Affine transformation of circular arc in 3D
6
$begingroup$
Start with a quarter-circle of radius 1 centered at the origin and lying in the xz-plane:
arc = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
I want to dilate this by a factor of 2 and shift the center to {3, 0, 0}, then show the result graphically.
The following does not work:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}]
Graphics3D[GeometricTransformation[arc3D, shiftAndDilate3D]]
The error I get is that "Graphics3DBox is not a Graphics3D primitive or directive."
What am I doing wrong?
graphics3d geometric-transform
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
$endgroup$
add a comment |
6
$begingroup$
Start with a quarter-circle of radius 1 centered at the origin and lying in the xz-plane:
arc = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
I want to dilate this by a factor of 2 and shift the center to {3, 0, 0}, then show the result graphically.
The following does not work:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}]
Graphics3D[GeometricTransformation[arc3D, shiftAndDilate3D]]
The error I get is that "Graphics3DBox is not a Graphics3D primitive or directive."
What am I doing wrong?
graphics3d geometric-transform
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
$endgroup$
$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago
add a comment |
6
6
6
$begingroup$
Start with a quarter-circle of radius 1 centered at the origin and lying in the xz-plane:
arc = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
I want to dilate this by a factor of 2 and shift the center to {3, 0, 0}, then show the result graphically.
The following does not work:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}]
Graphics3D[GeometricTransformation[arc3D, shiftAndDilate3D]]
The error I get is that "Graphics3DBox is not a Graphics3D primitive or directive."
What am I doing wrong?
graphics3d geometric-transform
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
$endgroup$
Start with a quarter-circle of radius 1 centered at the origin and lying in the xz-plane:
arc = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
I want to dilate this by a factor of 2 and shift the center to {3, 0, 0}, then show the result graphically.
The following does not work:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}]
Graphics3D[GeometricTransformation[arc3D, shiftAndDilate3D]]
The error I get is that "Graphics3DBox is not a Graphics3D primitive or directive."
What am I doing wrong?
graphics3d geometric-transform
graphics3d geometric-transform
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
97.3k10303463
asked 11 hours ago
murraymurray
6,2031835
asked 11 hours ago
murraymurray
6,2031835
asked 11 hours ago
murraymurray
6,2031835
6,2031835
$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago
add a comment |
$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago
$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:
ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:
ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
5
$begingroup$
You could work with regions instead. Your arc:
arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, [Pi]/2}}];
The transformed arc:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];
Visualization:
Show[
Region[arc, BaseStyle->Red],
Region[new, BaseStyle->Blue],
Axes->True
]
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
$endgroup$
$begingroup$
How should one know from the Mathematica documentation thatParametricRegionis a "region" suitable for the 1st argument toTransformedRegion? (The documentation involvingGeometricTransformatino,Region, etc., is sadly deficient. As withImage, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of theBaseStyleoption ofRegioninstead of the usualPlotStyleoption ofParametricPlot3Dand so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of aParametricRegionexpression with aRegion? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.
$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why doesParametricRegiontake as 2nd argument a list of lists (of parameters and their extent), whereasParametricPlot3Duses the parameter information as a list, then another list, etc.?
$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,ParametricRegion[{Cos[t], 0, Sin[t]}, t].
$endgroup$
– Carl Woll
6 hours ago
|
show 4 more comments
5
$begingroup$
You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:
arcLine =
First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]
Graphics3D[{
Red, arcLine,
Blue, GeometricTransformation[arcLine, shiftAndDilate3D]
}]
In red in the plot above is your original curve, in blue the transformed one.
answered 10 hours ago
MarcoBMarcoB
37k556113
$endgroup$
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "Graphicsprimitive", such as those discussed in this guide:Graphicsobjects. I fixed it in the answer.
$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, aParametricRegionseems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- theHeadof these objects doesn't tell you that they are "graphics objects"!
$endgroup$
– murray
7 hours ago
add a comment |
4
$begingroup$
Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:
arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
Show[arc3D,
Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
PlotRange -> All]
answered 10 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma beforePlotRange(fixed now).
$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
You could work with regions instead. Your arc:
arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, [Pi]/2}}];
The transformed arc:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];
Visualization:
Show[
Region[arc, BaseStyle->Red],
Region[new, BaseStyle->Blue],
Axes->True
]
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
$endgroup$
$begingroup$
How should one know from the Mathematica documentation thatParametricRegionis a "region" suitable for the 1st argument toTransformedRegion? (The documentation involvingGeometricTransformatino,Region, etc., is sadly deficient. As withImage, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of theBaseStyleoption ofRegioninstead of the usualPlotStyleoption ofParametricPlot3Dand so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of aParametricRegionexpression with aRegion? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.
$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why doesParametricRegiontake as 2nd argument a list of lists (of parameters and their extent), whereasParametricPlot3Duses the parameter information as a list, then another list, etc.?
$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,ParametricRegion[{Cos[t], 0, Sin[t]}, t].
