# Wikipedia:Reference desk/Archives/Science/2009 June 10

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# June 10

## Long term stability

From this site they said about :"long term stability time, what's the collisison of mercury, Earth, and Mars even mean, and what is the eccentricity of inner planets, and what the section mean about Pluto's eccentricity?--69.226.38.106 (talk) 01:01, 10 June 2009 (UTC)

The orbits of the planets (see orbital eccentricity) could change enough after billions of years for some of them to smack into each other (i.e. collide). Clarityfiend (talk) 02:29, 10 June 2009 (UTC)
In physics, there is a concept known as the n-body problem which basically states that for any value of n>2, the reliability of predicting the long-term motion of interacting bodies quickly drops to zero. In other words, where you have any group of objects which are all interacting with each other, if there are more than 2 objects in the group, their behavior is chaotic and their motion cannot be reliably predicted in the long term. In the short term, planets in our solar system behave like a 2-body problem (i.e. the planet itself and the Sun) since the gravitational effects of each planet on other planets is small, so on the order of, say, a few decades or a few hundred years, we can fairly accurately predict the motion of each planet by ignoring interplanetary gravitational effects. However, over time the system becomes less and less predictable because, over time, the planets gravitational effects on each other will sort of "add up" resulting in a system which diverges greatly from our predictions after several millions of years. What this all means for that article is that, while it looks like now the planets are in stable orbits, in 10-20 million years there is no way to predict how they will behave. We can be pretty sure the planets are not going to smash into each other simply by the fact that they have SO much space to move around in that even if their behavior becomes entirely unpredictable, it still only results in an infinitessimally small chance of collision between two of the major planets. However, what the chaotic behavior of the solar system means is that while we can be fairly certain the planets won't smash into each other, we can make no reliable predictions about exactly WHERE they will be in say another 50 million years. --Jayron32.talk.contribs 02:41, 10 June 2009 (UTC)
What I think they are saying is that while the average distance of these planets from the Sun may not change by much - the orbits may become more oval...less circular. If that's enough to make the orbit of one planet cross the orbit of another - then KAPOWW!!! - a very big mess! However, the timescale for these events is in the billions of years - more than you, I or humanity in general need worry about. Many of the events described are not due to happen for 5 billion years - and that's about as long as the Sun will last. So it's really kinda unimportant in the grand scheme of things. SteveBaker (talk) 02:54, 10 June 2009 (UTC)
Surely you aren't insinuating that five billion years and the ultimate fate of the sun and earth are unimportant in the grand "scheme of things", just because it's so much longer than a human lifespan timescale? I had no idea the grand scheme was so human-centric! Nimur (talk) 15:05, 10 June 2009 (UTC)
The long term stability of planetary orbits over a time period longer than the star they are orbiting will survive in its present form is pretty unimportant. When the sun throws off its outer layers and becomes a red giant the orbits will drastically change anyway, so stability is hardly an issue. --Tango (talk) 17:13, 10 June 2009 (UTC)
Here is a recent article from Space.com detailing the possibility of future collisions of inner planets as the Sun changes in size, with even the possibility of destabilizing the entire inner solar system. ~AH1(TCU) 20:27, 10 June 2009 (UTC)
This site say possible collision between Mercury and Earth, or Venus and Mars with Earth. Since Mercury is the smallest, then when it collides with Earth or Venus, then Mercury might be torn apart. If Mars collides with Earth or Venus, then Mars could be torn apart too. When collision comes, usually small one is gone, the big ones survvies.--69.226.38.106 (talk) 23:31, 10 June 2009 (UTC)
For the possible effects of a planet colliding with Earth, see this calculator. For other solar system objects, here is a less accurate and more informal calculation. Assuming the present average orbital speed of the planet, and neglecting that of the Earth's, as well as both planets' escape velocities, and assuming a 20-degree angle at collision, Mercury melts all of Earth and changes its rotation period/axial tilt, Venus destroys the Earth, and Mars melts just over half of Earth, disturbing it rotation and tilt. It appears the higher result for Mercury is caused by its density and speed. ~AH1(TCU) 01:30, 11 June 2009 (UTC)
They said in this article about Pluto and neptune orbital can change in 10-20 million years nobody will know what will happen to Pluto, but I don't think Pluto will ever crash into Neptune, that's Triton. It said Pluto and Neptune is 3:2 ratio, then Pluto might just get closer to Eris or I don't know. Since Moon and Earth is tidally lock, Galilean moons of Jupiter is also tidally lock, does that mean Galilean Moons is moving further away from Jupiter, sometime, Galilean moon will just eject from Jupiter. Is Titan also moving away from Saturn.--69.226.38.106 (talk) 02:06, 11 June 2009 (UTC)
• I honestly don't know anything about Long term stability of outer planets. I have absolutely no ides what they talking about Galilean moon vs. Jupiter. This is about the most clean I can ask question. I need some answers.--69.226.38.106 (talk) 23:03, 11 June 2009 (UTC)

Where's the mistake in this formula?

${\displaystyle i\hbar =i\hbar {\frac {dp}{dp}}=i\hbar {\frac {d}{dp}}p=xp=x(-i\hbar {\frac {d}{dx}})=-i\hbar (x{\frac {d}{dx}})=-i\hbar ({\frac {d}{dx}}x-1)=-i\hbar (1-1)=0}$

70.26.154.226 (talk) 04:48, 10 June 2009 (UTC)

Is dx/dx the first derivative of x with respect to x? It looks like you are confusing differentiation with multiplication in some places. --Jayron32.talk.contribs 05:49, 10 June 2009 (UTC)

If you make the substitution ${\displaystyle x\rightarrow i\hbar {\frac {d}{dp}}}$ you tacitly assuming to be working in the momentum space. Notice that here ${\displaystyle x\,}$ is an operator which therefore requires an state function ${\displaystyle \psi \,}$ to operate on. That factor is conventionally left out but you must keep in mind that it really is there. So, by ${\displaystyle {\frac {dp}{dp}}}$ we actually mean
${\displaystyle {\frac {d(p\psi )}{dp}}=({\frac {dp}{dp}})\psi +p{\frac {d\psi }{dp}}=\psi +p{\frac {d}{dp}}\psi }$
Or, after droping the ${\displaystyle \psi \,}$ according to convention
${\displaystyle {\frac {dp}{dp}}=1+p{\frac {d}{dp}}}$, and similarly we have
${\displaystyle {\frac {dx}{dx}}=1+x{\frac {d}{dx}}}$.
You used that last one correctly in your sixth step but than forgot about it in your seventh step. The right sequence is
${\displaystyle i\hbar =i\hbar 1=i\hbar ({\frac {dp}{dp}}-p{\frac {d}{dp}})=xp-px=x(-i\hbar {\frac {d}{dx}})-(-i\hbar {\frac {d}{dx}})x=-i\hbar (x{\frac {d}{dx}}-{\frac {dx}{dx}})=-i\hbar (-1)=i\hbar }$
Dauto (talk) 05:52, 10 June 2009 (UTC)
Oh, that makes sense. 70.27.198.174 (talk) 17:25, 10 June 2009 (UTC)

## Telecommunication in ships

Admiralty List of Radio Signals is a book from the HMSO which may be of use. Inmarsat could also be relevant. I will look for some more great sources later! Graeme Bartlett (talk) 07:17, 10 June 2009 (UTC)
What Wikipedia page did you go through? Here are some I found in Category:Maritime communication - Marine and mobile radio telephony, Maritime Mobile Service Identity, Vessel traffic service, Ship Security Alert System, Automatic Identification System, Global Maritime Distress Safety System, Long Range Identification and Tracking. Jay (talk) 07:21, 10 June 2009 (UTC)

Is a head on crash, at say 70 km/h, between 2 cars the same as one car hitting a solid stationary object at 140 km/h? I can't seem to get a definitive answer from myself. My preferred opinion is that it's not, because the change in momentum of the first situation car is half that of the second car, is it not? 124.169.131.68 (talk) 05:14, 10 June 2009 (UTC)

