{"content": "Title V of the URAA made several modifications to the Copyright law of the United States. It amended Title 17 (\"Copyrights\") of the United States Code to include a completely reworded article 104A on copyright restorations on foreign works and to include a new chapter 11, containing a prohibition of bootleg sound and video recordings of live performances. In Title 18 of the U.S. Code, a new article 2319A was inserted, detailing the penal measures against infringements of this new bootlegging prohibition.\n\nThe U.S. had joined the Berne Convention on March 1, 1989, when its Berne Convention Implementation Act of 1988 entered in force. Article 18 of the Berne Convention specified that the treaty covered all works that were still copyrighted in their source country and that had not entered the public domain in the country where copyright was claimed due to the expiration of a previously granted copyright there. Consequently, the U.S. would have had to grant copyright on foreign works that were never copyrighted before in the U.S. But the United States denied this retroactivity of the Berne Convention and applied the rules of the treaty only to works first published after March 1, 1989. Earlier foreign works that were not covered by other treaties and that had until then not been subject to copyright in the U.S. remained uncopyrighted in the United States.\n\nThe U.S. faced harsh criticism for its unilateral denouncement of the retroactivity of the Berne Convention defined in article 18, and ultimately reversed its position. The copyright changes implemented by the URAA in 17 USC 104A remedied the situation and brought the U.S. legislation in-line with the requirements of the Berne Convention.\n\n17 U.S.C. § 104A effectively copyrights many foreign works that were never before copyrighted in the U.S. The works are subject to the normal U.S. copyright term, as if they had never entered the public domain.\n\nThe affected works are those which were in the public domain either due to a lack of international copyright agreements between the U.S. and the country of origin of the work, or due to a failure to meet U.S. copyright registration and notification formalities. Also affected are works which did have previous U.S. copyright, but which entered the public domain due to a failure to renew the copyright. The law defines all of the affected works as \"restored works\" and the copyright granted to them as \"restored copyright\", even though many of the works never had U.S. copyright to restore.", "doc_id": "law_ex_111149_doc_2"}
{"content": "Legal traditions differ on whether a work in the public domain can have its copyright restored. In the European Union, the Copyright Duration Directive was applied retroactively, restoring and extending the terms of copyright on material previously in the public domain. Term extensions by the US and Australia generally have not removed works from the public domain, but rather delayed the addition of works to it. However, the United States moved away from that tradition with the Uruguay Round Agreements Act, which removed from the public domain many foreign-sourced works that had previously not been in copyright in the US for failure to comply with US-based formalities requirements. Consequently, in the US, foreign-sourced works and US-sourced works are now treated differently, with foreign-sourced works remaining under copyright regardless of compliance with formalities, while domestically sourced works may be in the public domain if they failed to comply with then-existing formalities requirements—a situation described as odd by some scholars, and unfair by some US-based rightsholders.", "doc_id": "law_ex_111149_doc_0"}
{"content": "Golan v. Holder, 565 U.S. 302 (2012), was a US Supreme Court case that dealt with copyright and the public domain. It held that the \"limited time\" language of the United States Constitution's Copyright Clause does not preclude the extension of copyright protections to works previously in the public domain. In particular, the case challenged the constitutionality of the application of Section 514 of the Uruguay Round Agreements Act of 1994, which implemented the provisions of trade agreements seeking to equalize copyright protection on an international basis. In the United States, the Act newly granted copyright status to foreign works previously in the public domain.\n\nThe two main arguments against the application of the Act in the case were that restoring copyright violates the \"limited time\" language of the United States Constitution's Copyright Clause, and that restoring to copyright works that had passed into the public domain interferes with the people's First Amendment right to use, copy and otherwise exploit the works and to freely express themselves through these works, thus also violating the Constitution's Copyright Clause.\n\nThe US Supreme Court held on January 18, 2012 that Section 514 of the Uruguay Round Agreements Act does not exceed Congress's authority under the Copyright Clause, and the court affirmed the judgment of the lower court by 6–2, with the opinion written by Justice Ginsburg. The practical effect of the decision is to confirm that works that were previously free to use, such as Prokofiev's Peter and the Wolf, are no longer in the public domain and are subject to use only with the permission of the copyright holder, such as", "doc_id": "law_ex_111149_doc_3"}
