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Just wondering what your take on this would be. — Preceding unsigned comment added by 96.238.137.168 (talk) 01:23, 20 February 2012 (UTC)[reply]

With safety gloves. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:04, 20 February 2012 (UTC)[reply]

If you know how much shrinkage to expect, you can make it larger than needed, to compensate for the shrinkage. As for excess heat, fans could be used, or other cooling methods. StuRat (talk) 05:42, 20 February 2012 (UTC)[reply]

Why does keeping a laptop plugged in constantly ruin the battery life? --108.225.117.174 (talk) 03:02, 20 February 2012 (UTC)[reply]

Who says it does? ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:03, 20 February 2012 (UTC)[reply]

"if you keep your laptop plugged in, you force your battery to remain at 4.2V continuously and these side reactions continue to happen and slowly kill the battery" "Laptop batteries may last a little longer if you let them fully discharge occasionally". Clarityfiend (talk) 03:15, 20 February 2012 (UTC)[reply]

See Memory effect. If you never discharge a battery all the way, it sort-of "forgets" what "all discharged" means, and you lose battery life. --Jayron32 03:44, 20 February 2012 (UTC)[reply]

The memory effect is a feature of NiCd batteries. Most laptops have a lithium-ion battery which has battery-management circuitry to prevent both overcharge and over-discharge. Both are harmful to battery life, with deep discharge making the battery unusable. The general advice seems to be to avoid leaving the power supply permanently connected, but many users leave laptops permanently plugged in with little noticeable effect on battery life. Ideally, the battery management circuitry should have a software link so that it can be adjusted to a "permanently plugged-in" setting where it stops charging just below the maximum power to extend battery life, and with a maximum charge setting for those who need many hours of use from battery alone. None of the laptops that I've used have this facility. I wonder why they don't. Dbfirs 13:46, 20 February 2012 (UTC)[reply]

I've read that lithium-ion batteries deteriorate faster when full, and also deteriorate faster the warmer they are. Operating a device while plugged in is the worst possible case: the battery is at full charge, and the device is warm.--Srleffler (talk) 18:30, 21 February 2012 (UTC)[reply]

Yes, that sounds accurate, but very few of us bother to disconnect our laptop batteries and store them in the fridge when the laptop is connected to the mains. I wonder if any research has been done to compare the shortening of life from "warm and fully charged" with the shortening of life from regularly charging and discharging. I suspect that the "warm and full" deterioration is measurable, but not a major factor in battery life, so most of us don't bother to allow the battery to discharge slightly, then move it to the fridge when we don't need it for a while. Dbfirs 09:48, 22 February 2012 (UTC)[reply]

From Marburg virus disease:

Marburgviruses are highly infectious, but not very contagious.

After looking through Infectious disease and Contagious disease, I can't figure out what this means — especially since the infectious disease article says "Infectious diseases that are especially infective are sometimes called contagious." How can a highly infectious disease not be very contagious? Or how can a less-contagious disease be characterised as highly infectious? Does it keep reinfecting hosts that already have it? Nyttend (talk) 04:44, 20 February 2012 (UTC)[reply]

Contagious is a measure of how easily it spreads in a human population. Ebola is another virus which is infectious but not very contagious, owing to the fact that victims quickly become incapacitated and die after infection before really having a chance to infect other people. On the other hand, HIV is given as an example highly contagious disease since victims can walk around for years without even knowing they're infected passing it on to people far away from where they initially caught the infection. Vespine (talk) 05:19, 20 February 2012 (UTC)[reply]

For completeness, diseases like Lyme disease are infectious but not contagious at all, since you can't catch them from other people. Vespine (talk) 05:20, 20 February 2012 (UTC)[reply]

So highly infectious basically means "if you get even a few pathogens, they have a high chance of infecting a substantial portion of the body and causing observable effects"? Nyttend (talk) 05:25, 20 February 2012 (UTC)[reply]

Well, I'm not a doctor but that's the way I understand it. I think it's more of a "threshold", as opposed to infecting "portions" of the body, but yeah... Also the Contagious disease article does say The boundary between contagious and non-contagious infectious diseases is not perfectly drawn. It contradicts me and says HIV is a non-contagious disease since it is not easily transmitted by physical contact, but then goes straight on to say In the present day, most sexually transmitted diseases are considered contagious. So I think the term has a slightly "loose" meaning. Vespine (talk) 05:28, 20 February 2012 (UTC)[reply]

I'm pretty sure "contagious" means "can be transmitted between individuals in a population". Malaria is infectious but not contagious -- you get malaria from mosquitos, not from other humans.--Atemperman (talk) 13:13, 20 February 2012 (UTC)[reply]

http://www.mediafire.com/imgbnc.php/cddd07abfa9b042f748ea3294428b71476d5fba1075598ef99d7079528e1c3a26g.jpg which insect is this ? — Preceding unsigned comment added by 182.178.249.145 (talk) 15:24, 20 February 2012 (UTC)[reply]

This will be tough without more info. Things that will help, if you know them (e.g. if you took the photo): Where was the photo taken? In what season? What kind of tree is that? How large is the specimen? Nevertheless, I'll start with some guesses:

