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2

I don't understand what's going on here. If the domain of the parametric parameter t is 0:100, then the function sin(2pi*t) should oscillate about 100 times, but it only oscillates a couple. What am I missing?

documentclass{article}

usepackage{tikz}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=40, smooth, variable=t] 

    plot ({t/100}, {sin(2*pi*t)*.5+.5});

  end{scope}

end{tikzpicture}

end{document}

tikz-pgf plot domain

share|improve this question

asked 4 hours ago

argentum2fargentum2f

1336

add a comment | 

2

I don't understand what's going on here. If the domain of the parametric parameter t is 0:100, then the function sin(2pi*t) should oscillate about 100 times, but it only oscillates a couple. What am I missing?

documentclass{article}

usepackage{tikz}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=40, smooth, variable=t] 

    plot ({t/100}, {sin(2*pi*t)*.5+.5});

  end{scope}

end{tikzpicture}

end{document}

tikz-pgf plot domain

share|improve this question

asked 4 hours ago

argentum2fargentum2f

1336

add a comment | 

I don't understand what's going on here. If the domain of the parametric parameter t is 0:100, then the function sin(2pi*t) should oscillate about 100 times, but it only oscillates a couple. What am I missing?

documentclass{article}

usepackage{tikz}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=40, smooth, variable=t] 

    plot ({t/100}, {sin(2*pi*t)*.5+.5});

  end{scope}

end{tikzpicture}

end{document}

tikz-pgf plot domain

share|improve this question

asked 4 hours ago

argentum2fargentum2f

1336

documentclass{article}

usepackage{tikz}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=40, smooth, variable=t] 

    plot ({t/100}, {sin(2*pi*t)*.5+.5});

  end{scope}

end{tikzpicture}

end{document}

share|improve this question

asked 4 hours ago

argentum2fargentum2f

1336

share|improve this question

asked 4 hours ago

argentum2fargentum2f

1336

asked 4 hours ago

argentum2fargentum2f

1336

asked 4 hours ago

argentum2fargentum2f

1336

1 Answer
1

active

oldest

votes

4

You got the domain right. The problem is that the trigonometric functions in TikZ are (oddly, in my opinion) in degrees by default. So 2*pi*t with t between 0 and 100 will give you the sine function (in degrees) between 0 and 628.3 degrees (which is about 10.96 radians) which almost 1.75 periods of the function. This is exactly what you see there: one full period and 3/4 of another.

You can tell TikZ to use radians by appending an r to the argument or using the rad function (see page 1005 of the TikZ-PGF manual, section 93.3.4 “Trigonometric functions”). I also added the FPU to allow the domain up to 100 and increased the number of samples to 400 following Kpym's suggestion (notice the aliasing):

documentclass{article}

usepackage{tikz}

usetikzlibrary{fpu}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=400, smooth, variable=t]

    plot ({t/100}, {sin(2*pi*t r)*.5+.5});

  end{scope}%                   ^

end{tikzpicture}

end{document}

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  with samples=400,smooth it looks better, but this is not the question ;)
  
  – Kpym
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kpym Certainly! I just removed the smooth to show that 100 cycles with 40 samples won't do (unless OP wants to show the aliasing). Thanks anyway :)
  
  – Phelype Oleinik
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Doh! Degrees... Yeah, don't know why that didn't occur to me. I'm just so used to thinking in radians. Usually trigonometric functions are defined assuming radians, but I guess it makes sense since everywhere else in latex assumes angles are degrees. I actually did try and turn up the samples at first - thought it was just aliasing.
  
  – argentum2f
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @argentum2f Yes, as I said “oddly”. All programming languages I know (which aren't that many, but...) have a sin function for argument radians and sind function for argument in degrees. Anyhow, Till must have had his reasons :)
  
  – Phelype Oleinik
  2 hours ago
  

  
  
  
  

add a comment | 

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4

You got the domain right. The problem is that the trigonometric functions in TikZ are (oddly, in my opinion) in degrees by default. So 2*pi*t with t between 0 and 100 will give you the sine function (in degrees) between 0 and 628.3 degrees (which is about 10.96 radians) which almost 1.75 periods of the function. This is exactly what you see there: one full period and 3/4 of another.