$endgroup$
– Carl Woll
6 hours ago
|
show 4 more comments
5
$begingroup$
You could work with regions instead. Your arc:
arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, [Pi]/2}}];
The transformed arc:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];
Visualization:
Show[
Region[arc, BaseStyle->Red],
Region[new, BaseStyle->Blue],
Axes->True
]
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
$endgroup$
$begingroup$
How should one know from the Mathematica documentation thatParametricRegionis a "region" suitable for the 1st argument toTransformedRegion? (The documentation involvingGeometricTransformatino,Region, etc., is sadly deficient. As withImage, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of theBaseStyleoption ofRegioninstead of the usualPlotStyleoption ofParametricPlot3Dand so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of aParametricRegionexpression with aRegion? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.
$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why doesParametricRegiontake as 2nd argument a list of lists (of parameters and their extent), whereasParametricPlot3Duses the parameter information as a list, then another list, etc.?
$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,ParametricRegion[{Cos[t], 0, Sin[t]}, t].
$endgroup$
– Carl Woll
6 hours ago
|
show 4 more comments
5
5
5
$begingroup$
You could work with regions instead. Your arc:
arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, [Pi]/2}}];
The transformed arc:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];
Visualization:
Show[
Region[arc, BaseStyle->Red],
Region[new, BaseStyle->Blue],
Axes->True
]
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
$endgroup$
You could work with regions instead. Your arc:
arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, [Pi]/2}}];
The transformed arc:
shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];
Visualization:
Show[
Region[arc, BaseStyle->Red],
Region[new, BaseStyle->Blue],
Axes->True
]
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
answered 10 hours ago
Carl WollCarl Woll
69.4k393179
69.4k393179
$begingroup$
How should one know from the Mathematica documentation thatParametricRegionis a "region" suitable for the 1st argument toTransformedRegion? (The documentation involvingGeometricTransformatino,Region, etc., is sadly deficient. As withImage, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of theBaseStyleoption ofRegioninstead of the usualPlotStyleoption ofParametricPlot3Dand so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of aParametricRegionexpression with aRegion? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.
$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why doesParametricRegiontake as 2nd argument a list of lists (of parameters and their extent), whereasParametricPlot3Duses the parameter information as a list, then another list, etc.?
$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,ParametricRegion[{Cos[t], 0, Sin[t]}, t].
$endgroup$
– Carl Woll
6 hours ago
|
show 4 more comments
$begingroup$
How should one know from the Mathematica documentation thatParametricRegionis a "region" suitable for the 1st argument toTransformedRegion? (The documentation involvingGeometricTransformatino,Region, etc., is sadly deficient. As withImage, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of theBaseStyleoption ofRegioninstead of the usualPlotStyleoption ofParametricPlot3Dand so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of aParametricRegionexpression with aRegion? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.
$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why doesParametricRegiontake as 2nd argument a list of lists (of parameters and their extent), whereasParametricPlot3Duses the parameter information as a list, then another list, etc.?
$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,ParametricRegion[{Cos[t], 0, Sin[t]}, t].
$endgroup$
– Carl Woll
6 hours ago
$begingroup$
How should one know from the Mathematica documentation that
ParametricRegion is a "region" suitable for the 1st argument to TransformedRegion? (The documentation involving GeometricTransformatino, Region, etc., is sadly deficient. As with Image, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)$endgroup$
– murray
7 hours ago
$begingroup$
How should one know from the Mathematica documentation that
ParametricRegion is a "region" suitable for the 1st argument to TransformedRegion? (The documentation involving GeometricTransformatino, Region, etc., is sadly deficient. As with Image, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.)$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region".
$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of the
BaseStyle option of Region instead of the usual PlotStyle option of ParametricPlot3D and so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of a ParametricRegion expression with a Region? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.$endgroup$
– murray
7 hours ago
$begingroup$
But I'm annoyed by the different method to style the resulting graphics display, namely, through use of the
BaseStyle option of Region instead of the usual PlotStyle option of ParametricPlot3D and so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of a ParametricRegion expression with a Region? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain.$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why does
ParametricRegion take as 2nd argument a list of lists (of parameters and their extent), whereas ParametricPlot3D uses the parameter information as a list, then another list, etc.?$endgroup$
– murray
7 hours ago
$begingroup$
Further rant: Why does
ParametricRegion take as 2nd argument a list of lists (of parameters and their extent), whereas ParametricPlot3D uses the parameter information as a list, then another list, etc.?$endgroup$
– murray
7 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,
ParametricRegion[{Cos[t], 0, Sin[t]}, t].$endgroup$
– Carl Woll
6 hours ago
$begingroup$
@murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g.,
ParametricRegion[{Cos[t], 0, Sin[t]}, t].$endgroup$
– Carl Woll
6 hours ago
|
show 4 more comments
5
$begingroup$
You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:
arcLine =
First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]
Graphics3D[{
Red, arcLine,
Blue, GeometricTransformation[arcLine, shiftAndDilate3D]
}]
In red in the plot above is your original curve, in blue the transformed one.
answered 10 hours ago
MarcoBMarcoB
37k556113
$endgroup$
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "Graphicsprimitive", such as those discussed in this guide:Graphicsobjects. I fixed it in the answer.