OK, I'll kick off the discussion. Yes, it is the same, if the 2 cars have the same speed and mass. Edison (talk) 05:31, 10 June 2009 (UTC)
If the collisions are perfectly elastic they are the same. In reality, car crashes, whether with other cars or stationary objects, are inelastic - lots of the energy goes into crumpling, heat, and other forms of energy loss. I think the essential question here though is: which scenario is more dangerous to the people in the car? Dcoetzee 05:43, 10 June 2009 (UTC)
(EC)Probably not; but the "solid stationary object" is vague. If a car at 140 km/h hits a parked car it would have roughly the same effect as two cars hitting head on at 70 km/h, asuming the parked car was immobilized, such as resting against a wall. However, if the car is hitting a wall directly at 140, the wall will deform less than a car would, so the car in this case would suffer more damage than hitting the parked car, since the deformation of the parked car would absorb some of the momentum of the collision; the wall will not. --Jayron32.talk.contribs 05:46, 10 June 2009 (UTC)
In fact, for either the head-on crash or the crash into a strong and solid wall, a totally inelastic collision is a much more realistic approximation than a totally elastic one. Which means that all the kinetic energy is converted to other forms, and a lot of it will go into damaging the car. Well, a car at 140 km/h has four times the kinetic energy of one at 70 km/h, so the 140 km/h crash releases twice as much energy as the head-on.
As Jayron says, a crash into a similar but parked car at 140 km/h is different -- that car will be put into motion by the collision, so it's not totally inelastic. --Anonymous, 05:55 UTC, June 10, 2009.
Thats not quite correct, assuming that the two collisions involve the same bodies then they will be the same. In accordance with galilean relativity, which is still accurate for small speed approximations such as this, all that matters is the two vehicles speed with respect to their centre of mass, which is the same in both cases. If however one car is held stationary (such as resting against a wall), then that will cause the collision to be different. This misunderstanding relates to the fact that the elasticity of a collision is not defined by the bodies final speed, or kinetic energy, in an arbitrarily chosen lab frame, but their seperation speed. Elocute (talk) 10:06, 10 June 2009 (UTC)
Another in the "no" category. Assuming that the two cars in the 70 kph head-on collision are the same (important for simplification as discussed above), let's consider what happens to car A: it's traveling 70 kph when its bumper hits car B, and at that point the bumper stops moving -- the crash occurs at that point where momentum cancels out. The equivalent single-car system is for car A to hit an ideal non-moving concrete wall at... 70 kph. The energy of car B in the original system can be safely and entirely ignored. If you're not persuaded by this, consider the original system but place an ultra-thin ideal concrete wall between the two cars -- the collision will be the same as if no wall existed. If you can place that ideal wall, though, then you can ignore what happens on the other side of the wall. — Lomn 11:27, 10 June 2009 (UTC)
It is not true to say that these situations are different, it is an extremely outdated idea to prescribe any significance to particular reference frames. Any apparent extra energy that you have in one frame cannot be used as it is recquired to keep the centre of mass of the system in constant motion. The laws of physics are the same for all reference frames and that is a tenet of modern science that is applicable to all situations. As long as the car approach speeds are the a constant the same collision will occur, regardless of the distribution of this approach speed between the car. Elocute (talk) 12:36, 10 June 2009 (UTC)
Er, what? You may want to clarify by tying your comments to a particular portion of mine. — Lomn 12:56, 10 June 2009 (UTC)
I'm just saying the two situations are the same. And then justifying that. I interpreted you post to say the situations were different, and so believed it recquired refuatation. Elocute (talk) 10:02, 11 June 2009 (UTC)
This is an old question - and we've discussed it here at least once before. The kinetic energy of the crash between two identical cars, hitting head on is twice that of one car hitting a brick wall. However, that energy is absorbed by crushing and crumpling the metal of the car(s) and with two cars, you have twice as much 'crumple zone' to absorb that energy - so we'd expect the damage to each car and the forces applied to the drivers to be exactly the same as with one car hitting a wall. That, however, assumes a totally solid wall which takes no damage whatever in the impact. That's really impossible - there are no infinitely rigid objects - so in practice, because the wall absorbs some of the energy in the single-car case, you're likely to be slightly better off hitting the wall than hitting another car. If the stationary obstacle is something that deforms or can be pushed out of the way (like a stationary car, for example) then the benefits of hitting the stationary object just get better still. SteveBaker (talk) 12:35, 10 June 2009 (UTC)
Note: the above is for a wall collision at 70 kph. Hitting a wall at 140 kph, for any substantially recognizable definition of "wall" (that is, not "a wall made of marshmallow creme"), is going to be worse. — Lomn 12:56, 10 June 2009 (UTC)
Steve hit my point. If the passengers were in perfectly rigid billiard balls with no way of seeing out, in a collision they would have no way of telling whether they hit a wall at 70 km/h or another ball going the other direction at 70 km/h, yes?124.169.131.68 (talk) 13:42, 10 June 2009 (UTC)
Yes, agreed, but in the original question that started this thread, they hit the wall at 140 km/h, which is very different from a 70 km/h head-on collision, even though relative velocities before the collision are the same in the two scenarios. Gandalf61 (talk) 13:47, 10 June 2009 (UTC)
Yes, but the difference is solely due to the difference between a car and a wall, not differences in velocities. A head on collision between a car at 70km/h and a wall (somehow) moving at 70km/h would be the same as a car hitting a stationary wall at 140km/h. --Tango (talk) 17:16, 10 June 2009 (UTC)
Gandalf, if the same objects have the same closing speed, the collision is the same. Elocute (talk) 10:00, 11 June 2009 (UTC)
Nonsense. True but only relevant if you assume the "solid stationary object" in the original question is another identical car, which isn't stated. See my response below. Gandalf61 (talk) 22:23, 11 June 2009 (UTC)

### Clarification

The appropriate scenarios for comparison are:-

• A: Two cars approaching each other on a straight line, each travelling at 70 kph.
• B: One car travelling at 140 kph directly towards the front end of a stationary car.

The damage taken by the cars depends on the impulse (change in momentum) applied to the cars. (Technically, the maximum force applied is probably more relevant, but this will be proportional to the impulse).

In scenario A, car 1 has momentum +70m (where m is the car's mass), and car 2 has momentum -70m (let's assume that our base unit for speed is kph). Due to crumpling of the cars, the collision is likely to have a low coefficient of restitution, leading to a (relatively) inelastic collision. Momentum must be conserved. The momentum from the two will tend to cancel each other out, causing them to come (approximately) to rest together. The impulse applied to car 1 is -70m. The impulse applied to car 2 is +70m.

In scenario B, car 1 has momentum +140m. Car 2 has momentum 0. Again, let's assume an inelastic collision. The combined mass of the two cars is 2m. Momentum must be conserved. Therefore the velocity of the combined cars is +140m divided by 2m, which is +70 kph. The impulse applied to car 1 is +140m - (70m), which is -70m. The impulse applied to car 2 is 0 + (70m), which is +70m.

In both scenarios, the impulses received by the cars are the same. Therefore the damage sustained should be the same.

In the third scenario, where a car travelling at 140 kph hits a stationary wall, we haven't defined the mass of the wall (and ground to which it is immovably attached). Let's assume that the wall's effective mass is a lot higher than that of the car (otherwise it wouldn't be an effective wall). The initial momentum of the car is +140m. The initial momentum of the wall is 0. Again, momentum is conserved. The final momentum of the combined car/wall is +140m. The velocity of the combined car/wall is +140m divided by a very high mass, which is close to zero. Therefore the change in velocity of the car is nearly -140 kph. Therefore the change in the momentum is nearly -140m, nearly twice as much damage as the collision in scenario A.

Kinetic energy is not a helpful concept in this problem because the amount of energy described varies according to the frame of reference. Axl ¤ [Talk] 20:51, 10 June 2009 (UTC)

I'll disagree on a couple of counts. First, the impulse is not the only relevant factor; the crushability (is there a proper term for this?) of both bodies matters significantly. Consider the difference between hitting a stationary wall at 70 kph versus hitting a mountain of marshmallow creme: in both cases, the car receives the same impulse, but with the marshmallow mountain it is spread over a much longer time, making for a much more gentle impact. Similarly, being in a car with larger (softer) crumple zones would spread out the impact, making for less damage to the occupants. Also, note that the relative crushability of the two bodies matters as to which one gets damaged more: if you hit a wall, the car is much softer, so it absorbs all of the energy of deformation; with the marshmallow mountain, it's much softer and hence takes almost all of the energy of deformation, leaving the car gooey but otherwise intact.
Speaking of energy of deformation, kinetic energy does matter ('cause it's what causes all that deformation), and it is helpful provided you do the math all the way through. In scenario A, each car has initial kinetic energy 2450m (that's .5*m*v^2), for a total of 4900m; their final KE is 0, so the whole 4900m goes into deforming the cars (and flinging bits off at high speed, etc). In scenario B, car 1 has initial KE of 9800m and car 2 zero, for a total initial KE of 9800m. This is twice the initial KE of scenario A, but watch where it goes: after the totally inelastic collision, both ex-cars are sliding along the roadway at 70 kph, giving each 2450m of KE, for a total final KE of 4900m. The difference, 4900m, goes to energy of deformation, part-flinging, etc, and it's exactly the same as in scenario A.
Finally, I'd like to quibble about one difference between the scenarios that everyone's been ignoring: what happens after the initial impact. In scenario A, there are two ex-cars at rest in the road. In scenario B, there are two ex-cars sliding down the road at 70 kph, maybe recieving additional damage as they come to a halt. Remember, they've still got 4900m of KE available to do further damage (e.g. if one of the ex-cars starts rolling over). -- Speaker to Lampposts (talk) 07:20, 11 June 2009 (UTC)
The two situations are exactly the same. Any increment in total kinetic energy is unusable as the Centre of Mass of the system must maintain a constant velocity. Changing reference frames (which is essentially what this problem is) will never cause the dynamics of the situation to be different. Both situations are exactly the same apart from the frame in which they are viewed. Elocute (talk) 09:58, 11 June 2009 (UTC)
Elocute, the two scenario are not identical apart from a change of reference frame. In the "head on" scenario, we are assuming that the second car has the same mass as the first. In the "wall" scenario we are assuming that the wall is immovable, so it must have a very large mass compared to the car (an "ideal" wall has an infinite mass). That is the key difference. The two scenarios have different outcomes (and the "stationary car" is a third scenario, which is different again if the stationary car has its hand brake on). Look at the motion of the centre of mass immediately after the collision in each case. Is the wall moving at 70 km/h immediately after the collision ? No it isn't.Gandalf61 (talk) 12:34, 11 June 2009 (UTC)
Who has claimed that hitting a wall is the same as hitting another car? The situation being discussed is two cars crashing into each other, either both moving or one moving and one stationary. The situation with a wall isn't interesting and was covered very quickly. --Tango (talk) 17:21, 11 June 2009 (UTC)
The original questioner said "a solid stationary object" which to me (and, clearly, to several other editors) means an immovable object such as a solidly built wall with foundations. If they had meant a stationary car I imagine they would have said so, and they would not have said "the change in momentum of the first situation car is half that of the second car", which is obviously not true for the stationary car scenario. Gandalf61 (talk) 22:23, 11 June 2009 (UTC)