{"content": "Sec. 2 · Declarations.\n\nThe Congress makes the following declarations:\n\n(1) The Convention for the Protection of Literary and Artistic Works, signed at Berne, Switzerland, on September 9, 1886, and all acts, protocols, and revisions thereto (hereafter in this Act referred to as the \"Berne Convention\") are not self-executing under the Constitution and laws of the United States.\n\n(2) The obligations of the United States under the Berne Convention may be performed only pursuant to appropriate domestic law.\n\n(3) The amendments made by this Act, together with the law as it exists on the date of the enactment of this Act, satisfy the obligations of the United States in adhering to the Berne Convention and no further rights or interests shall be recognized or created for that purpose.\n\nSec. 12 · Works in the public domain.\n\nTitle 17, United States Code, as amended by this Act, does not provide copyright protection for any work that is in the public domain in the United States.\n\nSec. 13 · Effective date: effect on pending cases.\n\n(a) Effective Date.—This Act and the amendments made by this Act take effect on the date on which the Berne Convention (as defined in section 101 of title 17, United States Code) enters into force with respect to the United States.\n\n(b) Effect on Pending Cases.—Any cause of action arising under title 17, United States Code, before the effective date of this Act shall be governed by the provisions of such title as in effect when the cause of action arose.", "doc_id": "law_ex_111149_doc_1"}
{"content": "I thought we shouldn't lubricate brushless motors?\n\nIt's true that we don't have to lubricate brushless motors with ball bearings. Ball bearings are sealed with grease inside, so applying oil doesn't really help much but actually could be counter-productive as well as making a mess because oil will attract dusts and dirt.\n\nThat's a different story when it comes to micro motors such as 0603 and 0703, as they don't have ball bearings but brass bushings (sintered bearings), which creates way more friction than ball bearings, not to mention they run at extremely high RPM and that only makes it worse. Oiling them reduces friction and help them run smoother! This makes more noticeable difference on lower quality motors.", "doc_id": "drones_ex_3025_doc_0"}
{"content": "A brushless motor used in RC is actually quite simple mechanically. There are only a few moving parts in the entire assembly. Those moving parts are limited to the rotor containing the permanent magnets and the bearings that support it.\n\nWhen considering lubrication, the only area that could possibly require it is the bearings, right? To understand the answer to this question, we have to get a good understanding as to how the bearings are protected. The typical bearings found in an RC motor have a shield that helps protect the bearings from foreign object debris AKA dirt. The key takeaway from this is that the bearings are not water proof and can certainly allow liquids to pass by the shield.\n\nCleaning the Bearings\n\nIf you happen to try and clean the bearing with a product similar to WD-40, the oil can actually pass by the shield and mix with the factory bearing lubrication ultimately leading to possible damage to the bearing. To clean the bearing, my preference is to take a dry cloth and wipe any dirt away from the visible bearing surface. I do not like to use any chemicals.\n\nLubricating the Bearings\n\nThe quick answer to lubricating or not is that I recommend not lubricating the bearings. First off, it's not as simple as just placing oil on to the bearings hoping to get by the shield of the bearing. Light oils may be able to make it easily past the shield of the bearing, but do not have the required lubrication characteristics. Light oils will more than likely not be able to withstand the concentrated heat and friction created within the bearing. The oil would burn off and subject the balls within the bearing to excessive frictional loads.\n\nTo correctly complete the process the shield must be removed, the balls (whole race) must be cleaned and then a proper bearing grease must be used prior to placing the shield back on to seal the deal.", "doc_id": "drones_ex_3025_doc_1"}
{"content": "Do NOT use a normal oil. It will attracts fine dust and sands. It is not visible to naked eye. Once it is sucked into the propeller bearings. It causes frictions to the bearings and eventually it impacts the high speed bearing. You will notice the weaker propeller spinning compare to the rest. Then, the drone will try to compensate by sending more power to the weaker propeller to ensure it is steady. This consumes a lot of battery.", "doc_id": "drones_ex_3025_doc_2"}
{"content": "The bearings are shielded- not sealed.\n\nThe reality is the bearings will probably outlast the service life of the drone without any additional oiling.\n\nDon't any magic \"salt absorbing\" or \"moisturising\" Magic oil in shielded bearings. The factory lube is sufficient for the service life.", "doc_id": "drones_ex_3025_doc_3"}
