Abstract:
An alignment system for aligning an acetabular cup insertion instrument utilizes pre-operative preferably standing x-rays and intra-operative x-rays to allow the surgeon to compensate for the position of a patient on an operating room table. The system uses a programmable computer connected to a digital x-ray system and an navigation tracking system to provide input for calculating the inclination and anteversion angles of an acetabular cup impactor based on a pre-operative plan developed from the standing x-rays. The system calculates changes in lengths and angles between anatomic landmarks on the pelvis to alter the intra-operative orientation of the acetabular reaming and impacting instruments to produce the desired inclination and anteversion angles when the pelvis is reoriented as the patient is placed on an operating table.

Description:
BACKGROUND OF THE INVENTION 
     Hip surgery requires the implantation of a femoral stem and an acetabular cup. The femoral stem has a spherical head that attaches to the neck of the stem and is free to articulate within a bearing insert that is fitted into the shell of an acetabular cup. Should the stem and cup not be positioned/aligned accurately, the neck of the stem may impinge on the lip of the insert resulting in a levering action that could allow the femoral head to cam out of the insert resulting in a permanent dislocation of the head. The impingement can also lead to excessive component wear and possibly failure. 
     In addition, a malpositioned cup can result in excessive liner wear even without impingement. A shell with a high abduction (inclination) angle can have joint forces concentrated near the cup liner rim, thereby increasing the wear rate due to concentrated forces. 
     Reducing or eliminating the chance for neck/insert impingement and high inclination angles is critical to eliminating dislocation and wear, which can ultimately result in a revision surgery to correct compound alignment (abduction and anteversion). 
     A recent study by H. Malchau et al., Clin. Orthop. Relat. Res. (2011) 469; 319-329 reviewed the implanted acetabular cup position post implantation in relation to the pelvic anatomy. The study of 1823 patients revealed that the cup position varied widely in reference to their described target zone. 
     One main reason for such variation is that the exact position of the patient&#39;s pelvis is not known in relation to the operating room (OR) table. Surgeons must rely on their experience to know how to position the cup, however the cup may not be implanted in the intended orientation. This is especially true with respect to less experienced surgeons. 
     Alignment of an acetabular cup can be achieved with an alignment guide that attaches to an insertion rod for facilitating the insertion of the acetabular cup into the acetabulum. The alignment guide preferably references the surgical table on which the patient rests. Conventionally, it is assumed that the patient&#39;s pelvis is parallel to the table, and that the surgical tables is parallel to the floor. Based on such assumptions, the ordinary position (in most patients) for the acetabular cup is 45° of inclination (abduction) and 20° of anteversion. For a discussion of angles of anteversion and also inclination or abduction of the acetabular cup when installed in the acetabulum, see, for example, U.S. Pat. No. 6,395,005, the disclosure of which is incorporated by reference herein in its entirety. 
     It has been found based on post-operative x-rays, however, that despite the alignment guide being parallel to the floor during insertion, of the acetabular cup, the resultant inclination or anteversion of the acetabulum in relation to the alignment guide is often different than expected and, thus, the acetabular cup has been installed at a less than ideal position. The pelvic position changes in relation to the operating room table which is not recognized during the procedure for example. 
     Presently most orthopedic companies offer instrumentation to direct reaming for acetabular cups and cup impaction which is an antenna-like device that attaches to either the reamer shaft or the cup impaction tool. Such cup alignment instruments are shown in, for example, U.S. Pat. Nos. 5,037,424, 5,571,111 and 6,395,005. When an x-shaped “antenna” is used a cup impactor that is oriented 45 degrees to the floor and 20 degrees to the long axis of the patient. The ‘X’ shape on the antenna is set parallel to the floor, and one leg of the ‘X’ set in line with the long axis of the body. One leg is for a left leg operation, and the other for a right leg operation. 
     Some of the drawbacks of this type of instrument are that the pelvis usually shifts when the patient is laid on the operating room table. If the pelvis does not shift, and the surgeon wants a 45/20 cup position, then the surgeon could use the instrument as is and get the perfect 45/20 alignment within the bone. However, most times the pelvis does shift in three possible planes: tilt, obliquity, and rotation. The surgeon does not know in which direction or by how much and therefore must use his experience or intuition to apply a correction factor to the direction of cup impaction. The actual cup orientation after impaction is not usually known until after the operation is complete and a post-operative x-ray is taken, and the patient is in recovery, and therefore at a time when changes to cup orientation are not possible without reoperation. 
     Another drawback is that the current antenna/impactor combination is set at set angles. For example a 45/20 degree abduction/anteversion orientation. If the surgeon determines that orientation of 40° to the floor and 15° to the long axis of the patient&#39;s femur is best for the patient, the set angles are of little use, or again the surgeon has to estimate a correct alignment. The orientation of the antenna/impactor combination in practice is set visually. The antenna shaft is set vertically, with the antenna ‘X’ cross bars parallel to the floor. The 20 degree orientation to the patient long axis is visual as well. Many surgeons do not use the antenna at all. 
