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Received 08/26/2015 Review began 09/01/2015 Review ended 09/21/2015 Published 10/02/2015
© Copyright 2015Brown. This is an open access articledistributed under the terms of theCreative Commons Attribution LicenseCC-BY 3.0., which permits unrestricteduse, distribution, and reproduction in anymedium, provided the original author andsource are credited.
The Mathematics of Three N-Localizers UsedTogether for Stereotactic NeurosurgeryRussell A. Brown
1.
Corresponding author: Russell A. Brown, [email protected]
AbstractThe N-localizer enjoys widespread use in image-guided stereotactic neurosurgery and radiosurgery. Thisarticle derives the mathematical equations that are used with three N-localizers and provides analogies,explanations, and appendices in order to promote a deeper understanding of the mathematical principlesthat govern the N-localizer.
Categories: Medical Physics, Radiation Oncology, NeurosurgeryKeywords: stereotactic neurosurgery, stereotactic radiosurgery, image guidance, image-guided, computedtomography, magnetic resonance imaging, positron emission tomography (pet), n-localizer, medical imaging, brainimaging
IntroductionThe N-localizer is a device that may be attached to a stereotactic frame (Figure 1) in order to facilitateimage-guided neurosurgery and radiosurgery using tomographic images that are obtained via computedtomography (CT), magnetic resonance (MR), or positron-emission tomography (PET) [1]. The mathematicsof the N-localizer have been discussed previously [2].
FIGURE 1: Three N-Localizers Attached to a Stereotactic FrameThree N-localizers are attached to this stereotactic frame and are merged end-to-end such that only sevenrods are present. The vertical rod at the right rear of the frame is larger in diameter than the other rods. Thislarge rod facilitates unambiguous interpretation of the fiducial circles and ellipses that the seven rods createin a tomographic image, as explained in the legend to Figure 5.
Technical ReportThe N-localizer comprises a diagonal rod that extends from the top of one vertical rod to the bottom of
Open Access TechnicalReport DOI: 10.7759/cureus.341
How to cite this articleBrown R A (October 02, 2015) The Mathematics of Three N-Localizers Used Together for Stereotactic Neurosurgery. Cureus 7(10): e341. DOI10.7759/cureus.341
another vertical rod (Figure 2). Assuming for the sake of simplicity that the two vertical rods areperpendicular to the tomographic section, the cross section of each vertical rod creates a fiducial circle andthe cross section of the diagonal rod creates a fiducial ellipse in the tomographic image, as shown in Figure2b. As the tomographic section moves from the top of the N-localizer towards the bottom of the N-localizer ,i.e. towards its point of attachment to the stereotactic frame (Figure 1), the ellipse will move away fromcircle and toward circle . The relative spacing between these three fiducials permits precise localizationof the tomographic section with respect to the N-localizer. The distance between the centers of circle and ellipse , and the distance between the centers of circles and are used to calculate the ratio
. This ratio represents the fraction of diagonal rod that extends from the top of vertical rod to the point of intersection of rod with the tomographic section. These linear geometric relationships existdue to the properties of similar triangles and are valid even if the vertical rods are not perpendicular to thetomographic section [3].
FIGURE 2: Intersection of the Tomographic Section with the N-Localizer Side view of the N-localizer. The tomographic section intersects the rods , , and . Tomographic
image. The intersection of the tomographic section with the rods , , and creates fiducial circles and and fiducial ellipse in the tomographic image. The distance between the centers of circle and ellipse
and the distance between the centers of circles and are used to calculate the ratio .This ratio represents the fraction of diagonal rod that extends from the top of rod to the point ofintersection of rod with the tomographic section.
It is convenient to ignore the thickness of the tomographic section and to approximate the tomographicsection as an infinitely thin plane. This "central" plane lies midway between the top and bottom halves ofthe tomographic section, analogous to the way that a slice of cheese is sandwiched between two slices ofbread. The central plane approximation is susceptible to error because of the partial volume effect thatderives from the several-millimeter thickness of the tomographic section [4-5]. The partial volume effectprevails because any structure that passes partially into the tomographic section, but does not span the fullthickness of that section, may be visible in the tomographic image. Hence, the position of that structure is
BA C
dAB AB dAC A C
f = /dAB dAC B AB
A B CA B C A C
B dAB AB dAC A C f = /dAB dAC
B AB
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determined to only a several-millimeter error that is a well-known limitation of tomographic imaging. In thefollowing discussion, the term "tomographic section" will be used as an abbreviation for the term "centralplane of the tomographic section."
The fraction is used to calculate the coordinates of the point of intersection between the longaxis of rod and the tomographic section (Figure 3). In this figure, points and represent thebeginning and end, respectively, of the vector that extends from the top of rod to the bottom of rod .This vector coincides with the long axis of rod . The coordinates of the beginning point andthe coordinates of the end point are known from the physical dimensions of the N-localizer.Hence, linear interpolation may be used to blend points and to obtain the coordinates ofthe point of intersection between the long axis of rod and the tomographic section
The vector form of Equation 1 shows explicitly the coordinates of points , , and
FIGURE 3: Calculation of the Point of Intersection Between the Rod Band the Tomographic SectionThe long axis of rod is represented by a vector that extends from point at the top of rod to point atthe bottom of rod . The coordinates of point and the coordinates of point areknown from the physical dimensions of the N-localizer. Hence, the ratio may be used to blend the
and coordinates of points and via linear interpolation as indicated by Equations 1and 2. This interpolation calculates the coordinates of the point of intersection between thelong axis of rod and the tomographic section.
