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November 13[edit]

Doing more damage with less recoil[edit]

I was wondering what the correlation between firing two different projectiles (of different masses) doing equal amounts of damage (in terms of kinetic energy), and their recoil. Lets say you have a 5.56×45mm NATO round fired at 940m/s. According to that article the bullet should have 1,767 joules of energy. The round itself is 4 grams approx. So lets say you have a projectile that is 4 miligrams instead. According to my calculations (I hope I didn't mess up my math), the projectile would have to move at 29,724m/s in order to have the same energy of 1,767 joules. My question is, if you had a weapon that could somehow fire this tiny projectile at such a high velocity, would the recoil be less? And if so, by how much? ScienceApe (talk) 02:10, 13 November 2011 (UTC)[reply]

...Can't you easily calculate the momentum? You know the mass and the velocity. If you want extensive analysis, the relevant article is elastic collision, because both energy and momentum are conserved in this scenario. There are other practical considerations - a tiny hypersonic projectile will not travel far before air resistance slows it considerably - or ablates it completely. Nimur (talk) 02:47, 13 November 2011 (UTC)[reply]

The article Recoil seems relevant. I am not highly confident about this area of physics as I would be with basic electricity, but here goes. Your high velocity and low mass scenarios as stated would have the same kinetic energy in the projectile. Now look at the momentum, mass times velocity, which governs recoil of a firearm: in case one, .004kg*940m/s=3.76 kg*m/s, while in the second case .000004kg*29,724m/s = .119 kg*m/s. The article does not clearly state that there is less recoil when there is less transfer of momentum, but it seems plausible. The Recoil article also talks about "recoil impulse," "recoil velocity," and "recoil energy." A separate question is the damage done to the target by the inelastic collision, when bullets with the same kinetic energy but different masses and velocities strike. Edison (talk) 03:03, 13 November 2011 (UTC)[reply]

In order for air resistance to be mitigated, the projectile would have to be made as small and dense as possible I believe. Or perhaps use plasma stealth to sheath it. 71.53.16.5 (talk) 15:16, 13 November 2011 (UTC)[reply]

Kinetic energy is p^2/2m, where p is momentum and m is mass. So two projectiles that have the same kinetic energy will have momenta that differ by the square root of the ratio of their masses, favoring the larger mass. So yes, the recoil would be less for the smaller projectile, by a factor of sqrt(1000). Nimur said that energy is conserved, but I don't see why--after all, the mass and bullet go from having a kinetic energy of 0 to a kinetic energy much higher than 0.

However, the damage dealt to the target is NOT proportional to the bullet's kinetic energy. Bullets from modern guns pass right through the victim's body if the gun is not too far away, and deliver only a fraction of its kinetic energy to the body. Your 4 mg bullet would have an even easier time passing through, and would deliver an even smaller percentage of its energy. Its smaller size also means that it's harder to manufacture, harder to load, more prone to breaking while inside the gun, and more prone to aerodynamic instability.

While we're on the subject of how to best kill our fellow human beings, expanding bullet are one way of making the bullet convert its energy into damage. I imagine they're also harder to manufacture for 4 mg bullets, but don't quote me on that. --140.180.3.244 (talk) 03:45, 13 November 2011 (UTC)[reply]

Good point about whether the event is elastic or inelastic. The correct model depends on which "event" we are modeling. I was referring only to the "recoil event," assuming a starting energy (from the chemical propellant) and no net initial momentum; then solve for the kinetic energy of the projectile and the recoil, constrained to have no net momentum. This is conceptually similar to solving an elastic collision system of equations - constraining both momentum and energy - except that the initial energy is chemical, not kinetic. 140.180.3.244 makes a valid point above; if we don't include the conversion of chemical energy into kinetic energy, we can't use this elastic collision model; the acceleration of the projectile is now an inelastic collision (therefore, we won't have enough information to solve the system of equations). The same is true for the terminal ballistic collision on the "receiving side" - when the projectile impacts the target, we probably want to use the inelastic collision model, which leaves a free parameter in the equation. Nimur (talk) 16:50, 13 November 2011 (UTC)[reply]

