Wikipedia:Reference desk/Archives/Science/2017 March 27

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March 27

Are there places with invasive species(es) that like its climate better than anywhere in their native range?

Are there any such species from continents instead of islands (which often have endemic clades that can't leave if a land bridge closes then climate goes downhill (end of the last ice age maybe?)) Of course they might also evolve to be weak against thing(s) they're now sheltered from (like losing fear of predators) and do even worse on the mainland even if the climate suits them better. Sagittarian Milky Way (talk) 17:42, 27 March 2017 (UTC)

Invasive species normally do better in their new environment due to a lack of predators, competitors and diseases or parasites that keep them under control in their native environment. I am unaware of any invasive species that are said to do better in a new climate, since most species already occupy most of the climate and environmental range they can tolerate (and where they are not excluded by competitors). The Gypsy moth became a widely problematic invasive pest in the Eastern US. It was controlled by spraying of Bacillus thuringiensis, a natural parasite of the moth imported from its native Eurasia. μηδείς (talk) 20:45, 27 March 2017 (UTC)

OR alert I'm not sure if the question relates to introduced species, but if it does, I wonder whether the Cane toad in Australia is a candidate. DrChrissy ^(talk) 21:29, 27 March 2017 (UTC)

Which, however, has to do with a lack of predators rather than with the climate. 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 00:24, 28 March 2017 (UTC)

What evidence do you have to support this statement? DrChrissy ^(talk) 19:10, 28 March 2017 (UTC)

Cane toads are lethally toxic, hence the lack of effective control by predation. μηδείς (talk) 15:57, 29 March 2017 (UTC)

Yes I know that - I lived in Australia for 12 years. My question referred to the suggestion their prevalence was a lack of predators rather than climate - this suggests there is a study that has examined and compared the 2 effects. This is a reference desk. DrChrissy ^(talk) 17:53, 29 March 2017 (UTC)

No, I don't think so. As long as you understand the concept of apex predator, and that this animal is one, you don't need a study to prove that they benefit by that status. I am not against someone producing a study to show whether they have reached their maximum range, which will depend on climate. But their success so far has been due to their toxicity, and there are plenty of documentaries and articles which discuss this. μηδείς (talk) 01:27, 30 March 2017 (UTC)

Hedgehogs in New Zealand. Kudzu and wisteria across North America. Rabbits in Australia. The "invasive" part of invasive species implies that the species does too well in the new environment. 71.85.51.150 (talk) 22:21, 27 March 2017 (UTC)

• BTW, the plural of species is simply species. μηδείς (talk) 01:29, 28 March 2017 (UTC)

• And to argue otherwise would be specious. StuRat (talk) 20:42, 28 March 2017 (UTC)

How about humans ? Human populations in Asia, and India and China, in particular, seem to have outstripped those in our native Africa. Part of the reason here is that parts of Africa have undergone desertification and can no longer support large human populations. StuRat (talk) 20:40, 28 March 2017 (UTC)

Humans are a rare case, insofar as our most unique adaptive qualities place us in what is sometimes called the "cognitive niche", ecologically speaking. The relatively rapid expansion of humans to occupy virtually every variety of terrestrial ecosystem on the planet has owed a lot to our ability to regulate body temperature through tools (everything from primitive fire building techniques and the construction of furs and clothing through to modern climate control technology). There has been a very slight effect upon our physiology by climate (ever-so-slightly shorter and stockier body types seem to correlatively predominate in colder regions), but for the most part, humans have rarely been discouraged from inhabiting a region on the basis of preferential physiological adaption to previous climates, if we are talking strictly about thermoregulation. Of course, the availability of resources in the climatological/ecological context is another matter entirely, as your comparison between Africa and Asia highlights. However, even then, note that human population densities in the modern world (and indeed before) are the product of a lot historical factors, and don't necessarily align 1-to-1 with what was available in terms of raw wild resources. Different continents provided different technological advantages to local populations, via the availability of basic materials, cultivatable crops, and domesticable animal species. Jared Diamond's most well-known known popular science work, Guns, Germs, and Steel, provides a good overview of this process as one of its main topics. Snow ^{let's rap} 23:57, 28 March 2017 (UTC)

Japanese knotweed was having a tough time in a volcanic area of its native country. Somebody brought it over to England's green and pleasant land and it's taken over. 86.169.56.176 (talk) 19:20, 29 March 2017 (UTC)

Yes, but again that is due to lack of natural control species in the UK rather than the different climate. However, there is a control insect that is being trialled for use in the UK.[1] Anyway, the original post seems to be asking about animals rather than plants. Richerman (talk) 19:47, 29 March 2017 (UTC)