$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, aParametricRegionseems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- theHeadof these objects doesn't tell you that they are "graphics objects"!
$endgroup$
– murray
7 hours ago
add a comment |
5
$begingroup$
You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:
arcLine =
First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]
Graphics3D[{
Red, arcLine,
Blue, GeometricTransformation[arcLine, shiftAndDilate3D]
}]
In red in the plot above is your original curve, in blue the transformed one.
answered 10 hours ago
MarcoBMarcoB
37k556113
$endgroup$
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "Graphicsprimitive", such as those discussed in this guide:Graphicsobjects. I fixed it in the answer.
$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, aParametricRegionseems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- theHeadof these objects doesn't tell you that they are "graphics objects"!
$endgroup$
– murray
7 hours ago
add a comment |
5
5
5
$begingroup$
You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:
arcLine =
First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]
Graphics3D[{
Red, arcLine,
Blue, GeometricTransformation[arcLine, shiftAndDilate3D]
}]
In red in the plot above is your original curve, in blue the transformed one.
answered 10 hours ago
MarcoBMarcoB
37k556113
$endgroup$
You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:
arcLine =
First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]
Graphics3D[{
Red, arcLine,
Blue, GeometricTransformation[arcLine, shiftAndDilate3D]
}]
In red in the plot above is your original curve, in blue the transformed one.
answered 10 hours ago
MarcoBMarcoB
37k556113
edited 9 hours ago
answered 10 hours ago
MarcoBMarcoB
37k556113
answered 10 hours ago
MarcoBMarcoB
37k556113
answered 10 hours ago
MarcoBMarcoB
37k556113
37k556113
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "Graphicsprimitive", such as those discussed in this guide:Graphicsobjects. I fixed it in the answer.
$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, aParametricRegionseems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- theHeadof these objects doesn't tell you that they are "graphics objects"!
$endgroup$
– murray
7 hours ago
add a comment |
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "Graphicsprimitive", such as those discussed in this guide:Graphicsobjects. I fixed it in the answer.
$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, aParametricRegionseems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- theHeadof these objects doesn't tell you that they are "graphics objects"!
$endgroup$
– murray
7 hours ago
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
Where in the documentation does it explain what is, and what is not, a "geometric object"?
$endgroup$
– murray
10 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "
Graphics primitive", such as those discussed in this guide: Graphics objects. I fixed it in the answer.$endgroup$
– MarcoB
9 hours ago
$begingroup$
@murray I was being loose with words there. I really meant a "
Graphics primitive", such as those discussed in this guide: Graphics objects. I fixed it in the answer.$endgroup$
– MarcoB
9 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, a
ParametricRegion seems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- the Head of these objects doesn't tell you that they are "graphics objects"!$endgroup$
– murray
7 hours ago
$begingroup$
Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, a
ParametricRegion seems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- the Head of these objects doesn't tell you that they are "graphics objects"!$endgroup$
– murray
7 hours ago
add a comment |
4
$begingroup$
Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:
arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
Show[arc3D,
Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
PlotRange -> All]
answered 10 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma beforePlotRange(fixed now).
$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
add a comment |
4
$begingroup$
Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:
arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
Show[arc3D,
Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
PlotRange -> All]
answered 10 hours ago
kglrkglr
187k10203421
$endgroup$
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma beforePlotRange(fixed now).
$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
add a comment |
4
4
4
$begingroup$
Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:
arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
Show[arc3D,
Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
PlotRange -> All]
answered 10 hours ago
kglrkglr
187k10203421
$endgroup$
Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:
arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
Show[arc3D,
Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
PlotRange -> All]
answered 10 hours ago
kglrkglr
187k10203421
edited 10 hours ago
answered 10 hours ago
kglrkglr
187k10203421
answered 10 hours ago
kglrkglr
187k10203421
answered 10 hours ago
kglrkglr
187k10203421
187k10203421
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma beforePlotRange(fixed now).
$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
add a comment |
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma beforePlotRange(fixed now).
$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string."
$endgroup$
– murray
10 hours ago
$begingroup$
@murray, forgot a comma before
PlotRange (fixed now).$endgroup$
– kglr
9 hours ago
$begingroup$
@murray, forgot a comma before
PlotRange (fixed now).$endgroup$
– kglr
9 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
$begingroup$
Sorry, I should have caught that!
$endgroup$
– murray
7 hours ago
add a comment |
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$begingroup$
At least for this case, it is much better to just apply the affine transformation to the parametric equations directly:
ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}].$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
@J.M.iscomputer-less: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer!
$endgroup$
– murray
1 hour ago