In a vacuum perhaps a smaller, faster projectile might make sense, but not with air resistance. The larger and slower the projectile, the less effect air resistance will have. However, there's also the effect of gravity to be considered. That is, too slow of a projectile may miss the target, due to it's curving path, or may fall short and expend it's energy on the ground. StuRat (talk) 05:56, 13 November 2011 (UTC)[reply]

You mean the more massive the projectile is. The larger the projectile, the more effect air resistance will have. Two objects of the same volume but of different mass will be affected by air resistance differently. The more massive (dense) object will travel further. Make a projectile smaller, but increase the mass that should improve ballistics. 71.53.16.5 (talk) 15:23, 13 November 2011 (UTC)[reply]

I'm assuming that the material is the same, so no change in density is possible. Thus, a larger projectile is also more massive, and would thus be slowed down less by air resistance. StuRat (talk) 16:57, 13 November 2011 (UTC)[reply]

Why are you assuming that? The material doesn't have to be the same. Also it's possible to just compress the thing and put a hard shell around it. 71.53.16.5 (talk) 18:48, 13 November 2011 (UTC)[reply]

Because the original Q was about mass versus velocity, with no mention of changing the density. Adding that variable only confuses matters unnecessarily. We could also discuss various bullet shapes, rotations, fluids other than air, etc., but none of this is what we were asked about. StuRat (talk) 03:53, 14 November 2011 (UTC)[reply]

To convert from energy to momentum, consider the formulae.

Energy kinetic = 0.5mv²

p = mv

Therefore, we can find the formula from Ek to p:

p (vector) = 2Ek/v. ~AH1 ^(discuss!) 18:59, 13 November 2011 (UTC)[reply]

why does Jupiter rotate round its axis so fast?[edit]

its period is the planet's shortest, at 9h 50m 30.0s --Akbarmohammadzade (talk) 09:49, 13 November 2011 (UTC)[reply]
we are preparing new results for way of explaining reason of Jupiter’s' fast rotation.--Akbarmohammadzade (talk) 10:42, 13 November 2011 (UTC)[reply]

The main reason is conservation of angular momentum. As the planets coalesced from the protoplanetary disc, they had to spin up to conserve overal angular momentum. As a rule, the larger the mass and the higher the density, the higher the spin rate. All the gas giants have short periods of rotation for their size - Jupiter is not unusual in that respect. See the table. --Stephan Schulz (talk) 11:36, 13 November 2011 (UTC)[reply]

Planet	Rotation period (h)	Radius (km)	Mass (kg)	Density (g/cm³
Sun	601.2	695,500	19,891×10²⁶	1.408
Jupiter	9.925	71,492	18.9886×10²⁶	1.326
Saturn	10.57	60,268	5.6846×10²⁶	0.687
Uranus	17.24	25,559	0.8681×10²⁶	1.27
Neptune	16.11	24,764	1.0243×10²⁶	1.638

As a rule of thumb, yes. But note that Uranus spins sideways (axial tilt 98°) and Venus spins backwards. Presumably, they spun up in the "right" direction initially, then got knocked out of whack. The point is, the angular momentum of a planet can change due to later events. (Adjusted formatting a bit, hope nobody minds.) Bobmath (talk) 15:04, 13 November 2011 (UTC)[reply]

Yes, later events can provide torques that will affect a planet's angular momentum, but Why would you presume that Uranus' sideways spin wasn't that way from the beginning? Dauto (talk) 15:13, 13 November 2011 (UTC)[reply]

It could have been, I suppose, but that's not the usual explanation, according to Uranus#Axial tilt. Bobmath (talk) 15:31, 13 November 2011 (UTC)[reply]

Because there is no plausible theory that would allow Neptune to have formed in situ with such a spin. Spinning Spark 15:40, 13 November 2011 (UTC)[reply]

