Wikipedia:Reference desk/Archives/Science/2009 October 4

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October 4

Help with clarifying what something I've added an article is....

I've just been doing an un-stubbing of the article for Departure Bay, British Columbia (Canada). I've added this (sourced and all). Below is a cut / paste from my additions to the article. I'm too tired to figure this out myself now; I've searched all over Wikipedia and can't find it (though I don't have or know of any good sources on chemistry / explosives). Can anyone out there help me clarify what "duraline stumping powder" is or could be? The original source I have consulted does not clarify. I'm not sure, but maybe it's designed for blowing up stumps. What gets me, though, is that it's included in a list of very common / general explosives (i.e., black powder and dynamite). Any help would be appreciated....

In 1892—after realizing that irrespective of the new plant in the Northfield area, demand was still increasing—the Hamilton Powder Company built an explosives (mainly black powder, dynamite, and duraline stumping powder) manufacturing plant on the shore of Departure Bay. The production of black powder was relatively dangerous, and the death of employees was frequently the result of accidents.

Peace and Passion ☮  ("I'm listening....") 05:07, 4 October 2009 (UTC)

Duraline is a brand name. For info about stumping powder, see Tree stump#Stump removal. It makes sense that a company that manufactured black powder would also manufacture stumping powder. The two products are similar in that they are both powdered mixtures containing potassium nitrate. Red Act (talk) 06:22, 4 October 2009 (UTC)

Wikipedia has an article on Tree Stumps? That's just awesome. Anyway, I thought "duraline" sounded like a brand name, but I decided it must not be given the context and the fact that "Duraline" just somehow sounds like a modern brand name, not a 1892 brand name from, relatively speaking, the middle of nowhere! But you must be right; maybe I'll just rephrase the paragraph to say solely "stumping powder" so the context is consistent. Thanks for your quick help,

Peace and Passion ☮  ("I'm listening....") 06:27, 4 October 2009 (UTC)

Hold on, I've just read the article and I'm not sure I completely understand. Do you mean to say that stumping powder is not an explosive, it's rather a rotting agent? So my whole phrasing of the sentence needs to be restructured? I just found a quote later on in the reference I'm using saying it's "less dangerous than either black powder or dynamite and was used mainly by farmers to blow up stumps." Now I'm confused, I have two ambiguous hints, seemingly contradictory! What should I do, lol ;) ...

Peace and Passion ☮  ("I'm listening....") 06:33, 4 October 2009 (UTC)

Whoops, it appears I've lead you astray! "Stumping powder" is for sure the term for an explosive used historically to blast stumps away, as a google search on the term will readily show. The modern chemical stump removal products (commonly potassium nitrate) used to accelerate the decay of the stump, as described in the tree stump article, is different. The modern products to accelerate decay of a stump are not called "stumping powder". Sorry! Red Act (talk) 19:08, 4 October 2009 (UTC)

Well, now we've figured it out. Thanks for adding it to the tree stump page; I'll just remove the "duraline" on the Departure Bay page and link it to tree stumps (it's a bit of a poor WP:EASTEREGG link, but it'll have to do).

Peace and Passion ☮  ("I'm listening....") 20:14, 4 October 2009 (UTC)

pH

Which of these has higher pH? 0.1M HCl(Hydrochloric Acid) or 0.1M CH3COOH(acetic acid)Why?.......Please answer....... —Preceding unsigned comment added by 59.92.242.163 (talk) 08:35, 4 October 2009 (UTC)

Acetic acid, because it is weaker. Tim Song (talk) 08:38, 4 October 2009 (UTC)

