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July 29[edit]

Calculating the distance between two planets on a given date[edit]

The title is pretty self-explanatory, I guess. I'm looking for an easy and simple way to do this, preferably online and preferably without installing any software. Thanks in advance. Evan ^{(talk|contribs)} 00:15, 29 July 2014 (UTC)[reply]

The JPL Solar System Simulator has the information you want. You will have to play around with the display settings and turn lots of things off to be able to actually read the distance figures. The information is obscured by a too-busy display on the default settings. There is a much simpler caculator at Wolfram but it is beta and doesn't give the date used so it's accuracy is probably suspect. SpinningSpark 01:10, 29 July 2014 (UTC) and 01:14, 29 July 2014 (UTC)[reply]

If you want high accuracy, use JPL's HORIZONS interface: [1]. For each planet, you can get the planet's right ascension, declination, and distance from Earth. This defines a spherical coordinate system, and you can calculate the distance using the planets' coordinates. I might write an app that automatically does this, if nobody finds a pre-existing one. --Bowlhover (talk) 01:32, 29 July 2014 (UTC)[reply]

Wolfram|Alpha will calculate it, for example distance between Venus and Mars on 31 December 2015 gives 0.9553 au. ---- CS Miller (talk) 08:46, 29 July 2014 (UTC)[reply]

I assume that it takes account that the planets' orbits are elliptical, rather than circular, However does it take account of perihelion precession, orbits slowly increasing, and other effects needed for calculating the planets' locations in the distant future/past? CS Miller (talk) 08:56, 29 July 2014 (UTC)[reply]

WolframAlpha looks accurate enough for my purposes. Thanks! Evan ^{(talk|contribs)} 16:13, 29 July 2014 (UTC)[reply]

Full body burns and infections[edit]

I've heard of people who have received full body burns being in the ICU and being unable to survive due to their skin having been peeled off and due to the resulting scabbing (all over their body), eventually die from infection. Is this have any basis in fact? I've researched but I couldn't find this specific scenario. Thanks. Tutelary (talk) 02:55, 29 July 2014 (UTC)[reply]

Burn#Prognosis explains that burns over more than 90% of the skin have an 85% fatality rate - and goes on to say that the most common complication is infection...so, what you've heard seems reasonably correct - although it's by no means the only possible outcome...after all, 15% survive - and of the 85% that die, some at least must die of the less common complications. SteveBaker (talk) 04:14, 29 July 2014 (UTC)[reply]

A quick Google search suggests that a complex condition called "burn shock" caused by "complex fluid, electrolyte, and protein shifts" is an important issue with major burns. It can be treated by "fluid resuscitation".[2] Alansplodge (talk) 18:03, 29 July 2014 (UTC)[reply]

Just to clarify, scabs are shields against infection, not causes of it. If your body is full of them, that's a sign of a problem, but only like how if your apartment is full of firefighters.

Inhaling fire really sucks for oxygen-rich lungs. There, and in other permeable places, scabs are more of a problem, but still not for infection reasons. InedibleHulk (talk) 03:17, 30 July 2014 (UTC)[reply]

To clarify, scabs would be a more major problem in your lungs and throat than on your skin. But they occur there far less frequently. What I meant by "more of a problem". InedibleHulk (talk) 23:47, 30 July 2014 (UTC) [reply]

I'll be very honest, I'm really really sleepy, I don't think I can adequately summarize my sources in a clear manner - at any rate, this [3] has a ton of good information, and is quite readable, these [4], [5], [6] are just abstracts, but may be of some interests. It does appear that a large number of deaths are caused from subsequent infection, sepsis seems to be a big one - the paper mentions that 75% perish from infection or inhalation related complications; I didn't see any clarification of how much each factor contributes, or how much overlap, etc. Sorry I could not be of more help.Phoenixia1177 (talk) 05:38, 30 July 2014 (UTC)[reply]

I believe dehydration is the leading cause of death from large burn areas that happen quickly. Infection doesn't kill very quickly. Without skin, fluids leave the body very quickly and hydration is difficult and death can occur quickly after the burn. --DHeyward (talk) 04:01, 31 July 2014 (UTC)[reply]

Thank you all for the responses, I think I understand now. ^^ Tutelary (talk) 20:42, 2 August 2014 (UTC)[reply]

"Red" and "blue" stars[edit]

Wnt (talk) 17:03, 29 July 2014 (UTC)[reply]

From the articles about stars, I get the impression that most "red" stars are even more infrared than red, and blue stars are usually more ultraviolet than blue.

