Ttdfjt

I want to make a lens aperture with different f numbers(sizes of the holes). I started with this code but it wont make it look like a complete aperture. Appreciate any help.

documentclass[aspectratio=43]beamer
usepackagetikz


begindocument

beginframeAperture image

begintikzpicture
clip (0,0) circle(1);
draw[thick] (0,0) circle(1);
foreach r in 0,40,...,360
filldraw[fill = black!80,draw = white,thick, rotate = r] (-3,-1) rectangle (-0.5,1);
endtikzpicture
endframe
enddocument

The current output is looking like this.

You could draw triangles instead of rectangles:

documentclassarticle
usepackagetikz
begindocument
begintikzpicture
 clip (0,0) circle (1);
 foreach r in 0,40,...,320
 fill[black!80,rotate=r] (-0.5,1) -- (-0.5,-0.13) --++ (130:1) -- cycle;
endtikzpicture
enddocument

A little late, but the following defines a pic that is more or less configurable (and I promptly used it to create an animation). The calculations are most likely inefficient and the overall code doesn't look as good as those posted by the usual TikZ experts.

documentclass[tikz]standalone

tikzset
 
 ,aperture segments/.initial = 6
 ,aperture radius/.initial = 3
 ,aperture closed/.initial = .5
 ,aperture/.pic=
 beginscope
 pgfkeysgetvalue/tikz/aperture segmentssegments
 pgfkeysgetvalue/tikz/aperture radiusrad
 pgfkeysgetvalue/tikz/aperture closedclosed
 pgfmathsetmacroang360/segments
 pgfmathsetmacroendang360-ang
 pgfmathsetmacroalphtild(180-ang)/2
 pgfmathsetmacrorp(1-closed)*rad
 pgfmathsetmacroccrad*sqrt(2*(1-cos(ang)))
 pgfmathsetmacrobpsqrt(rad*rad+rp*rp-2*rad*rp*cos(ang))
 pgfmathsetmacroalphprimasin(rp/bp*sin(ang))
 pgfmathsetmacroalphalphtild-alphprim
 pgfmathsetmacrobet180-ang-alph
 pgfmathsetmacrobbcc*sin(bet)/sin(ang)
 foreach r in 0,ang,...,endang
 
 filldraw[fill = black!80, draw = white, thick, rotate = r]
 (0:rad) ++(180-alphprim:bb) -- (0:rad)
 arc[start angle=0, end angle=ang, radius=rad]
 -- cycle;
 %
 endscope%
 
 

begindocument
foreachx in 0,0.025,...,1
 
 begintikzpicture
 pic [aperture segments=9, aperture closed=x] aperture;
 endtikzpicture
 
foreachx in 1,0.975,...,0
 
 begintikzpicture
 pic [aperture segments=9, aperture closed=x] aperture;
 endtikzpicture
 
enddocument

A small document which describes what is calculated above and why:

documentclass[]article

titleAperture drawing
authorSkillmon
date

usepackagetikz
usepackage[]amsmath
usepackagearray
usepackagecollcell
usepackagebooktabs
usepackagesiunitx

newcolumntypemathcol[1]>startmath#1<endmath
letstartmath(
letendmath)
newcolumntypemacrocol[1]>collectcellmakemacroname#1<endcollectcell
newcommandmakemacroname[1]
 %
 textttexpandafterstringcsname #1endcsname%
 

begindocument
maketitle

The geometry we use is shown in figure~reffig:geom.
In the equations below variables correspond to the names in the used
TitextitkZ code to draw the aperture. The correspondences are shown in
table~reftab:corres. In the following I don't care about the sign of the
angles denoted with $angle P_1P_2P_3$ and always only care for their absolute
value, so $angle P_1P_2P_3$ might actually be $angle P_3P_2P_1$.

