ParallelProjectionEllipse

Parallel Projection of an Ellipse David Eberly Geometric Tools, LLC http://www.geometrictools.com/ Copyright c 1998-2013. All Rights Reserved. Created: June 28, 2005 Last Modied: March 1, 2008 Contents 1 Discussion 2 1 1 Discussion This document describes an algorithm for projecting an ellipse in three dimensions onto a plane using a parallel projection. The plane of the ellipse is Ne
(XeCe) = 0, where Neis a unit-length normal vector, Ceis the plane origin, andXeis any point on the plane. Let UeandVebe vectors such that fUe;Ve;Negis an orthonormal set withWe=UeVe. Points on the plane are represented by Xe=Ce+ye;0Ue+ye;1Ve=Ce+h UeVei2 4ye;0 ye;13 5=Ce+JeYe (1) whereJeis the 32 matrix whose columns are the 3 1 vectors UeandVeand where Yeis the 21 vector whose rows are ye;0andye;1. The use of subscript eis to remind you these quantities are associated with the ellipse plane. Similarly, the projection plane is Np
(XpCp) = 0, where Npis a unit-length normal vector, Cpis the plane origin, and Xpis any point on the plane. Let UpandVpbe vectors such that fUp;Vp;Npgis an orthonormal set with Np=UpVp. Points on the plane are represented by Xp=Cp+yp;0Up+yp;1Vp=Cp+h UpVpi2 4yp;0 yp;13 5=Cp+JpYp (2) whereJpis the 32 matrix whose columns are the 3 1 vectors UpandVpand where Ypis the 21 vector whose rows are yp;0andyp;1. The use of subscript pis to remind you these quantities are associated with the projection plane. The projection is parallel to a specied unit-length direction vector D. For the projection to be invertible, it is required that Np
D6= 0 and Ne
D6= 0. The rst condition guarantees the projection actually exists and is onto the projection plane. The second condition guarantees that the projection is one-to-one (the ellipse does not project to a line segment). The two conditions, onto and one-to-one, mean the projection is invertible. The projection of Xeto a point Xpis the intersection of the line Xp=Xe+tDwith the projection plane. Substituting this into the projection plane equation Np
(XpCp) = 0 and solving for t produces the intersection point Xp=XeNp
(XeCp) D
NpD (3) The ellipse plane origin was chosen to be the ellipse center Ce. We have the freedom to choose the projection plane origin any way we like. Conveniently, let us choose it to be the projection of the ellipse center onto the projection plane, specically, Cp=CeNp
(CeCp) D
NpD (4) Consequently, equation (3) is equivalent to XpCp=XeCeNp
(XeCe) D
NpD= I3DNT p D
Np! (XeCe) (5) whereI3is the 33 identity matrix. 2 The plane origins in equations (1) and (2) may be subtracted from the equations to obtain XeCe=JeYe andXpCp=JpYp. Substitute these into equation (5) to obtain JpYp= I3DNT p D
Np! JeYe (6) Using the fact that JT pJp=I2, whereI2is the 22 identity matrix, we have Yp=JT p I3DNT p D
Np! JeYe=AYe (7) where the second equality denes the 2 2 matrixA. Since we know the projection is invertible, it must be the case that Ais invertible, in which case Ye=A1Yp (8) The ellipse is represented parametrically by Xe() =Ce+ (acos)Ue+ (bsin)Ve=Ce+JeYe() (9) where Ye() = [acos bsin]T, a 21 vector, and where 2[0;2). The positive values aandbare the ellipse major and minor axis lengths. Dene
to be the diagonal matrix whose diagonal entries are 1 =a and 1=b. Then
Ye=2 41=a 0 0 1=b3 52 4acos bsin3 5=2 4cos sin3 5 Since the right-hand side is a unit-length vector, we see that YT e
2Ye= 1 (10) Substituting equation (8) into this produces YT pMYp=YT p AT
2A1
Yp= 1 (11) whereATis the transpose of the inverse of A. The rst equality denes the matrix M. Equation (11) is the representation of the projected ellipse in the coordinate system of the projection plane. The projection is itself an ellipse. The matrix Mcan be factored into M=RDRT, whereRis a rotation matrix whose columns are eigenvectors for Mand whereD= Diag(d0;d1) are the eigenvalues, both positive. The projected ellipse major and minor axis lengths are 1 =pd0and 1=pd1. 3