Calculating sinusoidity of a line











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I have a line in a 2D plane. I wanted to calculate the sinusoidity of the line. What i thought was to calculate the arc length and then divide this with the eucledean distance between the initial and final points of the line. Is this a correct procedure?






geometry







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  • Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
    – Yves Daoust
    yesterday












  • how straight a curve is..
    – Harikrishnan R
    yesterday










  • If you have no precise requirement, this method is not worse than another.
    – Yves Daoust
    yesterday















up vote
0
down vote

favorite












I have a line in a 2D plane. I wanted to calculate the sinusoidity of the line. What i thought was to calculate the arc length and then divide this with the eucledean distance between the initial and final points of the line. Is this a correct procedure?






geometry







share|cite|improve this question






















  • Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
    – Yves Daoust
    yesterday












  • how straight a curve is..
    – Harikrishnan R
    yesterday










  • If you have no precise requirement, this method is not worse than another.
    – Yves Daoust
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a line in a 2D plane. I wanted to calculate the sinusoidity of the line. What i thought was to calculate the arc length and then divide this with the eucledean distance between the initial and final points of the line. Is this a correct procedure?






geometry







share|cite|improve this question













I have a line in a 2D plane. I wanted to calculate the sinusoidity of the line. What i thought was to calculate the arc length and then divide this with the eucledean distance between the initial and final points of the line. Is this a correct procedure?






geometry




geometry






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share|cite|improve this question











share|cite|improve this question




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asked yesterday









Harikrishnan R

32




32












  • Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
    – Yves Daoust
    yesterday












  • how straight a curve is..
    – Harikrishnan R
    yesterday










  • If you have no precise requirement, this method is not worse than another.
    – Yves Daoust
    yesterday


















  • Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
    – Yves Daoust
    yesterday












  • how straight a curve is..
    – Harikrishnan R
    yesterday










  • If you have no precise requirement, this method is not worse than another.
    – Yves Daoust
    yesterday
















Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
– Yves Daoust
yesterday






Define the sinusoidity. Currently, Google reports 61 occurrences of this word, which is a good sign that it doesn't exist.
– Yves Daoust
yesterday














how straight a curve is..
– Harikrishnan R
yesterday




how straight a curve is..
– Harikrishnan R
yesterday












If you have no precise requirement, this method is not worse than another.
– Yves Daoust
yesterday




If you have no precise requirement, this method is not worse than another.
– Yves Daoust
yesterday










1 Answer
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0
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accepted










You can define the radius of curvature at any point where the second derivative is defined by $$R=left|frac {(1+y')^{3/2}}{y''}right|$$



Your approach is a fine algorithm. Whether it matches what you are looking for is up to you. Note that it will not distinguish two sides of a triangle from a rather smooth curve of the same arc length and end points.






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    1 Answer
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    active

    oldest

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    active

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    active

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    up vote
    0
    down vote



    accepted










    You can define the radius of curvature at any point where the second derivative is defined by $$R=left|frac {(1+y')^{3/2}}{y''}right|$$



    Your approach is a fine algorithm. Whether it matches what you are looking for is up to you. Note that it will not distinguish two sides of a triangle from a rather smooth curve of the same arc length and end points.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      You can define the radius of curvature at any point where the second derivative is defined by $$R=left|frac {(1+y')^{3/2}}{y''}right|$$



      Your approach is a fine algorithm. Whether it matches what you are looking for is up to you. Note that it will not distinguish two sides of a triangle from a rather smooth curve of the same arc length and end points.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        You can define the radius of curvature at any point where the second derivative is defined by $$R=left|frac {(1+y')^{3/2}}{y''}right|$$



        Your approach is a fine algorithm. Whether it matches what you are looking for is up to you. Note that it will not distinguish two sides of a triangle from a rather smooth curve of the same arc length and end points.






        share|cite|improve this answer












        You can define the radius of curvature at any point where the second derivative is defined by $$R=left|frac {(1+y')^{3/2}}{y''}right|$$



        Your approach is a fine algorithm. Whether it matches what you are looking for is up to you. Note that it will not distinguish two sides of a triangle from a rather smooth curve of the same arc length and end points.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Ross Millikan

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