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December 25[edit]

Project management[edit]

In the UK, are project managers better respected and better known professionals than engineers? — Preceding unsigned comment added by 2A02:C7D:B91D:CC00:5146:5AF8:3AE6:5EDF (talk) 11:43, 25 December 2015 (UTC)[reply]

There's no way to answer that. Besides the problem of rating respect and/or fame, both professions are very widely dispersed, being present in all kinds of businesses. They're also completely different kinds of jobs, so it's like comparing apples to oranges. 99.235.223.170 (talk) 15:53, 25 December 2015 (UTC)[reply]

I like oranges better than apples. --Trovatore (talk) 21:29, 25 December 2015 (UTC) [reply]

Apples and oranges is the right article about comparing apples and organes.--Denidi (talk) 16:29, 26 December 2015 (UTC)[reply]

Project manager is a position that might be fulfilled by any profession, including engineers. Engineer is a profession. And as an engineer you might be at any level of development. Engineers in the UK are subject to the same stereotypes as abroad. It also appears that some engineering fields in the UK have high unemployment, which is definitely not good for getting respect. --Denidi (talk) 16:37, 26 December 2015 (UTC)[reply]

"Language Development"[edit]

[Moved to RD/L] Tevildo (talk) 18:25, 25 December 2015 (UTC)[reply]

Good introductory book on algorithms for a middle-school kid[edit]

cross-posting my question from the Math refdesk. Happy Holidays! --Dr Dima (talk) 20:22, 25 December 2015 (UTC)[reply]

Since no one has provided any answer I am going to suggest you read Lateral Thinking by Edward de Bono. Not algorithms as such, but thinking in such a way that you can come up with one. In theory our article on Algorithms should be understandable. If not comment on the talk page to see if improvements can happen. Graeme Bartlett (talk) 06:39, 27 December 2015 (UTC)[reply]

A more basic intro is at simple:Algorithm. Graeme Bartlett (talk) 06:40, 27 December 2015 (UTC)[reply]

Calculating joules without distances[edit]

Usually, the number of joules expended is calculated by multiplying the force in newtons by the distance in metres. But how do you calculate the expenditure of joules when there are no distances involved? E.g. how can I calculate the energy burned by holding a 1 kg laptop against the force of gravity for 60 seconds, but without moving it upwards or downwards? THank you.217.121.5.132 (talk) 20:44, 25 December 2015 (UTC)[reply]

In this case, you can still use the distance. The distance moved is zero, the work done is zero. However, your body will also generate heat just by standing there - this, unfortunately, can't be simply calculated. Tevildo (talk) 20:55, 25 December 2015 (UTC)[reply]

Well, there is in fact nonzero work done. Just not at such a macroscopically observable level. To hold the laptop against gravity, I believe you have to twitch muscle fibers (though I'm sketchy on the details), and that does work on the fibers themselves. You could compute this work if you knew how often the fibers twitch, how far they move, and under what force. As you say, this is not really a simple calculation of the sort that is usually embodied in a free-body diagram. --Trovatore (talk) 21:01, 25 December 2015 (UTC)[reply]

So there is no equation to calculate how many calories I'll burn if I hold a weight of X kg for Y seconds? 217.121.5.132 (talk) 21:13, 25 December 2015 (UTC)[reply]

No simple high-school-physics type equation, no. There might be (indeed, there almost certainly are) mathematical models that will estimate that value. But they will be approximate rather than exact, and (speculating here) they may involve inputs that have predictive power for your physiological makeup (e.g. age, sex, physical condition). --Trovatore (talk) 21:19, 25 December 2015 (UTC)[reply]

The rate of energy burn can be measured by measuring the amount of carbon dioxide exhaled as a measure of the metabolic rate. Less accurately, the amount of air intake can be measured and assume that all the oxygen taken in, or a fixed percentage of it, is burnt to CO₂. Even less accurately, the breathing rate can be recorded. There is a known relationship between energy and amount of oxygen consumed (its about 5 calories per litre of oxygen, but it does vary with type of food consumed). The quiescent metabolic rate (at rest) must be subtracted from this, and the answer is not going to be terribly precise because holding up a laptop is a very small part of the total energy consumption. However, a better question, as Trovatore hinted, is why does a static hold burn energy at all? It is certain that no one can indefinitely hold the dumbbell lift shown, yet no dynamic energy is being expended. SpinningSpark 00:42, 26 December 2015 (UTC)[reply]