My reasoning:

The peak wavelength is around 2.9mm divided by the surface temperature. So, 5000K - 580nm (yellow-orange), 4000K - 725nm (red), and many stars are way below 4000K and thus mainly infrared.
Going the other way, one finds 5800K - 500nm (green-blue), and 7250K - 400nm (violet-ultraviolet). The radiation of the latter is already 50% ultraviolet, and even A-type main-sequence stars are hotter.

So, is it correct that the 7250K star emits half its energy as UV, and a 3600K star half its energy as infrared? The figures look OK, but I'm not sure if I can take the peak as 50/50 mark.

Of course, names like "infrared dwarf" and "ultraviolet giant" are unnecessarily long; the dominant visible color avoids that problem nicely. I'm asking because somebody claimed that you couldn't see any stars in front of you when moving at 0.5c; their claim was that all colors would been blue-shifted out of the visible spectrum. Now, if most "red" stars emit lots of infrared, that issue doesn't even. - ¡Ouch! (^{hurt me} / _{more pain}) 09:55, 29 July 2014 (UTC)[reply]

Note that A-type main-sequence stars emit lots of infrared from the dust clouds surrounding them, so they would be visible too. 24.5.122.13 (talk) 10:04, 29 July 2014 (UTC)[reply]

Most of the energy radiated by a star is blackbody emission due to the hot gas. In supplement to that, the star also emits spectral radiation at various wavelengths - including visible, infrared, and elsewhere in the radio spectrum. Those spectra are emitted because of various atomic, nuclear, intermolecular, and bulk physical processes. In addition, there are very weak electromagnetic emissions at much lower frequency - say, a few microhertz for a typical star - due to the bulk movement of the plasma (the ionized gas inside the star), and its very complicated fluid interactions with its own electromagnetic radiation. All of these spectra are added together in a mostly linear fashion. Taking account of everything, we see that most of the energy spectrum still looks very much like a black-body spectrum - so it's radiating at all wavelengths. Astronomers then categorize as a "reddish" or "bluish" tint based on the peak temperature - a parameter that we might describe by its correlated color temperature (briefly, that's a mathematical formula to approximate the "equivalent" blackbody color). Nimur (talk) 16:55, 29 July 2014 (UTC)[reply]

Stars are a lot brighter-colored than I'd imagined - see the diagram from the end of color temperature I've reposted at top. If you say A-type main sequence star has color temperature 15,000, then that is accurate for what you see but not for the UV. Wnt (talk) 17:03, 29 July 2014 (UTC)[reply]

Sure! There are two blackbody curves. But this makes total sense: ionized plasmas commonly have per-species temperature. The best Wikipedia article may be nonthermal plasma (plasmas whose constituents are not in thermal equilibrium); but the best reference is (as always) Bittencourt's Fundamentals of Plasma Physics.

But for Vega in particular, things are even more complex: the temperature of the star varies with position, because of the complex relation between nuclear fusion reaction rates, plasma density (which is strongly affected by rotation), and convection of the bulk material to the stellar surface. Our article on Vega links to several research papers that discuss the temperature gradients, which are large enough to be observed from Earth. Nimur (talk) 18:37, 29 July 2014 (UTC)[reply]

Thanks Nimur and Wnt.

That A0V spectrum is quite remarkable; above 400nm, it fits the 15000K curve well, but in the UV range, all hell seems to break loose. Between 200 and 400nm, there is definitely too little output for an even remotely black body, but below 200,there is way too much (about 4 times the expected quantity around 50nm). If I didn't know better, I'd say the star is trolling the visible-light astronomers.

The 15000K curve looks like a good fit to the visible spectrum, but I wonder why the 9500K curve is in there. It doesn't fit either part of the curve well, except maybe the dim UV part between 240 and 370nm, and even then, the "peak" is far from pronounced. Was the 9500K chosen because the spectrum has its 50/50 point at the same wavelength, i.e. because half the emitted light has lower wavelength?

