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November 29

Causes of sudden death in dogs

Is there a list of causes of sudden death in dogs? One of my two dogs just died about an hour ago out of the blue.Uncle dan is home (talk) 02:39, 29 November 2016 (UTC)[reply]

It's going to be essentially the same as in humans: Cardiac arrest, arrythmia, heart attack, stroke, aneurysm, poison and pulmonary embolism. There are other, more rare circumstances, but they'll generally be some major defect in in breathing or circulation. Someguy1221 (talk) 02:48, 29 November 2016 (UTC)[reply]

My condolences. I looked at your user page and saw that your dog was 15 years old, so it was likely something related to old age. Our pet cat developed cancer at around the same age, and we had to have him euthanized. If you're in the Northern Hemisphere, it's possible the cold weather may have exacerbated an underlying health issue. That's why I went to your user page, to see if you stated your location. Let me clarify that I'm certainly not accusing you of neglect; no matter how warm and cozy you are, your body has to work harder to maintain its temperature in cold weather. Also as most people know infectious diseases spread more in cold weather. --47.138.163.230 (talk) 06:25, 29 November 2016 (UTC)[reply]

I'm sorry to hear it. We have an article on Dog health which lists common diseases of dogs as well as the foods and chemicals that commonly cause poisoning, and one on Aging in dogs. Neither makes for very pleasant reading though. Smurrayinchester 09:22, 29 November 2016 (UTC)[reply]

When I lost my last dog, it turned out that she had cancer throughout her body and in many vital organs. She'd shown no sign of it until a couple of days before she died. Our vet remarked that dogs are incredibly tough animals and can keep going despite amazing amounts of disease and injury - but when it finally overwhelms them, they go quickly. There is much to be said for that, I think.

FWIW: The only known cure for the loss you feel from losing an old friend is a new puppy...they have a way of providing the ideal distraction. SteveBaker (talk) 14:52, 29 November 2016 (UTC)[reply]

A new dog -- not necessarily a new puppy. We adopted a sweet little terrier who was maybe 10-12 years old at the time. Already housebroken, and old enough to be calm and not chew up the furniture. There's something special about older dogs, just like there's something special about puppies. Shock Brigade Harvester Boris (talk) 03:13, 30 November 2016 (UTC)[reply]

And by the way, her date of birth is September 11,2002, not 2001.Uncle dan is home (talk) 00:15, 30 November 2016 (UTC)[reply]

Sorry for the loss. The death of a little animal friend can cause long lasting grief that is hard for non-pet owners to understand. As for just getting a replacement, we may reach a point where we have just owned our last dog or cat, since we might not be around to tend them through their possible lifespan. As for sudden death, my vet said that a pet may start to feel pain and nausea, but conceal it to some extent, as a survival mechanism, since looking weak would not aid survival in the wild when with a group of other animals of the same species, and signs of weakness might draw the attention of predators. I knew an elderly pet who lived 6 months after the vet found a cancerous abdominal mass, which he said was inoperable due the the pet's age. The pet was seemingly cheerful and peppy until the last day when he suddenly couldn't get up. The vet said it was time to put him down. Edison (talk) 23:53, 30 November 2016 (UTC)[reply]

An animal's mother will always know if one of her newborn is sickly. If it is, when it comes to suck with the rest of the litter she will push it away, and it soon dies. 81.134.89.140 (talk) 01:12, 1 December 2016 (UTC)[reply]

Water temperature at the bottom of a kettle

Suppose you heat the bottom of the kettle to a temperature of 200 degrees Celsius, what will be the water temperature right near the metal layer? Will there be a sharp temperature difference or gradient? Gil_mo (talk) 09:59, 29 November 2016 (UTC)[reply]

When you boil a kettle the temperature inside eventually rises to 212 degrees F. As the water reaches that temperature it boils away. I guess this would happen here - so don't leave your kettle in contact with the heat source after the water is all gone. 86.145.54.170 (talk) 10:15, 29 November 2016 (UTC)[reply]

What would happen is this: the outside side of the bottom of the pan would be at 200°C while the inside side of the bottom of the pan would be at 100°C as long as the water lasted. If your pan has poor conductivity (thick stainless steel) you will need less energy to maintain that 100 degree difference. If your pan has good conductivity (thin copper) you will need a lot more energy -- likely more than you have available to you in any kitchen or even industrial setting, and a steady supply of water to replace what boils off. I doubt that any flame would be enough; you would likely need something like liquid sodium at 200°C (boiling point 882.8°C). --Guy Macon (talk) 10:31, 29 November 2016 (UTC)[reply]