• Could be a mealybug. Looks a lot like one. But we generally expect to see them aggregated on green plant tissue, not all alone on bark.
• Could also be some type of woodlice, which are actually crustaceans. They are more likely to be found alone on bark. The adults aren't generally white, but young ones are usually pale. SemanticMantis (talk) 15:50, 20 February 2012 (UTC)[reply]
  • Woodlouse would get my vote. Guettarda (talk) 20:55, 20 February 2012 (UTC)[reply]
    • The woodlouse for me, too. I see those all the time in Minnesota, only I call them "pillbugs." Stripey the crab (talk) 03:46, 23 February 2012 (UTC)[reply]
      • It was taken it in March 2011, Pakistan. It is spreading in my home garden. http://www.mediafire.com/imgbnc.php/2a135256a5246b1df1a662a4c4939f7ae23cacd5d43e04c7670dfcbfd1f77bfc6g.jpg and Second photo

Wow. I've never seen woodlice at that density, except maybe under a rotting log. Note that woodlice are detritovores, and will not harm plants. However, these additional pictures make me question both of my guesses above. The good news is, you can catch one and confirm/deny the woodlouse ID by counting the legs per segment, as described at woodlouse. I recommend you start a new question with all the pictures (questions here only attract good answers for a few days), and maybe also consult a more focused group here [1]. SemanticMantis (talk) 23:18, 24 February 2012 (UTC)[reply]

— Preceding unsigned comment added by Pendragon5 (talk • contribs) 23:16, 20 February 2012 (UTC)[reply] 

Can anyone helps me on number 16? Some drawing goes with the explanation will be highly appreciated! The answer for number 16 should be from 50-70 degrees (all they want is a rough calculation) so what matters is how to do it. Thanks!Pendragon5 (talk) 19:50, 20 February 2012 (UTC)[reply]

The redshift will give you the radial velocity (the component of velocity directly away from or towards you). You calculated the total velocity in the previous question. A bit of trigonometry will then give you the angle. --Tango (talk) 20:57, 20 February 2012 (UTC)[reply]

Can you this problem as an example for me please? I don't even know what angle are they looking for? If possible can you draw a picture to help me visualize and post it on here? It would help me a lot to solve a problem if i can visualize it.Pendragon5 (talk) 22:10, 20 February 2012 (UTC)[reply]

Draw a line between the observer and the star, then draw an arrow from the star pointing in the direction the star is moving. They want the angle between that line and that arrow. --Tango (talk) 23:09, 20 February 2012 (UTC)[reply]

No, not that angle. The star isn't going to move a significant distance in your lifetime, so considering it at two points isn't helpful. You're just interested in the instantaneous velocity. --Tango (talk) 23:35, 20 February 2012 (UTC)[reply]

I don't understand this at all. I can't even visualize what angle am i working on. I don't get what they mean by measured wavelength and laboratory wavelength. I have no knowledge of connecting the information they give to find the angle. Can you just do it as an example problem for me? So i can do the similar problem if i was given one. I'm totally blacked out in this problem sorry.Pendragon5 (talk) 23:55, 20 February 2012 (UTC)[reply]

I linked to an article, have you read it? --Tango (talk) 00:26, 21 February 2012 (UTC)[reply]

Yea i have been working on it and trying to understand it for the last few hours. Alright let just say that i'm being stupid on this ok.Pendragon5 (talk) 00:42, 21 February 2012 (UTC)[reply]

You know the projection of the radial velocity onto the direct line between you and the star. (You measured the red-shift, right? That's the velocity relative to you. And you calculated the orbital velocity, which is the total velocity of the star, right?) You're looking for the angle between the line that connects you to the star, and the line that the star is moving along. In physics parlance, the redshift tells you the projection of the total velocity on to your radial line. I'd use a dot product formula any time I have to calculate a projected vector; you can rearrange this formula. Have you studied much trigonometry yet? We can walk you through the details in a very step-by-step fashion, but even if we do, it might not be very helpful until you've got a really solid background in algebraic trigonometry. Nimur (talk) 01:52, 21 February 2012 (UTC)[reply]

i don't know how much is needed to be considered as having solid background in algebraic trigonometry. The highest math class i have done is precalculus, that's all. I remembered learn about dot product before but well we didn't apply it into complicated ideas, we basically do easy stuffs and obviously problems. And again i still can't picture what angle are we trying to find? Can someone just draw it on a piece of paper and upload it here please, that would be extremely helpful to me.

And no i still don't even know how to calculate the redshift. When i look at the formula, i don't know which one is which. Where should the 656.5386 nm measured value be in the formula? Where should the 656.3 nm laboratory value be in the formula? I still don't understand this statement "the line that connects you to the star, and the line that the star is moving along". The line that the star is moving along should be changing continuously since the star is always moving or what do you mean by moving along? What exactly angle is this? Pendragon5 (talk) 02:19, 21 February 2012 (UTC)[reply]

Calculation of redshift, ${\displaystyle z}$

Based on wavelength	Based on frequency
${\displaystyle z={\frac {\lambda _{\mathrm {obsv} }-\lambda _{\mathrm {emit} }}{\lambda _{\mathrm {emit} }}}}$	${\displaystyle z={\frac {f_{\mathrm {emit} }-f_{\mathrm {obsv} }}{f_{\mathrm {obsv} }}}}$
${\displaystyle 1+z={\frac {\lambda _{\mathrm {obsv} }}{\lambda _{\mathrm {emit} }}}}$	${\displaystyle 1+z={\frac {f_{\mathrm {emit} }}{f_{\mathrm {obsv} }}}}$

