New York Times article that is supposedly a reference for the 10 cm figure in fact says "Dr. Gross said his calculations indicated a shift of 6.5 inches in where the figure axis intersects the surface of the planet". The other ref for that sentence gives the 10 cm figure. Regardless, all these articles are from fairly soon after the quake, but it seems that the closer the date of the article to the actual quake, the bigger the figure given, suggesting a bit of early sensationalism or hyberbole. Wonder if anyone can find a genuinely good source from a month or more after the quake that would likely be more reliable. --jjron (talk) 13:30, 6 June 2011 (UTC)

Also worth pointin out that the tilt didn't change. What changed was the position of the axis on earth's surface. Dauto (talk) 17:35, 6 June 2011 (UTC)

Is it more correct that the earth's surface changed relative to the axis? Getting back to the original issue though, it would really be useful to see this as a before/after two sets of axes on the same globe. Even if the actual number could be verified, it still isn't clear to me whether the earth tilted more/less, or the solid vs axis precessed a bit. Spheres have lots of "axis" directions. DMacks (talk) 18:27, 6 June 2011 (UTC)

Changing the tilt of the earth's axis would imply a change in angular momentum which is not possible without an external torque. Dauto (talk) 19:44, 6 June 2011 (UTC)

Cats can rotate themselves in mid-air with no externally applied torque (Falling cat problem). DMacks (talk) 22:43, 6 June 2011 (UTC)

Yes, and they do that by changing the position of their bodies around a rotation axis which remains fixed because of the already pointed out conservation of angular momentum, which is what I was pointing out. Dauto (talk) 15:39, 7 June 2011 (UTC)

I'm not so sure about that. Picture you have a rotating cylinder with heavy iron balls clamped to each of four sides, which are also attached by dangling chains. You spin up the cylinder and suddenly one of the balls gets loose and starts swinging around several feet from it. Well that cylinder is going to start rotating around a new axis. (Something similar can be done with an ultracentrifuge when one or more of the sample tubes break, but there the rotating mass changes) In this case, part of Japan moved down, and so the axis should be a little further "down" from that spot also. The angular momentum and thus the axis of rotation may be fixed, but only relative to the total rotation, not relative to the planet surface. Wnt (talk) 18:02, 7 June 2011 (UTC)

What are the materials of Polyethylene terephthalate

Does it come from Petroleum?--Inspector (talk) 12:55, 6 June 2011 (UTC)

Yes. If you read down to the Recycling section of the article you linked (and the information should perhaps appear more prominently), you will find the statement

". . . recycling back to the initial raw materials purified terephthalic acid (PTA) or dimethyl terephthalate (DMT) and ethylene glycol (EG) where the polymer structure is destroyed completely . . . ."

If you then read the articles about those ingredients, linking through where necessary to articles about their production, you will find that they derive from petroleum. {The poster formerly known as 87.81.230.195} 90.201.110.217 (talk) 13:14, 6 June 2011 (UTC)

Most PET is probably made from oil and/or natural gas, as are other thermoplastics, but as our polyethylene article notes, it is possible to make such materials from sugar cane and similar cultivated products. In addition, the level of recycling in PET is relatively high compared to other materials. AndyTheGrump (talk) 13:17, 6 June 2011 (UTC)

Checking on the PET article talk page, I see that PepsiCo are apparently working on making PET bottles solely from plant materials: [38]. How 'green' this actually is may be open to question though. AndyTheGrump (talk) 13:28, 6 June 2011 (UTC)

So, How many steps are there between petroleum and PET?--Inspector (talk) 13:41, 6 June 2011 (UTC)

[39] might be a good read. --Stone (talk) 21:12, 6 June 2011 (UTC)

Global Warming – Why Predominantly Negative Effects?

It seems that global warming produces predominantly negative climatic effects, i.e. ones that are inconvenient to humans. Why is this? Would a cooler climate make for milder weather? I find it hard to believe that we've found ourselves in such a climatic sweet-spot — especially when you consider that average temperature has been continuously drifting one way or another throughout history. I was also wondering if there is any geographic location that might actually benefit from global warming.

Thanks!

Alfonse Stompanato (talk) 14:59, 6 June 2011 (UTC)

We have not "found ourselves in such a climatic sweet spot" by accident, we have adapted to it. The climate has actually been remarkably stable for the last 10000 or so years - i.e. for the whole time of human civilisation. We have adapted our agriculture to work well with the combination of climate, soil, and seeds known to us. We have grown large populations where a reliable water supply is available year round by rivers that are buffered by glaciers fed by winter snow that melt in the summer. We build our cities on the current shoreline for both commerce and fishing. The primary problem is not that different global temperatures are inherently better or worse (though they may well be, depending on what criteria you use), the problem is that we experience changes to the status quo at record speeds. On a larger scale, the same applies not only to humans, but to whole ecosystems. Given enough time, both humans and ecosystems will be able to adapt to a new steady state. But we are a long way from reaching a new steady state, and the disruption will be predominantly negative, simply because our lifestyle is optimised to current conditions. There are some other concerns, primarily that a higher temperature means more energy in the climate system, which likely leads to more or stronger extreme weather events. As for benefits, we may be able to reliably use the Northern Sea Route or the Northwest Passage for shipping. --Stephan Schulz (talk) 15:25, 6 June 2011 (UTC)

