Size image by area

Question

Sometimes I have to insert many relatively small different images, with different aspect ratio, which widths must be adjusted one by one in order to look approximately of the same size.

In such cases it would be useful to be able to size the image by area, that is choose the area and let LaTeX set image width or height so that width x height = area (keeping aspect ratio).

The ideal would be a new command built upon \includegraphics able to accept an area parameter (square root of area would be even better, because in order to get a sense of the size of the area you usually end up in calculating its square root). Such a command, given for example two images img_a (having aspect ratio W:H=2:1) and img_b (having aspect ratio W:H=2:3), with the following code

\includegraphicsbyarea[area=18]{img_a} % suppose area in square centimeters
\includegraphicsbyarea[area=18]{img_b}
\includegraphicsbyarea[area=32]{img_a}

would insert:

• img_a with width of 6 cm and height of 3 cm (6x3=18, 6:3=2:1)
• img_b with width of 3.464 cm and height of 5.196 cm (3.464x5.196=18, 3.464:5.196=2:3)
• img_a with width of 8 cm and height of 4 cm (8x4=32, 8:4=2:1)

Would it be possible?

Thanks in advance for any clue.

Answer 1

Here is an approach not redefining \includegraphics. However compared to scale option, I cut a branch because I don't have details of graphicx.sty in head, so possibly there would be a way to delegate to driver the final rescaling, which I am losing here. Ping @DavidCarlisle.

(I use xintexpr but xfp of course would do it as well as in Joseph's answer; also, up to some more cumbersome notations one could use only the macros of xintfrac, giving a tiny speed-up as expression parsing is skipped).

\documentclass{article}
\usepackage{graphicx}
\usepackage{xintexpr}
\makeatletter
\define@key{Gin}{sqrtofarea}{%
    \def\Gin@req@sizes{%
      \edef\Gin@scalex{\xinttheiexpr[5]% round fixed point to 5
                                       % fractional digits
                       \dimexpr#1\relax/
                            sqrt(\Gin@nat@height*\Gin@nat@width)
                       \relax}%
      \let\Gin@scaley\Gin@exclamation
      \Gin@req@height\Gin@scalex\Gin@nat@height
      \Gin@req@width\Gin@scalex\Gin@nat@width
      }%
  \@tempswatrue}
\makeatother
\begin{document}

\includegraphics[sqrtofarea=2cm]{example-image-a}

\includegraphics[sqrtofarea=3cm]{example-image-a}

\includegraphics[sqrtofarea=4cm]{example-image-a}

\includegraphics[sqrtofarea=5cm]{example-image-a}

\newbox\mybox
Equality only expected up to 5 digits of precision due to intrinsic
limitations of graphicx computations.

\setbox\mybox\hbox{\includegraphics[sqrtofarea=2cm]{example-image-a}}%

\xinttheiiexpr\ht\mybox*\wd\mybox\relax
?=
\xinttheiiexpr\dimexpr2cm\relax*\dimexpr2cm\relax\relax\ (4cm$^2$)

\setbox\mybox\hbox{\includegraphics[sqrtofarea=5cm]{example-image-a}}%

\xinttheiiexpr\ht\mybox*\wd\mybox\relax
?=
\xinttheiiexpr\dimexpr5cm\relax*\dimexpr5cm\relax\relax\ (25cm$^2$)
\end{document}

---

The sentence at bottom of image must be amended: in TeX all dimensions are integer multiples of 1sp. When we set the area square root as key, we automatically limit the achievable precision of the area. For example 5cm internally in TeX gives 9323399sp, hence a square equal to 86925768913201 as in image above. The previous square is 86925750266404 and the next one is 86925787560000, so they diverge in the 7th digit already and we can never overcome that possible imprecision when comparing a square with a produce height times width. Above we observe discrepancy already in 5th digit so the sentence is probably not completely wrong, but I felt I needed to add this mathematical precision.

