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November 29

How many SigFigs?

When performing the following simple addition:

    -260
   + 273.15

How many sigfigs should the answer, 13.15, carry? Intuitively, it should be 13. However, when this is inputted into the Science Tools Significant Figures calculator on a TI-84+, the answer says 10 (1 sigfig)

Which is correct? Thanks. Acceptable (talk) 00:09, 29 November 2007 (UTC)[reply]

Your calculator. The 260 has the tens place as its lowest significant figure, so the final answer should only be specified to the tens place. Someguy1221 (talk) 00:28, 29 November 2007 (UTC)[reply]

That's BOGUS. Sure, if the 'true' value is 261 and it was written as 260 then you have 2 digits of precision - but you don't know that. Perhaps the 'true' value was really 260.00000000 - it's written as 260 which is THREE significant digits. The fact is that with normal arithmetic symbolism, you simply can't tell whether it's 2 digits or 3. So you must calculate out to three digits in order to be sure you aren't losing information. It's not a "real" problem though - in any real situation, you know where the data came from and therefore how precise it is - and you know how much precision you need in the result. SteveBaker (talk) 01:16, 29 November 2007 (UTC)[reply]

Um, steve, that's what the period is for, dude. I'll agree with you though that this doesn't generally apply to "real" problems. Someguy1221 (talk) 01:22, 29 November 2007 (UTC)[reply]

OK - so you weaseled out of that one (damn!) - but I don't give up debunking bogosity that easily! How about this sum?

 -2600
+ 1234.45

Does 2600 have 2 or 3 significant digits? What about 26,000,000 ? You have no way to know - you are left guessing. If I tell you that the population of France is 26,000,000 people - you might guess that I gave you only two significant digits - but you don't know that there are really 26,012,345 people living there and I gave you 3 sig digits. In truth, if you aren't specifically told - you don't know. Teaching people that there are rules that tell you this stuff is BOGUS. SteveBaker (talk) 03:19, 29 November 2007 (UTC)[reply]

If it's possible for it to be ambiguous, you use scientific notation. :-p neener neener neeeeenerrrrrr. Anyway, I'm happy to say that I use plain old means and standard deviations when I do analytical experiments :-) Someguy1221 (talk) 03:23, 29 November 2007 (UTC)[reply]

Excuse me!?! I've never seen the population of France represented in scientific notation - NEVER! That's an utterly useless rule. SteveBaker (talk) 03:35, 29 November 2007 (UTC)[reply]

Population of France: 2.60×10⁷. Now you have!

Dragons flight (talk) 04:51, 29 November 2007 (UTC)[reply]

Oh no - now you've gone and done it. Here, let me fix that number for you: 02.60x10.00^07.00 ...there, much better! SteveBaker (talk) 06:22, 29 November 2007 (UTC)[reply]

In the scientific-notation version, the "10" and "7" are considered exact numbers, giving them an infinite number of significant digits. --Carnildo (talk) 01:16, 30 November 2007 (UTC)[reply]

So the '260' in the example up top had infinite precision according to you? You'll probably say "No" because you happen to know that 10 and 7 were integers and you happen to believe that the 260 is not. My point being that you have to somehow know. If you aren't told then you're guessing. Not good. SteveBaker 23:54, 1 December 2007 (UTC)[reply]

By the way...how many significant digits does '2' have? You're going to say "just one". So if I want to add 2 to 101. - the rule is that I have to do it to one significant digit so the answer is only 100 - unless I write 2.00 + 101. then I get 103. ?? That's something else I've NEVER seen anyone do? What's relevent in addition is the number of digits before/after the decimal point - not the number of significant digits. This supposed formalism is utterly useless in practice - and it's definitely going to cause horrible mistakes. Why are we even bothering to teach it to people? SteveBaker (talk) 03:44, 29 November 2007 (UTC)[reply]

For an amusing real-world conundrum involving trailing-zeros ambiguity, see the "Measurement" section of our article on Mt. Everest. —Steve Summit (talk) 03:53, 29 November 2007 (UTC)[reply]

Oh sorry, that'd my mistake, I guess I forgot to mention that. (-260) has 2 sigfigs, while (273.15) has 5. Acceptable (talk) 01:19, 29 November 2007 (UTC)[reply]

New question: how do I inform the calculator that this particuloar 260 accurate to three significat figures, if "260" is two significatn figures, and "260.0" is four significant figures? -Arch dude (talk) 01:05, 29 November 2007 (UTC)[reply]

And 260. has three sig figs ;-) Someguy1221 (talk) 01:06, 29 November 2007 (UTC)[reply]

AFTER EC: Oh, I thought when adding/subtracting sigfigs, the answer should have the decimal places of the value with the least decimal places. Acceptable (talk) 01:07, 29 November 2007 (UTC)[reply]