"Consider the difference between hitting a stationary wall at 70 kph versus hitting a mountain of marshmallow creme: in both cases, the car receives the same impulse, but with the marshmallow mountain it is spread over a much longer time, making for a much more gentle impact."

— Speaker to Lampposts

I agree, which is why I included the caveat "Technically, the maximum force applied is probably more relevant, but this will be proportional to the impulse". Axl ¤ [Talk] 14:12, 11 June 2009 (UTC)

## Bcf v Bcfe

Could some please provide an explanation of the differences betweeb Bcf and Bcfe?203.100.252.138 (talk) 05:29, 10 June 2009 (UTC)

Some context would help. See [1]. Billion cubic feet? Bioconcentration factor? For the energy field, see [2]. Edison (talk) 05:34, 10 June 2009 (UTC)
Some gas exploration and production companies report their reserves using BCF, Billions Cubic Feet for the gas and BCFE, Billions Cubic Feet Equivalent for any oil. Conversely, some oil companies do the reverse and quote their gas reserves as BBOE or billions of barrels oil equivalent, in fact I think that this is more common than using BCFE for oil (strictly that should be liquids as much of their non gas reserves may be condensate). Mikenorton (talk) 08:31, 10 June 2009 (UTC)
The companies do this so that they can discuss their performance using a single measure. There is a standard calculation to convert between them, approximately 1 barrel of oil = 6,000 cubic feet of gas, based on the amount of energy produced on burning. Mikenorton (talk) 09:14, 10 June 2009 (UTC)
An added benefit is that these conversions do not fluctuate with relative price changes. A barrel of oil equivalent is a specific quantity of natural gas, irrespective of the market price of gas or oil on any particular day. This makes it easier to compare quantities of gas production to oil production over a whole company - instead of specifying in dollars, which adds a whole economic variable into the comparison. Nimur (talk) 15:53, 10 June 2009 (UTC)

## geographical location 0°,0°

Which place on earth has a geographical location 0°,0°???Shraktu (talk) 10:01, 10 June 2009 (UTC)

A piece of ocean floor lying 614 km south of Accra, Ghana and 977 km west of Port Gentil, Gabon. Mikenorton (talk) 10:12, 10 June 2009 (UTC)
... which is in the Gulf of Guinea. It's where the middle red horizontal and vertical lines cross in this map projection. Gandalf61 (talk) 10:15, 10 June 2009 (UTC)
Why should it be the ocean floor and not the ocean surface? The "riddle" that I once read was "Where does 0 latitude meet 0 longitude at 0 altitude?"! Jay (talk) 11:40, 10 June 2009 (UTC)
Well, yes indeed - because otherwise I could perhaps answer our OP by saying "the center of the earth". However, Shraktu did say "What place on earth"...so even the most ridiculous nit-pick should allow Mikenorton's answer to be the correct one.
On a more useful note, one of my favorite sites on the Internet is the Degree Confluence Project which aims to have a photograph and description for every integer latitude/longitude intersection - and they are probably have about half of them done. Hence, if you head over to http://confluence.org/confluence.php?lat=0&lon=0 - you can see a photograph of 0°,0°. Predictably - it's a remarkably unremarkable piece of ocean - no visitors' center, no gift shop selling "I'm a zero!" t-shirts! The story behind the photograph is actually rather interesting - this particular group of visitors took with them a hollow steel sphere containing a hard vacuum to symbolize zero-ness and dropped it onto the ocean bed at (0,0). The ocean there is about 5,000m deep! SteveBaker (talk) 12:22, 10 June 2009 (UTC)
Water is part of Earth, so if he asked what point on Earth, you'd say on the water. Technically, he said on earth. Water isn't earth, dirt is. Even more nit-picky, he never said it couldn't be under any earth, and anywhere but the center of the Earth is going to be on some of it. — DanielLC 21:23, 10 June 2009 (UTC)
Haha, based on that site, the meaning of life is located here! Outlook doesn't look good...Drew Smith What I've done 09:12, 11 June 2009 (UTC)
The OP asked "on earth", and not "on land". Even otherwise, it may take a billion years for the ocean floor to become land. Jay (talk) 09:37, 11 June 2009 (UTC)
This particular bit of ocean floor is highly unlikely ever to become land, its likely fate is to be subducted back into the mantle along a destructive boundary at a location that we can only guess at. This will happen within the next couple of hundred million years for sure, possibly a lot sooner than that. Mikenorton (talk) 13:16, 11 June 2009 (UTC)

## Melting point of granite and silica

Granite, a rock composed largely of silica, has a fairly low melting point (for rocks); between 700 and 900 celsius, typically. However, silica has a melting point considerably higher; closer to 1650 C, according to silica. What is bringing the temperature down so much? I initially thought it might be because granite melting is measured at depth, not STP, but if it were to be melted at STP I'd expecting the melting point to be lower. —Preceding unsigned comment added by 157.203.42.175 (talk) 12:23, 10 June 2009 (UTC)

Eutectic point describes mineral eutectic mixtures - that is, certain combinations of minerals that may have a cumulative melting point lower than any of the constituent elements. The material properties of liquid-to-solid phase transitions, especially in the presence of crystal structure formation, is extraordinarily complex and is an active area of scientific research. Briefly summarized, it is quite possible that the presence of other molecules in the liquid mixture prevents crystallization until a much lower temperature (consider it like a "dopant" impurity). Even a small impurity in an otherwise crystalline structure can cause huge changes in macro-material properties. Nimur (talk) 14:26, 10 June 2009 (UTC)
For crustal materials, granite has the composition of the eutectic mixture, which is why it's so common. The precise composition and the melting temperature depend on the amount of water in the system; the more water, the lower the temperature at which it melts. Pressure (i.e. depth) also has an effect, but it's of less importance. Mikenorton (talk) 15:10, 10 June 2009 (UTC)

## DivX, avi, mp4, Xvid

Hi I put this on Divx but I should put it here:-

Divx, mp4 and avi are all over the internet from game trailers to home made camcorder stuff, iPod, keep going... Okay. When you read about them they are often grouped together, DivX and avi in particular. Anyways what this is all about is.. computers run avi. I have a load of avi and mp4 short films. I want to buy a dvd player capable of putting those on my bigger screen without turning my noisy computers on. Mp4 is easy to find. So, I am fairly convinced that a dvd player saying DivX will play avi because I check online and the talk is about compatibility but it's not clear. I just can't convince myself one way or another. Even here (DivX) it says "Partial backwards compatibility with AVI" but if there is a solid connection between DivX and avi, such as "As standard DivX compliant software/firmware can decode the infinitely popular en mass use avi format" or "DivX compliant has been known to be programmed with avi capabilities" or "DivX software has been produced with the capability of decoding it's strongest relative, avi, but as yet, avi decoding is rare in DivX compliant dvd machines." I will go ring a shop now about it but as for reading online ebay, wikipedia, none of those I couldn't satisfy myself if avi decoding was or not coming standard run in DivX. I was always sort of semi-convinced that DivX and avi were basically the same line but it's not written down like that. It's always in there but never exactly explained. Even avi provides us a sort of vague but un-solid connection between these two I would like to distinguish them properly if anyone can help at that pls. ~ R.T.G 13:04, 10 June 2009 (UTC)

