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\fancyhead[C]{ESTUDIO DE LA CINEMÁTICA DIRECTA Y CINEMÁTICA INVERSA DE UN MANIPULADOR PLANO DE 3 GRADOS DE LIBERTAD  $\bullet$ Junio 2015 $\bullet$} % Custom header text

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%----------------------------------------------------------------------------------------

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%	TITLE SECTION

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%----------------------------------------------------------------------------------------

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\title{\vspace{-15mm}\fontsize{24pt}{10pt}\selectfont\textbf{ESTUDIO DE LA CINEMÁTICA DIRECTA Y CINEMÁTICA INVERSA DE UN MANIPULADOR PLANO DE 3 GRADOS DE LIBERTAD}} % Article title

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\author{

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\large

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\textsc{Eduardo Vieira}\\% Your name 

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\normalsize Universidad Central de Venezuela \\ % Your institution

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\normalsize Facultad de Ingeniería \\

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\normalsize Escuela de Ingeniería Mecánica \\

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\normalsize eduardo.vieira@ucv.ve \\ % Your email address

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\normalsize Profesor: Arturo Gil \\

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\vspace{-5mm}

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}

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\date{}

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%----------------------------------------------------------------------------------------

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\begin{document}

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\maketitle % Insert title

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\thispagestyle{fancy} % All pages have headers and footers

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%----------------------------------------------------------------------------------------

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%	ABSTRACT

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%----------------------------------------------------------------------------------------

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%\begin{abstract}

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%\noindent \lipsum[1] % Dummy abstract text

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%\end{abstract}

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%----------------------------------------------------------------------------------------

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%	ARTICLE CONTENTS

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%----------------------------------------------------------------------------------------

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\begin{multicols}{2} % Two-column layout throughout the main article text

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\section{Introducción}

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\lettrine[nindent=0em,lines=3]{L} a cinemática del robot estudia el movimiento del mismo con respecto a un sistema de referencia sin considerar las fuerzas que intervienen. Así, la cinemática se interesa por la descripción analítica del movimiento espacial del robot como una función del tiempo, y en particular por las relaciones entre la posición y la orientación del extremo final del robot con los valores que toman sus coordenadas articulares. \\

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Existen dos problemas fundamentales a resolver en la cinemática del robot, el primero de ellos se conoce como el problema cinemático directo, y consiste en determinar cuál es la posición y orientación del extremo final del robot, con respecto a un sistema de coordenadas que se toma como referencia, conocidos los valores de las articulaciones y los parámetros geométricos de los elementos del robot; el segundo, denominado problema cinemático inverso, resuelve la configuración que debe adoptar el robot para una posición y orientación en su extremo conocidas. \\

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En el presente trabajo se estudiará el problema cinemático directo e inverso para un manipulador plano de 3 grados de libertad.  Además de desarrollará un algoritmo en python para la solución interactiva del problema.

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%------------------------------------------------

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\section{El Problema}

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\subsection{Se tiene}

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\textbf{Se tiene:} El siguiente manipulador plano \\

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\begin{center}

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 \includegraphics[width=200pt,keepaspectratio=true]{./Robot.png}

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 % P&ID_anti-suge.png: 0x0 pixel, 300dpi, 0.00x0.00 cm, bb=

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  \textit{Figura 1: Manipulador plano de 3 grados de libertad}

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\end{center}

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\subsection{Se pide}

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\begin{enumerate}

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\def\labelenumi{\arabic{enumi}.}

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\itemsep1pt\parskip0pt\parsep0pt

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\item

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  Definir la matriz de Denvit-Hatenberg.

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\def\labelenumii{\arabic{enumii}.}

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\setcounter{enumii}{1}

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\itemsep1pt\parskip0pt\parsep0pt

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\item

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  Estudiar la cinématica directa.

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\def\labelenumiii{\arabic{enumiii}.}

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\setcounter{enumiii}{2}

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\itemsep1pt\parskip0pt\parsep0pt

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\item

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   Estudiar la cinématica inversa.

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\end{enumerate}

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\subsection{Cinemática Directa}

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\subsubsection{Algoritmo de Denavit-Hartenberg}

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\begin{enumerate}

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\def\labelenumi{\arabic{enumi}.}

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\itemsep1pt\parskip0pt\parsep0pt

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\item

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  Numerar los eslabones comenzando con \(1\) (primer eslabón móvil dela

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  cadena) y acabando con \(n\) (último eslabón móvil). Se numerará como

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  eslabón \(0\) a la base fija del robot.