(I assume you meant Uranus.) Agreed. Specifically, solar systems form as a spinning disk, and planets therefore inherit that spin direction, while the speed is magnified, as mentioned, by the conservation of angular momentum, as the planet compacts. Note that while an impact after the planet is formed can change the direction of the spin, it's not likely to change the magnitude of the spin much. Any impact large enough to do that would also rip the planet apart. This may have actually happened in the early history of the Earth, with the Moon being the part which was ejected. Also note that the largest object in our Solar System, the Sun, has a much slower rotation of 25 days at the equator and 34.4 days at the poles. I attribute this slow rotation to the lower average density of the Sun. Once nuclear fusion stops and it collapses into a dwarf star, I'd expect it to rotate much faster. StuRat (talk) 16:45, 13 November 2011 (UTC)[reply]

There are two main models for planetary formation; I was going to create articles on the topic but forgot. One model, the most popular, suggests that the cores of giant planets formed first from hard objects, then the top gaseous layers coalesced (bottom-up approach), and yet the model gaining more from recent evidence is that the gaseous layers coalesced together initially (top-down), and the heavier materials sank to the bottom, fused and formed the planet. This has implications for the formation and evolution of the solar system, a field of study that is constantly gaining new research findings: [1]. ~AH1 ^(discuss!) 18:27, 13 November 2011 (UTC)[reply]

Apparently, there used to be a fifth gas giant up to 4 billion years ago, between Saturn and Uranus, it had a violent encounter with Jupiter, which caused it to be flung into deep space. It caused Neptune to have a highly irregular inclination, and caused Jupiter and the other gas giants to lose momentum and move further out from the sun. It also caused to the late heavy bombardment when disturbed the Oort cloud. Plasmic Physics (talk) 04:49, 14 November 2011 (UTC)[reply]

Have any links about that ? StuRat (talk) 05:11, 14 November 2011 (UTC)[reply]

Hypothetical fifth gas giant. PrimeHunter (talk) 05:22, 14 November 2011 (UTC)[reply]

While we are on the topic, is it purely coincidence that the gas giants follow a partial ROYGBIV sequence (O-Y,Y,G-B,B)? By speculation, the missing planet could be yellow-green or green in colour. Plasmic Physics (talk) 07:15, 14 November 2011 (UTC)[reply]

Yes, that's a complete coincidence. The colours are determined by what the surface layers are made up of. There certainly could be a trend in that make up, but it wouldn't result in a trend in the colours. The colours of different chemicals don't follow any nice patterns in relation to the EM spectrum. In fact, they often aren't monochromatic at all (which means they don't appear in a rainbow). --Tango (talk) 12:19, 14 November 2011 (UTC)[reply]

But the Sun has equal or higher density than Jupiter, Uranus and Saturn. --Sean 18:13, 14 November 2011 (UTC)[reply]

But it also has a 10 times larger radius than Jupiter. I've added it's data to the table. --Stephan Schulz (talk) 22:59, 14 November 2011 (UTC)[reply]

Why does Regulus spin so extemely fast? Count Iblis (talk) 17:25, 14 November 2011 (UTC)[reply]

It must have a broken regulator. :-) But seriously, perhaps it had a close encounter with another star ? It's in a multiple star system, so this seems quite possible. StuRat (talk) 22:48, 14 November 2011 (UTC)[reply]

Potential[edit]

Is potential a property of objects or is it a property of the space they are in, like a field? — Preceding unsigned comment added by 81.31.190.91 (talk) 13:08, 13 November 2011 (UTC)[reply]

Potential energy, potential difference, or some other potential? Plasmic Physics (talk) 13:13, 13 November 2011 (UTC)[reply]

Potential energy. — Preceding unsigned comment added by 92.42.55.141 (talk) 14:04, 13 November 2011 (UTC)[reply]

