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May 31[edit]

Are battery capacities going up or stagnant?[edit]

Does an AA battery of today hold more power than an AA battery of the 1970s? In millie amp hours please

I heard that battery technology has basically been stagnant since Telsa invented them Fishhungry (talk) 10:50, 31 May 2014 (UTC)[reply]

You might like to start with our article History of the battery, and the links and references there, before coming back with more specific questions. 86.146.28.105 (talk) 11:29, 31 May 2014 (UTC)[reply]

Is there something wrong with the question "have battery capacities gone up since the 1970s to the present day"? I'm not sure how to make it any more specific. Fishhungry (talk) 11:37, 31 May 2014 (UTC)[reply]

No, but you can find the answer (and ways to make your question more informative for yourself) by reading the article. Battery capacities have gone up. How, and what is meant by 'battery', can be answered by reading the provided reference. If you would like more specific references, having read that, please feel free to ask. 86.146.28.105 (talk) 11:41, 31 May 2014 (UTC)[reply]

See also, AA battery. 86.146.28.105 (talk) 11:43, 31 May 2014 (UTC)[reply]

So you're not going to answer the question, then? Why even bother to reply? I have dyslexia and I cannot read that entire article otherwise I would have done so myself. The point of asking the question was that I am unable to find the information myself and need help. If you don't want to help then don't respond. Fishhungry (talk) 11:46, 31 May 2014 (UTC)[reply]

He exactly answered the article by directing you to places where you can search for the answer yourself. Do you demand that the chef cut your food and feed it to you as well? --Jayron32 15:27, 31 May 2014 (UTC)[reply]

The answer is they have gone up. AA is the commonest standard size of electric cell; the word "battery" was extended to single electric cells in 1748 by Ben Franklin. Capacities of AA consumer batteries have increased since 1970 with introduction of new chemistries. Some dates of commercial availability are:

1970 Zinc-carbon, alkaline(Zn/Mn O₂) or rechargeable nickel-cadmium (Ni-Cd)
1989 Rechargeable Ni-MH
1991 Lithium or Rechargeable Li-ion
1997 Rechargeable Li-polymer

See History of the battery#20th century: new technologies and ubiquity. For a comparison table of AA battery capacities see Comparison. Nobody called "Telsa" or even the inventor Nikola Tesla 1856 - 1943 is credited with inventing the AA battery whose size was standardized 4 years after Tesla died. 84.209.89.214 (talk) 11:55, 31 May 2014 (UTC)[reply]

Thanks for the information. In terms of millie amp hours is it a big increase? AA batteries today come up to 3000 millie amp hours, what would an AA battery of the 1970s have been in millie amp hours? Many thanks for your help and sorry if I haven't been specific enough or worded it strangely, I do not have a full grasp on the English language although I try my best. Thanks Fishhungry (talk) 11:59, 31 May 2014 (UTC)[reply]

Click on this comparison table which shows that a 1970s Zinc-carbon battery gave only 400-1700 milliamp hours. By now it would be horribly corroded. 84.209.89.214 (talk) 12:14, 31 May 2014 (UTC)[reply]

Battery technology has improved in other ways as well. Energizer gives a lot of useful data on all of their products. For example, one can see that with conventional alkaline batteries, the capacity is highly dependent on the current draw. At 25 mA it has a capacity of nearly 3000 mAh, but at 500 mA operating current, the capacity is only around 1500 mAh. For lithium batteries however, the capacity is nearly independent of the current even up to 1 A. The alkaline batteries will also only last half as long in freezing conditions (0 C) than they will at room temperature while the lithium batteries have a nearly constant capacity from -20 C to 60 C Mr.Z-man 16:55, 31 May 2014 (UTC)[reply]

One comment, avoid "heavy duty" batteries. Despite the name, they are the weakest batteries on the market. At the very least, get alkaline batteries. StuRat (talk) 20:48, 31 May 2014 (UTC)[reply]

Another unfortunate trend I've seen is using absurdly small batteries where larger ones would be more appropriate. My bathroom scale, for example, uses a single watch batteries (CR2032), where 4 D-cell batteries would have fit just fine. The result is that I must change the battery ever few months. I have a flashlight, on the other hand, that has been running on the same 4 D-cells for years. StuRat (talk) 20:54, 31 May 2014 (UTC)[reply]