No it doesn't. It's one less. If you are really adding 100.6 to 100.6 but we do the sum accurate to three significant digits, you get 101. + 101. = 202. - but the real answer is 201.2 - so your answer isn't accurate to 3 significant digits anymore - it's now only accurate to 2. When you add two inaccurate things - the result is less accurate than the two numbers you added. Real math isn't done with 'significant digits' - it's done with error bars. So the right way to do this is to say 100.6 can be represented as 101 plus or minus 0.4, so when you add 101 (+/-0.4) to 101 (+/-0.4) you get 202 (+/-0.8) - which means that you know the true answer is between 202.8 and 201.2...which is good. This whole business of "significant digits" is just a VERY rough way to represent error bars and to make sure people don't over-specify the precision of the result. But to pretend that this is some kind of mathematical formalism is nonsense (but totally typical of the nonsense they fill school-kid's heads with when they could be teaching them REAL math). It's main purpose in practical math, science and engineering is to allow you to do 'back of envelope' calculations and not be ridiculously over-precise in your results. When it matters, you use proper error bars with all the formalism that entails. SteveBaker (talk) 03:33, 29 November 2007 (UTC)[reply]

If you do as scientists often do and treat your errors are normally distributed, then the standard error propagation tells you that 101 +/- 0.4 + 101 +/- 0.4 = 202 +/- (0.4*sqrt(2)) = 202 +/- 0.6. Though, for that matter, I can't imagine anyone looking at 100.6 and rushing to 101 +/- 0.4. The default position would be to ask the uncertainty, but if that is not available the customary assumption (at least back in my lab classes) is to assume the last digit is meaningful but on the threshold of not being so and go with 100.6 +/- 0.5, which would give the sum as 201.2 +/- 0.7. Dragons flight (talk) 04:48, 29 November 2007 (UTC)[reply]

Indeed, for many sorts of error, a normal distribution is an appropriate error model - and I agree that if you use that then no answer is "wrong" because it could simply be off in the long tail of the distribution. Hence if you calculate the answer to my little sum to be 202 +/- 0.6, you aren't saying that the true answer must lie between 201.4 and 202.6 (as indeed, it does not) - you are merely saying that the odds are much better that the true answer lies within that range than outside of it. However, it's not always the case that a normal distribution is appropriate. For example if you are using a computer to crunch your numbers, the error due to finite precision machine arithmetic is sharply delimited and equiprobable - it's certainly not gaussian.

But whatever - the point is that if you actually care about the precision of your results, you do something about it. The whole "significant digits" thing is vague, poorly determined and patchily implemented. It's simply a rule of thumb to prevent gross over-specification of precision. Once people start getting dogmatic about how it's treated, the system simply falls apart because of those weaknesses.

The computer geek in me is particularly upset because the rate at which precision is preserved, amplified or casually discarded by this so-called system depends entirely on what base of arithmetic you use! If you do your math in binary - then knowing how many significant binary-digits the number has preserves precision better than knowing how many significant decimal digits it has - and MUCH better than doing it in hexadecimal or radix 50 or counting the number of bytes (radix 256). Since the choice of a base-10 number scheme is ENTIRELY arbitrary, anything which bandies about "significant digits" is doomed to failure in the face of proper numerical analysis. SteveBaker (talk) 06:40, 29 November 2007 (UTC)[reply]

Significance arithmetic summarises quite a lot of what has been said here. 130.88.79.39 (talk) 14:30, 29 November 2007 (UTC)[reply]

One of the references in that article [1] says: The whole notion of significant digits is heavily flawed; see section 9 for more on this. Anything that can be done by means of significant digits can be done much better and more easily by other means. People who care about their data don’t use significant digits. There are plenty of important cases where following the usual "significant figures" rules would introduce large errors into the calculations. ...which is precisely my point (although I'll admit to VERY informally using the significant figures "rule" when I'm doing strictly back-of-envelope math. SteveBaker (talk) 20:31, 29 November 2007 (UTC)[reply]

This addition problem doesn't really involve the number of significant digits anyway. That's a multiplication/division factor. What is inolved here is that we don't know that the digits of one of the numbers is significant beyond the tens place (it might be significant in the units place, but not beyond there or it should have included significant digits after the decimal point. Gene Nygaard (talk) 22:02, 29 November 2007 (UTC)[reply]

Particle Accelerators use electric fields to accelerate and magnetic fields to bend particles?

Hi, in reading about the Large_Hadron_Collider and the article on Magnetism, I'm wondering some simple questions, like... 1) If a Proton at rest sees an electric field, the proton accelerates towards the negative voltage. As the proton attains the speed of Light (c), the electric field starts looking like a magnetic field, right? 2) The Large_Hadron_Collider has a round track 26.6 km long that bends the moving protons using magnetic fields. As the proton attains the speed of light (c), doesn't the proton see that magnetic field more and more as an electric field?

