Wikipedia:Reference desk/Archives/Science/2012 September 15

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September 15

continuity equation -- why partial, but not total derivative with respect to time ?

Mathematical proofs aside, I would like to intuitively grasp the meaning of the differential form. I can see flux lines going in and flux lines going out, and quantities building up, but I can't imagine how the partial differential of say charge density with respect to time relates better to charge balance than a total derivative of charge density. In fact, I have a hard time thinking about partial derivatives with respect to time versus total derivatives wrt time, whereas I can easily conceive the difference for the x,y,z counterparts. 128.143.102.189 (talk) 04:51, 15 September 2012 (UTC)

In the article I'm not convinced about the proof [1], it moves back and forth between partial and total derivatives. For instance it uses ${\frac {\partial q(t)}{\partial t}}$ but since q(t) is only a function of time it is the same as the total derivative, it then sticks with the partial when moving from charge to charge density but fails to discuss this. I'm not sure what the right course is though, sorry. Perhaps try math desk? 94.72.221.29 (talk) 08:03, 15 September 2012 (UTC)

How about saying $q(t)=\int _{V}\rho (\mathbf {r} ,t)dV$, then take time derivative:

${\frac {dq(t)}{dt}}={\frac {d}{dt}}\int _{V}\rho (\mathbf {r} ,t)dV$.

Now shrink the volume V to a very small blob centred on $\mathbf {r}$, we only want $\rho$ at the point $\mathbf {r}$ so we fix it with a partial deriv, hold $\mathbf {r}$ constant and just diff wrt to time:

${\frac {dq(t)}{dt}}={\frac {\partial }{\partial t}}\rho (\mathbf {r} ,t)\delta V$.

This is obviously very lazy, non-rigorous and probably wrong - you really need someone who knows what they are talking about. 94.72.221.29 (talk) 08:13, 15 September 2012 (UTC)

I tried to clarify continuity equation. Ruslik_Zero 13:12, 15 September 2012 (UTC)

Okay, what happens if you take the volume integral of the total time derivative of density? What would this generate and what would its physical meaning be? What would change about the integral and what could I add to the continuity equation to make it balance? 71.207.151.227 (talk) 22:28, 15 September 2012 (UTC)

Hmm, from the above explanation I kind of get it, but not completely. 71.207.151.227 (talk) 22:41, 15 September 2012 (UTC)

If you want co calculate the total time derivative of density you will need to specify how coordinates depend on time. The integral of this value does not have any physical meaning that I am aware of. Ruslik_Zero 11:48, 17 September 2012 (UTC)

Question about QM

I read this

Quantum mechanics is a result of human thought which is a mental analog to the microscope or telescope. It allows us to see external reality in a certain way due to the details of its construction but that very construction introduces distortions and filters that mean we are not seeing actual reality. It is a useful tool which produces "pictures" of the external world but again the pictures are not the world but a description of it.

Is it right? Bubba73 ^{You talkin' to me?} 05:05, 15 September 2012 (UTC)

When I was in the classroom teaching various high school science classes, I used to start every year in all of my classes with This painting by René Magritte. The lesson being that of course that isn't an apple. It is a picture of an apple. Scientific theories aren't apples. They are pictures of apples. Some are really nice pictures of apples. But at no point does a picture of an apple actually become an apple. I had an organic chemistry professor who said once "Atoms do what atoms do. If our models don't accurately portray what they do, it's because our models are not atoms." In a sense, all of the knowledge in your head consists of theories and models. You don't experience all of the world in reality, what you experience is the world filtered through your brain and made into models. It is all models. When I say "tree" you have a picture in your head of some sort of tree, and you easily recognize trees as trees, and not as, say, cats. That's because you have a working model of what a tree is, a "theory of a tree" in your head. When you see a never before seen object, you are able to instantly compare it to all of the models you have constructed for yourself, and decide "this is a tree" "this is a book of poetry" "this is a rock song I am hearing", because you atomatically create theories and models all the time to make your world comprehensible. Much of science is about extending our ability to make models beyond our own senses. You aren't equipped to experience an "atom" with your own senses. So what science does is develop those theories and models that we can't do automatically, because we can't use our own senses to probe nature to that level. That's all QM is: just as you have developed an internal "theory of tree" that is an idealized tree that allows you to identify an object as a tree, QM is an idealized portrait of physics that allows us to better elucidate exactly what is going on in, say, the atom. It is a tool for understanding the inner workings of matter. But just as Magritte's painting of the apple is not an apple, QM is not reality, but a picture of reality. All pictures are, by necessity, approximations of reality, and QM is an approximation. A really good approximation, but an approximation nonetheless. If you want to explore these ideas further, philosophy of science and philosophy of mind are good places to look for insight. --Jayron32 05:31, 15 September 2012 (UTC)