(after ec)

To answer your last, "benefit" is a subjective term. No doubt the colder regions would benefit from being made warmer, if you're a human being who likes warm conditions. But if you're a capercaillie who has evolved to fit one particular niche, warmer conditions would spell the end for your species in that area. --TammyMoet (talk) 15:28, 6 June 2011 (UTC)

Another point is that we aren't necessarily in a "sweet spot". A few thousand years ago much of the Sahara desert was grassland, and in other parts of the world many areas supported people that are now flooded by rising sea level. Looie496 (talk) 16:09, 6 June 2011 (UTC)

The life around us now has evolved for the last couple of million years to live at about the current temperature or a bit lower as during the ice ages. Dmcq (talk) 16:57, 6 June 2011 (UTC)

I have to admit, I can't shake off the feeling that the Ice Ages that mark our current geologic era are in some way "pathological". If there is anything to the Gaia hypothesis then it would appear that Gaia noticed that the environment was constantly changing for some reason and decided to put some thought into how to dig up the old carbon and get it back out into the atmosphere... While I recognize, of course, that the very rapid warming caused by humans has harmful effects on the ecology, the world has been through such drastic changes quite a bit over the past few million years. So I'm thinking that while controlling carbon emissions is necessary, I wouldn't want to see atmospheric CO2 rolled back to 1600 or even to 1950. I think this position is implicit in climate plans that speak of partial (even 80%) reduction in CO2 emissions, but I haven't really seen the ideal endpoint seriously discussed. Wnt (talk) 21:42, 7 June 2011 (UTC)

Is it right that.. (Talking birds)

All bird kinds in the world, could imitate human Speech Just like the Parrots, Myna birds, and others do so?...

is it right that the only "barricade" from all other bird kinds to do so, is just Neural? (Their pro-Larynx Areas in their nervous system).

have i understand correct?

Best blessings. — Preceding unsigned comment added by 109.67.42.106 (talk) 15:35, 6 June 2011 (UTC)

No, not all birds. Only birds from three groups, the Myna and the relatives of the Parrots, the Crows are known to mimic human speech. These birds are relatively intelligent, and have vocal tracts capable of imitating speech. See talking birds μηδείς (talk) 15:51, 6 June 2011 (UTC)

The question is whether more birds than just these varieties have vocal tracts capable of imitating speech, even if the birds themselves don't actually do so. The article does not address that (or much else). --Mr.98 (talk) 16:29, 6 June 2011 (UTC)

I now understand that it's also because of this 3 groups of Birds - unique Larynxal Anatomy (off course, in addition to their neural capability).

Just to ensure i understand correct,

blessings 109.67.42.106 (talk) 17:00, 6 June 2011 (UTC)

Haven't you asked this already up there at #A_beatiful_question_about_Parrots_and_their_voice_producing_system..? – b_jonas 18:17, 6 June 2011 (UTC)

He'd better be careful, or he'll be charged with contributing to the deliquency of a Myna. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:43, 6 June 2011 (UTC)

Just for the record the adjective from larynx is usually laryngeal Richard Avery (talk) 07:14, 7 June 2011 (UTC)

Note also birds use a Syrinx (bird anatomy). Wnt (talk) 21:34, 7 June 2011 (UTC)

radiation from magnets

Since photons are the force carrier for electromagnetism, do magnets give of any form of radiation from the electromagnetic spectrum? Bugboy52.4 ¦ =-= 17:25, 6 June 2011 (UTC)

If you hold a magnet in your hand and make fast movements, it will radiate electromagnetic radiation, albeit it not very much. Count Iblis (talk) 17:38, 6 June 2011 (UTC)

(ec) A static magnetic field does not produce any radiation, but a changing magnetic field does -- this is a consequence of Faraday's law of induction. Looie496 (talk) 17:41, 6 June 2011 (UTC)

Is the frequency constant, or is does it change? Bugboy52.4 ¦ =-= 18:15, 6 June 2011 (UTC)

The frequency radiated will be the same frequency of the shaking movements you apply on the magnet. Dauto (talk) 19:39, 6 June 2011 (UTC)

That means 4.3 x 10^13 (43,000,000,000,000) vibrations per second for red light. μηδείς (talk) 20:05, 6 June 2011 (UTC)

So it is possible to emit light in the visible spectrum? Bugboy52.4 ¦ =-= 19:47, 6 June 2011 (UTC)

Not with your handheld magnet, your hand does not shake that fast. What you ahve here are virtual photons. Graeme Bartlett (talk) 20:52, 6 June 2011 (UTC)

I'm not sure what you mean. Radiated photons are real photons, regardless of frequency. -- BenRG (talk) 22:52, 6 June 2011 (UTC)

EM radiation from a rapidly vibrating permanent magnet should be indistinguishable from that generated by , say an Alexanderson alternator(early 20th century radio transmitter via rotating generator) or from a solid state radio transmitter coupled to an antenna. except for the difficulty of achieving physical vibration of a permanent magnet at frequencies beyond a few tens of Hz. Edison (talk) 02:31, 7 June 2011 (UTC)