Just curious. Thanks! --InverseSubstance (talk) 02:43, 29 November 2007 (UTC)[reply]

(1) not quite, and (2) no. To be more specific, it isn't that an electric field "looks like" a magnetic field - put simply, when you have some charged particles and you are in a frame of reference where they are at rest, they have an electric field and no magnetic field, that will exert a certain force on another charged particle. When you are in a different (intertial) frame of reference, those particles will be moving, but by the rules of special relativity the force they exert on that other particle should be the same, even though length dilation means that the separation between them appears to have changed, and thus their electric field is different. The magnetic field is essentially the difference between the two possible electric fields. So yes, if you start with a "pure electric field", then as you accelerate you'll see some of the electric field "become" a magnetic field. However, if you start with a magnetic field, then as you accelerate the magnetic field will change, as will the associated electric field, but the one won't completely turn into the other. Confusing Manifestation(Say hi!) 02:53, 29 November 2007 (UTC)[reply]

One thing - the proton doesn't attain the speed of light - it can get close but it can never quite get there. SteveBaker (talk) 03:09, 29 November 2007 (UTC)[reply]

Extended/Extra Optical Zoom?

I bought Panasonic Lumix DMC-FZ18 recently, I have heard that its got "Extended Optical Zoom" but it works with lower resolution output only. http://panasonic.co.jp/pavc/global/lumix/fz18/18zoom.html . Does anyone know whats the difference between Extended/Extra Optical Zoom and Digital Zoom?

Thanks --Spundun (talk) 03:59, 29 November 2007 (UTC)[reply]

It's just precropped digital zoom. Uh, so basically, it's digital zoom :-) Someguy1221 (talk) 04:09, 29 November 2007 (UTC)[reply]

EOZ isn't truly digital zoom - it doesn't "stretch" pixels like digital zoom does. It is only used when you have chosen to take smaller pictures than your camera can take - for example, to save space on your memory card. Normally in this case the camera "squashes" pixels. However, if you wish to zoom in, the camera will use only pixels from the middle of your sensor - no "stretching" required. 138.38.151.57 (talk) 12:32, 29 November 2007 (UTC)[reply]

So in other words, it takes a full picture and crops it - that's "electronic zoom" - calling it "optical" is a nasty deception! Electronic zoom is by far inferior to optical zoom - but it's cheaper. If this were an optical zoom, the optics would have reconfigured to cover the entire sensor - then if a lower resolution were demanded by the user, "squashing" (as you put it) could be performed. However, by merging together a larger number of sensor pixels to get one stored pixel, you improve the quality of the image. SteveBaker (talk) 13:53, 29 November 2007 (UTC)[reply]

While I agree with you that "digital zoom" is vastly inferior (personally, in fact, I'd go farther and say that it's no zoom at all), based on the evidence at hand, I don't think we can quite prove malfeasance in this case.

Consider my camera (a 3.2 megapixel Canon PowerShot A510). I can shoot in four different resolutions:

L	2048x1536	(3.14MP)
M1	1600x1200	(1.92MP)
M2	1024x768	(0.78MP)
S	640x480	(0.30MP)

Normally I shoot in M1 mode, because my pictures aren't that great and mostly I use them on-line where anything bigger is just overkill. (And it doesn't even bother me that I've effectively castrated my already-puny 3.2 MP camera down to 1.9, because it's no contest for me.)

Obviously, the camera's internal CCD is 2048x1536, and when you shoot in one of the lower-resolution modes, the camera downsamples by a factor of 1.28, 2, or 3.2. Thus, the M1 images (for example) require 60% of the storage of an L image.

Now, suppose there was a way to divide the raw CCD image up into thirds, both horizontally and vertically. Suppose the camera downsampled the outer 8 of the resulting 9 subimages as usual, but left the center one at full CCD resolution, and stored this composite somehow. For M1, this would require 8/9 x 1.92MP + 1/9 x 3.14MP = 2.06 MP, or 65% of full-resolution storage. So for a 5% increase in storage (above normal M1, that is), you'd retain the ability to zoom back in to full CCD resolution, as long as you limited yourself to the image's fovea, so to speak.

I doubt you could store this "composite" image using a standard image file format, although I suppose you could arrange to use a higher JPEG quality factor on the center of the image than the periphery, and gain something. (Or perhaps you could store a 2048x1536 image as a 2048x1536 JPEG, with normal JPEG quality in the center and a much lower JPEG quality at the periphery, and then make a note to always display the image shrunk by 78%, so that the coarseness around the edges wouldn't show, but so that -- again -- you could digitally zoom in on the center without loss of sharpness.)

