Wikipedia:Reference desk/Archives/Science/2011 June 4

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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)