Read pH. As you will see, it is the negative logarithm of the hydrogen ion concentration. This means that a smaller concentration of hydrogen (hydronium) ions results in a large value in pH. Keeping this in mind, let us analyse this problem. HCl is a strong electrolyte, ie it dissociates completely to give Hydrogen cations and Chlorine anions. So, it gives a relatively higher concentration of Hydrogen ions. However, Acetic acid, being a weaker electrolyte, only partially dissociates into its respective ions. In fact, there exists an equilibrium between the ions and the acid itself. Hence, the concentration of hydrogen ions it can supply is considerably lower, and hence it has a higher pH. Rkr1991 ^{(Wanna chat?)} 09:23, 4 October 2009 (UTC)

irrelevent discussion removed (see talk page)83.100.251.196 (talk) 17:52, 4 October 2009 (UTC)

Fertilisation

fertilisation and activation of egg i.e. Sperm entry to ovum in exact sequence. —Preceding unsigned comment added by Humzaifat (talk • contribs) 10:27, 4 October 2009 (UTC)

See Human fertilization. Red Act (talk) 10:36, 4 October 2009 (UTC)

A question for (early) gun enthusiasts

I've read about some quite interesting contraptions used in a 16th century battle. In the 1552 siege of Eger, the greatly outnumbered defenders constructed a giant wooden wheel, filled it with explosives, and rolled it down the hill to break the ranks of their enemy. It's a pity no contemporary drawings exist, so we will never know how exactly it could have looked like, but based on what I've read, they had hundreds of "obsolete" guns, and they filled the wheel with them, setting the fuses so that they will be set off sequentially. (as in 1552 in Eastern Europe they still used bows and arrows and the most advanced gun of the time was probably just an arquebus, it think obsolete might mean handegonne). They filled some of them with sulfur, so it had some kind of flame thrower as well. So, my question would be:

• Can this be considered an early form of volley gun?
• Why was it never standard issued, if it was so effective, at least against enemy morale? Surely it would be limited to those on higher ground, but I never heard it being more than an improvised device. Or were there other uses of similar devices in other times and places I don't know about?

I found a movie on youtube [1] (at 6:30 to 7:30) containing it, however, it is just an artistic impression, it might or might not have been similar to it. --91.8.225.151 (talk) 11:20, 4 October 2009 (UTC)

• I don't think this could be considered a volley gun: I understand that term to mean a device firing multiple shots in the same direction in a controlled manner.
• It seems to me that such a device would only be effective in very limited circumstances, be too uncontrollable and subject to chance variations in terrain, shifting battlefield conditions or other factors, and so be more trouble than it was worth. A similar device called the Great Panjandrum was invented by the engineer (and novelist) Nevil Shute Norway during the Second World War, but in trials proved unsuccessful, as did a fictionalized radio-(un)controlled version of it in the TV show Dad's Army. 87.81.230.195 (talk) 14:22, 4 October 2009 (UTC)

The description sounds fanciful, like conjecture by someone of a later generation. People under seige usually do not have time to build something novel which uses up time, weapons, and gunpowder. How could they possible have timed the fuses so that the thing fired when it was aimed at the enemy as it rolled down the hill? What does a contemporaneous account say, exactly? Edison (talk) 23:29, 4 October 2009 (UTC)

It seems really unlikely to me. If these "obsolete" guns still fired - you can bet they'd have had a use for them up on the battlements. Our article: Siege of Eger says "...a water-mill wheel packed with gunpowder which he rolled into the Ottoman ranks. His secret lied in the gunpowder not simply exploding but sparking even more fire." - which sounds much more likely. They wouldn't have had to "construct" anything if the water-mill wheel already existed - and this description suggests that they just stuffed it full of explosives - no fancy triggering mechanism - no obsolete weapons - no "volley gun" - just a big rolling bomb. Some sites say that rusty gun-barrels (not entire guns) were packed with sulpher, flints and other 'junk' to add to the shrapnel. One presumes that the idea was that the force of the main explosion would force this junk outwards and cause more horrific injuries. Still no "volley gun" though. SteveBaker (talk) 02:10, 5 October 2009 (UTC)

grow

what way do horse nuts grow, with the white bit facing the sky or facing the earh. I need to know what way to plant them —Preceding unsigned comment added by WPRDQER01 (talk • contribs) 12:59, 4 October 2009 (UTC)

When these tree seed fall in the wild, the way they end up is probably random, so it's unlikely that it matters much. Plants are sensitive to gravity, so that roots and shoots quickly adopt the appropriate directions - see Gravitropism. You could try a small experiment by dropping some of them many times and seeing if they end up a particular way more often: if they do, that's the way to go.

