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?
trigonometry polynomials curvature
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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?
trigonometry polynomials curvature
Nobody wants to take a crack at it?
â Matt
May 21 '13 at 8:11
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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?
trigonometry polynomials curvature
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?
trigonometry polynomials curvature
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?
â Matt
May 21 '13 at 8:11
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41
add a comment |Â
Nobody wants to take a crack at it?
â Matt
May 21 '13 at 8:11
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41
Nobody wants to take a crack at it?
â Matt
May 21 '13 at 8:11
Nobody wants to take a crack at it?
â Matt
May 21 '13 at 8:11
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41
add a comment |Â
2 Answers
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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 âÂÂalpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe âÂÂalpha) 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
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Oct 28 '14 at 22:17
user188112
1
1
add a comment |Â
add a comment |Â
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 âÂÂalpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe âÂÂalpha) 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.
add a comment |Â
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 âÂÂalpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe âÂÂalpha) 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.
add a comment |Â
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 âÂÂalpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe âÂÂalpha) 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.
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 âÂÂalpha. Use the normal vector of the 2d plane and rotate normalized vector (x1,y1,z1) to (x2,y2,z2) by alpha (maybe âÂÂalpha) 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.
answered Mar 15 '15 at 23:42
tma
1
1
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Nobody wants to take a crack at it?
â Matt
May 21 '13 at 8:11
Is there any information that I am missing? Or is this not solvable?
â Matt
Jun 4 '13 at 10:41