So QM is really no different from other scientific theories in that regard, right? Bubba73 ^{You talkin' to me?} 05:44, 15 September 2012 (UTC)

No, that's what a theory is. A theory is a tool for studying a subject. It is an explanatory framework, a way of elucidating truth from some aspect of reality. Not all theories are equal, of course. QM is a particularly productive and useful theory, and it is very accurate, especially compared to the theories it supplanted, but it is a theory. It also isn't a complete picture of reality, for example it doesn't really deal with gravity in a satisfactory way. But what it does do well, it does very well, such as explaining the structures and interactions within and between atoms. --Jayron32 05:48, 15 September 2012 (UTC)

So the paragraph I quoted really applies to any scientific theory, not just QM, right? Bubba73 ^{You talkin' to me?} 16:00, 15 September 2012 (UTC)

Not to confuse you on this point: Yes. --Jayron32 03:13, 16 September 2012 (UTC)

I'd change that to "any model". A model, by definition, is a simplification of the real thing, such as a globe being a simplification of the Earth. StuRat (talk) 20:39, 15 September 2012 (UTC)

At the risk of politicizing this, I'd use the example of Global warming and the oft-heard retort that the predictions are "just the results of models". Of course, every scientific theory is, as you guys say, a model of a real-world process which is much more complicated. The belief that the sun will rise tomorrow is "just" the result of such a model, for another instance.Gzuckier (talk)

George Box is often quoted on this point: "All models are wrong, but some are useful." -- 205.175.124.30 (talk) 21:38, 15 September 2012 (UTC)

He was actually speaking of Kardashians. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:57, 16 September 2012 (UTC)

That'd be the some that aren't useful. --Jayron32 03:14, 16 September 2012 (UTC)

Depends on the use. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:01, 16 September 2012 (UTC)

* You get a copy of Medeis's star for that one Bugs. Definately the best answer of the day. --Jayron32 19:12, 16 September 2012 (UTC)

Resolved

Bubba73 ^{You talkin' to me?} 22:48, 16 September 2012 (UTC)

The return of the aether

So, just reading through some physics stuff, and let me state up front that I know next to nothing about real physics. But I was reading about the Superfluid vacuum theory, and it got me thinking, didn't I see this stuff before? Oh, yeah, Luminiferous aether. Near as I can tell, they are both ways of modeling a vacuum as a physical substance for explaining quantum-level behavior. So, any of the physics people out there. Using small words, can anyone explain how the Superfluid Vacuum Theory isn't just Luminiferous Aether wearing a new coat? --Jayron32 06:04, 15 September 2012 (UTC)

Not knowing much about the Superfluid Vacuum theory (SFV), I'll venture that the crucial difference is that the aether establishes an absolute referential frame while the SFV doesn't. Dauto (talk) 15:40, 15 September 2012 (UTC)

It appears that it does have an absolute rest state. More to the point, googling the phrase "superfluid vacuum theory" doesn't turn up any hits except copies of the Wikipedia article and a couple of random message board posts. I think the article should be deleted, or at least renamed to "superfluid models of the vacuum", edited to make it clear that hardly anyone believes in this stuff, and delinked from the many other articles that the page's creator added it to. -- BenRG (talk) 17:52, 15 September 2012 (UTC)

In otherwords, it is likely mostly bullshit. Thanks for that. I was skeptical, given that most mainstream physics texts don't light on this at all, which is why I asked the question. If BenRG hasn't heard of it, I take that as a good sign that it's hokum. Thanks Ben. --Jayron32 03:11, 16 September 2012 (UTC)

deer

can you get lyme disease from hunting and skinning deer? How big is the risk? Have there been any documented cases of this happening?--Wrk678 (talk) 11:13, 15 September 2012 (UTC)

I'll definitely say that yes. Deer ticks are directly related to Lyme disease, and you'll be hunting in the habitat of deers. Dying might also be the perfect occasion for ticks to leave it's host and search for a new, warmer host, like a hunter. OsmanRF34 (talk) 12:59, 15 September 2012 (UTC)

