November 6

Are there techniques for finding properties of a thermally unstable compound (TUS) beyond temperatures at which significant decomposition takes place? For example, even though mercury(II) hydride readily decomposes at approximately -125 degrees Celsius, is it practical to heat a sample faster than it can decompose and still be able to determine whether or not the decomposition temperature coincides with the melting point? How would I go about detecting a phase chance on the molecular scale under such dynamic conditions? Plasmic Physics (talk) 09:52, 6 November 2016 (UTC)[reply]

For higher temperatures this is certainly possible, and you may be able to overheat by some tens of degrees if you do it fast. However the decomposition products will contaminate your product and disturb the properties. Another issue is that heat is released during the decomposition, there will be positive runaway feedback, and it likely explodes. Perhaps you can subject the solid to high pressure, or high pressure under hydrogen to try to reverse the decomposition. Perhaps a thin layer on some substrate will be more stable. You may be able to use X-rays, Raman spectroscopy, optical dispersion, mechanical properties (eg the sound of the explosion), or even Neutron diffraction or Positron annihilation spectroscopy to find some properties. To see if the liquid is intrinsically unstable you could try dissolving your product in suitable liquids (liquid nitrogen, liquid xenon) and see if it can be recrystalised. Graeme Bartlett (talk) 10:52, 6 November 2016 (UTC)[reply]

What if I prepare microcrystals on a monolayer substrate, and heat the substrate in half-degree multiples, under hydrogen pressure? If I can get a good size distribution of microcrystals, I should be able to project behaviour for the bulk. I don't know if reversing the decomposition process is feasible, on account of the fact that atomic mercury requires excitation for insertion into the dihydrogen bond. Ergo, the pressure required may be too high. Yes... exposure of the compound to a pressure range could yield informative results. Recrystallization, if it works could improve the limited understanding of the solid structure. Plasmic Physics (talk) 10:26, 7 November 2016 (UTC)[reply]

Phase changes don't happen instantly. Whether this makes sense depends on a comparison between the time scale of decomposition and the time scale of the phase change. I believe that in most cases the decomposition will happen faster. Looie496 (talk) 14:37, 6 November 2016 (UTC)[reply]

• That exactly. Steel is "unstable" at room temperature (it demixes very very slowly into Fe and Fe3C) but there are tons of tables for its mechanical properties. Tigraan^{Click here to contact me} 17:20, 6 November 2016 (UTC)[reply]

HgH₂ mercury(II) dihydride.
Sleigh (talk) 05:01, 7 November 2016 (UTC)[reply]

The mechanism of decomposition appears to be loss of one hydrogen atom, which then reacts with other hydrogen to yield H²: HgH₂→H + HgH ; HgH → Hg + H (HgH is even less stable) ; H + HgH₂→H₂ + HgH; The overall reaction is exothermic, and once you start forming loose hydrogen atoms it will go out of control fast. This from http://pubs.rsc.org/en/Content/ArticleLanding/2005/CP/b412373e . The solid is made from molecules fairly weakly held together. So it is probably not stabilising the molecules much. This suggests that it is not reorienting or randomising the molecules that makes it unstable. So decomposition could be a long way from the theoretical melting point. The molecule is quite symmetrical, and so it can have a high melting point. cf

HgCl2 Graeme Bartlett (talk) 06:06, 7 November 2016 (UTC)[reply

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Perhaps the mercurophilic intermolecular forces have a stabalising tuning effect on the Hg-H bonding? You could be right regarding the high melting point, but the dipoles in HgCl2 is very much stronger than in HgH2, and what about carbon dioxide which is also a very symmetrical molecule? Plasmic Physics (talk) 10:26, 7 November 2016 (UTC)[reply]

The Hg to Hg bonding seems to very very weak, and you don't end up with a HHgHgH molecule at any point. That paper only says dispersion forces are holding the solid together, not any extra Hg-Hg stabilisation. Did you look at the Xuefeng Wang and Lester Andrews paper? Graeme Bartlett (talk) 10:47, 7 November 2016 (UTC)[reply]

