The nightingale’s song is one of the most celebrated sounds in the British countryside; rich, powerful and delivered under cover of darkness, when almost
Senza un nuovo trattato, i paesi saranno costretti a operare sulla base delle ipotesi peggiori su quali armi i loro avversari stanno costruendo, testando e schierando, avverte Nikolai Sokov, un ex negoziatore sovietico-russo sulle armi. “Una corsa agli armamenti non regolamentata renderebbe la situazione estremamente instabile.”
Il 5 febbraio scade l’ultimo Trattato sulla limitazione delle armi strategiche (Newstart) tra Russia e Stati Uniti. Se non verrà raggiunto un accordo, non ci saranno limiti all’espansione delle armi nucleari a lungo raggio dei due paesi.
Three days into a new conflict brewing in Ethiopia's Tigray region, the factions are either silent or somewhat cryptic thus far.
Today, in response to a call by the African Union for restraint, the TPLF, under whose command Tigrayan TDF rebels operate, wrote a response to the AU stating that they were open to engaging in dialogue. The letter does not acknowledge that fighting has broken out nor that the TPLF/TDF fired the first bullet. It's pretty vague and leaves more questions unanswered.
Är du i Lund med omnejd? Idag ses vi i caféet på stadsbiblioteket i Lund 14.00-16.00.
Digital självständighetsdag första söndagen varje månad. Om du redan kan lite , kom och tipsa om alternativ. Om du inte kan så mycket, kom och bli tipsad...
Behövligt initiativ! 💐 Även vi som inte kan ta oss fysiskt till bibliotek vill gärna bli självständiga. Önskan Att en fond skulle stödja er med pengar till en serie manualer och videos med aktuellt innehåll för oss funkisar.
du kan ju prova att byta till @Vivaldi 🇳🇴 och @Qwant 🇲🇫 idag och sedan posta med taggen #diday för att få hjälp, eller för att stolt "skryta" när det är gjort.
Det är en webbläsare. Men kul. Den kommer med ett gäng smarta funktioner inbyggda. Det är supersmidigt och spårar dig inte. Skaffa webbläsaren Vivaldi för stationära datorer, mobiler och din bil!
Tack! Qwant har som tillägg jag sen förra året. Lite sämre träffar på svenska. Jämförde sökresultat för - Immunförsvaret alla mot alla lyrik text - på 3 webbläsare. Qwant hittade ingen lyrik eller så är det ett användarfel :) Vivaldi nästa steg och hur jag importerar tiotusentals bokmärken
@Qwant @Vivaldi du kommer att klara det (installera Vivaldi med import av bokmärken). Jag tror på dig 🙌 🇪🇺 ✌️
Ja, Qwant är inte lika djup och bred. Men om 450 miljoner européer plötsligt använder Qwant får de resurser. Det är förstås något som tar tid är är långsiktigt. Men Qwant och Ecosia jobbar nu på ett europeiskt självständigt sökindex, så ju fler som använder dem ju mer resurser får de till det. Du får falla tillbaka på G när Qwant inte räcker till.
Kul med så mycket intresse och reboosts för #diday i Sverige. Vi behöver komma på en tag. SweDid eller något?
Alla kan vara med även om man inte är i Lund på stadsbiblioteket idag. Byt ut något från US-tech till europeisk tech och "rapportera in" det på Mastodon med #SweDiD och #diday.
Kan vara att byta sökmotor till 🇲🇫 Qwant eller 🇩🇪 Ecosia. Webbläsare till 🇧🇻 Vivaldi, 🇩🇪 Ecosia, 🇪🇺 LibreWolf eller 🇸🇪 Mullvad.
Eller e-post till 🇩🇪 Tuta, 🇨🇭 Proton. Eller något mer avancerat
När internationella brottmålsdomstolens (ICC) chefsåklagare, britten Karim Khan, kom till jobbet i den Haag en morgon i maj 2025, upptäckte han över morgonkaffet att hans Microsoft Outlook e-postkonto var avstängt som ett resultat av missnöje från de…
Shakur Stevenson delivered the most complete performance of his career on Saturday night, outmaneuvering, outthinking and ultimately outclassing Teófimo López
Sinds 1967 heeft Israël ruim achthonderdduizend Palestijnse mannen, vrouwen en kinderen van de Westoever, uit Oost-Jeruzalem en de Gazastrook gedetineerd. Dat is zo’n 20 procent van de bevolking, oftewel 40 procent van alle mannen, berekende B’Tselem. ‘Er zijn nauwelijks Palestijnse families waar geen familielid door het Israëlische gevangenissysteem is gehaald.’"
Israël houdt ruim 9300 Palestijnen vast. Veelal burgers die als politieke gevangenen zonder vorm van proces gevangen zitten. Ze worden vernederd en gefolterd. Het gevangenissysteem is een van Israëls wreedste onderdrukkingsmechanismen.
En opnieuw wordt duidelijk: er is geen volk dat niet in staat is tot misdaden tegen de menselijkheid, ondanks (of misschien mede dankzij) het feit dat zij er zelf ook slachtoffer van zijn geweest. Een gruwelijk en verontrustend besef omdat het ook onszelf betreft.
A procession of the world’s largest monumental bronze sculpture of St Charbel’s face moves through the streets of Punchbowl to St Charbel’s Church, before being placed permanently on the monastery building in#Sydney.
