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February 24

Weather control to prevent cold snaps and snowstorm s

Has anyone ever proposed weather control methods for preventing cold snaps and snowstorm s? — Preceding unsigned comment added by 73.118.181.38 (talk) 00:00, 24 February 2018 (UTC)

If by "cold snap" you're meaning the sort of unexpected late frost which damages seedlings and grape vines, then yes, they have. "Smudge pots" were a popular attempt, where orchards and vineyards would light smokey stoves between the trees. The smoke reduces radiative cooling, thus the temperature drop overnight. Andy Dingley (talk) 00:11, 24 February 2018 (UTC)

Andy Dingley's example and cloud seeding are about as far as this goes with our technology. Weather is pretty much impossible to control directly, as it is a system with many times the energy output that humanity can produce. Instead of spending energy in a futile effort to change the weather we typically tend to adapt our tools to let it do the work for us: e.g. windmills, wind turbines, hydroelectric power plants... Besides, cold snaps, snowstorms and other apparently extreme events can often be useful to the environment in unexpected ways. For example, sequoia relies on forest fires to procreate, while cold weather in the Arctic winter encases Alaskan and Siberian north coasts in ice, preventing coastal erosion by winter storms. 93.142.92.135 (talk) 01:27, 24 February 2018 (UTC)

I was about to reply with a supporting statement that cold snaps kill off insect pests, but a quick Googling has an amusing fifty-fifty split between articles supporting that it does and articles debunking that as a myth. Whether it kills them or not, my personal observation is that it does induce them to hibernate or otherwise quite being such pains. I quite look forward to it for that reason - and because a sharp frost typically stops my ragweed-induced hay fever. Matt Deres (talk) 15:08, 24 February 2018 (UTC)

Loss of cabin pressure

I was on a plane and they gave me a bread roll in a sealed plastic bag. I didn't eat it, but kept it.

During the trip, the bag got puffed up and under pressure, like a little football. Then it went back to normal.

I understand that the pressure in the cabin must have dropped. But is that normal and okay? Anna Frodesiak (talk) 00:58, 24 February 2018 (UTC)

Was the bag given to you on the ground? After you'd been in the air for a while? Nil Einne (talk) 01:03, 24 February 2018 (UTC)

The bag was given to me an hour into the flight, and it was baggy then. Then an hour or so later it was blown up. Then, hour(s) later, still in the middle of the flight, it went back to being baggy. Anna Frodesiak (talk) 01:08, 24 February 2018 (UTC)

AFAIK, barring stuff like damage, For a given commercial airline plane, cabin pressure will depend mostly on altitude, the higher the plane, the lower the cabin pressure. Nimur or someone would probably explain more, but I believe it's common for cruising altitude to increase over time as the plane gets lighter (burns up fuel), see [1] [2]. So it's not unexpected the cabin pressure will reduce, particularly if it's only been 1 hour. Altitude may also need to be adjusted for other reasons, particularly weather conditions. If you happen to have access (sorry I suspect you don't), you can see how the altitude and cabin pressure varied over one flight from London to Bangkok here [3]. Nil Einne (talk) 02:01, 24 February 2018 (UTC) 03:33, 24 February 2018 (UTC)

I don't think the pressure really depends much on altitude. As I understand it they pressurize the cabin to about 8000 ft for cruising, regardless of cruising altitude (which is obviously always higher than that). Used to be 7000, but 8000 saves on fuel, and makes the passengers slightly groggier and slightly less likely to cause trouble. --Trovatore (talk) 06:17, 24 February 2018 (UTC)

That mostly doesn't seem to be what the Respirology article found or says, unless I misunderstood it. Actually you don't need access to the whole source for this part although the graph may be useful, the abstract itself mentions "There was a linear fall in cabin pressure as the aircraft cruising altitude increased." That said, I did miss until now that this relationship only held during normal cruising, as the cruising altitude increased into the flight. On the 3 occasions the aircraft descended to a lower cruising altitude mid-flight, the cabin pressurisation was higher than predicted. (I'm pretty sure they mean in hPa, i.e. the pressurisation was at a lower altitude equivalent than expected.) It sounds like possibly they didn't have any occasions where the plane went to an unexpectedly high cruising altitude, so we don't know what would have happened then. And finally this was only during cruising, it doesn't apply during the descent and ascent phrase.

So my statement was at least a little wrong as the variation only seems to apply during normal cruising and probably won't apply during unexpected changes such as the weather example I mentioned. Also they only looked at 747-400 so it's possible other planes vary.

Note also in this study from 2004 to 2006, they don't seem to have generally reached the maximum allowed cabin pressure of 8,000 feet altitude equivalent. (Maximum seems to be what the sources use since they generally refer to the cabin pressure altitude equivalent.) The time frame suggests that this was probably mostly before high fuel prices really began to hit, and also before budget airlines really began to take off so these may have affected modern practices, even in full service airlines like those studied (Qantas, British Airways and a single Malaysian Airlines). And on that point, maybe only those airlines have/had these practices or even they only do/did it on the London to Bangkok route. Also it's possible that the airlines were aware that the study was being carried out and varied their practices. But I'm not sure as the study seemed to use a wristwatch altimeter and in-flight displays (which I take to mean any data shown on the entertainment systems).

That said, there also seems to be a push to higher Cabin pressurisation (i.e. lower altitude equivalents), e.g. the widely touted 6,000 feet for the 787. Our article also mentions something similar "A design goal for many, but not all, newer aircraft is to provide a lower cabin altitude than older designs. This can be beneficial for passenger comfort."

Our article also says "One study of 8 flights in Airbus A380 aircraft found a median cabin pressure altitude of 6,128 feet (1,868 m), and 65 flights in Boeing 747-400 aircraft found a median cabin pressure altitude of 5,159 feet (1,572 m)." referencing a 2010 study (albeit the link doesn't seem to work for me). So it doesn't seem that the findings of the earlier study on typical cabin pressurisations changed much in those 4-6 years. (I believe I looked into the issue of typical cabin pressurisations a bit more before, it's probably somewhere in the RD archives.) My memory is I found other stuff which supports the view the cabin pressurisation often doesn't reach the maximum allowed.

Anyway ultimately I don't know how well supported outside this study the finding on the variation of cabin pressurisation with cruising altitude during normal flight is either in the ~2005 period or nowadays. (My impression from vague memories of what I read before when looking into average or median cabin pressurisations is this is an area where there tends to be a lot of theory. What airliners actually do isn't well outlined or studied.)

Nil Einne (talk) 09:25, 24 February 2018 (UTC)

Hmm. Well, my phone can do both barometric pressure and GPS altitude, so maybe I'll start taking some data points and see what I can see. I suppose it's slightly less expensive to pressurize to 7000 ft if you're at 25000 than if you're at 35000, so if the airlines actually see any upside to pressurizing more than they have to, then I suppose they could. But I'm a little skeptical that they do — it's not something your average passenger tracks, so it probably doesn't give much competitive advantage. --Trovatore (talk) 09:58, 24 February 2018 (UTC)

Passengers rarely know it explicitly, but they feel more comfortable when cabins are pressurized to lower equivalent-altitudes. This is a major selling point for new airplanes like 787 and 777X: here's a blurb from Boeing.

Most of the time, the pilot configures the cabin pressure for "the lowest cabin altitude the plane can sustain," and that is basically limited by a few factors: how perfectly airtight is the cabin? How much energy can the engine bleed (e.g., the "air pumps") sustain? How much structural strength does the fuselage provide for the gauge-pressure difference between inside and outside?

As far as measuring this with your phone - I encourage the experiment, but I doubt it will work... I would not trust either the GPS altimeter and the baro-altimeter to accuracy of even +/- 1000 feet. Those kinds of consumer devices usually use software tricks and sensor-fusion to "improve" the data in a way that makes for very poor scientific instruments - especially when compared to a true altimeter. But if you collect data, please let us know! Nimur (talk) 17:46, 24 February 2018 (UTC)

I use the "GPS Status" app by MobiWIA, which reports the pressure in hPa, not as altitude. So it seems unlikely that it's doing any Kalman filtering or anything like that at the app level. I don't know enough about the internals to know whether the OS could be doing any such thing, but there's no obvious evidence of it as you watch the numbers change. --Trovatore (talk) 20:26, 24 February 2018 (UTC)

Don't pretend to know all the answers to this but here are some more likely possibilities. One doesn't want ones bread rolls to go pop. So when it was a football shape is was possibly out gassing (ie bags are not air tight). Second (you haven't mentioned which type of hairycraft you where in, nor the route) Most jet propelled aluminum cans in service today still use engine bleed air to pressurize the cabin. Bleed air is hot! So to increase the cabin temp, more air is vented out of the cabin (ie pressure is lowered) to allow more warm air to enter cabin from engine. To cool, venting is restricted and cabin pressure increases, slowing down the flow of warm air from engine. The battery in my slide-rule has gone flat so I can't use Log _e but fortunately there is this handy widget air-pressure-at-altitude-calculator. So by regulating the temperature, the cabin altitude pressure can fluctuate between 6000 to 8000 feet. Which equates to 11.78 to 10.92 PSI. Nearly one pound per square inch on the raper of said roll which has degassed whilst at lower pressure of 8000 ft. More than enough to deflate it. --Aspro (talk) 02:45, 24 February 2018 (UTC)

I once bought a soda at 700 feet and broke the seal at only 2 or 3 thousand feet and it burped fog. Air pressure's interesting stuff. (this was natural ground altitude) Sagittarian Milky Way (talk) 03:54, 24 February 2018 (UTC)

Very interesting indeed. Thank you all. I understand now that pressure changes within the cabin are nothing to worry about, and that strange things can happen. Many thanks to all. :) Anna Frodesiak (talk) 05:59, 24 February 2018 (UTC)

The flight attendants have to go through a canned routine set of announcements at the start of every flight, going through the "safety features of our Boeing 737 aircraft". One of the things they always say is ...although we never anticipate a change in cabin pressure.... Liars. Oh well. I know what they mean, and I guess that's what matters. --Trovatore (talk) 06:13, 24 February 2018 (UTC)

You will find they say something like "in the unlikely event of a cabin depressurization ..." The cabin pressure changes almost continuously throughout a flight, but uncommanded depressurization followed by deployment of the emergency oxygen system is indeed a rare event. Dolphin (t) 12:24, 24 February 2018 (UTC)

No, they don't. They say precisely "although we never anticipate a change in cabin pressure". I've heard it enough times that I can recite it in my sleep. This may vary by airline, but I guarantee you that that is word-for-word exactly what they say. --Trovatore (talk) 20:58, 24 February 2018 (UTC)

This seems to depend a lot on the airline and probably even the precise video. This American Airlines video for example says [4] "now, if the airplane loses pressure". This United Airlines video [5] "if necessary, an oxygen mask". (I came across another more standard safety video from United Airlines which said the same thing, but not linking to it because I'm not sure that Global Airline Safety Videos Youtube channel ensured they had the appropriate permissions for redistribution although I doubt UA will actually care. I had higher hopes for another video again with very similar wording albeit also fancy on Canal Plus Producciones's channel but looking a bit more they seem to be a relative small Venezuelan production company so I'm less certain. And I strongly doubt FrienldySKy made sure they had the appropriate permissions although the wording is slightly different on their old video.) This Virgin America one says [6] "if the cabin pressure's changin'". This KLM one says [7] "if there is a sudden decrease in cabin pressure". This Cathay Pacific one says [8] "oyxgen masks will drop automatically if they're needed". (An older one which coming from the production company which made it hopefully they know what they are doing [9] says "should oxygen be required". The smoking version says the same thing [10] although does remind you to extinguish your cigarette later.) Not sure what airline you are thinking of but I wonder if they still say that. With the modern trend of trying to produce fancy inflight safety videos, the would likely have a new one and perhaps the wording was changed. Nil Einne (talk) 02:14, 25 February 2018 (UTC)