You're not actually calculating work done by the laptop in that case, but the energy lost by biological inefficiency. The "work" done (which isn't) will be drastically different if you're holding it on your outstretched palm vs. if you're holding it like a suitcase or have it slung over a shoulder in a bag. You can calculate the waste heat but that's not work - for example, you cannot tap that to run a pendulum clock. Technically the waste heat isn't lost entirely to entropy yet; you could set up some kind of heat engine to recapture some of it. But if you count that you should count the "work" done by friction in a classical machine, since you could use that is a heat source and so forth. Bottom line: this is a feature, not a bug, of the formula. Wnt (talk) 12:51, 26 December 2015 (UTC)[reply]

Note that we have an article on isometric exercise -- the concept this question is about. It does not discuss energy consumption, though. Looie496 (talk) 13:49, 26 December 2015 (UTC)[reply]

The person holding a weight at arm's length is doing work, since he is resisting the downward acceleration that would result due to gravity. In other words the action of his arm is to keep moving the object upwards at an acceleration of -g, simply in order to keep the object at the same height. Turning things on their side, if a person pushed against a car to keep it stopped from running away on a slope, he would be expending energy and doing work simply in order to keep it stationary. Akld guy (talk) 21:06, 26 December 2015 (UTC)[reply]

Ah, no, this is a very common, but fundamental, misconception. The person holding the weight or the car stationary is not doing any work. See Work (physics). They'll be expending energy and generating heat, but, with no movement, there's no work being done. Tevildo (talk) 22:11, 26 December 2015 (UTC)[reply]

No work done at the gross level of looking at how the car or weight moves. As I said above, there is still work being done. You just can't see it without looking down at the level of muscle fibers or below. --Trovatore (talk) 22:14, 26 December 2015 (UTC)[reply]

I still think it's important to clarify that no _net_ work is being done on the laptop. If the OP were to adopt a more vigourous form of exercise by holding the laptop above his head and lifting it up and down, it would be meaningful to say that, in extending his arms, he is doing a calculable amount of work (m⋅g⋅h, where h is the height difference between the top of his head and his outstretched arms), and that he's not recovering this energy when he lowers it - but there's still no _net_ work being done. Tevildo (talk) 23:07, 26 December 2015 (UTC)[reply]

That's fair enough, as long as we keep in mind that "no work done on the laptop" is not the same as "no work done", period. --Trovatore (talk) 23:45, 26 December 2015 (UTC)[reply]

Yes, but why are they expending energy and generating heat. If I replace the person with a steel frame to hold the weight or the car, it won't get warm in the least no matter how long it holds it there. I understand Trovatore's point that muscle fibres are twitching as the hold goes on because the group that is contracting keeps changing as one group gets exhausted, but that still does not answer the question. Why should it be necessary for the fibres to take turns if they are not actually expending any energy? SpinningSpark 22:18, 26 December 2015 (UTC)[reply]

Well, it wouldn't be necessary, if the fibers were made of something rigid, say bone. But then you might have trouble using your arm for anything other than holding the weight.... --Trovatore (talk) 22:32, 26 December 2015 (UTC)[reply]

What difference does that make? A contracted muscle can be treated as a rigid body. Work is done while it is contracting, sure, but once there it is just a rigid member of the frame. Something is going on at the chemical level which, so far, no one has answered. SpinningSpark 22:46, 26 December 2015 (UTC)[reply]

Well, I don't know the chemistry in detail. Skeletal muscle might have some clues. At a higher level, you can certainly imagine a sort of ratchet system whereby muscles could be pulled into one configuration, then made rigid, and the rigor voluntarily released somehow at some later point. But that isn't how muscles work. Well, actually I think there is a sort of molecular ratchet, but not to the extent it that it can make the muscle rigid indefinitely without expenditure of energy.

