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3D Projection Onto a 2D Plane


Date: 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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College Higher-Dimensional Geometry

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