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April 11

Longest rotational period

Which object that orbits the Sun and does not orbit any other object orbiting the Sun, has the longest rotational period (prograde or retrograde, doesn't matter)? --Atethnekos (Discussion, Contributions) 04:49, 11 April 2014 (UTC)[reply]

Longer orbital periods are associated with greater distances from the Sun (see Kepler's second law), so your question is basically asking about the objects that are the furthest from the Sun while still within the Sun's gravitational potential well. I'm not sure which one specific object would be the answer to your question, but I think it'd be something within the Oort cloud. There are long-period comets (which are thought to originate from the Oort cloud) which have orbital periods in the thousands of years. For example, Comet McNaught has an orbital period of about 92,600 years. However, your criterion of an "object that orbits the Sun" is a bit fuzzy, because when you're talking about the objects with the longest orbital periods, that are the most weakly gravitationally bound to the Sun, there's fuzziness as to how long the object has to remain in the vicinity of the Sun before it gets lost as an interstellar comet for it to still fit your definition of an "object that orbits the Sun". Red Act (talk) 05:38, 11 April 2014 (UTC)[reply]

The OP asked about rotational period, not orbital period. --140.180.250.141 (talk) 05:57, 11 April 2014 (UTC)[reply]

Yes, I seem to have misread the question, so I have retracted my answer. Red Act (talk) 06:17, 11 April 2014 (UTC)[reply]

One could say it was redacted! Har! Dismas|^(talk) 15:36, 11 April 2014 (UTC)[reply]

Our article on Rotation period includes these figures for major bodies in the Solar System. Extreme rotation periods for some smaller bodies can be found here. Of the bodies known to Wikipedia, it looks like Venus is your answer. RomanSpa (talk) 06:56, 11 April 2014 (UTC)[reply]

Since you didn't specify a size, there are an almost infinite number of objects orbiting the Sun directly, when one includes the asteroid belt, Kuiper Belt, and Oort Cloud and allows for sizes down to a pebble. Of all those, I'd expect some have essentially no rotation at all. StuRat (talk) 13:03, 11 April 2014 (UTC)[reply]

It's infinite, there are neutral hydrogen atoms that orbit the Sun and they can have a total angular momentum of zero (electron in the 1s state and the electron-proton system in the total spin zero singlet state). Count Iblis (talk) 23:18, 11 April 2014 (UTC)[reply]

Probably Mercury (planet) is the correct homework answer as it is tide locked with the sun and Mercurians only get one "day" every 2 "years". (but I didn't check venus) --DHeyward (talk) 04:40, 12 April 2014 (UTC)[reply]

Please do so. —Tamfang (talk) 05:41, 12 April 2014 (UTC)[reply]

It's conflicting as Venus rotates clockwise. From the article Venus has a longer sidereal day and a shorter solar day than mercury. --DHeyward (talk) 15:31, 12 April 2014 (UTC)[reply]

Since the OP used the phrase "rotational period" rather than "solar day", I'm going with Venus, but the OP can make up its own mind. —Tamfang (talk) 05:53, 13 April 2014 (UTC)[reply]

Gas pipeline

How heavy must a piece of debris be (assuming it's travelling at the speed of a typical avalanche) to rupture a major aboveground gas pipeline? For example, would a good-sized boulder do the job? How about a snowmobile? Or a typical four-door sedan? Or a 3-ton truck? Thanks in advance! 24.5.122.13 (talk) 06:57, 11 April 2014 (UTC)[reply]

I guess we've won the War on Terror if Al Qaida is reduced to this. — Preceding unsigned comment added by 96.227.210.243 (talk) 12:47, 11 April 2014 (UTC)[reply]

HAHahaHAHahaHAHahaHAHahaHAHahaHAHaha μηδείς (talk) 22:04, 12 April 2014 (UTC)[reply]

Not a question about Al-Qaida, but a research question for a disaster novel. 24.5.122.13 (talk) 23:11, 11 April 2014 (UTC)[reply]

The shape and way it hits would make a huge difference. Imagine a block of masonry hitting it. If it hit corner first, it might well poke a hole, while a glancing blow on the side of the block would not. The difference is the pressure exerted at a given point. However, enough total force, even if evenly distributed, could still rip the pipeline off it's supports and rupture it. StuRat (talk) 12:53, 11 April 2014 (UTC)[reply]

So, a boulder could do it if it was big enough and/or it hit the "right" way, right? This is just what I hoped to hear -- the rupture of a gas pipeline (which is promptly ignited by a downed 39,000 volt power distribution line, incinerating an automobile with its occupants and threatening to do so to another in which a mother and her daughter are trapped and badly hurt) is an important plot element, and I'd be VERY disappointed if it turned out to be implausible. 24.5.122.13 (talk) 23:11, 11 April 2014 (UTC)[reply]