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\end{enumerate}

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\def\labelenumi{\arabic{enumi}.}

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\setcounter{enumi}{1}

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\itemsep1pt\parskip0pt\parsep0pt

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\item

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  Numerar cada articulación comenzando por \(1\) (la correspondiente al

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  primer grado de libertad) y acabando en \(n\).

342
\end{enumerate}

343


344
\begin{enumerate}

345
\def\labelenumi{\arabic{enumi}.}

346
\setcounter{enumi}{2}

347
\itemsep1pt\parskip0pt\parsep0pt

348
\item

349
  Localizar el eje de cada articulación. Si esta es rotativa, el eje

350
  será su propio eje de giro. Si es prismática, será el eje a lo largo

351
  del cual se produce el desplazamiento.

352
\end{enumerate}

353


354
\begin{enumerate}

355
\def\labelenumi{\arabic{enumi}.}

356
\setcounter{enumi}{3}

357
\itemsep1pt\parskip0pt\parsep0pt

358
\item

359
  Para \(i\) de \(0\) a \(n-1\), situar el eje \(Z_i\), sobre el eje de

360
  la articulación \(i+1\).

361
\end{enumerate}

362


363
\begin{enumerate}

364
\def\labelenumi{\arabic{enumi}.}

365
\setcounter{enumi}{4}

366
\itemsep1pt\parskip0pt\parsep0pt

367
\item

368
  Situar el origen del sistema de la base (\(S_0\)) en cualquier punto

369
  del eje \(Z_0\). Los ejes \(X_0\) e \(Y_0\) se situaran dé modo que

370
  formen un sistema dextrógiro con \(Z_0\).

371
\end{enumerate}

372


373
\begin{enumerate}

374
\def\labelenumi{\arabic{enumi}.}

375
\setcounter{enumi}{5}

376
\itemsep1pt\parskip0pt\parsep0pt

377
\item

378
  Para \(i\) de \(1\) a \(n-1\), situar el sistema (\(S_i\)) (solidario

379
  al eslabón \(i\)) en la intersección del eje \(Z_i\) con la línea

380
  normal común a \(Z_{i-1}\) y \(Z_i\). Si ambos ejes se cortasen se

381
  situaría (\(S_i\)) en el punto de corte. Si fuesen paralelos (\(S_i\))

382
  se situaría en la articulación \(i+1\).

383
\end{enumerate}

384


385
\begin{enumerate}

386
\def\labelenumi{\arabic{enumi}.}

387
\setcounter{enumi}{6}

388
\itemsep1pt\parskip0pt\parsep0pt

389
\item

390
  Situar $X_i$ en la línea normal común a $Z_{i-1}$ y $Z_{i}$.

391
\end{enumerate}

392


393
\begin{enumerate}

394
\def\labelenumi{\arabic{enumi}.}

395
\setcounter{enumi}{7}

396
\itemsep1pt\parskip0pt\parsep0pt

397
\item

398
  Situar $Y_i$ de modo que forme un sistema dextrógiro con $X_i$ y $Z_i$.

399
\end{enumerate}

400


401
\begin{enumerate}

402
\def\labelenumi{\arabic{enumi}.}

403
\setcounter{enumi}{8}

404
\itemsep1pt\parskip0pt\parsep0pt

405
\item

406
  Situar el sistema ($S_n$) en el extremo del robot de modo que $Z_n$ coincida

407
  con la dirección de $Z_{n-1}$ y $X_n$ sea normal a $Z_{n-1}$ y $Z_n$.

408
\end{enumerate}

409


410
\begin{enumerate}

411
\def\labelenumi{\arabic{enumi}.}

412
\setcounter{enumi}{9}

413
\itemsep1pt\parskip0pt\parsep0pt

414
\item

415
  Obtener $\theta_i$ como el ángulo que hay que girar en torno a $Z_{i-1}$ para que

416
  $X_{i-1}$ y $X_i$ queden paralelos.

417
\end{enumerate}

418


419
\begin{enumerate}

420
\def\labelenumi{\arabic{enumi}.}

421
\setcounter{enumi}{10}

422
\itemsep1pt\parskip0pt\parsep0pt

423
\item

424
  Obtener $D_i$ como la distancia, medida a lo largo de $Z_{i-1}$, que habría

425
  que desplazar ($S_{i-1}$) para que $X_i$ y $X_{i-1}$ quedasen alineados.

426
\end{enumerate}

427


428
\begin{enumerate}

429
\def\labelenumi{\arabic{enumi}.}

430
\setcounter{enumi}{11}

431
\itemsep1pt\parskip0pt\parsep0pt

432
\item

433
  Obtener $A_i$ como la distancia medida a lo largo de $X_i$ (que ahora

434
  coincidiría con $X_{i-1}$) que habría que desplazar el nuevo ($S_{i-1}$) para

435
  que su origen coincidiese con ($S_i$).