And i'm strongly doubt that they expect students to know advance math. All they required is basic math like algebra. Is there a faster way that can give a really rough answer? That's why the answer in the answer key is any answer from 50 degree to 70 degree is correct. So the rough calculation should be fine.Pendragon5 (talk) 02:26, 21 February 2012 (UTC)[reply]

You can either use math or memorization. The major tools that higher maths like calculus gives you is the ability to manipulate equations in complex ways to get new equations that tell you new things about the world. You can either start from one equation and do calculus to get all the equations you need, or you can memorize a whole bunch of equations. For example, you can essentially derive all of your Newtonian mechanics equations from a few basic starting points and a 1-semester calculus class. Or you can memorize like 50 equations and their applications. That's why people learn higher order math: it simplifies your life to know one equation and ways to manipulated into 50 new equations than to have to memorize all 50 without any idea how they relate to each other. --Jayron32 04:54, 21 February 2012 (UTC)[reply]

As with several of the other astronomy questions Pendragon5 has been asking about, this one implies a couple of unstated assumptions. First is that the binary system has no significant overall radial motion with respect to us, so that the entire redshift is a function only of the radial (with respect to our line of sight) component of the star's orbital velocity. If we want to come up with the actual orbital inclination and not just an upper limit on the inclination, then we must assume that the Hα wavelength given represents the maximum redshift observed during the star's 10 day orbit. Calculating the redshift yields a radial velocity of almost exactly half of the orbital velocity, so we know that at least the answer key is correct this time (though I don't know why they bracket the answer with a range of 50°-70°. If the second assumption doesn't hold, and this is just a single random spectral measurement of Star C, then all we can say is that 60° is the maximum inclination. If the first assumption doesn't hold, then all bets are off and nothing can be calculated. While I'm griping at the statement of the problem, I'll point out how strange it is to specify the observed Hα wavelength to 7 significant figures, but to round off the reference value to 4 significant figures. Hmph. -- ToE 13:40, 21 February 2012 (UTC)[reply]

Would it be possible to fly this RC quadrocopter indefinitely (or until mechanical failure) while having it connected to AC power via the use of a long extension cable? That is, can I remove the battery from the quadrocopter, make a long extension of the thinner wire before the power supply and fly the drone without a battery while tethered to its power supply? It says the payload is 700g, what are the implications as far as drawing out more wire? I mean, does this put more load on it and if so what is the relationship or rate?

As a side note, I do realize that it should include a small backup battery to stop it from crashing in case of disconnect and that I am not clear on how the drone will be autonomously hovering in the same spot indefinitely, but that doesn't matter yet. Also, I have seen a few videos of the Parrot AR.Drone being flown while tethered to its standard short-cabled AC adapter. I can see that the stability is affected even while flying indoors and that this is a potential problem. Even so, I would like to know the feasibility of hovering much further away whilst tethered, regardless of this potentially causing significant load on the copter.

In summary, I'd like to know:

1. Can the quadrocopter be flown without a battery while directly connected to the power supply?
2. Is it possible to extend the less weighty wire before the power brick?
3. What is the rate that (I assume) load increases as distance from the power supply increases?
4. I also found this pay load/power consumption chart. Lhcii (talk) 20:19, 20 February 2012 (UTC)[reply]

1) Your RC chopper is unlikely to work directly off A/C, without the battery, unless modified.

A few other thoughts:

A) 700 grams isn't much length for an electrical cord, but would be more if you eliminated the insulation (you would need minimal insulation to keep the positive from shorting to the negative, unless they were somehow kept physically apart). However, uninsulated wire would be unsafe, and would also be less effective as a tether, having only the strength of the copper to keep the chopper in place. So, overall, leaving the full insulation in place would make sense, which would only allow for a very short tether.

B) You might find that the time between mechanical failures is less than you might think, since it's designed for only short flights. It might overheat or loose lubrication rapidly.

C) I also share your concern over the stability of tethered flight, but attaching the tether directly to the CG point is the least likely to disrupt that stability.

D) Introducing a spring into the system (preferably at the bottom of the tether) would help prevent it from breaking it's tether on updrafts.

E) Constructing a tower to support most of the weight of the wire would allow the chopper to fly higher. StuRat (talk) 20:49, 20 February 2012 (UTC)[reply]

Thanks, StuRat, I appreciate the suggestions. I think that building a tower will be a perfect solution to the problem for many situations, attaching it with a spring is a great idea as well! I was thinking of using a gimbal, but that is much simpler and thus lighter. I am disappointed that 700g is so little wire, but I'm sure I could also increase my range by upgrading the chopper or adding additional choppers to ferry the power cables and possibly a camera as well(also increasing weight, I know).Lhcii (talk) 23:13, 20 February 2012 (UTC)[reply]

Something else to consider is ground effects. That is, your chopper will get more lift when closer to the ground. Thus, you could have a fairly stable spot where, if it drops lower the ground effects will lift it back up, or, of it rises up the lack of ground effects will bring it back down. To use this form of stability, it would need to be quite close to the ground though, no more than a few feet up. And, of course, the up/down stability is only one of many types to worry about (with pitching, rolling, moving forward/back or right/left, and twisting the cord being a few others). To address all of those, you might want 4 cables, one attached below each rotor, at maybe a 45 degree angle. Two of those cables could be the positive and negative wires. You'd want the cable weights to match, and the + and - wires to be on opposite sides, in case they don't match. This would mean 4 towers, and maybe a stepladder in the center to launch and retrieve the chopper (or maybe scaffolding with the cables tied off at each corner).