(Edited to add) I had intended to say "10's of kilohertz," imagining a small permanent magnet bolted to a loudspeaker or an ultrasonic transducer. A permanent magnet could also be bolted to a vibrating rod or wire in tension, for instance, which could easily be tuned to vibrate at frequencies above the audible range, which would in fact be in the low frequency radio frequency range. A magnet held in the hand cold be vibrated manually at 5 Hz or so. This low frequency EM wave could certainly be picked up by a suitable nearby antenna or pickup coil. I've tried the experiment, and viewed the somewhat distorted waveform on an oscilloscope. I agree that the transmitted energy would be negligible. Remind me how EM waves work: would a rapidly vibrating magnet moving left and right produce a transverse electrical field moving up and down, both fields moving away from the origin? If I took an electrically charged object and caused it to vibrate left and right similarly, would it produce a transverse magnetic wave moving up and down? Surely electromagnetic waves are not restricted to electronic oscillators.Edison (talk) 05:16, 8 June 2011 (UTC)

Our piezoelectricity article talks about speed and size of deformation, but not general commentary on frequency limits to its oscillation. However, one of the noted applications mentions 100 kHz–1 MHz. Put a lightweight magnet on it, and you're well into the AM radio range. DMacks (talk) 03:15, 7 June 2011 (UTC)

With a handheld magnet, you can only move it at frequencies up to 10Hz. Radiation efficiency from this would be extremely low with less than picowatts of power transmitted as traveling electromagnetic radiation. Almost all the energy would be in the near field which would extend over 1000 km. This would be dissipated by electric currents flowing in the earth or the ionosphere and so just about nothing would escape into a radiowave far field. Even at the AM frequencies mentioned by DMacks, the antennas are dozens of meters high to get any significant power out. So a small vibrating magnet would still not do much. I am probably addressing Count Iblis q rather than bugboy52.40. For the original question the virtual photons answer still stands. The size of these photons is enormous, thousands of kilometers, so they are not like your normal particle! Graeme Bartlett (talk) 11:26, 7 June 2011 (UTC)

What? Radiated photons are always real, and photons are always point particles, and these photons are most certainly 'ordinary photons' in the sense that no other kind of photons exist. The wavelength of a photon is not a measure of it's physical size!

Brightness of R136a1 when it goes hypernova

The article says it will have the luminosity of 100 typical supernovas. The brightness of supernova 1987A was magnitude +3, and its luminosity was about 1/10 of that of a typical supernova. So, does this mean that R136a1 when it explodes will become magnitude -4.5, making it barely visible at daytime? Count Iblis (talk) 17:44, 6 June 2011 (UTC)

They are about the same distance away from us, so yes. If those numbers are correct, then it would be about -4.5. --Tango (talk) 20:53, 6 June 2011 (UTC)

excited states / energy level diagram of nitrogen

If I look at the grotrian diagram of nitrogen in my book, I don't see a state 2²S_1/2, and the book says that it is not possible because of the Pauli principle. But if I assign the 3 valence electrons as follows, I don't see why this is wrong, where is my error:

• n₁=2, l₁=1, m_l1=-1, m_s1=1/2
• n₁=2, l₁=1, m_l1=0, m_s1=1/2
• n₁=2, l₁=1, m_l1=1, m_s1=-1/2

So the resulting atom should be S=1/2, L=0, J=1/2. ??

best thanks --helohe (talk) 21:01, 6 June 2011 (UTC)

Wow. It's been many years since I studied this stuff myself, but there are ideas about Selection rules like the Laporte rule knocking around in my head, though I am not fully sure how to make it work for you. Remember that monatomic nitrogen is a radical, which may introduce some weird symmetry problems and degeneracies, akin to what you see in odd-electron molecules and ligand complexes, vis-a-vis the Jahn–Teller effect. I've forgotten most of this stuff, and how it works, but that may give you some leads... --Jayron32 05:48, 7 June 2011 (UTC)

Desert basins

I was wondering what is distinguishable about desert basins which set them apart from deserts themselves. The article Kalahari Basin doesn't really help to explain what physical features separate it from the Kalahari itself.— Preceding unsigned comment added by Flaming Ferrari (talk • contribs) 21:08, 6 June 2011

I think the name Kalahari Basin refers to a drainage basin. The Kalahari Basin is an endorheic basin, which means that water flowing there does not reach the ocean. The same is true for the Great Basin in North America. In other contexts related to landforms, the word "basin" may mean something else; see Basin#Landforms. —Bkell (talk) 22:42, 6 June 2011 (UTC)

(Edit Conflict) So far as I understand it, there is no inherent connection between basins - large landform depressions of various sorts, and deserts - large areas with low precipitation: you can have either one without the other. However, if a basin happens to be in a desert, it's called a desert basin, and is likely to have certain features caused by the combination of shape and climate, namely seasonal water courses, salt lakes or salt flats (dried-up lakes), and perhaps subterranean water if the underlying geology is right. The Kalahari is a slightly different case in that the basin is larger than the desert, and the desert happens to be entirely contained within it; the Great Basin Desert in the USA is similar in this respect. Conversely, the Gobi Desert contains several basins. {The poster formerly known as 87.81.230.195} 90.201.110.217 (talk) 22:43, 6 June 2011 (UTC)

