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Normal vector, slope and aspect angle

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Old   February 2, 2011, 17:24
Default Normal vector, slope and aspect angle
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I have a set of (X,Y,Z) points representing different planar features. I need to calculate the slope and aspect angle of each plane using normal vectors.
For the slope, I want to compute angle between normal vector (NV) of each plane and NV of imaginary horizontal plane. Assume, the plane equation that I use is; Ax+By+c=z. hence, the normal vector of my plane is (a,b, -1). For my plane equation, what should be the equation of imaginary horizontal plane? I think equation of horizontal plane is z=c. Hence, the normal vector is (0,0,-1). Is this correct?
Also, I want to know the equation of plane which is parallel (1) to xz plane (2) to yz plane and (3) any vertical plane oriented in any direction (i.e. not parallel to xz or yz planes) with their corresponding normal vectors. After using the above two NVs, the slope angle of my plane with respect to horizontal plane is; acos{[a1.0+b1.0+(-1).(-1)]/[sqrt(a1.a1+b1.b1+1).(1)]}. I would like to know whether this equation is correct. Please comment me.

Other main thing I need to know is the way of computing aspect angle. I hope, I can compute the aspect angle by using the projected normal vector of each plane to the XY plane. Then, from the Y axis I can compute the aspect angle. But, I donít know how to get the projected NV once my vector is projected to XY plane. Then, can I apply the equation which gives angle between two vectors to compute angle of my desired vector from the y axis. (please mention the relevant equation)

On the other hand, I found, aspect angle is defined as the angle between any line which passes along the steepest slope of the plane and north direction (here, Y axis). Does this definition will follow, when taking normal vectos? I mean, does the projected normal vector always given along the steepest slope of the plane?
Please comment me. Thank you.
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