Equation 1 or 2 may be used to calculate the coordinates of the point of intersection between
f (x, y, z) P ′B
B P ′A P ′
C
A CB ( , , )xA yA zA P ′
A
( , , )xC yC zC P ′C
P ′A P ′
C ( , , )xB yB zB
P ′B B
= + f ( − ) = f + (1 − f) (1)P ′B P ′
A P ′C P ′
A P ′C P ′
A
(x, y, z) P ′A P ′
B P ′C
[ ] = f [ ] + (1 − f) [ ]xB yB zB xC yC zC xA yA zA
(2)
B P ′A
A P ′C
C ( , , )xA yA zA P ′A
( , , )xC yC zC P ′C
f = /dAB dAC
( , , )xA yA zA ( , , )xC yC zC P ′A
P ′C
( , , )xB yB zB P ′B
B
( , , )xB yB zB P ′B
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the long axis of rod and the tomographic section. The point , which lies on the long axis of rod in thethree-dimensional coordinate system of the N-localizer, corresponds to the analogous point , which liesat the center of ellipse in the two-dimensional coordinate system of the tomographic image (Figure 2b).Hence, there is a one-to-one linear mapping between a point from the N-localizer and a point from thetomographic image.
The attachment of three N-localizers to a stereotactic frame permits calculation of the , , and coordinates for the three respective points, , , and , in the
three-dimensional coordinate system of the stereotactic frame (Figure 4). These three points correspondrespectively to the three analogous points, , , and , in the two-dimensional coordinate system ofthe tomographic image. In the following discussion, the symbols, , , and , will be used as a shorthandnotation for , , and . The symbols, , , and , will be used as a shorthand notation for ,
, and .
FIGURE 4: Representation of the Tomographic Section in the Three-Dimensional Coordinate System of the Stereotactic FrameThe quadrilateral represents the tomographic section. The large oval depicts the circular base of thestereotactic frame (in perspective). The vertical and diagonal lines that are attached to the large ovalrepresent the nine rods. The centers of the six fiducial circles and the three fiducial ellipses that are createdin the tomographic image by these nine rods are shown as points that lie in the tomographic section. Thetomographic section intersects the long axes of the three diagonal rods at the points , , and thatcoincide with the respective centers , , and of the three ellipses (Figure 6). The , , and
coordinates of the respective points of intersection , , and are calculated in the three-dimensional coordinate system of the stereotactic frame using Equations 1 and 2. Because these threepoints determine the spatial orientation of a plane in three-dimensional space, the spatial orientation of thetomographic section is determined with respect to the stereotactic frame. The target point lies in thetomographic section. The coordinates of this target point are calculated in the three-dimensionalcoordinate system of the stereotactic frame using Equation 6.
The three points, , , and , lie on the three respective diagonal rods, , , and , and haverespective coordinates, , , and , in the three-dimensional coordinatesystem of the stereotactic frame (Figure 4). The analogous three points, , , and , lie at the centers ofthe three respective ellipses, , , and , and have coordinates, , , and , in the
B P ′B B
PB
B
( , , )xB1yB1
zB1
( , , )xB2yB2
zB2( , , )xB3
yB3zB3
P ′B1
P ′B2
P ′B3
PB1PB2
PB3
P ′1 P ′
2 P ′3
P ′B1
P ′B2
P ′B3
P1 P2 P3 PB1
PB2PB3
P ′1 P ′
2 P ′3
P1 P2 P3 ( , , )x1 y1 z1 ( , , )x2 y2 z2
( , , )x3 y3 z3 P ′1 P ′
2 P ′3
P ′T
( , , )xT yT zT
P ′1 P ′
2 P ′3 B1 B2 B3
(x, y, z) ( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3
P1 P2 P3
B1 B2 B3 (u, v) ( , )u1 v1 ( , )u2 v2 ( , )u3 v3
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two-dimensional coordinate system of the tomographic image (Figures 5-6).
FIGURE 5: CT Image with Three Sets of FiducialsCT image of a patient to whom a BRW CT localizer frame (Integra Radionics Inc., Burlington, MA), whichcomprises three N-localizers, is attached. The cross sections of the three N-localizers create three sets offiducials , , and in the CT image. The cursor indicates the target point .The large vertical rod allows it to be unambiguously distinguished from the other vertical rods andprovides a visual cue that this figure is rotated approximately 90 degrees clockwise relative to Figure 6 [6].
3 1 1 2 2 3 3
{ , , }A1 B1 C1 { , , }A2 B2 C2 { , , }A3 B3 C3 PT
A1
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FIGURE 6: Representation of the Two-Dimensional Coordinate Systemof the Tomographic ImageThe cross sections of the three N-localizers create three sets of fiducials , , and
in the tomographic image. Each set contains two circles and one ellipse that are collinear. Foreach set, the short double-ended arrows indicate the distance between the centers of circle andellipse and the long double-ended arrows indicate the distance between the centers of circles and .The centers , , and of the three ellipses coincide with the respective points of intersection , , and
of the long axes of the three diagonal rods with the tomographic section (Figure 4). The , , and coordinates of the centers , , and correspond respectively to the , , and
coordinates of the points of intersection , , and . The target point has coordinatesin the two-dimensional coordinate system of the tomographic image. The coordinates of theanalogous target point are calculated in the three-dimensional coordinate system of the stereotacticframe using Equation 6.