Well, as said in endorheic basin, they tend to be deserts because otherwise the water would break a channel flowing out. A desert is part of a system of classification of land that has a net moisture deficit. There's probably some slop here, I think, in that there are arid regions not classified as deserts that have a net moisture deficit, and lands with a net moisture deficit can still periodically overflow and erode a path to the ocean. Wnt (talk) 18:09, 7 June 2011 (UTC)

Rotational physics

Hello all. This is a problem that I made up but can't solve :( Imagine an unevenly weighted barbell, say with 20 kg on one end and 10 kg on the other, dropped from an airplane. Intuitively I know that if it is allowed to fall for long enough, the heavier end will be pointed towards the ground. However how do I prove this? I know that I have to use torques but I'm not sure how. Thanks for any help. 72.128.95.0 (talk) 21:08, 6 June 2011 (UTC)

The center of mass will be approximately one third of the bar length away from the 20 kg object. This is where the weight of the entire barbell can be considered to be acting. As the speed of the barbell increases there will be a significant drag force acting on it. The drag on the 20 kg object will be greater than the drag on the 10 kg object, but probably not twice as much so the drag force will not act at the center of mass. As a result, there will be a torque on the barbell caused by the weight (acting downwards) and the drag (acting upwards) and the fact that the two forces are not acting at the same point. This torque will cause the barbell to rotate so that the end with the 20 kg object is pointing downwards, and the end with the 10 kg object is pointing upwards. Dolphin (t) 22:26, 6 June 2011 (UTC)

(ec) There is also an unstable equilibrium if the barbell is exactly vertically oriented, with the heavier weight on the upper side. You can formally solve for the stable orientations by construction torque as a function of rotation angles τ = τ(φ,θ), and solve for zeros of the torque function by setting its gradient equal to zero; and to determine stability, solve for the concavity of the function. Nimur (talk) 22:35, 6 June 2011 (UTC)

Here's a related question: I expect that when you drop it, there is liable to be a little bit of spin, which is going to cause it to rotate or oscillate as it falls (connected with the principle you've just stated), and eventually the heavier end will be pointing downward. My question is: If it were dropped lighter-end first, and with such machine-like precision that it had no spin, wouldn't the light end of it hit the ground first? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:33, 6 June 2011 (UTC)

If it were dropped with the lighter-end first the weight and the drag would act through a common point so there would be no torque. However, this arrangement would be unstable and practical experience shows that even with machine-like precision, unstable arrangements don't persist for very long and they quickly move towards a more stable arrangement - in this case the barbell would invert and end up with the heavier end pointing downwards. Unstable arrangements can only be maintained with some sort of an active feedback control system - way beyond the scope of the simple barbell experiment. Dolphin (t) 22:40, 6 June 2011 (UTC)

Sure. The slightest variation in the air it's falling through would start it tumbling, I'm sure. I'm just talking theory. Yet another question: The barbell will oscillate and the heavier end will rotate to the downward. But it won't just stop there, right? It will kind of act like a pendulum, oscillating for some time, with less arc each time until it stabilizes. Right? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:02, 7 June 2011 (UTC)

It is dangerous to rely on intuition for things like this. The answer actually depends on how the weights are shaped. If you use two disc weights, a 20 kg on one side and a 10 kg on the other, then drag will be relatively greater on the lighter side, and you will end up heavy-side downward. But suppose you use three equal disc weights of 10 kg, with two on one side and one on the other. In that case the drag forces will balance, at least at the start, and there will be no intrinsic tendency to spin. However if there is any initial instability, things can get very complicated. Looie496 (talk) 22:54, 6 June 2011 (UTC)

@Looie496: Please explain drag will be relatively greater on the lighter side. Dolphin (t) 23:04, 6 June 2011 (UTC)

The surface-area-to-mass ratio will be larger for the lighter side. Looie496 (talk) 23:18, 6 June 2011 (UTC)

I agree that the surface-area-to-mass ratio will be larger for the lighter side. The larger object has greater surface area than that of the smaller so I would say the drag on the 20 kg object will be greater than on the 10 kg object, but by a factor less than 2 for the reason you have given. The factor is less than 2 so a torque acts on the barbell to cause it to fall heavy-side down. If the factor was exactly 2 there would be no torque and the barbell would fall with the bar horizontal. (If the factor was greater than 2 the torque would cause the barbell to fall light-side down.) Dolphin (t) 23:53, 6 June 2011 (UTC)

Any equations we specify to estimate drag based on the geometry and mass distribution will be just that - estimates. As every high-school physics student knows, force due to air resistance is a very difficult parameter to quantify and model. But, for any specified model of air resistance, we can construct the equations of motion and estimate the dynamics of the object as it falls and tumbles. As I described above, one such approach is to seek the steady-state, where net torque is zero. This is not the only way we could model the dynamics; but it's simple and allows us to find a "preferred" orientation. In reality, with turbulence, velocity-dependent drag, and so on, the actual model should account for all kinds of higher-order effects. When I need more sophisticated models, I toss out the elementary physics and use MDATCOM for modeling airflow and estimating turbulence. Nimur (talk) 00:23, 7 June 2011 (UTC)