Some links for you: deer tick, Lyme disease. StuRat (talk) 14:33, 15 September 2012 (UTC)

Given that the geographic range of the deer tick is only the eastern USA, can one say that the answer to the OP's question is "No" if outside that area? HiLo48 (talk) 16:57, 15 September 2012 (UTC)

To the hunting part, yes, but a deer could conceivable be taken outside that region before being skinned. I'm not sure how long the ticks remain after the deer's death. StuRat (talk) 20:33, 15 September 2012 (UTC)

The insect hangs around the dead deer for only about 4 seconds. Hence the old saying, "... 5, 6, pick up ticks." ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:00, 16 September 2012 (UTC)

Not to contradict the other answers, but perhaps to clarify them, the only recognized mode of transmission is by the bite of a tick, or (much less likely) some other biting insect. So the only way it could happen is if your activities led you to pick up ticks. Looie496 (talk) 17:26, 15 September 2012 (UTC)

Keep also in mind that even if you never see a deer while hunting, running around in deer habitats will likely expose you to deer ticks. Whether you get bitten or not depends on whether you've taken proper precautions in your attire and etc. (I once brushed against some low-hanging tree branches in the northeast and managed to pick up no fewer than 10 ticks. Psychologically disturbing to keep finding them on me, but none of them bit me.) --Mr.98 (talk) 23:19, 15 September 2012 (UTC)

Not only that, but as the article Lyme disease linked to above shows, one can get it anywhere in the Northern Hemisphere, not just the Eastern US from eastern deer ticks. See http://www.lymedisease.org/california/california_map.html To answer the question as worded, you cannot get the disease simply from hunting or skinning a deer per se, there needs to be blood or bodily fluid transfer, the chance of that from a dead deer is miniscule, but this being biology, anything is imaginable. μηδείς (talk) 01:09, 16 September 2012 (UTC)

PS, the link "deer tick" Hilo linked to above was a misleading redirect to Ixodes scapularis when he posted it. I have changed it to a disambiguation page. In California the deer tick is Ixodes pacificus. μηδείς (talk) 04:31, 16 September 2012 (UTC)

Are there no ticks in Australia? In Europe, Ixodes ricinus is sometimes called the "deer tick" or "sheep tick". I'll add this to the disambiguation page. Dbfirs 07:02, 16 September 2012 (UTC)

how likely is it a deer tick will bite you when skinning a deer? is it true the "insect hangs around the dead deer for only about 4 seconds"  ? i live in the northeast--Wrk678 (talk) 17:05, 17 September 2012 (UTC)

According to the article, males mostly just nest in deer to be near the females, and aren't aggressive feeder. They would probably stick around until the body temperature dropped. Females will probably continue to feed so long as the blood is warm enough and, especially, not coagulated. The notion they would leave the host in four seconds is absurd, death itself is a long process, consider resuscitation. That's pretty much magical thinking. That doesn't mean some ticks might not change hosts regardless of their "knowledge" of the deer's death. μηδείς (talk) 18:03, 17 September 2012 (UTC)

Conduction of heat and electricity.

Is there any material that does not conduct electricity but conducts heat and is easily available? — Preceding unsigned comment added by Daivatb (talk • contribs) 11:29, 15 September 2012 (UTC)

Aluminium nitride. -- Finlay McWalterჷTalk 11:38, 15 September 2012 (UTC)

Pure water (must be really pure, though, as even a tiny contamination makes it electrically conductive). Being a fluid, water also has the ability to transfer heat by convection, which can be even more effective than conduction. StuRat (talk) 14:31, 15 September 2012 (UTC)

Diamond. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:49, 15 September 2012 (UTC)

Glass (moderate conductor of heat). StuRat (talk) 15:03, 15 September 2012 (UTC)

The material of choice with such properties is sapphire, which is much less exotic than diamond. Ruslik_Zero 16:36, 15 September 2012 (UTC)

Diamond is not "exotic"! Whoop whoop pull up ^{Bitching Betty | Averted crashes} 17:08, 15 September 2012 (UTC)

Unless you only need tiny sizes, diamond isn't likely to be "easily available". StuRat (talk) 20:27, 15 September 2012 (UTC)

By those standards, neither is sapphire or ultrapure water. And glass is a mediocre conductor of heat. Whoop whoop pull up ^{Bitching Betty | Averted crashes}