{"content": "Diversity[edit]\nThere are a large number of different odor receptors, with as many as 1,000 in the mammalian genome which represents approximately 3% of the genes in the genome.  However, not all of these potential odor receptor genes are expressed and functional.  According to an analysis of data derived from the Human Genome Project, humans have approximately 400 functional genes coding for olfactory receptors, and the remaining 600 candidates are pseudogenes.\nThe reason for the large number of different odor receptors is to provide a system for discriminating between as many different odors as possible.  Even so, each odor receptor does not detect a single odor.  Rather each individual odor receptor is broadly tuned to be activated by a number of similar odorant structures.  Analogous to the immune system, the diversity that exists within the olfactory receptor family allows molecules that have never been encountered before to be characterized. However, unlike the immune system, which generates diversity through in-situ recombination, every single olfactory receptor is translated from a specific gene; hence the large portion of the genome devoted to encoding OR genes. Furthermore, most odors activate more than one type of odor receptor. Since the number of combinations and permutations of olfactory receptors is very large, the olfactory receptor system is capable of detecting and distinguishing between a very large number of odorant molecules.\nDeorphanization of odor receptors can be completed using electrophysiological and imaging techniques to analyze the response profiles of single sensory neurons to odor repertoires. Such data open the way to the deciphering of the combinatorial code of the perception of smells.\nSuch diversity of OR expression maximizes the capacity of olfaction. Both monoallelic OR expression in a single neuron and maximal diversity of OR expression in the neuron population are essential for specificity and sensitivity of olfactory sensing. Thus, olfactory receptor activation is a dual-objective design problem. Using mathematical modeling and computer simulations, Tian et al proposed an evolutionarily optimized three-layer regulation mechanism, which includes zonal segregation, epigenetic barrier crossing coupled to a negative feedback loop and an enhancer competition step\n\n. This model not only recapitulates monoallelic OR expression but also elucidates how the olfactory system maximizes and maintains the diversity of OR expression.", "doc_id": "RGB_equivalent_for_smells/Olfactory_receptor_3.txt"}
{"content": "Physiological basis in vertebrates[edit]\nMain olfactory system[edit]\n\nMain article: Olfactory system\nIn humans and other vertebrates, smells are sensed by olfactory sensory neurons in the olfactory epithelium. The olfactory epithelium is made up of at least six morphologically and biochemically different cell types. The proportion of olfactory epithelium compared to respiratory epithelium (not innervated, or supplied with nerves) gives an indication of the animal's olfactory sensitivity. Humans have about 10 cm (1.6 sq in) of olfactory epithelium, whereas some dogs have 170 cm (26 sq in). A dog's olfactory epithelium is also considerably more densely innervated, with a hundred times more receptors per square centimeter. The sensory olfactory system integrates with other senses to form the perception of flavor. Often, land organisms will have separate olfaction systems for smell and taste (orthonasal smell and retronasal smell), but water-dwelling organisms usually have only one system.\nMolecules of odorants passing through the superior nasal concha of the nasal passages dissolve in the mucus that lines the superior portion of the cavity and are detected by olfactory receptors on the dendrites of the olfactory sensory neurons. This may occur by diffusion or by the binding of the odorant to odorant-binding proteins. The mucus overlying the epithelium contains mucopolysaccharides, salts, enzymes, and antibodies (these are highly important, as the olfactory neurons provide a direct passage for infection to pass to the brain). This mucus acts as a solvent for odor molecules, flows constantly, and is replaced approximately every ten minutes.\nIn insects, smells are sensed by olfactory sensory neurons in the chemosensory sensilla, which are present in insect antenna, palps, and tarsa, but also on other parts of the insect body. Odorants penetrate into the cuticle pores of chemosensory sensilla and get in contact with insect odorant-binding proteins (OBPs) or Chemosensory proteins (CSPs), before activating the sensory neurons.\nReceptor neuron[edit]\nThe binding of the ligand (odor molecule or odorant) to the receptor leads to an action potential in the receptor neuron, via a second messenger pathway, depending on the organism. In mammals, the odorants stimulate adenylate cyclase to synthesize cAMP via a G protein called Golf. cAMP, which is the second messenger here, opens a cyclic nucleotide-gated ion channel (CNG), producing an influx of cations (largely Ca with some Na) into the cell, slightly depolarising it. The Ca in turn opens a Ca-activated chloride channel, leading to efflux of Cl, further depolarizing the cell and triggering an action potential. Ca is then extruded through a sodium-calcium exchanger. A calcium-calmodulin complex also acts to inhibit the binding of cAMP to the cAMP-dependent channel, thus contributing to olfactory adaptation.\nThe main olfactory system of some mammals also contains small subpopulations of olfactory sensory neurons that detect and transduce odors somewhat differently. Olfactory sensory neurons that use trace amine-associated receptors (TAARs) to detect odors use the same second messenger signaling cascade as do the canonical olfactory sensory neurons. Other subpopulations, such as those that express the receptor guanylyl cyclase GC-D (Gucy2d) or the soluble guanylyl cyclase Gucy1b2, use a cGMP cascade to transduce their odorant ligands. These distinct subpopulations (olfactory subsystems) appear specialized for the detection of small groups of chemical stimuli.