I have no idea if this is actually what the DMC-FZ18 is doing, but it's not completely outside the bounds of possibility. —Steve Summit (talk) 01:30, 30 November 2007 (UTC)[reply]

Yes, EOZ still isn't as good a thing as true optical zoom. It's still much better than standard digital zoom - it lets you zoom in without upsampling. 138.38.149.208 (talk) 08:56, 30 November 2007 (UTC)[reply]

Time to fall

If a point mass of mass m is released at some distance d above the surface of a perfect sphere of uniform density, with mass M and radius r, then, using Newtonian universal gravitation and assuming there are no other forces acting within the system and ignoring the effect of the point mass' gravity on the sphere, how long does it take the point mass to fall to the surface of the sphere? This question occurred to me a while back and I have been trying to figure it out, but I have no idea how it would be calculated, since the distance travelled at any time depends on the velocity/acceleration, but the acceleration is constantly changing and depends on the distance from the centre of the sphere. --superioridad ^(discusión) 04:57, 29 November 2007 (UTC)[reply]

The answer is calculus (mostly integral) - consider acceleration, velocity and radius/position all as functions of time, use calculus to find an expression for the position, and solve to find the two time points. As for an actual answer ... give me a second (or perhaps someone else will take care of it). Confusing Manifestation(Say hi!) 05:55, 29 November 2007 (UTC)[reply]

We do have an article on this. It's not the most apparent method of approaching this, but it starts with more general conditions. Someguy1221 (talk) 06:07, 29 November 2007 (UTC)[reply]

And I suppose you could use that solution, then transfer to a co-ordinate system where the second mass is stationary, if you wanted the "Earth/bigger object doesn't move" solution. Confusing Manifestation(Say hi!) 06:12, 29 November 2007 (UTC)[reply]

(e/c)... OK, time for some TeX. Acceleration is given by Newton as

$a=-{\frac {GM}{r^{2}}}$

But $a={\frac {dv}{dt}}$ , and using a little fiddling we get

$-{\frac {GM}{r^{2}}}={\frac {dv}{dt}}={\frac {dv}{dr}}{\frac {dr}{dt}}=v{\frac {dv}{dr}}$

And if we integrate with respect to r, we have

${\begin{aligned}\int \limits _{r_{1}}^{r_{2}}-{\frac {GM}{r^{2}}}dr&=\int \limits _{r_{1}}^{r_{2}}v{\frac {dv}{dr}}dr\\\left[{\frac {GM}{r}}\right]_{r_{1}}^{r_{2}}&=\int \limits _{v_{1}}^{v_{2}}v\ dv\\{\frac {GM}{r_{1}}}-{\frac {GM}{r_{2}}}&=\left[{\frac {1}{2}}v^{2}\right]_{v_{1}}^{v_{2}}\\&={\frac {1}{2}}{v_{1}}^{2}-{\frac {1}{2}}{v_{2}}^{2}\end{aligned}}$

... and then you use the fact that $v={\frac {dr}{dt}}$ and do some more calculus, and plug in your known values, and get some kind of answer. I have to go now, but I'm sure you or someone else can happily work away at this step. Confusing Manifestation(Say hi!) 06:12, 29 November 2007 (UTC)[reply]

${\sqrt {d^{3} \over {2GM}}}$ - Dragons flight (talk) 06:28, 29 November 2007 (UTC)[reply]

Drunken vampire bats

If a vampire bat sucks the blood of a drunk person, does the bat become drunk too? --Candy-Panda (talk) 09:55, 29 November 2007 (UTC)[reply]

I don't believe it would because of the way it can quickly excrete it.--58.111.143.164 (talk) 12:50, 29 November 2007 (UTC)[reply]

Also consider how the body dilutes alcohol that's ingested. You can drink 95% pure alcohol, which your body will dilute down -- half a percent of alcohol in your blood has a 50% chance of killing you. So if the bat is drinking blood, chances are good that it's drinking at most a 0.5% alcoholic drink, which in the U.S. is the upper limit to be considered a non-alcoholic drink. So, given a lot of other unfounded assumptions about a bat's metabolism and response to alcohol, the answer is: the bat becomes about as drunk as you could get on O'Douls. jeﬀjon (talk) 14:34, 29 November 2007 (UTC)[reply]

Using carbon nanotubes to transport electrons?

I read on the article on CNT's that certain CNT structures are excellent conductors of electricity, while others are semi conductors. Could CNT's be used instead of metal wires to transport electricity? Are there any advantages to this over what we currently use? 64.236.121.129 (talk) 14:43, 29 November 2007 (UTC)[reply]

You might enjoy our article about field emission displays for one such use of carbon nanotubes. I think a lot of the possible applications haven't yet been explored because of the relative shortage of long nanotubes.