AVI is a container format, which is a specified packing order for the bytes that describe the video, audio, and any other multimedia (like subtitles, metadata about the media clip, etc.) You can put almost any type of compressed multimedia (any codec) inside an AVI file. To make it worse, many encoders will put non-standard codecs into AVI formats anyway (like variable-bit-rate-audio), and because of sloppy software, these files "work" in many media players, so nobody complains. MP4 is an approximate name for a container format, designed only for use with a specific family of codecs. It has also come to refer to that family of codecs as well. Codecs are rules for compressing and expanding streams of bytes into full images to reconstruct a movie. Codecs do not specify a file-format or a specific byte order. DivX is a brand-name for a commercial version of several different codecs, including an implementation of H.264, (roughly speaking, a type of MP4). If a device claims it can play DivX files, it probably implements the DivX decoder, and probably expects the file to be in an AVI container. This does not mean that it can play all MP4 files, which might be encoded with any of a wide number of other standard codecs. To play a file, a player must understand the container format and have a compatible decoder for any given file. Nimur (talk) 14:33, 10 June 2009 (UTC)
Aaah you are right. In the 'Properties->Summary' there is info on wether the files are Xvid or what. I does seem that most of the files are DivX or Xvid so I am just taking the word now that these formats would accept the avi container naturally, once they are Xvid and DviX. tnks! ~ R.T.G 15:00, 10 June 2009 (UTC)
Note that the H.264 version of DivX (DivX 7) was only released in January 2009 and the H.264 DivX codec is called DivX Plus HD. While it's possible some files marked DivX are encoded with the H.264 codec, I've never come across any personally. Generally speaking, the vast majority of DivX files, and DivX players are MPEG-4 ASP compatible only. XviD is also an MPEG-4 ASP codec although the features enabled by default may vary from that supported by DivX. Generally speaking though, most DivX players will be able to play a XviD encoded file fine, at worst perhaps requiring a difference fourCC (once you started talking HD things may start to get more problematic). In any case, this isn't really a science desk question. Note that as DivX and XviD are MPEG-4 ASP, it's possible to store them in the MP4 container, as well as ogm and Matroska. Nil Einne (talk) 07:21, 11 June 2009 (UTC)
I guess the thing is to buy the one that takes the firmware upgrades and to check in the shop that it plays most of your stuff. It's a bit like when they raed the cpu's to 1gz, a new one every few weeks. ~ R.T.G 13:22, 12 June 2009 (UTC)

## Explosive friction

There are "weak" elements to such a degree that on a mathematical symetrical chart the there is a space where in all probability the very weakest are there but too weak to detect (I saw a TED talks about it very good). There are detectable wavelengths that just cut through our little crust like rain and we never even notice them. According to our science they are a kind of light and therefor a kind of mass. So, what I am saying is there is a grain to the fabric that is infinitely fine to our perception. So, when an explosion goes off a rapid expansion occurs. Let's say you are watching a movie and the guys put some C4 plastic around a door. Enough force is produced to break steel as big as you like. The only opposing friction is in the air. Now I know little about it but it seemed to me that the force against the door would be represnted opposingly as compressed air with a mass and velocity enough to break the door open. i.e. not only to move the air as fast as a truck heavy enough to break the door but to instantly cause compression in the air the mass-size of that truck. But, the amount of air in the mass of a truck is enormous, is there really enough friction in free floating air alone to produce that kind of opposing force? Wouldn't the same charge work in space? If any charge worked in space like that (and would it?), where would the opposing friction come from then? Is it possible that large friction is produced against radiated cosmic mass whose natural home/origin is right in the middle of explosive (star) forces? If you were feet first on a steel plate (in space) and an explosion occured beneath the plate, apparently you would fly in the opposite direction but there must be friction right? If there was no opposing friction a space-walker could happily sleep through a huge explosion right beside him and just be pushed off at speed, right? It must grip something and in an area of open space the largest mass is the radiation? Also, if you can answer some of that, what effect, with a view to friction, do explosions have on magnetism? ~ R.T.G 14:16, 10 June 2009 (UTC)

It's not friction, RTG, it's momentum. In your steel plate example, gases are forced away from the plate by the explosion. This gives them momentum, which is equal to mass times velocity. In order to balance that momentum so that the universe doesn't fall over, they need to push the plate in the opposite direction. That is why the plate experience a force. The law of conservation of momentum means that the momentum of the gases is equal and opposite to the momentum of the plate (plus you standing on it). As an equation: (mass of gases) x (speed of gases) = (mass of plate etc.) x (speed of plate etc.). The gases might not appear to have enough mass to produce much of a force on the plate, but when you multiply their small mass by their huge speed, you get a large momentum. --Heron (talk) 16:23, 10 June 2009 (UTC)
Well I didn't think that explosives produced a signifigant amount of gases. Maybe I should read up a bit about C4 and TNT. ~ R.T.G 17:19, 10 June 2009 (UTC)
No it is absolutely the function of explosives to produce as much gas as possible - with as much velocity as possible. SteveBaker (talk) 18:18, 10 June 2009 (UTC)
This is a bit off the wall but friction on stellar scales can be quite impressive. But in such cases it is not explosive force but massive, massive gravitational forces at play. --98.217.14.211 (talk) 17:21, 10 June 2009 (UTC)
What would be the blast effect at normal atmospheric pressure versus the blast effect in space, of the same explosive set off at the same distance from the same surface? Gas effects only, neglecting bomb casing fragments. Edison (talk) 21:07, 10 June 2009 (UTC)
The explosion pushing against the plate is effectively a very short-lived rocket. The thrust equation for a rocket includes a term for the reduction in thrust due to ambient pressure, but in an explosion the exhaust pressure ought to be vastly greater than atmospheric pressure, so I doubt that atmospheric pressure would make much difference to the thrust. --Heron (talk) 09:52, 11 June 2009 (UTC)
The explosion which cuts steel is also quite different to the stick of dynamite used for mining, for example. To cut steel or blow a doorway into a wall one would use a shaped charge which uses the explosionto force a material (typically copper) through the thing to be cut, so you have high velocity gasses in one direction, and a high velocity "blade" of lining material going in the other direction. For the explosion+plate in space, see also Project Orion --Polysylabic Pseudonym (talk) 05:20, 12 June 2009 (UTC)
Thats a good article "Even compared to today's best chemical rockets or even other nuclear designs, Orion's potential performance was stunning." ~ R.T.G 13:43, 12 June 2009 (UTC)

## A problem of Doppler effect

The article about emission theory points out an odd result as a problem, that, 'a radiant star moves across our field of vision, light given off by differently-moving atoms in its atmosphere should take different amounts of time to reach us. Since the retreating atoms would have a "red" Doppler shift, and the approaching ones a "blue" Doppler shift, the passing star might be expected to appear as a "rainbow streak".' If Doppler effect (relativistic or non-relativistic) does occur, similar problems seems inevitable. If "rainbow streak" is unobservable, why is it? Doppler effect surely occurs, doesn't it? Like sushi (talk) 15:55, 10 June 2009 (UTC)

Yes, relativistic doppler effect does occur but the conclusion that light given off by different atom should take different amounts of time to reach us is faulty. They don't. Dauto (talk) 16:46, 10 June 2009 (UTC)
IF emission theory was true - then we might expect some of these strange things...but it's not true - so we don't see them.
Out in the real universe (the one in which relativity won this argument) atoms moving through the atmosphere at different speeds and directions would indeed be producing doppler-shifted light. That has nothing to do with travel time though because the speed of light is constant no matter the relative speed of source and . But the speed that atoms move relative to one-another is so tiny that it's negligable compared to the motion of the star relative to us...and in any case there are insanely large numbers of atoms and the various colors are going to average out. SteveBaker (talk) 18:52, 10 June 2009 (UTC)
Putting aside the travel time of light and motion of atoms, just assuming that Doppler effect occurs still seems to cause "rainbow streak" if "a radiant star moves across our field of vision". I thought the motion of the star causes "rainbow streak".
I may have been misunderstanding what is written. It says 'when a "star moves across"' and "the passing star...", so I thought motion of the star was included in the cause. Is the motion of the star not taken into account here? Does it cause Doppler effect but it's not what is called "rainbow streak" here?
Like sushi (talk) 01:06, 11 June 2009 (UTC)
Perhaps you don't understand how the doppler effect works: The amount of color shift depends on the speed of the light source relative to us. If the star is moving in more or less a straight line at more or less constant speed over the period we're watching it for - then the redshift will be constant. The light would look a little reddish - or a little blueish - but it can't VARY in color unless it's accelerating...and accelerating pretty rapidly too! I guess you might see a rainbow redshift as a star fell into a black hole or something. SteveBaker (talk) 03:09, 11 June 2009 (UTC)
I thought of a star moving side to side, initially from a relatively far point, through a nearest point, then to a relatively far point. But thinking about it, the effect may be far too small. Do you think so?
Like sushi (talk) 09:28, 11 June 2009 (UTC)
And if the speed of the star does not change much, and if the prediction of shift by emission theory is also dependent on the speed of the source, does it not turn out that "rainbow streak" does not practically occur by emission theory either?
Like sushi (talk) 09:50, 11 June 2009 (UTC)
That's true. But stars are so far away - and move so amazingly slowly - that it would take hundreds of human generations for the star to go from rushing more or less towards us to rushing more or less away. So in the case you describe, the color of the star would indeed change from a blueish tint to a reddish tint as it went past - just as the sound of a car engine goes "wiiiiiiiooooooo" as it goes past you. But the star takes so long to do that and the speed is so slow that the color shift is fairly small and the time it takes to change from blueish to reddish is measured in thousands to millions of years (at best). So I suppose technically there could be a 'rainbow streak' - but you'd have to be watching a star that's doing this specific manouver for millions of years in order to see it. SteveBaker (talk) 14:26, 11 June 2009 (UTC)