In order to make that question meaningful you first have to answer the question: "what difference does it make?" If there is no physically measurable difference between the two possibilities, than the question is not physically meaningful. Dauto (talk) 15:03, 13 November 2011 (UTC)[reply]

There are a number of different things that are called "potential energy". Gravitational potential energy and electrostatic potential energy are described as being due to an object's location in a field. Elastic potential energy (like a stretched spring) and chemical potential energy (like a charged battery) are usually seen as being inherent in the object itself. However, if you consider the microscopic scale, the last two are special cases of electrostatic potential (the electrons in a stretched spring are being pushed into a configuration where they're too close to each other (though this is somewhat of an oversimplification)). I think potential energy should be seen as a property of systems (groups of objects) rather than individual objects, but maybe someone will set me straight. Bobmath (talk) 15:27, 13 November 2011 (UTC)[reply]

What I meant is(I'm the OP, I was in a library) it's common to draw a diagram with the vertical axis representing U and the horizontal one representing place (in 1-dimension), and while explaining the meaning of potential energy, it's often said that the potential energy of an object depends on where it is (in that diagram, for example).does the potential energy depend on the object, too, or it just depends on where it is? — Preceding unsigned comment added by Irrational number (talk • contribs) 16:12, 13 November 2011 (UTC)[reply]

I'm still not sure what kind of potential energy you're thinking of, but maybe it's Gravitational potential energy#Local approximation. In that case, m stands for the mass of the object, so the answer is, yes, the energy does depend on the object. Bobmath (talk) 16:22, 13 November 2011 (UTC)[reply]

Another simple example is the electric potential energy - and it depends on the charge of the test particle.

In elementary physics, we use a test particle that is assumed to be so small that it doesn't modify the potential energy function as it moves around (the value of the object's potential energy changes - but not the graph of the potential energy); we use this simplification because the math is easier. Once you can easily solve equations for static potential energy functions, you will graduate to more difficult mathematical tasks that require recalculating the potential energy field as a function of the test-particle's position. Nimur (talk) 17:03, 13 November 2011 (UTC)[reply]

Elastic potential energy is yet another. What the OP seems to be hinting at is the similarity between the different sorts of "potential", for example voltage is potential difference. See also quantum energy, zero-point energy and field lines. ~AH1 ^(discuss!) 18:22, 13 November 2011 (UTC)[reply]

Potential is a property of the field and the object in question. 71.53.16.5 (talk) 18:45, 13 November 2011 (UTC)[reply]

Somewhat confusingly, Gravitational potential includes only the effect of the central mass, not the test mass being acted on. 69.234.123.255 (talk) 19:51, 13 November 2011 (UTC)[reply]

That's because the potential and the potential energy are not the same thing. The gravitational potential energy is the the potential times the mass of the test particle and the electric potential energy is the electric potential times the charge of the test particle. That way the potential is independent from the test particle properties. From that point of view, the potential can be seen as field. Dauto (talk) 20:23, 13 November 2011 (UTC)[reply]

Diode and clipper circuit[edit]

I have 2 questions regarding diodes. 1. If i were to connect a resistor with a diode, would i get a DC output voltage across the resistor which would be equal to the barrier potential of the diode? In other words, would the diode act as a (low voltage) battery? 2. If i were to add an AC supply in parallel to the resistor across which the output voltage is being measured, i would get a simple clipper circuit. My question is, what would be the shape of the output voltage graph? Supposing the diode is forward biased(hence conducting) in the positive cycle, and also in parallel with the resistor, would i get a straight line at or a curve with a peak at the barrier potential of the diode. I believe i should get a peak, since the diode is turned on because of the AC supply, and since the diode and resistor are in parallel, the voltage across the diode will be replicated across the output. However, a colleague said i should get a straight line across the output for this phase as the voltage is being provided by the diode. I hope the question is easily understood. Thanks. --175.110.202.85 (talk) 13:33, 13 November 2011 (UTC)[reply]

Regular diodes don't provide any significant energy. Except for a little thermal noise, the voltage across them is 0 when there is no energy source in the circuit.