---

Here is same with xfp (basically copied from Joseph's way of using it):

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Gin}{sqrtofarea}{%
    \def\Gin@req@sizes{%
      \edef\Gin@scalex{\fpeval{#1/sqrt(\Gin@nat@height*\Gin@nat@width)}}%
      \let\Gin@scaley\Gin@exclamation
      \Gin@req@height\Gin@scalex\Gin@nat@height
      \Gin@req@width\Gin@scalex\Gin@nat@width
      }%
  \@tempswatrue}
\makeatother
\begin{document}

\includegraphics[sqrtofarea=2cm]{example-image-a}

\includegraphics[sqrtofarea=3cm]{example-image-a}

\includegraphics[sqrtofarea=4cm]{example-image-a}

\includegraphics[sqrtofarea=5cm]{example-image-a}

\end{document}

About this:

• no wrapping of #1 in \dimexpr #1\relax needed here; xintexpr could easily be extended to recognize cm, in, pt, etc ... units so that e.g. 2cm is understood automatically, but the problem is that it would then do an exact conversion to a fractional number of sp units, whereas TeX process is more complex than simply using a proportionality factor and proceeds with rounding and truncating in various directions at various stages; so using exact conversion factor means not doing same operations as TeX itself. For this reason, no such units are defined yet in xintexpr and user must go via \dimexpr #1\relax; the xintexpr parser will apply \number to this, triggering TeX's own way to convert dimensional units.

• I am not expert in xfp so I don't know if sometimes such computation could result in a scientific notation which would break TeX later; in the xintexpr solution I applied a transformation to fixed point value with 5 fractional digits. I don't know how one does that in xfp and whether it could be needed here in some cases. In the MWE above it works fine.

Answer 2

Scaling to an area is an unusual request: normally one knows the target height or width. The following takes the image and scales such that the area is that given in the optional argument, interpreted in square centimetres:

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Garea}{area}{\def\Garea@area{#1}}
\define@key{Garea}{areaunit}{\def\Garea@unit{#1}}
\define@key{Gin}{area}{} % So we can pass through easily
\define@key{Gin}{areaunit}{}
\newcommand*{\Garea@area}{}
\newcommand*{\Garea@unit}{cm}
\newcommand{\includegraphicsbyarea}[2][]{%
  \setkeys{Garea}{#1}%
  \ifx\Garea@area\@empty
    \gdef\Garea@scale{scale = 1}%
  \else
    \begingroup
      \setbox0=\hbox{\includegraphics[#1]{#2}}%
      \xdef\Garea@scale{scale =
        \fpeval{sqrt((\Garea@area * 1 \Garea@unit * 1 \Garea@unit)
          /(\the\ht0 * \the\wd0))}}%
    \endgroup
  \fi
  \expandafter\includegraphics\expandafter[\Garea@scale,#1]{#2}%
}
\makeatother
\begin{document}
\includegraphicsbyarea[area=100]{example-image-a}
\includegraphicsbyarea[area=100,areaunit=pt]{example-image-a}
\end{document}

It still applies any options to the graphic: I've ordered such that the area scale is first, but one might want to put it last.

---

An alternative interface uses the square root of the area of the image: conveniently this has units we can work with directly

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Garea}{sqrtarea}{\def\Garea@sqrtarea{#1}}
\define@key{Gin}{sqrtarea}{}
\newcommand*{\Garea@sqrtarea}{}
\newcommand{\includegraphicsbyarea}[2][]{%
  \setkeys{Garea}{#1}%
  \ifx\Garea@diag\@empty
    \gdef\Garea@scale{scale = 1}%
  \else
    \begingroup
      \setbox0=\hbox{\includegraphics[#1]{#2}}%
      \xdef\Garea@scale{scale =
        \fpeval{\Garea@sqrtarea /(sqrt(\the\ht0 * \the\wd0))}}%
    \endgroup
  \fi
  \expandafter\includegraphics\expandafter[\Garea@scale,#1]{#2}%
}
\makeatother
\begin{document}
\includegraphicsbyarea[sqrtarea = 5cm]{example-image-a}
\includegraphicsbyarea[sqrtarea = 10cm]{example-image-a}
\end{document}

(See the answer by mmj for an alternative name for the key.)

Answer 3

Update 3

This is the last revision, I promise.