...while the inside side of the bottom of the pan would be at 100°C as long as the water lasted... - not true, even under the (possibly wrong) assumption that there is no film of gas causing Leidenfrost effect here. The layer of fluid next to the solid will have a finite temperature difference with the solid just next to the fluid; in other words, such an interface is described by a (finite) heat transfer coefficient rather than by a thermal diffusivity coefficient. Tigraan^{Click here to contact me} 12:31, 29 November 2016 (UTC)[reply]

So to answer Gilmo's question, there will be a "sharp temperature difference" (i.e. a discontinuity in temperature across the interface), but how much exactly (and whether it is negligible compared to conduction inside the kettle, convective effects etc.) cannot be said without more precisions. As Jayron32 mentioned, this is a complex subject. Tigraan^{Click here to contact me} 12:31, 29 November 2016 (UTC)[reply]

What you are asking to describe is a complex physical phenomenon; local temperature will vary throughout a substance due to various effects, including specific heat capacity, temperature differential (i.e. Newton's law of cooling), heat flux, the effect of convection in fluids, etc. There are entire higher-university level courses that are dedicated to teaching the complex relationships and mathematics involved in calculating what you are asking. You might want to read up on things like Non-equilibrium thermodynamics as well. --Jayron 32 12:08, 29 November 2016 (UTC)[reply]

In short, will or will not the water just at the hot solid surface be above 100 Celsius, or not? Should we also take into account that the water at the very bottom is under pressure of the water above, thus raising the boiling point? Gil_mo (talk) 13:49, 29 November 2016 (UTC)[reply]

This image helps to visualize the water in contact with the kettle bottom. The lowest layer of molecules are water vapour i.e. Steam, superheated over 100 °C (212 °F). Water vapour is a gas and a thermal insulator so there is a steep thermal gradient through this layer. However it is usually turbulent and its depth is unstable, except in the special case of the Leidenfrost effect which can occur where there are only small water drop(s) on a very hot surface: a stable cushion of vapour forms under the droplet and insulates it from rapid evaporation. Most of the temperature difference 200°C to 100 °C is is across the vapour layer. Above this is a relatively thick layer of liquid water at 100 °C which absorbs conducted heat from below, the latent heat of vapourization that transforms it to gas. Whether there is a further temperature variation higher in the water depends on how recently the kettle was filled; the liquid may soon all be at 100 °C with streams of vapour bubbling to the surface. Blooteuth (talk) 20:15, 29 November 2016 (UTC)[reply]

If this layer of gas existed at the bottom of an ordinary boiling pot, we would see it as being silvery at certain angles. We don't see the silvery effect, therefore the water is still wetting the metal. Of course once the conditions are right for Leidenfrost effect (as happens when you pour water into an overheated cast iron pan), the physics change radically. --Guy Macon (talk) 22:01, 29 November 2016 (UTC)[reply]

The water will be at 100°C. under normal conditions, water above 100°C isn't water -- it is steam. The water at the very bottom will indeed be under pressure from the water above, thus raising the boiling point a bit, but for a normal-sized pot the effect will be negligible, and smaller than that caused by changes in altitude or barometric pressure. (Note that I said 100°C, not 100.0000°C.) And as long as that water is still wetting the metal, the top layer of atoms in the metal will be at 100°C. --Guy Macon (talk) 21:54, 29 November 2016 (UTC)[reply]

No. Water is a common chemical substance that can exist in three states: liquid, solid (ice), and gas (invisible water vapor in the air). Do not neglect the fact that Gil_mo's kettle will advance from nucleate to transition boiling. Blooteuth (talk) 18:56, 30 November 2016 (UTC)[reply]

Let's denote the cross section of the bottom of the kettle by $A$ , the temperature on the outside $T_{1}$ , the inside temperature by $T_{2}$ , and the thickness of the kettle by $d$ . Then the heat flux into the kettle is:

J_{1}=\lambda {\frac {T_{1}-T_{2}}{d}}A

where $\lambda$ is the thermal conduction coefficient of the metal. This must be balanced by the heat escaping from the kettle which will be primarily due to the latent heat of vaporization. If the cross section of the opening is $B$ , then this is:

J_{2}=L\rho vB

where $L$ is the latent heat of vaporization, $\rho$ the density of the escaping steam and $v$ the flow velocity of the steam moving through the opening. Equating $J_{1}$ to $J_{2}$ allows you to solve for the mass flux $\rho v$ (note that the latent heat does not depend strongly on the temperature). Conservation of mass implies that the product of the mass flux times the cross section this flux moves through, is conserved. So, this yields the flux of escaping steam from the surface of the water. The next step is to estimate the required overpressure in the kettle that would support this flux, this is not so straightforward it depends on the frictional losses particularly at the opening. This overpressure then yields the temperature via the Clausius–Clapeyron relation. The higher the overpressure, the higher the temperature will be above the normal boiling point of 100°C. Count Iblis (talk) 22:53, 29 November 2016 (UTC)[reply]

