sexta-feira, 14 de janeiro de 2011

How to Run Android Applications on Ubuntu - Softpedia

How to Run Android Applications on Ubuntu - Softpedia: "

How to Run Android Applications on Ubuntu

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June 25th, 2009, 18:04 GMT| By Marius Nestor

Android 1.5 Emulator on Ubuntu 9.10 Alpha 2

When Google announced and released Android, back in October 2008, everyone knew that it would become the best operating system for mobile devices. Not only is Android open source, but it also comes with a Software Development Kit, which offers the necessary APIs and utilities for developers to easily build powerful applications for Android-powered mobile devices. The following tutorial was created especially for those of you who want to test the Android platform and install various applications, on the popular Ubuntu operating system. OK, so let's get started... shall we?

Grab the Android SDK from Softpedia and save the file on your home folder.

Editor's note: The tutorial was rewritten for the new Android 2.0 or later, which provides a graphical user interface to setup a virtual device and the SD card. This makes everything a lot easier. No more command-line madness!

Step 1 - Installing the requirements

Until the download is over, make sure that you have Java installed and the 32-bit libraries (for the x86_64 users ONLY). If you don't have Java (or the 32-bit libraries), go to System -> Administration -> Synaptic Package Manager...

...search for openjdk and double-click on the openjdk-6-jre entry...

...then, search for ia32-libs (ONLY if you are on a x86_64 machine), and double-click on the ia32-libs entry...

Now, click the 'Apply' button to install the packages. Wait for the packages to be installed and close Synaptic when the process is finished.

Step 2 - Android Setup

When the Android SDK download is over, right-click on the file and choose the 'Extract Here...' option...

Enter the extracted folder, then enter the tools folder and double click the android file. Click on the "Run" button when you will be asked what you want to do, and the Android SDK and AVD Manager interface will appear...

Go to the "Settings" section and make sure you check the "Force https://..." box. Click the "Save & Apply" button....

Now go to the 'Installed Packages' section and click the 'Update All' button. A window will appear with all the available updates. Click the 'Install Accepted' button...

...and wait for the packages to be downloaded and installed. It will take a while if you have a slow bandwidth, so go see a movie or something until it finishes...

Close the update window when it's done and you will see all the installed SDKs in the 'Installed Packages' section.

And now, let's create the virtual device. Go to the 'Virtual Device' section and click the 'New' button. In the new window do the following:

- put a name to the device;
- select a target (Android system);
- put the size for the SD Card;
- add the hardware you want have in the emulator.

It should look something like this...

Click the 'Create AVD' button when you're done setting up the virtual device and wait for it to finish. It takes about 1 minute, and you'll be notified by a pop-up...

Note: In the above setup, we've created a virtual device for Android 2.0.1 with a 2 GB SD card and the following hardware components: SD Card, GPS, Accelerometer, Track-ball and touch-screen.

Now click the 'Start' button, and the 'Launch' button from the next dialog, and the emulator will start...

To make things a lot simpler let's create a desktop shortcut, so you won't have to open the terminal every time and type some command, in order to start the Android emulator. Therefore, right-click on your desktop and choose the 'Create Launcher...' option...

In the Create Launcher window, type 'Android Emulator' (without quotes) in the Name field, and paste the below line in the Command field. Optionally, you can also put a nice icon if you click the icon button on the left...

/home/YOURUSERNAME/android-sdk-linux_86/tools/emulator @softpedia

Note: Please replace YOURUSERNAME and the name of the Android Virtual Device (softpedia in our case) with your USERNAME and the name you gave to the virtual device. DO NOT REMOVE the @ sign.

Step 3 - Run applications in Android

All you have to do now is double-click that desktop shortcut you've just created. The Android emulator will start. Wait for the operating system to load...

When the Android operating system has loaded, you can install and test applications. If you are used with the Android platform, you already know how to do that, but if this is your first time... follow the next instructions.

Android 1.1

Android 1.5

Click the Browser icon, wait for the browser to load and click Menu -> Go to URL. Enter the address from where you can download an Android application with the apk extension. For example, we've easily installed Android's Fortune from Launchpad...

...all you have to do is follow the on-screen instructions!

Have fun, and do not hesitate to comment if you want to know more about Android, or if you're stuck somewhere in the tutorial.

Follow the editor on Twitter @mariusnestor

Get Softpedia News on your mobile by pointing your browser to m.softpedia.com. Bookmark and enjoy!

segunda-feira, 27 de dezembro de 2010

3D dissecado

http://www.gizmodo.com.br/conteudo/giz-explica-como-o-3d-funciona/

Published with Blogger-droid v1.6.5

quinta-feira, 23 de dezembro de 2010

Descoberta mostra que sólitons, além de ondas, são partículas complexas

Descoberta mostra que sólitons, além de ondas, são partículas complexas: "Parece um contra-senso falar em estrutura interna de uma onda. Mas os sólitons - ondas solitárias - não são solitários como se pensava; eles possuem estruturas internas intricadas. As aplicações vão da comunicação aos músculos artificiais."

sexta-feira, 10 de dezembro de 2010

Ensaios de Futurologia.