## calculating gas viscosity

I have 89% H2, 8% CH4 1% C2H6, and 1% N2 with a remainder in higher MW gasses. Is there a formula I would use to calculate the gas viscosity? 65.121.141.34 (talk) 16:02, 10 June 2009 (UTC)

The constant of viscosity of a gas can be calculated as follows. Just realised I have no idea how to write a formula in here, I'll try the cumbersome method of trying to write it in words.
The constant of viscosity of a perfect gas is equal to one third of the mean free path multiplied by the mean speed and the gas's molar mass and concentration. These could all be worked out from the information you are given and a periodic table.
The other method you could use is to conduct a Pouiseuille flow experiment, which will also allow to you work out the viscosity.

Hope this helps a little bit Alaphent (talk) 16:23, 10 June 2009 (UTC)

## Suspend a weight between two poles - how strong must the cable and poles be?

Hello, I asked a question above and I think I confused the issue by including too much information. What I am looking for is information on a physics question. Suppose I have two steel poles set vertically into the ground. The distance between the two poles is Y. Between the two poles is stretched a steel cable. I want to hang weight X in the middle of that cable. There are no other supports or guy wires involved. How do I calculate the needed size/strength of the cable and poles? Thanks! JohnMGarrison (talk) 17:14, 10 June 2009 (UTC)

Can we assume X is significantly greater than the weight of the cable itself? --Tango (talk) 17:24, 10 June 2009 (UTC)
Yes, we can assume that. I want to treat this as a very simple problem - no wind, no weight of materials, etc. Any ideas? JohnMGarrison (talk) 17:38, 10 June 2009 (UTC)
We need to know the length of the cable - is it taut - or does it sag in the middle? In your previous question you talked about a sign as your weight 'X' - be aware that if you treat the weight as a 'point mass' then the answer will be radically different if the weight were suspended at a couple of points along the wire instead of just one. These cases are all different (even if the poles are the same distance apart and the mass of 'X' is the same in each case):
    |\       /|        |\        /|      |_____________|
| \     / |        | \      / |      |      |      |
|  \   /  |        |  \____/  |      |      X      |
|   \ /   |        |  |XXXX|  |      |             |
|    X    |        |          |      |             |
|         |        |          |      |             |

SteveBaker (talk) 18:14, 10 June 2009 (UTC)
However, if the distance between the poles is much greater than the size of the sign, we can reasonably approximate it as a point mass. --Tango (talk) 18:47, 10 June 2009 (UTC)
That's certainly true - but our OP was originally talking about hanging some kind of banner across the street - so the span of the sign could easily be 80% of the distance between the poles. Depending on the stiffness of the sign - that could make a massive difference. When the wires a loose, (as in my left-hand diagram) more of the weight is acting downwards on the poles. Most poles are extremely strong in compression - so that would be a good thing. However, if the wire is more or less straight (as in the right-hand diagram) - then all of the force would be applied to bending the poles inwards - which would be very likely to bend, snap or uproot them. The difference between the middle diagram and the left-hand one is that the diagonal parts of the cable are at an even steeper slope than in the left-most case. SteveBaker (talk) 18:59, 10 June 2009 (UTC)
Photo 1
• Great ASCII art Steve - thank you. I see what you mean about compression vs bending. You are right that what I really have in mind is not a point mass, but something that spans a significant fraction of the distance. I am thinking about a neighborhood sign such as in the photo to the right:
The poles don't show in that picture but my guess is that the sign spans about 60% of the distance between the signs. The cables supporting the sign are angled like in your first 2 drawings. I think your middle drawing is closest to this actual case, except that in the actual case the sign is suspended by 4 cables: 2 attach to the center, while 2 attach near either end. (and then 2 more cables below the sign for stabilization).
This is a sign hanging over a street in San Diego, California. I'd like to understand how to size the cables and poles so that they safely support a sign like this. Again, we need to assume no additional guy wires are possible, just the poles. JohnMGarrison (talk) 19:42, 10 June 2009 (UTC)
Update: In Photo 2 and Photo 3 you can see the supporting poles. JohnMGarrison (talk) 19:46, 10 June 2009 (UTC)
If you don't have guy wires you'll need to either have wide bases on the poles or have them sunk a significant distance into the ground. Is one of those an option? --Tango (talk) 19:49, 10 June 2009 (UTC)
Hmm at a guess I'd say the sign was designed so that the middle wires (the ones that meet at a point) and the tops of the poles corresponding to them can support all the weight of the sign. That would seem to be the conservative way to calculate it. The other wires seem to be there just to keep the sign stable and provide a little more peace of mind for people standing under it (also a redundancy if one of the middle cables fails due to corrosion or whatever). It looks from the pictures that the poles are much bigger than they "need" to be and metal cables like that can hold much more weight than that sign. Checking to make sure would involve making a free body diagram of the sign, using it to make sure the tension on the cables is less than what they're rated at, checking to see that the bending of the poles is not beyond their strength (you could do that from fundamentals or you can likely find a rating for it being bent like that) and check that the pole is buried deep enough in concrete at the base to support the moment without moving. At least that's what I think :) TastyCakes (talk) 20:01, 10 June 2009 (UTC)
• Tango: Yes, the poles could be sunk very deep into the ground if required. Thickening the base is also a possibility but the desire would be to not take up too much of the sidewalk as shown in photo 2.
• TastyCakes: I think you are right that the middle cables probably support all the weight. In that case, can we treat this like a point mass? In the case of this sign, the poles do bend in slightly. I don't think they were built that way so that must have happened over time.
1. Suppose for a moment that someone was worried that the poles lean in a little, and they want to calculate if the sign was safe. Is there a simple formula they could use to get at least a rough idea?
2. If this design were to be strengthened while keeping the same basic design (in other words, not added bars, trusses, cables), what would be the simplest way to do that? Bigger cable? Thicker post? Bury the post more deeply? Taller post (to ensure more compresion and less sideways force)?
Thanks very much! JohnMGarrison (talk) 20:15, 10 June 2009 (UTC)
Yes, I think you can treat it as a point mass. As for the poles being strong enough, this appears to be a fairly straight forward Bending question. An engineering book on statics or mechanics of materials should have the necessary formulas. To make the structure overall stronger, you'd have to figure out the weakest part of it now - that is what part of it is closest to its failure point. I don't know if it would be the poles or the cables in this case, you'd have to get at least a rough estimate of the numbers. If the cables are the weaker link, put bigger ones in, if the pole is the weakest link, put a bigger one in or support it somehow (with guide wires or something). TastyCakes (talk) 20:42, 10 June 2009 (UTC)
Also note that the fastener that the two middle cables are attached to could be the weakest point and should be analysed to check. TastyCakes (talk) 20:46, 10 June 2009 (UTC)
I've got a whole bunch of stuff here on Wire Rope Safe Working Load Table. Sling Angle SWL, Wire Rope Clips/Number and Spacing but no way am I going to disclose that here, nor am I going to check my Machinery Handbook for column loading. JMG, you blew your cover long ago - you seem to be asking about how to suspend an actual sign on an actual rope above actual human beings. There's only one answer - consult a licensed professional engineer. It's a simple enough problem, maybe $400 to find the solution, max. It's nice that you try to devolve this to a simple physics problem, but you still aren't phrasing it right. You should be asking for a formula to calculate the stress in the various components. Instead you ask for the "needed size". This is a no-no. We have no way of knowing the needed factors for wind-loading, ice-loading, nor any other safety factors in your jurisdiction. You need to specify the precise design case and ask an engineer to work it out. You're playing with fire if you really want to hang a sign. No-one here carries liability insurance. Franamax (talk) 21:42, 10 June 2009 (UTC) • Franamax. thanks for the input but you really assume too much. I am not about to hang a sign like this myself. I'm just trying to learn more about it to satisfy my curiosity. That isn't worth a$400 consultation. Also, as I said more than 1 time, I'm not interested at this point in all the intricacies such as wind (we have some in San Diego) or ice (which we rarely have). I'm just trying to learn the basics right now. I'll follow TastyCakes suggestion and read up on bending. Thanks, TastyCakes. If anyone else knows a simple formula I could use to get the approximate stress, I would be grateful. JohnMGarrison (talk) 22:09, 10 June 2009 (UTC)
JMG have you contacted your local DOT yet?71.236.26.74 (talk) 22:03, 10 June 2009 (UTC)
With regards to the sign shown here in Photo 1, the City of San Diego says the sign meets code and is safe. I have heard a local resident say that "national building codes have been revised since Hurricane Katrina and that signs can no longer be suspended by cables." I don't think building codes are set at a *national* level. That is why I asked at one point about building codes. I emphasize again I am not looking for legal advice and I am not looking for a structural engineer to give me a hard and fast answer. I'm just looking to learn a bit about the issues involved, particularly the physics. JohnMGarrison (talk) 22:17, 10 June 2009 (UTC)
Update: I think my neighbor was thinking of International Building Code, which does attempt to bring some standardization across the United States. It seems, however, that local jurisdictions may adopt it completely or with changes or not at all. JohnMGarrison (talk) 23:38, 10 June 2009 (UTC)
In that case (and sorry if my alarm bells started ringing), the critical factors seem to be stress in the cable, stress at the connectors, and side-loading stress on the vertical column. Stress in the wire is dependent on the angle between the support and the load (see SB's ASCII diagrams, the lesser the angle between load and support, the greater the stress in the wire). My handbook for SWL for a Two Part Bridle Sling with a 120-degree included-angle (roughly analogous if you draw the sling upside down) is 920 lbs for 1/4" wire rope - but note that the stress rises rapidly as the included angle approaches 180-degrees, and lift figures are based on still air, since no crane operator fires up on a windy day. The side-loading on the column is basically one-half of the suspended weight acting horizontally, figure out the cosines for yourself but these are dwarfed by the safety factors. Machinery Handbook or Mark's Engineering will give you those calculations, be aware though that column calculations are subject to dispute. That's why a guy wire is so often used on the other side, then the column strength can be calculated as a simple compression/buckling problem, you can use Euler's formula or the AISC standard for that. The notable item in all this is safety-factor x safety-factor x safety-factor. You're asking a question which science doesn't necessarily answer, but engineering does - the difference being that engineering is concerned with which choices result in the least number of people dying as a result of the "oopsies". Franamax (talk) 22:56, 10 June 2009 (UTC)