I'm presuming, by the way, that by "horse nuts" you mean the seeds or 'conkers' of the Horse chestnut tree: which hasn't been much modified by human cultivation; seeds of plants that have been more heavily modified by selection, cross-breeding and other techniques may be less robust, so that the sowing orientation matters more.

However, there may be other factors than orientation that affect whether or not a particular tree's seeds start to grow, like whether they're buried or not (and how deep, and in what), or the temperatures they're exposed to - the article on Germination mentions some of these factors. Some seeds will only grow if they've been exposed to frost, or to fire, or even if they've passed through the digestive tract of a particular animal or bird! I don't think any of these apply in this case, but I haven't found any reference that says so definitely. Googling on 'Horse chestnut germination' brings up quite a few relevant references and discussions (do explore them yourself), but overall conkers seem to be unfussy and I don't see any critical factors. Be aware though, that Horse Chestnut trees are quite slow growing (and can live for centuries), so you're not going to have a spectacular, flowering tree anytime soon. 87.81.230.195 (talk) 14:04, 4 October 2009 (UTC)

Built in GPS Mobile Phones

Lately I was searching for a mobile that serves all important features and couldn't find any matching device. I was wondering for the reasons, companies don't provide such services together in one device:

• Dual mode :GSM/CDMA
• GPS asisted and built-in.
• OS: Windows Mobile
• Camera
• SD card

Is it because there is no much request for such gadgets (in my opinion these aren't gadgets), or complications of manufacturing both networks? Perhaps, the question is about technology rather than science. --Email4mobile (talk) 15:08, 4 October 2009 (UTC)

A website like CNET allows you to add requirements to narrow down a list of phones that match your criteria. I couldn't find any such phones on that site, but other comparison-shopping sites have similar feature filters. Nimur (talk) 16:31, 4 October 2009 (UTC)

Because anyone building such a nice phone wouldn't put such a pile of stinking pooh on it as Windows Mobile? Heck - even Steve Ballmer admits that. SteveBaker (talk) 15:26, 5 October 2009 (UTC)

How hard and where did you look? Without intending to disrespect Nimur, I'm somewhat doubtful cnet is particularly complete when it comes to this sort of thing. I doubt it is even mostly complete with the major international brands. And there are a whole bunch of Chinese phones that it definitely doesn't contain. There are quite a few Chinese dual mode phones and also some smart phones usually with Windows Mobile (probably because until the launch of Google Android there was little other ready made option and companies are still getting their heads around Google AndroidO). It wouldn't surprise me if there's one that combines both. I did find [2] although I'm not sure if they're a Chinese company. You're generally unlikely to hear about these phones as they're not sold (or approved by regulators) in most developed countries, in fact China is usually their primary market. In fact it may be difficult to find anything about some of these phones sometimes from my experience only the cheap phones with MTK chipsets and Nucleus Plus OS and a cloned iPhone (particularly) or Android or whatever interface get much attention (and no I don't consider these smartphones because they usually aren't powerful or featured enough to be called smart phones IMHO). Nil Einne (talk) 17:13, 5 October 2009 (UTC)

No disrespect taken : ) Nimur (talk) 00:07, 6 October 2009 (UTC)

There are not many dual mode GSM/CDMA phones at all. (If CDMA was supposed to be CDMA 2000 or similar earlier standart). -Yyy (talk) 07:30, 6 October 2009 (UTC)

