How to specify a point belong to an ellipse?
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
degrees[36]
begin{pspicture}[showgrid](-3,-3)(4,4)
defa{2}
defxa{pscalculate{a*cos(5)}}
defya{pscalculate{(a/4)*sin(5)}}
psellipse(0,0)(+a,a/4)
pstGeonode[PosAngle=-45](xa,ya){A}
pscircle{2}
multido{i=0+1}{36}{psdot(2;i)}
end{pspicture}
end{document}
Question: How to specify a point belong to an ellipse? Ps: The syntax (r;angle) help to specify a point belong to a circle.
pstricks
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
add a comment |
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
degrees[36]
begin{pspicture}[showgrid](-3,-3)(4,4)
defa{2}
defxa{pscalculate{a*cos(5)}}
defya{pscalculate{(a/4)*sin(5)}}
psellipse(0,0)(+a,a/4)
pstGeonode[PosAngle=-45](xa,ya){A}
pscircle{2}
multido{i=0+1}{36}{psdot(2;i)}
end{pspicture}
end{document}
Question: How to specify a point belong to an ellipse? Ps: The syntax (r;angle) help to specify a point belong to a circle.
pstricks
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
add a comment |
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
degrees[36]
begin{pspicture}[showgrid](-3,-3)(4,4)
defa{2}
defxa{pscalculate{a*cos(5)}}
defya{pscalculate{(a/4)*sin(5)}}
psellipse(0,0)(+a,a/4)
pstGeonode[PosAngle=-45](xa,ya){A}
pscircle{2}
multido{i=0+1}{36}{psdot(2;i)}
end{pspicture}
end{document}
Question: How to specify a point belong to an ellipse? Ps: The syntax (r;angle) help to specify a point belong to a circle.
pstricks
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
degrees[36]
begin{pspicture}[showgrid](-3,-3)(4,4)
defa{2}
defxa{pscalculate{a*cos(5)}}
defya{pscalculate{(a/4)*sin(5)}}
psellipse(0,0)(+a,a/4)
pstGeonode[PosAngle=-45](xa,ya){A}
pscircle{2}
multido{i=0+1}{36}{psdot(2;i)}
end{pspicture}
end{document}
Question: How to specify a point belong to an ellipse? Ps: The syntax (r;angle) help to specify a point belong to a circle.
pstricks
pstricks
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
edited Feb 23 at 9:29
chishimutoji
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
asked Feb 23 at 9:22
chishimutojichishimutoji
8251320
8251320
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
x = a cos t
y = b sin t
is the parameterization of an ellipse but the t does not correspond to the angle of position vector (x,y). Let Θ be the angle of position vector.
It is easy to show that tan t = (a sin Θ) / (b cos Θ).
The remaining will be self-explanatory. :-)
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 2 copy exch THETAspace sin mul exch THETAspace cos mul atan PtoCab){2pt}
end{pspicture}}
end{document}
Explanation
3 2 2 copyproduces3 2 3 2exchproduces3 2 2 3THETAspace sin mulproduces3 2 2 3*sin(Θ)exchproduces3 2 3*sin(Θ) 2THETAspace cos mulproduces3 2 3*sin(Θ) 2*cos(Θ)atanproduces3 2 tPtoCabproducesx yPtoCabneeds 3 operandsa b tthat will be converted toa*cos(t) b*sin(t).atanneeds 2 operandsy xto produces a quadrant-dependent angle.
Final Release
documentclass[border=15pt,pstricks]{standalone}
usepackage{pst-eucl}
pstVerb{/P2EC {3 copy sin 3 -1 roll mul 3 -1 roll cos 3 -1 roll mul atan PtoCab} bind def}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 THETAspace P2EC){2pt}
end{pspicture}}
end{document}
I introduce a new macro P2EC (Polar to Elliptical Cartesian) that will convert a b Θ to a*b*cos Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ) a*b*sin Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ).
