Distance from point to plane

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Nicolás
Nicolás 2014-12-7
回答： Thorsten 2014-12-15
Hi everywhere!
A question; How can I compute the distance between a plane (in ellipsoidal coordinates) to a point (in ellipsoidal coordinates too)?
The plane have 4 points (the borders points), and I need calculate the closest distance from this plane to a point.
For example;
%Lon. %Lat. %Depth (km)
Plane= [-74.65180 -37.74230 1.62620
-72.86330 -33.14430 1.62620
-70.88940 -33.64880 62.19350
-72.67800 -38.24670 62.19350
-74.65180 -37.74230 1.62620]
Point=[-72.7060 -37.7950]
Thank you so much!

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回答（3 个）

Matt J
Matt J 2014-12-7
编辑：Matt J 2014-12-7
Well, the fact that you have ellipsoidal coordinates shouldn't matter, because you can convert them to Cartesian coordinates PlaneCart and PointCart. Once you've done so
c=mean(PlaneCart);
N=null(bsxfun(@minus, PlaneCart,c));
N=mean(N,2);
distance=abs(dot(N,c-PointCart));
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Nicolás
Nicolás 2014-12-8
Yes, but obtain similar "d" values with my script. This "d" values are wrong.
Matt J
Matt J 2014-12-9
Let's do this together with the following sample data
PlaneCart=eye(3); PointCart=[0,0,0];
When I run with the above inputs, I obtain
distance =
0.5774
Do you not obtain the same thing? And do you disagree that this is the correct distance from the origin to the plane containing the rows of eye(3)? If you disagree, then you have given us the wrong description of your problem in some way.

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Thorsten
Thorsten 2014-12-8
This is a math question rather than a Matlab question. First convert your plane from 3-point-form to Hessian normal form http://mathworld.wolfram.com/Plane.html. Second compute the point-plane distance http://mathworld.wolfram.com/HessianNormalForm.html
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Nicolás
Nicolás 2014-12-8
Thank you for answered.
The mathematical problem is clear, the use of Hessian normal is useful for define the signs if the point is on the same side of the plane as the normal vector and negative if it is on the opposite side ( refer ). Therefore the next script is similar to use the Hessian normal form (but not work);
Plane= [-37.7423 -74.6518 1.6262
-33.1443 -72.8633 1.6262
-33.6488 -70.8894 62.1935
-38.2467 -72.6780 62.1935];
Point=[ -33.3210 -71.4110];
for i=1:4
[x(i),y(i),z(i)]=ell2xyz(deg2rad(Plane(i,1)),deg2rad(Plane(i,2)),-Plane(i,3)*1000);
end
P1=[x(1),y(1),z(1)]/1000; %en km
P2=[x(2),y(2),z(2)]/1000; %en km
P3=[x(3),y(3),z(3)]/1000; %en km
normal = cross(P1-P2, P1-P3);
d=dot(normal, P1); d=-d;
%Equation form: a*x+b*y+c*z+d=0
%Traspaso a coordenadas cartesianas
[xPo,yPo,zPo]=ell2xyz(deg2rad(Point(1)),deg2rad(Point(2)),0);
xPo=xPo/1000; yPo=yPo/1000; zPo=zPo/1000; %en km
d1=norm(normal(1)*xPo+normal(2)*yPo+normal(3)*zPo+d);
d2=sqrt(normal(1)^2+normal(2)^2+normal(3)^2);
d=(d1/d2);
The problem is to implement in Matlab, because, the result "d" is wrong.
Regards.

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Thorsten
Thorsten 2014-12-15
Plane= [-37.7423 -74.6518 1.6262
-33.1443 -72.8633 1.6262
-33.6488 -70.8894 62.1935
-38.2467 -72.6780 62.1935];
Point=[ -33.3210 -71.4110];
for i=1:4
[x(i),y(i),z(i)]=ell2xyz(deg2rad(Plane(i,1)),deg2rad(Plane(i,2)),-Plane(i,3)*1000);
end
P1=[x(1),y(1),z(1)]/1000; %en km
P2=[x(2),y(2),z(2)]/1000; %en km
P3=[x(3),y(3),z(3)]/1000; %en km
P4=[x(4) y(4) z(4)]/1000;
n = cross(P1-P2, P1-P3);
d = dot(-P1, n)
nnorm = n/norm(n);
p = d/norm(n);
Distance = dot(nnorm, P4) + p
Distance =
2.8967
That means that Point 4 is not in the plane, but has a distance of 2.89 km.
[xPo,yPo,zPo]=ell2xyz(deg2rad(Point(1)),deg2rad(Point(2)),0);
Po = [xPo yPo zPo];
Distance = dot(nnorm, Po) + p
Distance =
-6.0717e+006
So point Po lies 6.07 km low the plane. Does this make sense?

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