beginfigure
 centering
 begintikzpicture
 defsegments9
 defrad3
 defclosed0.3
 pgfmathsetmacroang360/segments
 pgfmathsetmacroendang360-ang
 pgfmathsetmacroalphtild(180-ang)/2
 pgfmathsetmacrorp(1-closed)*rad
 pgfmathsetmacroccrad*sqrt(2*(1-cos(ang)))
 pgfmathsetmacrobpsqrt(rad*rad+rp*rp-2*rad*rp*cos(ang))
 pgfmathsetmacroalphprimasin(rp/bp*sin(ang))
 pgfmathsetmacroalphalphtild-alphprim
 pgfmathsetmacrobet180-ang-alph
 pgfmathsetmacrobbcc*sin(bet)/sin(ang)
 draw 
 (0,0) coordinate(O) circle [radius=1.5pt] node [below]$O$
 (0:rad) coordinate(A) circle [radius=1.5pt] node [below right]$A$
 ++(180-alphprim:bb)
 coordinate(C) circle [radius=1.5pt] node [above left]$C$
 (ang:rad) coordinate(B) circle [radius=1.5pt] node [above right]$B$
 (ang:rp) coordinate(D) circle [radius=1.5pt] node [below]$D$
 ;
 draw[blue]
 (A) arc[start angle=0, end angle=40, radius=3cm] -- (C) -- cycle;
 draw[gray, dashed]
 (O) -- (A)
 (A) -- (B)
 (O) -- (B)
 ;
 endtikzpicture
 caption
 %
 The geometry in which we want to calculate the position of point $C$%
 labelfig:geom%
 
endfigure

begintable
 centering
 begintabularmathcolc macrocoll mathcoll
 toprule
 multicolumn1lVariable & multicolumn1lMacro name
 & multicolumn1lMeaning \
 midrule
 barp & closed & overlineDB/overlineOB \
 n & segments \
 r & rad & overlineOA=overlineOB \
 gamma & ang & angle AOB \
 tildealpha & alphtild & angle OAB \
 r_p & rp & overlineOD \
 c & cc & overlineBA \
 b_p & bp & overlineAD \
 alpha' & alphprim & angle OAC \
 alpha & alph & angle CAB \
 beta & bet & angle ABC \
 b & bb & overlineAC \
 bottomrule
 endtabular
 caption
 %
 Variable-Macro-Correspondence and their geometrical meaning in
 figure~reffig:geom.%
 labeltab:corres%
 
endtable

The variables we know the values of are $n$, $r$, and $barp$. $barp$ is in
the range $[0,1]$ and describes how closed the aperture is. $n$ is the number of
aperture segments and $r$ is the outer radius of the aperture. From $n$ we get
the angle $gamma = angle AOB$ straight forward:

beginequation
 gamma = fracang360n
endequation
Also relatively easy to calculate are the value of $p$ and $r_p$:
beginalign
 p &= 1 - barp \
 r_p &= rp
endalign
The next thing we want to know is the angle $angle ACB$. We know how many edges
the polygon of the aperture will have ($n$), so we know the sum
of internal angles and $angle ACB$ is the adjacent angle of one of the internal
angles:
beginequation
 angle ACB = ang180 - ang180 frac(n - 2)n
 = ang180 cdot (1 - 1 + frac2n) = fracang360n = gamma
endequation
The distances $c$ and $b_p$ can be calculated using the law of cosines:
beginalign
 c &= sqrt2r^2 - 2r^2cos gamma = r sqrt2(1-cosgamma) \
 b_p &= sqrtr^2 + r_p^2 - 2rr_pcosgamma labeleq:bp
endalign
We could further simplify eq.~refeq:bp, but this should suffice.
$tildealpha$ can be calculated with the sum of internal angles in the
triangle $OAB$ since it is isosceles. With the sine theorem we can calculate
$alpha'$ and therefore $alpha$ and $beta$:
beginalign
 tildealpha &= fracang180-gamma2 \
 alpha' &= arcsin ( fracr_pb_psingamma ) \
 alpha &= tildealpha - alpha' \
 beta &= ang180 - gamma - alpha
endalign
Using the sine theorem again we calculate the value of $b$ and with the position
of $A$ and the angle $alpha'$ we get the position of $C$:
beginequation
 b = c fracsinbetasingamma
endequation
In TitextitkZ the point $C$ is now positioned at
verb|(0:rad) ++(180-alphprim:bb)|. Now we can draw our aperture segments.
enddocument