436
\end{enumerate}

437


438
\begin{enumerate}

439
\def\labelenumi{\arabic{enumi}.}

440
\setcounter{enumi}{12}

441
\itemsep1pt\parskip0pt\parsep0pt

442
\item

443
  Obtener $\alpha_i$ como el ángulo que habría que girar entorno a $X_i$ (que ahora

444
  coincidiría con $X_{i-1}$), para que el nuevo ($S_{i-1}$) coincidiese totalmente

445
  con ($S_i$).

446
\end{enumerate}

447


448
\begin{enumerate}

449
\def\labelenumi{\arabic{enumi}.}

450
\setcounter{enumi}{13}

451
\itemsep1pt\parskip0pt\parsep0pt

452
\item

453
  Obtener las matrices de transformación $^{i-1}A_i$.

454
\end{enumerate}

455


456
\begin{enumerate}

457
\def\labelenumi{\arabic{enumi}.}

458
\setcounter{enumi}{14}

459
\itemsep1pt\parskip0pt\parsep0pt

460
\item

461
  Obtener la matriz de transformación que relaciona el sistema de la

462
  base con el del extremo del robot $T = ^0A_i  ^1A_2\ldots{} ^{n-1}A_n$.

463
\end{enumerate}

464


465
\begin{enumerate}

466
\def\labelenumi{\arabic{enumi}.}

467
\setcounter{enumi}{15}

468
\itemsep1pt\parskip0pt\parsep0pt

469
\item

470
  La matriz $T$ define la orientación (submatriz de rotación) y posición

471
  (submatriz de traslación) del extremo referido ala base en función de

472
  las n coordenadas articulares.

473
\end{enumerate}

474


475
Al aplicar el algorítmo quedaría

476
\begin{center}

477
 \includegraphics[width=200pt,keepaspectratio=true]{./RobotDH5.png}

478
 % P&ID_anti-suge.png: 0x0 pixel, 300dpi, 0.00x0.00 cm, bb=

479
  \textit{Figura 2: Algoritmo de Denavit-Hartenberg aplicado al manipulador}

480
\end{center}

481
Los paŕametros para cada articulación son

482


483
\begin{center}

484
 \includegraphics[width=240pt,keepaspectratio=true]{./Tabla_DH.png}

485
 % P&ID_anti-suge.png: 0x0 pixel, 300dpi, 0.00x0.00 cm, bb=

486
  \textit{Tabla 1: Parámetros de Denavit-Hartenberg}

487
\end{center}

488


489
La matriz de transformación \(^{0}T_{1}\) será

490
\begin{equation}

491
 \left[\begin{matrix}\cos{\left (q_{1} \right )} & - \sin{\left (q_{1} \right )} & 0 & 4 \cos{\left (q_{1} \right )}\\\sin{\left (q_{1} \right )} & \cos{\left (q_{1} \right )} & 0 & 4 \sin{\left (q_{1} \right )}\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]

492
\end{equation} 

493


494
$^{1}T_{2}$

495
\begin{equation}

496
 \left[\begin{matrix}- \cos{\left (q_{2} \right )} & \sin{\left (q_{2} \right )} & 0 & - 3 \cos{\left (q_{2} \right )}\\- \sin{\left (q_{2} \right )} & - \cos{\left (q_{2} \right )} & 0 & - 3 \sin{\left (q_{2} \right )}\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]

497
\end{equation}

498


499
Y finalmente $^{2}T_{3}$

500
\begin{equation}

501
 \left[\begin{matrix}- \cos{\left (q_{3} \right )} & \sin{\left (q_{3} \right )} & 0 & - 2 \cos{\left (q_{3} \right )}\\- \sin{\left (q_{3} \right )} & - \cos{\left (q_{3} \right )} & 0 & - 2 \sin{\left (q_{3} \right )}\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]

502
\end{equation} 

503


504
La matriz de transformación que relaciona el extremo de la herramienta del robot con el sistema definido en la base $T$ será $T = ^{0}T_{1} \cdot ^{1}T_{2} \cdot ^{2}T_{3}$

505


506
\begin{equation}

507
\left[\begin{matrix}\alpha & \beta & 0 & \gamma\\\ \delta & \epsilon & 0 & \kappa \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]

508
\end{equation} 

509
Donde $\alpha=- \left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \cos{\left (q_{3} \right )} - \left(\sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} + \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \sin{\left (q_{3} \right )}$ \\