You can tell TikZ to use radians by appending an r to the argument or using the rad function (see page 1005 of the TikZ-PGF manual, section 93.3.4 “Trigonometric functions”). I also added the FPU to allow the domain up to 100 and increased the number of samples to 400 following Kpym's suggestion (notice the aliasing):

documentclass{article}

usepackage{tikz}

usetikzlibrary{fpu}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=400, smooth, variable=t]

    plot ({t/100}, {sin(2*pi*t r)*.5+.5});

  end{scope}%                   ^

end{tikzpicture}

end{document}

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  with samples=400,smooth it looks better, but this is not the question ;)
  
  – Kpym
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kpym Certainly! I just removed the smooth to show that 100 cycles with 40 samples won't do (unless OP wants to show the aliasing). Thanks anyway :)
  
  – Phelype Oleinik
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Doh! Degrees... Yeah, don't know why that didn't occur to me. I'm just so used to thinking in radians. Usually trigonometric functions are defined assuming radians, but I guess it makes sense since everywhere else in latex assumes angles are degrees. I actually did try and turn up the samples at first - thought it was just aliasing.
  
  – argentum2f
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @argentum2f Yes, as I said “oddly”. All programming languages I know (which aren't that many, but...) have a sin function for argument radians and sind function for argument in degrees. Anyhow, Till must have had his reasons :)
  
  – Phelype Oleinik
  2 hours ago
  

  
  
  
  

add a comment | 

4

You got the domain right. The problem is that the trigonometric functions in TikZ are (oddly, in my opinion) in degrees by default. So 2*pi*t with t between 0 and 100 will give you the sine function (in degrees) between 0 and 628.3 degrees (which is about 10.96 radians) which almost 1.75 periods of the function. This is exactly what you see there: one full period and 3/4 of another.

You can tell TikZ to use radians by appending an r to the argument or using the rad function (see page 1005 of the TikZ-PGF manual, section 93.3.4 “Trigonometric functions”). I also added the FPU to allow the domain up to 100 and increased the number of samples to 400 following Kpym's suggestion (notice the aliasing):

documentclass{article}

usepackage{tikz}

usetikzlibrary{fpu}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=400, smooth, variable=t]

    plot ({t/100}, {sin(2*pi*t r)*.5+.5});

  end{scope}%                   ^

end{tikzpicture}

end{document}

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  with samples=400,smooth it looks better, but this is not the question ;)
  
  – Kpym
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kpym Certainly! I just removed the smooth to show that 100 cycles with 40 samples won't do (unless OP wants to show the aliasing). Thanks anyway :)
  
  – Phelype Oleinik
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Doh! Degrees... Yeah, don't know why that didn't occur to me. I'm just so used to thinking in radians. Usually trigonometric functions are defined assuming radians, but I guess it makes sense since everywhere else in latex assumes angles are degrees. I actually did try and turn up the samples at first - thought it was just aliasing.
  
  – argentum2f
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @argentum2f Yes, as I said “oddly”. All programming languages I know (which aren't that many, but...) have a sin function for argument radians and sind function for argument in degrees. Anyhow, Till must have had his reasons :)
  
  – Phelype Oleinik
  2 hours ago
  

  
  
  
  

add a comment | 

You got the domain right. The problem is that the trigonometric functions in TikZ are (oddly, in my opinion) in degrees by default. So 2*pi*t with t between 0 and 100 will give you the sine function (in degrees) between 0 and 628.3 degrees (which is about 10.96 radians) which almost 1.75 periods of the function. This is exactly what you see there: one full period and 3/4 of another.

You can tell TikZ to use radians by appending an r to the argument or using the rad function (see page 1005 of the TikZ-PGF manual, section 93.3.4 “Trigonometric functions”). I also added the FPU to allow the domain up to 100 and increased the number of samples to 400 following Kpym's suggestion (notice the aliasing):

documentclass{article}

usepackage{tikz}

usetikzlibrary{fpu}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=400, smooth, variable=t]

    plot ({t/100}, {sin(2*pi*t r)*.5+.5});

  end{scope}%                   ^

end{tikzpicture}

end{document}

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

documentclass{article}

usepackage{tikz}

usetikzlibrary{fpu}

begin{document}

begin{tikzpicture}

  begin{scope}[x=.6textwidth,y=.6textwidth]

    pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}

    draw[very thin,color=gray, step=.1] (0.0,0.0) grid (1,1);

    draw [ thick,  domain=0:100, samples=400, smooth, variable=t]

    plot ({t/100}, {sin(2*pi*t r)*.5+.5});

  end{scope}%                   ^

end{tikzpicture}

end{document}

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688

answered 4 hours ago

Phelype OleinikPhelype Oleinik

24.2k54688