Why the function ScalingFunctions does not work?
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$begingroup$
Why highlighted in red and does not work function ScalingFunctions -> {None, "Reverse"} in ParametricPlot?
ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1, ScalingFunctions -> {None, "Reverse"}]
graphics parametric-functions
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
$endgroup$
add a comment |
$begingroup$
Why highlighted in red and does not work function ScalingFunctions -> {None, "Reverse"} in ParametricPlot?
ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1, ScalingFunctions -> {None, "Reverse"}]
graphics parametric-functions
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
$endgroup$
$begingroup$
The option is not implemented forParametricPlot. CompareOptions[Plot, ScalingFunctions]withOptions[ParametricPlot, ScalingFunctions].
$endgroup$
– Bob Hanlon
May 20 at 13:55
add a comment |
$begingroup$
Why highlighted in red and does not work function ScalingFunctions -> {None, "Reverse"} in ParametricPlot?
ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1, ScalingFunctions -> {None, "Reverse"}]
graphics parametric-functions
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
$endgroup$
Why highlighted in red and does not work function ScalingFunctions -> {None, "Reverse"} in ParametricPlot?
ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1, ScalingFunctions -> {None, "Reverse"}]
graphics parametric-functions
graphics parametric-functions
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
edited May 20 at 13:44
Bob Hanlon
63.8k3 gold badges36 silver badges99 bronze badges
63.8k3 gold badges36 silver badges99 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
asked May 20 at 13:28
ArtArt
233 bronze badges
233 bronze badges
$begingroup$
The option is not implemented forParametricPlot. CompareOptions[Plot, ScalingFunctions]withOptions[ParametricPlot, ScalingFunctions].
$endgroup$
– Bob Hanlon
May 20 at 13:55
add a comment |
$begingroup$
The option is not implemented forParametricPlot. CompareOptions[Plot, ScalingFunctions]withOptions[ParametricPlot, ScalingFunctions].
$endgroup$
– Bob Hanlon
May 20 at 13:55
$begingroup$
The option is not implemented for
ParametricPlot. Compare Options[Plot, ScalingFunctions] with Options[ParametricPlot, ScalingFunctions].$endgroup$
– Bob Hanlon
May 20 at 13:55
$begingroup$
The option is not implemented for
ParametricPlot. Compare Options[Plot, ScalingFunctions] with Options[ParametricPlot, ScalingFunctions].$endgroup$
– Bob Hanlon
May 20 at 13:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
While ScalingFunctions is not documented to work with ParametricPlot, I think that is only because it fails when using it together with a PlotRange option. Take a look at the result of your ParametricPlot:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
Notice that the vertical plot range is actually {0, -3} and not {0, 3}. So, either remove the PlotRange option:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
or modify it to reflect the transform being used:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,-3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None, "Reverse"}
]
Another possibility is to use PlotRange -> All.
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
$begingroup$
ScalingFunctions is not an option ParametricPlot.
To get the desired look, you can post-process the output of ParametricPlot to vertically flip the coordinates of line objects (using ScalingTransform[{1,-1}]) and reverse the vertical axis tick labels using Charting`ScaledTicks["Reverse"]:
Show[ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1] /. Line[a_] :> Line[ScalingTransform[{1, -1}]@a],
PlotRange -> All,
Ticks -> {Automatic, Charting`ScaledTicks["Reverse"]}]
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
While ScalingFunctions is not documented to work with ParametricPlot, I think that is only because it fails when using it together with a PlotRange option. Take a look at the result of your ParametricPlot:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
Notice that the vertical plot range is actually {0, -3} and not {0, 3}. So, either remove the PlotRange option:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
or modify it to reflect the transform being used:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,-3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None, "Reverse"}
]
Another possibility is to use PlotRange -> All.
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
$begingroup$
While ScalingFunctions is not documented to work with ParametricPlot, I think that is only because it fails when using it together with a PlotRange option. Take a look at the result of your ParametricPlot:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
Notice that the vertical plot range is actually {0, -3} and not {0, 3}. So, either remove the PlotRange option:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
or modify it to reflect the transform being used:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,-3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None, "Reverse"}
]
Another possibility is to use PlotRange -> All.