\nThis mechanism of transduction is somewhat unusual, in that cAMP works by directly binding to the ion channel rather than through activation of protein kinase A. It is similar to the transduction mechanism for photoreceptors, in which the second messenger cGMP works by directly binding to ion channels, suggesting that maybe one of these receptors was evolutionarily adapted into the other. There are also considerable similarities in the immediate processing of stimuli by lateral inhibition.\nAveraged activity of the receptor neurons can be measured in several ways. In vertebrates, responses to an odor can be measured by an electro-olfactogram or through calcium imaging of receptor neuron terminals in the olfactory bulb. In insects, one can perform electroantennography or calcium imaging within the olfactory bulb.\nOlfactory bulb projections[edit]\nSchematic of the early olfactory system including the olfactory epithelium and bulb. Each ORN expresses one OR that responds to different odorants. Odorant molecules bind to ORs on cilia. ORs activate ORNs that transduce the input signal into action potentials. In general, glomeruli receive input from ORs of one specific type and connect to the principal neurons of the OB, mitral and tufted cells (MT cells).\nOlfactory sensory neurons project axons to the brain within the olfactory nerve, (cranial nerve I). These nerve fibers, lacking myelin sheaths, pass to the olfactory bulb of the brain through perforations in the cribriform plate, which in turn projects olfactory information to the olfactory cortex and other areas. The axons from the olfactory receptors converge in the outer layer of the olfactory bulb within small (≈50 micrometers in diameter) structures called glomeruli. Mitral cells, located in the inner layer of the olfactory bulb, form synapses with the axons of the sensory neurons within glomeruli and send the information about the odor to other parts of the olfactory system, where multiple signals may be processed to form a synthesized olfactory perception. A large degree of convergence occurs, with 25,000 axons synapsing on 25 or so mitral cells, and with each of these mitral cells projecting to multiple glomeruli. Mitral cells also project to periglomerular cells and granular cells that inhibit the mitral cells surrounding it (lateral inhibition). Granular cells also mediate inhibition and excitation of mitral cells through pathways from centrifugal fibers and the anterior olfactory nuclei. Neuromodulators like acetylcholine, serotonin and norepinephrine all send axons to the olfactory bulb and have been implicated in gain modulation, pattern separation, and memory functions, respectively.\nThe mitral cells leave the olfactory bulb in the lateral olfactory tract, which synapses on five major regions of the cerebrum: the anterior olfactory nucleus, the olfactory tubercle, the amygdala, the piriform cortex, and the entorhinal cortex. The anterior olfactory nucleus projects, via the anterior commissure, to the contralateral olfactory bulb, inhibiting it. The piriform cortex has two major divisions with anatomically distinct organizations and functions. The anterior piriform cortex (APC) appears to be better at determining the chemical structure of the odorant molecules, and the posterior piriform cortex (PPC) has a strong role in categorizing odors and assessing similarities between odors (e.g. minty, woody, and citrus are odors that can, despite being highly variant chemicals, be distinguished via the PPC in a concentration-independent manner). The piriform cortex projects to the medial dorsal nucleus of the thalamus, which then projects to the orbitofrontal cortex. The orbitofrontal cortex mediates conscious perception of the odor. The three-layered piriform cortex projects to a number of thalamic and hypothalamic nuclei, the hippocampus and amygdala and the orbitofrontal cortex, but its function is largely unknown. The entorhinal cortex projects to the amygdala and is involved in emotional and autonomic responses to odor. It also projects to the hippocampus and is involved in motivation and memory. Odor information is stored in long-term memory and has strong connections to emotional memory. This is possibly due to the olfactory system's close anatomical ties to the limbic system and hippocampus, areas of the brain that have long been known to be involved in emotion and place memory, respectively.\nSince any one receptor is responsive to various odorants, and there is a great deal of convergence at the level of the olfactory bulb, it may seem strange that human beings are able to distinguish so many different odors. It seems that a highly complex form of processing must be occurring; however, as it can be shown that, while many neurons in the olfactory bulb (and even the pyriform cortex and amygdala) are responsive to many different odors, half the neurons in the orbitofrontal cortex are responsive to only one odor, and the rest to only a few. It has been shown through microelectrode studies that each individual odor gives a particular spatial map of excitation in the olfactory bulb. It is possible that the brain is able to distinguish specific odors through spatial encoding, but temporal coding must also be taken into account. Over time, the spatial maps change, even for one particular odor, and the brain must be able to process these details as well.