If a diode, a resistor, and an AC supply are connected in parallel, something will be destroyed. Perhaps the AC supply, perhaps the wire, or perhaps the diode. Jc3s5h (talk) 13:59, 13 November 2011 (UTC)[reply]

Ok so lets add a resistor in series as an electrical ballast in the 2nd circuit.--175.110.202.85 (talk) 14:26, 13 November 2011 (UTC)[reply]

The two resistors and the AC supply are a linear network, so a Thevenin equivalent can be formed of a supply voltage in series with a single combined resistor. A diode connected to this circuit will have the same behaviour as the original circuit. Assuming the current delivered is within the capabilities of the diode, then while the voltage is applied reverse biased to the diode, the diode voltage will be almost equal to the open-circuit voltage of the Thevenin equivalent (ie a sinusiodal half cycle). However, during the forward half-cycle the voltage will be limited by the forward voltage of the diode. For an ideal diode with a perfectly square transfer function that voltage would be constant with time. However, real diodes have a transfer function that follows an exponential law, so yes, you would see a peak, but it will be a much flatter and lower peak than obtained in the reverse direction. In many design applications this is insignificant and is simply ignored and the forward voltage is assumed to be flat - either zero volts or a small constant. Spinning Spark 15:26, 13 November 2011 (UTC)[reply]

Subliminal distraction exposure?[edit]

Hi. Apparently in the 1960's, psychological researchers did something regarding cubicle-level protection and suggested that movement or sensory stimuli at the far edge of our perception could induce psychosis. This is the main website for a spattered collection of research and seemingly unconnected topics, though there are other sources also. Is this a potentially valid theory, that merits articles? Thanks. ~AH1 ^(discuss!) 18:30, 13 November 2011 (UTC)[reply]

The website fails several criteria for establishing itself as a reliable source. If I had to enumerate the shortcomings, I would bring up at least the following problems:

The author makes no effort to self-identify as an individual or an affiliate of any institution with any degree of credential or expertise in the field.
Almost the first line of the page accosts me, imploring me to save the page to my desktop for future reference. If the material had merit, I would decide for myself how best to reference it in the future.
The website is poorly formatted, exhibiting little organization and very amateurish web design.
Regarding the actual content - there's essentially zero summary or introduction or abstract of any type. The author immediately begins delving into specific examples to illustrate his theory.
The theory is pretty outlandish. As a scientifically curious individual, I keep an open mind to new ideas, but the ensemble presentation of this website almost immediately identifies the author as a cruft-peddler. In the first few seconds after I started to read the site, I had already decided this. Perhaps I am making up my mind too quickly; but I've seen enough pseudoscience on the internet to identify it pretty fast.
In the absence of a reputation or credential of any kind, the website should be discounted entirely. There are too many sources of information out there to lend credit to everyone's idea. Save yourself time - go grab any textbook on neuropsychology out of your local library or bookstore - it is almost a certainty that they will provide better reading material than the site you linked. I'm sure one of the other regulars will be by momentarily to post links to their favorite "how to spot crank science" rundowns. Nimur (talk) 19:37, 13 November 2011 (UTC)[reply]

____________

Site author's reply:

The site provides sources to verify this problem was discovered and solved forty years ago, "Letters" page. (I literally tell you who to call.) No effort was made for a professional presentation. In nine years of emails and letters only one doctor has admitted seeing cases of Subliminal Distraction psychotic episodes. He like everyone else aware of it believes the problem can only cause a harmless temporary episode of confusion.

The site is designed to gather evidence not disseminate it. If a critic has not followed simple instructions to verify the problem exists and a multimillion dollar industry exists to prevent the believed harmless episode in business offices, what is that critic's opinion worth?

Anyone with a computer at home or a child in high school and college should be aware of this normal feature of our physiology.