Why isn't there? Presumably it is too difficult to make it work, and doesn't confer enough evolutionary advantage. That's just a boilerplate answer, but that's what I'd go with pending contrary information. --Trovatore (talk) 22:54, 26 December 2015 (UTC)[reply]

Oh, and the fibers are doing work. They are not doing work on the weight, per se. But they are doing work, moving against a resistive force. --Trovatore (talk) 22:41, 26 December 2015 (UTC)[reply]

Yes, but why is it necessary for the fibres to move? SpinningSpark 22:46, 26 December 2015 (UTC)[reply]

I found this book that contains the sentence "This is the reason why muscles get tired supporting a static load." The answer isn't simple, otherwise this thread would be much shorter. The basic idea seems to be that a muscle fibre has two options: consume a molecule of ATP and contract temporarily, or don't consume a molecule of ATP and relax. ATP is our fuel, so our muscles consume fuel to stay still under load. Note that there is no "hold your current position regardless of the load" option. We aren't made of struts and ratchets so we can't do that. --Heron (talk) 11:37, 27 December 2015 (UTC)[reply]

Thanks for finding that, but ATP is merely the form in which muscles consume energy, so saying a muscle must consume ATP to stay contracted is just paraphrasing the question. Why must a muscle consume energy to stay contracted? However, the previous page, if I have understood it correctly, seems to give the answer, it is needed to overcome thermal fluctuations that tend to release the contraction. That would seem to suggest that there is going to be a definite relationship between static force applied and energy consumed, so in principle, there should be a simple and direct answer to the OP's orginal question (if we are given the weight of the laptop, and any mechanical (dis)advantage, such as the lever action of the dumbbell lift, is taken into account.) SpinningSpark 13:00, 27 December 2015 (UTC)[reply]

No, I don't think you will ever find a simple definite relationship because there are too many variables, including how many muscles are involved and how hard they have to work. If the laptop is balanced on the head, then muscles will not need to work hard because the weight is mainly supported by fairly rigid structures, but if the laptop is held between fingers and thumb at the end of an outstretched arm, then lots of muscles are under strain, so will need to expend energy. Dbfirs 15:54, 27 December 2015 (UTC)[reply]

Well, the implication of that source is that there is a simple relationship between the force exerted and ATP expended, and that is not dependent on which muscles are involved, or the specific configuration of the load. Those things might affect the total force required though, so it comes down to just a knowledge of the mechanical arrangement in a specific case (length of lever arms, distance to muscle attachment points etc, otherwise known as anatomy). For holding up a laptop, I would think that if one held it close to one's body then the force being applied would be pretty close to the force of gravity acting on the laptop. SpinningSpark 16:43, 27 December 2015 (UTC)[reply]

Yes, if one could take account of each of the dozens of muscles involved and the force provided by each, then it might be possible to calculate the "work" done (energy expended) by the muscles for each position of the laptop, though it would differ between individuals because everyone's skeleton and musculature is different. Dbfirs 16:57, 27 December 2015 (UTC)[reply]

There are some minor errors in the statement: "consume a molecule of ATP and contract temporarily, or don't consume a molecule of ATP and relax". ATP is use to relax a muscle fiber. The contraction is generated when the potential energy in the sarcomere is released. Rigor mortis occurs because there is no energy to relax. Additionally, it is a lot more than one molecule of ATP involved per fiber or even sarcomere.

As for the main point, as pointed out before, the work done on the laptop is identical to the work done by a shelf holding up the laptop: zero. The amount of energy expended, which could be expressed as work on the molecules in the environment which is a fancy way of saying increasing the temperature, would highly variable with a variety of factors including muscle mass, distribution of fiber types (slow twitch vs. fast twitch), and the the weight of the arm itself. Any such model would be fairly inaccurate relative to real world conditions. Breathing and circulation have to increase to support the effort, so there is energy not directed to the laptop involved. It would be easier to measure metabolism by an indirect measure such as oxygen consumption/carbon dioxide production and compare results when the arm is outstretched with and without the weight.BiologicalMe (talk) 17:31, 27 December 2015 (UTC)[reply]

