3D Projection Onto a 2D PlaneDate: 7/9/96 at 15:22:57 From: Anonymous Subject: 3D Projection Onto a 2D Plane I'm dealing with a theoretical geological problem where I consider a roughly planar orebody as being 2D. I have drill-hole values and 3D coordinates for a number of points. Fitting a plane to these points is simple. What I want to do is project these points onto the plane, estimate values in 2D and then project the estimates back onto the original orebody. Many thanks. Date: 7/10/96 at 13:34:48 From: Doctor Jerry Subject: Re: 3D Projection Onto a 2D Plane I'm not sure I understand what you want. Here's what I understand. 1. You have several points (x1,y1,z1), (x2,y2,z2), . . . ,(xn,yn,zn). 2. You have fitted a plane, say z = Ax+By+C, to these points. 3. You want to "project these points onto the plane...". I take this to mean that you want to find where a vertical line through, say, (xi,yi,zi), passes through the plane. If this is what you want, then all you have to do is substitute xi and yi into the equation of the plane. The projection of (xi,yi,zi) onto the plane would be (xi,yi,Axi+Byi+C). 4. I don't know what you mean by estimate values in 2D and then project the estimates back onto the original orebody. I'd appreciate a little help with what you want. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 7/14/96 at 22:43:28 From: Anonymous Subject: Re: 3D Projection Onto a 2D Plane Thank you for your reply. I really haven't been clear at all have I? The problem is as follows. I am attempting to deal with an almost flat 3D ore-body in 2D. Given a set of sample values and thicknesses throughout the body I want to fit the best plane to the points. We assume that the 3D locations are placed at the centre of lines drawn perpendicular to the plane. Fitting the plane z = Ax+By+C to these points I can do. In order to reduce the problem to a 2D one I intend to work with accumulations (grade * thickness). For each point (x,y,z) I need a transformation to the fitted plane so that the new coordinates would be (x'=f(x,y,z), y'=g(x,y,z) z'=0). Each point should be projected onto the plane along a line passing through the point and perpendicular to the fitted plane. Once I have (x',y') I can use 2D geostatistics to evaluate the ore- body. Through the process of semi-variogram modeling and kriging I should be able mark off the plane into rectangles and estimate values and statistical variances for each of them. Once I have done this I will need to transform the centres of each rectangle back into the original 3D coordinates. The reason for all this is that geostatistics is much easier in 2D. 3D geostatistics tends to take the form of evaluating multiple horizontal slices, somewhat messy and each slice is evaluated independently from the next slice. Since this is almost a planar situation 2D evaluation would be the best option. Many thanks for your help. Date: 7/29/96 at 13:32:0 From: Doctor Jerry Subject: Re: 3D Projection Onto a 2D Plane Perhaps I'm beginning to understand what you want. Let me state my understanding and a reply. If you haven't given up on me and this still isn't what you want, please feel free to try again. It appears to me that the (x',y') values are coordinates of the projected points, but relative to a system of coordinates in the plane. This requires that we somehow choose a set of coordinate axes in the fitted plane. In my earlier comments, I stated how to obtain the coordinates (x,y,z) of the projection of a sample point into the fitted plane. With this as a given, so that z = Ax+By+C, we need to obtain (x',y'), which gives the position of this point relative to a set of axes in the plane. Here's an example. Suppose the plane is z=2x-3y+5 and we are considering the point (4,7,-8) in this plane. A normal vector to this plane is (-2,3,1). So, by inspection, a vector perpendicular to the plane is (3,2,0). We may cross (-2,3,1) and (3,2,0) to get the vector (-2,3,-13). Let e1=(1/sqrt(13))(3,2,0) and e2=(1/sqrt(182))(-2,3,-13). These are orthogonal unit vectors in the fitted plane. Now choose a fixed point in the plane as origin, say, (0,0,5). To calculate x' and y' for (4,7,-8), we calculate the vector from (4,7,-8) to (0,0,5), which is (4,7,-8)-(0,0,5) = (4,7,-13). This vector lies in the plane and, hence, can be written as a linear combination of e1 and e2, which also lie in this plane. So, we solve (4,7,-13)=x'*e1+y'e2, obtaining x'=2 sqrt(13) and y'=sqrt(182). These are the 2D coordinates of the point (4,7,-8). -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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