This is going off-topic but has anyone any idea why they do this? My mum's doorbell could fit two AAs but they chose to design it with a place for a CR2032. The result is more expensive and frequent battery changes. I think AAs are the most prolific and cheap batteries available. 78.148.110.113 (talk) 17:16, 2 June 2014 (UTC)[reply]

It is perplexing. All I can think of is that they share common components with portable devices, where of course weight and size are far more important. Also presumably few people think about the batteries when they make the purchase decision. (I actually did, but my bathroom scale was marked down 90%, from $30 to $3, so I decided to buy it anyway, as that $27 will buy a lot of batteries.) StuRat (talk) 01:24, 3 June 2014 (UTC)[reply]

fractals[edit]

Is it possible that there could exist a fractal seed for anything imaginable? I know that fractals expand indefinitely from the starting point. So could the entire universe be created from a fractal? Theoretically I mean Armesh1997 (talk) 16:01, 31 May 2014 (UTC)[reply]

Please don't post requests for opinions on the reference desk. This is not a forum for the discussion of original research... Sebastian Garth (talk) 17:06, 31 May 2014 (UTC)[reply]

A Fractal is a set of image data that can be generated by a simple mathematical formula; by "expand indefinitely" one means that fractal sets have a property of Self similarity at unlimited increasing magnifications. The short article Fractal compression explains the difficulty of synthesizing a fractal generator formula for just an arbitrary single 2-D image. Approximating anything more complex with three dimensions and time is beyond any present practical capability. The question "is the Universe a fractal?" belongs to Metaphysics (for which we have no reference desk) because the purported Universe includes ourselves conceiving its possibility. The intriguing thought may been inspired by the book The Fractal Geometry of Nature (1982) by Benoit Mandelbrot, renowned for the fractal image shown at right. 84.209.89.214 (talk) 18:25, 31 May 2014 (UTC)[reply]

moons affect on sea level[edit]

If Earths moon suddenly disappeared would the sea level rise of fall? — Preceding unsigned comment added by 203.223.33.101 (talk) 16:09, 31 May 2014 (UTC)[reply]

yes. --Jayron32 16:59, 31 May 2014 (UTC)[reply]

We answered this in January 2009 and again in September 2011. If you seek a scientific answer, you need to formulate the question more formally. Objects do not simply disappear, and if you can provide a mechanism to explain a way that objects could disappear, then we could follow that logic and determine consequences. If you simply rely on "magic" to make events happen, or to change the properties of the universe, then anything is a plausible consequence. Nimur (talk) 17:26, 31 May 2014 (UTC)[reply]

The cycle of tidal rise and fall of sea level, which is an effect (not "affect") of the Moon's gravity, is unaffected by a Lunar eclipse. 84.209.89.214 (talk) 18:47, 31 May 2014 (UTC)[reply]

To be specific, the moon cannot simply disappear because it would violate all the fundamental conservation laws (of linear momentum, angular momentum, mass-energy, and everything else).--Jasper Deng (talk) 22:39, 31 May 2014 (UTC)[reply]

We could sanitize the question easily if we assumed that the Moon is split in half (sort of) and that the forward half is propelled fast enough to exceed escape velocity (that'd take a delta v of 41.5% of its current orbital v – make that 1.2km/s) while the other half experiences recoil and exceeds escape v in the other direction. The latter would be at a delta v of about 241.5% (so, ~7km/s), so that "half" should be much smaller actually. The Moon would disappear far from suddenly, but it would disappear (escape). The Helium 3 contents of the Moon should be sufficient for that kind of acceleration; we had a topic once that stated that solar energy was insufficient, though. - ¡Ouch! (^{hurt me} / _{more pain}) 06:58, 3 June 2014 (UTC)[reply]

Old topic about how to lose the Moon. I estimated the mass ratio there, and it looks I got the order of magnitude right (although it's about 15:85 rather than 20:80). It's in the "late" reply. - ¡Ouch! (^{hurt me} / _{more pain}) 07:11, 3 June 2014 (UTC)[reply]

Ouch indeed, I used the wrong orbital v figures.

