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Polar Plots - extra loops appear

Posted 10 years ago
When plotting under Graphics, 
I get the expected cloverleaf of 3 loops.
However, surprisingly,  in each loop appears another smaller loop.
How can the presence of these smaller loops be explained ?
POSTED BY: Jaak Bouckaert
5 Replies
Try this to show how 3 parts of the plot fit together
Show[PolarPlot[{1 - 2 Cos[3 t]}, {t, 0, 2 Pi/3}, PlotStyle -> Black],
PolarPlot[{1 - 2 Cos[3 t]}, {t, 2 Pi/3, 4 Pi/3}, PlotStyle -> Red],
PolarPlot[{1 - 2 Cos[3 t]}, {t, 4 Pi/3, 2 Pi}, PlotStyle -> Blue]]
POSTED BY: S M Blinder
Posted 10 years ago
When limits are set to [0,2Pi/3] ,indeed only one big loop is shown and no small loop inside of it.
And now arises a new question: why small loops with limits [0,2Pi] and none with limits [0,2Pi/3] ?
The presence of  small loops seems odd: for every given t in [0,2Pi] , there is only 1 value for r = 1-2Cos[3t]
POSTED BY: Jaak Bouckaert
You can also get a sense sense of how this function unfolds by exploring it with a Manipulate:
    PolarPlot[{1 - 2 Cos[3 t]}, {t, 0, angle},
     PlotRange -> {{-3, 3}, {-3, 3}}, Frame -> True,
     PlotLabel -> N@angle],
    Plot[{1 - 2 Cos[3 t]}, {t, 0, angle},
     PlotRange -> {{0, 2 Pi}, {-1.1, 3}}, Frame -> True,
     PlotLabel -> N@angle]
{{angle, 0.01}, 0.01, 2 Pi}

Just drag the slider to see Professor Blinder's separate expressions unnfold.
POSTED BY: David Reiss
Posted 10 years ago
After a night's sleep, I think I found myself a rather simple explanation:

Normally, negative values of r ( = 1-2Cos3t  in this case) are not considered in polar coordinates.
However, they indeed seem to not have been discarded here, but presented in the opposite direction of the pole.
So for instance ( when plotting with limits [0 , 2pi] ) , for t = 4pi/3 , an angle in the third quadrant, r = -1 , and this value is given as the top of the small loop in the first quadrant.

No wonder then, that limiting t to [0, 2Pi/3] , no small loop appears in the big one, because inside this latter, r is only negative for values of t outside this region of t.

Another interesting way to see and understand this is asking for  r = 1-2Cos3t  with limits for t : [-pi/9 , 7pi/9]                                                                
POSTED BY: Jaak Bouckaert
For 3 simple loops try
PolarPlot[Sin[3 t], {t, 0, Pi}]
PolarPlot[Cos[3 t], {t, 0, Pi}]
POSTED BY: S M Blinder
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