I already stated that about glass. Both large sapphires and pure water are easier to obtain than large diamonds. StuRat (talk) 23:29, 15 September 2012 (UTC)

Without knowing the context or the application, it is impossible to properly answer this question. In many applications, the actuall degree or heat conduction is unimportant. An example: In electronics, power transistors need cooling - this is most often done by bolting them on to a heatsink (an aluminium piece with a large surface area), with a mica washer between the transistor and the sink. The washer conducts the heat from the transistor, but electrically insulates it. Mica is NOT a good conductor of heat, but it is a VERY good electrical insulator. In this application it works very well heat-wise, because it is very thin, typically about 0.2 mm over an area of 400 mm². Distilled water has been used as a coolant in very high power radio transmitters, exposed to high voltages. As said above, the electrical properties with commonly available distilled water are let's say rather poor. But when you are shifting 100 kW of heat, a few watts lost due to water electrical conduction doesn't matter. Keit120.145.29.139 (talk) 03:06, 16 September 2012 (UTC)

No one has mentioned Transformer oil?? Vespine (talk) 23:05, 16 September 2012 (UTC)

Good call. Any mineral oil is an electrical insulator. However it still depends on the application. The electrical insulation properties of transformer oil are excellent. But is not as good as water at conducting heat. It's good at DC and power frequencies, but at radio frequecies the dielectric loss in transformer oil is worse. That's why transformer oil is used in power transformers, but water has been used in high power radio transmitters. Ratbone58.164.237.102 (talk) 00:26, 17 September 2012 (UTC)

And the silicone grease sold as heat sink compound. Gzuckier (talk) 04:03, 17 September 2012 (UTC)

Actually, no. Heatsink grease is intended to fill microscopic voids between two hard surfaces. These microscopic voids arise because of normal "mill finish". Without the grease there would be in effect small islands of metal contact surrounded by airgaps, thermal conductivity 0.024 W/m.K - so the thermal conductivity does not need to be very good to make a usefull improvement in the intended application. It is a common misconception among electronics hobbyists and radio hams that heatsink greases are good conductors of heat. A typical grease is about 0.9 to 2.5 W/m.K, contrasted with aluminium at 250 W/m.K, or carbon steel at 40 to 50 W/m.K. Instructions for using heatsink grease emphasise the need to apply a THIN smear so you just fill the voids. Keit121.215.23.108 (talk) 04:41, 17 September 2012 (UTC)

I remember this test from years ago where Dan discovered that toothpaste and Vegimite perform just about as well as most commercial heatsink compounds. Of course their problem is they dry out quite quickly, but it's still very funny. Vespine (talk) 22:39, 17 September 2012 (UTC)

Height of the nail

Secondary growth results in an increase in diameter. Obstructions, both foreign objects such as this metal post, and parts of the plant, such as stubs of limbs, can be "swallowed" by continued growth.

A nail is inserted in the trunk of a tree at a height of 1 meter from the groung level. After three years the nail will remain at the same position or it will rise above. Why? Sunny Singh (DAV) (talk) 16:29, 15 September 2012 (UTC)

I believe it depends on the type and age of the tree. Many mature trees grow mainly at the tips of branches, while younger trees can also grow elsewhere. StuRat (talk) 16:34, 15 September 2012 (UTC)

Husband and Wife crab apple (Malus sylvestris) at Lynncraigs farm, Dalry, North Ayrshire, Scotland.

In almost all cases (I can't think of an exception, unless you are dealing with the tree's first season of growth) it will remain at the same height from the ground, but will appear to have sunk into the tree. This is because of the difference between primary and secondary growth. Primary growth is the (normally ) yearly extension of the growth of the tips of the trunk and then branches. Only this yearly growth at the tips contributes to an increase in height and length. Secondary growth in rings adds girth to the previous years' primary and secondary growth. This secondary growth will cause the growth of three new rings external to the point to which the nail had originally been sunk. Eventually the nail will be engulfed by new rings and disappear, which is why the terrorist practice of "spiking" trees is so dangerous. See also meristem on the growth of plants and especially intercalary meristem which is found in grasses and very few other plants and is the only mechanism by which a stem could grow lengthwise in its middle. μηδείς (talk) 20:53, 15 September 2012 (UTC)

Witness the unfortunately common sight around my yard of the tree which has grown up under a chain link fence and engulfed it; the fence is not pulled up in the process. Gzuckier (talk) 04:12, 17 September 2012 (UTC)