Calculating electrical resistance

Calculating the electrical resistance of a body such as a wire, with a length much greater than its diameter, is straightforward, knowing the resistivity of the material, e.g. 1.59×10⁻⁸ ohm-metres for silver. Equally easy is the resistance of several wires in parallel, even if they're uninsulated and touching - they can be regarded as separate, or as a single wire with the combined CSA. But how would the resistance be determined of something more complicated, say a thin square lamina between opposite corners, or a thin circular one between points diametrically opposite? Each shape could be considered as consisting of many wires of different lengths in parallel, but it all looks nastily infinite and bound to require integral calculus.→86.148.186.208 (talk) 15:32, 4 October 2009 (UTC)

You might find sheet resistance helpful. The correct unit to use here is "ohms per square" (Ω/☐) (That is "ohms per square", not "ohms per square inch", not "ohms per square meter" - it has units of ohms). This has huge application in ASIC design - in semiconductor processing, there is always a need to calculate the effective areal resistance between two ports, and even with CAD it is not usually suitable to perform a full nasty integral every time. For a non-rectangular area (which is very uncommon in semiconductor processing), you can approximate as a sum of rectangular regions. Two or three regions can form a pretty reasonable polygon for estimation purposes. Keep in mind that geometric tolerances in silicon manufacture are often on the order of 10-20% - so very accurate estimation of the resistance is less important than designing for a wide range of acceptable values! Nimur (talk) 16:20, 4 October 2009 (UTC)

The integrals don't seem that difficult, and I can't imagine how to get the resistance without doing them; for the circular one the resistance would be muʃt r² sin²(a) da (from a = 0 to pi) (where mu is the resistance in omhs per unit length, t is the thickness, r is ...) - just by summing the cross sectional areas from one side to another. The resistance accross a diagonal of rectangle is also doable.83.100.251.196 (talk) 17:48, 4 October 2009 (UTC)

Would you like to set up the integral to estimate resistance between two points on this standard-cell? Nimur (talk) 18:41, 4 October 2009 (UTC)

It might not be a good idea, seeing the mess I made of the above ^^

This manual from UC Berkeley sets up the theory (including an integral derivation) and relates it to the practical lab methods. Nimur (talk) 18:54, 4 October 2009 (UTC)

Correction for the disc - using R=muL/A R=muʃr sin(a) / t r sin(a) dx = muʃdx/t (between 0 and pi) = pi mu/t - interestingly the resistance accross the disc is unrelated to its radius! that is for edge to edge resistance between opposite points, not inner points...in a linear field..83.100.251.196 (talk) 18:48, 4 October 2009 (UTC)

I see what you're trying to do with that integral, but I'm fairly certain it's incorrect. You're basically making the assumption that the voltage is constant along a line perpendicular to the line going through the two end points. To see that it's incorrect, think about what equipotential lines should really look like in an infinitesimal region near one of the end points. Red Act (talk) 22:20, 4 October 2009 (UTC)

Yes I aggree (quote me "for a linear field") to get this I'd need the disc sandwiched between two capacitive plates.. Not really what was originally asked.83.100.251.196 (talk) 23:54, 4 October 2009 (UTC)

Another thing to bear in mind is that the resistance between two points is theoretically infinite because the contact areas are theoretically zero. In practice, the answer depends on the size of the contact areas. To get a meaningful answer you need to measure the resistance between two well-defined contact areas, such as two parallel wires bonded to opposite sides of a square sheet, or two metal discs bonded to the ends of a cylinder. I don't see how you're going to get a meaningful answer to the case involving the opposite corners of a rectangle. --Heron (talk) 18:46, 4 October 2009 (UTC)

In practice, you measure this with a four point probe. This is a standard piece of equipment in a semiconductor lab. Here is one from University of Pennsylvania. The "infinitesimal point-of-contact" is solved by making a slight scratch into the surface with the needle-like tip of the probe; this makes a good approximation to a "point"-contact. The measurement of a rectangular sheet-resistance can be seen in these photographs from Unisoku (a commercial vendor). Nimur (talk) 18:52, 4 October 2009 (UTC)