Most of my flights are on Southwest Airlines. Yes, they definitely still say it. --Trovatore (talk) 03:17, 25 February 2018 (UTC)

Can't find any safety videos. But it seems unlike many other airlines who are competing to have some viral hit safety video, Southwest Airlines instead known for their flight attendants spicing up the safety presentations. So I can find videos of the flight attendants saying this [11] [12] [13] or similar [14]. Although not all seem to say it [15] [16] [17] [18] (while some of these look to be semi spontaneous or invented by the flight attendant, this one at least looks scripted probably by the airline since other than the fact she seems to be reading it, it sounds very similar to this [19] [20]). Nil Einne (talk) 11:51, 25 February 2018 (UTC)

Southwest doesn't use "videos", at least not on their bread-and-butter short-haul flights. The 737s have no video screens. --Trovatore (talk) 20:21, 25 February 2018 (UTC)

Loss of pressure can carry effects, ranging from "nothing" to "passenger discomfort" to "potentially serious physiological response," to "serious medical emergency"... Commercial airplanes never (intentionally) expose passengers to hazardous cabin pressure. In aviation, the two types that are of interest are rapid cabin pressure loss, and slow pressure loss. In the event of a sudden depressurization, most passengers will lose "useful consciousness" within a moment or two. (During my High Altitude Hypoxia training, I lasted almost 300 seconds at equivalent of 29000 feet altitude, with a little fuzziness on the "useful" bit - in this case "useful consciousness" is a technical term meaning that I had enough energy to put on my oxygen mask - barely - I had some blackout periods in there, and could not do basic aviation arithmetic correctly). If you're interested in developing a sense for what altitude hypoxia does, have a look at the famous Four of Spades video. While watching that training video, bear in mind that you are watching a high-IQ, high-adrenaline, college-educated Air Force officer describing his symptoms. Absent oxygen, your brain does things differently. If you want to survive a depressurization event, the only thing your brain must do is to remember is to put the oxygen mask on. If you can not remember to don the mask, it will not matter if you are awake. If the depressurization occurs slowly, you might not notice the onset of hypoxia, and you may lose useful consciousness without noticing it. If the crew informs you to do so, put the mask on immediately. Nimur (talk) 17:59, 24 February 2018 (UTC)

By now we've established that the ambient cabin pressure does change - and this is safe and normal on a commercial airliner. You'll typically lose about 5 or 8 inches of mercury worth of pressure (25 kPa) by the time you hit cruise altitude. At that point, the snack bag (which has sea-level air inside it) will tend to puff up. For me, the most interesting detail is that the snack-bag eventually re-equilibrates to ambient pressure, but only after some delay! Those little bags of cookies or peanuts are frequently bags made from plastic or metallized plastic - sort of like mylar - and sealed using hot-bar pressed seams. Mylar is amazing aerospace material - compared to our best engineered materials, it's darned near airtight - but not actually perfectly airtight! We don't know whether air leaks through the seams or through the material itself. But you can sort of develop an intuition about how fast air flows into and out of the bag based on how effectively it withstands its internal pressure against the ambient cabin pressure! Nimur (talk) 17:46, 24 February 2018 (UTC)

As I understood the OP's comments, the bag was not puffed up when they received it 1 hour into the flight, so the plane was likely already at some level of cruising altitude. This puffing up happened due to changes in cabin pressure (probably due to changes in cruising altitude) after the ascent phase. Nil Einne (talk) 02:14, 25 February 2018 (UTC)

February 25

A low viscosity liquid which will harden after 20+ minute working time?

I have a small hobby project in which I'm printing some keyboard keycaps with an SLA 3D printer with the intention of filling in an inset legend with some kind of paint or resin which will harden in a contrasting colour (white since the SLA resin is black) but I don't know exactly what to use. The spaces I fill will be really small so I need to use something with low viscosity which I can inject with a (blunt, narrow needle) syringe and it will self-level and harden very slowly so I have time to inject about 70 characters in total. I know two-part resins are available but I don't know their viscocity or if there are more suitable alternatives. Can anyone advise a suitable material available in small quantities for this purpose? 185.222.217.213 (talk) 14:34, 25 February 2018 (UTC)

There'll be lots of low-viscosity potting epoxy on the web. But I'd have thought some pot of enamel paint for hobbyists would do the job just as well. I think it would probably be worthwhile painting the flat area of the keycap with some masking fluid too. Dmcq (talk) 14:58, 25 February 2018 (UTC)

• Stupid-sounding idea, but what about solder, if you are OK with a metal-silvery finish? The idea is that you might already have some training and equipment for soldering. (I am assuming that since you hobby-print 3D pieces there's a good chance you have standard electronics material already. If not, follow Dmcq's advice instead.)

I would try the standard tin-lead at first on a test cap (to check that you can reliably apply the solder without melting the plastic too much; you may need to thicken the plastic cap though). If you can 3D-print freely, that is easy enough to test. If this does not work, you could switch to using a low-temperature solder (Solder#Solder_alloys lists many; I could find some Cerrolow 117 for about $40 on ebay), which should be easier to use (you can overheat the fusion temp by more, so you have more time to wipe mistakes, and viscosity is lower, while being at lower absolute temperatures for the plastic cap). Tigraan^{Click here to contact me} 13:40, 26 February 2018 (UTC)

Doesn't he need something that adheres very well to the plastic? If the label comes out of even one key it will look like a really shoddy piece of workmanship. But I don't understand why 20+ minute hardening time is needed - can't the keys be done one by one and left to harden progressively? Wnt (talk) 00:22, 1 March 2018 (UTC)

Feynman Lectures. Exercises. Exercise 19-17 JPG

. .

...

A yo-yo like spool consists of two uniform discs, each of mass M and radius R, and an axle of radius r and negligible mass. A thread wound around the axle is attached to the ceiling, and the spool is released from rest a distance D below the ceiling.

a) If there is to be no pendulum -- like swinging motion, what angle should the thread make with the vertical as the spool is released?

b) What is the downward acceleration of the center of the spool?

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

I have found the acceleration with assumption the motion is vertical (with zero angle ).On time 6:16 of a video youtube.com/watch?v=kdTLZ6-hVq8 it seems there is no sidewise motion. But I don't understand why the angle is zero. The acceleration of the yo-yo is $g<a<0$. So it has apparent weight = $2M(g-a)$ like in the going down elevator. This net force must create a torque and shift the yo-yo so that the center of mass comes to be under the attachment point on the ceiling . Username160611000000 (talk) 16:05, 25 February 2018 (UTC)

The tension in the thread and the true weight of the spool (acting vertically) cannot be collinear otherwise there would be no torque to provide the angular acceleration. From the moment of inertia and the angular acceleration, you can calculate the perpendicular distance of the line of action of the tension in the thread from the centre of the spool, and hence the initial angle of the thread. From the order of the questions, I assume that you were supposed to set up simultaneous equations for linear and angular acceleration. If you have used an energy method for the acceleration, then you have probably calculated the vertical component of the tension. (I can see now that you didn't.) Dbfirs 17:13, 25 February 2018 (UTC)

"the perpendicular distance of the line of action of the tension in the thread from the centre of the spool" = r. How from this can I find the angle? Username160611000000 (talk) 18:08, 25 February 2018 (UTC)

I have calculated the acceleration form next formulas.

$\alpha ={\tfrac {\tau }{I}}={\tfrac {2Mgr}{{\tfrac {2MR^{2}}{2}}+2Mr^{2}}}$

$a=r\alpha ={\tfrac {gr^{2}}{{\tfrac {R^{2}}{2}}+r^{2}}}$

The moment of inertia $I$ and the torque $\tau$ were calculated around a point A (see image PNG) where the thread touched the spool. From answers it is correct jpg. The tension is then $T=2M(g-a)$, but again using the assumption that the spool is moving vertically.

In The Solutions is said:

...

The spool is acted upon by gravity, directed vertically downward, and the tension force T along the thread. The spool will not swing if there are no horizontal forces, i.e. if the thread is vertical.

—  MEPhI , Solutions (Google Translate)

But it does not explain the case when the thread is fixed , then the spool will go so that the center of mass is under the point on the ceiling PNG . I.e. we hold the spool with fixed thread by hand in position, showed in the exercise PNG; then the spool is let fall. The spool then 100% will go to the right, but in the absence of horizontal forces at starting moment. I can understand it like next: the spool starts rotation and pulls the thread away from the vertical. The inclined thread generate a horizontal force. It proves that absence of horizontal forces CAN produce horizontal motion and it demolishes the arguments from The Solutions. Username160611000000 (talk) 18:20, 25 February 2018 (UTC)

• • From assumption that the center of mass must be under the ceiling fixing point I can calculate the angle: $\sin(\theta )={\tfrac {r}{OC}}$ (see image). It is not clear from the statement of the exercise is $D=OC$ or $D=AC$, but it is not a problem. The problem is how to prove that the center of mass must be under the point on the ceiling, moreover the experiment shows that it's not the case. Username160611000000 (talk) 08:18, 26 February 2018 (UTC)

Now I'm confused. It's too long since I did this type of problem. Perhaps someone else can help? I had assumed that the spool was released with the thread vertical, but there will be horizontal motion in this case as the spool rotates about the lower end of the thread. Perhaps you were meant to assume that the centre of mass is moved to be under the point of suspension before release, but then the tension in the string will cause a swing the other way. Dbfirs 10:11, 26 February 2018 (UTC)

@Dbfirs:but then the tension in the string will cause a swing the other way. The tension will create a torque about point O only in O - reference frame (ref. frame in which point O is at rest). First, the tension will simply increase angular velocity about point O, there is no guarantee that the point O will move horizontally. Second, O - ref. frame is an accelerating frame, so there may be some complications. On the other hand point A is at rest in ceiling ref. frame (at least during small time at start). Username160611000000 (talk) 17:09, 26 February 2018 (UTC)

Hmmm. I think the thread, wherever unspooled, does not actually move; thus all downward force on it is opposed by upward force. And at the moment of release, the spool has no downward velocity. Therefore, it applies a torque according to the newtons (M * g) multiplied by the lever arm = radius r. To oppose the torque, the string can be non-vertical, applying a contrary lever arm. That puts the center of mass of the spool directly under the string, AFAICT - in other words, the center of mass is stable when directly below the suspension point, like with anything else. That makes the angle sin-1(r/D).

Now as for the downward acceleration, we know it can move only r * however many radians it turns by. From the list of moments of inertia, I = 1/2 mr^2 for a disk. We use the quation from moment of inertia that tau = I alpha, where tau (the torque) is that M*g*r thing. So alpha = M*g*r / (1/2 MR^2) given that R is the disk radius and M is I think the same M, or alpha = 2*g*r/R^2. The distance, velocity, and acceleration should all be r* the angular versions I think, so I get 2*g*r^2/R^2 ... hmmm, units should be acceleration, it tends to zero for a very thin axle -- but, I get double speed falling if you wrap a string around a soup can, which is wrong. Sigh. This time I'll post my detritus in hope someone finds a fix, with apologies. Wnt (talk) 02:11, 27 February 2018 (UTC)

@Wnt: With the non-vertical thread, I'm pretty sure the spool would swing in the direction of the horizontal component of the string tension. Compare this with picture frame wires, which typically have two attachments with opposing tensions that can be adjusted (they tend to get misaligned anyway). Take away one of the attachments and the frame swings. Assuming an infinite thread and fall of the spool (with gravity not changing), the spool's mass should oscillate about the thread's vertical position as the horizontal tension is a restoring force towards vertical with each swing. -Modocc (talk) 19:14, 27 February 2018 (UTC)

So with the picture frame we have something like this , where last position is the one when oscillations stop. I'm not sure the frame center of mass will go left.

