Timeline for Interpretation of PolarPlot
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2018 at 16:08 | vote | accept | Gennaro Arguzzi | ||
| Nov 14, 2018 at 16:04 | vote | accept | Gennaro Arguzzi | ||
| Nov 14, 2018 at 16:08 | |||||
| Nov 14, 2018 at 12:56 | answer | added | Alex Trounev | timeline score: 5 | |
| Nov 14, 2018 at 11:44 | comment | added | Kuba | I'm deleting previous comment as I don't have time for discussion and it was off topic anyway. Please explain clearly what do you want to achieve in Mathematica and where are you stuck. | |
| Nov 14, 2018 at 11:40 | comment | added | Gennaro Arguzzi | @Kuba please tell me what is in my case $x^2$ anf $f$. | |
| Nov 14, 2018 at 11:18 | comment | added | Gennaro Arguzzi | Hi @Kuba, by looking the plot I see that the right angle is $\pi+(\pi/4)$, that is almost 3.9 (check it by using Get Coordinates please). Is there a way to get the right angle through math formulas? | |
| Nov 14, 2018 at 10:59 | comment | added | Kuba |
t = -r/2 and r = -Pi/2 which means t = Pi / 2 /2 = Pi /4?
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| Nov 14, 2018 at 10:42 | comment | added | Gennaro Arguzzi | @Kuba Ok, your example is clear. I have the inverse problem: given $r=-1.57$, what is the angle (without knowing x and y)? | |
| Nov 14, 2018 at 10:36 | comment | added | Kuba | If you have a point (1,1) cartesian, is it (sqrt(2), Pi/4) or (-sqrt(2), 3Pi/4)? My point is that information that you use negative radius is gone so you have to think about it yourself. | |
| Nov 14, 2018 at 10:33 | comment | added | Lotus | I think the confusion here is what exactly does GetCoordinates give for PolarPlot | |
| Nov 14, 2018 at 10:30 | comment | added | Gennaro Arguzzi | Hello @Kuba, can you explain your comment please? If the angle is unknown, how can I do the reflection? Maybe the angle is $\pi/4$? | |
| Nov 14, 2018 at 10:27 | comment | added | Kuba | That is your problem, the point you measured has r = 1.57. If you want to get an angle for r = -1.57 you need to reflect that point through the origin. | |
| Nov 14, 2018 at 10:26 | history | edited | Gennaro Arguzzi |
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| Nov 14, 2018 at 10:01 | history | asked | Gennaro Arguzzi | CC BY-SA 4.0 |