In order to facilitate calculation of the coordinates of the target point , it is convenient toproject the , , and coordinates of the three centers, , , and , of the ellipses ontothe plane in three-dimensional space by appending a third coordinate to create ,
, and coordinates. The -coordinate may be set arbitrarily to any non-zero value, e.g., 1,so long as same value of is used for each of the three -coordinates. The equations that are presented inthe remainder of this article assume that a value of has been used to project the , , and
coordinates. If a value of were used instead of to project these coordinates, theequations that are presented in the remainder of this article would no longer apply and would requirerevision so that the calculations that these equations describe may produce correct results.
Because three points determine the orientation of a plane in three-dimensional space, thethree coordinates, , , and , together with the three coordinates, ,
, and , determine the spatial orientation of the tomographic section with respect to thestereotactic frame. This spatial orientation or linear mapping is specified by the matrix elements
{ , , }A1 B1 C1 { , , }A2 B2 C2
{ , , }A3 B3 C3
dAB AB dAC A C
P1 P2 P3 P ′1 P ′
2P ′
3 ( , )u1 v1 ( , )u2 v2
( , )u3 v3 P1 P2 P3 ( , , )x1 y1 z1 ( , , )x2 y2 z2
( , , )x3 y3 z3 P ′1 P ′
2 P ′3 PT ( , )uT vT
( , , )xT yT zT
P ′T
( , , )xT yT zT P ′T
( , )u1 v1 ( , )u2 v2 ( , )u3 v3 P1 P2 P3
w = 1 w = 1 ( , , 1)u1 v1
( , , 1)u2 v2 ( , , 1)u3 v3 w
w w
w = 1 ( , )u1 v1 ( , )u2 v2
( , )u3 v3 w ≠ 1 w = 1
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 ( , )u1 v1
( , )u2 v2 ( , )u3 v3
m11
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through in the matrix equation
Equation 3 represents concisely a system of nine simultaneous linear equations that determine the spatialorientation of the tomographic section with respect to the stereotactic frame. This equation transforms the
, , and coordinates from the two-dimensional coordinate system of the tomographicimage to create , , and coordinates in the three-dimensional coordinatesystem of the stereotactic frame.
Equation 3 assumes a linear mapping from the two-dimensional coordinate system of the tomographicimage to the three-dimensional coordinate system of the stereotactic frame. Magnetic resonance (MR)images are susceptible to nonlinear distortion that invalidates this linear mapping and nullifies theapplicability of Equation 3. For this reason, the Brown-Roberts-Wells (BRW) stereotactic frame [7] that wasused initially with computed tomography (CT) required modification to eliminate nonlinear distortion ofMR images. The CT-compatible BRW frame comprised an aluminum ring in which the magnetic field that theMR scanner generated to acquire MR images induced eddy currents. Those eddy currents distorted the MRimages. Replacing one section of the aluminum ring with a nonmetallic insert prevented magneticallyinduced circumferential eddy currents and eliminated nonlinear distortion of the MR images [8].
An analogy provides insight into how the transformation of Equation 3 operates. Consider the tomographicimage to be an elastic membrane. The transformation describes the process of stretching the membrane inthe plane of the tomographic image, rotating the membrane about an axis that is normal to the plane of thetomographic image, tilting the membrane, if necessary, so that it is not parallel to the base of thestereotactic frame, and lifting the membrane into place upon the scaffold of the three N-localizers, such thatthe three points, , , and , from the tomographic image precisely coincide with the respectivethree points, , , and , from the stereotactic frame. Then, any other point that lies on themembrane, e.g., the target point , is transformed by the same stretching, rotating, tilting, and liftingprocesses that transformed the three points, , , and . In this manner, the coordinates of thetarget point may be transformed from the two-dimensional coordinate system of the tomographic imageinto the three-dimensional coordinate system of the stereotactic frame to produce the
coordinates of the analogous target point .
Equation 3 may be rewritten in more compact form as
In Equation 4, represents the matrix of , , and coordinates in the coordinatesystem of the stereotactic frame. represents the matrix of , , and coordinatesin the coordinate system of the tomographic image. represents the matrix of elements, through ,that defines the transformation from the two-dimensional coordinate system of the tomographic image tothe three-dimensional coordinate system of the stereotactic frame.
The elements of and are known, but the elements of are unknown. It is possible to solve Equation 4 forthe elements of
In this equation, represents the inverse of matrix . The inverse of is guaranteed to exist so long as the, , and coordinates of the centers of the three ellipses , , and are not
collinear. This non-collinearity is enforced by careful design of the stereotactic frame, as will be explainedbelow in the Discussion.
