Wikipedia:Reference desk/Archives/Science/2016 September 6

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September 6

Coords

Rule of thumb: the antipodes is almost never anywhere interesting. Excepting some bits of southern South America, and Antarctica which happens to fall mostly in the Arctic Ocean, and Australia which happens to fall in the Atlantic, the Pacific + Indian Ocean is nearly half the globe.

What would be the furthest point from coordinates 34°S 20°E? Thanks. Anna Frodesiak (talk) 07:54, 6 September 2016 (UTC)

Because of the way latitude and longitude work, to find the antipodes of a particular point you reverse the direction for both coordinates and subtract the longitude number from 180°. So that's 34°N 160°W, which is in the ocean about 1,400 km north of Honolulu. --69.159.61.86 (talk) 08:05, 6 September 2016 (UTC)

• (edit conflict) @Anna Frodesiak: The fish: 34°N 160°W (somewhere north of Honolulu in the Pacific Ocean).

How to fish: see latitude, longitude. The "furthest point" on a sphere is the point on the other side of a diameter of the sphere. To find it, take a symmetry around the equator (exchange North and South) and rotate 180° ("200°E" = 160°W). Tigraan^{Click here to contact me} 08:09, 6 September 2016 (UTC)

Thank you so much folks! Actually, the reason I asked is quite trivial. I just want to know the furthest distance from the blue cheese capital of the world. But I know, you guys secretly get a bit of a kick out of figuring out science questions. :) Thank you again! Anna Frodesiak (talk) 08:24, 6 September 2016 (UTC)

Not from Clayton, Polk County, Wisconsin? -- ToE 16:40, 6 September 2016 (UTC)

See, and I thought it would have been Roquefort-sur-Soulzon... --Jayron32 16:49, 6 September 2016 (UTC)

Stilton, surely. Alansplodge (talk) 23:38, 8 September 2016 (UTC)

I would have thought the antipode of a place famous for cheese that can be made into a sauce would be a place with a distinct shortage of cheese sauce. StuRat (talk) 21:27, 6 September 2016 (UTC)

• The diagram caption at right is correct, but it does say "almost". One of the exceptions is that there are places where one city is antipodal to another. For example, Neiva and Palembang. --69.159.61.86 (talk) 18:43, 6 September 2016 (UTC)
  • New Zealand is over Spain as well. Rmhermen (talk) 01:30, 7 September 2016 (UTC)

Thank you all again! And I like the image caption. Very interesting. Anna Frodesiak (talk) 02:09, 7 September 2016 (UTC)

Coriolis force. Centrifugal force

As we know, the Coriolis force appears in a noninertial reference frame, only when the body moves in this r.f. Centrifugal force does not depend on whether the body is moving. Explain please, if we have a noninertial r.f. e.g. a carousel and a body hanged over the carousel, then from the carousel r.f. what forces will act on the body? From the inertial r.f. the body will stay on the same point, from the carousel r.f. the body will describe a circle path. I think only the Coriolis force will act. Is it correct? — Preceding unsigned comment added by 12.215.81.250 (talk) 08:59, 6 September 2016 (UTC)

Well, according formulas $F_{c}=2mv\omega$, but a force needed to move the body along a circle path $=mv\omega$. — Preceding unsigned comment added by 12.215.81.250 (talk) 09:51, 6 September 2016 (UTC)

In the non-inertial frame there is a centrifugal force ${\vec {F}}_{ce}=m{\vec {r}}\omega ^{2}$. The Coriolis force is ${\vec {F}}_{co}=-2m{\vec {\omega }}\times {\vec {v}}$. As it is given that the object is stationary in the inertial frame, we can substitute Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \vec{v}=\vec{r}\times\vec{\omega}} , leading to ${\vec {F}}_{co}=-2m{\vec {\omega }}\times \left({\vec {r}}\times {\vec {\omega }}\right)=-2m\left(\left({\vec {\omega }}\cdot {\vec {\omega }}\right){\vec {r}}-\left({\vec {\omega }}\cdot {\vec {r}}\right){\vec {\omega }}\right)=-2m{\vec {r}}\omega ^{2}$, as we also know that ${\vec {\omega }}\cdot {\vec {r}}=0$ (the object is in a plane perpendicular to the axis of rotation). Combining these two we get ${\vec {F}}_{sum}={\vec {F}}_{ce}+{\vec {F}}_{co}=-m{\vec {r}}\omega ^{2}$. So there is the centrifugal force acting outward and the Coriolis force acting twice as strong inward, combined delivering exactly the necessary centripetal force to make the object go in circles. PiusImpavidus (talk) 10:04, 6 September 2016 (UTC)