Engineering and non technical project management career path

Per the note at the top of the page, we cannot answer requests for opinion. Matt Deres (talk) 00:35, 29 March 2017 (UTC)
The following discussion has been closed. Please do not modify it.
If a junior engineer goes to a non engineering organisation for 2 years to gain some general project management skills, would he easily be able to go back to engineering after? 2A02:C7D:B91F:6A00:C840:DAC4:FAF4:C6D9 (talk) 18:02, 27 March 2017 (UTC) Maybe? You'd have to talk to people who have done so, likely someone in human resources where decisions about hiring and firing are done. --Jayron32 18:07, 27 March 2017 (UTC) That would depend on what branch of engineering. Project management is very sensitive to trends and fashion, but these aren't consistent across disciplines. If the engineer went somewhere that gave them experience of a narrow skill they would find useful in the future, and whichever discipline that was, then this would be valuable. But production line management, or software project management, would help far less if they went into civil engineering. Andy Dingley (talk) 19:34, 27 March 2017 (UTC) What about high level programme management of public policy projects, related to infrastructure, to civil engineering? 2A02:C7D:B91F:6A00:7C29:82C:CF1D:82B5 (talk) 19:48, 27 March 2017 (UTC) Depends on the local fashions, but that sounds worthwhile. Often it's about the size of budget - projects with similar budgets share similar approaches and the experience is transferrable. Andy Dingley (talk) 20:19, 27 March 2017 (UTC) Budgets would be a lot lower for public policy compared to civil engineering though. 82.132.216.32 (talk) 23:11, 27 March 2017 (UTC) I wouldn't have thought it would help with civil engineering. A large part of the work in civil engineering is project management anyway and the person would have a gap in their civil engineering experience. In other types of engineering the work tends to be more with other engineers rather than supervising non-engineers, it might be possible to use the project management experience to get a leg up but one would have to have done very well at the project management. Otherwise yet again it would just be lost experience. Dmcq (talk) 11:19, 28 March 2017 (UTC)

Center of mass

How to prove this formula below? I do know it works for 2 masses because of torque. יהודה שמחה ולדמן (talk) 20:34, 27 March 2017 (UTC)

$x={\frac {m_{1}x_{1}+\cdots +m_{n}x_{n}}{m_{1}+\cdots +m_{n}}}$

That is actually just the definition of the center of mass (for discrete masses). There's no way to prove something that is defined without using the definition itself, which is circular. - Lindert (talk) 20:47, 27 March 2017 (UTC)

Bummer.

But this formula does represent the physical reality, something here doesn't add up. יהודה שמחה ולדמן (talk) 21:02, 27 March 2017 (UTC)

You can't prove the definition, but you can prove that it is useful. Using ordinary mathematics, you can prove that the center of mass formulation can be made to appear in calculations of the effects of gravity or torque on a system of rigidly connected masses, simply based on first principles. Someguy1221 (talk) 21:04, 27 March 2017 (UTC)

(edit conflict) I agree you can't prove it, since it's the definition of center of mass. But - depending on why you want to "prove" it, and how rigorous you need to be - if you accept it's true for 2 masses, you can convince yourself it's true for 3 masses, m1, m2, and m3, by using (a) the center of mass for masses m1 and m2 as a first term of a new equation, and (b) mass m3 as the second term of that equation, and then substituting. This can be extended as often as needed. --Floquenbeam (talk) 21:08, 27 March 2017 (UTC)

I'll tell you the truth - now that you mention it, all I really know is the lever formula $m_{1}d_{1}=m_{2}d_{2}$ . So how did we get to $x={\frac {m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}}$ ? יהודה שמחה ולדמן (talk) 08:32, 28 March 2017 (UTC)

If the lever is in balance, the pivot of the lever is the center of mass, and d₁ and d₂ are the distances from that point right? Consider the following 'diagram':

---O--------m1------P---------------m2----

where O is the origin of the x-axis, m₁ and m₂ are the two masses, and P is the pivot point, or the center of mass. Counting from the reference point O, the positions of the masses m₁ and m₂ are denoted x₁ and x₂ respectively and the position of P, or the center of mass, we will call x_cm. From the diagram we can see that $d_{1}=x_{cm}-x_{1}$ and $d_{2}=x_{2}-x_{cm}$. Now we substitute those in the formula $m_{1}d_{1}=m_{2}d_{2}$, so we get $m_{1}(x_{cm}-x_{1})=m_{2}(x_{2}-x_{cm})$. Getting rid of the brackets, we get $m_{1}x_{cm}-m_{1}x_{1}=m_{2}x_{2}-m_{2}x_{cm}$. Moving terms with x_cm to the left, and x₁ and x₂ to the right of the equal sign gives $(m_{1}+m_{2})x_{cm}=m_{1}x_{1}+m_{2}x_{2}$, which leads to $x_{cm}={\frac {m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}}$. I hope this is helpful. - Lindert (talk) 10:04, 28 March 2017 (UTC)

