1

I wanted to wrap my figure and just write text around it but this happens: enter image description here

Here's my code:

\subsection{Príklad č.2}
\textbf{Zadanie:} Na obrázku \ref{koleno} je znázornené koleno.\\

\begin{wrapfigure}{r}{0.40\textwidth}
\includegraphics[width=0.9\linewidth]{uloha2.png} 
\caption{Koleno (zdroj: Zadania bonusových príkladov )}
\label{koleno}

\end{wrapfigure}

\begin{tabular}{l l}
 $D_1 = 20 \quad mm$   & $p= 49 900 \quad Pa $\\
 $D_2 = 10 \quad mm$   & $V_k =1.8 \cdot 10^{-5} \quad m^3$\\ 
 $H = 0.1 \quad m$ & $\rho = 998 \quad kg\cdot m^{-3}$ \\
 $h = 25 \quad mm$ &  $g = 9.81 \quad m \cdot s^{-1}$\\
\end{tabular}

\\ Vypočítať: 
\begin{itemize}
   \item Silu $\Vec{F_K}$ pôsobiacu na koleno (vzťah odvodiť z vety o zmene hybnosti)
   \item Odvodiť vzťah závislosti priemeru paprsku na súradnici z. Predpokladá sa, že paprsok je celú dobu kompaktný a nerozkladá sa.
   \item Silu pôsobiacu na dosku.
\end{itemize}
\subsubsection{BR 2 (výstup z kolena) - 3 (vrchol paprsku)}
\begin{equation}
\frac{p}{\rho} + \frac{{v_2}^2}{2} + gh_2 = \frac{p_3}{\rho} + \frac{{v_3}^2}{2} + gh_3
\end{equation}
Tlak okolia je konštantný, takže mi vypadne z rovnice tlaková potencionálna energia na vrchole paprsku. Taktiež mi vypadne kinetická merná energia na vrchole paprsku, pretože tam je už všetka spotrebovaná. 
\begin{equation*}
   \frac{p}{\rho} + \frac{{v_2}^2}{2} + gh_2 = gh_3
\end{equation*}
\begin{equation*}
\frac{{v_2}^2}{2} = g (h_3 - h_2 ) - \frac{p}{\rho}
\end{equation*}
\begin{equation*}
v_2 = \sqrt{2 \cdot (g \cdot H - \frac{p}{\rho}) }
\end{equation*}

Do you know what's the problem ? thanks

2
  • 1
    Unrelated to the issue itself, but you might want to take a look at the siunitx package in order to improve the typesetting of numbers and their units.
    – leandriis
    Apr 17 at 7:38
  • thank you so much, I am still learning LaTex. Good to know !
    – Michaela
    Apr 17 at 9:26
2

When things that are not plain text are present, the automatic calculations of the figure space is often wrong. You can do two things:

  1. In this case, you have a lot of white space above. Try adding a \vspace*{-1.5cm} (or whatever) before the \includegraphics.

  2. say explicitly to wrapfig how many lines you want to skip. You have to use the optional argument:

    \begin{wrapfigure}[16]{r}
    

    adjusting the 16 more or less by try and error (LaTeX is not really though for wrapfigures, I fear).

If you had posted a complete minimal working example (MWE), I could have tried it...

3
  • Thank you, it worked. but now the rest of the page is like this. Like if there would be a figure there or something
    – Michaela
    Apr 16 at 16:53
  • oh I cant post an image to a commment. like the text is all just on the left side of the page
    – Michaela
    Apr 16 at 16:53
  • Yes, change the 16 until you got your result... it's difficult to tell if you don't prepare a MWE. Also, there is very interesting info here: tex.stackexchange.com/questions/238680/…
    – Rmano
    Apr 16 at 16:57
2

Probably this alternative, that uses two minipages instead of wrapfig also suits your needs:

enter image description here

\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{siunitx}
\usepackage{caption}
\usepackage[export]{adjustbox}
\begin{document}

\subsection{Príklad č.2}

\begin{minipage}[t]{0.55\linewidth}
\textbf{Zadanie:} Na obrázku \ref{koleno} je znázornené koleno.
\[
\begin{aligned}
 D_1 &= \SI{20}{\mm}   & p    &= \SI{49 900}{\Pa}          \\
 D_2 &= \SI{10}{\mm}   & V_k  &= \SI{1.8e-5}{\cubic\m}     \\ 
 H   &= \SI{0.1}{\m}   & \rho &= \SI{998}{\kg\per\cubic\m} \\
 h   &= \SI{25}{\mm}   & g    &= \SI{9.81}{\m\per\s}       \\
\end{aligned}
\]
 Vypočítať: 
\begin{itemize}
   \item Silu $\Vec{F_K}$ pôsobiacu na koleno (vzťah odvodiť z vety o zmene hybnosti)
   \item Odvodiť vzťah závislosti priemeru paprsku na súradnici z. Predpokladá sa, že paprsok je celú dobu kompaktný a nerozkladá sa.
   \item Silu pôsobiacu na dosku.
\end{itemize}
\end{minipage}\hfill
\begin{minipage}[t]{0.4\linewidth}
\includegraphics[width=\linewidth,valign=t]{example-image-10x16} 
\captionof{figure}{Koleno (zdroj: Zadania bonusových príkladov )}
\label{koleno}
\end{minipage}

\subsubsection{BR 2 (výstup z kolena) - 3 (vrchol paprsku)}
\begin{equation}
\frac{p}{\rho} + \frac{{v_2}^2}{2} + gh_2 = \frac{p_3}{\rho} + \frac{{v_3}^2}{2} + gh_3
\end{equation}
Tlak okolia je konštantný, takže mi vypadne z rovnice tlaková potencionálna energia na vrchole paprsku. Taktiež mi vypadne kinetická merná energia na vrchole paprsku, pretože tam je už všetka spotrebovaná. 
\begin{align*}
\frac{p}{\rho} + \frac{{v_2}^2}{2} + gh_2 &= gh_3    \\
\frac{{v_2}^2}{2} &= g (h_3 - h_2 ) - \frac{p}{\rho} \\
v_2 &= \sqrt{2 \cdot (g \cdot H - \frac{p}{\rho}) }
\end{align*}
\end{document}

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