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
$begingroup$
While ScalingFunctions is not documented to work with ParametricPlot, I think that is only because it fails when using it together with a PlotRange option. Take a look at the result of your ParametricPlot:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
Notice that the vertical plot range is actually {0, -3} and not {0, 3}. So, either remove the PlotRange option:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
or modify it to reflect the transform being used:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,-3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None, "Reverse"}
]
Another possibility is to use PlotRange -> All.
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
$endgroup$
While ScalingFunctions is not documented to work with ParametricPlot, I think that is only because it fails when using it together with a PlotRange option. Take a look at the result of your ParametricPlot:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
Notice that the vertical plot range is actually {0, -3} and not {0, 3}. So, either remove the PlotRange option:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
AspectRatio->1,
ScalingFunctions->{None,"Reverse"}
]
or modify it to reflect the transform being used:
ParametricPlot[
{
0.06677273831511694*(1-E^(-8.145*t)),
-0.10917030600597447*(1-E^(-8.145*t))+1.2031921424186618*t
},
{t,0,2.6},
PlotRange->{{0,0.07},{0.,-3.0191292643518968`}},
AspectRatio->1,
ScalingFunctions->{None, "Reverse"}
]
Another possibility is to use PlotRange -> All.
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
answered May 20 at 14:19
Carl WollCarl Woll
88.3k3 gold badges117 silver badges227 bronze badges
88.3k3 gold badges117 silver badges227 bronze badges
add a comment |
add a comment |
$begingroup$
ScalingFunctions is not an option ParametricPlot.
To get the desired look, you can post-process the output of ParametricPlot to vertically flip the coordinates of line objects (using ScalingTransform[{1,-1}]) and reverse the vertical axis tick labels using Charting`ScaledTicks["Reverse"]:
Show[ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1] /. Line[a_] :> Line[ScalingTransform[{1, -1}]@a],
PlotRange -> All,
Ticks -> {Automatic, Charting`ScaledTicks["Reverse"]}]
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
$endgroup$
add a comment |
$begingroup$
ScalingFunctions is not an option ParametricPlot.
To get the desired look, you can post-process the output of ParametricPlot to vertically flip the coordinates of line objects (using ScalingTransform[{1,-1}]) and reverse the vertical axis tick labels using Charting`ScaledTicks["Reverse"]:
Show[ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1] /. Line[a_] :> Line[ScalingTransform[{1, -1}]@a],
PlotRange -> All,
Ticks -> {Automatic, Charting`ScaledTicks["Reverse"]}]
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
$endgroup$
add a comment |
$begingroup$
ScalingFunctions is not an option ParametricPlot.
To get the desired look, you can post-process the output of ParametricPlot to vertically flip the coordinates of line objects (using ScalingTransform[{1,-1}]) and reverse the vertical axis tick labels using Charting`ScaledTicks["Reverse"]:
Show[ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1] /. Line[a_] :> Line[ScalingTransform[{1, -1}]@a],
PlotRange -> All,
Ticks -> {Automatic, Charting`ScaledTicks["Reverse"]}]
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
$endgroup$
ScalingFunctions is not an option ParametricPlot.
To get the desired look, you can post-process the output of ParametricPlot to vertically flip the coordinates of line objects (using ScalingTransform[{1,-1}]) and reverse the vertical axis tick labels using Charting`ScaledTicks["Reverse"]:
Show[ParametricPlot[{0.06677273831511694*(1 - E^(-8.145*t)),
-0.10917030600597447*(1 - E^(-8.145*t)) + 1.2031921424186618*t},
{t, 0, 2.6}, PlotRange -> {{0, 0.07}, {0., 3.0191292643518968`}},
AspectRatio -> 1] /. Line[a_] :> Line[ScalingTransform[{1, -1}]@a],
PlotRange -> All,
Ticks -> {Automatic, Charting`ScaledTicks["Reverse"]}]
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
answered May 20 at 13:59
kglrkglr
207k10 gold badges237 silver badges470 bronze badges
207k10 gold badges237 silver badges470 bronze badges
add a comment |
add a comment |
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$begingroup$
The option is not implemented for
ParametricPlot. CompareOptions[Plot, ScalingFunctions]withOptions[ParametricPlot, ScalingFunctions].$endgroup$
– Bob Hanlon
May 20 at 13:55