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Is there an algorithms to determine correct angles

  1. Jun 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello!
    I am struggling with plotting polar graphs manually (without any help of the calculator). My main unresolved issue is with finding correct values of theta in a given range.
    For example, I have an equation:
    $r = cos(5\theta)$

    2. Relevant equations
    and I know that the range I have to work with is [-π/5 , π/5]. Here are the values I get when arbitrary choose values of theta to compute r, then x and y.

    Screen Shot 2017-06-04 at 13.25.18.png

    But the problem is that I don't know of any algorithm that I could apply in any such case to choose correct values of theta - here I started with -π/5, took π/6, next π/10, and 0 for the first half.
    I suppose there should be some algorithm or rules applicable for such choices.
    Please, help me to fill that gap in my knowledge.

    3. The attempt at a solution
    Thank you very much!
     
  2. jcsd
  3. Jun 4, 2017 #2

    Nidum

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    Increment the angle by fixed amounts starting at the lowest range value and ending at the highest range value . You have to decide what increment is appropriate for your application .

    For example we could decide on 0.1 of π/5 as the increment .

    Then starting from the lowest range value and incrementing the angle sequentially by 0.1 of π/5 we get :

    - π/5 , - 0.9 x π/5 , - 0.8 x π/5 , ............. , 0 , 0.1 x π/5 , 0.2 x π/5 , .......... , π/5

    Note though that it may be easier to get or make some proper polar graph paper and plot results directly rather than convert to Cartesian coordinates .
     
  4. Jun 4, 2017 #3

    FactChecker

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    If you want to do quick plots from memory, you should use the angles inside sin() and cos() whose values you can remember. You should memorize the sin() and cos() of 0°, 30°, 45°, 60°, and 90°. Pick theta so that you are taking sin() and cos() of those angles.

    (The sin() of 0°, 30°, 45°, 60°, and 90° is √0/2=0, √1/2=1/2, √2/2, √3/2, √4/2=1. And use √2/2 ≅ 1.414/2 = 0.707 and √3/2 ≅ 1.732/2 = 0.866)
     
    Last edited: Jun 4, 2017
  5. Jun 4, 2017 #4
    Thank you very much. I will try to follow your advice, and see where it brings me to. :) As to direct plotting, well, no, I need to do the fundamental cycle first on r theta graph, and then convert to Cartesian; all manually, without any machines. :-)
     
  6. Jun 4, 2017 #5
    Yes, I know all these values pretty well - seems the only topic I definitely don't have issues with. I did try using these "standard" angles, but this is wrong in this case because the range of the given equation is 2π/5. So, I wondered what algorithm should be used that works in all cases.
     
  7. Jun 4, 2017 #6

    FactChecker

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    The equation is cos(5θ), so for instance √2/2 = cos(5θ) when 5θ = π/4 or θ = π/20. π/20 is well within the range ±π/4. So you can get a lot of points to plot within the range of ±π/4.
     
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