Yes, I did look at their paper. Metallophilic forces are dispersion forces. I also looked at their paper published one year earlier, titled: Solid Mercury Dihydride: Mercurophilic Bonding in Molecular HgH2 Polymers. Plasmic Physics (talk) 18:11, 7 November 2016 (UTC)[reply]

Carbon dioxide melts at -56° which is quite elevated compared to other molecules with similar or higher mass, but less neat shapes. eg

ClF. HHgH has a somewhat similar mass to Polonium hydride which melts at -35°. But HHgH being symmetrical would have a higher theoretical melting point. A theoretical boiling point could be comparable between the two, but is made more complex because of dimers and trimers. Graeme Bartlett (talk) 21:56, 7 November 2016 (UTC)[reply

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The

bomb calorimeter is used to get the enthalpy of decomposition on heating of samples of compounds. Heating is supplied by a wire electric heater inside the bomb, so I suppose you could try, depending on the sample's thermal stability, to superheat the sample and get some of the information you're looking for. loupgarous (talk) 02:33, 9 November 2016 (UTC)[reply

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That article describes a plant with yellow or cream flowers, growing in Colombia. When you eliminate the species with red flowers, and those which don't grow as far north as Colombia, you are reduced to either B. aurea or B. arborea. Wymspen (talk) 11:48, 6 November 2016 (UTC)[reply]

Thanks. I don't understand what's with the red flowers though. —  Ark25  (talk) 16:57, 6 November 2016 (UTC)[reply]

There are seven species of Brugmansia - several of which have reddish coloured flowers, rather than the yellow flowers of B. aurea. Have a look at Brugmansia sanguinea - the name gives a good clue to the blood red colour. Wymspen (talk) 09:16, 7 November 2016 (UTC)[reply]

I saw in this video on youtube (and also on facebook) a group of Indian people who makes some things which look like dangerous things (such as: getting a lot of beats which normal body can not survive after them and getting an heavy hammer into the chest or one of them was run over by car and motorcycle at the same time - on the torso and the head and laying on standing nails with naked upper bodies), but after that in the past I asked here about phenomenon of putting hand into boiled oil by Indian you linked me the trick behind it and then I saw that every one can actually do that, so then I started to cast doubt in any exceptional phenomenon. What is the explanation in this case? 93.126.88.30 (talk) 15:18, 6 November 2016 (UTC)[reply]

The usual explanation for many of these things is the difference between force and pressure. The human body can withstand a great deal of force, so long as it is distributed over a wide area. The bed of nails provides many not-so-sharp nails, to spread the force over a wide area. The hammer also spreads the force out. Replace that with a pick, then apply the same force, and it would cause major damage. StuRat (talk) 15:35, 6 November 2016 (UTC)[reply]

See also our

Bed of Nails article. CodeTalker (talk) 16:31, 6 November 2016 (UTC)[reply

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And see Leidenfrost effect for the "boiling oil" phenomenon. Tevildo (talk) 22:13, 6 November 2016 (UTC)[reply]

You need strong core muscles to be able to do this safely. Your core muscles will absorb the shock waves, and deflect them so that your organs don't get injured. To deal with more static pressures, as StuRat points out, what matters is that the force exerted on your body will be spread out, but to make sure this indeed happens when a car is running over your belly, you need to be able to keep your core muscles tight. If these exercises are not too easy for you, you probably won't be able to do that. Count Iblis (talk) 03:12, 7 November 2016 (UTC)[reply]

Hello , what is the highest speed of a target that tanks with the best fire control system can hit ? 149.200.231.84 (talk) 16:21, 6 November 2016 (UTC) In other words can tanks hit the opponent tanks driving at 45 km/hr speed ? 149.200.231.84 (talk) 16:45, 6 November 2016 (UTC)[reply]