The peak of the 9500K curve is right where I expected it (around 300nm - I don't even know where I found the "divide 2.9mm by temperature" rule), but the 15000K curve doesn't even seem to have a peak, nor does it come down when wavelengths approach zero. Why? - ¡Ouch! (^{hurt me} / _{more pain}) 08:09, 30 July 2014 (UTC)[reply]

The 15000K curve is the spectrum of the black body radiation that comes from the lower parts of the star (its maximum is outside the plot limits). It is modified by absorption in the star's atmosphere into the observed spectrum. For A stars this absorption is mostly due to atomic hydrogen - the Balmer lines are clearly visible in the visible part of the spectrum (starting from Hα at 656 nm), converging to the Balmer limit at 365 nm. At shorter wavelength, photons are absorbed at all wavelengths, not just in separated absorption lines. 9500 K is the star's effective temperature, it is defined by the total luminosity of the star, i.e. by the integral of the observed spectrum. The 9500 K curve should have the same integral as the observed spectrum, although not the same shape. --Wrongfilter (talk) 12:25, 30 July 2014 (UTC)[reply]

¡OOPS! — I didn't see that the wavelength doesn't approach zero in the graph; the left margin is 120nm, which leaves enough space for the 15000K curve (which should have its peak around 193nm) to approach the origin.

The 15000K curve "enters" the graph area around 360nm, just short of twice its peak wavelength. If I look at the corresponding wavelength for the 9500K curve, I get about 570nm, and the other point with the same Y coordinate is around 180nm. Which means that the 15000K curve "left" the graph area between 110 and 120nm.

The y coordinate is around 1, about half the peak value, so the 15000K curve should, by a similar kind of educated guess, peak around y=6 to 7.

I found Wien's displacement law in the color temperature article; it is basically my peak formula, with a constant b = 2.89777×10⁻³ m K which agrees quite well with the 2.9mm I used. - ¡Ouch! (^{hurt me} / _{more pain}) 06:29, 31 July 2014 (UTC)[reply]

Amelia[edit]

When embarking on her solo flight across the Pacific in 1935, did Amelia Earhart take off from Wheeler Army Airfield with a less-than-maximum fuel load? Because I've tried it in Microsoft Flight Simulator, and I found that when taking off from Wheeler in a Lockheed Vega with the full fuel load of 650 gallons, it's almost impossible to clear the high ground to the east. So how did she manage to do it (other than by being a better pilot than me, which she no doubt was, but I don't think that alone could have made a difference)? 24.5.122.13 (talk) 23:57, 29 July 2014 (UTC)[reply]

Are you correctly simulating wind, temperature, and density altitude? Just like the textbook says: aircraft performance depends on atmospheric conditions. A lot. The effects are so important that atmospheric conditions constitute the first half of the chapter on aircraft performance!

If you fly long enough, you'll eventually meet a pilot who forgets to account for wind and temperature, and ends up in a tragic fireball. He was, by the numbers and the ratings, a better pilot than I... but still could not execute a climb at three times the calculated maximum aircraft performance. Maybe it would have been possible, had there been a stronger headwind or if the day weren't so hot... because the same aircraft had executed the same departure many times before ... Nimur (talk) 05:29, 30 July 2014 (UTC)[reply]

Yes, I'm simulating weather conditions as accurately as possible -- I took the weather info straight out of Last Flight (where she wrote, among other things, that the winds for that takeoff were very weak). If anything, the temperature I've inputted into the weather engine might be too low (which would make the air density too high, which in turn would make it easier to gain altitude). 24.5.122.13 (talk) 05:46, 30 July 2014 (UTC)[reply]

(After more close review of the Lockheed Vega performance, and the terrain at HHI): The PHHI AF/D entry makes no remark about terrain. You're departing Runway 6? What altitude are you turning crosswind?

Something about your simulation is screwy. Our article lists a 1300 foot per minute rate of climb - which would be specified at maximum gross takeoff weight. The sectional shows no terrain or obstacles higher than 3,500' east of the field. The strip itself is 5600 feet long. Whatever terrain - or aircraft performance - that your simulator is using appears to contradict the actual data. Terrain should not be even remotely hazardous. There isn't really even any justification for using a short field takeoff procedure.

Or... and I'm reluctant to even mention it... but perhaps you don't know how to fly a simulated taildragger very well at all! Review the airplane flying handbook, and be sure you're rotating at liftoff speed, and flying a normal pattern, climbing at Vy, ...