Once the elements of matrix have been calculated via Equation 5, it is possible to transform the coordinates of the target point from the two-dimensional coordinate system of the tomographic image tothe three dimensional coordinate system of the stereotactic frame to obtain the coordinates ofthe analogous target point . In order to accomplish this transformation, the coordinates of areused to form the vector that is post-multiplied by matrix to obtain the vector thatcontains the coordinates of
Moreover, it is possible to calculate the inverse of matrix
The inverse matrix may be used perform a transformation analogous to the transformation of Equation6 but in the reverse direction. This reverse transformation transforms the coordinates of a point
from the three-dimensional coordinate system of the stereotactic frame to the two-dimensional
m33
=x1
x2
x3
y1
y2
y3
z1
z2
z3
u1
u2
u3
v1
v2
v3
111
m11
m21
m31
m12
m22
m32
m13
m23
m33
(3)
( , )u1 v1 ( , )u2 v2 ( , )u3 v3
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3
P1 P2 P3
P ′1 P ′
2 P ′3
PT
P1 P2 P3 ( , )uT vT
PT
( , , )xT yT zT P ′T
= (4)
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3
( , , 1)u1 v1 ( , , 1)u2 v2 ( , , 1)u3 v3
m11 m33
= (5)−1
−1
( , , 1)u1 v1 ( , , 1)u2 v2 ( , , 1)u3 v3 B1 B2 B3
( , )uT vT
PT
( , , )xT yT zT
P ′T ( , )uT vT PT
[ 1]uT vT [ ]xT yT zT
( , , )xT yT zT P ′T
[ ] = [ 1] (6)xT yT zT uT vT
= (7)−1 −1
−1
( , , )xQ yQ zQ
P ′Q
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coordinate system of the tomographic image to obtain the coordinates of the analogous point . Inorder to accomplish this reverse transformation, the coordinates of are used to form thevector that is post-multiplied by matrix to obtain the vector that contains the
coordinates of
Equation 8 yields coordinates for instead of coordinates. The -coordinate equals 1 if and only if the point lies in the tomographic section that corresponds to the plane inthree-dimensional space [2]. Similarly, appears in the two-dimensional tomographic image if and only if
. In the case that does not lie in the tomographic section, so does not appear in thetomographic image.
One case where does not appear in the tomographic image occurs when the point and a second point define the intended trajectory of a surgical probe but neither nor lies in an intermediate
tomographic section (Figure 7). In this case, and , so neither nor appears in theintermediate tomographic image.
However, in this case, the neurosurgeon may wish to know where the intended probe trajectory wouldintersect the intermediate tomographic section. In order to provide this information, the points and
are used to define the vector from to . This vector is then used to calculate the coordinatesof a third point for which (Figure 7). Because , appears in the intermediate tomographicimage; hence, a mark may be superimposed on that tomographic image at the coordinates toindicate where the intended probe trajectory would intersect the intermediate tomographic section [2]. It ispossible to distinguish two configurations of and relative to an intermediate tomographic image:
and . All other configurations can be made to conform to one of these twoconfigurations via interchange of and and/or inverting the signs of both and . Theconfiguration specifies that and are located on opposite sides of an intermediatetomographic image; thus, linear interpolation may be used to calculate (Figure 7). The configuration
specifies that and are located on the same side of a non-intermediate tomographicimage; thus, linear extrapolation may be used to calculate (Figure 8).
( , )uQ vQ PQ
( , , )xQ yQ zQ P ′Q
[ ]xQ yQ zQ−1 [ ]uQ vQ wQ
( , , )uQ vQ wQ PQ
[ ] = [ ] (8)uQ vQ wQ xQ yQ zQ−1
( , , )uQ vQ wQ PQ ( , , 1)uQ vQ w wQ
P ′Q w = 1
PQ
= 1wQ P ′Q ≠ 1wQ PQ
PQ P ′Q
P ′R P ′
Q P ′R
≠ 1wQ ≠ 1wR PQ PR
PQ
PR PQ PR ( , )uS vS
PS = 1wS = 1wS PS
( , )uS vS PS
PQ PR
> 1 >wQ wR > > 1wQ wR
PQ PR wQ wR
> 1 >wQ wR PQ PR
PS
> > 1wQ wR PQ PR
PS
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FIGURE 7: Interpolation Within the Vector from to in Order toObtain the Point that Appears in the Tomographic ImageThe points and are located on opposite sides of an intermediate tomographic image for which .The distances and are used to obtain the interpolant
This interpolant is used to calculate the coordinates of the point that appears in the tomographicimage.
PQ PR
PS
PQ PR = 1wS
− 1wQ − = ( − 1) − ( − 1)wQ wR wQ wR
t =− 1wQ
−wQ wR
( , )uS vS PS
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FIGURE 8: Extrapolation Beyond the Vector from to in Order toObtain the Point that Appears in the Tomographic ImageThe points and are located on the same side of a non-intermediate tomographic image for which
. The distances and are used to obtain the extrapolant
This extrapolant is used to calculate the coordinates of the point that appears in the tomographicimage.
For either interpolation or extrapolation, the term
is used to calculate the coordinates of by blending the and coordinates of and
The vector form of Equation 10 shows explicitly the , , and coordinates ofthe respective points , and
It is necessary to calculate only the coordinates of using Equation 11 because due to thedefinition of in Equation 9. It is possible to prove that by substituting Equation 9 into Equation 11then expanding the resulting expression in the -coordinate to obtain
DiscussionThe above mathematical formulation imposes some constraints on the physical design of the stereotacticframe and on the mathematical model of that frame. Specifically, Equations 5 and 7 require that themathematical model of the frame permit the inverse matrices, and , to exist.
