Can you help me reverse the Minimum Curvature Method?

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The minimum curvature method is used in oil drilling to calculate positional data from directional data. A survey is a reading at a certain depth down the borehole that contains measured depth, inclination, and azimuth. Two consecutive surveys (directional data) can then be used to calculate the position of the second survey in x,y,z terms (northing, easting, true vertical depth).



This part is fairly straight-forward. However, I would like to reverse the Minimum Curvature method in order to take two consecutive points, and calculate the directional values for one of the points.



The equations for the Minimum Curvature method are here:
http://www.relps.com/faq/MinimumCurvatureEquations.pdf



My knowns are A1, I1, North, East, and TVD. I am trying to solve for A2, I2, and MD.



I have tried to use a number of tools to solve systems of equations, but without any luck. Can you help?







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  • Nobody wants to take a crack at it?
    – Matt
    May 21 '13 at 8:11










  • Is there any information that I am missing? Or is this not solvable?
    – Matt
    Jun 4 '13 at 10:41














up vote
1
down vote

favorite
1












The minimum curvature method is used in oil drilling to calculate positional data from directional data. A survey is a reading at a certain depth down the borehole that contains measured depth, inclination, and azimuth. Two consecutive surveys (directional data) can then be used to calculate the position of the second survey in x,y,z terms (northing, easting, true vertical depth).



This part is fairly straight-forward. However, I would like to reverse the Minimum Curvature method in order to take two consecutive points, and calculate the directional values for one of the points.



The equations for the Minimum Curvature method are here:
http://www.relps.com/faq/MinimumCurvatureEquations.pdf



My knowns are A1, I1, North, East, and TVD. I am trying to solve for A2, I2, and MD.



I have tried to use a number of tools to solve systems of equations, but without any luck. Can you help?







share|cite|improve this question






















  • Nobody wants to take a crack at it?
    – Matt
    May 21 '13 at 8:11










  • Is there any information that I am missing? Or is this not solvable?
    – Matt
    Jun 4 '13 at 10:41












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





The minimum curvature method is used in oil drilling to calculate positional data from directional data. A survey is a reading at a certain depth down the borehole that contains measured depth, inclination, and azimuth. Two consecutive surveys (directional data) can then be used to calculate the position of the second survey in x,y,z terms (northing, easting, true vertical depth).



This part is fairly straight-forward. However, I would like to reverse the Minimum Curvature method in order to take two consecutive points, and calculate the directional values for one of the points.



The equations for the Minimum Curvature method are here:
http://www.relps.com/faq/MinimumCurvatureEquations.pdf



My knowns are A1, I1, North, East, and TVD. I am trying to solve for A2, I2, and MD.



I have tried to use a number of tools to solve systems of equations, but without any luck. Can you help?







share|cite|improve this question














The minimum curvature method is used in oil drilling to calculate positional data from directional data. A survey is a reading at a certain depth down the borehole that contains measured depth, inclination, and azimuth. Two consecutive surveys (directional data) can then be used to calculate the position of the second survey in x,y,z terms (northing, easting, true vertical depth).



This part is fairly straight-forward. However, I would like to reverse the Minimum Curvature method in order to take two consecutive points, and calculate the directional values for one of the points.



The equations for the Minimum Curvature method are here:
http://www.relps.com/faq/MinimumCurvatureEquations.pdf



My knowns are A1, I1, North, East, and TVD. I am trying to solve for A2, I2, and MD.



I have tried to use a number of tools to solve systems of equations, but without any luck. Can you help?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 22 '13 at 14:20

























asked May 16 '13 at 13:11









Matt

62




62











  • Nobody wants to take a crack at it?
    – Matt
    May 21 '13 at 8:11










  • Is there any information that I am missing? Or is this not solvable?
    – Matt
    Jun 4 '13 at 10:41
















  • Nobody wants to take a crack at it?
    – Matt
    May 21 '13 at 8:11










  • Is there any information that I am missing? Or is this not solvable?
    – Matt
    Jun 4 '13 at 10:41















Nobody wants to take a crack at it?
– Matt
May 21 '13 at 8:11




Nobody wants to take a crack at it?
– Matt
May 21 '13 at 8:11












Is there any information that I am missing? Or is this not solvable?
– Matt
Jun 4 '13 at 10:41




Is there any information that I am missing? Or is this not solvable?
– Matt
Jun 4 '13 at 10:41










2 Answers
2






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0
down vote













How accurate do you need to be. Have you tried using average angle calculations? As long as a couple of centimetres one way or the other don't matter the average angle calculation will work. It is going to get messy but you've got 3 equations and so you should be able to solve for your 3 unknown values.






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    In order to get a better picture you should look at the problem in 2d. Your arc from (x1,y1,z1) to (x2,y2,z2) lives in a 2d plane, also in the same pane the tangents (a1,i1) and (a2, i2). The 2d plane is given by the vector (x1,y1,y3) to (x2,y2,z2) and vector converted from polar to Cartesian of (a1, i1). In case their co-linear is just a straight line and your done. Given the angle between the (x1,y1,z2) and (a1, i1) be alpha, then the angle between (x2,y2,z2) and (a2, i2) is –alpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe –alpha) and converter back to polar coordinates, which gives you (a2,i2). If d is the distance from (x1,y1,z1) to (x2,y2,z2) then MD = d* alpha /sin(alpha).



    That said, in a directional drilling setting you may get Cartesian coordinates which do not perfectly follow minimum curvature. Furthermore, to reconstruct reasonable well the inclination and the azimuth the input requires accuracy of close to float, which is typically much smaller than you actual can measure. Just calculate the Cartesian coordinates of your trajectory, round them to 0.01m and try to reconstruct inclination and azimuth.






