Weblog on Physics

Tuesday, 30 July 2013

HD 189733: An Eclipsing Exoplanet Seen In X-rays For First Time

Friday, 12 July 2013

Chords in trigonometry

Source: Wikipedia.

Chords were used extensively in the early development of trigonometry. The first known trigonometric table, compiled by Hipparchus, tabulated the value of the chord function for every 7.5 degrees. In the second century AD, Ptolemy of Alexandria compiled a more extensive table of chords in his book on astronomy, giving the value of the chord for angles ranging from 1/2 degree to 180 degrees by increments of half a degree. The circle was of diameter 120, and the chord lengths are accurate to two base-60 digits after the integer part. The chord function is defined geometrically as in the picture to the left. The chord of an angle is the length of the chord between two points on a unit circle separated by that angle. The chord function can be related to the modern sine function, by taking one of the points to be (1,0), and the other point to be (cos $\theta$ , sin $\theta$ ), and then using the Pythagorean theorem to calculate the chord length:

$\mathrm{crd}\ \theta = \sqrt{(1-\cos \theta)^2+\sin^2 \theta} = \sqrt{2-2\cos \theta} =2 \sin \left(\frac{\theta }{2}\right).$

The last step uses the half-angle formula. Much as modern trigonometry is built on the sine function, ancient trigonometry was built on the chord function.

The chord function satisfies many identities analogous to well-known modern ones:

Name	Sine-based	Chord-based
Pythagorean	$\sin^2 \theta + \cos^2 \theta = 1 \,$	$\mathrm{crd}^2 \theta + \mathrm{crd}^2 (180^\circ - \theta) = 4 \,$
Half-angle	$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos \theta}{2}} \,$	$\mathrm{crd}\ \frac{\theta}{2} = \pm \sqrt{2-\mathrm{crd}(180^\circ - \theta)} \,$
Apothem (a)	$c=2 \sqrt{r^2- a^2}$	$c=\sqrt{D ^2-4 a^2}$
Angle (θ)	$c=2 r \sin \left(\frac{\theta }{2}\right)$	$c=D \sin \left(\frac{\theta }{2}\right)$

Tuesday, 18 June 2013

Pendolo fisico , piccole oscillazioni

Moti alternati

Dal Progetto Glues del Prof. M. Baronti
http://www.diptem.unige.it/baronti/GLUES/Glues_negli%20anni.htm

Adattato dal progetto sul moto alternato

The Remarkable Properties of Mythological Social Networks | MIT Technology Review

"Today, P J Miranda at the Federal Technological University of Paraná in Brazil and a couple of pals study the social network between characters in Homer’s ancient Greek poem, the Odyssey. Their conclusion is that this social network bears remarkable similarities to Facebook, Twitter and the like and that this may offer an important clue about the origin of this ancient story."
The Remarkable Properties of Mythological Social Networks | MIT Technology Review

Monday, 17 June 2013

Festival Beethoven

Dal 24 al 30 giugno 2013, Piazza San Carlo

Le 9 Sinfonie con l’Orchestra Sinfonica Nazionale della RAI e il Coro del Teatro Regio. I Concerti con l’Orchestra Filarmonica di Torino e grandi interpreti.

Tutte le sere, ore 21, Ingresso libero
http://www.festivalbeethoven.it/

Immagine con non-photorealistic rendering; vedi A.C. Sparavigna, B Montrucchio,

Non-photorealistic image rendering with a labyrinthine tiling,
http://arxiv.org/abs/cs/0609084

Friday, 14 June 2013

Moving sand dunes on the Google Earth

Several methods exist for surveying the dunes and estimate their migration rate. Among methods suitable for the macroscopic scale, the use of the satellite images available on Google Earth is a convenient resource, in particular because of its time series. Some examples at http://arxiv.org/abs/1301.1290 (arXiv: January 2013)

Thursday, 13 June 2013

Positive or negative? Nanoparticle surface charge affects cell-membrane interactions - physicsworld.com

Monopoles unwind magnetic whorls - physicsworld.com

Bob e Joe, lavavetri

Two window washers, Bob and Joe, are on a 3.00-m-long, 345-N scaffold supported by two cables attached to its ends. Bob weighs 750 N and stands 1.00 m from the left end, as shown in Figure. Two meters from the left end is the 500-N washing equipment. Joe is 0.500 m from the right end and weighs 1 000 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?

Risolviamo con equilibrio di forze e di momenti usando lo schema seguente per masse, distanze e tensioni funi.

T_o + T_d = m_a g + m_b g + m_c g

a m_a + b m_b + c m_c - d T_d = 0 (polo in o)

Dato che le masse e le distanze sono note, le due equazioni mi danno le due tensioni funi.