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Math Help - Dip angle and direction of a plane described by 3 points

  1. #1
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    Dip angle and direction of a plane described by 3 points

    I'm trying to make a model in excel which calculates the dip angle
    and direction (azimuth or bearing) of a plane descriped by 3 points in a coordinate
    system.
    The planned use is a quick-look geomodel where I would like to
    calculate formation dip based on the observation of a specific
    geological feature in three separate wells. The points where the
    well intersects the formation in each well is described by a UTM
    latitude/longitude and depth below a fixed datum. Point 1 (x1, y1,
    z1), point 2 (x2, y2, z2), point 3 (x3,y3, z3).
    By definition, the dip angle is perpendicular to the strike (a
    horizontal line through the plane) 0 degrees meaning a horizontal
    plane/90 degrees meaning a vertical plane, and the dip azimuth is
    the compass direction/bearing of the dip angle.
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  2. #2
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    Quote Originally Posted by prevaricator View Post
    I'm trying to make a model in excel which calculates the dip angle
    and direction (azimuth or bearing) of a plane descriped by 3 points in a coordinate
    system.
    The planned use is a quick-look geomodel where I would like to
    calculate formation dip based on the observation of a specific
    geological feature in three separate wells. The points where the
    well intersects the formation in each well is described by a UTM
    latitude/longitude and depth below a fixed datum. Point 1 (x1, y1,
    z1), point 2 (x2, y2, z2), point 3 (x3,y3, z3).
    By definition, the dip angle is perpendicular to the strike (a
    horizontal line through the plane) 0 degrees meaning a horizontal
    plane/90 degrees meaning a vertical plane, and the dip azimuth is
    the compass direction/bearing of the dip angle.
    Given the three points, you can construct two vectors <x_2-x_1,y_2-y_1,z_2-z_1> and <x_3-x_1,y_3-y_1,z_3-z_1>. The cross product of those two vectors is normal to the plane. I'm not going to do the calculation myself, but call that <v_x,v_y,v_z>. The "dip angle", the angle between that normal vector and the horizontal plane which is the angle between <v_x,v_y,v_z> and <v_x,v_y,0>. The angle, \theta, is given by cos(\theta)= \frac{u\cdot v}{|u||v|}.
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