\nInputs from the two nostrils have separate inputs to the brain, with the result that, when each nostril takes up a different odorant, a person may experience perceptual rivalry in the olfactory sense akin to that of binocular rivalry.\nIn insects, smells are sensed by sensilla located on the antenna and maxillary palp and first processed by the antennal lobe (analogous to the olfactory bulb), and next by the mushroom bodies and lateral horn.\nCoding and perception[edit]\nThe process by which olfactory information is coded in the brain to allow for proper perception is still being researched, and is not completely understood. When an odorant is detected by receptors, they in a sense break the odorant down, and then the brain puts the odorant back together for identification and perception. The odorant binds to receptors that recognize only a specific functional group, or feature, of the odorant, which is why the chemical nature of the odorant is important.\nAfter binding the odorant, the receptor is activated and will send a signal to the glomeruli  in the olfactory bulb. Each glomerulus receives signals from multiple receptors that detect similar odorant features. Because several receptor types are activated due to the different chemical features of the odorant, several glomeruli are activated as well. The signals from the glomeruli are transformed to a pattern of oscillations of neural activities of the mitral cells, the output neurons from the olfactory bulb. Olfactory bulb sends this pattern to the olfactory cortex. Olfactory cortex is thought to have associative memories, so that it resonates to this bulbar pattern when the odor object is recognized. The cortex sends centrifugal feedback to the bulb. This feedback could suppress bulbar responses to the recognized odor objects, causing olfactory adaptation to background odors, so that the newly arrived foreground odor objects could be singled out for better recognition. During odor search, feedback could also be used to enhance odor detection. The distributed code allows the brain to detect specific odors in mixtures of many background odors.\nIt is a general idea that the layout of brain structures corresponds to physical features of stimuli (called topographic coding), and similar analogies have been made in smell with concepts such as a layout corresponding to chemical features (called chemotopy) or perceptual features. While chemotopy remains a highly controversial concept, evidence exists for perceptual information implemented in the spatial dimensions of olfactory networks.\nAccessory olfactory system[edit]\nMany animals, including most mammals and reptiles, but not humans, have two distinct and segregated olfactory systems: a main olfactory system, which detects volatile stimuli, and an accessory olfactory system, which detects fluid-phase stimuli. Behavioral evidence suggests that these fluid-phase stimuli often function as pheromones, although pheromones can also be detected by the main olfactory system. In the accessory olfactory system, stimuli are detected by the vomeronasal organ, located in the vomer, between the nose and the mouth. Snakes use it to smell prey, sticking their tongue out and touching it to the organ. Some mammals make a facial expression called flehmen to direct stimuli to this organ.\nThe sensory receptors of the accessory olfactory system are located in the vomeronasal organ. As in the main olfactory system, the axons of these sensory neurons project from the vomeronasal organ to the accessory olfactory bulb, which in the mouse is located on the dorsal-posterior portion of the main olfactory bulb. Unlike in the main olfactory system, the axons that leave the accessory olfactory bulb do not project to the brain's cortex but rather to targets in the amygdala and bed nucleus of the stria terminalis, and from there to the hypothalamus, where they may influence aggression and mating behavior.", "doc_id": "RGB_equivalent_for_smells/Sense_of_smell_5.txt"}
{"content": "Machine-based smelling[edit]\n\nMain article: Machine olfaction\nScientists have devised methods for quantifying the intensity of odors, in particular for the purpose of analyzing unpleasant or objectionable odors released by an industrial source into a community. Since the 1800s industrial countries have encountered incidents where proximity of an industrial source or landfill produced adverse reactions among nearby residents regarding airborne odor. The basic theory of odor analysis is to measure what extent of dilution with \"pure\" air is required before the sample in question is rendered indistinguishable from the \"pure\" or reference standard. Since each person perceives odor differently, an \"odor panel\" composed of several different people is assembled, each sniffing the same sample of diluted specimen air. A field olfactometer can be utilized to determine the magnitude of an odor.\nMany air management districts in the US have numerical standards of acceptability for the intensity of odor that is allowed to cross into a residential property. For example, the Bay Area Air Quality Management District has applied its standard in regulating numerous industries, landfills, and sewage treatment plants. Example applications this district has engaged are the San Mateo, California, wastewater treatment plant; the Shoreline Amphitheatre in Mountain View, California; and the IT Corporation waste ponds, Martinez, California.\n", "doc_id": "RGB_equivalent_for_smells/Sense_of_smell_8.txt"}