Why are you all saying that the drag depends on the mass? Drag depends on surface area, velocity and, in very complicated ways, shape and texture, but it shouldn't depend on mass. The resultant force will depend on the mass, since it's the sum of the weight and the drag, but those are two separate forces. If we assume the weights on either end are identical except for their mass, then the drag will be the same on both, so can be considered to act through the geometric centre. That is not the same as the centre of gravity, which will be nearer the heavier end, so you have a torque and the object rotates. It will eventually reach a stable equilibrium when the heavier weight is at the bottom (since then the drag acts along the length of the object, so goes through the centre of gravity) and stay there. How long it takes to get to that equilibrium will depend on how big the drag is compared to the mass and how far from the geometric centre the centre of gravity is. A barbell would take a while to reach equilibrium, while a piece of paper with a blob of blutac on one edge would reach it very quickly. --Tango (talk) 00:31, 7 June 2011 (UTC)

We are all making the obvious assumption that the objects on either end of the bar are made of the same material so have the same density. Therefore their volumes are directly proportional to their masses. Dolphin (t) 00:37, 7 June 2011 (UTC)

Yes. Now here's a different question: What if the two ends of the barbell were identical in size, but one was made of iron and one made of aluminum. Does "drag" still figure into it? ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:01, 7 June 2011 (UTC)

See Tango's answer just above that you're indenting from. The forces will act through the geometric center because the shapes are symmetric, but will not act through the center of mass because of the density difference. This creates torque that will eventually leave the heavier end at the bottom. — Lomn 12:19, 7 June 2011 (UTC)

Another way to approach this problem is to look at the location of the center of mass (where "weight" forces balance) versus the location of the center of pressure (where "drag" forces balance). The object will be stable when the center of pressure is behind the center of mass (which is why arrows have feathers--the feathers move the center of pressure backwards without appreciably affecting the center of mass). In essence, that's no different than what everyone else is saying, but I think it might be easier to visualize. Andrew Jameson (talk) 12:40, 7 June 2011 (UTC)

Surely at terminal velocity we are way above the critical reynolds number for steady flow, and the barbell will tumble as it sheds vortices. Supposing that the lyapunov exponents are greater than one (which for such a flow they almost certainly are) then there is simply nothing that can be known about the orientation of the system on impact, based on its orientation at release. — Preceding unsigned comment added by 129.67.37.190 (talk) 15:00, 7 June 2011 (UTC)

Ionic liquids

Why don't the intermolecular forces between the positive and negative ions crystallize them into a solid? --75.40.204.106 (talk) 23:13, 6 June 2011 (UTC)

Our fairly detailed article entitled "ionic liquid" says that it's just an ionic compound with a low melting point ("liquid at room temp" for example), not an intrinsically "liquid" chemical. DMacks (talk) 23:29, 6 June 2011 (UTC)

Sleep and endurance athletes

This 2007 NYTimes article explores why endurance athletes sleep so much. I tried searching the scientific literature about this, but came up empty. Although I'm a bit groggy from a workout this morning. Can someone explain why long, intense workouts seem to cause sleepiness (preferably cited to review articles)? Yet short workout seem to have the opposite effect? I looked at our sleep article, but was unimpressed. Thanks. -Atmoz (talk) 23:46, 6 June 2011 (UTC)

For short workouts, you get the benefit of the adrenaline, which wakes you up. For longer workouts, this wears off and fatigue starts to take it's place. This means that damage, like tiny tears in muscles and cracks in bones, builds up and needs to be repaired during sleep. StuRat (talk) 22:37, 7 June 2011 (UTC)

The short answer is that there isn't really much to add to the NYT article. (Gina Kolata is very good, and doesn't miss much.) The literature confirms that prolonged intense exercise leads to increased sleep (e.g., PMID 9140908), but doesn't explain why. It isn't that hard, of course, to come up with more or less plausible speculations. Looie496 (talk) 23:31, 7 June 2011 (UTC)

Thanks to you both. ...but doesn't explain why. It isn't that hard, of course, to come up with more or less plausible speculations. Yup. The most common I've seen is actually StuRat's example that sleep is needed to repair tissue. But why would you need to sleep for that and not just rest? What are the biological processes that occur during sleep that facilitate this healing? I don't expect the ref desk to answer that, because based on my searching the last couple days, it doesn't appear that anyone knows the answer, or everyone is satisfied with the plausible speculations. I guess I'll just have to be satisfied with that for now. :) -Atmoz (talk) 02:05, 8 June 2011 (UTC)

Well, there actually is some information about how sleep promotes recovery -- our article Sleep#Restoration summarizes the evidence. Basically the sleep state induces hormonal and immune system changes that don't happen during a simple resting state. What is not clearly understood is how tissue damage produces an increase in sleep -- which I saw as the question that was being asked. Looie496 (talk) 03:09, 8 June 2011 (UTC)

June 7

Bowel and bladder control with ALS

Can people with ALS usually detect a #1 or #2 coming on and, if so, do anything to hold it in? 76.27.175.80 (talk) 00:22, 7 June 2011 (UTC)

Did you read ALS? if you look at the end of the 'disease progression' section your question is answered, to some extent. Richard Avery (talk) 07:10, 7 June 2011 (UTC)

The original question was as follows:

Can people like name redacted usually detect a #1 or #2 coming on and, if so, do anything to hold it in?