Thank you all for your enlightening answers! Gil_mo (talk) 06:41, 30 November 2016 (UTC)[reply]

At a microscopic scale, the temperature of water/steam in bubbles can be surprising at times - see sonoluminescence. Wnt (talk) 12:13, 30 November 2016 (UTC)[reply]

Water has a high heat capacity and a high Enthalpy of vaporization. A fascinating phenomenon which I think sheds some light on this question: you can boil water over an open fire in a plastic bottle and for the same reason you can put a cigarette lighter under (a filled) water balloon and it will not pop. An open flame is 1000+ degrees, but the water on the inside of the bottle / balloon will NOT get over 100 degrees which is not enough to melt the plastic, well very small parts or layers of the water might get a bit over 100 but no significant portion of it will get anywhere near the melting point of the plastic / rubber. Convection is very efficient, the "hot" parts of the water will carry the heat up and away and be replaced by 'colder' parts. You can find many demonstrations of this on youtube. Vespine (talk) 22:05, 30 November 2016 (UTC)[reply]

Helicopter (in)stability

Is it true that twin-rotor helicopters are more stable than single-rotor ones? Of all the commonly used rotor systems -- Sikorsky system (conventional single-rotor), NOTAR (if different from Sikorsky system), Piasecki system (tandem rotors), Kamov system (coaxial rotors), and Flettner system (synchropter), how do they rank in the order from least stable to most stable (or if you prefer, from most stable to least stable)? 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 10:56, 29 November 2016 (UTC)[reply]

According to our Tandem rotors article: "Advantages of the tandem-rotor system are a larger centre of gravity range and good longitudinal stability. Disadvantages of the tandem-rotor system are a complex transmission, and the need for two large rotors".

According to our Coaxial rotors article: "Because of the mechanical complexity, many helicopter designs use alternate configurations to avoid problems that arise when only one rotor is used". Alansplodge (talk) 21:51, 1 December 2016 (UTC)[reply]

Helicopter yaw control

How is yaw controlled in helicopters with tip jets? 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 10:57, 29 November 2016 (UTC)[reply]

If they don't have a tail rotor (like the NHI H-3 Kolibrie and the American Helicopter XH-26 Jet Jeep), I would expect that they need a rudder (like the Sud-Ouest Djinn), which would depend on the forward motion. However, I don't see one in the image of the Hiller YH-32 Hornet (unless it uses its two tail planes for that), so I don't see how that would work.

A brake could allow turning in one direction, but I don't think that would be used. Rmvandijk (talk) 14:12, 29 November 2016 (UTC)[reply]

Seems to me that because there is no engine inside the body of the helicopter, the torque that normally tries to spin the body of the helicopter in the opposite direction of the rotors is vastly less (presuming the friction in the shaft is kept low). So it might not take very much at all to control the direction of the helicopter. In forward flight, a small rudder would be all that you'd need...it might also be possible to route some of the thrust that goes to the tip jets to a couple of small jets on the tail of the helicopter. However, it's not clear what they *actually* do...these kinds of craft are rare and exotic beasts! SteveBaker (talk) 14:57, 29 November 2016 (UTC)[reply]

Indeed, they are experimental aircraft. This means that they're "non-standard" and can be modified irregularly. Among the few famous tip-jet helicopters, some used bleed exhaust or a drive shaft from the main powerplant to power the tail rotor; some used a redundant smaller powerplant to drive the tail rotor; some did not have any tail rotor at all, like the Hiller Hornet. The point is, these were "skunk works" projects, metaphorically and literally. The designs changed any time the mission planners demanded they needed to change, sometimes bypassing regulatory testing, paperwork, and documentation.

If you get a chance to visit San Carlos, California, the Hiller Aviation Museum has tons of neat stuff. How did these weird aircrafts really work? We shall know soon - it's been nearly fifty years since most of them flew, and the details have begun to emerge... Some say that a band of rowdy helicopter pilots meet on Monday nights to dispel rotorcraft myths and share experiences.

Last week's discussion was on LAX00FA306 ([1], [2]), which has incredible explanatory power for the mandatory Special Awareness Training in 14 CFR 61. Certain helicopters - particularly a few conventional models like R-22 and the UH-1 - do not behave well in zero-G condition. If the pilot attempts to correct this unusual condition by application of cyclic or adverse yaw, the rotors flutter and flap and just ... fly away. We will probably never know if tipjet rotorcraft suffered that problem... but if you visit the CIA Memorial Wall, you can imagine whatever you want about those unmarked stars.