Atlant (talk) 16:42, 29 November 2007 (UTC)[reply]

CNT's could certainly do that (and they are also very strong - which would be handy). However, the longest ones we can make are a few millimeters long - and the longest ones we can make in industrial quantities are utterly microscopic in length. So the technology just isn't there yet. However, there is huge interest in making long buckytubes - they have HUGE implications for materials science and a lot of very smart people are trying very hard to crack the problem of making them. I think our kids are going to have a lot of fun with these things. SteveBaker (talk) 20:16, 29 November 2007 (UTC)[reply]

If you fill an airtight car with helium, will it become lighter?

^Topic 64.236.121.129 (talk) 14:52, 29 November 2007 (UTC)[reply]

No, you can only pump Helium in if the air can be displaced. With the car airtight, there is no place for the air to go, so no more room inside for the helium. - 131.211.161.119 (talk) 14:54, 29 November 2007 (UTC)[reply]

That's not what I meant. I meant the car is already filled with helium, and is sealed airtight to prevent the gas from escaping. 64.236.121.129 (talk) 15:05, 29 November 2007 (UTC)[reply]

A car filled with helium would weigh less than a car filled with air, all else being equal. From a practical point of view, helium is a very small molecule and can often escape from containers that are "air-tight" (for example, certain types of balloons). ike9898 (talk) 15:00, 29 November 2007 (UTC)[reply]

I see. If this is true, why not fill compartments on airplanes (seperate from the cabin which contains people) with helium to make the plane lighter. 64.236.121.129 (talk) 15:05, 29 November 2007 (UTC)[reply]

As stated - the helium will escape. Also, to get any substantial effect, you would need to create a helium cavity the size of a zeppelin - which would greatly inhibit the aircraft's ability to fly. All in all - the benefit is not worth the effort. You'd get a much better benefit by banning all luggage from aircraft. -- kainaw ™ 15:08, 29 November 2007 (UTC)[reply]

Ok lets say the airplane is a military fighter. We have a couple of lightweight tanks filled with helium. Once the helium escapes, it can drop the tanks. Wouldn't this help? 64.236.121.129 (talk) 15:11, 29 November 2007 (UTC)[reply]

To get any significant benefit you would very large helium tanks with very lightweight walls. Maybe some sort of fabric or plastic walls. Then you want to make sure your centre of lift is above your centre of gravity, for stability - so put your crew cabin underneath your big helium tanks. No need for a fuselage or those heavy wings - just stick a rudder and elevators onto your big helium tanks. What you now have is called a dirigible. Gandalf61 (talk) 15:25, 29 November 2007 (UTC)[reply]
No. It's very unlikely that you can build tanks that can withstand military-style maneuvers and still have positive buoyancy even if you assume they contain a vacuum. Also, you would still increase mass and, presumably, air resistance. The lift you can get from the surrounding air is fairly low - about 1.2 kg/m³ at sea level. --Stephan Schulz (talk) 15:34, 29 November 2007 (UTC)[reply]

Ahh! Is it time for our next round regarding "vacuum balloons"? If we went with the version inflated by photons, perhaps they could perform double-duty by both providing buoyancy and, later, allowing them to be dropped on the enemy as a sort of photon torpedo? This would be especially effective if they were inflated with some nasty photons such as gamma rays.

Atlant (talk) 16:45, 29 November 2007 (UTC)[reply]

Hmmm, here is an idea: A balloon made from a very thin film with a very high static charge. If you can charge it enough, it should keep the balloon expanded even against air pressure. Of course it needs a bit of material science handwaving ;-). --Stephan Schulz (talk) 12:46, 30 November 2007 (UTC)[reply]

On the topic of vacuum balloons, I saw one when I was watching an episode of Pushing Daisies last night on television. The protagonist had a flashback to a childhood boarding school's science lab, and he recalled an incident with a friend who had been an expert at constructing paper airplanes. One of the characters filled a balloon from a benchtop gas hose barb; once inflated and tied off, the balloon floated away. (Appropriately enough for this thread, I believe the balloon was actually used to provide lift to a paper airplane.) There was a very quick cut that showed the gas valve as it was being turned on, it bore the legend VAC. I thought that it was a very clever touch. TenOfAllTrades(talk) 17:47, 29 November 2007 (UTC)[reply]

Imagine a helium balloon. Think of how little a helium balloon can lift. Those little plastic tags they put on the end of the string are usually sufficient to keep the balloon from flying away. Even if your helium 'tanks' are made of the same lightweight material as the balloon you got at the fair the 'tanks' would only make the plane lighter by the weight of that little plastic tag. Not very useful. The problem is that air is already so incredibly light that replacing it with something even lighter does not achieve much. That is why even small, four passenger blimps have gas-bags the size of an office building. 72.10.110.107 (talk) 15:50, 29 November 2007 (UTC)[reply]

Also - aircraft flies at an altitude where there is almost no buoyancy offered by helium. -- kainaw ™ 16:07, 29 November 2007 (UTC)[reply]