1. From the geometry of the cable length, pole separation Y and the way the weight is attached, calculate the angle w by which each cable deviates from horizontal.

2. When the tension in a cable is T the upward component of the force is T.sine(w)

3. When there are two cable spans as in SteveBaker's sketches we have X = 2.T.sine(w) where X and T are in the same units such as pounds or kilograms. Then by manipulating: T = X/(2.sine(w)) gives the tension T in the cables.

4. When there are more than two cable spans at different angles from horizontal it is not obvious how the weight X is shared among them. It would be safer to allow for a cable to break, i.e. the possibility of each cable bearing the whole weight X as calculated above.

5. The pole tops are pulled horizontally by a force T.cos(w) tending to bend them inwards. The downward force on the poles is simply X/2.

6. The sign in the photographs has additional cables holding the bottom corners down. Any tension in those cables adds to the effective weight X of the sign and to the horizontal force on the poles. Without wind there is no need for those cables, but their presence implies that there can be wind. The calculations in 2. to 5. above take no account of wind. Cuddlyable3 (talk) 23:01, 10 June 2009 (UTC)

• Dear Franamax, Cuddlyable3, thank you very much. This is exactly the kind of info I was looking for. I originally knew enough to realize I had to care about both downwards force and sideways force, but I didn't know exactly how to start. This will get me started in the right direction. Thanks again, JohnMGarrison (talk) 23:13, 10 June 2009 (UTC)
The thing is that hanging a sign is an engineering problem. That is only partly physics. A large part in engineering is (still) trial and error. So you get people in labs that pull cables till they break. You get guys in another lab that dangle weights from cables and kick the weight with various speeds and frequencies till the cable breaks. Next door they're doing the same thing, but they're freezing or heating the cable, too. The results end up in materials data sheets and engineering handbooks or software. Just for starters I don't think your calculations include the elasticity of your cables yet. Even without wind your cables are going to stretch when you apply tension. Wire rope is also going to twist, so you are getting an elastic spring effect and some torsion. All without wind and still not even thinking of California's earth quaking. The reason the person mentioned the revised building codes is that when there's a notable incident engineers go and do forensics afterward. So you find that the signs with the X gauge cable hung by N number of wires got blown off by Andrew and after Katrina they found that only the signs on cross beams survived. They write a report that gets reviewed by lots of committees dealing with standards. If enough committee members agree the standard gets changed. 71.236.26.74 (talk) 23:47, 10 June 2009 (UTC)
Thank you, anon. I know about these issues in a general sense. I have relatives who are engineers. I know that they do a lot of testing, and I know that there is *always* a large safety factor applied to whatever they *think* is a sufficient solution. I also know that our understanding of these things evolves with experience. Thanks for your help. JohnMGarrison (talk) 23:52, 10 June 2009 (UTC)

## Matter and antimatter would be a subset of what?

In other words, France and China are both countries. Matter and antimatter are both... what? (as in a term that exclusively englobes only those two terms -not something too vast as universe, which also includes "energy and momentum, and the physical laws and constants that govern them"-) Not sure if such a term exists, but worth a try, I guess. Thanks, TomasBat 17:38, 10 June 2009 (UTC)

I just checked the Wiktionary definitions for matter and antimatter. The defs "(physics) Matter made up of normal particles, not antiparticles. (Non-antimatter matter)" and "(physics) Matter that is composed of the antiparticles of those that constitute normal matter" imply that matter is both a general term, referring to compositions of particles and compositions of anti-particles, as well as a term referring specefically to that composed of particles. Thus matter would be the term I would be looking for, right?

Matter: (physics) The basic structural component of the universe. Matter usually has mass and volume.

• Matter: (physics) Matter made up of normal particles, not antiparticles. (Non-antimatter matter).
• Antimatter: (physics) Matter that is composed of the antiparticles of those that constitute normal matter.

TomasBat 17:46, 10 June 2009 (UTC)

Sounds good to me. Just keep in mind you shouldn't take any of the above definitions very seriously. There is no scientifically based definition of what should be considered a particle and what should be considered an anti-particle. We chose to define an electron as a particle and a positron as an antiparticle simply because ordinary matter is composed of lots of electrons (among other particles) but no positrons. But that's just a convention. the oposite convention would work just fine. Dauto (talk) 18:03, 10 June 2009 (UTC)
Our matter article says, "in discussions of matter and antimatter, normal matter has been referred to by Alfvén as koinomatter", which implies that what you say above is correct: normal matter and antimatter are both a kind of matter, and you have to distinguish from context which "matter" you mean. --Sean 18:14, 10 June 2009 (UTC)
The word you are all looking for is Substance. --Shaggorama (talk) 05:26, 13 June 2009 (UTC)

## Relativity, exponential curves, infinity of time(ish)

All natural things outside of pure mathematics would seem to have a solid portion of infinite exponential curvature in their structure. Apparently instruments were available at one time with the sensitivity to record and calculate the Einstein theory of time dilation on a supersonic craft and found dilation theory to be bang on, definite fluctuations in time/motion in the supposed direction. Einsteins theories also, if I am quoting correctly, viewed the uppermost light wave we can detect to be more or less the uppermost and fastest matter there is. So, given that the exponential curve of improbability is part of the fabric of everything off paper, would it be delicate to assume that further up the ladder of detection, as is the nature of light, a certain amount of exponentiality occurs in the dilation curve so as light would not only be younger than any other entombed and ageing matter, but so little aged that in our perception hasn't even aged at all? What is the likelyhood, as with quark and quantum theory, that light goes far beyond our scales in terms of substance? ~ R.T.G 18:06, 10 June 2009 (UTC)