Also I checked cylinder with fixed thread : 12345678 Username160611000000 (talk) 12:20, 28 February 2018 (UTC)

If the tension is very small or negligible compared to the mass so is the lateral motion of the swing (or the rocking) of that mass. With the picture frame you are simply showing the very end result... a stationary frame and not the oscillations that had to first be dampened due to the spring tension that was on the wire and became free to set the frame in motion. I can't quite make out what your dropped spool test shows. If it is swinging rather than dropping straight down the string is not spooling very well, too tightly wound or not enough spool mass etc. Try again by wrapping it around a heavy jar and see if you get the same result. There are different factors that will influence any experiment or test. But to get an idea of the magnitude of the forces involved, consider a bicycle. If one lifts it at the handle bars straight up the bike rotates about its rear wheel which will stay put, but if one does this at the slightest angle toward the center of the bike it will roll backward even with a slight pull. -Modocc (talk) 13:29, 28 February 2018 (UTC)

• • • • • I can't quite make out what your dropped spool test shows. As I told, experiment with the wound thread gives strictly vertical thread during unwinding. Snapshots show that the cylinder CM goes sidewise only on 6th snapshot (each snapshot time = 1/24 sec). So the inclined thread doesn't imply horizontal motion and the vertical thread doesn't imply absence of horizontal motion. Username160611000000 (talk) 17:10, 28 February 2018 (UTC)

I see. You placed the spool in such a way that is reversed from what was expected; thus the spool is not spun initially for half a turn without any significant tension (other than to remove the thread), thus it gets a large pendulum kick to the right when it undergoes tension... which should be expected because the string happens to be no longer vertical at that point when it actually starts unspooling normally. -Modocc (talk) 17:46, 28 February 2018 (UTC)

I still do not trust The Solution reasoning (and so I can't explain the experiment theoretically). How to model the motion of the spool (or disk) by numerical methods shown in lecture 9? To do this, we need to know the expression for the tension force (in terms of coordinates or velocities and independent from 2nd Newton's law), but it is unknown. Username160611000000 (talk) 19:02, 28 February 2018 (UTC)

The picture isn't the same as the spool, because the thread's attachment can *rotate* freely. Of course, the thread is not rigidly attached to the spool... but so long as it supports the spool as it unrolls, it has to remain at a fixed angle that meets almost perpendicularly at one end. Its attachment point can never be right on top of the spool, unless it runs out and reaches the point where it's tacked on to something. Wnt (talk) 14:49, 28 February 2018 (UTC)

True, the thread's vertical component of the spring tension supports the spool or counters gravity and that is true for the frame's wire. The horizontal component of the tension in both do not. The point I am making with the picture is that its horizontal tension is divided between left and right tensions that balance each other, thus preventing lateral motion and swing of the mass center. Replace the wires with springs and the ensuing oscillations become very apparent. With the spool, without any lateral force acting on it, it drops straight down and this should be the cased as diagrammed in the problem statement with that shows a vertical thread. Add an unbalanced lateral force and there is lateral movement that has to change the angle of the thread (in part, because the thread remains attached to the ceiling) which changes the lateral force which is always a component of the tension vector. The magnitude of the changing horizontal component then has to oscillate as pendulums tend to do and it doesn't take much force to set them off. --Modocc (talk) 16:17, 28 February 2018 (UTC)

Maybe I'm missing what you're saying, but if you pull up a picture that is hanging off a bent wire coat hanger, then (barring bending of the hanger...) you expect to be able to pull it straight up even though the hanger makes an extreme angle. In this case the lower leg of the hanger is the near-horizontal radius from where the thread separates from the spool to the center of the spool. And the upper leg of the bent hanger is the string deviating from vertical to meet the spool. Wnt (talk) 00:01, 1 March 2018 (UTC)

The coat hanger is rigid, but the spool is free to separate from the thread. Elastic tension is all that is needed here anyway and you calculated the string tension with the string vertical, but then moved the contact point from vertical to non-vertical. However, the string tension is zero if the thread starts out horizontal and the spool has to drop and swing a bit before the torque increases significantly. Since the spool is free to separate from the thread, the unopposed horizontal component of that tension accelerates the spool's mass center laterally as the spool continues to move downward. Modocc (talk) 15:50, 1 March 2018 (UTC)

Where the problem statement states "no pendulum -- like swinging motion" is a bit unclear, however the solution that the string (or thread) needs to be vertical such that the net horizontal force acting on the disks' axle is zero is a reasonable interpretation of that... any angle from vertical gives a measurable horizontal force component from the string's total tension and without any other horizontal force also acting on the mass center, that lateral string tension results in its horizontal acceleration. Now if the string is "fixed at point A" as in your diagram, the spool thread is essentially stuck as the spool rotates without the thread being released, creating a pendulum with lateral string tension acting on the spool and which you didn't account for and the motion of which is to be avoided. --Modocc (talk) 17:30, 27 February 2018 (UTC)

The on-line handouts and examples from major educational establishments (Browns, MIT etc) all seem to assume that the thread remains vertical. I'm trying to convince myself that this will be the case for all values of moment of inertia, then it does make the calculations much simpler. Maybe it can be proved that the moment of the weight about the (moving) point of contact of the lower end of the string always produces a rotation at exactly the rate to match the downward speed of the centre of mass. If so, then careful release with the string vertical will always produce a pure unrolling motion without any swing. Dbfirs 11:37, 1 March 2018 (UTC)

I think you meant a pure rolling motion. Gears, wheels, balls or tires can also be set rolling down a vertical wall with minimal contact. Of course when the wall is not vertical, the angle is set opposite of vertical than that of the spool here since the force acting on their mass centers is compression and not tension. For a given angle from vertical, the component force due to gravity that accelerates an object down an incline contributes to the thread tension. In particular, if they were to oppose one another there is no acceleration and the object is held stationary on the incline and the force of that tension is mg*the sine or cosine of which ever angle one is referring to (calculated here). For the spool, there is no incline thus no compression force, but there is also centripetal acceleration to consider too. In any case, the tension vector can also be broken down into two components, vertical and horizontal, to determine the horizontal acceleration of the bob at any given moment. -Modocc (talk) 18:35, 1 March 2018 (UTC)

See this reference for calculating an ordinary pendulum rod's tension. --Modocc (talk) 19:43, 1 March 2018 (UTC)

Modocc, I was discussing the original question. I've no problem with the rotational dynamics of rolling down a rough slope where the acceleration depends on the moment of inertia. I'm also fully familiar with the dynamics of a pendulum for a small angle of swing. I'm just trying to convince myself that this problem can be treated in the same way. Dbfirs 07:49, 2 March 2018 (UTC)

The pendulum reference is valid for large angles not just small angle swings and the original problem involves leverage of an object while under the influence of gravity and these other cases should help the OP, Wnt and others understand how to work it and why, especially since the OP asked about how the thread tension can be calculated. In addition, the applied force on the spool acts on its mass center. An example is the firing of a projectile from an unmoored ship's cannon. It does not matter where on board one places the cannon, conservation of momentum requires the ship's mass center to be acted on by the cannon recoil in a direction opposite that of the projectile even when the cannon is somewhere in a far corner such that it sets the ship spinning too. Here, the thread tension does the same, torquing the spool while it acts on the spool's mass center. -Modocc (talk) 13:06, 2 March 2018 (UTC)

True about the pendulum, except for the usual calculation of the period of swing. Your ship example is a good illustration. The rotational behaviour of the ship varies with the positioning of the canon, and in this case the swinging behaviour varies with the initial release configuration. I would like to be able to prove that if the thread starts vertical then it remains vertical, assuming no snagging. I used to prove every year from first principles that the motion can be separated into the linear acceleration of the centre of mass plus the rotation about the centre of mass, and this principle applies here in exactly the same way if we are allowed to assume that the thread remains vertical. I'm just worried about the assumption. Dbfirs 14:35, 2 March 2018 (UTC)

• In the simple pendulum problem the tension can be easily found from centripetal acceleration formula $v^{2}/R$, coordinates and velocity of a bob. In the spool problem constraint is still present and as I can see I can find one tension component from another , e.g. $T_{y}$ from $T_{x}$ , but I cannot find $T_{x}$ (so I should use $T_{x}(t+\Delta t)=T_{x}(t)$ ).Username160611000000 (talk) 19:42, 2 March 2018 (UTC)

How well do Tanks and other tracked vehicles coast?

I was talking to a buddy of mine about tanks the other week, and we were wondering how well tanks coast and if it is possible to put regen braking on a tracked vehicle. Does anyone have any experience with tanks here? Our Tank steering article wasn't much help, and neither was Google. Thanks, L3X1 ◊distænt write◊ 16:36, 25 February 2018 (UTC)

I can't find any in service, but this paper discusses the concept and notes that there are some hybrid-electric tracked vehicle development programs in the works. I don't know how well tanks coast, but regenerative braking definitely seems to be seen as a non-laughable concept. TenOfAllTrades(talk) 17:49, 25 February 2018 (UTC)

Considering that the M1 Abrams gets only .6 miles per gallon, I doubt "coasting" is even the right word; it barely budges without eating a lot of fuel. Matt Deres (talk) 21:05, 25 February 2018 (UTC)

Well, the thing weighs 60-70 tons, so unless there is some jake or automatic engine braking going on, if you let off the throttle at 30 or 45mph inertia says it will keep on going for a little bit. I've driven some hydraulic transmission construction vehicles, but they always would come to a halt pretty quick if you let off the gas. Thanks, L3X1 ◊distænt write◊ 22:21, 25 February 2018 (UTC)

Common Tanks are very ineffective vehicles because of their weight and Caterpillar tracks. They also lose their speed fast, even on streets, because the traks are very sturdy and stiff. A recuperation system would not be worth the extra weight and besides it would have to be absurdly huge to take up and feed back the braking energy of 60+ tons at 60-70 km/h. Because of their ineffectiveness most modern Armoured personnel carrier constructions went back to wheels. --Kharon (talk) 05:39, 27 February 2018 (UTC)

This is kind of an aside, but why hasn't anyone come out with wheels covered in retractable spikes or more sophisticated specialized grabbers or pseudopods? I'd think each could have its own little computer brain and (at least at military funding levels) its own camera. I'm picturing whizzing down the road on spikes that drive into ice for traction, stop short to sail over speed bumps without a jiggle, and go the extra two inches to cross a problem pothole. But conceivably each one could drive its own little pitons this way and that, hundreds of times a second, and allow you to drive straight up the wall, maybe even along the ceiling. One of the few things you could actually do with all this cheap computing/surveillance technology. Wnt (talk) 00:28, 1 March 2018 (UTC)

10-4 to that, I like the way you think! L3X1 ◊distænt write◊ 02:23, 1 March 2018 (UTC)

Why do humans have body hair?