510


511
\vspace*{1\baselineskip}

512
$\beta=\left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \sin{\left (q_{3} \right )} - \left(\sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} + \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \cos{\left (q_{3} \right )}$ \\

513


514
\vspace*{1\baselineskip}

515
$\gamma=- 2 \left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \cos{\left (q_{3} \right )} - 2 \left(\sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} + \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \sin{\left (q_{3} \right )} + 3 \sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - 3 \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )} + 4 \cos{\left (q_{1} \right )}$\\

516


517
\vspace*{1\baselineskip}

518
$\delta=- \left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \sin{\left (q_{3} \right )} - \left(- \sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} - \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \cos{\left (q_{3} \right )} $\\

519


520
\vspace*{1\baselineskip}

521
$\epsilon=- \left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \cos{\left (q_{3} \right )} + \left(- \sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} - \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \sin{\left (q_{3} \right )} $\\

522


523
\vspace*{1\baselineskip}

524
$\kappa=- 2 \left(\sin{\left (q_{1} \right )} \sin{\left (q_{2} \right )} - \cos{\left (q_{1} \right )} \cos{\left (q_{2} \right )}\right) \sin{\left (q_{3} \right )} - 2 \left(- \sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} - \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}\right) \cos{\left (q_{3} \right )} - 3 \sin{\left (q_{1} \right )} \cos{\left (q_{2} \right )} + 4 \sin{\left (q_{1} \right )} - 3 \sin{\left (q_{2} \right )} \cos{\left (q_{1} \right )}$

525
\vspace*{1\baselineskip}

526
Sean $x_r$, $y_r$ y $z_r$ las coordenadas de un punto medidas desde el sistema de referencia ubicado en el extremo del robot.

527
\begin{equation}

528
 P_r = \left[\begin{matrix}x_{r}\\y_{r}\\z_{r}\\1\end{matrix}\right]

529
\end{equation} 

530
Entonces, el mismo punto medido en el sistema de referencia de la base será $T \cdot P_r$

531


532
 

533
$x = x_{r} (- (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \cos{ (q_{3}  )} \\ - (\sin{ (q_{1}  )} \cos{ (q_{2}  )} + \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \sin{ (q_{3}  )}) + y_{r} ((\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \sin{ (q_{3}  )} - (\sin{ (q_{1}  )} \cos{ (q_{2}  )} + \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \cos{ (q_{3}  )}) - 2 (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \cos{ (q_{3}  )} - 2 (\sin{ (q_{1}  )} \cos{ (q_{2}  )} + \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \sin{ (q_{3}  )} + 3 \sin{ (q_{1}  )} \sin{ (q_{2}  )} - 3 \cos{ (q_{1}  )} \cos{ (q_{2}  )} + 4 \cos{ (q_{1}  )}$\\

534


535
\vspace*{1\baselineskip}

536
$y = x_{r} (- (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \sin{ (q_{3}  )} - (- \sin{ (q_{1}  )} \cos{ (q_{2}  )} - \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \cos{ (q_{3}  )}) + y_{r} (- (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \cos{ (q_{3}  )} + (- \sin{ (q_{1}  )} \cos{ (q_{2}  )} - \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \sin{ (q_{3}  )}) - 2 (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \sin{ (q_{3}  )} - 2 (- \sin{ (q_{1}  )} \cos{ (q_{2}  )} - \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \cos{ (q_{3}  )} - 3 \sin{ (q_{1}  )} \cos{ (q_{2}  )} + 4 \sin{ (q_{1}  )} - 3 \sin{ (q_{2}  )} \cos{ (q_{1}  )}$

537
\vspace*{1\baselineskip}

538


539
Si colocamos el sistema de referencia en la base ($x_r$ y $y_r$ iguales a cero) el punto que determina la ubicación de la

540
herramienta se obtiene de la siguiente manera \\

541
$x=- 2 (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \cos{ (q_{3}  )} - 2 (\sin{ (q_{1}  )} \cos{ (q_{2}  )} + \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \sin{ (q_{3}  )} + 3 \sin{ (q_{1}  )} \sin{ (q_{2}  )} - 3 \cos{ (q_{1}  )} \cos{ (q_{2}  )} + 4 \cos{ (q_{1}  )}$