The inverse matrix exists if and only if the points , , and are neither collinear nor lie on a planethat passes through the origin of the coordinate system. Similarly, the inverse exists if and onlyif the points , , and are neither collinear nor lie on a plane that passes through the origin of the
PQ PR
PS
PQ PR
= 1wS − 1wQ − = ( − 1) − ( − 1)wQ wR wQ wR
t =− 1wQ
−wQ wR
( , )uS vS PS
t = (9)− 1wQ
−wQ wR
( , )uS vS PS ( , )uQ vQ ( , )uR vR
PQ PR
= + t ( − ) = t + (1 − t) (10)PS PQ PR PQ PR PQ
( , , )uQ vQ wQ ( , , )uR vR wR ( , , )uS vS wS
PQ PR PS
[ ] = t [ ] + (1 − t) [ ]uS vS wS uR vR wR uQ vQ wQ
(11)
( , )uS vS PS = 1wS
t = 1wS
w
= + 1 −wS
− 1wQ
−wQ wR
wR
− 1wQ
−wQ wR
wQ
= = 1 (12)( − 1) + (1 − )wQ wR wR wQ
−wQ wR
−1 −1
−1 P1 P2 P3
(u, v, w) −1
P ′1 P ′
2 P ′3
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coordinate system.
The collinearity requirement is satisfied for both and by judiciously choosing the positions of thethree N-localizers relative to the stereotactic frame. Because for contemporary stereotactic frames the N-localizers are positioned either around the circumference of a circle or on the faces of a cube, neither thepoints , , and nor the points , , and can possibly be collinear.
The requirement that the points , , and do not lie on a plane that passes through the origin of the coordinate system is satisfied by choosing to project the , , and
coordinates to create , , and coordinates.
The requirement that the points , , and do not lie on a plane that passes through the origin of the coordinate system may be satisfied by judiciously defining the coordinate system of the
stereotactic frame, such that the -coordinate cannot equal zero anywhere along the diagonal rods. Figure 3demonstrates that Equations 1 and 2 will never produce along the diagonal rods so long as the intervalfrom to does not contain zero. One way to satisfy this requirement is to define the origin of the
coordinate system of the stereotactic frame to lie below the base of the N-localizers, such that the -coordinate of the origin is always less than the -coordinate of point .
When all three of the above requirements are satisfied, the matrix will correctly transform the , , and coordinates of points , , and from the two-dimensional coordinate system of
the tomographic image to create the , , and coordinates of points , , and in the three-dimensional coordinate system of the stereotactic frame as indicated by Equation 4. Also, the
inverse matrix will correctly perform the inverse of that transformation.
ConclusionsThe N-localizer is a simple yet powerful tool for image-guided stereotactic neurosurgery andradiosurgery. The N-localizer enables the transformation of coordinates from the two-dimensionalcoordinate system of the computed tomography (CT), magnetic resonance (MR) or positron-emissiontomography (PET) image to the three-dimensional coordinate system of the stereotactic frame to obtain
coordinates. The matrix that accomplishes this transformation may be inverted; the resulting inversematrix enables the transformation of coordinates from the three-dimensional coordinate system ofthe stereotactic frame to the two-dimensional coordinate system of the computed tomography or magneticresonance image to obtain coordinates.
AppendicesAppendix 1: Derivation of Equation 3Equation 3 transforms coordinates from the two-dimensional coordinate of the tomographic image tothe three-dimensional coordinate system of the stereotactic frame to produce coordinates. Prior touse in Equation 3, the coordinates are projected onto the plane in three-dimensional space byappending a third coordinate to create coordinates. Equation 3 is derived as follows.
Transformation of coordinates from one three-dimensional coordinate system to another three-dimensionalcoordinate system may be accomplished via matrix multiplication that operates in a four-dimensional space[9]. However, in order that this four-dimensional space may be used to transform the two-dimensional
coordinates into three-dimensional coordinates, it is necessary first to create three-dimensional coordinates by projecting the coordinates onto the plane in three-dimensional space by appending a third coordinate . Then it is necessary to create four-dimensional
coordinates by projecting the coordinates onto the hyperplane in four-dimensionalspace by appending a fourth, homogenous [10] coordinate . The coordinates may betransformed to obtain coordinates using a four by four transformation matrix that contains thematrix elements through
In Equation 13, the third row of the transformation matrix includes elements , , and and thefourth row includes elements , , and . This non-standard numbering convention for these matrixelements is convenient to the remainder of this derivation of Equation 3. Also, the matrix elements in thefourth column of this transformation matrix have the values of 0, 0, 0 and 1 because Equation 13 expressesan affine transformation that comprises only scale, rotate and translate operations [9]. Theseoperations accomplish the stretching, rotating, tilting and lifting processes that were described for the
(x, y, z)
−1 −1
P1 P2 P3 P ′1 P ′
2 P ′3
P1 P2 P3
(u, v, w) w = 1 ( , )u1 v1 ( , )u2 v2 ( , )u3 v3
( , , 1)u1 v1 ( , , 1)u2 v2 ( , , 1)u3 v3
P ′1 P ′
2 P ′3
(x, y, z) (x, y, z)z
z = 0zA zC
(x, y, z) z
zC PC
( , , 1)u1 v1
( , , 1)u2 v2 ( , , 1)u3 v3 P1 P2 P3
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 P ′1 P ′
2P ′
3−1
(u, v)
(x, y, z)(x, y, z)
(u, v)
(u, v)(x, y, z)
(u, v) w = 1w = 1 (u, v, 1)
(u, v) (x, y, z)(u, v, 0) (u, v) h = 0
h = 0(u, v, 0, 1) (u, v, 0) w = 1
w = 1 (u, v, 0, 1)(x, y, z, 1)
m11 m43
x1
x2
x3
y1
y2
y3
z1
z2
z3
111
= (13)u1
u2
u3
v1
v2
v3
000
111
m11
m21
m41
m31
m12
m22
m42
m32
m13
m23
m43
m33
00
01
m41 m42 m43
m31 m32 m33
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 11 of 21
membrane analogy in association with Equation 3.