Why?86.8.202.234 (talk) 22:10, 25 February 2018 (UTC)

Because mammal! "It [ hair ] is a definitive characteristic of the class." {The poster formerly known as 87.81.230.195} 90.220.212.253 (talk) 03:16, 26 February 2018 (UTC)

For the lack of body hair in humans, see What is the latest theory of why humans lost their body hair? Why are we the only hairless primate? from Scientific American. Alansplodge (talk) 09:10, 26 February 2018 (UTC)

We're not hairless. ←Baseball Bugs ^{What's up, Doc?} carrots→ 10:34, 26 February 2018 (UTC)

Indeed, but a lot less hairy than all other primates (most people anyway). Alansplodge (talk) 11:20, 26 February 2018 (UTC)

"Hairless" does not simply mean "less hair". I'm reminded of Richard Armour's comment about the wireless, i.e. the radio: "Although it had a great many wires, it had less than it might have." ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:21, 26 February 2018 (UTC)

Still, the concept of the human as a hairless ape is a common enough trope, see, for example, the sentiment in the title of the famous book by Desmond Morris, The Naked Ape. --Jayron32 17:41, 26 February 2018 (UTC)

This is the bloke who wrote the offending article. Alansplodge (talk) 18:05, 26 February 2018 (UTC)

FWIW, the Aquatic ape hypothesis mentioned in that article is not really supported by any paleontological or archaeological evidence and is generally dismissed by most anthropologists (outside of a few proponents) as an interesting thought experiment. In terms of body hair, of course, we have about the same number of hairs per square inch as a chimp, it's just that the hairs are shorter and finer. Matt Deres (talk) 14:36, 26 February 2018 (UTC)

Thank you. Yes, some of us have particularly fine hair. Drmies (talk) 17:42, 26 February 2018 (UTC)

The problem about the lack of paleontological or archaeological evidence for the Aquatic ape hypothesis is that it would be extremely difficult to find any even if it were true. The hypothesis supposes the aquatic behaviour (leading to various physical adaptations) was practiced on marine shores, but since the postulated period in question sea levels have risen by hundreds of feet, so any such evidence will have been almost if not entirely obliterated, and any that remains would be almost impossible to find and excavate, even if we were looking in the right places (which we're not). In this case, the maxim "absence of evidence is not evidence of absence" is particularly apt. {The poster formerly known as 87.81.230.195} 90.220.212.253 (talk) 23:28, 26 February 2018 (UTC)

I think a "mud ape" hypothesis is more interesting. After all, polar bears can swim with thick body hair, but we are constantly reading about well-meaning humans pulling things out of mud pits. In earlier times they would also have meant well... meant to eat well. The Okavango delta in particular is known for periods of flooding and periods of mud and periods of wildfire, and seems like an interesting possible ancestral situation. The feet of humans, like lechwe, seem elongated -- could it have been for crossing that mud? Wnt (talk) 01:46, 27 February 2018 (UTC)

So you're saying it's unfalsifiable? Then I'm afraid it's not science. We have found and studied numerous human settlements that are now underwater, having been submerged by rising sea levels, so the implication that "evidence may be underwater so we just have to throw up our hands" is nonsense. All conclusions in science are provisional, so it's possible in the future we could unearth some stunning evidence for the hypothesis, and if we do, it will presumably be revisited. Until then, it fails in explanatory power relative to other hypotheses. RationalWiki has a good summary of its problems. Although as noted, a problem is that people often seem to mean different things when talking about the "aquatic ape hypothesis". The notion that marine environments may have played some role in human evolution is, I don't think, very crazy, though we need stronger evidence to state that more definitely. The idea that human ancestors at some point lived entirely in the water like whales or dolphins is ludicrous. --47.146.60.177 (talk) 08:23, 27 February 2018 (UTC)

Yes, that site is a good overview of the problems with the AAH, but the problem it goes deeper than that. We all have hooded noses, we all have sebaceous glands, we all have all those items that proponents of the theory say were likely derived from a semi-aquatic existence. In order to be true, aquaticism would have to have been a massively important evolutionary feature where divergence from that adaptation meant complete lack of procreation. It would require a bottleneck where all non-aquatic people were wiped out to the point where even recessive non-aquatic genes were virtually eliminated. Matt Deres (talk) 13:33, 27 February 2018 (UTC)

That's not really a valid argument at the end. Many racial features and even more substantial cosmetic alterations in the lineage leading to humans (such as hooded noses) can be seen as cosmetic features that potentially might have been basically under no selection at all. If on average a person had to dive into a creek once a generation to avoid a barrage of rocks, spears, or arrows, that would be enough selection to make big changes in even a thousand years. Wnt (talk) 15:14, 27 February 2018 (UTC)

I'm sorry, but that's just ridiculous - you don't develop subcutaneous fat and a "hairless body" because you jump in a lake every few months. We're talking about a massive evolutionary pressure here - one that hasn't seen itself reset despite millennia since it proved advantageous. We're all different shapes and sizes and colours, but whatever selected for hooded noses did so with such survival differentiation that essentially nobody gets born without them any more. You can't have it both ways; it might hang on if there's no longer a survival bias, but something hugely significant must have selected for it at some point. Matt Deres (talk) 14:37, 28 February 2018 (UTC)

[In answer to 47.146.60.177] "The idea that human ancestors at some point lived entirely in the water like whales or dolphins is ludicrous": yes, I agree with that entirely, but I'm not aware that anybody has ever suggested such an extreme version of AAH. As for the inaccessibility of evidence making it unfalsifiable and therefore unscientific: no, not in principle, just extremely difficult (like, for example, Einstein's theorised gravity waves which eventually were detected against his own expectations). And yes, we have found and excavated submerged human settlement sites, but not, I think, any sites a million or so years old, and also some 300 feet below current sea levels, and also many miles from current coastlines (as most potential sites would now be). However, we're now just exchanging opinions without reliable sources, so we perhaps ought to draw the discussion to a close. {The poster formerly known as 87.81.230.195} 90.220.212.253 (talk) 06:48, 28 February 2018 (UTC)

February 26

Static discharge and lightning igniting fuel

According to the article Pan Am Flight 214, the plane was damaged due to lightning igniting fuel vapors. The response was to install Static discharger on aircraft. I read the article on that, and it seems to be focused on preventing static electricity build up from interfering with cabling and electronic equipment. What I don't get is how discharging static electricity would prevent a lightning strike from igniting fuel vapor. Can someone explain? RudolfRed (talk) 00:29, 26 February 2018 (UTC)

From reading our articles, Static discharger and that on the flight, I don't think it would have helped in the situation with Flight 214. If you look at the report, this was only one of a number of the recommendations made to FAA Wikisource:Aircraft Accident Report: Pan Am Flight 214/Attachment II in response to the accident which the FAA then implemented Wikisource:Aircraft Accident Report: Pan Am Flight 214/Attachment III. I'm fairly sure it's common the CAB/NTSB and FAA will feel that what was discovered from an accident suggests the need for something, even if that probably wouldn't have helped in the specific accident. This may have been even more common in the past, when there were still a lot of unknowns. See also the report itself, especially the end Wikisource:Aircraft Accident Report: Pan Am Flight 214. The modern FAA's lessons learnt also doesn't mention static dischargers but does suggest the accident was one of the key catalysts for further research etc that lead to improved protection against lightning strikes [21]. This, while maybe not the best source (NY Daily News) is also perhaps of interest [22]. As may be obvious from Inerting system, something which arose, at least in part from flight 214, is still an ongoing issue. Nil Einne (talk) 02:29, 26 February 2018 (UTC)

@Nil Einne: Thanks for the explanation and detailed answer. RudolfRed (talk) 02:43, 26 February 2018 (UTC)

I suspect an indirect lightning strike could create a static charge (flying through where a bolt passed, or by electromagnetic induction as the bolt increased and decreased in intensity). Wnt (talk) 02:15, 27 February 2018 (UTC)

February 27

Isn't the anthropic principle a tautology?

Isn't the anthropic principle a tautology? Doesn't it basically say that we contemplate our existence, well, because we can? Or does it relate to the old cogito ergo sum (with which I incidentally disagree)? I feel I'm missing something important here. 93.142.74.255 (talk) 09:11, 27 February 2018 (UTC)

I don't see how that would make it a tautology? Have you read Anthropic principle?--Shantavira|^{feed me} 09:20, 27 February 2018 (UTC)

• The observation behind the "anthropic principle" is by itself a fairly mundane observation (though not strictly a tautology). Our article mentions that Others[ref] have criticised the word "principle" as being too grandiose to describe straightforward applications of selection effects. Its point is to explain apparent paradoxes by selection bias.

For instance, a classic creationist argument is to say that humans have unique characteristics among animals, Earth has unique climatic conditions and is the only place to host life, physical constants of the universe are tuned extremely precisely to allow matter to exist as we know it, etc.; all this would have had a ridiculous chance of happening by random chance, and a single deviation would have caused sentient life not to exist; hence, there must be some deity who fine-tuned all of this. The AP is in effect saying that we wouldn't be here to observe it if it was not the case, which invalidates the "hence" (because in both the deity-created and the random-dice universes, only the outcomes with a sentient species to observe it are sampled).

Of course, only the precise formulation is new, not the idea itself - the very similar "shipwrecked crew make no offerings" existed in the Antiquity. Tigraan^{Click here to contact me} 09:53, 27 February 2018 (UTC)

The anthropic principle is notable mostly in the use, and mostly in the use to dispel "it won't happen because it hasn't happened before" type arguments. For example, the Earth didn't collide with another planet a million years ago because then we wouldn't be here -- which means that we don't know (at least from Earth alone) whether such collisions happen all the time. Or, more practically, we know the Earth didn't overheat and form a Venus-like greenhouse because we're alive to look, but that means we don't know whether the odds of that happening are low in the future. In the extreme case, I recall a Nature paper from some decades ago arguing that because the position of a human today is random within the order of all humans ever to exist, humans probably will not produce more offspring than that in the future; hence an end of the world can be predicted within a few generations. But I didn't track it down just now. Wnt (talk) 13:01, 27 February 2018 (UTC)

@Wnt: You are describing the Doomsday argument. Tigraan^{Click here to contact me} 19:05, 27 February 2018 (UTC)

• The anthropic principle is tautological in the week sense (of course we have to live in a universe that supports our sort of life in order to support our sort of life) and is entirely arbitrary and unfalsifiable in the strong sense. In the strong sense, the claim is that the universe had to be the ways it is (or very close) because otherwise there would be no intelligent life to observe it.

We have no way of knowing that there being intelligent life to observe it is a necessary property of existence, and we have no way of knowing that in other universes without matter as we know it there might not be some sort of non-material life as we don't know it. We cannot enumerate these possible universes, we have no idea of the unknown unknowns or any way of actually modelling these existences other than in very crude parochial terms.

"week sense": is that when you sense how weeks pass? --Hofhof (talk) 18:37, 27 February 2018 (UTC)

Seems like they pass more quickly every year. --47.146.60.177 (talk) 19:00, 27 February 2018 (UTC)

Linking relevant article... Hofhof (talk) 19:25, 27 February 2018 (UTC)

To New Yorkers or isolated Pacific islanders, life is only possible in Manhattan or on a coral atoll. See the famous parochial New Yorker Magazine cover, and translate that into a cosmological model if you will. μηδείς (talk) 16:52, 27 February 2018 (UTC)

Said cover being titled View of the World from 9th Avenue. --47.146.60.177 (talk) 18:59, 27 February 2018 (UTC)

• Thnaks to IP 47's image link and key nigh. I flame WP for having an antic spell-checker with know glamour funk shin. As for the numerology Wnt alluded to, there was also an argument before 2001 that, since statistically, any entity is statistically 5% likely to be in its first %5 of lifespan (this assumes a very unlikely linear relation) and 5% likely to be in its last 5% of it's existence, we can be sure that the Catholic church, if assumed to be a round 2,000 years old, has existed that long, there is a 90% chance that it won't live fewer than another hundred and five years or longer than another 38,000 years. This is, of course, yet more meaningless gibberish, funded at taxpayer dollars. μηδείς (talk) 03:31, 28 February 2018 (UTC)

• Here[23] are some old Scott Aaronson lecture notes with more conundrums like the ones Medeis mentioned. Aaronson mentions Nick Bostrom a few times and likes Bostrom's book[24] that I haven't looked at. 173.228.123.121 (talk) 08:52, 1 March 2018 (UTC)

I don't agree with this. There's way too many assumptions, e.g. that (ignoring our personal existence) Doom Soon and Doom Late scenarios have similar probabilities, or even that we're equally likely to be born as any being in the same world (I see no reasons against, but I see no reasons for, either). And the Dice Room paradox can well be explained by the fact that the expected number of people put in the room as well as the expected number of victims is unbounded. When dealing with infinities a lot of paradoxes arise in the probability theory, and since there's no evidence yet for either an infinite amount of people or infinite amount of universes/multiverses, this doesn't open up a new line of reasoning for me. Although, dealing on a daily basis with, essentially, applied mathematics, I might be biased here... --OP 93.139.80.133 (talk) 23:32, 1 March 2018 (UTC)

• OP here, thanks for the answers, folks. I kinda understand now that the principle is useful for dealing with theological fine tuning arguments, although it reminds me of Pascal's wager, which rests on the same problem as the Dice Room paradox mentioned in the Scott Aaronson link. I think this is probably why I'm so uncomfortable with it. 93.139.80.133 (talk) 23:32, 1 March 2018 (UTC)

Also: Medeis, your calculation about the Catholic Church's expected longevity is reminiscent of Laplace's solution to the sunrise problem. 173.228.123.121 (talk) 00:40, 2 March 2018 (UTC)

• Um, the fact that I mentioned a bullshit numerological "principle" don't make it mine! Bejeezus! μηδείς (talk) 05:54, 2 March 2018 (UTC)

Deflating tyres to drive over snow or sand

When you deflate the tyres to drive in the desert or over snow, how much bigger does the surface in contact with the ground get? How meaningful is that (let's suppose that you could inflate the tyres as soon as you drive over a different terrain)? How much more friction can be expected? --Hofhof (talk) 18:20, 27 February 2018 (UTC)

From the basic definition of pressure, (pressure equals force times area): if the entire weight of a vehicle is supported by its pneumatic tires, then the area in contact with the ground is inversely related to the tire-pressure. Of course, the simple model assumes a perfect, nonrigid wheel. Automobile tire dynamics constitutes an entire field of advanced science and engineering.