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      2 Answers
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      2 Answers
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      up vote
      0
      down vote













      How accurate do you need to be. Have you tried using average angle calculations? As long as a couple of centimetres one way or the other don't matter the average angle calculation will work. It is going to get messy but you've got 3 equations and so you should be able to solve for your 3 unknown values.






      share|cite|improve this answer
























        up vote
        0
        down vote













        How accurate do you need to be. Have you tried using average angle calculations? As long as a couple of centimetres one way or the other don't matter the average angle calculation will work. It is going to get messy but you've got 3 equations and so you should be able to solve for your 3 unknown values.






        share|cite|improve this answer






















          up vote
          0
          down vote










          up vote
          0
          down vote









          How accurate do you need to be. Have you tried using average angle calculations? As long as a couple of centimetres one way or the other don't matter the average angle calculation will work. It is going to get messy but you've got 3 equations and so you should be able to solve for your 3 unknown values.






          share|cite|improve this answer












          How accurate do you need to be. Have you tried using average angle calculations? As long as a couple of centimetres one way or the other don't matter the average angle calculation will work. It is going to get messy but you've got 3 equations and so you should be able to solve for your 3 unknown values.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 28 '14 at 22:17









          user188112

          1




          1




















              up vote
              0
              down vote













              In order to get a better picture you should look at the problem in 2d. Your arc from (x1,y1,z1) to (x2,y2,z2) lives in a 2d plane, also in the same pane the tangents (a1,i1) and (a2, i2). The 2d plane is given by the vector (x1,y1,y3) to (x2,y2,z2) and vector converted from polar to Cartesian of (a1, i1). In case their co-linear is just a straight line and your done. Given the angle between the (x1,y1,z2) and (a1, i1) be alpha, then the angle between (x2,y2,z2) and (a2, i2) is –alpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe –alpha) and converter back to polar coordinates, which gives you (a2,i2). If d is the distance from (x1,y1,z1) to (x2,y2,z2) then MD = d* alpha /sin(alpha).



              That said, in a directional drilling setting you may get Cartesian coordinates which do not perfectly follow minimum curvature. Furthermore, to reconstruct reasonable well the inclination and the azimuth the input requires accuracy of close to float, which is typically much smaller than you actual can measure. Just calculate the Cartesian coordinates of your trajectory, round them to 0.01m and try to reconstruct inclination and azimuth.






              share|cite|improve this answer
























                up vote
                0
                down vote













                In order to get a better picture you should look at the problem in 2d. Your arc from (x1,y1,z1) to (x2,y2,z2) lives in a 2d plane, also in the same pane the tangents (a1,i1) and (a2, i2). The 2d plane is given by the vector (x1,y1,y3) to (x2,y2,z2) and vector converted from polar to Cartesian of (a1, i1). In case their co-linear is just a straight line and your done. Given the angle between the (x1,y1,z2) and (a1, i1) be alpha, then the angle between (x2,y2,z2) and (a2, i2) is –alpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe –alpha) and converter back to polar coordinates, which gives you (a2,i2). If d is the distance from (x1,y1,z1) to (x2,y2,z2) then MD = d* alpha /sin(alpha).



                That said, in a directional drilling setting you may get Cartesian coordinates which do not perfectly follow minimum curvature. Furthermore, to reconstruct reasonable well the inclination and the azimuth the input requires accuracy of close to float, which is typically much smaller than you actual can measure. Just calculate the Cartesian coordinates of your trajectory, round them to 0.01m and try to reconstruct inclination and azimuth.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  In order to get a better picture you should look at the problem in 2d. Your arc from (x1,y1,z1) to (x2,y2,z2) lives in a 2d plane, also in the same pane the tangents (a1,i1) and (a2, i2). The 2d plane is given by the vector (x1,y1,y3) to (x2,y2,z2) and vector converted from polar to Cartesian of (a1, i1). In case their co-linear is just a straight line and your done. Given the angle between the (x1,y1,z2) and (a1, i1) be alpha, then the angle between (x2,y2,z2) and (a2, i2) is –alpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe –alpha) and converter back to polar coordinates, which gives you (a2,i2). If d is the distance from (x1,y1,z1) to (x2,y2,z2) then MD = d* alpha /sin(alpha).



                  That said, in a directional drilling setting you may get Cartesian coordinates which do not perfectly follow minimum curvature. Furthermore, to reconstruct reasonable well the inclination and the azimuth the input requires accuracy of close to float, which is typically much smaller than you actual can measure. Just calculate the Cartesian coordinates of your trajectory, round them to 0.01m and try to reconstruct inclination and azimuth.






                  share|cite|improve this answer












                  In order to get a better picture you should look at the problem in 2d. Your arc from (x1,y1,z1) to (x2,y2,z2) lives in a 2d plane, also in the same pane the tangents (a1,i1) and (a2, i2). The 2d plane is given by the vector (x1,y1,y3) to (x2,y2,z2) and vector converted from polar to Cartesian of (a1, i1). In case their co-linear is just a straight line and your done. Given the angle between the (x1,y1,z2) and (a1, i1) be alpha, then the angle between (x2,y2,z2) and (a2, i2) is –alpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe –alpha) and converter back to polar coordinates, which gives you (a2,i2). If d is the distance from (x1,y1,z1) to (x2,y2,z2) then MD = d* alpha /sin(alpha).



                  That said, in a directional drilling setting you may get Cartesian coordinates which do not perfectly follow minimum curvature. Furthermore, to reconstruct reasonable well the inclination and the azimuth the input requires accuracy of close to float, which is typically much smaller than you actual can measure. Just calculate the Cartesian coordinates of your trajectory, round them to 0.01m and try to reconstruct inclination and azimuth.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 15 '15 at 23:42









                  tma

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