BTW, why do you want to do this ? (Your answer might affect our suggestions.) StuRat (talk) 23:20, 20 February 2012 (UTC)[reply]

I'm a bit confused by your proposal. Are you saying you're planning to make the extension on the AC/line voltage side? If so this seems to carry obvious safety implications you need to consider. As StuRat has said, you want properly insulated wire. Also have you considered this means you will have to be carrying the AC adapter which will add more weight, probably significantly. Even with a switching mode PSU, it's likely to be heavy given the amount of power the device appears to need (the fact that a 2S2000mAh battery only lasts about 12 minutes tells you it's a lot) and will be another safety concern (since you're potentially going to drop it if things go wrong). All in all, while lithium polymer batteries when used in RC devices aren't exactly the safest thing, the plan to carry around an line voltage extension cord and an adapter on a copter seems a worse idea.

If you're thinking of making the extension on the DC side, are you aware what kind of current you're going to need to support? [2] suggests it needs a minimum of 50W (and I'm guessing that's the average meaning it will likely need more for peaks) up to 170W with a full load. That does of course demonstrate another problem with your proposal, the more weight the more power you need. But more importantly, if you look at the graph it seems like you'll need to supply at least 18A (at about 11V). But it has peaks of up to 30A, so really you'll should support that. (Perhaps you don't need to go quite that high, but you probably want to support more then 18A.) [3] [4] gives you an idea of what sort of wire gauge you'll be needing depending on the length of the wire, it seems to me you'll want at least 8 AWG probably 6 or even 4. Raising the voltage, should help, the pages suggest it supports 2S and 3S batteries, suggest it does support somewhat of a voltage range. If it support 4S or even higher without being very inefficient this would be good.

P.S. I somewhat doubt the AC adapter supplied with the device has the capacity to power it live. Also while the RD is great, you may want to consider a specialised forum. While I don't know that much about RC, I have come across some forums on searches for other things and the people there usually seem to be somewhat informed. For example [5] appears to have over 708 pages apparently on the copter you linked above and I noticed some discussion of 4S. [6] is another one I've come across before and also [7].

Nil Einne (talk) 20:07, 21 February 2012 (UTC)[reply]

Some other thoughts:

a) If you use the 4 tower system I proposed above, I suggest a net between them to catch the chopper if it falls.

b) Make sure the wire lengths are sufficient to reach the net without them being pulled taught.

These steps may make it possible to reuse the setup. However, the wires may still get chopped by the blades, so have backup wires. The net may also be damaged, so have extra netting. StuRat (talk) 21:36, 21 February 2012 (UTC)[reply]

Thank you StuRat and Nil Einne for your input. I have a lot to think about from your ideas. Nil Einne, the forums that you linked to me are full of information, I am sure I will find some nice tips on the Gaui and its different batteries. StuRat, I do like the four tower system. Perhaps it is more viable for testing purposes than it would be in actual application, but helpful regardless. The reason I asked the question in the first place is because I thought it would be a good way to take time-lapse or at least long duration video from a bird's eye view. I thought this would be effective particularly for youth sports games as it would be much more portable than a camera on a pole or ladder and would give a better perspective on the game to the viewers (more like a broadcast TV association football match). It seems like it won't be possible to spool out enough wire to let the chopper gain significant height, especially if it is going to be carrying a camera as well. — Preceding unsigned comment added by Lhcii (talk • contribs) 17:44, 22 February 2012 (UTC)[reply]

In that case I think a camera on the end of a telescoping pole would be the way to go, perhaps with a spike for the ground and 3 guy-wires to secure it. Here's a 25 foot tall telescoping flagpole for around the same price as your chopper: [8]. The helicopter would introduce unwanted vibrations which would blur the images, and the flagpole should be able to support a better camera. StuRat (talk) 18:54, 22 February 2012 (UTC)[reply]

Ah, you are very right, that would be much more suitable and less costly to maintain to boot. I guess my idea doesn't work out for my original intention, then. In any case, I thank you all for your input and assure you that it wasn't a waste. I'm sure that I'll be able to apply some of your ideas for testing purpose or otherwise, especially if I find a use for a tethered camera drone.Lhcii (talk) 01:07, 23 February 2012 (UTC)[reply]

OK, I'll mark this question resolved. Incidentally, when I was trying to figure out why you wanted a tethered chopper, my best guess was to demonstrate the chopper itself, say at a science fair or hobby store that sells them. StuRat (talk) 02:14, 23 February 2012 (UTC)[reply]

Hello. I would like to synthesize triphenylmethanol via a Grignard reaction. A common side-product is biphenyl, which prefers to dissolve in hexanes over ether. How can adding hexanes purify a mix of triphenylmethanol, biphenyl, and ether? The boiling point of biphenyl is higher than of triphenylmethanol. Rotary evaporation will not separate biphenyl from my desired product, eh? Thanks in advance. --Mayfare (talk) 22:19, 20 February 2012 (UTC)[reply]