Nimur (talk) 06:05, 30 July 2014 (UTC)[reply]

Another thing to consider: her Lockheed Electra was a "...highly modified Model 10E" (emphasis mine) —which undoubtedly meant that it was lighter than the standard model. 71.20.250.51 (talk) 06:26, 30 July 2014 (UTC)[reply]

That might well be something to consider if it was relevant. It isn't because Earhart wasn't flying the Electra at the time. AndyTheGrump (talk) 06:34, 30 July 2014 (UTC)[reply]

Okay, roger that. — Although she "sold her 5B Vega to Philadelphia's Franklin Institute in 1933" [she purchased "a new Lockheed 5C Vega"] [before] "In July 1936 Amelia took delivery of a Lockheed L-10E Electra" ~My bad, ~:71.20.250.51 (talk) 07:43, 30 July 2014 (UTC)[reply]

(un-indent) Really? The Vega climbs out at 1300 fpm? That must be with much less than 650 gallons of gasoline -- I've flown the Vega many times, and with the full 650 gallons on board, even at maximum sustained power (30 in. manifold pressure, 2200 rpm) and with a perfect climb at V(y) (110-120 mph), it barely gets 250-300 fpm. (Now once you get down to 75% fuel, it's a different story altogether...) So the only conclusion I can draw here is, the performance data given in Wikipedia for the Vega is for a much lower fuel loading than the maximum 650 gallons -- with the full 650 gallons, the Vega is much heavier than normal, with the corresponding reduction in performance. (The Electra, as far as I can tell, was even worse -- with the full 1100 gallons of fuel, it was not only grossly overloaded, but also dangerously nose-heavy so it would only rotate with great difficulty and a generous use of nose-up trim.) 24.5.122.13 (talk) 09:00, 30 July 2014 (UTC)[reply]

BTW, has anyone noticed that the performance data given for the Vega in the article lists the maximum range as only 725 miles -- nowhere near the distances involved here? 24.5.122.13 (talk) 09:04, 30 July 2014 (UTC)[reply]

Also, even though the performance data might have been given for maximum gross weight, the article lists this weight as 4500 lbs. -- which, with an empty weight of 2565 lbs. and (I presume) a 130-pound pilot, would be reached with less than 300 gallons of fuel. (With the full 650 gallons, the actual takeoff weight would be more like 6600 pounds!) 24.5.122.13 (talk) 09:16, 30 July 2014 (UTC)[reply]

A couple points:

Do not overload the aircraft. The aircraft must be within its weight and balance envelope. That means fuel plus baggage plus passenger weight plus engine oil plus everything else has to be below maximum gross takeoff weight, and the moment arm also needs to be calculated. Aircraft performance envelopes aren't exaggerated. They don't include a built-in "engineering fudge factor." When loaded and fueled, the aircraft needs to be below that value. Need a reference? The official textbook addendum, The Aircraft Weight and Balance Handbook. Want another reference? Weight and Balance (Chapter 9 of the Pilot's Handbook of Aeronautical Knowledge).
Performance numbers are typically quoted at maximum gross takeoff weight. I wouldn't trust the 1300 foot-per-minute, or any other parameter, at face value, though, because...
The Wikipedia article is not a reliable source for aircraft performance numbers - it's just a useful summary in a convenient form suitable for an encyclopedia. If you're really flying, or even if you're just serious about simulating, there is one and only one reference for aircraft performance: the legal and approved Pilot's Operating Handbook for the aircraft. That will provide performance numbers and charts for the aircraft at various conditions. Anything else is bogus.

Nimur (talk) 14:43, 30 July 2014 (UTC)[reply]

True -- except that for long-range overwater flying of the kind that Amelia (and other pioneering flyers like Lindberg, Post, Kingsford-Smith, etc.) had done, there's no choice but to overload the aircraft with fuel, because that's the only way you can reach the other side of the pond (without in-flight refueling, which was not available in the 1930s and is not supported in Flight Simulator). Which brings up the original question once again: Just how much fuel did Amelia have on board so as to reach the mainland but do so without crashing on takeoff? 24.5.122.13 (talk) 00:01, 31 July 2014 (UTC)[reply]

I can't be quite sure, but I've re-read her bio by Susan Wells, and that book seems to mention that she had 470 gallons on board for the Pacific flight. I'll try taking off with that amount and see if I make it over the ridge. (Should be doable without major problems -- I've flown the Vega with different fuel loads, and with 450 gallons it still takes off quite briskly.) 24.5.122.13 (talk) 10:42, 31 July 2014 (UTC)[reply]