Os displays de papel em baixa qualidade serão uma realidade nos próximos cinco anos e a China usará como adesivos inteligentes em diversos produtos.

Já os displays de papel de alta qualidade (alta resolução e taxa de atualização) serão amplamente utilizados no setor de decoração a um custo bastante acessível nos próximos dez anos.

Tá registrado.

Published with Blogger-droid v1.6.5

sexta-feira, 22 de outubro de 2010

No Retreat, No Surrender !!!

Leiam em: http://www.dedalus-atlas.blogspot.com/

O que faz uma pessoa tomar uma atitude dessas ? Posso citar vários motivos, mas não aceito nenhum. Desistir nunca pode ser uma opção. Ontem estava muito triste e sentido, hoje estou tremendamente chateado com este meu amigo. Estas fases as quais estou passando são esperadas quando sofremos tamanha perda.

Valeu Sandro. Valeu nossa discussões que não levavam a nada a não ser satisfazer nossos próprios egos. Valeu pelos papos de cultura inútil, mas que fazíamos rir a beça. Tá bom ... Agora só depois !!!

Abraços !!!

HULK

quinta-feira, 21 de outubro de 2010

Remote desktop only on local network

Remote desktop only on local network: "From: http://ubuntuforums.org/showthread.php?t=1143079&page=2

had the same problem with 9.04, no advanced tab, server was not listening to the port (even default one), but I just found out how to set all this.

To use an alternative port:
run gconf-editor from terminal
in the open window go to:
/desktop/gnome/remote_access
Change the alternative_port value (mine was 5900) to the one you want
Select (check) use_alternative_port

And now, run:
/usr/lib/vino/vino-server
(you can make it run at login by addin the previous line in the: System->Preferences->Startup Applications)

Good Luck!
zensys
June 17th, 2009, 03:01 PM
I had the same problem but was able to solve it by unchecking the option 'configure network automatically to accept connections' This option requires the UPnP feature in your router to be enabled.

Surprisingly (I find computers have a strong resemblance to women) now I also have outside connectivity with the automatically accept connections option checked and UPnP disabled.

Hope this helps more than it confuses!"

segunda-feira, 4 de outubro de 2010

Método de integração

Método de integração: "

From Wikibooks, the open-content textbooks collection Calculus | Integration techniques

Jump to: navigation,

A Wikibookian suggests that Solving Integrals by Trigonometric substitution be merged into this book or chapter.

Discuss whether or not this merger should happen on the discussion page.

← Integration techniques/Partial Fraction Decomposition	Calculus	Integration techniques/Tangent Half Angle →
	Integration techniques/Trigonometric Substitution

If the integrand contains a single factor of one of the forms $\sqrt{a^2-x^2} \mbox{ or } \sqrt{a^2+x^2} \mbox{ or } \sqrt{x^2-a^2}$ we can try a trigonometric substitution.

If the integrand contains $\sqrt{a^2-x^2}$ let x = asinθ and use the identity 1 − sin²θ = cos²θ.

If the integrand contains $\sqrt{a^2+x^2}$ let x = atanθ and use the identity 1 + tan²θ = sec²θ.

If the integrand contains $\sqrt{x^2-a^2}$ let x = asecθ and use the identity sec²θ − 1 = tan²θ.

[hide]

[edit] Sine substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a piece of the form $\sqrt{a^2-x^2}$ we use the substitution

$x=a\sin \theta \quad dx=a \cos \theta d\theta$

This will transform the integrand to a trigonometic function. If the new integrand can't be integrated on sight then the tan-half-angle substitution described below will generally transform it into a more tractable algebraic integrand.

Eg, if the integrand is √(1-x²),

$\begin{matrix} \int_0^1 \sqrt{1-x^2} dx & = & \int_0^{\pi/2} \sqrt{1-\sin^2 \theta} \cos \theta \, d\theta \\ & = & \int_0^{\pi/2} \cos^2 \theta \, d\theta \\ & = & \frac{1}{2} \int_0^{\pi/2} 1+ \cos 2\theta \, d\theta \\ & = & \frac{\pi}{4} \end{matrix}$

If the integrand is √(1+x)/√(1-x), we can rewrite it as

$\sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{1+x}{1+x}\frac{1+x}{1-x}} =\frac{1+x}{\sqrt{1-x^2}}$