Have you tried chromatography methods to seperate your products? If thin layer chromatography indicates a reasonable difference in Rf values (i.e. if the spots seperate enough), then you stand a good chance of Column chromatography working to seperate the products from each other. Collect all of your factions, identify which fractions contain your desired product using a combination of the TLC results (the compounds come off of the column in the same order they moved up the TLC plate) and analytical techniques (NMR and/or Mass spectrometry should work) and then collect those fractions and isolate your desired products, perhaps by distillation or crystalization or something like that. --Jayron32 23:02, 20 February 2012 (UTC)[reply]

This isn't something I know well, but a simple Google search turns up many hits - apparently it's a common classroom exercise.^[9] For example [10],[11],[12], and [13]. To quote this last (with a little imperfection of grammar), "The petroleum ether will dissolve the non-polar biphenyl, while the product triphenylmethanol does not." (The point is, the product is an alcohol) Using chromatography... someone's thinking like a biochemist. ;) Wnt (talk) 04:01, 21 February 2012 (UTC)[reply]

I did, in my mispent youth, spend about 18 months working in an organic synthesis lab. It was the most mispent part of my youth, in hindsight. But hey, it prepared me to answer this question. --Jayron32 04:48, 21 February 2012 (UTC)[reply]

I have an aluminum pot which, when filled with water and allowed to sit, gets white nodules forming on the bottom of the interior. Are these some type of salt ? I use this pot to replace water in my "humidifier" (a larger pot I leave on the stove all day). I don't think I want to eat anything out of this pot. Interestingly, I have another aluminum pot which doesn't seem to do this. What might be different in their constructions ? StuRat (talk) 22:20, 20 February 2012 (UTC)[reply]

Pure aluminium quickly develops an impervious, transparent oxide layer which prevents further oxidization. It seems likely that your pot was formed form aluminium which contained impurities. These (probably iron) prevent this protective film of oxides from forming. Never-the-less the aluminium -which is very reactive- still insists on oxidizing but in a different form. --Aspro (talk) 23:37, 20 February 2012 (UTC)[reply]

If one looks on page 33 one will see Fig 7 showing such an iron inclusion.[14]--Aspro (talk) 00:14, 21 February 2012 (UTC)[reply]

Sounds like calcification deposits. I expect the one pot has tiny roughnesses that serve as nucleation centers to allow the dissolved solids in the water (which is probably pretty hard) to come out of solution. Looie496 (talk) 00:11, 21 February 2012 (UTC)[reply]

I second this explanation (calcification). Aluminum, as noted above, is highly reactive. This means that it forms its protective oxide layer instantly, pretty much regardless of inclusions or impurities. This makes aluminum, in practice, almost inert. David Spector (user/talk) 02:27, 21 February 2012 (UTC)[reply]

We have city water, which isn't particularly hard. Also, why does it form nodules, instead of a continuous film ? StuRat (talk) 03:25, 21 February 2012 (UTC)[reply]

Q:why does it form nodules. Ans: OK, lets construct a rough pons asinorum. As I said, “Pure aluminium quickly develops an impervious, transparent oxide layer which prevents further oxidization”. Yet in the real would things are different AND we are not just talking about gaseous oxygen in this 'case study' regarding your pot. Some aluminium pots are made from high grade aluminium sheet (often thin and flexible) . These tend to get a good protective coat of oxides (two) which are self repairing every time they get scratched. Some pots are cast. These might use lower grade and recycled aluminium alloys with impurities. The latter are more prone to getting what you appear to describe but both can suffer from it. Now, what I need to know from you is -when the nodules have been removed are there little pits in the surface under them? If so you have quite ordinary and common pit- nodules of aluminium hydroxide. See here:[15]. Now. If what you have is pit-nodules, then the reason why they form as they do: As the last reference implies -there is natural stuff in your tap water that makes the original passive oxide layer permeable. Just being permeable though is not going to do much unless there is something to force the cations of aluminium out. So, back to my first reference of Fig 7 pg.33 An iron impurity can cause that area to become cathodic. In turn, it will make any near by weakness in the oxide skin anodic. So out flow the Al cations. Now. They are not flowing out (by way of perhaps mainly Cl electrolyte in your case) not into gaseous oxygen but water. As I said in my first post on this “aluminium -which is very reactive- still insists on oxidizing but in a different form” and so it uses the first thing it comes across -in this case water. Aluminium hydroxide has a volume many times the size of the pit it leaves behind and so it grows out and above the surface. This can happen in just a few days. Other cations in the taps water may be attracted to the cathodic spots (depending on their relative nobility to the Al) and after a few days equalise the charge so weakening the galvanic force and slowing the corrosive effect but there are too many variable here for me to guess at. --Aspro (talk) 19:00, 21 February 2012 (UTC)[reply]

A beautiful answer. Forget "calcification". David Spector (user/talk) 23:33, 21 February 2012 (UTC)[reply]

Thanks, I like to charge £70 per hour for consultations, so if anyone has any other questions please let me know how many pounds worth of my time they would like and where to send them my pro-former invoice (actually I'm a little embarrassed, as that wasn't a very good explanation since I wanted it to be in plain and simple English. Yet, my mind is getting a little fuddled. My quack say's it old age but I think its too much beer) (no, second thoughts -how can one drink too much beer?).--Aspro (talk) 23:57, 21 February 2012 (UTC)[reply]