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:a b c d e 3 1 rollproducesa b e c d.a b c d e 3 -1 rollproducesa b d e c.a b c d e 3 2 rollproducesa b d e c.
– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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active
oldest
votes
x = a cos t
y = b sin t
is the parameterization of an ellipse but the t does not correspond to the angle of position vector (x,y). Let Θ be the angle of position vector.
It is easy to show that tan t = (a sin Θ) / (b cos Θ).
The remaining will be self-explanatory. :-)
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 2 copy exch THETAspace sin mul exch THETAspace cos mul atan PtoCab){2pt}
end{pspicture}}
end{document}
Explanation
3 2 2 copyproduces3 2 3 2exchproduces3 2 2 3THETAspace sin mulproduces3 2 2 3*sin(Θ)exchproduces3 2 3*sin(Θ) 2THETAspace cos mulproduces3 2 3*sin(Θ) 2*cos(Θ)atanproduces3 2 tPtoCabproducesx yPtoCabneeds 3 operandsa b tthat will be converted toa*cos(t) b*sin(t).atanneeds 2 operandsy xto produces a quadrant-dependent angle.
Final Release
documentclass[border=15pt,pstricks]{standalone}
usepackage{pst-eucl}
pstVerb{/P2EC {3 copy sin 3 -1 roll mul 3 -1 roll cos 3 -1 roll mul atan PtoCab} bind def}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 THETAspace P2EC){2pt}
end{pspicture}}
end{document}
I introduce a new macro P2EC (Polar to Elliptical Cartesian) that will convert a b Θ to a*b*cos Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ) a*b*sin Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ).
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:a b c d e 3 1 rollproducesa b e c d.a b c d e 3 -1 rollproducesa b d e c.a b c d e 3 2 rollproducesa b d e c.
– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
add a comment |
x = a cos t
y = b sin t
is the parameterization of an ellipse but the t does not correspond to the angle of position vector (x,y). Let Θ be the angle of position vector.
It is easy to show that tan t = (a sin Θ) / (b cos Θ).
The remaining will be self-explanatory. :-)
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 2 copy exch THETAspace sin mul exch THETAspace cos mul atan PtoCab){2pt}
end{pspicture}}
end{document}
Explanation
3 2 2 copyproduces3 2 3 2exchproduces3 2 2 3THETAspace sin mulproduces3 2 2 3*sin(Θ)exchproduces3 2 3*sin(Θ) 2THETAspace cos mulproduces3 2 3*sin(Θ) 2*cos(Θ)atanproduces3 2 tPtoCabproducesx yPtoCabneeds 3 operandsa b tthat will be converted toa*cos(t) b*sin(t).atanneeds 2 operandsy xto produces a quadrant-dependent angle.
Final Release
documentclass[border=15pt,pstricks]{standalone}
usepackage{pst-eucl}
pstVerb{/P2EC {3 copy sin 3 -1 roll mul 3 -1 roll cos 3 -1 roll mul atan PtoCab} bind def}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 THETAspace P2EC){2pt}
end{pspicture}}
end{document}
I introduce a new macro P2EC (Polar to Elliptical Cartesian) that will convert a b Θ to a*b*cos Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ) a*b*sin Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ).
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:a b c d e 3 1 rollproducesa b e c d.a b c d e 3 -1 rollproducesa b d e c.a b c d e 3 2 rollproducesa b d e c.
– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
add a comment |
x = a cos t
y = b sin t
is the parameterization of an ellipse but the t does not correspond to the angle of position vector (x,y). Let Θ be the angle of position vector.
It is easy to show that tan t = (a sin Θ) / (b cos Θ).
The remaining will be self-explanatory. :-)
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 2 copy exch THETAspace sin mul exch THETAspace cos mul atan PtoCab){2pt}
end{pspicture}}
end{document}
Explanation
3 2 2 copyproduces3 2 3 2exchproduces3 2 2 3THETAspace sin mulproduces3 2 2 3*sin(Θ)exchproduces3 2 3*sin(Θ) 2THETAspace cos mulproduces3 2 3*sin(Θ) 2*cos(Θ)atanproduces3 2 tPtoCabproducesx yPtoCabneeds 3 operandsa b tthat will be converted toa*cos(t) b*sin(t).atanneeds 2 operandsy xto produces a quadrant-dependent angle.