If we compare low Earth orbit to the Moon, we have an orbital radius of about 1:60, and thus, by Kepler's 3rd law, an orbital v ratio of about 8:1. That would reduce the delta-v figures to 0.4 and 2.4km/s respectively. The mass ratio is unchanged, though, because we can treat the Moon orbit as roughly circular with an error of less than 10%. Also, using helium3 as a fuel is not efficient enough yet. It only shows that the energy needed is there. - ¡Ouch! (^{hurt me} / _{more pain}) 07:41, 3 June 2014 (UTC)[reply]

Of course this is a hypothetical question But I can't believe that the science desk has not answered hypothetical questions in the past.

If the moon just quietly disappeared from the solar system without any cataclysmic events, so that the amount of water on the earth was unchanged, wouldn't it be a fairly safe answer to say that in the short term, the AVERAGE sea level would be the same? CBHA (talk) 01:55, 1 June 2014 (UTC)[reply]

Its fairly safe to say it won't happen, and if it did, as long as the reasons for it occurring are unexplained, the consequences cannot be predicted. AndyTheGrump (talk) 02:05, 1 June 2014 (UTC)[reply]

Don't forget that the sun also causes tides...not as big as moon-induced tides...but definitely still there. But the total volume of water wouldn't change, so the average sea level wouldn't change directly from the loss of the moon. However, the moon does things to stabilize the axis of rotation of the earth - and in the longer term, that shift in the position of the poles would cause complicated interactions with ice at the poles - which in turn could easily result in either higher or lower mean sea levels. It's a really complicated thing. SteveBaker (talk) 16:05, 1 June 2014 (UTC)[reply]

Transmitting information faster than the speed of light[edit]

It has long been a principle in science, based on the Special Theory of Relativity, that information could not be transmitted from one place to another faster than the speed of light, but a press report about a piece in Science states that it has been done over 10 feet, reliably, and that it should work for thousands of feet, in a case where the spin of an electron is changed and this changes the spin of another remote electron due to quantum entanglement. The press report says "Researchers from the Kavli Institute of Nanoscience of Delft University of Technology in the Netherlands have created a new system capable of reliably transporting information." Our related articles on quantum teleportation say that such demonstrations do not actually transmit information. I did not see in our articles a description of what would each observer would detect when the event happened. Some explanation of how it misses being the transmission of information would be informative, and what the minimum change would be in the process for there to be information transmitted. "Something just happened" at a remote location would be somewhat informative, but less than "event one" versus "event two" happened. What would be the theoretical distance such quantum entanglement could encompass, with a spin change at one locatio being instantly mirrored at the other location, and why couldn't it convey information? (other than "that would violate the rulz"Edison (talk) 19:25, 31 May 2014 (UTC)[reply]

I haven't read the piece in Science; I don't know if there's some genuine novelty here. However, the usual quantum-entanglement thingies, like the Aspect experiment, don't give any obvious way to transmit information faster than light. What happens is that observer A makes a measurement, and observer B makes a measurement, and observer A's measurement causes "wavefunction collapse" (looking at it from the Copenhagen viewpoint) whose effects seem to proceed superluminally to B. But A doesn't get to choose which way it goes, so A can't use it to transmit information. --Trovatore (talk) 19:35, 31 May 2014 (UTC)[reply]

Anything that did have the ability to detemine the outcome would, definitionally, be a hidden variable. Whether you believe that such a state-variable exists is irrelevant, because in either case, it still isn't transmitting information from A to B. Nimur (talk) 20:35, 31 May 2014 (UTC)[reply]

Yea, it's logically like you sent an envelope with a photo of your room to Mars, and then an astronaut opens it to find out what your room looks like. The time to send that info to Mars was not how long it took to open the envelope, it was the entire trip. StuRat (talk) 20:43, 31 May 2014 (UTC)[reply]

However, quantum tunneling does indeed appear to allow particles to move faster than light. See Quantum_tunneling#Faster_than_light. StuRat (talk) 20:40, 31 May 2014 (UTC)[reply]

As I recall there's some trickery involved with that. To begin with, any particle can move "faster than light" in the sense that it is a wave of probability that may be found at any point, so when observed at two times very close together, it can be found in very different places. But the amplitude of the wave doesn't move faster than light. In the case of the tunneling though, there's also the issue that AFAIK the amplitude splits into two parts; you might say that the "frontmost" part goes through the barrier and the "rearmost" part gets reflected backward, more or less. Wnt (talk) 20:52, 31 May 2014 (UTC)[reply]