Not to mention that all your passengers would suffocate. shoy (^words _words) 17:24, 29 November 2007 (UTC)[reply]

No they wouldn't. In my first question, I didn't specify people were in the car, and in my second question, the helium would be in either compartments seperate from the cabin, or in tanks. But I agree with the other users that they prolly wouldn't contribute much lift to the airplane. 64.236.121.129 (talk) 17:42, 29 November 2007 (UTC)[reply]

This is definitely a non-starter. The trouble with the aircraft idea is that very little volume can be enclosed within the plane and the additional weight required to get it airtight (seals and whatnot) would easily outweigh the benefits. One liter of helium can only lift just over a gram. Take an F16 fighter - it's 14m long and the fuselage is about 3x3 m - the wings are 27 sq.m in area and about 20cm thick. So we have a total volume of about 130 cubic meters - which is 130,000 liters. If the ENTIRE plane was filled with nothing but helium (ie we take out the engine, fuel tanks, instruments, landing gear, internal bracing...the lot), it would be lighter by 130,000 grams - 130kg. Fully fuelled, the plane weighs 12,000kg. So at VERY best, you'd get about a 1% weight saving. But in truth there is very little free space inside an F16 - perhaps a hundedth of it might be utilised giving us an 0.01% weight reduction overall. The hassle involved is simply not worth it. Worse still - if that gas is pressurised to sea level then as you take off and the air gets less and less dense, the helium produces less and less lift. In a zepplin or a blimp, the lift bags expand to take account of that - but you can't do that in these confined little spaces. So in the end, this is certainly a lot more trouble than it's worth - and in all likelyhood would actually make the plane heavier - not lighter. A similar argument applies to a car. My MINI Cooper'S has 1.47 cubic meters of interior volume, that's 1470 liters - so 1.47kg of lift from filling it with helium. Sadly, it weighs 1200kg - so now we are looking at 0.1% reduction in weight. But its hard to imagine you could seal up all of the little holes and escape routes with less than 1.47kg of material - so in practice, you wouldn't save anything. SteveBaker (talk) 20:07, 29 November 2007 (UTC)[reply]

Indeed hmm. Is it possible to just shove more gas in there by force, and increase the density of helium in the tanks. Just hypothetically, assume the tanks are invincible, and won't explode. 64.236.121.129 (talk) 20:17, 29 November 2007 (UTC)[reply]

Helium atoms still have positive mass (just lighter than the molecules that make up air). If you shove more helium into the tank, increasing its density, you make it heavier, not lighter. -- Coneslayer (talk) 20:21, 29 November 2007 (UTC)[reply]

Going by what Alant said, if you had tanks that were filled with nothing, there's a vacuum in them, they would provide more lift than helium? 64.236.121.129 (talk) 20:27, 29 November 2007 (UTC)[reply]

Yes, vacuum (nothing) is lighter than helium (something). But it requires a much stronger, and hence more massive, container. -- Coneslayer (talk) 20:36, 29 November 2007 (UTC)[reply]

Yep - vacuum balloons sound like a great idea - but air pressure is 15psi - that's fifteen pounds pushing in on every square inch of the surface. It takes a pretty heavy construction to resist that much force - and that's going to be heavier than a fairly small amount of helium. SteveBaker (talk) 00:25, 30 November 2007 (UTC)[reply]

That's why we have to inflate them with an enormous flux of massless photons! (In case anyone's wondering, yes, I'm joking, at least until perfectly-reflective unobtainium becomes available.)

Atlant (talk) 12:53, 30 November 2007 (UTC)[reply]

Steve said 1 liter of helium can lift 1 gram. How much can 1 liter of vacuum lift? 64.236.121.129 13:56, 30 November 2007 (UTC)[reply]

Based on the densities listed in Lighter than air, one liter of helium has a buoyancy of 1.11 grams, minus the weight of its container, if immersed in sea-level air. A liter of vacuum has a buoyancy of 1.29 grams, minus the weight of its (by necessity much heavier) container. The difference is very small, since helium is already very much lighter than air. --mglg_(talk) 17:22, 30 November 2007 (UTC)[reply]

I see. But the buoyancy will increase if the air density is higher right? So if we were on Venus, it would produce more lift? 64.236.121.129 18:55, 30 November 2007 (UTC)[reply]

Yes. Closer to home than Venus, see Density altitude.