Yup. Even to a hopeless failure at chemistry/physics like me, Aspro's answer makes sense. There has been a rather pointless debate about 'Aluminium' vs 'Aluminum' as the proper name for this substance, whereas it should most probably have been named 'LookingForAnExcuseToReactWithOxygenAtTheFirstOpportunityium'. Alumin(i)um as a metal isn't something you'd expect to find in nature for a very good reason - there is far too much oxygen about, and it will steal it from any source available - often with spectacular results. An aluminium pot is nothing more than a localised hiccup in entropy, and liable to find a way to restore disorder at every opportunity. From the perspective of the universe, it isn't the behaviour caused by 'inclusions of iron' that have to be explained, but instead the much less plausible 'inclusions of metallic aluminium'. (and in answer to Aspro's final question, I'd say not at the time, no, but come the morning after, probably ;-) ). AndyTheGrump (talk) 06:47, 22 February 2012 (UTC)[reply]

UPDATE: I scraped a few of the nodules off, and no pits are visible. However, they might be microscopic. I also got brave enough to taste the nodules, and they seemed slightly salty (but definitely not all salt). StuRat (talk) 06:15, 22 February 2012 (UTC)[reply]

Interesting observation (and like a true scientist did you grab a bottle distilled vinegar out the kitchen to see if it both dissolved the nodule and my argument?) . As I said towards the end “there are too many variable here for me to guess at” yet I too have wondered about this 'saltiness', (In cleaning out a badly corroded heat exchangers, some of it inevitably ends up on one's lips) . The following is just idle speculation so don't take it as gospel: It reminds me a little of Salty liquorice. So, Perhaps a Al cation gets the chance to gab a couple of passing Chlorates (present in the tap water). Strip off their oxygen, leaving the Cl's to in turn grab a passing nitrogen molecule. This 'might' result in creating Ammonium chloride which will taste something like salt but not table salt. Again, this is just speculation that I can not give any real reasoning to as why this reaction should happen and it might very well be the result of another chemical compound that I can not guess at (ie. I' am clutching at straws). Never-the-less, it is obviously not what pure aluminium hydroxide tastes like. It would really need a proper experiment to discover and explain the sensation of saltiness of this crud. So in other words, I don't think your having Gustational hallucinations.

--Aspro (talk) 18:26, 22 February 2012 (UTC)[reply]

I tried some balsamic vinegar, and it didn't seem to have any effect. I'll buy some distilled vinegar when I go to the store. StuRat (talk) 18:49, 22 February 2012 (UTC)[reply]

Vinegar is vinegar - apart from the price. The balsamic may have made the nodules taste better but if there was no reaction -that’s it. I just said distilled because that's what I mostly use for cleaning windows, de-calcifying kettles and other domestic chores that I can't get either of the wives to do. Buy it by the gallon and its really cheap - and we get eight pints to the gallon where I live!.. --Aspro (talk) 19:17, 22 February 2012 (UTC)[reply]

Wives ? I take it you're a Mormon, then ? :-) StuRat (talk) 19:21, 22 February 2012 (UTC) [reply]

No ! Just very gullible, and unable to change the locks faster than they can get new keys cut - so just given up and accepted it... --Aspro (talk) 19:53, 22 February 2012 (UTC)[reply]

I recall not being able to dissolve the exudate with vinegar, too. Can't be a hydroxide, probably not calcium carbonate. My intuition says an aluminum salt. Aluminum chloride? Also, many salts taste salty without being a chloride. Would be a nice 'unknown substance' problem for a school chemistry lab class. David Spector (user/talk) 01:08, 23 February 2012 (UTC)[reply]

OK, thanks everyone, I will mark this Q resolved. StuRat (talk) 19:22, 24 February 2012 (UTC)[reply]

Hi. Our tabby female cat recently gave birth to three kittens who have since been purchased and rehomed. A short time after our other cat became pregnant and this evening gave birth to five kittens. The other cat seems to be taking a motherly interest in this new batch of kittens. The new mother doesn't seem to mind but we're concerned that this will have an adverse effect on the connection between the mother and her litter? If so how do we prevent the other cat from exhibiting this maternal behaviour? --Hadseys (talk) 22:31, 20 February 2012 (UTC)[reply]

I imagine some type of hormone therapy could remove the "maternal" hormones from the first momcat, but, of course, no vet would do that. Keeping them separated would work, but really, I don't see the danger. It's far better to have two mothers than have one trying to kill them, after all. The only physical risk I see is if the first momcat has stopped producing milk, but prevents the kittens from getting milk from their mother. If this is the case, then yes, keep them locked in different rooms until the kittens are weaned. If you do have to separate them, I suggest some surrogate kitten for the kittenless one. Rolled up socks seem to work. StuRat (talk) 23:44, 20 February 2012 (UTC)[reply]

Don't worry about the other female kitten taking a maternal interest, from my experience with cats and kittens it's perfectly normal. In feral groups, female cats will raise the kittens between them. The time to worry is if an entire male cat takes interest in kittens as he will regard them as a threat and potential source of food. Oh and I'd get all your cats neutered as soon as they are old enough. The last thing the world needs is more cats (and I'm a cat lover). --TammyMoet (talk) 11:10, 21 February 2012 (UTC)[reply]