If you have an experiment in which you localize the particle (detect for its presence) in two places that are further apart than light could reach in the time difference, only one of the detectors will trigger, never both under these conditions, rather like quantum entanglement. So I fail to see how quantum tunnelling can be interpreted, in any sense, to allow tunnelling of a particle at faster than the speed of light. As in BenRG's waveguide example below, the wavefunction will not become established there at faster than the speed of light. I expect that the referenced portion of that article presents controversial research, and as such, does not belong there. —Quondum 14:34, 1 June 2014 (UTC)[reply]

Tunneling happens in classical wave theories too; see evanescent-wave coupling. It doesn't violate the speed-of-light limit. The evanescent wave is outside the waveguide (i.e. in the "classically forbidden region") whether or not you've put another wave guide there to pick it up. The wave outside the waveguide doesn't lag behind the wave inside (much as field lines don't). Thus the second wave guide will get a (blurred) copy of the signal with no distance-dependent time delay. This confuses some people into thinking the wave is jumping instantaneously across the intermediate space. But if you alter the signal inside the first wave guide, the evanescent wave will update to reflect the change at a speed of at most c, just like the electric field lines if you change the velocity of the charge. -- BenRG (talk) 06:48, 1 June 2014 (UTC)[reply]

The quantum teleportation protocol requires transmitting two (classical) bits from point A to point B. The qubit isn't transferred from A to B until those classical bits are acted upon at the receiving end. The only advantage of quantum teleportation over just sending the qubit from A to B is that you can use a classical channel (like the Internet) instead of a quantum channel. Against that there's the serious disadvantage that you have to create a Bell pair and send the halves to A and B, which itself requires a quantum channel connecting A and B (though it needn't still exist at the time the "teleportation" is done). It's also very difficult to store Bell pairs for any significant length of time without losing the entanglement, but if you have any qubits worth teleporting in the first place then presumably you've solved that problem somehow. -- BenRG (talk) 06:48, 1 June 2014 (UTC)[reply]

Calculating the top speed of a car given its power, drag coefficient, and cross-sectional area[edit]

I take the motion to be one-dimensional, assume that the engine power is constant (since if I understand it correctly, full throttle means operating at maximum power), and ignore all dissipative forces except air drag (for now). The net power on the system is given by $P=H-D$ , where H is the engine power (constant) and $D=kv^{3}$ is the power of the drag force, where $k={\frac {1}{2}}\rho C_{D}A$ rolls these coefficients into a single constant. This sets up the nonlinear ordinary differential equation $mv{\frac {dv}{dt}}=H-kv^{3}$ . I separated variables as in the drag equation article, but the resulting expression is very hard to solve for v:

$mv{\frac {dv}{dt}}=H-kv^{3}$

$\int _{0}^{v}{\frac {mv}{H-kv^{3}}}dv=\int _{0}^{t}\ dt$

$-m{\frac {2\ln {(H^{\frac {1}{3}}-k^{\frac {1}{3}}v)}-\ln {(H^{\frac {2}{3}}+(Hk)^{\frac {1}{3}}v+k^{\frac {2}{3}}v)}+2{\sqrt {3}}\arctan {\frac {2({\frac {k}{H}})^{\frac {1}{3}}v}{\sqrt {3}}}}{6(Hk^{2})^{\frac {1}{3}}}}=t$

Perhaps I could change the arctangent to the (complex) hyperbolic arctangent, which is expressible in terms of the natural logarithm, and then apply laws of logarithms, but it gets really messy afterwards (what is amazing is that on the left side, that expression is indeed 0 when v=0). Perhaps we don't need to solve for v, but letting t→∞ is not really useful unless we do.

I do realize that if P=0, then the maximum velocity is reached, from which it follows that $v_{max}=({\frac {H}{k}})^{\frac {1}{3}}$ , but I'd like to see that from the equation above too.