L K Tucker 108.206.18.197 (talk) 02:37, 30 December 2011 (UTC)[reply]

Speed[edit]

How fast is the average squirrel, maximum? Eta-theta 18:57, 13 November 2011 (UTC)[reply]

Wolfram Alpha has no data for squirrel, but for rat, it estimates 6 mph (9.6 km/hr). Both are rodents, but they are not even in the same family - so, use caution with that estimate; in any case, I think it's a reasonable ballpark. The problem is confounded by definition - are we talking endurance-running velocity, or sprinting top-speed? Squirrels aren't known as long-distance runners. Again, use caution comparing sprint velocities between different animals under different conditions.

I found this journal article, The influence of distance to refuge on flight initiation distance in the gray squirrel (Sciurus carolinensis), quantitatively analyzing squirrel velocity in bursty sprints, as a function of distance they run, and with some analysis of their psychology. Nimur (talk) 19:12, 13 November 2011 (UTC)[reply]

That seems a bit slow, to me. Sprinting, they seem about as fast as a person, so maybe twice that fast. StuRat (talk) 19:46, 13 November 2011 (UTC)[reply]

Yes, I agree. I have seen a squirrel in the park outrun a large dog over a considerable distance (the chase started well away from trees). I am pretty sure the squirrel was going a lot faster than 6 mph. Maybe two or three times that. Of course, the squirrel had an incentive... Spinning Spark 20:05, 13 November 2011 (UTC)[reply]

This site says 12mph for "squirrel", this site says 12.4 mph (rather precisely) for Eastern Gray Squirrel. Spinning Spark 20:16, 13 November 2011 (UTC)[reply]

Interesting... Do they also have info about the airspeed of an unladen swallow? 67.169.177.176 (talk) 23:48, 13 November 2011 (UTC)[reply]

No, but this has a go. Why do I never get the easy questions like "what...is your name?" and "what...is your favourite colour?". It it it it it. Spinning Spark 00:44, 14 November 2011 (UTC)[reply]

Don't you mean an unladen flying squirrel? Clarityfiend (talk) 01:45, 14 November 2011 (UTC)[reply]

As 20km/h = 12.43mph, this seems to be a case of false precision. 109.150.130.187 (talk) 23:49, 14 November 2011 (UTC)[reply]

acidic alpha-proton in Ascorbic acid?[edit]

I note that the chiral hydrogen seems potentially acidic, because the conjugate base can form an aromaticity-stabilised compound. Does it have a pKa lower than that of water? I'm guessing this alpha-carbon would less acidic than the second hydroxyl group? Or is this carbon the second pKa? Would this be one route to interconvert between L-ascorbic acid and D-ascorbic acid? elle _{vécut heureuse} ^{à jamais} (be free) 23:14, 13 November 2011 (UTC)[reply]

Nope, the chiral hydrogen is bound directly to the carbon and therefore is not significantly acidic -- it's the vinylogous hydrogen on the second hydroxyl group that's responsible for the acidity, and the other vinylogous hydrogen gives the second pKa. FWiW 67.169.177.176 (talk) 23:46, 13 November 2011 (UTC)[reply]

What do you mean "bound directly to the carbon? The alpha-proton of ethyl acetoacetate is "bound directly to the carbon" but I its pKa is 10. elle _{vécut heureuse} ^{à jamais} (be free) 00:03, 14 November 2011 (UTC)[reply]

This Paper did some electronic calculations, and it appears that the dianion with the second deprotonation at either of the two hydroxyl groups on the side chain are of similar energy (lowest of all the dianion tautomers they calculated). Buddy431 (talk) 00:34, 14 November 2011 (UTC)[reply]

Thanks. The side-chain hydroxyls, then, not the second vinilogous hydroxyl. I stand corrected. 67.169.177.176 (talk) 01:54, 14 November 2011 (UTC)[reply]

Hmm, but they don't discuss the alpha-proton. I'll try running some calculations myself. elle _{vécut heureuse} ^{à jamais} (be free) 01:18, 14 November 2011 (UTC)[reply]