542


543
\vspace*{1\baselineskip}

544
$y=- 2 (\sin{ (q_{1}  )} \sin{ (q_{2}  )} - \cos{ (q_{1}  )} \cos{ (q_{2}  )}) \sin{ (q_{3}  )} - 2 (- \sin{ (q_{1}  )} \cos{ (q_{2}  )} - \sin{ (q_{2}  )} \cos{ (q_{1}  )}) \cos{ (q_{3}  )} - 3 \sin{ (q_{1}  )} \cos{ (q_{2}  )} + 4 \sin{ (q_{1}  )} - 3 \sin{ (q_{2}  )} \cos{ (q_{1}  )}$

545


546
\end{multicols}

547
\subsubsection{Código en python}

548
En la siguiente entrada determinaremos la posición de la herramienta modificando los valores de 

549
$q_1$, $q_2$ y $q_3$ en los widgets interactivos

550


551
    \begin{Verbatim}[commandchars=\\\{\}]

552
{\color{incolor}In [{\color{incolor}1}]:} \PY{k+kn}{import} \PY{n+nn}{matplotlib.pyplot} \PY{k+kn}{as} \PY{n+nn}{plt}

553
         \PY{c}{\PYZsh{}plt.style.use(\PYZsq{}bmh\PYZsq{})}

554
         \PY{k+kn}{from} \PY{n+nn}{numpy} \PY{k+kn}{import} \PY{n}{sin}\PY{p}{,} \PY{n}{cos}\PY{p}{,} \PY{n}{pi}\PY{p}{,} \PY{n}{arctan2}

555
         \PY{k+kn}{from} \PY{n+nn}{IPython.html.widgets} \PY{k+kn}{import} \PY{n}{interact}\PY{p}{,} \PY{n}{FloatSlider}\PY{p}{,} \PY{n}{interactive}

556
         \PY{k+kn}{from} \PY{n+nn}{IPython.display} \PY{k+kn}{import} \PY{n}{display}

557
         \PY{o}{\PYZpc{}}\PY{k}{matplotlib} inline

558
         \PY{n}{fact} \PY{o}{=} \PY{l+m+mi}{180} \PY{o}{/} \PY{n}{pi}

559
         \PY{k}{def} \PY{n+nf}{punto}\PY{p}{(}\PY{n}{q1}\PY{p}{,}\PY{n}{q2}\PY{p}{,}\PY{n}{q3}\PY{p}{)}\PY{p}{:}

560
             \PY{n}{x} \PY{o}{=} \PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{p}{(}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q3}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2} \PY{o}{*} \PYZbs{}

561
             \PY{p}{(}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{+} \PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q3}\PY{p}{)} \PY{o}{+} \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{\PYZhy{}} \PYZbs{}

562
             \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{+} \PY{l+m+mi}{4}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

563
             

564
             \PY{n}{y} \PY{o}{=} \PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{p}{(}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q3}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2} \PY{o}{*} \PYZbs{}

565
             \PY{p}{(}\PY{o}{\PYZhy{}}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q3}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)} \PY{o}{+} \PYZbs{}

566
             \PY{l+m+mi}{4}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

567
             

568
             \PY{n}{x1} \PY{o}{=} \PY{l+m+mi}{4}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

569
             \PY{n}{y1} \PY{o}{=} \PY{l+m+mi}{4}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

570
             \PY{n}{x2} \PY{o}{=} \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{4}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

571
             \PY{n}{y2} \PY{o}{=} \PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{4}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{sin}\PY{p}{(}\PY{n}{q2}\PY{p}{)}\PY{o}{*}\PY{n}{cos}\PY{p}{(}\PY{n}{q1}\PY{p}{)}

572
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{n}{x}\PY{p}{,}\PY{n}{y}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{r+}\PY{l+s}{\PYZsq{}}\PY{p}{)}

573
             \PY{n}{plt}\PY{o}{.}\PY{n}{ylim}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mf}{9.5}\PY{p}{,}\PY{l+m+mf}{9.5}\PY{p}{)}

574
             \PY{n}{plt}\PY{o}{.}\PY{n}{xlim}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mf}{9.5}\PY{p}{,}\PY{l+m+mf}{9.5}\PY{p}{)}

575
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{x1}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{y1}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{k\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

576
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{n}{x1}\PY{p}{,}\PY{n}{x2}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{n}{y1}\PY{p}{,}\PY{n}{y2}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{r\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

577
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{n}{x2}\PY{p}{,}\PY{n}{x}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{n}{y2}\PY{p}{,}\PY{n}{y}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{b\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

578
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{k\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

579
             \PY{n}{plt}\PY{o}{.}\PY{n}{grid}\PY{p}{(}\PY{n+nb+bp}{True}\PY{p}{)}

580
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{x [m] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{x}

581
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{y [m] = }\PY{l+s}{\PYZdq{}}\PY{p}{,}\PY{n}{y}