I re-worded the question to the version you see above, out of respect for the person named. Dolphin (t) 12:59, 7 June 2011 (UTC)

Mechanism of rotating gas cloud round gravity center

question :how dos any interstellar gas cloud rotate round gravity field ?(replay by intering all dynamic and termodynamic and statistical physics factors)--[[

Special:Contributions/78.38.28.3|78.38.28.3]] (talk) 03:00, 7 June 2011 (UTC)—Preceding unsigned comment added by 78.38.28.3 (talk) 03:00, 7 June 2011

Why don't you create an account? Your questions often look like homework but for those who know you it is clear that they are not, and that you ask out of genuine interest. It would also make it easier to put some advice or help on your discussion page. And, when you sign your posts with ~~~~ the user name you had chosen would automatically appear and you don't have to do all this signing by hand each time. 93.132.188.46 (talk) 10:50, 7 June 2011 (UTC)

And it would not be as easy for others to put misuse to your questions, pretending it came from you. 93.132.188.46 (talk) 10:53, 7 June 2011 (UTC)

I have to apologize for my silly proposal. I thought for myself: why not create an account for myself? But I failed for every try because all user names I could think of are already in use or rejected because they are too similar to some already existing ones. 93.132.188.46 (talk) 20:22, 7 June 2011 (UTC)

many thanks of your guid I have account and this is my signature(I am online now)--Akbarmohammadzade (talk) 04:53, 8 June 2011 (UTC) (talk) 04:47, 8 June 2011 (UTC)

To get to the actual Q, the cloud likely has a tiny rotation inherited from larger galactic rotation, which in turn comes from galactic cluster eddies, ultimately going back to the Big Bang (although what caused the initial rotation right after the Big Bang is a mystery). As gravity pulls the gas in closer, it spins faster, due to the conservation of angular momentum. StuRat (talk) 20:42, 7 June 2011 (UTC)

I want to know can any gas cloud or nebula be able to have momentum of inerthia , and center of mass?--Akbarmohammadzade (talk) 04:56, 8 June 2011 (UTC)

Yes, a cloud of gas has both. StuRat (talk) 08:09, 8 June 2011 (UTC)

grouping and cross-matching in blood bank

can plasma be used (instead of serum) for blood grouping and cross-matching in blood bank? — Preceding unsigned comment added by Goldenvikie (talk • contribs) 17:07, 7 June 2011 (UTC)

Yes. Blood anticoagulated with EDTA is often used. Details here. --NorwegianBlue ^talk 20:55, 7 June 2011 (UTC)

How healthful is this bread?

I want to buy low-sodium, low-sugar 100% whole wheat bread, but it doesn't seem to exist. For some reason this bread [40] is not labeled as 100% whole wheat. What's your verdict? (The link includes a list of ingredients as well as the Nutrition Facts. 66.108.223.179 (talk) 19:14, 7 June 2011 (UTC)

I'm surprised it can be sodium free, since I thought yeast needed some sodium to grow. That bread looks to be healthy, though. StuRat (talk) 20:34, 7 June 2011 (UTC)

Not at all: you can make bread without salt very easily: it rises quicker, but tastes blander. Less of a problem if you spread it with salted butter ;) The yeast don't need the salt: the salt actually slows the rising down, which is handy if the weather is warm, you don't have a cool enough spot, and you want a slow rise for taste or convenience's sake. You can only take that so far before the bread is inedible, which generally hits well before you risk killing the yeast completely. 86.163.0.72 (talk) 20:46, 7 June 2011 (UTC)

You all are talking about different things. The OP is asking about low-sodium, StuRat reads this as "sodium free" and 86.~ talks about salt, and it reads as if he is talking about additional table salt (NaCl) that was not already present in the wheat. 93.132.188.46 (talk) 21:25, 7 June 2011 (UTC)

No, I think StuRat is refering to the bread in the link which is listed as sodium free. Richard Avery (talk) 22:18, 7 June 2011 (UTC)

How would it be possible to make bread literally sodium free? Not adding salt and using distilled water would sure be necessary but still there would be the sodium from the wheat. Plants, like all living things (live as we known it), contain essential parts of sodium. 93.132.188.46 (talk) 22:31, 7 June 2011 (UTC)

Yes, I refer to the 0 mg of sodium listed under the "Nutrition" tab on the link. Perhaps it's a case of "rounding down", and it really has a small amount of sodium. I still think that yeast must require some sodium to grow, but perhaps the small amount already present in the wheat and water is sufficient, with no need for added sodium. StuRat (talk) 08:02, 8 June 2011 (UTC)

This bread is healthful and widely distributed. Their "Yoga Bread" is very good but so are their other breads. The company is called "The Baker". They make granola which isn't bad either, though I prefer udi's granola—also widely distributed. Bus stop (talk) 23:40, 7 June 2011 (UTC)

Physics Is Phun!