Possible, sure. Make sure a particularly sharp protrusion from the boulder hits the pipeline first. Also, you'd need to explain why they would build a pipeline where there's an obvious avalanche danger. My suggestion, explain it with acoustic lubrication. This is a situation where the sound/vibrations of the landslide hit just the right frequency, providing a type of lubrication that allows the landslide to go much further than predicted. When you see film of such a landslide, it looks like rushing water, due to the greatly reduced friction. To explain it to the readers, the sound frequency makes the boulders vibrate, so they are only in contact with the ground and each other part of the time, significantly lowering the average friction versus if they were inconstant contact. The larger the objects, the lower frequency is needed for this type of resonance, so boulders require quite a low frequency sound. StuRat (talk) 23:30, 11 April 2014 (UTC)[reply]

In the Canadian Rockies, there's often no choice but to build stuff despite a known avalanche hazard. A more important question is why that part of the pipeline wasn't built underground -- perhaps it would be to save on construction costs, or maybe it was supposed to be built underground but they decide to turn on the gas first and only then cover the trench because the project is running badly behind schedule? 24.5.122.13 (talk) 23:45, 11 April 2014 (UTC)[reply]

It would be nice to actually cite references when stating that something is "possible, sure". Here's a South African report that talks (on page 5) about how impacts are a common cause of pipeline ruptures, although it's talking about underground pipelines. Also note the following sections where it talks about fireballs and such, which may be useful for the novel. Here are US NTSB reports on pipeline incidents, but the table does not classify them by cause, so you might have to read a number of them to see to find ones that are relevant to your scenario. --50.100.193.30 (talk) 00:53, 12 April 2014 (UTC)[reply]

Makes it so the avalanche derails a train into the pipeline. The build those things together :). --DHeyward (talk) 03:24, 13 April 2014 (UTC)[reply]

You can't have a train without a railroad line (the disaster scene is NOT on the Canadian Pacific or Canadian National) -- but I could try making the avalanche push a truck into the pipeline (which should do the job just as well), and then have the resulting fireball and jet fire incinerate the truck. In fact, this would be PERFECT for my story! 24.5.122.13 (talk) 04:29, 13 April 2014 (UTC)[reply]

Bug ID

Should I be worried that this dropped down onto my desk? -- Zanimum (talk) 12:10, 11 April 2014 (UTC)[reply]

Wood louse (Isopoda) suborder, pill bug (Armadillidiidae) family. Very common here. I've never known them to be harmful. In fact, they are kind of fun. Poke them on the back and they will roll up into a little ball like an armadillo, to defend themselves. You can then play marbles with them, until they get tired of it and walk away. :-) StuRat (talk) 12:30, 11 April 2014 (UTC)[reply]

A solitary one is not a problem, but; "When large numbers of woodlice are found indoors, perhaps clustered in wall crevices or under skirting boards etc., it is always worth checking for excessive dampness in these places - just in case there is a structural problem with the damp proofing or damp course".[1] Alansplodge (talk) 13:03, 11 April 2014 (UTC)[reply]

Right, but they still aren't a problem (other than a cosmetic one), but rather they just indicate that a problem exists. StuRat (talk) 13:56, 11 April 2014 (UTC)[reply]

Well, the question was "should I be worried". Referenced answer: "Maybe". Alansplodge (talk) 22:43, 11 April 2014 (UTC)[reply]

Please do not worry unduly because it looks like Armadillidium vulgare has survived the fall. Let's not be negative about his lifestyle. 84.209.89.214 (talk) 12:44, 11 April 2014 (UTC)[reply]

'Round these parts, we call those roly polys. Justin15w (talk) 14:38, 11 April 2014 (UTC)[reply]

I had an interesting experience with one once. I wasn't in the mood to play with it when it crawled into my house, so I flushed it. At the moment it disappeared down the toilet, everything went black and silent, so I thought "Damn, that pill bug must have been God, and now the universe has ended" ... at least until the power came back on a minute later. :-) StuRat (talk) 12:48, 11 April 2014 (UTC)[reply]

:) -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]

In Kent, according to a former colleague, pill woodlice are known colloquially as "monkey peas",[2] presumably because monkeys are supposed eat them like peas? The origin of this seems to be completely unknown - there are no wild monkeys in the south east of England. Alansplodge (talk) 12:56, 11 April 2014 (UTC)[reply]

:) -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]

In Norse areas of northern England they were known as "thuslice" /θʊslaɪs/ (originally Thor's lice). Dbfirs 21:16, 11 April 2014 (UTC)[reply]

Thanks all! I've seen pillbugs a lot outdoors, never in. I work in a 148-year-old building, so we frequently have surprises. -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]