Atlant 22:27, 30 November 2007 (UTC)[reply]

I just wanted to chyme in that there are certain plastics that contain helium well and the helium won't leak for a long time. We've all seen those silver foil baloons. They are normally made of mylar and they hold helium for about 2 weeks. Ultimately you would have to get a specification on your meterial, to find out the oxygen and helium permeability rating. Rfwoolf (talk) 13:06, 5 December 2007 (UTC)[reply]

ionic,covalent,coordinate bonds

Which one of the following compounds contains all the three ionic, covalent and coordinate compounds?? A-hydrogen cyanide B-ammonium nitrate C-potassium permanganate D-sulphuric acid —Preceding unsigned comment added by Knowiteverything (talk • contribs) 15:54, 29 November 2007 (UTC)[reply]

This a really easy homework question!--Stone (talk) 16:53, 29 November 2007 (UTC)[reply]

One of three relevant wiki articles gives the right answer in a nice picture!!!!--Stone (talk) 16:57, 29 November 2007 (UTC)[reply]

ornamental millet

Yesterday I saw a plant at a Nursery named as an ornamental millet. It was a feathery gras like plant with a deep puple head. The staff was unable to give me any information on it and I can't find any reference to ornamental millets in Western Garden Book or on line. Can you help? —Preceding unsigned comment added by 69.3.233.29 (talk) 17:54, 29 November 2007 (UTC)[reply]

Tall, dark and handsome describes this purple-leaved millet. Young plants are green-leaved, direct sunlight induces the purple leaf color. Capable of growing 3 to 5 feet tall, the plants are embellished with 8 to 12 inch flower spikes. The immature spikes can be cut and used dramatically in floral arrangements. Left on the plant, the millet seed spike attracts birds that snack on seed. 'Purple Majesty' is very easy to grow and is very tolerant of heat and low moisture. The purple leaf blades and spike are distinctly different from all other ornamentals.

Purple Majesty is best started outdoors when temperatures are consistently above 60 degrees or indoors in warmed seed trays. Chilly weather will stop plant growth or weaken its stems, so don't start your seeds too far in advance. Germination is very quick -- just 3 days! -- and plant growth is rapid and vigorous under good conditions. If you begin seeds indoors, the plants will remain green until set outside, then burnish a lovely violet within several days! You might like to buy some from here, [2]Richard Avery (talk) 18:15, 29 November 2007 (UTC)[reply]

Largest marine animal in captivity

Quick question: what is the largest marine animal ever kept in a zoo or aquarium? 68.23.172.215 (talk) 18:36, 29 November 2007 (UTC)[reply]

I think Japan has an aquarium with a whale shark in it. Whale sharks are larger than orcas which should come in 2nd. 64.236.121.129 (talk) 19:45, 29 November 2007 (UTC)[reply]

The Georgia Aquarium also has whale sharks. --Sean 23:45, 29 November 2007 (UTC)[reply]

What an excellent question. I would be one of the first in line if there was something exciting like a giant squid (extremely rare) or a blue whale. Truth is you'd need an extremely large area to hold these behemoths. Rfwoolf (talk) 13:08, 5 December 2007 (UTC)[reply]

How big would your arms have to be in order to fly/glide?

Lets assume you weigh 140 pounds, and you want to glide by stretching out your arms like a plane and running forward off of a cliff. If your arm width is normal, how long would they have to be in order for the air moving around it to provide enough lift to glide? 64.236.121.129 (talk) 20:24, 29 November 2007 (UTC)[reply]

I doubt you'd ever glide, no matter how long they are. Ignore your body for a moment, and just consider the arms. If you make your arm a foot longer, then that extra foot of arm has to produce at least enough lift to support itself, or you're no better off. But an arm, being roughly cylindrical and fairly dense, is not a very good airfoil. If you think about objects similar to a piece of arm, like a baseball bat or slender log, they don't "glide" if you throw them. That is, they don't produce much lift compared to their own weight. Thus, I don't think that making your arms longer is going to produce enough lift to support your arms, let alone your body. -- Coneslayer (talk) 20:34, 29 November 2007 (UTC)[reply]

I guess, we will have to change the shape of the arms also in addition to changing the size only. Airplave wings have very specific shapes, which provide the airplane enough lift to fly. Once the shape based on small prototypes has been determined, we will accordingly fix the size depending upon our weight. But our lower body also has to be in accordance with the aerodynamic structure. So, I guess if you have distended belly for example, you will have more problem than those having 6 pack :). DSachan (talk) 20:37, 29 November 2007 (UTC)[reply]

Shape helps a lot, but I think air moving over any object should eventually provide lift, depending on the speed of the airflow, and the weight of the object. Hmm, that's why cars have rear spoilers. When they go fast, the air moving around it can cause lift, which makes the tires lose traction with the ground. The spoiler keeps the car in contact with the ground. 64.236.121.129 (talk) 20:43, 29 November 2007 (UTC)[reply]

I guess, this is a wrong statement to make that air moving over any object should eventually provide lift, depending on the speed of the airflow, and the weight of the object. It will be possible only when the body itself is such that it is able to create the pressure difference on its two sides. This will eventually depend on the shape, and texture of the body. Of course, speed of the airflow plays a role in it, but only a part of the whole. DSachan (talk) 20:51, 29 November 2007 (UTC)[reply]