The extent to which contact patch size affects friction and traction and rolling resistance is nontrivial, because the wheels are nonrigid and the surface contact dynamics are complicated. (For starters, read Slip (vehicle dynamics)). So, you cannot accurately use a simple model derived from first principles; there are so many confounding variables. Instead, empirical laws are derived for automobiles; for specific modes of operation, on specific road surfaces, and for specific vehicle models, you can obtain a table of tire performance.

Our article on SAE J2452 - an automotive industry standard for tire performance - has links to more detailed technical resources.

Nimur (talk) 19:24, 27 February 2018 (UTC)

• This is complicated, and in general don't do it.

It used to work, back in the days of crossply tyres. Tyres were also of a taller cross section. As practically all modern tyres are radials, they're a different matter: there aren't many modern tyres where you can deflate them sufficiently to make a difference to their shape where this isn't also going to cause so much bending in the sidewall as to damage the tyre.

In principle, the idea is to let the tyre flex more, thus increasing the size of the contact patch. Ideally this should increase lengthwise, but not sideways - if it increases much sideways, you're killing those sidewalls. Another problem is that reducing the pressure reduces the load carried pneumatically as a balloon tyre and increases the proportion being carried by the sidewalls - which are also getting bent out of their ideal shape.

There's no increase in friction as such - the "grip" of tyres on soft ground is far more complicated than such a simple term. In this case, it's more about the structural integrity of the surface layer of the ground and nothing to do with the friction to the tyre. A hard tyre with a small patch will have a greater loading on the surface (force/area) and this tends to break up the integrity of the mud or snow - the limit for the unbroken surface is higher than the force available once it starts to crumble.

Where deflation is used today it's mostly with CTIS and tyres designed specifically to cope. They might also need a beadlock system to stop the tyres pulling off the rims when there's not enough pressure to hold them otherwise. Tyres can be re-inflated as soon as you hit tarmac, which is important for allowing higher road speeds and avoiding damage. Andy Dingley (talk) 20:49, 27 February 2018 (UTC)

Tires are much more abused in high speed curves. Todays sidewalls or tires in general can take an awesome lot of "abuse", so its a minor problem to run them partly deflated for some hours to give them more contact on sand, mud and alike where there is danger of sinking in to a degree you get stuck. This does not work on snow and ice tho because you actually will get more grip by having your given weight on a very small contact area. --Kharon (talk) 21:42, 27 February 2018 (UTC)

Tyres "in high speed curves" aren't abused, they're run at proper pressures, so that the sidewalls aren't excessively flexed as they are when the same tyre is run under-inflated. If they were run like this, then they'd fail very rapidly. Tyres in high speed curves are heavily loaded, but that's not the same as abuse - it's what they're designed for -and their sidewalls are in the correct shape. Andy Dingley (talk) 22:24, 27 February 2018 (UTC)

(irony on) Ah, is that why "under-inflated" tires always squeal so loud compared to the silence of tires in sharp curves at high speed?(irony off) I guess the logic dictates. --Kharon (talk) 09:09, 28 February 2018 (UTC)

Instead of speculating wildly on topics I don't even know about, I've been reading a book this month called Race Car Vehicle Dynamics (Milliken, 1995); you can even buy it on Amazon. It has a whole chapter on tire dynamics, including proprietary data from Goodyear for tires at various pressure points.

If you like vehicle physics, it is something of a great read, albeit extremely technical.

My insight for this morning, after reviewing the tire physics chapter, is that aerodynamic drag on the tire is sufficiently large that it cannot be ignored in a discussion of net rolling resistance. Automotive engineers have models to estimate that - just like how aerospace engineers measure the induced drag on an airfoil!

Nimur (talk) 16:55, 28 February 2018 (UTC)

Tyrrell P34 Andy Dingley (talk) 11:59, 1 March 2018 (UTC)

Probably does not matter in some years when all vehicles drive on Audi RSQ-style ballon tires. --Kharon (talk) 22:16, 1 March 2018 (UTC)

Bison fur pattern

Why American bison has only that partial fur covering his upper parts, neck and head instead of the entire body? Perhaps some other animals have the same. 212.180.235.46 (talk) 18:34, 27 February 2018 (UTC)

Compared with the European bison, for example? And note that they both have plenty of fur. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:13, 27 February 2018 (UTC)

I don't know about the southern states, but in the northern states and in Canada, the fur over the shoulder hump grows long and thick as it gets colder. In the winter, the fur is very thick and helps them handle the cold (as cold as -40°). When it warms, the fur molts in large clumps and the difference is mainly in texture during the summer rather than length. 209.149.113.5 (talk) 20:46, 27 February 2018 (UTC)

This is the science desk. Please be more precise, since we can't tell whether you meant -40 F or C. μηδείς (talk) 19:58, 2 March 2018 (UTC)

Compared with more or less uniform fur distribution in other animals. Also, it it the same mechanism involved in uneven hair distribution in humans compared to apes who have more uniform fur? Thanks. 212.180.235.46 (talk) 21:23, 27 February 2018 (UTC)

It is hard to say if their hair has changed from behavior or if their behavior is based on their hair. Bison are known for facing into cold winds. Growing up on a reservation in Montana, we had a saying that you should stand and face harsh conditions the way the bison stand and face the cold bitter winter winds. They don't need thick hair on their rears because they take the wind to their face. 156.143.240.137 (talk) 23:44, 27 February 2018 (UTC)

• Horses, lions, and humans all have manes, see also sexual selection. μηδείς (talk) 03:17, 28 February 2018 (UTC)
  • As do some other cats, such as Maine Coons. ←Baseball Bugs ^{What's up, Doc?} carrots→ 10:39, 28 February 2018 (UTC)
    • I apologize for having given an exclusive-seeming list. For goshsakes, see sexual dimorphy at bird. μηδείς (talk) 04:56, 2 March 2018 (UTC)

February 28

Chemical purification, part 2

Of the many chemicals used for the liquid-liquid extraction of iron from phosphoric acid, which ones have the highest flash points? 2601:646:8E01:7E0B:0:0:0:64DA (talk) 05:07, 28 February 2018 (UTC)

If you have a list of chemicals, this sort of information can usually be found on the SDS, on the packages themselves, or our wikipedia article for each. But we can't read your mind to know what specific chemicals you mean or even what ref you are reading. DMacks (talk) 17:16, 28 February 2018 (UTC)

From what I have read on liquid-liquid extractions, the common liquid used is kerosene, with some other chellating substance dissolved in it. Usually this is optimised to be effective and cheap. If you do your own research you may find high flash point non-polar solvents that may work eg perfluorocarbons or chlorocarbons, but they are probably undesirable for other reasons. Graeme Bartlett (talk) 23:15, 28 February 2018 (UTC)

Reference: https://patents.google.com/patent/US3694153 look for kerosene. Graeme Bartlett (talk) 23:27, 28 February 2018 (UTC)

Interesting! I had another process in mind, where the acid is first mixed with table salt to complex the iron, and then the iron chloride complex is extracted (I found it in an old book in the university library, I don't remember its name but I can give it to you tomorrow), but this MIGHT be even better! Will it work on 85% acid, or will I have to dilute it first? 2601:646:8E01:7E0B:0:0:0:64DA (talk) 08:05, 1 March 2018 (UTC)

If you read the patent, it says that the acid become too viscous if it is too concentrated. Graeme Bartlett (talk) 12:43, 1 March 2018 (UTC)

All right, that's not a big deal -- I'll have to dilute the acid before use anyway, so I can just purify it after dilution. And kerosene is not too flammable for me, so I guess that's what I'll be using as my solvent (although, if diesel fuel would work, I'll use that because it's more readily available). 2601:646:8E01:7E0B:0:0:0:64DA (talk) 06:06, 2 March 2018 (UTC)

Chemical purification, part 3

Is it possible to remove iron from aluminum hydroxide and copper oxide to levels of 1 ppm or less by selectively dissolving the material to be purified (in, say, sodium hydroxide for the aluminum hydroxide or ammonium carbonate solution for the copper oxide), physically removing the undissolved iron oxide/hydroxide (e.g. by filtration), and then recovering the purified aluminum hydroxide or copper oxide from the solution? Or do other techniques have to be used for this level of purity? 2601:646:8E01:7E0B:0:0:0:64DA (talk) 05:16, 28 February 2018 (UTC)

For your ammonia solution see https://pubs.acs.org/doi/abs/10.1021/i260041a028 where the answer is no, some iron would probably dissolve in your solution. Graeme Bartlett (talk) 23:42, 28 February 2018 (UTC)

Thanks for the bad news! And for the mini-Bayer process? 2601:646:8E01:7E0B:0:0:0:64DA (talk) 08:06, 1 March 2018 (UTC)

I couldn't easily find any sources for that. But if you really oxidise your iron to the red mud stage it is really insoluble. But then it depends on how good your filtration is. Even the dust blowing over your solution could add a part-per-million of iron! But for your copper extraction there appears to be more complex solutions and processes that will get the purity. Graeme Bartlett (talk) 12:21, 1 March 2018 (UTC)

OK here it says it will contain a few hundredths of one percent of iron, ie a few 100 parts per million in the alumina at the precipitation stage. But this page http://sourcedb.ipe.cas.cn/zw/lwlb/200909/P020090909606185656884.pdf says that you can precipitate more iron by adding methanol. So there will be ways around the problem. Graeme Bartlett (talk) 12:41, 1 March 2018 (UTC)

Whoa, methanol, I'd rather not go there -- I'll be working near gas appliances, using this stuff would be an accident waiting to happen! 2601:646:8E01:7E0B:0:0:0:64DA (talk) 06:08, 2 March 2018 (UTC)

... but your kerosene has a much lower autoignition temperature (though I agree that flashpoint of methanol could be a problem). You could always precipitate the iron in a safe environment, then evaporate the methanol safely. Dbfirs 11:48, 2 March 2018 (UTC)

...except there is no other place for me to do the purifications -- the only other available area is off-limits to any work with hazardous chemicals because there's food being prepared there. And autoignition temperature is mainly a problem if I actually spill the kerosene on the gas appliances (which most emphatically WILL NOT happen -- my work area is a fair distance away), whereas the flash point and/or fire point is a much bigger hazard -- the way the whole thing is set up, the main fire hazard is if the methanol vapors get sucked into the gas appliances and ignite! 2601:646:8E01:7E0B:59E2:B6:A6B8:FAC7 (talk) 09:41, 3 March 2018 (UTC)

Calculating the activation energy of a redox reaction

The reaction in question involves a galvanic cell setup. If I know the electrode potentials of the reactants, can I multiply that by the total charge flow (given by the area under a current-time graph) to get the activation energy? I think this because potential*charge = potential energy which I suspect is the energy needed for the reaction to proceed. The Average Wikipedian (talk) 15:10, 28 February 2018 (UTC)

This set of lecture slides describes how to calculate the activation energy of an electrochemical reaction quantitatively, and seems to directly answer your query. --Jayron32 15:14, 28 February 2018 (UTC)

Why is estradiol cypionate so much more potent than estradiol valerate?