Could you say that again but using phrases that make sense so normal people will understand it? Dauto (talk) 18:35, 10 June 2009 (UTC)
I don't really understand what you are saying, but I can tell you that time dilation has been measured using atomic clocks on aeroplanes and spacecraft. I can also tell you that light, in a sense, experiences infinite time dilation, so experiences no time passing at all. Proper time for a photon is not well defined, but can be thought of as not progressing at all. So, all light has zero age from its own point of view. Light is not generally considered to be matter, although it can be thought of as being made up of particles, photons. There is a very big difference between matter travelling close to the speed of light and light itself, due to the matter having non-zero rest mass. Does any of that help? --Tango (talk) 18:45, 10 June 2009 (UTC)
Hmmm... I said most stuff seems to be what you would call an expression of a 3 dimensional fractal:- a ratio with a mathematical formula appied that goes on forever (1 + 1 = 2 + 1 = 3 + 1 = 4.... 1 atom + 2 atom = 3 + 2 atom = 5 + 2 atom = 7 atom... 1 X mass + 2 X speed = 3 X time + 1 mass + 2 speed = 6 time + 1 and 2 = 9 x time...) Apparently travelling at the speed of light can alter time to a ratio of 8:1 (an object at the speed of light ages 1 year for every 8 of our normal years, time dilation). Physical evidence has been shown of time dilation. If light has mass then would it be possible that light itself is an object affected by time dilation and if so, as they smashed the atom and found smaller peices called quarks and smashed them and found smaller peices again, could there be light in the upper range beyond our detection, and, if there is light going faster than we can comprehend, could the ratio of time dilation be infinite (could a beam of light we haven't detected be going so fast that for every 1 year that passes at speed much more than 8 years passes in normal time, even hundreds, thousands of years until there was a mass or energy to which the concept of time passing was minor or immeasurable?). Surely, if all forms of light are structures with different mass or with different wavelengths they travel a different speeds. For instance, what is the speed of a quark or a Preon in relation to visible light? What is the speed and mass of electricity? I don't know how to simplify or which terms are wrong. If it is possible to move time dilation from a ratio of 1:1 to 2:1, 3:1 up to 8:1... is it not natural to assume ratios of 10:1, 100:1, 1000:1 and so on? If it is conceivable, what are the hypothesis? Is it fair to say that the 8:1 dilation ratio and the speed of light as we know it may be like a 'hard limit' to us in all practicality, but, more reasonable to assume the possibility of an infinite and possibly exponential curve on the acceptable values of light speed and time dilation? If that is reasonable, is there even some failed experiments in studying it? I'm sure the like of Einstein or those he inspires must have worked extensively at it, no? ~ R.T.G 20:25, 10 June 2009 (UTC)
I've already told you, time dilation at the speed of light in infinite. An object travelling at the speed of light doesn't age at all. I have no idea where you got the figure of 8 from, but it is completely wrong. --Tango (talk) 21:06, 10 June 2009 (UTC)
I got 8:1 off a movie or something to be honest but the idea that light speed was finite I find in general ("nothing faster than light : The speed of light = x") I guess I am driving at a sort of flat light (light diamond?) which would probably possess that immunity to time. It does help. Perhaps light is invisibly balanced with a dark-matter anti-light and finite speed would make sense. Flat light could be a balancing point between that (something is bound to be stationary or it really does go bang!). OK, thanks tango :D ~ R.T.G 23:42, 10 June 2009 (UTC)
The speed of light is finite, it's about 300,000,000 m/s, but the time dilation at the speed of light is infinite. The factor is ${\displaystyle {\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$. If v=c you end up dividing by zero, which (in this context, where we are only interested in v approaching c from below) gives us zeroinfinity. I really have no idea what you are talking about with flat light and anti-light, you seem to be just making up words. --Tango (talk) 23:52, 10 June 2009 (UTC)
You seem to have a lot of technical terms mixed up and confused. I suggest you might try reading over our article on special relativity a bit more carefully. Time dilation is relatively straightforward and contains the actual equations and derivations for coming up with it. It is not as complicated as you seem to think it is—it is wonderfully, elegantly simple if you just reason through it. --98.217.14.211 (talk) 23:53, 10 June 2009 (UTC)
I'm sorry, variance in light speed is a possibility. Even here on Wikipedia (Speed of light) it says, "...observations show the variation of the speed of light as the universe ages is less than two parts in 10 to the power of 16 per year, for both microwaves and visible light." As for saying anti-light... in a universe of dark-matter and anti-matter on faith, words like anti-light and dark-light are as 1 is to 2, a to b etc. Light is in waves, if it was flatter/less wavy/more streamlined, would it not go faster (obviously that will get me a "no it doesn't change speed")? According to Tango, dilation is considered infinite as things stand but I've never heard it described as infinite and we are bombarded with titbits about dilation sometimes. In our perception light is a force comparable to gravity and magnetism. All natural things we know are rough around the edges with potential spikes not smooth and constant (maybe that is just what light is, the only smooth and constant matter there is but who knows, just as likely not, isn't it?). ~ R.T.G 15:16, 12 June 2009 (UTC)
You need to read more basic science before you can seriously think about such things. I'll try parsing your last paragraph and I hope that will help you understand why.
You said: I'm sorry, variance in light speed is a possibility. Even here on Wikipedia (Speed of light) it says, "...observations show the variation of the speed of light as the universe ages is less than two parts in 10 to the power of 16 per year, for both microwaves and visible light."
Yes, that is a possibility but it is considered a highly speculative possibility. There no evidence that it actually has happened and your quote of less than two parts in 10 to the power of 16 means that if there is any variation, it is so small to the point of being unobservable with current technology. That does not mean that it is impossible but it does mean that all observations are consistent with the current scientific consensus that there is no variation. We should stick to scientific consensus unless there is some strong reason to do otherwise.
You said: As for saying anti-light... in a universe of dark-matter and anti-matter on faith, words like anti-light and dark-light are as 1 is to 2, a to b etc.
unlike anti-matter and dark-matter, anti-light and dark-light aren't standard terms within science jargon. If you're going to introduce new terms, you need to be very specific about what you mean by those terms, otherwise they become meaningless pseudo-scientific babble. Your attempt at an explanation was words like anti-light and dark-light are as 1 is to 2, a to b etc. Sorry but that is woefully inadequate definition. Anti-matter and Dark-matter have very specific definitions that do not lend themselves to such analogies. You most definitely can (and should) do better than that. By the Way, Anti-matter and Dark-matter are not based on faith. Anti-matter has been produced and observed in labs and cosmic rays for more than 70 years now and its existence is a very well stablished fact of nature. Dark-matter existence has also become stablished on very strong observational grouds throughout the last decades. I can go into more detail about their eperimental evidence later if you desire.
You said Light is in waves, if it was flatter/less wavy/more streamlined, would it not go faster (obviously that will get me a "no it doesn't change speed")?
You missunderstand the statement that light is a wave. That does not mean that light follows a wavy path down the road. In fact light follows a staight path. By wave it is meant that something is oscillating and that oscillation propagates. In the case of light the electric field and the magnetic field are both oscillating and propagating and that's what's meant by light is a wave. That clearly underscores your lack of basic understanding about the things you are talking about.
You said According to Tango, dilation is considered infinite as things stand but I've never heard it described as infinite and we are bombarded with titbits about dilation sometimes.
Tango is right. You're also right to complain about being bombard by titbits of disconnect information that is often misleading or even downright wrong. That's why my advice is to stop paying attention to all that misinformation and to read a good book about the subject form your local library.
You said In our perception light is a force comparable to gravity and magnetism.
Your perception is misleading. light is definitaly related to magnetism (As I said, it is the oscillation of the electric and magnetic fields) but it is not a force properly speaking.
You said All natural things we know are rough around the edges with potential spikes not smooth and constant (maybe that is just what light is, the only smooth and constant matter there is but who knows, just as likely not, isn't it?).
I don't know what to make of that. I think you have expressed yourself extremely poorly here. Dauto (talk) 17:34, 12 June 2009 (UTC)
That is all I am looking for:- what is that speculation, even if highly unlikely, with some variance there must have been speculation. If I am correct dark-matter is matter of enormous mass beyond that we can even analyse and anti-matter is a speculation, with some proof, that matter as we know it could exist as an anti-matter that would be destructive to it's counter parts. Perhaps a light we have not detected exists with zero oscillation at an even greater speed (flat-light?/dark-matter). Perhaps an anti-light exists similar to how anti-matter exists but how would you define that? I don't know. Don't even have any ideas about it beyond the relationship between matter and anti-matter. "All natural things we know..." - Even light is various. Absolute exactness is only true on paper/mathematics. Obviously it is outside my sphere but at any rate it is hard to beleive that radio waves do not have a standard difference in speed to gamma radiation even if that is so minute that we never detected it (although of course not much good if never detected). Is there truly nothing in a possible upper spectrum to gamma radiation even if undetected or inconclusive? Hardly out of the realm. ~ R.T.G 20:51, 12 June 2009 (UTC)
RTG: Please - stop talking...just stop. These words that you are typing are all just nonsense - babble - any old scientific-sounding words strung together at random. Dauto did a masterful job of trying to dissect your stream-of-consciousness BS - and I agree - you have absolutely not the first clue about what you're talking about - it's not that you are a little bit wrong - you aren't even using words that make sense in a scientific context! Go and read some text books...heck just tune into the science channel once in a while!
You say things like "If I am correct dark-matter is matter of enormous mass beyond that we can even analyse and anti-matter is a speculation". What kind of arrogance makes you think you have a snowball's chance in hell of being right? The odds of your (frankly) wild-assed guess about dark matter being within a million miles of being right are very, very close to zero. Antimatter is certainly not "speculation". Let me just give you a clue about how inane that statement is. Positrons are anti-electrons...genuine, honest-to-goodness anti-matter. They were discovered in 1932 for chrissakes - this isn't something new and revolutionary. People actually USE anti-electrons all the time in their day-to-day lives. If you get cancer and have a "PET" scan at your local hospital they are using Positron emission tomography to look at your tissues. Yeah - positrons...antimatter...the stuff you think is "speculative". We use antimatter every day in thousands of hospitals around the world! The pretty spinning image at right was taken using antimatter in a PET scanner. Speculation!?! Have any clue how dumb that statement sounds?
If you keep up this kind of behavior, everyone here on the science desk is going to write you off as a crank or a troll - we'll start to ignore your questions - and if you keep it up, we'll probably start deleting them on sight. SteveBaker (talk) 23:02, 12 June 2009 (UTC)
SteveBaker has already hit all the points here. BTW steve, nice picture. I just want to reinforce the point that there is no royal road to science, which is another way of saying that you will never achieve true understanding of those things you're talking about unless you sit down and do the hard work of studying science. It is a long road, but every step of it is worth the effort. The first thing you must do is to stop doubting (for the now at least) what scientists are telling you. Believe me, the speed of light really is constant. Dauto (talk) 04:32, 13 June 2009 (UTC)