Equation 13 may be rewritten in more compact form as
In Equation 14, represents the matrix of , , and coordinates. represents the matrix of , , and coordinates. represents the
transformation matrix of elements through that defines the transformation from the two-dimensional coordinate system of the tomographic image to the three-dimensional coordinate system of thestereotactic frame.
A comparison of Equations 3 and 13 reveals that both equations produce identical results for the , , and coordinates. The third column of and the third row of do not affect the , , and coordinates. The fourth column of affects only the fourth column of
but does not affect the , , and coordinates. Hence, these columns and thisrow may be removed from , , and without affecting the result of Equation 13. Their removal yieldsEquation 3, thus completing the derivation of Equation 3.
There is a significant difference between Equations 3 and 13. None of the matrices in Equation 13 have aninverse because neither nor is a square matrix. In contrast, the matrices , , and in Equation 3potentially have inverses because these matrices are square matrices. Equations 5, 7, and 8 require thatthese matrices have inverses. Hence, in order to express the transformation from the two-dimensionalcoordinate system of the tomographic image to the three-dimensional coordinate system of the stereotacticframe and vice versa, Equation 3 must be used instead of Equation 13.
Appendix 2: Derivation of the Distance Equation 9 calculates the interpolant or extrapolant in the coordinate system of the tomographicimage. This interpolant or extrapolant is calculated in terms of the perpendicular distance from a point
to the plane of the tomographic image. The distance is derived as follows.
In the three-dimensional coordinate system of the stereotactic frame, the equation for the centralplane of the tomographic section is given by the following equation that involves a determinant [11-12]
Expanding this determinant using the cofactors [12] of the elements , , , and 1 in the first row of thedeterminant yields
Equation 16 may be rewritten in more compact form as
where , , , and represent the determinants in Equation 16. The determinants , , and may beexpanded using the cofactors of the elements in their third columns as follows
The normalized perpendicular distance from a point , which has coordinates , to the central planeof the tomographic section may be calculated as [11]
= (14)
( , , , 1)x1 y1 z1 ( , , , 1)x2 y2 z2 ( , , , 1)x3 y3 z3
( , , 0, 1)u1 v1 ( , , 0, 1)u2 v2 ( , , 0, 1)u3 v3
m11 m43
( , , )x1 y1 z1
( , , )x2 y2 z2 ( , , )x3 y3 z3
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3
w − 1t (u, v, w)
w − 1P w − 1
(x, y, z)
= 0 (15)
∣
∣
∣∣∣∣
x
x1
x2
x3
y
y1
y2
y3
z
z1
z2
z3
11
11
∣
∣
∣∣∣∣
x y z
x − y + z
∣
∣
∣∣
y1
y2
y3
z1
z2
z3
111
∣
∣
∣∣
∣
∣
∣∣
x1
x2
x3
z1
z2
z3
111
∣
∣
∣∣
∣
∣
∣∣
x1
x2
x3
y1
y2
y3
111
∣
∣
∣∣
− = 0 (16)∣
∣
∣∣
x1
x2
x3
y1
y2
y3
z1
z2
z3
∣
∣
∣∣
Ax + By + Cz + D = 0 (17)
A B C D A B C
A = = − +∣
∣
∣∣
y1
y2
y3
z1
z2
z3
111
∣
∣
∣∣
∣∣∣y2
y3
z2
z3
∣∣∣
∣∣∣y1
y3
z1
z3
∣∣∣
∣∣∣y1
y2
z1
z2
∣∣∣
B = − = − + −∣
∣
∣∣
x1
x2
x3
z1
z2
z3
111
∣
∣
∣∣
∣∣∣x2
x3
z2
z3
∣∣∣
∣∣∣x1
x3
z1
z3
∣∣∣
∣∣∣x1
x2
z1
z2
∣∣∣
C = = − +∣
∣
∣∣
x1
x2
x3
y1
y2
y3
111
∣
∣
∣∣
∣∣∣x2
x3
y2
y3
∣∣∣
∣∣∣x1
x3
y1
y3
∣∣∣
∣∣∣x1
x2
y1
y2
∣∣∣
D = −∣
∣
∣∣
x1
x2
x3
y1
y2
y3
z1
z2
z3
∣
∣
∣∣
(18a)
(18b)
(18c)
(18d)
d P ′ (x, y, z)
d = (19)Ax + By + Cz + D
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 12 of 21
This equation for the normalized perpendicular distance will be compared to the equation for the distance that is derived below.