In theory a tank can hit a target going at any speed - even a jet aircraft flying past at low level. However, the faster the target, the greater the element of chance in actually scoring a hit. Wymspen (talk) 16:52, 6 November 2016 (UTC)[reply]

Quantitative details about the capabilities of modern weapon systems are not generally released to the public. It is not likely that you, or anyone, will be able to find a reliable published source that provides very exact information. For example, read about how the term "technical data" is defined and used in this context; and read all about the rules that pertain to it. You can find out, for example, that a commercially-available motor exists which can accurately track some angular velocity ...but you won't reliably be able to determine that weapon-system-X uses motor-model-number-Y ... and so on.

You can try to "guess" at the limitations by estimating engineering parameters from first principles of physics and from generic engineering concepts, but if you're honest with yourself and your math, you'll probably end up with an answer "somewhere in the range between DC to light" - or, to use a little less jargon, somewhere between negative-infinity and positive-infinity.

Some great resources to read include Jane's numerous publications, and the website of the Federation of American Scientists. Both resources publish perspectives on defense systems, and often link to specific programs provided by contractors like Raytheon's AFATDS and to public data from

DTIC, like this public program review (2005) of the technology development during the development of the Abrams tank

. That report, written by some very well-credentialed authors, includes the startling and fascinating claim that the upgrade to the M1A2 digital computer fire control system quantitatively degraded fire rate and fire accuracy compared to the earlier M1A1.

To quote one other author who holds notoriously strong opinions on the efficacy of digital computers, "I have not yet made up my mind on how I feel about relying on a digital computer during critical phases..."

Nimur (talk) 16:03, 7 November 2016 (UTC)[reply]

Thank you very much 188.247.77.185 (talk) 20:21, 7 November 2016 (UTC) Great information , but I supposed an idea and I tried to test it , that if battle tanks can protect themselves by driving at high speed , so we can say that the differences in fire range and accuracy between different types of tanks can be cured by this technique .92.253.23.28 (talk) 04:36, 8 November 2016 (UTC)[reply]

Well, fire control systems function by taking the following and predicting where you need to aim the gun: Gunner's speed and heading; target's speed and heading; round velocity; distance to target; weather conditions; elevation difference. So the question of accuracy is a question of how well these parameters can be measured or (in the case of round velocity) predicted. And I completely agree with Nimur, that how accurately this can be done is going to be a state secret. Driving faster and faster will decrease the gunner's accuracy, since discrepancies between measured and actual velocity translate to larger distances between the target and the round's trajectory as target velocity increases. But without technical information about an MBT's fire control system and the fidelity of a round's nominal velocity, we can't know how fast you'd have to go to seriously undermine a tank's accuracy. I'd imagine a more effective technique, if possible, would be a jinking maneuver, which we don't seem to have an article on. Basically, this consists of making random changes to your direction of travel, and is (or at least was) taught to combat pilots as a method of avoiding direct-fire anti-aircraft weapons. Imagine I'm fleeing from an M1 Abrams at a range of 3000m, which is a bit over 2 seconds of travel time for its APFSDS round. If I'm traveling at 30m/s, and I turn 90 degrees at the moment the tank fires, then by the time that 2 seconds has passed, I'll be about 85 meters from where I would have been, had I continued in a straight line, and that round will almost certainly miss. There's nothing an unguided weapon can do about this maneuver if my movements are truly unpredictable - given a certain opponent, there is always a range and speed I can stay at where I'll always be able to dodge his projectiles. Of course, if I screw up once my vehicle is destroyed. Or my opponent can lob high explosives and hope the shrapnel gets me. Or if he has a lot of friends, they can spread out their fire to take into account my possible choices of direction. But I suppose if you're just trying to reach cover, or you're faster than the tank and trying to get away, this might be your best chance. Someguy1221 (talk) 05:05, 8 November 2016 (UTC)[reply]

Very great , thank you all 37.202.116.92 (talk) 04:09, 9 November 2016 (UTC)[reply]