Interesting links, Nimur, but I'm not convinced. I think you're all answering the wrong question. The 4-point probe that you refer to is just a Kelvin probe. It solves the practical problem of contact resistance, but no amount of experimental ingeniousness will get around the underlying geometrical fact that you can't measure the resistance of an object that terminates in a point, which is what the OP wanted to do. If a material has finite conductivity then a perfectly sharp corner of the stuff must have infinite resistance. And if the corner is not a mathematical point but a real-life slightly blunt one, then its resistance depends entirely on its imperfect geometry. You may be able to measure resistivity using your devilish 4-point method, but you can't measure resistance, because the latter is determined by the geometry of the object. All your calculations are doomed to fail if you ignore this distinction. --Heron (talk) 20:04, 5 October 2009 (UTC)

Okay, but there is no better real-world approximation to an infinitesimal-point-of-contact than a needle-point. That is why it is used in industrial processes - to measure resistance and resistivity. Nimur (talk) 00:09, 6 October 2009 (UTC)

To accurately calculate the resistance of a complicated object (as opposed to measuring that resistance), you'd generally need to use the finite element method, except in cases where symmetry enables you to simplify the problem enough to solve it, or at least closely approximate it, analytically. Red Act (talk) 22:40, 4 October 2009 (UTC)

I have a wire Frame cube. The resistance of each edge of the cube is 1 ohm. What is the resistance from diagonally opposite corners of the cube? This is not home work but something been puzzling me for years. How does one work it out using normal circuit theory?--SpiceJar (talk) 22:33, 4 October 2009 (UTC)

See here or Answer 15 here. Gandalf61 (talk) 23:15, 4 October 2009 (UTC)

Temparature in Space

As far as I know temperature is the numerical expression of average kinetic energy of motion of molecules.So what will the temperature of space where there are no molecules? —Preceding unsigned comment added by 113.199.141.152 (talk) 16:28, 4 October 2009 (UTC)

Temperature is not simply the average kinetic energy of motion of molecules -- for a precise definition you would have to read the Entropy article, but it isn't easy going. Basically temperature is a number that determines which way the net flow of energy will go when two systems are brought into contact with each other -- it always goes from higher temperature to lower temperature. In outer space with no massive particles, the temperature is set by the cosmic microwave background radiation. If you shield a portion of space so that no particles whatsoever can penetrate, the temperature is undefined. Looie496 (talk) 16:44, 4 October 2009 (UTC)

Also, there is a small amount of stuff out there. The average is about one atom per cubic centimeter, I believe. --Sean 16:57, 4 October 2009 (UTC)

The mass-density and energy-density depend where you are, of course. A temperature for a perfect vacuum can be defined, but it is quite different from the conventional usage of "temperature" (which is exactly as the OP stated - the average kinetic energy of the molecules). Nimur (talk) 18:37, 4 October 2009 (UTC)

Interestingly, the Boomerang Nebula is actually colder than the cosmic microwave background, and is the coldest known natural environment. Gandalf61 (talk) 21:34, 4 October 2009 (UTC)

I am a body of mass 1kg floating in deep space. My surface area is 1 m^2. My surface emissivity is 0.95. I have been here many millions of years. What is my temperature? (Not homework just curiosity) —Preceding unsigned comment added by SpiceJar (talk • contribs) 22:49, 4 October 2009 (UTC)

Assuming that by deep space you mean away from any potential source of radiation such as stars, your temperature is the same as the cosmic microwave background which is about 2.7 K. all the info you gave about size mass, and composition is unecessary. Dauto (talk) 00:35, 5 October 2009 (UTC)