Thank you. יהודה שמחה ולדמן (talk) 10:31, 28 March 2017 (UTC)

Without any reference to "center of mass", the original formulation could be translated to english as "the weighted mean position of the masses". This seems like a fairly intuitive *definition* of "center of mass" although as noted, its applicability depends on context. 92.90.16.42 (talk) 14:06, 28 March 2017 (UTC)

By itself the formula for the center of mass is not a definition. It is obtained as a result of separation of the angular momentum of a system of $N$ particles into the internal and external parts, which are conserved separately. We can write for the total momentum and the corresponding velocity

${\vec {P}}=\sum \limits _{k=1}^{N}{\vec {p}}_{k}=\sum \limits _{k=1}^{N}m_{k}{\vec {v}}_{k},\quad {\vec {V}}={\frac {\vec {P}}{\sum \limits _{k=1}^{N}m_{k}}}$,

The later velocity is called the center of mass velocity. For the total angular momentum

${\vec {M}}=\sum \limits _{k=1}^{N}{\vec {x}}_{k}\times {\vec {p}}_{k}=\sum \limits _{k=1}^{N}{\vec {x}}_{k}\times m_{k}({\vec {V}}+\Delta {\vec {v}}_{k}),$

where $\Delta {\vec {v}}_{k}$ are particle velocities in the frame of reference where ${\vec {V}}=0$ (center of mass frame of reference). Then

${\vec {M}}=\left(\sum \limits _{k=1}^{N}{\vec {x}}_{k}m_{k}\right)\times {\vec {V}}_{k}+\sum \limits _{k=1}^{N}{\vec {x}}_{k}\times m_{k}\Delta {\vec {v}}_{k}={\vec {X}}\times {\vec {P}}+{\vec {M}}_{int},$

where ${\vec {M}}_{int}$ is the internal angular momentum of the system and

${\vec {X}}={\frac {\sum \limits _{k=1}^{N}m_{k}{\vec {x}}_{k}}{\sum \limits _{k=1}^{N}m_{k}}}.$

Ruslik_Zero 19:19, 28 March 2017 (UTC)

You can prove it experimentally. That is, you can balance an object on it's center of mass. StuRat (talk) 20:45, 28 March 2017 (UTC)

This looks like it is ripe for a proof by mathematical induction. If we assume the center of mass of x₁ to x_n is at center_n, and we introduce a new mass at x_n+1 with mass m_n+1, then at a new center_n+1 the torque in one direction from the preceding summed mass * distance will precisely equal the torque in the other from the new mass m_x+1 in the other. That means you take the distance x_n+1 - center_n and divide it in proportion to the ratio of the new mass vs. all the old masses. This math converts your denominator from n masses to n+1 masses, and gives you a new-minted term for the numerator to match. You still have to assume that you are confident of how torque works. Wnt (talk) 23:01, 28 March 2017 (UTC)

How close are nerves to the skin?

I have a mole on my ankle that a doctor wants to remove, I know from previous ankle surgery stories that the sural nerve is close by and that if it is jabbed by the needle from the numbing injection or the scapel it will cause nerve damage. I am wondering how close to the skin the sural nerve is. --Sara203040 (talk) 20:54, 27 March 2017 (UTC)

Have you reviewed Sural nerve? ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:57, 27 March 2017 (UTC)

Yes I did read that and it doesn't answer my question. Does anyone know?--Sara203040 (talk) 22:22, 28 March 2017 (UTC)

I am not clear on whether you're talking about a very superficial anaesthesia hitting it by accident, or an intentional sural nerve block. For the latter see [2]; note it says 5 or 6 ml is injected, so I assume the needle doesn't really have to be directly touching the nerve. Wnt (talk) 23:12, 28 March 2017 (UTC)

I am talking about very superficial anaesthesia hitting it by accident. There doses not seem to much if any muscle under the area the mole is in, so my concern in the needle hitting the nerve by accident. I am wondering how many millimeters below the skin the sural nerve is.--Sara203040 (talk) 14:06, 29 March 2017 (UTC)

This page discusses procedures for giving a proper sural nerve block, including locations and depths of injections. This page seems to discuss the anatomy of the sural nerve specifically. Perhaps one of those can help you find the information you seek. --Jayron32 14:24, 29 March 2017 (UTC)

The jamanetwork link says the nerve depths were measured near the "posterior crural intermuscular septum". Where exactly is that on the leg? I am not sure where that is but from the pictures in the pdf if looks more like it is on the calf rather than the ankle where my mole is and where there is less muscle. Is the nerve depth the same? It looks like it could be as little as 5mm higher up on the leg, which makes me rather nervous to be sticking a needle in that area.--Sara203040 (talk) 21:10, 29 March 2017 (UTC)