{"content": "part5 -------------------\nApparent paradox[edit]\nAn equation similar to that of Kelvin can be derived for the solubility of small particles or droplets in a liquid, by means of the connection between vapour pressure and solubility, thus the Kelvin equation also applies to solids, to slightly soluble liquids, and their solutions if the partial pressure \n\n\n\np\n\n\n{\\displaystyle p}\n\n is replaced by the solubility of the solid (\n\n\n\nc\n\n\n{\\displaystyle c}\n\n) (or a second liquid) at the given radius, \n\n\n\nr\n\n\n{\\displaystyle r}\n\n, and \n\n\n\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p_{\\rm {sat}}}\n\n by the solubility at a plane surface (\n\n\n\n\nc\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle c_{\\rm {sat}}}\n\n). Hence small particles (like small droplets) are more soluble than larger ones.  The equation would then be given by:\nln\n⁡\n\n\nc\n\nc\n\n\ns\na\nt\n\n\n\n\n\n=\n\n\n\n2\nγ\n\nV\n\nm\n\n\n\n\nr\nR\nT\n\n\n\n.\n\n\n{\\displaystyle \\ln {\\frac {c}{c_{\\rm {sat}}}}={\\frac {2\\gamma V_{\\text{m}}}{rRT}}.}\nThese results led to the problem of how new phases can ever arise from old ones. For example, if a container filled with water vapour at slightly below the saturation pressure is suddenly cooled, perhaps by adiabatic expansion, as in a cloud chamber, the vapour may become supersaturated with respect to liquid water. It is then in a metastable state, and we may expect condensation to take place. A reasonable molecular model of condensation would seem to be that two or three molecules of water vapour come together to form a tiny droplet, and that this nucleus of condensation then grows by accretion, as additional vapour molecules happen to hit it. The Kelvin equation, however, indicates that a tiny droplet like this nucleus, being only a few ångströms in diameter, would have a vapour pressure many times that of the bulk liquid. As far as tiny nuclei are concerned, the vapour would not be supersaturated at all. Such nuclei should immediately re-evaporate, and the emergence of a new phase at the equilibrium pressure, or even moderately above it should be impossible. Hence, the over-saturation must be several times higher than the normal saturation value for spontaneous nucleation to occur.\nThere are two ways of resolving this paradox. In the first place, we know the statistical basis of the second law of thermodynamics. In any system at equilibrium, there are always fluctuations around the equilibrium condition, and if the system contains few molecules, these fluctuations may be relatively large. There is always a chance that an appropriate fluctuation may lead to the formation of a nucleus of a new phase, even though the tiny nucleus could be called thermodynamically unstable. The chance of a fluctuation is e, where ΔS is the deviation of the entropy from the equilibrium value.\nIt is unlikely, however, that new phases often arise by this fluctuation mechanism and the resultant spontaneous nucleation. Calculations show that the chance,  e, is usually too small. It is more likely that tiny dust particles act as nuclei in supersaturated vapours or solutions. In the cloud chamber, it is the clusters of ions caused by a passing high-energy particle that acts as nucleation centers. Actually, vapours seem to be much less finicky than solutions about the sort of nuclei required. This is because a liquid will condense on almost any surface, but crystallization requires the presence of crystal faces of the proper kind.\nFor a sessile drop residing on a solid surface, the Kelvin equation is modified near the contact line, due to intermolecular interactions between the liquid drop and the solid surface.  This extended Kelvin equation is given by\nln\n⁡\n\n\nc\n\nc\n\n\ns\na\nt\n\n\n\n\n\n=\n\n\n\nV\n\nm\n\n\n\nR\nT\n\n\n\n\n(\n\n\n\n\n2\nγ\n\nr\n\n\n+\nΠ\n\n)\n\n.\n\n\n{\\displaystyle \\ln {\\frac {c}{c_{\\rm {sat}}}}={\\frac {V_{\\text{m}}}{RT}}\\left({\\frac {2\\gamma }{r}}+\\Pi \\right).}\nwhere \n\n\n\nΠ\n\n\n{\\displaystyle \\Pi }\n\n is the disjoining pressure that accounts for the intermolecular interactions between the sessile drop and the solid and \n\n\n\n\n(\n\n2\nγ\n\n/\n\nr\n\n)\n\n\n\n{\\displaystyle \\left(2\\gamma /r\\right)}\n\n is the Laplace pressure, accounting for the curvature-induced pressure inside the liquid drop. When the interactions are attractive in nature, the disjoining pressure, \n\n\n\nΠ\n\n\n{\\displaystyle \\Pi }\n\n is negative. Near the contact line, the disjoining pressure dominates over the Laplace pressure, implying that the solubility, \n\n\n\nc\n\n\n{\\displaystyle c}\n\n is less than \n\n\n\n\nc\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle c_{\\rm {sat}}}\n\n. This implies that a new phase can spontaneously grow on a solid surface, even under saturation conditions.\n", "doc_id": "humidity_and_rain/Kelvin_equation5.txt"}