582
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{Ángulo [deg] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{fact} \PY{o}{*} \PY{n}{arctan2}\PY{p}{(}\PY{p}{(}\PY{n}{y}\PY{o}{\PYZhy{}}\PY{n}{y2}\PY{p}{)}\PY{p}{,}\PY{p}{(}\PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{x2}\PY{p}{)}\PY{p}{)}

583
         \PY{n}{q1\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{pi}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,} \PYZbs{}

584
                                 \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{Angulo \PYZdl{}q\PYZus{}1\PYZdl{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

585
         \PY{n}{q2\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{n}{pi}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,} \PYZbs{}

586
                                 \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{Angulo \PYZdl{}q\PYZus{}2\PYZdl{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

587
         \PY{n}{q3\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{pi}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,} \PYZbs{}

588
                                 \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{Angulo \PYZdl{}q\PYZus{}3\PYZdl{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

589
         \PY{n}{w}\PY{o}{=}\PY{n}{interactive}\PY{p}{(}\PY{n}{punto}\PY{p}{,}\PY{n}{q1}\PY{o}{=}\PY{n}{q1\PYZus{}slider}\PY{p}{,}\PY{n}{q2}\PY{o}{=}\PY{n}{q2\PYZus{}slider}\PY{p}{,}\PY{n}{q3}\PY{o}{=}\PY{n}{q3\PYZus{}slider}\PY{p}{)}

590
         \PY{n}{display}\PY{p}{(}\PY{n}{w}\PY{p}{)}

591
\end{Verbatim}

592


593


594
\begin{center}

595
 \includegraphics[width=500pt,keepaspectratio=True]{./cap_cin_dir.png}

596
 % P&ID_anti-suge.png: 0x0 pixel, 300dpi, 0.00x0.00 cm, bb=

597
  \textit{Figura 3: Captura de la ejecución del código en un Notebook de IPython}

598
\end{center}

599


600


601


602
\begin{multicols}{2}

603


604
\subsection{Cinemática Inversa}

605
Realizaremos la cinemática inversa mediate métodos geométricos. Sea Pher el punto desado de la herramienta con coordenadas

606
$x_h$ y $y_h$ ($z_h=0$ al tratarse de un robot plano) y formando un ángulo $\theta_h$ con la horizontal. \\

607
Por trigonometría las coordenadas de la articulación 3 serán

608
\(x_3 = x_h - l_3 * cos(\theta)\) y \(y_3 = y_h - l_3 * sin(\theta)\).

609
Podemos obtener \(q_1\) y \(q_2\) con

610
\(q_2 = arctg({{\pm \sqrt{1-cos^2 (q_2)}} \over {cos (q_2)}})\) donde

611
\(cos (q_2) = {{x_3^2 + y_3^2 - l_1^2 - l_2^2} \over {2 l_1 l_2}}\) y

612
\(q_1 = arctg({{y_3}\over{\pm x_3}})-arctg({{l_2 sin(q_2)} \over {l_1 + l_2 cos(q_2)}})\)

613
(Barrientos, Fundamentos de Robótica, 2007).

614


615
\end{multicols}

616
\subsubsection{Código en python}

617


618
\begin{Verbatim}[commandchars=\\\{\}]

619
{\color{incolor}In [{\color{incolor}2}]:} \PY{k+kn}{from} \PY{n+nn}{numpy} \PY{k+kn}{import} \PY{n}{sqrt}\PY{p}{,} \PY{n}{arcsin}

620
         \PY{k}{def} \PY{n+nf}{cin\PYZus{}inv}\PY{p}{(}\PY{n}{x}\PY{p}{,}\PY{n}{y}\PY{p}{,}\PY{n}{theta}\PY{p}{)}\PY{p}{:}

621
             \PY{n}{theta} \PY{o}{=} \PY{n}{theta} \PY{o}{/} \PY{n}{fact}

622
             \PY{n}{x\PYZus{}3} \PY{o}{=} \PY{n}{x} \PY{o}{\PYZhy{}} \PY{n}{l\PYZus{}3} \PY{o}{*} \PY{n}{cos}\PY{p}{(}\PY{n}{theta}\PY{p}{)}

623
             \PY{n}{y\PYZus{}3} \PY{o}{=} \PY{n}{y} \PY{o}{\PYZhy{}} \PY{n}{l\PYZus{}3} \PY{o}{*} \PY{n}{sin}\PY{p}{(}\PY{n}{theta}\PY{p}{)}