The only change is the use of pgfmath.sty in place of fp-eval.sty which, in turn, required a little special handling to avoid dimension too large errors. As well, the use of pgfmath.sty allowed for considerable simplification. The new output is identical to the old.

%===8><-----%

I was unhappy with the way that I left the \scaletoarea macro. It has been altered to be, hopefully, a little more user-friendly. I have made it possible to use the options to \includegraphics for all of the macros. Take a look at the examples.

%===8><---%

I take it that you want all of the graphics to have the same area regardless of the aspect ratio. The following code does that. There are two approaches outlined here.

The first scales subsequent graphics based on the size of the first graphic Use \fpic for the first graphic; using \spic for the subsequent graphics will size them to have the same area as the first graphic. Use the options to \fpic (the same as those to \includegraphics) to size the first graphic accordingly.

Second, if you know the target area, you can use \scaletoarea{<unitless area>}{<linear unit of area>}{<name of graphic>}. For example, if you were to scale a graphic, foo, so that it is 5 square inches in area, you would write: \scaletoarea{5}{in}{foo}. Any of the units recognized by TeX can be used.

    \documentclass{article} 
\usepackage{graphicx,calc,pgfmath} 
\usepackage[margin=0.5in]{geometry}

\newlength\fpicw
\newlength\fpich
\newlength\spicw
\newlength\spich
\newsavebox{\firstpic}
\newsavebox{\nextpics}
\newlength{\targetarea}

%% \fpic[<options to \includegraphics>]{<name of graphic>}
\newcommand{\fpic}[2][]{% First picture, use \includegraphics, and options
    \sbox{\firstpic}{\includegraphics[#1]{#2}}% 
    \settoheight{\fpich}{\usebox{\firstpic}}% 
    \settowidth{\fpicw}{\usebox{\firstpic}}%
    \usebox{\firstpic}%
} 

%% \spic[<options to \includegraphics>]{<graphic name>}
\newcommand{\spic}[2][]{% Second and succeeding pictures
    \sbox{\nextpics}{\includegraphics[#1]{#2}}%
    \settoheight{\spich}{\usebox{\nextpics}}% 
    \settowidth{\spicw}{\usebox{\nextpics}}%
    \pgfmathsetmacro{\scaling}{sqrt((\fpicw/\spicw)*(\fpich/\spich))}%
    \scalebox{\scaling}{\usebox{\nextpics}}% 
    \typeout{**The scaling (#2) = \scaling}% Comment-out if not needed
}

%% \scaletoarea[<options to \includegraphics>]{<unitless area>}{<unit of area (linear)>}{<graphic name>}
\newcommand{\scaletoarea}[4][]{%
    \pgfmathsetmacro{\mytmp}{sqrt(#2)}%
    \setlength{\targetarea}{\mytmp #3}%
    \sbox{\nextpics}{\includegraphics[#1]{#4}}%
    \settoheight{\spich}{\usebox{\nextpics}}% 
    \settowidth{\spicw}{\usebox{\nextpics}}%
    \pgfmathsetmacro{\scaling}{sqrt((\targetarea/\spicw)*(\targetarea/\spich))}%
    \scalebox{\scaling}{\usebox{\nextpics}}% 
    \typeout{++The scaling (#4) = \scaling}% Comment-out if not needed
}

%% \scaletoareab[<options to \includegraphics>]{<square root of desired area>}{<name of graphic>}
\newcommand{\scaletoareab}[3][]{%
    \setlength{\targetarea}{#2}%
    \sbox{\nextpics}{\includegraphics[#1]{#3}}%
    \settoheight{\spich}{\usebox{\nextpics}}% 
    \settowidth{\spicw}{\usebox{\nextpics}}%
    \pgfmathsetmacro{\scaling}{sqrt((\targetarea/\spicw)*(\targetarea/\spich))}%
    \scalebox{\scaling}{\usebox{\nextpics}}% 
    \message{++The scaling (#3) = \number\scaling}%
}

\begin{document} 

\fpic[width=1.25in]{Peppers}\spic{Pasta}\spic{OldImage}\spic{BethanyDrawing}

\scaletoarea{3}{in}{Peppers}\scaletoarea{12}{cm}{Pasta}\scaletoarea{2}{in}{OldImage}\scaletoarea{61}{pc}{BethanyDrawing}