Here's an Army Agency for Aviation Safety video hosted on YouTube: Mast Bumping - Causes and Prevention.

Nimur (talk) 15:27, 29 November 2016 (UTC)[reply]

So, how did yaw control work in the Hornet? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 21:14, 29 November 2016 (UTC)[reply]

"Barely." From the exhibit page of the Smithsonian's National Air and Space Museum, you can read about the various modifications: in the first incarnation, a rudder (only) was used, and it was controlled by the collective. (I imagine the yaw control wires were bungeed to the stick like some early Bonanzas and the Ercoupe). Later, a (very unusual asymmetric) tail rotor was added to the aircraft, driven by the starter motor (...a battery-powered electric propulsion system). Though its performance is not explicitly described, if you have any familiarity with aircraft design, you can imagine that this probably didn't result in great yaw authority, nor great duration of flight. The Air and Space Museum lists fuel consumption rates, and computes a maximal flight time of 30 minutes; in actual truth, the helicopter probably could not fly at all. (In the United States, it is generally illegal to begin a flight in an aircraft that is carrying less than 30 minutes of fuel remaining on board, 14 CFR 91.151 - well, twenty minutes for rotorcraft, so ...).

"The flames coming out of the ramjets produced an incredibly bright white halo when the HOE-1 was flown at night. This was a considerable disadvantage in the military environment, and the effect prompted a number of UFO sightings when operated in the vicinity of populated areas.... The noise generated by the ramjets was also quite considerable, and did not endear the United Helicopter's Palo Alto, California facility to its neighbors." (Smithsonian Institution).

The interested reader will be happy to know that the Palo Alto test facility, declassified by the National Reconnaissance Office in year 2006, was not a Hiller Helicopter facility, as it was loudly and widely advertised and publicly known; but the location was in fact a CIA "General Participant" development facility for Project CORONA, America's first observation and surveillance satellite. You can find some great historical photographs of the facility during the 1940s and 1950s in this book, Palo Alto Over Time, featured in the Palo Alto Daily News. Interestingly, the location of this facility was on Willow Road, which is today the mailing address of a more modern ~~surveillance program~~ corporation.

When one wants to keep a secret project quiet, one has to find some excuse for moving so many engineers into and out of a building each day.

Nimur (talk) 21:40, 29 November 2016 (UTC)[reply]

Mostly it doesn't need to be. There's no major torque reaction to deal with, so no need to counteract it. Yaw is only needed for literally yawing the helicopter. As helicopters also have a cyclic pitch which can translate them sideways, then there's not even as much need for that.

Tip jets can be hot or cold. Cold tip jets are powered from a compressor bleed on a gas turbine, implying that there's a supply of pressurised air available and so puffer jets could be used (as worked for the Hawker Harrier), although I can't think offhand of a tipjet helicopter which used them. A few helicopters, particularly the French designs, but also the Fairey Ultra-light used a tailplane with conventional rudder and their gas turbine exhaust over this was enough to give very good yaw control, even in the hover.

Hot jets, such as the British fuel-burning jets supplied with air from a central compressor, or the US designs with a self-contained jet on each rotor, have a harder time of it. The Hiller Hornet was tested as a gunship (claimed to be the first, although the Germans had tried it during WWII) but its poor yaw control put paid to that. Various Gyrodynes, such as the Fairey Rotodyne and its precursors, had good yaw control at speed, by controlling their propulsion props, but were limited in the hover. A few, such as the Jet Jeep and the (un)Flying Ocarina, took a geared drive from the main rotor hub and used this to drive a conventional helicopter tail rotor, although much undersized from the usual. Andy Dingley (talk) 16:36, 29 November 2016 (UTC)[reply]

Do you know if the control surfaces at the tail of the Hiller Hornet (Are those inverted ruddervators?) gave some yaw control even while hovering by deflecting downwash? -- ToE 13:15, 30 November 2016 (UTC)[reply]

The original Hornet HJ-1 had a vertical stabiliser with a rudder effect, by the whole vertical stabiliser moving. This worked with forward airspeed, but as the hinge axis was at something like 45°, this could also deflect pure rotorwash sideways in the hover.

The military Hornet YH-32, as illustrated, had the two ruddervators. As I understand it (but don't have good docs for) these moved in two axes: "pitch", as elevators, and also in "roll" to control yaw. This is why they were mounted in the V shape, rather than flat, so that rolling them gave an asymmetric side thrust and thus yawing moment. This allowed the whole helicopter to be pitched and yawed, for aiming the guns. They didn't work terribly well as a yaw control though. Andy Dingley (talk) 18:50, 30 November 2016 (UTC)[reply]