Why would driving spikes into trees (that's what I assume it means) be at all dangerous? Objects driven into a tree, unless they completely surround the trunk and girdle the tree, are harmless to the tree, unless they disrupt the sap flow immediately below a branch, in which case the only thing that dies is that particular branch. Eventually, the spike or nail or whatever will be completely engulfed by the tree, and it will become an inclusion in the tree, doing no more harm to the tree than a lithopedion does to a human. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 22:22, 15 September 2012 (UTC)

Environmental activists put spikes in trees so they will destroy the chainsaws of anyone trying to cut them down. Rather than risk this, it is presumed that the lumberjacks will leave those trees alone. However, if the lumberjack doesn't know they are there, it could also cause the chainsaw to "buck", hit the lumberjack, and injure or kill him. Not likely, but possible. StuRat (talk) 22:30, 15 September 2012 (UTC)

(After EC, but with references) The danger is not to the tree, but to loggers. The idea is to make logging more dangerous than it already is, and thereby potentially stop logging operations due to safety risks. When a large chainsaw hits a hidden spike in a tree, the chainsaw reacts violently, and can harm the operator. I think this practice came to national attention in the USA during controversy surrounding protecting spotted owl habitat from timber harvesting. See Tree_spiking, eco-terrorism, Spotted_owl#Conservation, or google similar terms. SemanticMantis (talk) 22:32, 15 September 2012 (UTC)

According to the tree spiking article, only one injury has ever resulted from tree spiking, and even that injury is dubious. A lumberjack is many orders of magnitude more likely to win the lottery than to die from tree spiking. --140.180.247.208 (talk) 22:28, 16 September 2012 (UTC)

Now that you mention it, my admittedly limited chainsaw experience would suggest that if the nail is imbedded enough to not be visible the saw will just cut through it while making a lot of fuss and getting dull; if the nail were loose enough to get dislodged and fired into space it would be visible to the observant chainsawist.Gzuckier (talk) 04:12, 17 September 2012 (UTC)

Oh. I thought Medeis meant dangerous to the tree. Apparently not. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 22:34, 15 September 2012 (UTC)

They always told us it was harmful to a tree to put nails into it, peel bark off it, etc. I don't know about that. But it's safe to say that applying a chainsaw to a tree is liable to endanger it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:00, 16 September 2012 (UTC)

Odd, there doesn't seem to be any good schematic drawing of this at all on the web. Much of the imagery of secondary growth seems to be related to pot plants, which don't really exhibit it in the sense of wood. But see figure three here for a hint at how a tree grows: http://www.cmg.colostate.edu/gardennotes/771.html μηδείς (talk) 21:12, 15 September 2012 (UTC)

The caption to the image raises an interesting question: can a growing tree engulf a living part of another plant, such as a branch from another tree that was pressed against the trunk of the tree in question? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 22:39, 15 September 2012 (UTC)

Yes, and some vines/trees intentionally engulf others, even going so far as to smother them. StuRat (talk) 23:35, 15 September 2012 (UTC)

For which see strangler fig and husband and wife trees. μηδείς (talk) 23:38, 15 September 2012 (UTC)

Thanks for the spiking reference, had dinner not intervened I meant to come back and find the WHAAOE link. Interestingly there is an herbicide called Spike 80DF that has famously been used to murder trees, but I was talking about the saw-sabotaging practice mentioned above. μηδείς (talk) 23:38, 15 September 2012 (UTC)

One can kill a tree, but one cannot "murder" a tree. Murder is the unlawful taking of human life. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:53, 16 September 2012 (UTC)

Remind me not to let you water my plants while I'm away. μηδείς (talk) 02:57, 16 September 2012 (UTC)

It's deja vu all over again. In any case, I try to keep my plants well watered. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:00, 16 September 2012 (UTC)

Yes, I can easily imagine you murdering my Euphorbias with your careless oversaturation. Or at least negligently manslaughtering them. μηδείς (talk) 03:08, 16 September 2012 (UTC)

That would be plantslaughtering them. -- ♬ Jack of Oz ♬ ^{[your turn]} 03:34, 16 September 2012 (UTC)

Even if they end up in a persistent vegetative state ? StuRat (talk) 04:01, 16 September 2012 (UTC)