{"content": "part2 -------------------\nFormulation[edit]\nThe original form of the Kelvin equation, published in 1871, is: \n\n\n\n\np\n(\n\nr\n\n1\n\n\n,\n\nr\n\n2\n\n\n)\n=\nP\n−\n\n\n\nγ\n\n\nρ\n\n\nv\na\np\no\nr\n\n\n\n\n\n(\n\nρ\n\n\nl\ni\nq\nu\ni\nd\n\n\n\n−\n\nρ\n\n\nv\na\np\no\nr\n\n\n\n)\n\n\n\n\n(\n\n\n\n1\n\nr\n\n1\n\n\n\n\n+\n\n\n1\n\nr\n\n2\n\n\n\n\n\n)\n\n,\n\n\n{\\displaystyle p(r_{1},r_{2})=P-{\\frac {\\gamma \\,\\rho _{\\rm {vapor}}}{(\\rho _{\\rm {liquid}}-\\rho _{\\rm {vapor}})}}\\left({\\frac {1}{r_{1}}}+{\\frac {1}{r_{2}}}\\right),}\n\n\nwhere:\np\n(\nr\n)\n\n\n{\\displaystyle p(r)}\n\n = vapor pressure at a curved interface of radius \n\n\n\nr\n\n\n{\\displaystyle r}\n\n\n\n\n\n\nP\n\n\n{\\displaystyle P}\n\n = vapor pressure at flat interface (\n\n\n\nr\n=\n∞\n\n\n{\\displaystyle r=\\infty }\n\n) = \n\n\n\n\np\n\ne\nq\n\n\n\n\n{\\displaystyle p_{eq}}\n\n\n\n\n\n\nγ\n\n\n{\\displaystyle \\gamma }\n\n = surface tension\n\n\n\n\n\nρ\n\n\nv\na\np\no\nr\n\n\n\n\n\n{\\displaystyle \\rho _{\\rm {vapor}}}\n\n = density of vapor\n\n\n\n\n\nρ\n\n\nl\ni\nq\nu\ni\nd\n\n\n\n\n\n{\\displaystyle \\rho _{\\rm {liquid}}}\n\n = density of liquid\n\n\n\n\n\nr\n\n1\n\n\n\n\n{\\displaystyle r_{1}}\n\n , \n\n\n\n\nr\n\n2\n\n\n\n\n{\\displaystyle r_{2}}\n\n = radii of curvature along the principal sections of the curved interface.\nThis may be written in the following form, known as the Ostwald–Freundlich equation:\n\n\n\n\nln\n⁡\n\n\np\n\np\n\n\ns\na\nt\n\n\n\n\n\n=\n\n\n\n2\nγ\n\nV\n\nm\n\n\n\n\nr\nR\nT\n\n\n\n,\n\n\n{\\displaystyle \\ln {\\frac {p}{p_{\\rm {sat}}}}={\\frac {2\\gamma V_{\\text{m}}}{rRT}},}\n\n\nwhere \n\n\n\np\n\n\n{\\displaystyle p}\n\n is the actual vapour pressure,\n\n\n\n\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p_{\\rm {sat}}}\n\n is the saturated vapour pressure when the surface is flat,\n\n\n\n\nγ\n\n\n{\\displaystyle \\gamma }\n\n is the liquid/vapor surface tension, \n\n\n\n\nV\n\nm\n\n\n\n\n{\\displaystyle V_{\\text{m}}}\n\n is the molar volume of the liquid, \n\n\n\nR\n\n\n{\\displaystyle R}\n\n is the universal gas constant, \n\n\n\nr\n\n\n{\\displaystyle r}\n\n is the radius of the droplet, and \n\n\n\nT\n\n\n{\\displaystyle T}\n\n is temperature.\nEquilibrium vapor pressure depends on droplet size.\nIf the curvature is convex, \n\n\n\nr\n\n\n{\\displaystyle r}\n\n is positive, then  \n\n\n\np\n>\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p>p_{\\rm {sat}}}\n\n\nIf the curvature is concave, \n\n\n\nr\n\n\n{\\displaystyle r}\n\n is negative, then \n\n\n\np\n<\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p<p_{\\rm {sat}}}\nAs \n\n\n\nr\n\n\n{\\displaystyle r}\n\n increases, \n\n\n\np\n\n\n{\\displaystyle p}\n\n decreases towards \n\n\n\n\np\n\ns\na\nt\n\n\n\n\n{\\displaystyle p_{sat}}\n\n, and the droplets grow into bulk liquid.\nIf the vapour is cooled, then \n\n\n\nT\n\n\n{\\displaystyle T}\n\n decreases, but so does \n\n\n\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p_{\\rm {sat}}}\n\n. This means \n\n\n\np\n\n/\n\n\np\n\n\ns\na\nt\n\n\n\n\n\n{\\displaystyle p/p_{\\rm {sat}}}\n\n increases as the liquid is cooled. \n\n\n\nγ\n\n\n{\\displaystyle \\gamma }\n\n and \n\n\n\n\nV\n\nm\n\n\n\n\n{\\displaystyle V_{\\text{m}}}\n\n may be treated as approximately fixed, which means that the critical radius \n\n\n\nr\n\n\n{\\displaystyle r}\n\n must also decrease.\nThe further a vapour is supercooled, the smaller the critical radius becomes. Ultimately it can become as small as a few molecules, and the liquid undergoes homogeneous nucleation and growth.\nA system containing a pure homogeneous vapour and liquid in equilibrium. In a thought experiment, a non-wetting tube is inserted into the liquid, causing the liquid in the tube to move downwards. The vapour pressure above the curved interface is then higher than that for the planar interface. This picture provides a simple conceptual basis for the Kelvin equation.\nThe change in vapor pressure can be attributed to changes in the Laplace pressure. When the Laplace pressure rises in a droplet, the droplet tends to evaporate more easily.\nWhen applying the Kelvin equation, two cases must be distinguished: A drop of liquid in its own vapor will result in a convex liquid surface, and a bubble of vapor in a liquid will result in a concave liquid surface.\n", "doc_id": "humidity_and_rain/Kelvin_equation2.txt"}
{"content": "part4 -------------------\nDerivation using the Gibbs free energy[edit]\nThe formal definition of the Gibbs free energy for a parcel of volume \n\n\n\nV\n\n\n{\\displaystyle V}\n\n, pressure \n\n\n\nP\n\n\n{\\displaystyle P}\n\n and temperature \n\n\n\nT\n\n\n{\\displaystyle T}\n\n is given by:\nG\n=\nU\n+\np\nV\n−\nT\nS\n,\n\n\n{\\displaystyle G=U+pV-TS,}\nwhere \n\n\n\nU\n\n\n{\\displaystyle U}\n\n is the internal energy and \n\n\n\nS\n\n\n{\\displaystyle S}\n\n is the entropy. The differential form of the Gibbs free energy can be given as\nd\nG\n=\n−\nS\nd\nT\n+\nV\nd\nP\n+\n\n∑\n\ni\n=\n1\n\n\nk\n\n\n\nμ\n\ni\n\n\nd\n\nn\n\ni\n\n\n,\n\n\n{\\displaystyle dG=-SdT+VdP+\\sum _{i=1}^{k}\\mu _{i}dn_{i},}\nwhere \n\n\n\nμ\n\n\n{\\displaystyle \\mu }\n\n is the chemical potential and \n\n\n\nn\n\n\n{\\displaystyle n}\n\n is the number of moles. Suppose we have a substance \n\n\n\nx\n\n\n{\\displaystyle x}\n\n which contains no impurities. Let's consider the formation of a single drop of \n\n\n\nx\n\n\n{\\displaystyle x}\n\n with radius \n\n\n\nr\n\n\n{\\displaystyle r}\n\n containing \n\n\n\n\nn\n\nx\n\n\n\n\n{\\displaystyle n_{x}}\n\n molecules from its pure vapor. The change in the Gibbs free energy due to this process is\nΔ\nG\n=\n\nG\n\nd\n\n\n−\n\nG\n\nv\n\n\n,\n\n\n{\\displaystyle \\Delta G=G_{d}-G_{v},}\nwhere \n\n\n\n\nG\n\nd\n\n\n\n\n{\\displaystyle G_{d}}\n\n and \n\n\n\n\nG\n\nv\n\n\n\n\n{\\displaystyle G_{v}}\n\n are the Gibbs energies of the drop and vapor respectively. Suppose we have \n\n\n\n\nN\n\ni\n\n\n\n\n{\\displaystyle N_{i}}\n\n molecules in the vapor phase initially. After the formation of the drop, this number decreases to \n\n\n\n\nN\n\nf\n\n\n\n\n{\\displaystyle N_{f}}\n\n, where\nN\n\nf\n\n\n=\n\nN\n\ni\n\n\n−\n\nn\n\nx\n\n\n.