A curious thing happened today. I got up from my seat on the bus before it stopped moving and as the velocity decreased exponentially I felt that I were going to fall forward. I compensated by walking faster. Now my question is, was I really walking faster, or was it an illusion? What are the pertinent functions and how can I notate this action? Schyler (one language) 23:27, 7 June 2011 (UTC)

A very qualitative discussion: Let's imagine the bus was going a constant speed, say, 25 mph. While you're sitting on the bus, or even walking around on it, you won't feel like you're moving at all — basic Galilean invariance. When the bus starts to slow down, your body is still going 25 mph, but you are slowing down along with the bus. (If the bus slows down immediately, you'll fly through the window at 25 mph, because you didn't adjust to the speed of the bus.) If you're up and walking around, you're going to be potentially not slowing down at the same rate as the bus itself, and thus feel an acceleration (a falling feeling). The illusion is that when you walk forward, you feel like you're speeding up. You're not — the bus is slowing down, but your relative speed vs. the relative speed of the bus is probably faster than the bus (e.g. the bus is going 23 mph, but you're still going 25 mph). --Mr.98 (talk) 02:10, 8 June 2011 (UTC)

Yes, you were walking faster. When you are seated, friction is equal and opposite to the stopping force causing the bus to lose forward momentum. Therefore, when you stand up during this time, friction is drastically decreased, meaning the stopping force is also decreased for you compared to the bus. This difference in stopping force inturn results in a decreased deceleration for you, i.e. you lose forward momentum at a slower rate than the bus. Plasmic Physics (talk) 02:26, 8 June 2011 (UTC)

Friction has nothing to do with this. It is a question of relative acceleration, as Mr.98 explains, though it is also a matter of how you 'perceive' this. If you are standing still, and start to tip forward, you put a foot in front of you. If you are walking, and start to tip forward, you compensate by walking faster. Try standing still in the bus as it brakes - you will step backwards to avoid falling. Essentially, you are trying to keep the vector of forces generated by your weight (vertically downwards) and the deceleration of the bus (horizontally forwards) acting through your feet - you have to move them forward of your centre of gravity to avoid falling over. AndyTheGrump (talk) 02:35, 8 June 2011 (UTC)

Sorry, not friction, but torque. Plasmic Physics (talk) 07:23, 8 June 2011 (UTC)

June 8

Red/white poppy flowers

Recently in south east england I saw some poppies in someone's garden where the petals were both red and white, with the two colours irregularly mixed together. Does anyone know the species and or variety please? They were undoubtedly cultivated garden flowers and not the wild ones. Thanks 92.29.122.28 (talk) 00:25, 8 June 2011 (UTC)

Interesting. I wonder if it's a chimera (plant). StuRat (talk) 07:58, 8 June 2011 (UTC)

Stomach and food poisoning

I thought that bacteria, apart from Heliobacter, could not survive being in the stomach. If that is true, then how do people catch food poisoning? 92.29.122.28 (talk) 00:29, 8 June 2011 (UTC)

(a) The stomach is a harsh environment for bacteria but it doesn't infallibly kill them. (b) However, lots of food poisoning comes from toxins created by bacteria rather than from ingesting live bacteria; for example botulism, usually. Looie496 (talk) 00:37, 8 June 2011 (UTC)

Thanks, but news about the current E. coli food poisening scare in Germany suggests that the bacteria do infect people. 92.29.122.28 (talk) 00:39, 8 June 2011 (UTC)

Well yes, no one said otherwise Nil Einne (talk) 01:02, 8 June 2011 (UTC)

(e/c)There are many bacteria (including quite a lot of the more normal variants of E. coli) that routinely hang out in the gut. Once you get past the stomach it's quite habitable, and the stomach acid might kill 99.99% of the bacteria and still leave enough to establish an infection. EHEC inovlves an actual infection, so something has to survive, but it's more on the lines of toxicoinfection - the bacteria produces a toxin that causes the issue, the mass of bacteria isn't the problem. SDY (talk) 01:10, 8 June 2011 (UTC)

One method is endospores, which are the cellular equivalent of a bank vault. To quote the article, "Endospores can survive without nutrients. They are resistant to ultraviolet radiation, desiccation, high temperature, extreme freezing and chemical disinfectants. Common anti-bacterial agents that work by destroying vegetative cell walls do not affect endospores." Reputedly botulism endospores are killed by stomach acid,^[41] but Clostridium difficile (a diarrhea bug) is not.^[42] (But I don't get why people on Prilosec OTC don't get botulism from honey then...) Wnt (talk) 01:34, 8 June 2011 (UTC)

Exercise and rest

If for example someone ran a marathon or half-marathon one day, how do sports coaches calculate how many days they should wait before doing the same thing again? I understand that it is possible to over-train. Thanks 92.29.122.28 (talk) 00:38, 8 June 2011 (UTC)

I believe these are mostly rules of thumb based on experience, not anything derived from science. Looie496 (talk) 01:10, 8 June 2011 (UTC)

Animal Testing

The PETA article says that they "promote alternatives [to animal testing], including embryonic stem cell research and in vitro cell research". Would these be viable alternatives to animal testing? 65.92.5.252 (talk) 01:05, 8 June 2011 (UTC)

I think in general cell culture studies have a fairly poor reputation - different cell lines tend to do different things. In part that's because to this day people still fairly often use cell lines from tumors that have diverged greatly from the natural tissue they are supposed to represent - even having different karyotypes. I think that careful immortalization of cell lines with the smallest amount of genetic change possible might improve their usefulness a little.

That said, there's no substitute for the real animal - and the real animal is scarcely a substitute for the human. In vitro, dextran sulfate cures HIV. In vitro, immortality is easy, and gene therapy can target more than 90% of cells. There's not much you can do with pharmacokinetics and bioavailability in a tissue culture dish. And it makes poor practice for medical students learning surgery...