Then we can make the substitution

$\begin{matrix} \int_0^a \frac{1+x}{\sqrt{1-x^2}} dx & = & \int_0^\alpha \frac{1+\sin \theta}{\cos \theta} \cos \theta \, d\theta & 0 <a < 1 \\ & = & \int_0^\alpha 1+ \sin \theta \, d\theta & \alpha = \sin^{-1} a \\ & = & \alpha + \left[ - \cos \theta \right]_0^\alpha & \\ & = & \alpha + 1 - \cos \alpha & \\ & = & 1+ \sin^{-1} a - \sqrt{1-a^2} & \\ \end{matrix}$

[edit] Tangent substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

When the integrand contains a piece of the form $\sqrt{a^2+x^2}$ we use the substitution

$x = a \tan \theta \quad \sqrt{x^2+a^2} = a \sec \theta \quad dx = a \sec^2 \theta d\theta$

E.g, if the integrand is (x²+a²)^-3/2 then on making this substitution we find

$\begin{matrix} \int_0^z \left( x^2+a^2 \right)^{-\frac{3}{2}}dx & = & a^{-2} \int_0^\alpha \cos \theta \, d\theta & z>0 \\ & = & a^{-2} \left[ \sin \theta \right]_0^\alpha & \alpha = \tan^{-1} (z/a) \\ & = & a^{-2} \sin \alpha & \\ & = & a^{-2} \frac{z/a}{\sqrt{1+z^2/a^2}} & = \frac{1}{a^2} \frac{z}{\sqrt{a^2+z^2}} \\ \end{matrix}$

If the integral is

$I= \int_0^z \sqrt{x^2+a^2} \quad z>0$

then on making this substitution we find

$\begin{matrix} I & = & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & & & \alpha = \tan^{-1} (z/a) \\ & = & a^2 \int_0^\alpha \sec \theta \, d\tan \theta & & & \\ & = & a^2 [ \sec \theta \tan \theta ]_0^\alpha & - & a^2 \int_0^\alpha \sec \theta \tan^2 \theta \, d\theta & \\ & = & a^2 \sec \alpha \tan \alpha & - & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & a^2 \sec \alpha \tan \alpha & - & I & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ \end{matrix}$

After integrating by parts, and using trigonometric identities, we've ended up with an expression involving the original integral. In cases like this we must now rearrange the equation so that the original integral is on one side only

$\begin{matrix} I & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \left[ \ln \left( \sec \theta + \tan \theta \right) \right]_0^\alpha \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \ln \left( \sec \alpha + \tan \alpha \right) \\ & = & \frac{1}{2}a^2 \left( \sqrt{1+\frac{z^2}{a^2}} \right) \frac{z}{a} & + & \frac{1}{2}a^2 \ln \left( \sqrt{1+\frac{z^2}{a^2}}+\frac{z}{a} \right) \\ & = & \frac{1}{2}z\sqrt{z^2+a^2} & + & \frac{1}{2}a^2 \ln \left(\frac{z}{a} + \sqrt{1+\frac{z^2}{a^2}} \right) \\ \end{matrix}$

As we would expect from the integrand, this is approximately z²/2 for large z.

[edit] Secant substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a factor of the form $\sqrt{x^2-a^2}$ we use the substitution

$x = a \sec \theta \quad dx = a \sec \theta \tan \theta d\theta \quad \sqrt{x^2-a^2} = a \tan \theta.$

[edit] Example 1

Find $\int_1^z \frac{\sqrt{x^2-1}}{x}dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x}dx & = & \int_1^\alpha \frac{\tan \theta }{\sec \theta }\sec \theta \tan \theta \,d\theta & z>1 \\ & = & \int_0^\alpha \tan^2 \theta \, d\theta & \alpha = \sec^{-1} z \\ & = & \left[ \tan \theta -\theta \right]_0^\alpha & \tan \alpha = \sqrt{\sec^2 \alpha -1} \\ & =& \tan \alpha -\alpha & \tan \alpha = \sqrt{z^2-1} \\ & =& \sqrt{z^2-1} - \sec^{-1} z & \\ \end{matrix}$

[edit] Example 2

Find $\int_1^z \frac{\sqrt{x^2-1}}{x^2} dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & \int_1^\alpha \frac{\tan \theta}{\sec^2 \theta}\sec \theta \tan \theta \, d\theta & z>1 \\ & = & \int_0^\alpha \frac{\sin^2 \theta}{\cos \theta} d\theta & \alpha = \sec^{-1} z \\ \end{matrix}$

We can now integrate by parts

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & -\left[ \tan \theta \cos \theta \right]_0^\alpha + \int_0^\alpha \sec \theta \, d\theta \\ & = & -\sin \alpha +\left[ \ln (\sec \theta + \tan \theta ) \right]_0^\alpha \\ & = & \ln (\sec \alpha + \tan \alpha ) - \sin \alpha \\ & = & \ln (z+ \sqrt{z^2-1} ) - \frac{\sqrt{z^2-1}}{z}\\ \end{matrix}$

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sexta-feira, 14 de janeiro de 2011