Simple beam calculations

I have a beam with a simple support at one end and a roller support at the other end. The beam has a 30kN load 1m from the left support, a 50kN load 2m from the left and a uniform load 3m from the left, which spans 2m. The overall span of the beam is 6m. I've simplified this to a simply supported beam, which I'm assuming is correct. In order to find the reactions at either end I've done the following. R1+R2=w1+w2+(w3*length of uniform load/2), and for the moments R2*L=w1*x1+w2*x2+w*length of uniform load*distance to centre of uniform load from left of beam. I've basically done this by simplifying the uniform load to a concentrated load acting at the centre of the length of the uniform load. But apparently this is incorrect as my instructor has used W*l instead of w*l/2 to find the contribution of the uniform load to shear force equilibrium, and wl^2/2 instead of wl^2/8 for uniform load contribution to moment equilibrium, which in my opinion treats the uniform load section as a cantilever. Where have I gone wrong in my calculations? Why is that section a cantilever? Clover345 (talk) 13:32, 11 April 2014 (UTC)[reply]

We need a diagram. Is this correct (not quite drawn to scale) ?

                          UNIFORM LOAD
         30kN       50kN /            \
          |          |  /              \
          |          | /                \
<-- 1m -->V<-- 1m -->V/<------ 2m ------>\<------ 2m ------> 
============================================================
o                                                          ^

I assumed the uniform load is centered 3m from the left. Also, is the magnitude of the uniform load specified ? StuRat (talk) 13:51, 11 April 2014 (UTC)[reply]

The uniform load is centred 4m from the left and it's value is 10kN/m. Also the simple support is at the left and the roller support is at the right. Clover345 (talk) 14:46, 11 April 2014 (UTC)[reply]

Like this ?

                                    UNIFORM LOAD
         30kN       50kN           /   10kN/m   \
          |          |            /              \
<-- 1m -->|<-- 1m -->|<-- 1m --> /<----- 2m ----->\ <-- 1m --> 
          V          V          vvvvvvvvvvvvvvvvvvvv
==============================================================
^                                                            o

StuRat (talk) 16:31, 11 April 2014 (UTC)[reply]

Ye, that's it exactly. That is a simply supported beam right? According to my instructors calculations that's a cantilever but I can't understand why. Clover345 (talk) 16:48, 11 April 2014 (UTC)[reply]

Unfortunately, it's been too many years since I've done such a problem to answer reliably, but now that the Q is clear and we have a diagram, hopefully others can answer. (Your approach of dealing with it as two simple supports seems right to me, too, so either I'm missing something or the instructor made a mistake.) StuRat (talk) 17:08, 11 April 2014 (UTC)[reply]

If the uniform load is specified as 10kN/m then its (not spelled "it's") total force is 10k/m times 2m. This is what the instructor means by W*I and the OP has no reason to divide by 2 in the expression for R1+R2 which comes out to 100 kN. The sum of the clockwise moments about the left end is 30 + 50*2 + 10*2*4 kN m = 210 kN m which is in equilibrium with 35 kN force on the support at the right end. That leaves 65 kN force at the left end. The beam is not a cantilever unless it extends (not shown) beyond the right hand support. It's commendable that contributor StuRat shows correct spelling here. 84.209.89.214 (talk) 19:38, 11 April 2014 (UTC)[reply]

Thanks but why is this? I thought that in a simple beam with a uniform load, the moment is given by wx/2*(1-x), whilst the shear force is given by w(l/2-x). Clover345 (talk) 20:07, 11 April 2014 (UTC)[reply]

This may help. 84.209.89.214 (talk) 21:42, 11 April 2014 (UTC)[reply]

Weather/travel

I have heard there is a possibility of a storm on the east coast early next week what is the chance that incoming flights into airports in the northeast will be affected — Preceding unsigned comment added by 75.69.246.118 (talk) 14:55, 11 April 2014 (UTC)[reply]

OP geolocates to Massachusetts, so that's probably the coast he's interested in. Rojomoke (talk) 16:24, 11 April 2014 (UTC)[reply]

Looking up the weather for Boston, I get rain and winds up to 25 mph on Tuesday, which doesn't seem like enough to cancel flights, to me. If you can be more specific as to the exact arrival date, time and location, we can give a better answer. StuRat (talk) 16:27, 11 April 2014 (UTC)[reply]

The OP should call his airline. We don't have some secret knowledge they don't have. μηδείς (talk) 20:11, 11 April 2014 (UTC)[reply]

The National Weather Service's area forecast for the Boston region, for April 11, 1745 UTC, includes notification for an AIRMET (AIRMET Sierra) with IFR conditions, mountain obscuration, overcast skies, low ceilings, thunderstorms, severe or greater turbulence, severe ice, low level wind shear, anf IFR conditions. You can check the Terminal Aerodrome Forecast for most major commercial airports to get specific forecast probabilities for specific types of weather. It is at the discretion of the airline whether they ought to delay or cancel flights ahead of time when such weather is forecast. Nimur (talk) 23:41, 11 April 2014 (UTC)[reply]