\n\n\n{\\displaystyle N_{f}=N_{i}-n_{x}.}\nLet \n\n\n\n\ng\n\nv\n\n\n\n\n{\\displaystyle g_{v}}\n\n and \n\n\n\n\ng\n\nl\n\n\n\n\n{\\displaystyle g_{l}}\n\n represent the Gibbs free energy of a molecule in the vapor and liquid phase respectively. The change in the Gibbs free energy is then:\nΔ\nG\n=\n\nN\n\nf\n\n\n\ng\n\nv\n\n\n+\n\nn\n\nx\n\n\n\ng\n\nl\n\n\n+\n4\nπ\n\nr\n\n2\n\n\nσ\n−\n\nN\n\ni\n\n\n\ng\n\nv\n\n\n,\n\n\n{\\displaystyle \\Delta G=N_{f}g_{v}+n_{x}g_{l}+4\\pi r^{2}\\sigma -N_{i}g_{v},}\nwhere \n\n\n\n4\nπ\n\nr\n\n2\n\n\nσ\n\n\n{\\displaystyle 4\\pi r^{2}\\sigma }\n\n is the Gibbs free energy associated with an interface with radius of curvature \n\n\n\nr\n\n\n{\\displaystyle r}\n\n and surface tension \n\n\n\nσ\n\n\n{\\displaystyle \\sigma }\n\n. The equation can be rearranged to give\nΔ\nG\n=\n(\n\nN\n\ni\n\n\n−\n\nn\n\nx\n\n\n)\n\ng\n\nv\n\n\n+\n\nn\n\nx\n\n\n\ng\n\nl\n\n\n+\n4\nπ\n\nr\n\n2\n\n\nσ\n−\n\nN\n\ni\n\n\n\ng\n\nv\n\n\n=\n\nn\n\nx\n\n\n(\n\ng\n\nl\n\n\n−\n\ng\n\nv\n\n\n)\n+\n4\nπ\n\nr\n\n2\n\n\nσ\n.\n\n\n{\\displaystyle \\Delta G=(N_{i}-n_{x})g_{v}+n_{x}g_{l}+4\\pi r^{2}\\sigma -N_{i}g_{v}=n_{x}(g_{l}-g_{v})+4\\pi r^{2}\\sigma .}\nLet \n\n\n\n\nv\n\nl\n\n\n\n\n{\\displaystyle v_{l}}\n\n and \n\n\n\n\nv\n\nv\n\n\n\n\n{\\displaystyle v_{v}}\n\n be the volume occupied by one molecule in the liquid phase and vapor phase respectively. If the drop is considered to be spherical, then\nn\n\nx\n\n\n\nv\n\nl\n\n\n=\n\n\n4\n3\n\n\nπ\n\nr\n\n3\n\n\n.\n\n\n{\\displaystyle n_{x}v_{l}={\\frac {4}{3}}\\pi r^{3}.}\nThe number of molecules in the drop is then given by\nn\n\nx\n\n\n=\n\n\n\n4\nπ\n\nr\n\n3\n\n\n\n\n3\n\nv\n\nl\n\n\n\n\n\n.\n\n\n{\\displaystyle n_{x}={\\frac {4\\pi r^{3}}{3v_{l}}}.}\nThe change in Gibbs energy is then\nΔ\nG\n=\n\n\n\n4\nπ\n\nr\n\n3\n\n\n\n\n3\n\nv\n\nl\n\n\n\n\n\n(\n\ng\n\nl\n\n\n−\n\ng\n\nv\n\n\n)\n+\n4\nπ\n\nr\n\n2\n\n\nσ\n.\n\n\n{\\displaystyle \\Delta G={\\frac {4\\pi r^{3}}{3v_{l}}}(g_{l}-g_{v})+4\\pi r^{2}\\sigma .}\nThe differential form of the Gibbs free energy of one molecule at constant temperature and constant number of molecules can be given by:\nd\ng\n=\n(\n\nv\n\nl\n\n\n−\n\nv\n\nv\n\n\n)\nd\nP\n.\n\n\n{\\displaystyle dg=(v_{l}-v_{v})dP.}\nIf we assume that \n\n\n\n\nv\n\nv\n\n\n≫\n\nv\n\nl\n\n\n\n\n{\\displaystyle v_{v}\\gg v_{l}}\n\n then\nd\ng\n≃\n−\n\nv\n\nv\n\n\nd\nP\n.\n\n\n{\\displaystyle dg\\simeq -v_{v}dP.}\nThe vapor phase is also assumed to behave like an ideal gas, so\nv\n\nv\n\n\n=\n\n\n\nk\nT\n\nP\n\n\n,\n\n\n{\\displaystyle v_{v}={\\frac {kT}{P}},}\nwhere \n\n\n\nk\n\n\n{\\displaystyle k}\n\n is the Boltzmann constant. Thus, the change in the Gibbs free energy for one molecule is\nΔ\ng\n=\n−\nk\nT\n\n∫\n\n\nP\n\ns\na\nt\n\n\n\n\nP\n\n\n\n\n\nd\nP\n\nP\n\n\n,\n\n\n{\\displaystyle \\Delta g=-kT\\int \\limits _{P_{sat}}^{P}{\\frac {dP}{P}},}\nwhere \n\n\n\n\nP\n\ns\na\nt\n\n\n\n\n{\\displaystyle P_{sat}}\n\n is the saturated vapor pressure of \n\n\n\nx\n\n\n{\\displaystyle x}\n\n over a flat surface and \n\n\n\nP\n\n\n{\\displaystyle P}\n\n is the actual vapor pressure over the liquid. Solving the integral, we have\nΔ\ng\n=\n\ng\n\nl\n\n\n−\n\ng\n\nv\n\n\n=\n−\nk\nT\nln\n⁡\n\n\n(\n\n\n\n\nP\n\nP\n\ns\na\nt\n\n\n\n\n\n\n)\n\n\n.\n\n\n{\\displaystyle \\Delta g=g_{l}-g_{v}=-kT\\ln {\\Bigl (}{\\frac {P}{P_{sat}}}{\\Bigr )}.}\nThe change in the Gibbs free energy following the formation of the drop is then\nΔ\nG\n=\n−\n\n\n4\n3\n\n\nπ\n\nr\n\n3\n\n\n\n\n\nk\nT\n\n\nv\n\nl\n\n\n\n\nln\n⁡\n\n\n(\n\n\n\n\nP\n\nP\n\ns\na\nt\n\n\n\n\n\n\n)\n\n\n+\n4\nπ\n\nr\n\n2\n\n\nσ\n.\n\n\n{\\displaystyle \\Delta G=-{\\frac {4}{3}}\\pi r^{3}{\\frac {kT}{v_{l}}}\\ln {\\Bigl (}{\\frac {P}{P_{sat}}}{\\Bigr )}+4\\pi r^{2}\\sigma .}\nThe derivative of this equation with respect to \n\n\n\nr\n\n\n{\\displaystyle r}\n\n is\n∂\n\n\n(\n\n\nΔ\nG\n\n\n)\n\n\n\n\n∂\nr\n\n\n\n=\n−\n4\nπ\n\nr\n\n2\n\n\n\n\n\nk\nT\n\n\nv\n\nl\n\n\n\n\nln\n⁡\n\n\n(\n\n\n\n\nP\n\nP\n\ns\na\nt\n\n\n\n\n\n\n)\n\n\n+\n8\nπ\nr\nσ\n.\n\n\n{\\displaystyle {\\frac {\\partial {\\bigl (}\\Delta G{\\bigr )}}{\\partial r}}=-4\\pi r^{2}{\\frac {kT}{v_{l}}}\\ln {\\Bigl (}{\\frac {P}{P_{sat}}}{\\Bigr )}+8\\pi r\\sigma .}\nThe maximum value occurs when the derivative equals zero. The radius corresponding to this value is:\nr\n=\n\n\n\n2\n\nv\n\nl\n\n\nσ\n\n\nk\nT\nln\n⁡\n\n\n(\n\n\n\n\nP\n\nP\n\ns\na\nt\n\n\n\n\n\n\n)\n\n\n\n\n\n.\n\n\n{\\displaystyle r={\\frac {2v_{l}\\sigma }{kT\\ln {\\Bigl (}{\\frac {P}{P_{sat}}}{\\Bigr )}}}.}\nRearranging this equation gives the Ostwald–Freundlich form of the Kelvin equation:\nln\n⁡\n\n\n(\n\n\n\n\nP\n\nP\n\ns\na\nt\n\n\n\n\n\n\n)\n\n\n=\n\n\n\n2\n\nv\n\nl\n\n\nσ\n\n\nr\nk\nT\n\n\n\n.\n\n\n{\\displaystyle \\ln {\\Bigl (}{\\frac {P}{P_{sat}}}{\\Bigr )}={\\frac {2v_{l}\\sigma }{rkT}}.}\n", "doc_id": "humidity_and_rain/Kelvin_equation4.txt"}
{"content": "part1 -------------------\nThe Kelvin equation describes the change in vapour pressure due to a curved liquid–vapor interface, such as the surface of a droplet. The vapor pressure at a convex curved surface is higher than that at a flat surface. The Kelvin equation is dependent upon thermodynamic principles and does not allude to special properties of materials. It is also used for determination of pore size distribution of a porous medium using adsorption porosimetry.  The equation is named in honor of William Thomson, also known as Lord Kelvin.\n", "doc_id": "humidity_and_rain/Kelvin_equation1.txt"}