Final Release
documentclass[border=15pt,pstricks]{standalone}
usepackage{pst-eucl}
pstVerb{/P2EC {3 copy sin 3 -1 roll mul 3 -1 roll cos 3 -1 roll mul atan PtoCab} bind def}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 THETAspace P2EC){2pt}
end{pspicture}}
end{document}
I introduce a new macro P2EC (Polar to Elliptical Cartesian) that will convert a b Θ to a*b*cos Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ) a*b*sin Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ).
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
x = a cos t
y = b sin t
is the parameterization of an ellipse but the t does not correspond to the angle of position vector (x,y). Let Θ be the angle of position vector.
It is easy to show that tan t = (a sin Θ) / (b cos Θ).
The remaining will be self-explanatory. :-)
documentclass[border=15pt,pstricks,12pt]{standalone}
usepackage{pst-eucl,pst-calculate}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 2 copy exch THETAspace sin mul exch THETAspace cos mul atan PtoCab){2pt}
end{pspicture}}
end{document}
Explanation
3 2 2 copyproduces3 2 3 2exchproduces3 2 2 3THETAspace sin mulproduces3 2 2 3*sin(Θ)exchproduces3 2 3*sin(Θ) 2THETAspace cos mulproduces3 2 3*sin(Θ) 2*cos(Θ)atanproduces3 2 tPtoCabproducesx yPtoCabneeds 3 operandsa b tthat will be converted toa*cos(t) b*sin(t).atanneeds 2 operandsy xto produces a quadrant-dependent angle.
Final Release
documentclass[border=15pt,pstricks]{standalone}
usepackage{pst-eucl}
pstVerb{/P2EC {3 copy sin 3 -1 roll mul 3 -1 roll cos 3 -1 roll mul atan PtoCab} bind def}
begin{document}
foreach THETA in {60,150,240,330}{%
begin{pspicture}[showgrid](-4,-4)(4,4)
psline[linecolor=red](3;THETA)
psellipse(0,0)(3,2)
qdisk(!3 2 THETAspace P2EC){2pt}
end{pspicture}}
end{document}
I introduce a new macro P2EC (Polar to Elliptical Cartesian) that will convert a b Θ to a*b*cos Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ) a*b*sin Θ/sqrt(a^2 * sin^2 Θ + b^2 * cos^2 Θ).
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
edited Feb 23 at 13:28
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
answered Feb 23 at 12:18
The Inventor of GodThe Inventor of God
4,93211041
4,93211041
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:a b c d e 3 1 rollproducesa b e c d.a b c d e 3 -1 rollproducesa b d e c.a b c d e 3 2 rollproducesa b d e c.
– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
add a comment |
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:a b c d e 3 1 rollproducesa b e c d.a b c d e 3 -1 rollproducesa b d e c.a b c d e 3 2 rollproducesa b d e c.
– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Yes, I see. Thanks for your explanation about PS language...
– chishimutoji
Feb 23 at 12:44
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
Hi, with your final code, what is the roll and bind? :-)
– chishimutoji
Feb 23 at 14:31
@chishimotoji:
a b c d e 3 1 roll produces a b e c d. a b c d e 3 -1 roll produces a b d e c. a b c d e 3 2 roll produces a b d e c.– The Inventor of God
Feb 23 at 14:59
@chishimotoji:
a b c d e 3 1 roll produces a b e c d. a b c d e 3 -1 roll produces a b d e c. a b c d e 3 2 roll produces a b d e c.– The Inventor of God
Feb 23 at 14:59
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
Yes, I will consider it carefully.
– chishimutoji
Feb 23 at 15:06
add a comment |
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