In order to calculate the distance , the coordinates of the point that corresponds to the point are obtained by transforming the coordinates of the point via Equation 8 then by substituting
the definitions of matrices , , and from Equation 3
Substituting the inverse of the matrix , which is defined as its adjoint [12] divided by its determinant, intoEquation 20 yields
Transformation of the coordinates of the point to obtain only the -coordinate of the point requires only the vector from the third column of . Hence, keeping only the third column of the matrixthat results from the post-multiplication of by produces the following expression for that contains athree-element column vector
Rewriting Equation 22 in more compact form using the definitions of , , , and from Equation 18 yields
Performing the vector multiplication of Equation 23 produces the -coordinate of the point
The perpendicular distance from the point to the plane of the tomographic image is given by
Comparison of Equation 25 to Equation 19 reveals that the numerators of these equations are identical buttheir denominators differ, as can be demonstrated by expanding the determinants , , , and thenshowing that . Thus, the distance that is calculated using Equation 25 differs by a factor of
from the normalized distance that is calculated using Equation 19. However, this factoris not relevant to the interpolant or extrapolant that is calculated via Equation 9 because Equation 9calculates a ratio of distances that eliminates this factor. Hence, may be used to construct theinterpolant or extrapolant according to Equation 9.
Appendix 3: Transformation of the Target Point and the Analogous
d = (19)Ax + By + Cz + D
+ +A2 B2 C 2− −−−−−−−−−−√
w − 1
w − 1 (u, v, w) P
P ′ (x, y, z) P ′
[u v w] = [x y z] = [x y z]−1 −1
= [x y z] (20)x1
x2
x3
y1
y2
y3
z1
z2
z3
−1u1
u2
u3
v1
v2
v3
111
[u v w]
= [x y z]
∣∣∣y2
y3
z2
z3
∣∣∣
− ∣∣∣x2
x3
z2
z3
∣∣∣
∣∣∣x2
x3
y2
y3
∣∣∣
− ∣∣∣y1
y3
z1
z3
∣∣∣
∣∣∣x1
x3
z1
z3
∣∣∣
− ∣∣∣x1
x3
y1
y3
∣∣∣
∣∣∣y1
y2
z1
z2
∣∣∣
− ∣∣∣x1
x2
z1
z2
∣∣∣
∣∣∣x1
x2
y1
y2
∣∣∣
∣
∣
∣∣
x1
x2
x3
y1
y2
y3
z1
z2
z3
∣
∣
∣∣
u1
u2
u3
v1
v2
v3
111
(21)
(x, y, z) P ′ w P−1
−1 w
w = [x y z]
− +y2
y3
z2
z3
∣∣∣y1
y3
z1
z3
∣∣∣
∣∣∣y1
y2
z1
z2
∣∣∣
− + −∣∣∣x2
x3
z2
z3
∣∣∣
∣∣∣x1
x3
z1
z3
∣∣∣
∣∣∣x1
x2
z1
z2
∣∣∣
− +∣∣∣x2
x3
y2
y3
∣∣∣
∣∣∣x1
x3
y1
y3
∣∣∣
∣∣∣x1
x2
y1
y2
∣∣∣
∣
∣
∣∣
x1
x2
x3
y1
y2
y3
z1
z2
z3
∣
∣
∣∣
(22)
A B C D
w = [x y z] (23)
A
B
C
−D
w P
w = (24)Ax + By + Cz
−D
P w − 1
w − 1 = (25)Ax + By + Cz + D
−D
A B C D
≠ + +D2 A2 B2 C 2
− /D+ +A2 B2 C 2− −−−−−−−−−−√
t
w − 1t
PT
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 13 of 21
Target Point Equation 3 expresses the transformation of points , , and from the two-dimensional coordinatesystem of the tomographic section to obtain the analogous points , , and in the three-dimensionalcoordinate system of the stereotactic frame. However, the above discussion of the membrane analogy assertsthat any point that lies in the plane of points , , and may be transformed from the two-dimensionalcoordinate system of the tomographic section to the three-dimensional coordinate system of the stereotacticframe via the same transformation that transforms points , , and . This assertion obtains due to theprinciples of linear algebra and is proved as follows.
Equation 3 transforms en masse the , , and coordinates of the three points , ,and to create the , , and coordinates of the three points , , and . Analternative is to transform separately the , , and coordinates of the three points
, , and to obtain the , , and coordinates of the three points , , and , respectively
Equation 26 produces the same result for the , , and coordinates of points , , and as does Equation 3.
It is possible to represent any point that lies in a plane defined by three other points as a linear combinationof those three points using the barycentric coordinates , , and that satisfy the condition
[13]. For example, with reference to Figure 9, the target point may be represented as alinear combination of the three points , , and
P ′T
P1 P2 P3
P ′1 P ′
2 P ′3
P1 P2 P3
P1 P2 P3
( , , 1)u1 v1 ( , , 1)u2 v2 ( , , 1)u3 v3 P1 P2
P3 ( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 P ′1 P ′
2 P ′3
( , , 1)u1 v1 ( , , 1)u2 v2 ( , , 1)u3 v3
P1 P2 P3 ( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 P ′1 P ′
2P ′
3
= [ ] = [ ] =P ′1 x1 y1 z1 1u1 v1
m11
m21
m31
m12
m22
m32
m13
m23
m33
P1
(26a)
= [ ] = [ ] =P ′2 x2 y2 z2 1u2 v2
m11
m21
m31
m12
m22
m32
m13
m23
m33
P2
(26b)
= [ ] = [ ] =P ′3 x3 y3 z3 1u3 v3
m11
m21
m31
m12
m22
m32
m13
m23
m33
P3
(26c)
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 P ′1
P ′2 P ′
3
b1 b2 b3
+ + = 1b1 b2 b3 P ′T
P ′1 P ′
2 P ′3
= + + (27)P ′T b1P ′
1 b2P ′2 b3P ′
3
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 14 of 21
FIGURE 9: Representation of the Tomographic Section in the Three-Dimensional Coordinate System of the Stereotactic FrameThe quadrilateral represents the tomographic section. The large oval depicts the circular base of thestereotactic frame (in perspective). The vertical and diagonal lines that are attached to the large ovalrepresent the nine rods. The centers of the six fiducial circles and the three fiducial ellipses that are createdin the tomographic image by these nine rods are shown as points that lie in the tomographic section. Thetomographic section intersects the long axes of the three diagonal rods at points , , and that coincidewith the respective centers , , and of the three ellipses (Figure 10). The target point lies in the planeof the triangle . Hence, its coordinates may be expressed as a linear combination of the
, , and coordinates of the points , , and using barycentric coordinates asindicated by Equation 27.