What if we now consider internal friction forces? Since friction is directly proportional to the normal forces of the surfaces, and those normal forces are basically constant (I don't know how if I had to assume otherwise), then the friction forces are constant and the sum is denoted as f. This adds another term on the right:

$mv{\frac {dv}{dt}}=H-fv-kv^{3}$

$\int _{0}^{v}{\frac {mv}{H-fv-kv^{3}}}dv=\int _{0}^{t}\ dt$

Unfortunately things are even messier (integrating the left side involves partial fraction decomposition using the cumbersome cubic formula to find roots... ugh). Even the "simple" approach of letting the power equal 0 leads to $v_{max}=({\sqrt {{\frac {f^{3}}{27k^{3}}}+{\frac {H^{2}}{4k^{2}}}}}+{\frac {H}{2k}})^{\frac {1}{3}}+({\frac {H}{2k}}-{\sqrt {{\frac {f^{3}}{27k^{3}}}+{\frac {H^{2}}{4k^{2}}}}})^{\frac {1}{3}}$ , which (as expected) reduces to $v_{max}=({\frac {H}{k}})^{\frac {1}{3}}$ when f=0. Has anyone previously solved these equations for v and derived the velocities by then letting t→∞?--Jasper Deng (talk) 22:34, 31 May 2014 (UTC)[reply]

I suspect that this would be a purely academic exercise, as many other factors will affect the maximum speed, such as the gearing. Also, the drag coefficient isn't constant, and would be expected to change when flow transitions from laminar to turbulent. StuRat (talk) 23:54, 31 May 2014 (UTC)[reply]

Well, I was certain that it being constant would be a valid approximation because terminal velocity makes that approximation. For simplicity, I'm assuming a continuously variable automatic transmission so gear shifts aren't a factor. The only other factors would be other dissipative forces such as viscosity of exhaust or gasoline, but they shouldn't be that hard to deal with if I can already fold them into the fv or drag terms on the right side of the equation.--Jasper Deng (talk) 02:29, 1 June 2014 (UTC)[reply]

What are you trying to learn from the entire velocity as a function of time? For a qualitative feel of the time behaviour of the system as it approaches terminal velocity, surely it would be sufficient to approximate the cubic with its tangent where the power is balanced? I'd expect essentially an exponential decay towards the limiting velocity. —Quondum 02:49, 1 June 2014 (UTC)[reply]

Oh I'm essentially looking for a quantitative approach to this. By tangent, do you mean a linear approximation? You might be right that the decay is in the form of

v_{max}-e^{-G(t)}

for some increasing function G of time, but based on the form of my friction-free (drag-only) equation's solution above, it would seem like it is instead in the form of

e^{-{\frac {G(t)}{H(t}}}

where

-\infty <\lim _{t\to \infty }{\frac {G(t)}{H(t)}}<\infty

.--Jasper Deng (talk) 05:12, 1 June 2014 (UTC)[reply]

Yes, I meant linear approximation, and I was expecting from this approximation to get specifically

v_{max}-e^{-\alpha t}

, where alpha is a constant. —Quondum 14:25, 1 June 2014 (UTC)[reply]

Letting

t\rightarrow \infty

is useful even without solving for v. In order for the LHS to diverge, the numerator must; the denominator is a constant. The third term in it is bounded, and the second can only diverge when v does (since the argument of the logarithm is bounded away from 0). So the first must diverge, which occurs exactly when the terminal velocity you derived from

P=0

. --Tardis (talk) 15:26, 1 June 2014 (UTC)[reply]

The OP asks, "Has anyone previously solved these equations?" The answer is, almost certainly: between Carl Friedrich Gauss and Leonhard Euler and Pierre-Simon Laplace, every closed-form differential equation whose parameter-space has cardinality aleph 0 has been solved analytically... and thanks to Joseph-Louis Lagrange, the same is also true for equations whose parameter space is continuous! And now that we have computational mathematics, we know whether the solution exists a priori, and we even have a method to pick the algorithm to numerically approximate the solution. With fast computers and good free software, you can approximate the solution numerically at no cost. So, if the solution exists, then its form is already known to humans... somebody has solved it. But you probably won't like the form in which such solutions are presented: the math gets a little bit abstract!