Best...comeback...ever. (at least in this thread) μηδείς (talk) 04:24, 16 September 2012 (UTC)

Keep in on life support, as with the famous Boston I.V. ←Baseball Bugs ^{What's up, Doc?} carrots→

So, Sunny, who won the bet? μηδείς (talk) 19:17, 16 September 2012 (UTC)

I think you. Sunny Singh (DAV) (talk) 17:17, 17 September 2012 (UTC)

(I assumed your question was a bar-bet type one. μηδείς (talk) 02:14, 18 September 2012 (UTC))

Speed on one point

To measure speed, do you really need to measure where you are in two points? I mean, on point p you also have speed, so could you measure this speed that you have just with data from this point? OsmanRF34 (talk) 21:36, 15 September 2012 (UTC)

Speed is by definition distance travelled over time so if you just have one point you have no way of knowing your speed; in other words, if you only know you are at exactly 20 m from your front door at exactly noon, what is your speed? Who knows? You could be in the middle of just standing there, or running away, or driving away, or walking in circles, or any other action. However, if you know that at ten to noon you were at your front door, then you could calculate your average speed to be 20 m / 10 min or 2 m/min or 0.12 kph - there is still a problem with this, as in the intervening time you could have walked the 20 m directly, if you are a very elderly person or some kind of small animal, but you could just as well have gone around the house, or gone down the block, or gone halfway to the sun and come back if you happen to be a photon under the influence of some weird gravitational field - but at least you will have some idea of your "speed".

Mathematically speaking, given a well-defined, differentiable position function p(t) of time t it is possible to find instantaneous speed exactly at any time, by taking the derivative p'(t) with respect to t. This is because in such a case you have or can calculate an infinite number of data points, so calculus "works". If you can model a physical phenomenon with some such function, such as the falling of an apple from a tree, then indeed you can find the speed with just one data point, the time after it has fallen. However, you will probably be slightly off due to imprecision in determining air resistance and other variables.

In real life, you don't have a magical formula that can tell you exactly where you are at any arbitrarily precise given time, so you must use an average, hence you need more than one data point. 24.92.74.238 (talk) 21:47, 15 September 2012 (UTC)

I don't buy that. I can theoretically measure your speed by (for example) placing an immovable obstacle in your path. When you splatter into the object, I can (in principle) calculate the rise in temperature, and knowing your mass and the mass of the object you struck, deduce the amount of kinetic energy that was involved and use that to deduce your speed at the instant of impact. This is a somewhat destructive approach - but it points the way to doing the measurement at a single point. In practical terms, a police "speed gun" does a measurement at a single point by measuring the doppler shift of reflected radio or light waves from a moving object. SteveBaker (talk) 22:06, 15 September 2012 (UTC)

The measure doppler shift you need to measure the signal over a period of time in order to determine its frequency. Indirectly measuring the kinetic energy of an object will also take some time. To measure the speed at a single point, you need to measure it in a single instant, which you can't do. --Tango (talk) 23:18, 15 September 2012 (UTC)

IP 24 and Tango are correct, point in space equates with instant in time. Doppler shift and splatter both imply change over time. I am not quite sure how to reference this other than suggesting a basic course in physics. μηδείς (talk) 23:27, 15 September 2012 (UTC)

Uh—Speed#Instantaneous speed? Isn't it easy to find the speed of an object at one instant in time by using calculus? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 23:46, 15 September 2012 (UTC)

Yes, but the calculation is still based on a measurement over time/distance. μηδείς (talk) 00:04, 16 September 2012 (UTC)

Medies seems right on this. If you think about the practical method of finding "speed" of a moving object, even instantaneous speed, involves making a large number of discrete time-and-distance measurments to give you points on a graph, and then you fill in the "gaps" via interpolation. You can then take the derivative of that curve, but to build the curve you still have to connect the dots at some point. Or you could just use a camera and a speedometer, which measures speed continuously. --Jayron32 03:06, 16 September 2012 (UTC)

Depends on how seriously you mean "measure". One could calculate the speed by the integral of the acceleration, for instance; you could use acceleration data over all previous points but you could also use the "timestamp" at the one particular time and integrate over all previous times, thereby avoiding the "measure" of more than one point, at least directly. Or a theoretical speed based on some measure of forces involved. But a direct measurement in the real world always means just an average calculated from the difference in time and the difference in space between two points. Gzuckier (talk) 04:21, 17 September 2012 (UTC)