624
             \PY{n}{cos\PYZus{}q\PYZus{}2} \PY{o}{=} \PY{p}{(}\PY{n}{x\PYZus{}3}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{o}{+} \PY{n}{y\PYZus{}3}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{o}{\PYZhy{}} \PY{n}{l\PYZus{}1}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{o}{\PYZhy{}} \PY{n}{l\PYZus{}2}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)} \PY{o}{/} \PY{p}{(}\PY{l+m+mi}{2} \PY{o}{*} \PY{n}{l\PYZus{}1} \PY{o}{*} \PY{n}{l\PYZus{}2}\PY{p}{)}

625
             \PY{n}{q\PYZus{}2} \PY{o}{=} \PY{n}{arctan2}\PY{p}{(}\PY{n}{sqrt}\PY{p}{(}\PY{l+m+mi}{1} \PY{o}{\PYZhy{}} \PY{p}{(}\PY{n}{cos\PYZus{}q\PYZus{}2}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{,}\PY{n}{cos\PYZus{}q\PYZus{}2}\PY{p}{)}

626
             \PY{n}{q\PYZus{}1} \PY{o}{=} \PY{n}{arctan2}\PY{p}{(}\PY{n}{y\PYZus{}3}\PY{p}{,}\PY{n}{x\PYZus{}3}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n}{arctan2}\PY{p}{(}\PY{p}{(}\PY{n}{l\PYZus{}2} \PY{o}{*} \PY{n}{sin}\PY{p}{(}\PY{n}{q\PYZus{}2}\PY{p}{)}\PY{p}{)}\PY{p}{,}\PY{p}{(}\PY{n}{l\PYZus{}1} \PY{o}{+} \PY{n}{l\PYZus{}2} \PY{o}{*} \PY{n}{cos\PYZus{}q\PYZus{}2}\PY{p}{)}\PY{p}{)}

627
             \PY{n}{l} \PY{o}{=} \PY{n}{sqrt}\PY{p}{(}\PY{n}{l\PYZus{}1}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{o}{+} \PY{n}{l\PYZus{}2}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2} \PY{o}{*} \PY{n}{l\PYZus{}1} \PY{o}{*} \PY{n}{l\PYZus{}2} \PY{o}{*} \PY{n}{cos\PYZus{}q\PYZus{}2}\PY{p}{)}

628
             \PY{n}{alpha} \PY{o}{=} \PY{n}{arcsin}\PY{p}{(}\PY{n}{l\PYZus{}1} \PY{o}{*} \PY{n}{sin}\PY{p}{(}\PY{n}{q\PYZus{}2}\PY{p}{)} \PY{o}{/} \PY{n}{l}\PY{p}{)}

629
             \PY{n}{beta} \PY{o}{=} \PY{n}{arctan2}\PY{p}{(}\PY{n}{y\PYZus{}3}\PY{p}{,}\PY{n}{x\PYZus{}3}\PY{p}{)}

630
             \PY{n}{k} \PY{o}{=} \PY{n}{pi} \PY{o}{\PYZhy{}} \PY{n}{theta}

631
             \PY{n}{q\PYZus{}3} \PY{o}{=} \PY{p}{(}\PY{n}{alpha} \PY{o}{+} \PY{n}{k} \PY{o}{+} \PY{n}{beta}\PY{p}{)}

632
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{n}{x}\PY{p}{,} \PY{n}{y}\PY{p}{,} \PY{l+s}{\PYZsq{}}\PY{l+s}{r+}\PY{l+s}{\PYZsq{}}\PY{p}{)}

633
             \PY{n}{plt}\PY{o}{.}\PY{n}{ylim}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mf}{9.5}\PY{p}{,} \PY{l+m+mf}{9.5}\PY{p}{)}

634
             \PY{n}{plt}\PY{o}{.}\PY{n}{xlim}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mf}{9.5}\PY{p}{,} \PY{l+m+mf}{9.5}\PY{p}{)}

635
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{k\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

636
             \PY{n}{x\PYZus{}2} \PY{o}{=} \PY{n}{l\PYZus{}1} \PY{o}{*} \PY{n}{cos}\PY{p}{(}\PY{n}{q\PYZus{}1}\PY{p}{)}

637
             \PY{n}{y\PYZus{}2} \PY{o}{=} \PY{n}{l\PYZus{}1} \PY{o}{*} \PY{n}{sin}\PY{p}{(}\PY{n}{q\PYZus{}1}\PY{p}{)}

638
             \PY{n}{x\PYZus{}1} \PY{o}{=} \PY{l+m+mi}{0}

639
             \PY{n}{y\PYZus{}1} \PY{o}{=} \PY{l+m+mi}{0}

640
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{n}{x\PYZus{}1}\PY{p}{,}\PY{n}{x\PYZus{}2}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{n}{y\PYZus{}1}\PY{p}{,}\PY{n}{y\PYZus{}2}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{r\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