AFAIK, estradiol cypionate is not active in itself and neither is the valerate ester: they're both prodrugs for estradiol. However, 5 mg of estradiol cypionate apparently is just as effective as 20 mg of estradiol valerate and estradiol cypionate is not manufactured at higher dosages. What is the reason behind this increased efficacy? Yanping Nora Soong (talk) 21:10, 28 February 2018 (UTC)

According to cypionic acid:

The lipophilicity of the cypionate group allows the prodrug to be sequestered in fat depots after intramuscular injection.[2] The ester group is slowly hydrolyzed by metabolic enzymes, releasing steady doses of the active ingredient.

So maybe it's not so much potency as time-release? --21:15, 28 February 2018 (UTC)

To elaborate a bit on the answer; this is a pharmacokinetics issue rather than a pharmacodynamics issue. The way that the cypionate is metabolized affects how the drug is released into the body; the drug itself has the same effects, but the body's action on the two molecules is different. --Jayron32 00:23, 1 March 2018 (UTC)

• This is the sort of thing one takes a full course in organic chemistry, plus a full course in psychopharmacology for. I am not sure the ref desk is certified to teach such courses, and I have had both, but at an Ivy-League university, and I had to repeat Organic because a mere one out of ten of my syntheses blew up, virtually guaranteeing a D grade, which would count as an F toward the Bio major. Enrole in a good STEM school, then get back with the stuck point. μηδείς (talk) 03:02, 1 March 2018 (UTC)

Actually our questioner sounds to be the same kind of student. And so probably has access to better experts on campus. Graeme Bartlett (talk) 12:15, 1 March 2018 (UTC)

Honestly, I think the graph in estradiol cypionate sums it up very well, and an expert would bring little to the table -- at least on this narrow issue of dosage -- other than by finding data of that type. Medical dosage recommendations are often quite arbitrary, really - consider the history of the birth control pill. [25] To call dosage empirical is too generous - empirical means that people trying it have found that X works; but medical testing means more like "somebody tried it once and found that X works and it will cost more than you can possibly spend to have the bureaucrats revisit the issue." Wnt (talk) 03:09, 2 March 2018 (UTC)

Yes, it is a good thing that the two above users have learned to imitute me exalctly. μηδείς (talk) 04:53, 2 March 2018 (UTC)

March 1

How "fine tuned" is the exponent of E=mc²?

No one knows if there's other Big Bangs with other exponents like E=mc^1.9 but if there are what's the anthropic principle range? Sagittarian Milky Way (talk) 04:07, 1 March 2018 (UTC)

The c² simply makes the units work out — see dimensional analysis. In natural units, the equation is just E=m.

Personally I can't make any sense of the idea of a possible world where E=mc^1.9, but maybe someone else can. --Trovatore (talk) 04:27, 1 March 2018 (UTC)

• It came as a shock to me when I finally realized (after the age of 60) that for equations describing the real world, dimensional analysis pretty much restricts exponents to being integers, except for exponents of dimensionless numbers. Furthermore, the exponent is also dimensionless. -Arch dude (talk) 06:34, 1 March 2018 (UTC)

  Per archdude, the exponent just means E = m * c * c , it's not a magical constant or anything, it's just a function of basic mathematical notation. --Jayron32 12:04, 1 March 2018 (UTC)

Exactly as Jayron wrote its all just simple math. Its not "fine tuned" but absolute and exact by definition because every part of the equation is exactly defined. Energy mass and speed are the same in your home, somewhere a million lightyears away in space and even close to a black hole tho our understanding of physics reportedly gets odd the closer you go in. Luckily you stayed away from that with your question. --Kharon (talk) 19:01, 1 March 2018 (UTC)

• Not quite. The speed of sound is square root of (gamma*R*T) in the correct units. I am sure there are other examples.Greglocock (talk) 23:15, 1 March 2018 (UTC)

$c_{\mathrm {ideal} }={\sqrt {\gamma \cdot R_{*}\cdot T}}$ at Speed of sound#Speed of sound in ideal gases and air is only possible because the multiplication under the square root produces units which can be expressed as squares, e.g. giving ${\sqrt {\tfrac {m^{2}}{s^{2}}}}$ = m/s. We could also write the relationship as $c_{\mathrm {ideal} }^{2}=\gamma \cdot R_{*}\cdot T$, just like E = mc² is equivalent to $c={\sqrt {\tfrac {E}{m}}}$. In practice we just tend to write formulas so the value we are most likely to want to compute is isolated. PrimeHunter (talk) 02:15, 2 March 2018 (UTC)

Agreed. In which case the statement at the top of this sub-thread is meaningless. If you exponentiate (?) both sides of the equation enough times you can get rid of square roots, always, and if you take the square root of both sides you can always get fractional powers. Greglocock (talk) 04:09, 2 March 2018 (UTC)

I'm confused by the claim that the exponent has to be exactly 2, at least as a question of physics. E=mc² comes from a certain mathematical model where the exponent is exactly 2, and the model apparently closely matches experimental observation. But maybe the model is only a close approximation. I don't see an inherent reason why the exponent couldn't be 1.9998 or whatever. Is it somehow different from the exponent 2 in the inverse-square law of gravity, which could conceivably be approximate?[26] I also remember something about Gauss trying to confirm the gravitational inverse-square law using astronomical observations.

Regarding Einstein's discovery of the exponent 2 in his famous equation: see here. 00:32, 2 March 2018 (UTC) — Preceding unsigned comment added by 173.228.123.121 (talk)

What exactly would it mean for the exponent to be 1.998? In (say) SI units, you would be saying that the energy in kg*m^2/s^2 would be equal to the mass in kg times c to the power 1.998, which gives you units on the right-hand side of kg*m^1.998/s^1.998. What on Earth is that supposed to even mean? --Trovatore (talk) 00:40, 2 March 2018 (UTC)

We might say that energy more fundamentally is a conserved quantity with certain physical characteristics, from which experimental observation leads to that formula you gave. But the formula itself could also be an approximation rather than an exact specification of what energy really is (i.e. the quantity in the formula might turn out to not have the exact properties we want energy to have in the physical universe). "Spacetime swimming"[27] is an example of how energy and momentum conservation are more subtle than they seem.

Semi-related: we're taught in school that F=ma, alternatively written F=m dv/dt, implicitly assuming m is constant. But I've heard that Newton's Principia actually says (translated to modern notation) something likeF=d(mv)/dt, i.e. Newton left open the possibility that m varies with velocity, as later predicted by relativity. I haven't checked out the claim though. 173.228.123.121 (talk) 01:22, 2 March 2018 (UTC)

You haven't responded to the request for clarification. What would it even mean for the exponent to be something other than 2? On the face of it, it's like you're asking whether the mass of the Sun could be Omaha, Nebraska. --Trovatore (talk) 01:27, 2 March 2018 (UTC)

Trovatore, maybe there's a philosophical disconnect. I'm positing that the physical universe has various symmetries, which by Noether's theorem lead to corresponding conservation laws. Those laws will hold even if there are no humans in the universe trying to guess at what they are. One of those symmetries might pertain to a quantity I'll call "X", which is conserved in physical systems like planetary orbits or elastic collisions. Now humans come along and make observations, and eventually develop a theory (i.e. a mathematical model) in which planetary motion conserves something they call "E" (for energy). That theorized E is very close to but not exactly equal to the quantity X that's conserved in the actual physical universe.

So what is the intended referent of the word "energy"? If energy is conceptually the thing specified in the human formula, then energy is by definition exactly E. But if it's conceptually the physically conserved quantity such that yada yada, we should say that energy is actually X, where E is just an approximation to X because the human mathematical model is slightly off: to be perfectly accurate the formulas we have for energy might have to be tweaked by a small correction factor, like changing 2 to 1.999998 and maybe a few other things.

I may be missing something but I don't see an obvious argument that the exponent should be have to be 2. Sure, the 2 is necessary to make the (human-theorized) units work, but the same can be said about the gravitation law $F=G{\frac {m_{1}m_{2}}{r^{2}}}\$. Yet we already know (see citation above) that the 2 in the gravitation law has not always been considered exact. Am I missing something here, that gives the 2 in the energy equation a more fundamental character than the 2 in the gravitation law? Maybe I am, but it doesn't seem obvious. I don't see a way to write the gravitation law with exponent 1.999998 while still having the units work, but physicists apparently still aren't certain that the exponent is exactly 2. So I'm left open to the possibility that we live in a universe in which the units don't quite work. Assuming that they work and concluding that the exponent is 2 is circular reasoning. 173.228.123.121 (talk) 02:52, 2 March 2018 (UTC)

That's completely different. Your r is a variable here. It's at least logically possible that you could have a force depending on r^2.02; you would fix up the units in the proportionality constant G.

The c in E=mc² is not a variable. It's a constant. There's no other constant around to fix up the units with. --Trovatore (talk) 03:11, 2 March 2018 (UTC)

Couldn't there be a multiverse with a universe where c is 101% this universe's c? All other things being the same antimatter should make a bigger boom there even though c is still 1 in real units. Sagittarian Milky Way (talk) 03:38, 2 March 2018 (UTC)

Maybe there is another universe that only has 1 dimension so there is no time and everything is stuck in a line with no chance to ever move because there is no time for that. You ask like if it is really a line or maybe a giant circle but there are no circles in 1 dimension because circles need 2 dimensions. So my offer is you draw me a 1 dimensional circle then i will show you that c is always and forever exactly c. Pinky Promise! --Kharon (talk) 04:11, 2 March 2018 (UTC)

A 1-D circle is a degenerate circle aka a point. Sagittarian Milky Way (talk) 04:31, 2 March 2018 (UTC)

Does not make sense to call a point a circle, nomatter how many dimensions. I could aswell call a question an answer and tag this section as solved. --Kharon (talk) 05:01, 2 March 2018 (UTC)

This is off-topic, of course, but actually a one-dimensional circle is an ordinary circle. Moving along a circle, you can go only forwards or backwards, same as on a line. A two-dimensional circle is a sphere, on which you can go east-west or north-south, same as on a plane. --Trovatore (talk) 20:23, 2 March 2018 (UTC)

So line and circle are just the next lower dimensional analogues of topological spheres and mugs? Sagittarian Milky Way (talk) 21:13, 2 March 2018 (UTC)

I'm not totally sure what you're asking. The circle is the one-dimensional analogue of both the sphere and the torus ("coffee mug"); you'll see it notated S¹ or T¹, depending on what the author wants to emphasize. --Trovatore (talk) 21:28, 2 March 2018 (UTC)

Oh I see that "number of holes" is far less fundamental than the S or T categories and is called genus (mathematics). Sagittarian Milky Way (talk) 01:21, 3 March 2018 (UTC)

(ec) Oh ok, I see what you mean: in relativity theory c is constant and m as the mass of a multi-particle system is additive between particles, etc. So sure, in the math of that theory, the exponent must be 2. Physically though, maybe we have something like E=mc^f(m) where f(m) is extremely close to 2 in the regimes we've been able to experimentally observe, but there's still some potential discrepancy. And relativity itself is the subject of many experimental verifications, some of whose purpose is to validate the hypothesis that c is constant, up to some amount of precision. I don't claim to be smart about this stuff, but I don't see a way to be sure that we'll eventually stop being able to find more wiggle room at the "non-crazy physics" level (i.e. the level where we are now, where maybe we're open to a r^2.02 gravitational force but can experimentally rule out r^2.05). At the purely mathematical level, of course, eliminating the wiggle room is impossible even in principle. 173.228.123.121 (talk) 04:01, 2 March 2018 (UTC)

Since in the right units the equation is just E=m, as already pointed out, what you're essentially asking is if it's possible that E is not exactly m. Or in other words, does E=km (or E=kmc²) where k is a proportionality constant that is not quite 1. That is, can the inertial and/or gravitational mass of a closed and isolated system change due to a mass-energy conversion? Our article on mass-energy equivalence describes a variety of experiments done to test this theory, and the best data cited on Wikipedia would bound k at 1±0.00004. Someguy1221 (talk) 04:14, 2 March 2018 (UTC)

(ec) No, you're still not following. It's not about c being constant or not in theory. People have proposed theories with a variable speed of light, though it's tricky to say what you actually mean (in SI, for example, the meter is defined in terms of the second and c, so it's actually logically impossible for c to be anything other than what it is).