Of course, even animals can be a poor substitute for humans ... especially when studied by stupid people. See TGN1412, for example. That study would have been better off if they'd been allowed to go straight from the mouse to the human, because testing the human monoclonal antibody in a monkey made them think it had a low affinity ... when in fact there were a few 'minor' differences in the primate sequence. The results weren't pretty.

Much of this has more to do with "ethics", such as it is, than science. If we were serious about fighting disease - serious in the way that we've been about fighting wars - then we could line up a million test subjects, take every drug before it's ever entered the "pipeline", and test it on them flat out. Give the casualties medals, the survivors money and veteran health benefits, and jump medicine 15 years into the future - saving more lives than were lost. I don't know why that is only a noble thing to do when you're trying to kill people. Wnt (talk) 01:21, 8 June 2011 (UTC)

Not doing the basic research first would be ridiculous. The actual success rate for new chemical entities is dismal. The military equivalent would be like a human wave attack. Read up on WWI for how effective those are. For serious problems like HIV and malaria, it's not the lack of drugs that fails us, it's that the patients can't afford them, so mass suicide in the name of drug research isn't exactly justified. SDY (talk) 01:36, 8 June 2011 (UTC)

To be clear, I was not suggesting to try drugs before some rationale for activity had been devised. Nor trying them at once in large doses - rather, working up the dose gradually in a few subjects, like Sasha Shulgin's psychonauts. As for affordability, one aspect of a military-like campaign is that the drugs thus invented would be public property, and therefore quite inexpensive. Wnt (talk) 01:41, 8 June 2011 (UTC)

The way it's done currently is actually fairly sensible, though sometimes slow: they try and throw out the losers (the vast majority of candidates) in pre-clinical (mostly animal) testing, which is cheap compared to human testing because the standard of care is much lower (no one expects a $20,000 hospital stay for mice, except maybe PETA). Initial testing for things like genotoxicity (i.e. "is the drug mutagenic/carcinogenic?") are done in bacteria, which is dirt cheap (no one cares if you slaughter a million bacteria). Only 10-20% of drugs that enter clinical trials are approved, but most drugs don't make it to clinical trials at all, and approval isn't the final word on whether a drug is worth using. If the chances were higher, it might be worthwhile to go to human testing earlier, but the chances are really, really poor, and you'd just be throwing away people. PETA's point is that we're basically just throwing away animals with the same callous disregard. There are some ethical standards in animal testing, but they're mostly focused on easily avoided pain and suffering as well as ensuring the scientific validity of the study so that the deaths are not meaningless. As for "inexpensive" that a government pays for it doesn't make it free. SDY (talk) 07:03, 8 June 2011 (UTC)

With all forms of "intellectual property" there is a trade-off between the price and the number of copies sold. In the case of the newest pharmaceuticals, this is becoming rather absurd, with drugs routinely introduced that cost over $100,000 per patient, even though they are ordinary monoclonal antibodies any passable technician can make.

I don't deny the usefulness of screening in bacteria, cell culture and mice in many situations - it's more the bureaucratic requirement for a certain sequence of studies before a human test can be done. The primate studies bother me the most, but in general, once a compound has some promising indication, it would be nice to see if it can help those in great need. Instead companies are supposed to go through a gauntlet of intermediate tests, anteing up the venture capital ahead of time, paying the bureaucrats' salaries as they go, in the hope of a payout ten years later that depends more on marketing than medicine. There has to be a better way to do things. Wnt (talk) 08:05, 8 June 2011 (UTC)

Four wheel drive..... and two wheel drive

Hi. I know that four whell drive vehicle is more powerful then two drive vehicle . but which one consumes more fuel two wheel drive or four wheel drive.......... I have doubt because I think both carry equal weight so both should consume equal fuel. — Preceding unsigned comment added by 220.225.96.217 (talk) 05:48, 8 June 2011 (UTC)

Four wheel drive vehicle has at least one extra differential and a set of driving axles to wheels. Thes increases both weight and mechanicla losses (friction), so it will consume more fuel. -Yyy (talk) 06:52, 8 June 2011 (UTC)

Agreed. Also, 4-wheel drive vehicles are not inherently more powerful than 2WD. However, in order to be able to still move at a reasonable speed despite the increase in weight and friction, they do tend to put more powerful engines in 4WD vehicles. Such engines are often diesels and/or mated to transmission geared lower, though, meaning 4WD vehicles typically have more towing and climbing torque, but less horsepower, and therefore don't typically accelerate very quickly. StuRat (talk) 07:51, 8 June 2011 (UTC)

A king cobra question

I saw a TV show recently about a king cobra tracking project. One finding that apparently wasn't expected (although I'm not sure why) was that males sometimes kill pregnant females. Unfortunately, having an attention span of a puppy, I may have missed the bit where they explained how a king cobra can kill a king cobra. Seems puzzling given that they use their venom to kill other snakes. What does this mean ? King cobra venom is toxic to king cobras ? King cobras are only immune (I assume they are immune) to their own venom ? It wasn't the venom that killed the females ? I can't find anything in our article or in google. Thoughts ? Sean.hoyland - talk 08:12, 8 June 2011 (UTC)