Although the term isn’t one that would be used by scientists, you could say that “anti-light” does indeed exist. However, “anti-light” is the exact same thing as “light”. Light consists of elementary particles called photons. You can annihilate a photon with the photon’s antiparticle. However, the photon’s antiparticle is just another photon. The situation is different with, say, electrons, because an electron has a nonzero charge, namely, –e. The charge of a particle and its antiparticle must sum to zero, so the antiparticle of an electron, the positron, must have a charge of +e, which is different from the –e charge of an electron. So the positron is a different kind of particle from an electron (unless you start talking about time reversal) . But the charge of a photon is 0, the negative of which is still zero, so you can’t distinguish between a photon and its antiparticle based on charge. There don’t exist “antiphotons” which are distinguishable from photons.

“Flat light” is another phrase that a scientist wouldn’t use, but the concept can be viewed as fitting in with current scientific theory, at least as a limit. The photons that light consists of come in various wavelengths. A photon with a longer wavelength can sort of be thought of as being “flatter” than a photon with a shorter wavelength, although “flatness” brings to mind a kind of transverse “thickness” to a photon that isn’t really accurate. But as pointed out by others, in any experiment ever performed, “flatter” light (i.e., photons with a longer wavelength) has the exact same speed as “wavier” light (i.e., photons with a shorter wavelength). Even the tiniest difference in speed between the two would result in a violation of special relativity, among other things.

The statement “If I am correct dark-matter is matter of enormous mass beyond that we can even analyse” may be a perfectly reasonable guess, depending on how you interpret the phrase “beyond that we can even analyse”. The most widely discussed models of nonbaryonic dark matter are based on the cold dark matter hypothesis. One of the two proposed categories of what cold dark matter consists of is weakly interacting massive particles, of which the leading candidate is the neutralino. The proposed neutralinos do indeed have an enormous mass, compared to other elementary particles. Red Act (talk) 09:50, 13 June 2009 (UTC)

Thanks for that answer Red Act. Well, Steve's just a prick with every paragraph comprising personal attack (I don't know him..!?) so, I removed the whole thing. If people think I am just asking questions as an attempt to insult or... I don't know. I had this thought that maybe something classed as light could move at an infinite speed and experience an infinite time dilation. I knew that light speed is considered constant. I thought it would be great to find some study of that idea or get a decent speculation that wasn't just, "No, light speed is constant and we don't discuss it." You guys might think it is the biggest load of crap but I thought it was great to see the answers. I didn't even complete a year of secondary school. What somebody thinks I am hiding might make me laugh, so what. Faster light is an interesting idea and so is anti- anything. ~ R.T.G 13:41, 13 June 2009 (UTC)
RTG, you're not supposed to delete other people's posts (nomatter how much you deslike them). I added it back
The problem with your posts is not that they are too speculative. speculation is fine. The problem is that your post show a lack of undertanding of the current scientific knowlege and you show no signs of even trying to learn it. For instance, I wrote a careful post in which (among other things) I pointed out that anti-matter is not a guess or speculation, that it is a scientific fact. On your very next post you said If I am correct dark-matter is matter of enormous mass beyond that we can even analyse and anti-matter is a speculation . Obviously either you didn't read my post or you didn't care that much for it. Either way it was quite a turn off for me. Dauto (talk) 19:47, 13 June 2009 (UTC)

## Top Speed

Basically, if I know the power and torque of my engine at a specific rpm. As well as the mass and drag coefficient of the car/bike, is it possible to work out a theoretical top speed? Alaphent (talk) 18:43, 10 June 2009 (UTC)

No, that is insufficient. You also need details about the transmission and the wheels - basically, which engine RPM corresponds to which speed of the outer rim of the tire. It is rarely the case that the engine provides maximal shaft power at the highest attainable speed. And even then you will only get an approximation. --Stephan Schulz (talk) 20:52, 10 June 2009 (UTC)
No - you don't need that. But you DO need to know the rolling resistance. If you have all of the numbers you can get a pretty close approximation that way. If you know the peak power output - it's pretty safe to assume that the car will be producing that amount of power at it's top speed. Then you can compute the drag force in Newtons and the know that:
 Power = ( Drag + RollingResistance ) x Speed

Of course drag is proportional to speed squared - so you wind up with a quadratic equation for the maximum speed. The only other things you need to know is the peak power of the motor in Watts, the coefficient of drag, the density of the air and the frontal area of the vehicle.
If you don't know the rolling resistance or the drag coefficient - you can do it experimentally. Get a friend to sit in the car with you with a stopwatch. Find a long, straight, level, empty, stretch of typical road on a windless day - get the car up to (say) 70mph - then put the car in neutral and without touching the brakes - let it coast until it stops. (Please be VERY careful doing this!) Have your copilot note the times at which you reach 65mph, 60mph, 55mph...all the way down to 5mph. Armed with those times & speeds - you can calculate the acceleration (well, 'deceleration') every 5mph from 70mph to zero. Knowing the mass of the car (and that Force = Mass x Acceleration) - you can calculate the force due to a combination of drag and rolling resistance that caused that deceleration - and you can do that for every 5mph of speed. Multiplying each number by that speed gives you the amount of power the engine would have to provide in order to counteract the combined drag/friction forces (Power = Force x Velocity). Get some graph paper and plot that graph. Now you have a graph of engine power needed to maintain a constant speed versus speed. Since drag is proportional to the speed squared and rolling resistance is proportional the speed you should get a curve that indicates a steeply increasing power demand at higher speeds. Now - all you have to do is extrapolate that curve until the amount of power required equals the power that the engine can produce at the wheels. That's your top speed!
SteveBaker (talk) 23:52, 10 June 2009 (UTC)

## Bacterial Infections

Why are Gram Negative infections supposed to be so much worse for the human body than Gram Positive ones? Is this actually true? 87.115.3.165 (talk) 22:02, 10 June 2009 (UTC)

Have you read our pages yet? Gram-negative bacteria, Gram-positive bacteria "worse for the human body" isn't a very qualifying term. More drug resistant, higher number of infectious species, shorter incubation time, higher reproduction rate, primarily targets essential organs, etc.? The Immune system has a lot to do with it. 71.236.26.74 (talk) 22:19, 10 June 2009 (UTC)
See Endotoxin. --NorwegianBlue talk 00:15, 11 June 2009 (UTC)

I'm not a microbiologist (I just happen to have taken college-level biochem), but as far as I remember, Gram-negative bacteria may be somewhat harder for the immune system to kill because they have a thick protective layer of glycoproteins on the outside of their cell walls. It doesn't make all that much of a difference, though, so there are probably other factors involved as well. FWiW 76.21.37.87 (talk) 04:28, 14 June 2009 (UTC)

## Grandma's cold treatment

I got a cold, and a friend advised me to put salted water into my nose. Question: has anybody heard about this remedy? I am not necessarily asking for a medical advice about its validity, I just would like to make sure it's not another :-( practical joke on me, before trying out --pma (talk) 22:42, 10 June 2009 (UTC)

It's not new, see [3] [4] [5] and also nasal irrigation. Nanonic (talk) 22:48, 10 June 2009 (UTC)
It may or may not help - but it's not a practical joke - my wife also swears by it. However - we're not allowed to offer medical advice here - so we can't tell you whether it works or not - or whether to try it or not. SteveBaker (talk) 02:47, 11 June 2009 (UTC)
I can say from firsthand experience that it feels weirder than shit the first time you try it. --jpgordon::==( o ) 15:52, 11 June 2009 (UTC)
Why did you try stuffing shit up your nose? O_o Vimescarrot (talk) 18:26, 11 June 2009 (UTC)
it is not a treatment for a cold, its a way to clear your sinus, like when you go swimming in the ocean you will often end up with snot coming out of your nose - its the salt water that causes it.

--58.111.132.138 (talk) 12:02, 13 June 2009 (UTC)

Thanks everybody for the opinions and the useful links! I had a great time in experimentation, with the exception of the first trial --it went directly in the brain, I suppose. Anyway, the lesson of our fathers proved useful: carefully examine your failure; do not give up; correct your method and go ahead! Now it works smoothly. --pma (talk) 07:53, 15 June 2009 (UTC)