That's not the point. The point is that c is a constant in the equation in question. The equation does not express E as a function of m and c. It expresses E as a function of m, period. It can be right or it can be wrong, but it can't be wrong in terms of the exponent of c. The c² is just a conversion constant, not a variable. --Trovatore (talk) 04:17, 2 March 2018 (UTC)

Trov, I interpreted the OP question as being about the natural universe, while the SI is just a way to organize the numbers in a particular mathematical model that aims to describe the universe but might not do so exactly. It could be that there's no way to make the math exactly describe the universe while still keeping the SI intact. So "changing the exponent is inconsistent with the SI" might only mean that the SI is unphysical. Sure, c is constant in the equation, but the question was about the universe and not about the equation. 173.228.123.121 (talk) 05:38, 2 March 2018 (UTC)

Well, there's certainly plenty we don't know about the universe, or that we think we know but we could be wrong. However, I don't see how you can express any of that by asking about how precisely we know the exponent in E=mc². If it's anything other than exactly 2, then the whole underlying theory makes no sense at all and you have to start over from scratch. It's not at all like asking about empirically measured quantities like the fine structure constant. --Trovatore (talk) 06:10, 2 March 2018 (UTC)

You know what, I expressed myself too softly in the struck-out part above. It simply doesn't make any sense to talk about the exponent being anything other than exactly 2. Forget changing the underlying theory. An exponent other than 2 simply does not state any claim. If the units are wrong, then it's like an ungrammatical utterance; it doesn't have any meaning whatsoever. --Trovatore (talk) 06:14, 2 March 2018 (UTC)

Well, I guess we're in dead horse territory and I'm still not totally convinced, but maybe I'm just clueless. Thanks and I'll stop posting about this. 173.228.123.121 (talk) 09:24, 2 March 2018 (UTC)

Here is a brief version: E is energy but mc^1.9 is not a measure of energy (or of anything meaningful) so it doesn't make sense to discuss whether it's equal to E, or could be equal in another universe.

It would be like discussing whether 5 kg can be equal to 3 s. A measure of mass cannot be compared to a measure of time. A formula like E = 0.9mc² could be discussed since 0.9mc² is a measure of energy, just 10% less energy than Einstein says. PrimeHunter (talk) 14:38, 2 March 2018 (UTC)

• I think that, while all of the responses here are correct, most of you are getting lost in the weeds in trying to explain the OP's misconceptions here, which are largely due to a misunderstanding of mathematics rather than of physics. You're all providing physics explanations for what is a fundementally mathematical mistake. Let me try this again: As with all equations, there are two different bits: there are values and there are mathematical operations. A value is something like energy or speed or mass. Some of these values may be constants, and some may be variables, but that's not really relevent to our discussion; the speed of light is a constant, but for the purpose of this discussion, it needn't be. All that matters is the equation has three actual values: energy, mass, and speed. What we do with those values is set up a relationship using mathematical operations, and the relationshop is this: energy is equal to the mass multplied by the speed of light multiplied by the speed of light again. For notation purposes, when we multiply something two times we right a little "2" as a superscript next to that value in the equation. But, and I can't stress this enough which is why I'm going to put it in bold, that little 2 is not a value, it's a mathematical operation that means multiply by this value twice. It's completely nonsensical to change the value to anything except a whole number here (there are times when non-integer exponents are meaningful and useful; this is not one of them). If we change the number to anything except 2, that's a completely different equation. If there is "fine tuning" to be done, it's with the "c" value. Changing that little "2" is not even wrong. --Jayron32 15:28, 2 March 2018 (UTC)

The problem is I never decomposed it all the way down to (kg·(m/(s·s)))·m = kg·m/s·m/s or realized it's basically a tautology before Trovatore mentioned dimensional analysis. I just thought of it as the joules to kg conversation rate of this universe (or ergs to gram). Maybe it's an integer in all universe(s) cause reality seems to like elegance but the most fundamental deep reasons for everything are unknown so fractional or irrational isn't ruled out right? But it is. Sagittarian Milky Way (talk) 21:13, 2 March 2018 (UTC)

Well it could change but so much of physics would have to change it would be a total and complete overhaul. The question is a bit like asking if speed went up as distance divided by time to the power of 0.9. I believe it would require factional dimensions so we lived in something like a 3.1 dimensional world for instance and what that would be like I haven't the foggiest idea. Dmcq (talk) 18:57, 2 March 2018 (UTC)

One issue brought up in the thread is that dimensions don't have to use integer powers of the units in the measuring system. E.g. if the inverse-square gravitational force turns out to really have exponent 2.02 rather than 2, then (per Trovatore) we'd have to redefine the gravitational constant G and it would fractional units. Or in the era before SI, the electrostatic force was F=q₁q₂/r² where the q's are charges (note the absence of a multiplicative constant like ¹/_4πε₀). Knowing that force=mass*distance/time², that meant the units of charge were mass^1/2*distance^3/2*time⁻¹. But the OP has now clarified they meant the question in a simpler way, so everything's cool. 173.228.123.121 (talk) 01:39, 3 March 2018 (UTC)

Cold air, colder water

Apologies if this has been asked before - I skimmed the archive but couldn't find it. Imagine it's a warmish summer day of about 20°C. You jump into a swimming pool with the water at about 27°C, and the first thing you do is feel cold. Why? Is it because your body warms a thin layer of air around itself like a blanket, which is stripped away when you enter the water, or is it simply that water (or liquid in general perhaps) is better able to take heat from a warm solid than air (or gas in general)? Thanks in advance, Grutness...wha? 13:51, 1 March 2018 (UTC)

Because water is about 3 orders of magnitude more dense than air, so it is better at conducting heat away from your body. It isn't just the difference in temperature, it is how fast the energy is transferred from one material to another that matters here. In order to lower the temperature of your body, molecules of something else have to bump into your body and have some of that energy transferred to them. There's about 1000 times more collisions with the water molecules than the air molecules. Of course, when you do finally equilbrate with the surroundings (well, when your corpse does) the final temperature of your body in 20 degree air is colder than in 27 degree water. However, your body will lose heat faster in the water, resulting in you feeling colder. Also, water has a very high heat capacity, meaning that it is very efficient at soaking up heat energy. Qualitatively, this is all pretty intuitive if you understand what heat energy is (it is just the kinetic energy of individual molecules), quantitatively this is very complex, however, and requires some understanding of Heat transfer, which requires far more calculus than I'm prepared to go into. --Jayron32 13:58, 1 March 2018 (UTC)

Some of each: which is why both wet-suits and dry-suits are used, for example. All the best: Rich Farmbrough, 14:02, 1 March 2018 (UTC).

Another data point here and a couple of references to back this up: water can feel cold, but obviously not always, for my family prefers swimming in the mid-Alantic ocean surf in the late summer and early fall when the water is warmer than the air, for after the initial sensation of coldness when entering the water by diving in our bodies quickly acclimate to the water to the point that it is like taking a warm bath because the water is warm (lukewarm starts at about 27°C and can feel cold and the water temps we swim in are between 27°C and 28°C) and we towel off in a hurry because the evaporation makes us colder anytime the air is cooler than room temp. That the water feels between cold or lukewarm but can feel warmer over time (if it's not too cold to start with) is because the body adapts to the different temperatures, see table 3 of this extensive reference. For instance, at around 28°Celsius the skin vasodilation increases. On page 17 it states "When water temperatures rise between 30°C to 34°C, there is intense vasodilatation[sic], strong blood flow, an increase in the peripheral and core temperatures, and a decrease in [athletic] performance. These temperatures are appropriate foe[sic] exercise activities for infants and children, as well as for the elderly (65, 55). When temperatures drop between 25°C to 20°C, there is pronounced vasoconstriction, an increase in oxygen consumption, a decrease in peripheral and core temperatures, and a drop in maximum performance. These temperatures are appropriate for endurance swimmers (12, 57, 45)". As you can see, YMMV (your mileage may vary) because there are different responses to different water temperatures for different individuals. -Modocc (talk) 23:54, 1 March 2018 (UTC)

Thanks all. @Modocc:, I take it that also explains why your pulse rises when you get into a hot bath? I've noticed that my pulse is often around 120 when I'm taking a bath, compared with my normal resting rate in the high 60s. Grutness...wha? 04:00, 2 March 2018 (UTC)

Er, I am not free to speculate as to why your pulse increases as it does, per the reference desk guidelines at the top of the page which state "We don't answer (and may remove) questions that require medical diagnosis or legal advice." Thus I cannot answer your question regarding your experience and medical condition. In addition, I have not read the entire source that I provided above nor its references nor do I have any additional references which adequately address the actual quantitative magnitude of the effects of hot water on individuals so as to know or understand what is considered a normal change or not. However, I did find a vague description of the general effects of hot water on people from Cleveland Clinic here. My very limited understanding from this one source in particular is that vasodilation of the peripheral blood vessels moves blood out of the body's core which lowers the blood pressure causing the heart to pump faster and harder to keep it near normal. -Modocc (talk) 14:14, 2 March 2018 (UTC)

:) I wasn't asking for a diagnosis! I've just looked it up elsewhere online and yes, it seems to be the reason and is perfectly normal and - as I suspected - is caused by the same phenomenon. Grutness...wha? 01:15, 3 March 2018 (UTC)

What do you call...

...those cards that you use in photographs to give an index to colour and scale.

All the best: Rich Farmbrough, 13:58, 1 March 2018 (UTC).

Like this: [28]? I can't find any standardized name other than "calibration chart" or "calibration card". --Jayron32 14:25, 1 March 2018 (UTC)

I suspect you want the Munsell color system, but it may be another color model. Wnt (talk) 15:20, 1 March 2018 (UTC)

• Photographic scale[29][30] (archaeological) or ABFO scale (criminal) [31]
• Kodak gray card

Andy Dingley (talk) 15:59, 1 March 2018 (UTC)

Thanks both, looks like "color scale card" is close to what I was thinking of, but the one that fits in a wallet seems ideal. I also found these nice 1cm cubes. [32] All the best: Rich Farmbrough, 16:57, 1 March 2018 (UTC).

It is also common to refer to the color calibration chart by the vendor name - as there is at least some merit to the idea that a vendor's proprietary quality-control methodology makes their chart different from any other particular "standard" color chart. After all, these are meant to be used for calibration - one hopes that every aspect of the chart is calibrated - from the ink and printing process all the way down to the paper's optical and surface properties! One hopes that the optical properties of the calibration chart is more accurate and precise than the optical equipment that is photographing or measuring the chart!

So, for example, you can buy a Macbeth chart, or a Munsell chart, or a Pantone chart, or a Kodak chart; these are among the most commonly-used (and commonly-copied) proprietary form of the color chart. As it turns out, economic forces are at play, and as of 2018, almost all of these proprietary calibration methodologies are now owned by one conglomerate, X-Rite Pantone, an American optical instrument company. They sell, for example, the ColorChecker Classic, which you may know as the GretagMacbeth chart (named for the original conglomerate who bought the conglomerate who bought the rights to the chart and its methodology...)