Note that the work done on the laptop actually doesn't depend on the particulars of myosin contraction, because it can be dissipated before reaching it. If you put a vibrator under a sofa cushion, it's going to do a lot less mechanical work on the woman sitting on it than if it were in direct contact with her. And generally speaking, the body strives to have coordination that suppresses tremor - if you had Parkinson's, you'd do a lot more work on the laptop than otherwise! It is work to wiggle the laptop back and forth, even if some is positive work and some is negative work. When the laptop is accelerating downward, you're doing less negative work (because the force is less) than when it is accelerating upward and you are doing positive work against a greater-than-g force; thus, if you have one of those shake-em-up flashlights it can actually get power by your wiggling it back and forth in place, because you're delivering work to it. I say this tentatively, since I could have fouled something up. Wnt (talk) 19:04, 29 December 2015 (UTC)[reply]

Is there H2O (water) in milk?[edit]

23:47, 25 December 2015 (UTC) — Preceding unsigned comment added by 92.249.70.153 (talk)

Have you read Milk it might help. MilborneOne (talk) 23:50, 25 December 2015 (UTC)[reply]

And according to the article, the answer is "Yes." ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:06, 26 December 2015 (UTC)[reply]

Yes, Baseball Bugs is right. Average whole unprocessed cow's milk is 87.8% water. Even more shockingly, whiskey is roughly half water. Cullen³²⁸ Let's discuss it 06:39, 26 December 2015 (UTC)[reply]

Since life is basically a bunch of stuff dissolved in water, things produced by organisms are generally water solutions: blood, sweat, tears, milk, mucus, urine, whatever. (I guess toil would be the exception.) --71.119.131.184 (talk) 10:11, 26 December 2015 (UTC)[reply]

I meant to ask if the water in the form of free H2O molecules in the milk. Second, when I read about something that dissolved in water, does it mean that the water in this solution is in form of free H2O molecules? (I always thought that when something dissolved in water it means that the water molecules create new bonds with the water molecules) 17:53, 26 December 2015 (UTC) — Preceding unsigned comment added by 92.249.70.153 (talk)

I think this is a bit complicated. If I understand correctly (and I may not), even pure liquid water is not necessarily made up of "free H₂O molecules", depending on what you mean by that. They organize themselves into groups connected by weak hydrogen bonds. Our water article has a little on this but not very much. --Trovatore (talk) 21:29, 26 December 2015 (UTC)[reply]

When a substance is dissolved in water, there is a mixture of free H2O molecules and molecules of the solute. The bonding between the solvent (water) and the solute is not so strong that the molecules of the solvent become something else. Anyway, with milk 87.8% water, even if some of the water was bound, some of it won't be bound. Robert McClenon (talk) 17:58, 26 December 2015 (UTC)[reply]

Also, most of the non-aqueous content of milk is not dissolved but emulsified (dispersed), so that the H2O molecules do not bond with it. The fact that milk is colloidal in nature is why it may separate under some conditions and can be separated into cream and skim milk. Robert McClenon (talk) 18:04, 26 December 2015 (UTC)[reply]

Thank you! 92.249.70.153 (talk) 01:24, 27 December 2015 (UTC)[reply]

Substances that dissolve do tend to form bonds with the water - these are hydrogen bonds. They're chemical bonds sort of the way Pluto is a planet. Since lots of things will interfere with hydrogen bonds (like dissolving a substance in water) we don't think of them as defining the chemical. Also, a solvation shell isn't as consistent an object as a chemical compound that has just one structural formula. The interaction is strongest when you have something charged like hydronium - the free H+ interacts so strongly with water that it can't really be said to be outside the molecule, and it is surrounded by a shell of other water molecules that strongly interact. Other compounds interact less, but they have to have some sort of charge to interest the water molecules in interaction, or else they would be hydrophobic. I should add that I actually have no idea whether there is some compound that is so aberrantly shaped that it interacts so poorly with itself that it will dissolve in water despite not having any polar interaction, simply because nothing holds it together. (I suppose noble gases dissolve somewhat in water to fulfill that criterion, but it seems like cheating; I want something with multiple atoms per molecule to be properly amused) Wnt (talk) 18:54, 29 December 2015 (UTC)[reply]