I think a more pointed question is due: is this method actually used by experts in the field of automotive engineering? Any half-trained ape can vomit out thirty pages of equations (assuming the ape is half-trained in mathematical physics). But after the novelty of multiple-pages-of-math begins wearing off, we need to start grounding our efforts in reality. Though we can come up with our own equations, this is impractical. If you know anything about the involved math, even a simple equation with a few terms can be dramatically difficult to solve. Worse yet: even if you do solve it - there's a really good chance that your answer has no connection to physical reality.

So, instead of creating your own equation, why not start by reading a book on automotive engineering? You can learn how other experts have derived top speeds for automobiles; you can see which models are useful, and which parameters are determined from first principles, as opposed to experiment.

If you haven't got anywhere to start, begin at the Society of Automotive Engineers website. They have a library of free- and non-free publications, including Automotive Engineering Fundamentals. Nimur (talk) 17:40, 1 June 2014 (UTC)[reply]

Oh I know I'm making a lot of idealizations; the "car" in my problem is any object with constant mass (approximately) that moves with constant power in one dimension and obeys the drag force law. This is primarily of academic interest to me, as I'm essentially interested in how we treat a constant power rather than a constant force (as with terminal velocities with gravity close enough to the surface such that the force is essentially mg in magnitude).--Jasper Deng (talk) 19:26, 1 June 2014 (UTC)[reply]

As an automotive engineer I solve it by noting that at maximum speed P=kv^3, so v=(P/k)^(1/3). However that is for optimal gearing and ignores rolling resistance, driveline losses etc. You may be looking for this https://iversity.org/my/courses/vehicle-dynamics-i-accelerating-and-braking/lesson_units/16028 registration is required. Greglocock (talk) 22:07, 1 June 2014 (UTC)[reply]

Many years ago, I did the math for my car - and the numbers came out pretty much on the money. See http://sjbaker.org/telamom/speed_math.html - but getting good numbers for friction and rolling resistance by calculation alone is unlikely...I had to find "typical" numbers for those things. SteveBaker (talk) 20:34, 3 June 2014 (UTC)[reply]

horseshoes[edit]

why do ppl think horseshoes are lucky? are they? — Preceding unsigned comment added by 201.210.249.61 (talk) 22:44, 31 May 2014 (UTC)[reply]

I think this belongs at another reference desk.--Jasper Deng (talk) 22:48, 31 May 2014 (UTC)[reply]

Well horses that have horseshoes, and hence owners, tend to live much longer than wild horses with neither. StuRat (talk) 23:50, 31 May 2014 (UTC)[reply]

Lucky W Amulet Archive by Catherine Yronwode claims that the crescent form of the horseshoe links the symbol to pagan Moon goddesses of ancient Europe such as Artemis and Diana, and that the protection invoked is that of the goddess herself, or, more particularly, of her sacred vulva. As such, the horseshoe is related to other magically protective doorway-goddesses, such as the Irish sheela-na-gig, and to lunar protectresses such as the Blessed Virgin Mary, who is often shown standing on a crescent moon and placed within a vulval mandorla or vesica pisces.

In most of Europe, the Middle-East, and Spanish-colonial Latin America protective horseshoes are placed in a downward facing or vulval position... but in some parts of Ireland and Britain people believe that the shoes must be turned upward or "the luck will run out." Americans of English and Irish descent prefer to display horseshoes upward; those of German, Austrian, Italian, Spanish, and Balkan descent generally hang them downward.

See also Horseshoe#Folklore. 84.209.89.214 (talk) 23:58, 31 May 2014 (UTC)[reply]

To hang a horseshoe on a wall with the ends pointing upwards takes two nails. Hanging downwards needs only one. The latter is obviously more efficient. HiLo48 (talk) 00:34, 1 June 2014 (UTC)[reply]

Taking short-cuts can result in bad luck. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:36, 1 June 2014 (UTC)[reply]

Apocryphally, Niels Bohr had a lucky horseshoe in his office. When asked if such a great scientist could believe in good luck charms, Niels said "Of course not ... but I am told it works even if you don't believe in it." [1] SemanticMantis (talk) 02:08, 1 June 2014 (UTC)[reply]

My guess is that in the old days, if you had a horse that threw a shoe, it would be lucky if you had a horseshoe on the wall. Wnt (talk) 04:44, 2 June 2014 (UTC)[reply]