Personally, I'm a fan of the more interesting angular resolution charts, which are not proprietary; but beware! There's no value in a poorly-manufactured optical calibration knock-off! (Tip: whether you're calibrating color, intensity, or resolution, don't simply download a JPEG and print it ... calibration of a calibration system is... a bit involved)!

Nimur (talk) 17:30, 1 March 2018 (UTC)

Coloured cubes are pretty handy, but mostly for setting up lighting, rather than the camera itself. It can be a nuisance for some jobs if your lights from different directions have different colour temperatures (typically artificial fill-in under directional natural light).

Resolution test targets, like the USAF one, are a different thing.[33] They're easy enough to print yourself, if you want a large one to test some telescopic system - it's microscopic one which test the limits of your production system. I did one myself nearly 30 years ago, as an exercise in learning PostScript coding.

The USAF one is a bit limited though. It's no use for demonstrating astigmatism, or basic setup of an optical system. For that you're better with something like the IEEE target [34] or just something with rosette targets on it [35] Andy Dingley (talk) 20:02, 1 March 2018 (UTC)

Very true! There are proverbial truck-loads of special-purpose calibration charts for specific use-cases, lenses, adaptive-optics systems, and so on! As I frequently do, I refer the interested reader to Applied Photographic Optics ... if you actually want to get into the meat of real, modern lens design! Nimur (talk) 23:21, 1 March 2018 (UTC)

I was kind of hoping Lenna would lead somewhere, but the closest I get is Standard test image, neither of which are really what's needed. Matt Deres (talk) 17:46, 1 March 2018 (UTC)

@Rich Farmbrough. You mentioned cubes. At first, I was a little disappointed when I revived my Spydercube. It was smaller than the illustration suggested. I thought I had wasted my money on a bit of plastic. In use though, for ' when one just wants to take a quick photograph ' and perhaps correct it after in Photoshop or GIMP. It provides a good and quick datum. The little silver ball on top, even indicates any highlight that may burn out. Take it that you know how to use your camera's gammer display in order to expose to the right. This cube makes it easier, without having to think too much about what one is doing. --Aspro (talk) 21:25, 1 March 2018 (UTC)

March 2

Chemical grades

What's the difference between "lab grade" and "reagent grade" cupric oxide in terms of impurities (and especially iron content)? Also, for ACS grade phosphoric acid, what's the maximum allowable iron content? 2601:646:8E01:7E0B:0:0:0:64DA (talk) 06:10, 2 March 2018 (UTC)

Reagent grade will mean it meets the standard of "Committee on Analytical Reagents of the American Chemical Society". Perhaps 95% pure. Lab grade is not so good, and you had better check the manufacturer specs. ACS grade is likely the same as reagent grade. Graeme Bartlett (talk) 10:06, 2 March 2018 (UTC)

ACS grade phosphoric acid allows up to 30 ppm of iron in phosphoric acid. So if you want 1 part per million this is not the standard you need. The reference is ISBN 9780841230460. ACS grade cupric oxide can contain up to 500 ppm of iron. Graeme Bartlett (talk) 11:22, 2 March 2018 (UTC)

These phphoric acid products have less than 10 ppm iron[36][37]. This one is extremely pure: [38], but you will pay heavily. Graeme Bartlett (talk) 11:40, 2 March 2018 (UTC)

500 ppm of iron in the copper oxide?! Mother of God! Are there any higher grades of the stuff??? (Incidentally, the acid is less of a problem, because I now know a few tricks for purifying it.) Also, how about the standards for copper carbonate and copper hydroxide (my Plan B if I can't get copper oxide of decent quality)? 2601:646:8E01:7E0B:59E2:B6:A6B8:FAC7 (talk) 09:43, 3 March 2018 (UTC)

These two do not have a reagent standard from ACS. The list of what there are standards for is available here: https://pubs.acs.org/isbn/9780841230460 Graeme Bartlett (talk) 11:02, 3 March 2018 (UTC)

Is there a health advantage in skimming cooked watery food?

I'm not sure how common it is in nowadays world widely but I'm sure that it's most common in Arabic countries to skim foods while cooking, it says to remove the "scum" from the surface of these dishes while cooking. (That is the definition that I found for "skim" on the internet: "To remove scum, fat or other impurities from the surface of a liquid, such as stock or jam, while it is cooking."). It's mainly common in meat dishes but not only, it's commonly used also also for fabaceae. Now my question is if there is/ are health advantage/s / beneficials for this act? The rumors claim that this "scum" of the surface is not healthy to the body. Furthermore I read in the past that there are in the "scum" some proteins that the body doesn't like 93.126.116.89 (talk) 18:04, 2 March 2018 (UTC)

Fabacaea is a fancy way of saying “legumes.”Edison (talk) 18:19, 2 March 2018 (UTC)

Legumes is a fancy way of saying "beans". --Jayron32 18:44, 2 March 2018 (UTC)

"Beans" is a fancy way of saying testicles? What is the source of this "Arabic countries" claim? My dad is not Arabic, and he boils a lot of his meats and discards the fat. It makes his chili taste like dry dog food (yes, I have eaten dog food, although it's been decades, and was for fun), and you have to add fat (olive oil or sour cream) to make it edible. μηδείς (talk) 19:27, 2 March 2018 (UTC)

The source for "Arabic countries" is likely the OP. I.E. they are going by their own personal experience and knowledge. It's true that this isn't a good source, but it's also largely irrelevant. The only reason they appear to have brought this up is to explain the background as to why they have this question. Even if it isn't as common in Arabic countries as suggested, the question itself is still valid. Nil Einne (talk) 13:37, 3 March 2018 (UTC)

This seems to be an WP:ENGVAR issue. A peanut is a legume. I would not call it a bean. In fact, I would not even call many peas beans. Our Bean article also talks about the complexity of defining what a bean is. Nil Einne (talk) 13:43, 3 March 2018 (UTC)

I think this comes up in western cooking frequently when talking about making stock. Here, the consensus is that skimming the foam off simmering stock is for reasons of aesthetics and taste rather than health. - Nunh-huh 19:37, 2 March 2018 (UTC)

Indeed I'm not a native English speaker and by fabacaea I meant to any kind of legumes (such as beans, peas, chick-pea, lentils etc.). I'm afraid that my question regarding to the health advantages of skimming will become a conversion about the definition of fabacaeae. If it complicates my question, then you can remove the part which talks about fabaceae and refer the rest:) Thank you! 93.126.116.89 (talk) 19:41, 2 March 2018 (UTC)

Burn Rates

I'm trying to find a table of average Burn rates for materials which might or might not be considered explosive. Burn rates should be specified by comparative numbers, not ordinals; for example, meters/second (or fps). I would like common substances, such as:

Substance	Burn Rate (m/s)
Cigarette embers	0.5 - 3.0
Kitchen match	40 - 110
Campfire	200 - 300
Fireworks	450 - 1500
Black powder	1100 - 2000
TNT	7500 - 9000

All the burn rates in the above table are made up; that's what I need.

I don't need a table. A long discussion (or many articles) of various materials that has the numeric burn rates embedded in the paragraph(s) will do just fine. And I know that burn rates are subject to discussion. What I want is an average range that covers 75 to 90 percent of the possible values.

Anything you can provide me will be greatly appreciated.

--RoyGoldsmith (talk) 21:08, 2 March 2018 (UTC)

No idea about low-velocity stuff like cigarettes but this stuff is measured carefully for rocket propellants (see articles related to specific impulse). I click around a little and couldn't find any tabulation, which surprised me. But fancy rockets use difficult-to-handle cryogenic liquid hydrogen and LOX because of its high velocity (resulting from low molecular weight). People even tried to use the even nastier hydrogen fluoride but that luckily was considered unmanageable. 173.228.123.121 (talk) 01:50, 3 March 2018 (UTC)

Look for a publication of the Lawrence Livermore National Laboratory Properties of Chemical Explosives[39] which has detonation velocity for many compounds. They use mm/μsecond (which is km/sec) Graeme Bartlett (talk) 04:33, 3 March 2018 (UTC)

Considering how long a cigarette lasts, a better estimate would be 0.5 m/h 78.0.254.206 (talk) 05:43, 3 March 2018 (UTC)

Note that the OP is asking about cigarette embers, not cigarettes. Nil Einne (talk) 15:13, 3 March 2018 (UTC)

Note that the OP explicitly states that the numbers are made up. A campfire that burned at 300 m/s would be quite a sight. μηδείς (talk) 19:15, 3 March 2018 (UTC)

Neanderthal, rats

Did Neanderthals have rats or mice as pests?144.35.114.28 (talk) 21:36, 2 March 2018 (UTC)

• Word is that house mice Mus musculus arose in the Middle East at the dawn of agriculture, long after the Neanderthals were gone. Rats are older, but one wonders how well they would do as commensals with hunting-gathering ice age Neanderthals. Probably not in any kind of permanent association. As an example, the Canadian province of Alberta is rat-free. Heaviside glow (talk) 22:46, 2 March 2018 (UTC)

They can survive Norway cold but not Alberta? Sagittarian Milky Way (talk) 01:29, 3 March 2018 (UTC)

Not for lack of trying... [40] 78.0.254.206 (talk) 05:41, 3 March 2018 (UTC)

There are still parts of the world where a nice fat rat is prey, rather than pest. Animals only become pests when they are competing with humans for a common food source - which wasn't a significant issue before the start of agriculture. Wymspen (talk) 23:03, 2 March 2018 (UTC)

Homo neanderthalensis lived in Germany which was completely covered by Forest in that time and which is still home to predators like Owls, Weasels, Foxes and Lynxes which all hunt small Rodents. That is probably why Rodents propagation is so high. Actually they became a "pest" because their predators where thinned out as homo sapiens rose to higher culture, hunted excessively for their fur to make luxury clothing. Homo neanderthalensis most likely had little interest in all these small predators and so the rodent population was likely much smaller in the wild compared to their numbers in medieval and todays cities. --Kharon (talk) 00:41, 3 March 2018 (UTC)

The wood mouse is native to European temperate woodland and will enter buildings to find food given the chance. Not sure how long they've been around though. Alansplodge (talk) 12:07, 3 March 2018 (UTC)

March 3

CMB is remained from past age of universe or recent

  I have  three difficulties  in cosmology :
    1.in the case so we really judge about the world , then what means just now in our universe ?
    2. rotation mode of galaxy members , why we believe both orbital rotation of galaxy members and stability of galaxy shape and arms?
    3.  Either it is so that CMB is recent thermal Image of universe or it refers  very past time when world age was 380000 years?--Akbarmohammadzade (talk) 09:43, 3 March 2018 (UTC)

1. We count time in cosmology using our reference frame as determined on earth. So that means that we are not moving at close to the speed of light compared to the rest of the universe. Barycentric Coordinate Time can be used, which cuts out gravitational time dilation of the Earth and Sun. In the past cosmology still described the known universe. So centuries ago, Earth and planets made up a much bigger component. If you see time as "Before Present", it means before 1 January 1950, so you could say that "now" is that date.

2.Our article mentioning spiral arm fails to say much at all. See also Scutum–Centaurus Arm and Perseus Arm and Orion Arm. But these do not mention how they were discovered or anything about long. term stability. The arms are probably due to density waves in the gas, resulting in star formation, with bright new big stars marking the arms. Graeme Bartlett (talk) 11:20, 3 March 2018 (UTC)

3.Cosmic microwave background is believed to be ancient light. Graeme Bartlett (talk) 11:20, 3 March 2018 (UTC)

Eyelid anatomy

What is the name for the small flesh colored bump that can be seen on the lateral commissure when you pull the eye lid away from the eye horizontally? It can be seen touching both the upper and loser eyelids. This http://slideplayer.com/8275868/25/images/3/%28a%29+Surface+anatomy+of+the+right+eye.jpg is the most detailed chart I could find. --User777123 (talk) 18:19, 3 March 2018 (UTC)