641
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{n}{x\PYZus{}2}\PY{p}{,}\PY{n}{x\PYZus{}3}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{n}{y\PYZus{}2}\PY{p}{,}\PY{n}{y\PYZus{}3}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{b\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

642
             \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{p}{[}\PY{n}{x\PYZus{}3}\PY{p}{,}\PY{n}{x}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{n}{y\PYZus{}3}\PY{p}{,}\PY{n}{y}\PY{p}{]}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{k\PYZhy{}}\PY{l+s}{\PYZsq{}}\PY{p}{)}

643
             \PY{n}{plt}\PY{o}{.}\PY{n}{grid}\PY{p}{(}\PY{n+nb+bp}{True}\PY{p}{)}

644
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{q1 [deg] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{q\PYZus{}1} \PY{o}{*} \PY{n}{fact}

645
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{q2 [deg] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{q\PYZus{}2} \PY{o}{*} \PY{n}{fact}

646
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{q3 [deg] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{q\PYZus{}3} \PY{o}{*} \PY{n}{fact}

647
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{l1 [m] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{sqrt}\PY{p}{(}\PY{p}{(}\PY{n}{x\PYZus{}2}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{+}\PY{p}{(}\PY{n}{y\PYZus{}2}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{,} \PY{n}{l\PYZus{}1}

648
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{l2 [m] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{sqrt}\PY{p}{(}\PY{p}{(}\PY{p}{(}\PY{n}{x\PYZus{}2}\PY{o}{\PYZhy{}}\PY{n}{x\PYZus{}3}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{+}\PY{p}{(}\PY{p}{(}\PY{n}{y\PYZus{}2}\PY{o}{\PYZhy{}}\PY{n}{y\PYZus{}3}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{,} \PY{n}{l\PYZus{}2}

649
             \PY{k}{print} \PY{l+s}{\PYZdq{}}\PY{l+s}{l3 [m] = }\PY{l+s}{\PYZdq{}}\PY{p}{,} \PY{n}{sqrt}\PY{p}{(}\PY{p}{(}\PY{p}{(}\PY{n}{x\PYZus{}3}\PY{o}{\PYZhy{}}\PY{n}{x}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{+}\PY{p}{(}\PY{p}{(}\PY{n}{y\PYZus{}3}\PY{o}{\PYZhy{}}\PY{n}{y}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{,} \PY{n}{l\PYZus{}3}

650
         \PY{n}{x\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{9}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{9}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,} \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{X}\PY{l+s}{\PYZsq{}}\PY{p}{)}

651
         \PY{n}{y\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{9}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{9}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{l+m+mi}{4}\PY{p}{,} \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{Y}\PY{l+s}{\PYZsq{}}\PY{p}{)}

652
         \PY{n}{theta\PYZus{}slider} \PY{o}{=} \PY{n}{FloatSlider}\PY{p}{(}\PY{n+nb}{min}\PY{o}{=}\PY{l+m+mi}{0}\PY{p}{,} \PY{n+nb}{max}\PY{o}{=}\PY{l+m+mi}{360}\PY{p}{,} \PY{n}{step}\PY{o}{=}\PY{l+m+mf}{0.1}\PY{p}{,} \PY{n}{value}\PY{o}{=}\PY{l+m+mi}{200}\PY{p}{,} \PY{n}{description}\PY{o}{=}\PY{l+s}{\PYZsq{}}\PY{l+s}{Theta}\PY{l+s}{\PYZsq{}}\PY{p}{)}

653
         \PY{n}{w}\PY{o}{=}\PY{n}{interactive}\PY{p}{(}\PY{n}{cin\PYZus{}inv}\PY{p}{,}\PY{n}{x}\PY{o}{=}\PY{n}{x\PYZus{}slider}\PY{p}{,}\PY{n}{y}\PY{o}{=}\PY{n}{y\PYZus{}slider}\PY{p}{,}\PY{n}{theta}\PY{o}{=}\PY{n}{theta\PYZus{}slider}\PY{p}{)}

654
         \PY{n}{display}\PY{p}{(}\PY{n}{w}\PY{p}{)}

655
\end{Verbatim}

656


657
\begin{center}

658
 \includegraphics[width=480pt,keepaspectratio=True]{./cap_cin_inv.png}

659
  \textit{Figura 4: Captura de la ejecución del código en un Notebook de IPython}

660
\end{center}

661


662


663
\end{document}

664


665