I think you mean "intact" male. Not entire. Dismas|^(talk) 11:13, 21 February 2012 (UTC)[reply]

I've seen both used. --TammyMoet (talk) 22:09, 21 February 2012 (UTC)[reply]

I've seen male cats take a keen interest in others' kittens. It's often perfectly harmless. Kittens like to play, and will naturally try to interest other cats in playing with them. Some cats (including adult males) are perfectly happy to oblige. Unless you sense aggression from the adult male, I wouldn't worry. And as to your situation, I highly doubt it will have any "adverse effect on the connection between the mother and her litter" in a detrimental way. Be thrilled that they're all on good terms. 58.111.178.170 (talk) 15:58, 21 February 2012 (UTC)[reply]

• Dont worry we will spay her, the other one's already been done. They just both escaped out of the house one time and got knocked up =/.The kittens though, if anyone's interested, are beautiful. Four with white fur with random black tortoiseshell markings scattered about it and a pure black one

Have you got pictures? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:24, 21 February 2012 (UTC)[reply]

is this an observable fact? 203.112.82.128 (talk) 23:15, 20 February 2012 (UTC)[reply]

Yes. You can observe the sun, moon and planets moving in the sky and they move in exactly the way the heliocentric model of the solar system predicts. --Tango (talk) 23:38, 20 February 2012 (UTC)[reply]

You can distinguish the modern heliocentric model from the Ptolemaic model by the phases of Venus, from the Tychonic model by the stellar parallax, and from Newtonian orbits by careful observation of the perihelion of Mercury, among many other careful observations that align with the modern understanding (I have picked out only a few historically significant ones). Observations do not prove theories, but they do disprove them, leaving only those theories which fit the data and remain falsifiable. --Mr.98 (talk) 23:47, 20 February 2012 (UTC)[reply]

And, of course, we've sent ships to all the planets and the Sun. I don't think they would hit their targets if our heliocentric model of the solar system was wrong. StuRat (talk) 23:49, 20 February 2012 (UTC)[reply]

I agree with the above conclusions, but I think the above answers are not quite accurate. The pre-Copernican Ptolemaic system was amply verified by observations and made pretty much the same predictions. The main problem with the ptolemic system was that it assumed circular orbits, which the copernican model also assumed to begin with. From the geocentric model article: The geocentric system was still held for many years afterwards, as at the time the Copernican system did not offer better predictions than the geocentric system, and it posed problems for both natural philosophy and scripture. The Copernican system was no more accurate than Ptolemy's system, because it still used circular orbits. This was not altered until Johannes Kepler postulated that they were elliptical (Kepler's first law of planetary motion). If Kepler's laws were applied to the ptolemic system, there is no reason it couldn't have been used to send space craft to the sun and planets. Fundamentally it's a question of relativity and perspective. Also, strictly speaking, Heliocentrism is the theory that the sun is at the center of the UNIVERSE, which we obviously no longer believe to be the case, but locally to our solar system, it holds true. (EDIT: I posted this before I saw Mr.98's reply which I don't have any issues with)Vespine (talk) 00:01, 21 February 2012 (UTC)[reply]

Would going with elliptical orbits improve the geocentric model beyond what epicycles did ? For example, could they explain the apparent retrograde motion of Mars ? Or do you mean a geocentric model with both elliptical orbits and epicycles ? Also, I don't think they would be accurate enough for space shots, unless you are assuming local corrections to navigation. StuRat (talk) 00:09, 21 February 2012 (UTC)[reply]

I'll admit I have no idea:) I could be wrong, but you COULD make an accurate working model from geocentrism, even if you had to add "planetary constants" for each orbit or whatever, it might make it exceedingly complex, but the point is that it doesn't necessarily make it false and conversely doesn't necessarily make heliocentrism a fact. I'm probably reading WAY too much into probably just a casual question. If the OP is interested I think this is a question about Philosophy of science, Scientific realism and even Model dependent realism is an interesting article. Vespine (talk) 00:15, 21 February 2012 (UTC)[reply]

OP here, i think we could still send ships to the sun if we have a good geocentric model, what im interested in is do we establish heliocentrism as fact analogous to the fact that birds can fly or a basketball is sphere. 203.112.82.128 (talk) 00:45, 21 February 2012 (UTC)[reply]

In which case, as I suspected, this is a philosophical question the answer of which depends on which stand point you accept. The articles I linked above are a good starting point. I'm only a novice philosopher but for example if you subscribe to Scientific realism I think you would argue that heliocentrism can be a fact, if however you subscribe to Model dependent realism then, if I understand correctly, you don't really believe in facts, just the usefulness of the models we adopt. Vespine (talk) 01:18, 21 February 2012 (UTC)[reply]

Maybe what im really interested in is if we have a probe that can actually see the planets move around the sun or something like that.203.112.82.128 (talk) 01:33, 21 February 2012 (UTC)[reply]

We can see the planets move around the Sun from Earth! For example, almost every year I watch Mars undergo apparent retrograde motion. This is something I can see as plainly as snow in winter. This particular motion makes more sense if you realize that Mars is exterior to Earth and orbits the Sun, consistent with an orbit dominated by Newtonian gravity. I watch Jupiter and Venus orbit the Sun, and I watch Moon orbit the Earth. You don't have to look very hard to see these things: they're big, bright, and move slowly. Nimur (talk) 01:43, 21 February 2012 (UTC)[reply]