I read that the pressure difference thing is a fallacy when it comes to wings and lift. The articles here say that something as simple as a flat board, angled up will cause lift. 64.236.121.129 (talk) 20:56, 29 November 2007 (UTC)[reply]

A flat board can certainly generate lift. I've seen a couple of radio controlled planes that flew reasonably well with flat 'planks' of foam polystyrene for wings. The pressure difference thing isn't exactly a fallacy - it's just a very minor part of what's going on. Many aerobatic aircraft have symmetrical cross-sections so they'll fly just as well inverted as the right way up. They get their lift because of the angle of attack of the wing to the air flow. SteveBaker (talk) 04:39, 30 November 2007 (UTC)[reply]

(edit conflict) Why do you say "any object"? You're contradicting yourself. If air moving over "any object" caused lift, then the air moving over the spoiler (or, say, a Formula 1 wing) would cause lift, rather than downforce. If you have a perfect cylinder in air (as a model for your arm), why should airflow cause lift? You could flip the whole scene upside down, and it would look the same. -- Coneslayer (talk) 20:52, 29 November 2007 (UTC)[reply]

Right, I should say any object that isn't shaped at an angle so that the airflow pushes the object down. 64.236.121.129 (talk) 20:56, 29 November 2007 (UTC)[reply]

To elaborate, another example on wiki also says that if you stick your arm out of a moving car and angle it up, the air flow will push your arm back and up, causing lift. Factors that affect lift are airflow, weight of the wing, shape of the wing, and wing area. 64.236.121.129 (talk) 20:59, 29 November 2007 (UTC)[reply]

'Glide' can have many meanings—how much lift will you settle for? Take the Space Shuttle orbiter—it glides in to a landing, but it lands very steeply and very quickly compared to a commercial airliner, owing to its great weight and short, stubby wings. (It's been said that the Shuttle 'glides like a toolbox'.) See also our article on lifting body. TenOfAllTrades(talk) 21:48, 29 November 2007 (UTC)[reply]

Take a look at a hang glider - now you have your answer. SteveBaker (talk) 00:04, 30 November 2007 (UTC)[reply]

Actually, Wingsuit flying is just about...kinda/sorta gliding. It's all a matter of what glide ratio you anticipate getting! SteveBaker (talk) 01:55, 30 November 2007 (UTC)[reply]

Why doesn't buoyancy work in solids?

It works in liquids, and gas... Why not solids? 64.236.121.129 (talk) 20:40, 29 November 2007 (UTC)[reply]

It does. Dragons flight (talk) 20:43, 29 November 2007 (UTC)[reply]

Ok lets say, we cover a person in lead, then let it solidify. Since he weighs less than the lead, his corpse should come to the top? 64.236.121.129 (talk) 20:47, 29 November 2007 (UTC)[reply]

Most solids have enough internal rigidity to resist the bouyant force, but on very large scales or long timescales, the same principle applies. For example, salt diapirs. Dragons flight (talk) 20:54, 29 November 2007 (UTC)[reply]

Or the ballast in the basement of the Kansai Airport to prevent it from floating, perhaps? Or am I remembering it wrong... --Mdwyer (talk) 22:27, 29 November 2007 (UTC)[reply]

If you vibrate a container full of balls made of wood and others made of lead, all the same size, presumably the wooden ones will float to the top due to buoyancy. --Sean 23:51, 29 November 2007 (UTC)[reply]

This is probably irrelevant, but worth a mention: Didn't they find that glass -- although being a solid -- behaves like a liquid because over time they've found glass gets thinner on the top and thicker at the bottom - gravity has an affect. Is this relevant at all? Rfwoolf (talk) 13:13, 5 December 2007 (UTC)[reply]

Wrong; the existence of antique windows that are thinner at the top than at the bottom is the result of medieval glass production techniques being imperfect. If you have a big plate of glass that's thin at one end and thick at the other end (a not-uncommon occurrence back then), and you install it with the thick end up, it's top-heavy, and far more likely to fall out and break. So you install it with the thick end on the bottom (although some do still exist with the thick end on top). Another point to consider is that, if medieval glass had flowed that much, Roman and pre-Roman glass artifacts should show even more signs of flow; they do not (and antique telescopes would be useless; they likewise are not). DS (talk) 15:10, 5 December 2007 (UTC)[reply]

Natural and Phenomena and Natural Resources

People from Europe came to the New World, noe the United States, for the natural resources.What kinds of resources or products did they ship back to Europe? —Preceding unsigned comment added by Felicia25 (talk • contribs) 22:44, 29 November 2007 (UTC)[reply]

Colonization_of_America#Economic_immigrants. Someguy1221 (talk) 23:35, 29 November 2007 (UTC)[reply]