Photograph: Christopher Khoury/APA/Zuma Press Wire/Shutterstock
The Israeli Gov't is currently fêting Swedish politician Jimmy Åkesson, leader of an extreme-rightwing party founded by literal Nazis. In case you were confused about what the alliances look like now. 🧵/1 www.dn.se/varlden/sver...
Two pairs of ants set off from the same starting point, walking side-by-side. One pair takes the red path, the other pair takes the blue path.
The red and blue paths are geodesics (the straightest possible paths, like great circles on a sphere), and each ant in a pair follows a geodesic that lies a small distance to either side of the marked one, initially pointing in the same direction.
If this ellipsoid has semi-axes of length 1 and 3, the ants taking the blue path will bump into each other at the “pole”, where all meridians converge, after travelling a distance of 3.34122.
How far will the ants following the red path travel before they bump into each other? More, less, the same?
To answer this, we can start with the fact that on a *sphere* of radius R, any two ants following neighbouring great circles would travel a distance of (π/2) R before their paths met. That behaviour comes from the intrinsic curvature of the sphere, which is measured by its Gaussian curvature, K = 1/R^2.
But what is the Gaussian curvature of our ellipsoid, at its equator? The equator itself has a curvature of 1, but the meridians are ellipses that are 3 times wider. What is the curvature of an ellipse with semi-axes a and b, at the point where the minor semi-axis meets the ellipse?
The ellipse takes the form:
y(x) = b √[1-(x/a)^2]
For small x:
y(x) ≈ b (1-½(x/a)^2)
The second derivative is:
y''(x) = -b/a^2
If we set a = b = R this would be -1/R, so the curvature here is just the opposite of the second derivative.
The Gaussian curvature is the product of the curvatures along two perpendicular directions where it achieves its maximum and minimum values. For our ellipsoid, these curvatures are 1/b for the equa
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To answer this, we can start with the fact that on a *sphere* of radius R, any two ants following neighbouring great circles would travel a distance of (π/2) R before their paths met. That behaviour comes from the intrinsic curvature of the sphere, which is measured by its Gaussian curvature, K = 1/R^2.
But what is the Gaussian curvature of our ellipsoid, at its equator? The equator itself has a curvature of 1, but the meridians are ellipses that are 3 times wider. What is the curvature of an ellipse with semi-axes a and b, at the point where the minor semi-axis meets the ellipse?
The ellipse takes the form:
y(x) = b √[1-(x/a)^2]
For small x:
y(x) ≈ b (1-½(x/a)^2)
The second derivative is:
y''(x) = -b/a^2
If we set a = b = R this would be -1/R, so the curvature here is just the opposite of the second derivative.
The Gaussian curvature is the product of the curvatures along two perpendicular directions where it achieves its maximum and minimum values. For our ellipsoid, these curvatures are 1/b for the equator and b/a^2 for the meridian, so we have:
K = 1/a^2
In other words, the Gaussian curvature at the equator of the ellipsoid is completely independent of b (the radius of the equator) and is the same as that of a sphere of radius a (the polar radius)!
That in turn means that the ants following the red path will bump into each other after travelling a distance:
(π/2) a = 3π/2 ≈ 4.71239
So, perhaps a bit counterintuitively, the ants following the small red circle will travel *farther* before they converge than the ants following the larger blue ellipse.
@slashdottir My intuition didn't immediately figure out what to do with "each ant in a pair follows a geodesic that lies a small distance to either side of the marked one, initially pointing in the same direction." (and in fact the illustration only shows one path per pair, I guess because figuring out where the other one lies is basically what the problem is about). @gregeganSF
@soaproot @slashdottir I could use an explanatory illustration of the other paths then because I think I must've totally misunderstood the problem statement
For the blue path, which is a meridian, the ants that walk side-by-side and “follow the blue path” just walk along two meridians with longitudes that are a tiny bit west, and a tiny bit east, of the one that is drawn. All three meridians meet at the pole, so the ants bump into each other at the pole.
For the red path, again the ants walk side-by-side, each of them a very small distance on either side of the red path (i.e. one of them very slightly north of the equator, the other one very slightly south of the equator). But circles of latitude are not geodesics, so their latitudes do not stay constant.
I did not draw the individual paths for the ants because they're meant to be an “infinitesimal" distance on either side of the paths that are drawn.
@soaproot @slashdottir hmm. If the red paths are so close to the red circle (which is of radius 1, right?) as to be indistinguishable, I don't understand how the path length is so different from the (half? since the paths meet at ascending and descending nodes?) circumference of the unit circle. Except that that must be the point of the exercise, but if it's so visually similar...hm.
Sorry for being the thick student but I get concerned when I don't understand ellipses
On a sphere of radius 1 the ants following the red path would meet after they have gone 90 degrees around the equator, so one quarter of the circumference. If you extended their tracks into complete great circles, the distance between their two crossing points would be half the circumference.
On the ellipsoid, their paths still stay very close to the equator (the absolute value of their latitude is a maximum when they start, and I have said that is very small). But being close to the same circle, on a very different surface, doesn't mean they should meet up after a similar distance.
For example, if the red circle was on a cylinder, the ants would never bump into each other.
Electric Sheep
in reply to Quinn Norton • • •