It is necessary to work out how long it would take to reach thermal equilibrium, but it is not sufficient - the initial temperature is obviously needed. I would expect something that small to have reached equilibrium after millions of years even if it started of extremely hot, though - the square-cube law is important here, small objects exchange heat very quickly. --Tango (talk) 02:22, 5 October 2009 (UTC)

Indeed. A white dwarf is just such an object, only a bit heavier than 1kg. However, none has reached thermal equilibrium yet - the universe is too young (and "too young" at 14 billion years, not 6000 ;-). --Stephan Schulz (talk) 14:39, 5 October 2009 (UTC)

Um, our white dwarf article gives a range of 0.17-1.33 solar masses, average of 0.6, this corresponds to an estimate of 1kg being off by about 30 orders of magnitude. Googlemeister (talk) 18:18, 5 October 2009 (UTC)

Well, I did say "a bit heavier". --Stephan Schulz (talk) 21:46, 5 October 2009 (UTC)

Yes, a white dwarf is rather big so takes a long time to cool down. A 1kg mass with a surface area of 1m² would cool down much faster. --Tango (talk) 22:24, 5 October 2009 (UTC)

Energy transformation and space-time

I went through some interesting subjects today as in "speed of photons/light" and the "current source voltage source" discussions. I was wondering of the answers; though, accidentally I had a relevant question to ask. Does energy require time and/or space to be transformed from one form to another or can this transformation reduce to power terms at all cases?--Email4mobile (talk) 17:13, 4 October 2009 (UTC)

I think you are asking whether any such form of energy transfer can occur instantaneously. The most complete answer is that all physical interactions appear to be quantized on some level. This means that the smallest/fastest/most-infinitesimal version of any particular interaction must occur on a discrete time scale. Notably, though, we have not fully explained the quantum nature of gravity, for example; there are other complex interactions which are not necessarily best-modeled with a quantum theory. But, it is probably safe to say that there is no process which can occur faster than one interaction per Planck time. Nimur (talk) 18:48, 4 October 2009 (UTC)

Indeed. Time periods less than the Planck time are essentially meaningless within the framework of our current theories. It is possible (I'd even say it's likely) that we'll come up with a theory in the future that has a better explanation of what happens on small time-scales, but even then it is likely to still be a quantum theory, so there will be a minimum amount of time that any process must take. --Tango (talk) 19:06, 4 October 2009 (UTC)

International Space Station Experiments

Is there a comprehensive fully itemised list of all the experiments that have been and are currently being conducted, and how and where can I access it? —Preceding unsigned comment added by Hardy6273 (talk • contribs) 18:56, 4 October 2009 (UTC)

I don't know about a comprehensive list, but there is some information on the subject on the NASA site here. --Tango (talk) 19:00, 4 October 2009 (UTC)

Not only is there a comprehensive list - the data collected resides in the public domain (with few caveats and exceptions). You can access the data, perform original research, and submit your results for peer review. It's surprising to me that few "crank scientists" take advantage of this - instead, they seem to prefer to speculate about perpetual motion machines instead of freely accessing the best scientific data this side of the Van Allen belts. Nimur (talk) 19:05, 4 October 2009 (UTC)

Yes but it's not really surprising to you, is it? --Mr.98 (talk) 21:32, 4 October 2009 (UTC)

I hope not. For those not sufficiently familiar with cranks to understand, allow me to explain: The data is only useful to people with significant scientific knowledge and understanding, which is precisely what cranks lack (and often pride themselves in lacking). --Tango (talk) 21:43, 4 October 2009 (UTC)

Sclerophyllous plant communities

Besides the Mediterranean biome and in Oceania, is there any other region of the world that presents a mostly sclerophyllous vegetation cover, or one that this vegetation can be considered to play a significant part in? —Preceding unsigned comment added by 201.21.180.57 (talk) 23:58, 4 October 2009 (UTC)

Have you looked at the Sclerophyll article? It mentions California, Chile and South Africa for starters. 87.81.230.195 (talk) 02:34, 5 October 2009 (UTC)