\scaletoarea[width=0.5in,height=3in,keepaspectratio=false]{3}{in}{Peppers}
\scaletoareab[width=0.5in,height=3in,keepaspectratio=false]{1.732050807568877in}{Peppers}

\end{document} 

Answer 4

Here is an MCVE that fulfills your requirements provided th. I have also added commented the code and added a full explanation with regards to the logic of the program.

Sample output:

Now the explanation:

The code!:

\documentclass[12pt]{article} 
\usepackage{graphicx, color}
\usepackage[margin=0.5in]{geometry}
\usepackage{calculator}         % Very important package. Provides
                                % basic mathematical operations like
                                % addition, subtracting, multiplying
                                % dividing, rounding, truncating, and
                                % finding the GCD (greatest common divisor).
\usepackage{caption, amsmath, amssymb}

\makeatletter
\newcommand*{\getlength}[1]{\strip@pt#1}
\makeatother

% For the area, and ratio (width:height) = (\xRatio:\yRatio)
\newlength{\areaGiven}
\newlength{\xRatio}
\newlength{\yRatio}

% Accepts 4 argments: area, width ratio, height ratio, and name of image (respectively).
\newcommand*{\adjustImage}[4]{%
  \setlength\areaGiven{#1 cm}
  \setlength\xRatio{#2 cm}
  \setlength\yRatio{#3 cm}

  \noindent Given information~~~~~~~~~~~: \textbf{area} = #1cm, 
  \textbf{ratio} $(width:height) = #2:#3$.

  % Convert the unit from pt to cm. Note, 1pt = 0.0352778cm.
  % If you really care about exact numbers, 1.0000008pt = 0.035277806cm.
  % If inches is the unit to focus on, then change 0.0352778 to 0.0138889.
  \MULTIPLY{\getlength{\areaGiven}}{0.0352778}{\areaInCm}
  \MULTIPLY{\getlength{\xRatio}}{0.0352778}{\xRatioInCm}
  \MULTIPLY{\getlength{\yRatio}}{0.0352778}{\yRatioInCm}

  \noindent Converting from pt to cm : \textbf{area} = \areaInCm cm,
  \textbf{ratio} $(width:height) = \xRatioInCm cm:\yRatioInCm cm$.

  % Keep in mind that there will be rounding errors. The answer
  % we will obtain is a decimal answer. Since pixels are atomic, we need to 
  % truncate (or round) the decimal values to secure a whole number.
  % I used the TRUNCATE command, you can comment the next three line and 
  % uncomment the ROUND commands if you want to round.
  \TRUNCATE[0]{\areaInCm}{\areaInCm}
  \TRUNCATE[0]{\xRatioInCm}{\xRatioInCm}
  \TRUNCATE[0]{\yRatioInCm}{\yRatioInCm}
  % OR use round instead. It is better for images with areas 100+.
  % \ROUND[0]{\areaInCm}{\areaInCm}
  % \ROUND[0]{\xRatioInCm}{\xRatioInCm}
  % \ROUND[0]{\yRatioInCm}{\yRatioInCm}

  \noindent After truncating~~~~~~~~~~~~~: \textbf{area} = \areaInCm,
  \textbf{ratio} $(width:height) = \xRatioInCm cm:\yRatioInCm cm$.

  % The following are the steps of find the result mathematically.
  % This is explained in details at the end of the docmument.
  \MULTIPLY{\xRatioInCm}{\yRatioInCm}{\k}   % 1) finds (a*b) in ak*bk = area.
  \DIVIDE{\areaInCm}{\k}{\k}                % 2) finds (area/(a*b)) in k^2 = (area/(a*b)).
  \SQUAREROOT{\k}{\kFinal}                  % 3) finds of k in k = sqrt((area/(a*b))).
  \MULTIPLY{\xRatioInCm}{\kFinal}{\x}       % 4) multiplies the x-ratio by k.
  \MULTIPLY{\yRatioInCm}{\kFinal}{\y}       % 5) multiplies the y-ratio by k.