So that means nothing can disprove heliocentrism? 203.112.82.129 (talk) 01:58, 21 February 2012 (UTC)[reply]

It's not that nothing can disprove it, but that it's consistent with all observations, which the previous models were not. And with all the observational data we now have, it would take a truly elaborate and complicated geocentric model to also match those observations. The modern heliocentric model also benefits from the fact that it can be constructed directly from universal laws of physics, which have themselves been extensively demonstrated to agree with observations unrelated to planetary motion. You reach a point where asking if something is a fact just becomes meaningless. See brain in a vat, for example. It's trivial to construct a theory centered on that concept that is consistent with all human observations. If you're asking it's safe to consider heliocentrism to be a fact, the answer is yes. If you're asking what it means to be a fact, you have wandered into the realm of philosophy. Someguy1221 (talk) 02:20, 21 February 2012 (UTC)[reply]

Philosophically, we could still have a geocentric system - just take all of our standard equations, convert to a coordinate system centered on the Earth's surface (or to be precise, some specific part of it), and in that coordinate system they still revolve around the Earth. Either way you just work the numbers; what they mean is merely "an interpretation", like the many-worlds interpretation or the Copenhagen interpretation of quantum mechanics. That does seem kind of absurd though, doesn't it? Wnt (talk) 02:49, 21 February 2012 (UTC)[reply]

Of course, to accept a geocentric system, we would need some mechanism by which the far more massive Sun would orbit the Earth. Does the Earth have hidden mass, like a black hole inside a hollow sphere ? If so, why isn't the force of gravity more for us ? Does the Sun have much less mass than we think ? If so, how can it support nuclear fission ? We would have to toss out much of physics to come up with a universe where geocentrism is still possible. StuRat (talk) 03:20, 21 February 2012 (UTC)[reply]

There are plenty of observations that could have disproven it, but in fact, they disproved other alternatives. --Mr.98 (talk) 13:15, 21 February 2012 (UTC)[reply]

Also note that heliocentrism isn't 100% right. The Sun does wobble a bit as it is also affected by the gravitational attraction of the planets. In a two-body system it's more correct to say that both objects orbit about the barycenter of the system. With three or more bodies, it gets even more complex. StuRat (talk) 03:23, 21 February 2012 (UTC)[reply]

Is Heliocentrism an observable fact? No! The center of the universe is not a well defined place. To Aristotle the center of the world was the point towards heavy objects are attracted, which is the center of the Earth. To Aristarchos of Samos and Copernicus the center was the Sun. To Kepler the Sun is not in the center of the elliptical planetary orbit, but rather at a focal point. To Newton any point may serve as the center, and there is not even an absolute zero velocity - any constant velocity displaced coordinate system will do - but there is an absolute zero point for acceleration, which is the acceleration of a particle that is subject to no force. To Einstein even the absolute zero acceleration is undefined, as nonzero acceleration may be identified with a gravitation. The Cosmic microwave background radiation#CMBR dipole anisotropy seems to indicate a zero point of velocity, but no geometrical center of the universe can be identified. Bo Jacoby (talk) 12:51, 21 February 2012 (UTC).[reply]

It all depends what one really means by "heliocentrism." --Mr.98 (talk) 13:15, 21 February 2012 (UTC)[reply]

One thing here: the OP is referring to "facts" like "birds can fly" or "a basketball is a sphere" as the gold standard of truth. But this is misleading. There's a preference for visual epistemology here which is very problematic: "seeing" something is not necessarily better or more iron-clad evidence than many other forms of observation. Secondarily, as with those examples, life gets complicated. Basketballs are not perfect spheres; not all birds can fly; and the definition of "bird" and "flight" can vary quite a bit once one starts hashing out very precise definitions (is gliding flying? if we put an ostrich on an airplane, can it fly?). Establishing even basic "facts" from apparently raw and unadulterated sense data is more tricky than it seems — aside from the fact that your sense data may be flawed or unreliable (you could be crazy, or dreaming, or inhibited in some way), you also have to make the initial selection of what sense data to retrieve in the first place, which shapes your entire outlook (relying solely on a visual spectrum of light will cause you to miss quite a lot of observable phenomena). If you want to start parsing out interesting and complicated epistemological questions about how we know anything and what is a fact, there is a rich history of these issues in the philosophy of science that one might explore. --Mr.98 (talk) 13:15, 21 February 2012 (UTC)[reply]

... and remember that, in the words of the Galaxy Song, "The sun and you and me and all the stars that we can see, Are moving at a million miles a day" around the centre of our galaxy. Actually, according to orders of magnitude (speed), it is more like 10 million miles a day. But the point is that is several times greater than the Earth's orbital speed relative to the Sun. Gandalf61 (talk) 15:25, 21 February 2012 (UTC)[reply]

And the Milky Way is moving away from wherever the Big Bang started, right? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:35, 22 February 2012 (UTC)[reply]

Actually Big Bang did not happen at any single point, it happened everywhere. Space is expanding which results in growing distance among galaxies. - manya (talk) 03:55, 22 February 2012 (UTC)[reply]

It is histories and philosophies of science all the way down. Fifelfoo (talk) 04:06, 22 February 2012 (UTC)[reply]