Similarly, with reference to Figure 10, the analogous target point may be represented as a linearcombination of the three points , , and .
P ′1 P ′
2 P ′3
P1 P2 P3 P ′T
P ′1P ′
2P ′3 ( , , )xT yT zT
( , , )x1 y1 z1 ( , , )x2 y2 z2 ( , , )x3 y3 z3 P ′1 P ′
2 P ′3
PT
P1 P2 P3
= + + (28)PT b1P1 b2P2 b3P3
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 15 of 21
FIGURE 10: Representation of the Two-Dimensional Coordinate Systemof the Tomographic ImageThe cross sections of the three N-localizers create three sets of fiducials , , and
in the tomographic image. Each set contains two circles and one ellipse that are collinear. Thecenters , , and of the three ellipses coincide with the respective points of intersection , , and ofthe long axes of the three diagonal rods with the tomographic section (Figure 9). The , , and
coordinates of the centers , , and correspond respectively to the , , and coordinates of the points of intersection , , and . The target point lies in the plane of the
triangle . Hence, its coordinates may be expressed as a linear combination of the , , and coordinates of the points , , and using barycentric coordinates as indicated by
Equation 28.
Because the matrix that transforms into via Equation 4 describes a linear transformation, thebarycentric coordinates , , and apply to both and . Hence, these barycentric coordinates may beused in both Equation 27 and Equation 28. These equations describe interpolation in a plane that isanalogous to the interpolation along a line that is expressed by Equation 1.
Using the matrix to transform the point as shown in Equation 6 and substituting Equations 26-28yields
Eliminating the intermediate steps from Equation 29 and showing explicitly the coordinates of and the coordinates of yields
{ , , }A1 B1 C1 { , , }A2 B2 C2
{ , , }A3 B3 C3
P1 P2 P3 P ′1 P ′
2 P ′3
( , )u1 v1 ( , )u2 v2
( , )u3 v3 P1 P2 P3 ( , , )x1 y1 z1 ( , , )x2 y2 z2
( , , )x3 y3 z3 P ′1 P ′
2 P ′3 PT
P1P2P3 ( , )uT vT ( , )u1 v1
( , )u2 v2 ( , )u3 v3 P1 P2 P3
b1 b2 b3 PT P ′T
PT
(29)
PT = ( + + )b1P1 b2P2 b3P3
= + +b1P1 b2P2 b3P3
= + +b1P ′1 b2P ′
2 b3P ′3
= P ′T
( , , )xT yT zT
P ′T ( , )uT vT PT
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 16 of 21
Equation 30 proves that any point that lies in the tomographic image may be transformed from the two-dimensional coordinate system of that image to the three-dimensional coordinate system of the stereotacticframe to obtain the analogous point .
Using the inverse matrix to transform the point as shown in Equation 8 and substituting Equations26-28 yields
Eliminating the intermediate steps from Equation 31 and showing explicitly the coordinates of and the coordinates of yields
Equation 32 proves that any point that lies in the plane of the tomographic section may be transformedfrom the three-dimensional coordinate system of the stereotactic frame to the two-dimensional coordinatesystem of the tomographic image to obtain the analogous point .
Additional InformationDisclosuresHuman subjects: All authors have confirmed that this study did not involve human participants or tissue.Animal subjects: All authors have confirmed that this study did not involve animal subjects or tissue.Conflicts of interest: In compliance with the ICMJE uniform disclosure form, all authors declare thefollowing: Payment/services info: All authors have declared that no financial support was received fromany organization for the submitted work. Financial relationships: All authors have declared that they haveno financial relationships at present or within the previous three years with any organizations that mighthave an interest in the submitted work. Other relationships: All authors have declared that there are noother relationships or activities that could appear to have influenced the submitted work.
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= [ ] = [ ] = (30)P ′T xT yT zT 1uT vT PT
PT
P ′T
−1 P ′T
(31)
P ′T
−1 = ( + + )b1P ′1 b2P ′
2 b3P ′3
−1
= + +b1P ′1
−1 b2P ′2
−1 b3P ′3
−1
= + +b1P1 b2P2 b3P3
= PT
( , , )xT yT zT
P ′T ( , )uT vT PT
= [ ] = [ ] = (32)PT 1uT vT xT yT zT−1 P ′
T−1
P ′T
PT
2015 Brown et al. Cureus 7(10): e341. DOI 10.7759/cureus.341 17 of 21