  \noindent \textbf{Final dimensions are}~~~: 
  \textbf{width} = \x cm, \textbf{height} = \y cm.

  \begin{center}
    \includegraphics[width = \x cm, height = \y cm]{#4} % 6*3 cm
    \captionof{figure}{
      \textbf{area} = #1cm, \textbf{ratio} 
      $(width:height) = #2:#3$ $\longrightarrow$
      \textbf{width} = \x cm, \textbf{height} = \y cm.
    }
  \end{center}  
}

\begin{document}
  \noindent
  \textcolor{red}{\textbf{Example A:}}\newline
  \adjustImage{18}{2}{1}{example-image-a} % area = 18cm, ratio (width:height) = 2:1
  \textcolor{red}{\textbf{Example B:}}\newline
  \adjustImage{18}{2}{3}{example-image-b} % area = 18cm, ratio (width:height) = 2:3
  \textcolor{red}{\textbf{Example C:}}\newline
  \adjustImage{32}{2}{1}{example-image-c} % area = 32cm, ratio (width:height) = 2:1
  % Maximum area is 575 units! 
  % \adjustImage{575}{2}{1}{TestImage.png} % area = 575cm, ratio (width:height) = 2:1

  \newpage
  \newgeometry{margin=1in}

  % Here is the explanation of the logic.
  \noindent\textbf{Explaining the logic with an example:}\newline
  The logic is a simple mathematical manipulation.\newline
  First, we need our three variables, 
  the total area we will call $area$ and the ratio $x : y$.\newline\newline
  Let $area = 18$ and the ratio $x : y = 2 : 1$.\newline
  We know that $x \times y = area$ or in other words, $2x \times y = 18$.\newline
  For simplicity, let's only use one variable, $k$. Substitute $k$ \newline
  in our previous equation for both $x$ and $y$, $2k \times k = 18$.\newline
  Solve for $k$:
  \begin{align*} 
    2k \times k &= 18\\
    2k^{2}      &= 18\\
    k^{2}       &= 9\\
    k           &= 3
  \end{align*}
  Now, we multiply $k = 3$ in our original ratio $x : y = 2 : 1$.\newline
  Hence, our desired ratio is :
  \begin{align*} 
    x : y &= 2(k) : 1(k)\\
    x : y &= 2(3) : 1(3)\\
    x : y &=    6 : 3\\
  \end{align*}\bigskip

  % Formal definition.
  \noindent\textbf{Formal definition:}\newline
  Let $area$ represent the total area of the image and the ratio $x : y$ respectively
  represent the $width : height$ of the image.\newline
  Find a value $k$ such that $area = kx \times ky$:
  \begin{align*} 
    area &= (k)x \times (k)y\\
    area &= k^{2}\times (xy)\\  % Step 1 (in the preamble) stores x*y.
    \frac{area}{xy} &= k^{2}\\  % Step 2 (in the preamble) stores area/(x*y)
    \sqrt{\frac{area}{xy}} &= k\\% Step 3 (in the preamble) stores sqrt((area/(x*y)))
  \end{align*}
  Now, multiply $k$ by the our ratio given, $x : y$. % Steps 4 and 5 (in the preamble).
  \[kx : ky\]
  Which yields to $area = kx \times ky$. \hfill $\blacksquare$
\end{document}

Answer 5

\documentclass[12pt]{article}
\usepackage{graphicx}    
\begin{document}

\begin{figure}% only needed if you want captions
\includegraphics[width=\linewidth]{example-image-a}

\bigskip
\makebox[\linewidth]{% only needed if the textwidth is smaller than the images
 \includegraphics[width=6cm,height=3cm,keepaspectratio=false]{example-image-b}
 \includegraphics[width=3.464cm,height=5.196cm,keepaspectratio=false]{example-image-b}
 \includegraphics[width=8cm,height=4cm,keepaspectratio=false]{example-image-b}%
}
\end{figure}

\end{document}