     Some major prosthetic hip joint companies offer a navigation option to surgeons. This system uses cameras in the operating room and optical trackers on instrumentation. From a clinical perspective, the major drawbacks for navigation are that the technique involves placing invasive pins having the tracker thereon in the patient pelvis and femur. The pins are placed in the pelvis and the femur, through the skin and screwed into the bone. Using pins results in multiple separate wounds and increases the possibility of infection. This technique is also time intensive. Pins must be placed and pointers with trackers on them are used multiple times to register anatomy. This technique has a learning curve. The software and technique require extensive training and practical experience. Some systems require a pre-op CT scan which is costly. 
     A more recent development is digital imaging which produces an x-ray like image on a digital receiver. Once digitized, the digital image can be used to identify points in order for the system to calculate lengths and angles which could be used by the surgeon to help to identify how the pelvis is oriented to the operating room table. The surgeon can take pre-operative and intra-operative x-rays and pick points on the screen to calculate lengths and angles. This is a relatively new technology with a relatively small amount of users. 
     The x-rays can be visually observed for comparison. The system can aid the user by allowing the user to plan by designating the desired cup inclination and version angles pre-operatively as well as taking dimensions that will help to designate leg length and femoral stem offset corrections. Taking dimensions that will measure the cup inclination and version angles of the actual implanted cup intra-operatively as well as taking dimensions that show that actual leg length and offset of the trials or implants. 
     However, they do not aid the user by figuring out how the pelvis has shifted on the table as there is no algorithm to do this. A pelvis may have tilted by 15 degrees, yet the cup angle measurement only measures the cup angle at the plane that the x-ray is taken. If the pelvis has tilted by 15 degrees, then the correct cup placement would not be 45°/20° to the intra-op x-ray image, but should be adjusted to account for the tilt. 
     These current digital imaging systems do not have an algorithm that tries to compare pre-operative and intra-operative images to calculate how the intra-operative pelvic position changes in orientation to a pre-operative image. Furtheremore, the current digital systems don&#39;t calculate the cup impaction angles that would account for these changes. Instead the cup needs to be first impacted into the bone prior to taking the image, and reoriented if not in the desired position. Reorienting the shell could compromise the fit and security of the cup to the acetabular bone cavity. Multiple reorientations could possibly compromise the fit to the point that a secure fit is no longer achieved. In this situation, the surgeon may have to remove the shell, ream up to the next size shell, and start over. The removal of further acetabular bone is not ideal as this could compromise the overall strength of the remaining bone, and reduce the amount of bone for any future revisions. Current digital imaging techniques require successive intra-operative images, exposing the patient and the surgical team to higher levels of radiation than with a single image. 
     BRIEF SUMMARY OF THE INVENTION 
     The present invention, in the preferred embodiment, uses a combination of digital imaging and orientation technology to improve upon the limitations described above, along with an algorithm that determines the amount of pelvic movement in the three planes (obliquity, tilt, rotation) that is used as input for the orientation technology. The preferred embodiment to calculate pelvic tilt obliquity and rotation inoperatively which no other system does. Most pre-operative x-ray images are taken lying down, and hence placing the pelvis in an unnatural position. A standing x-ray is the gold standard as the amount of pelvic tilt when standing is what is right for that individual. This invention serves to recreate the natural standing x-ray tilt and, if desired, obliquity and rotation amount intra-operatively by adjusting the orientation of reaming and cup impaction, and performing these functions at the pre-op plan angles determined by the user (e.g. at 40° inclination and 15° anteversion). 
     The connection uses a method for aligning an acetabular cup including: taking a pre-operative preferably standing anterior/posterior view digital x-ray image and a lateral view standing digital x-ray image of the pelvis of a patient. A desired cup abduction and anteversion angle is determined based on two standing x-ray images. At least three points on the digital anterior/posterior digital x-ray image and at least two points on the lateral digital x-ray image are identified. The lengths and angles between each of the points on both the anterior/posterior digital image and the lateral digital image are calculated. A patient is positioned on an operating table in an operating room. The preferred operating table has a reference element thereon. In the preferred embodiment, the operating room has a navigation system therein, the preferred operating table has a navigation tracker mounted in a known position with respect to the operating table. Alternately, the reference system could reference the floor or other fixed point including the operating room table. At least one pelvic digital x-ray image of the patient positioned on the operating table is taken with the x-rays including the reference element. At least three points are identified on the intra-operative x-ray image. In the preferred embodiment, the points corresponding to the points on the pre-operative x-ray image. The lengths and angles between each of the at least three points on the intra-operative digital images are then calculated. An intra-operative angular deviation of the acetabular cup insertion instrument from the desired abduction and anteversion angle i.e. calculated by comparing the dimensional differences between the points on the pre-operative standing x-ray images and the at least one intra-operative x-ray image. The insertion instrument is then aligned to a calculated angular position, based on the intra-operative deviation, using the navigation camera and a navigation tracker mounted on the insertion instrument. 
     The at least three points on the anterior/posterior pre-operative and intra-operative images are preferably the right and left promontory points and the pubic symphysis. The at least two points on the intra-operative lateral image are preferably the left or right promontory and the public symphysis. However, other points can be used. Additional points may be selected from the group consisting of acetabular teardrops, the obturator foramen and the base of the left and right ischial rings, the points being identified on the pre-operative and intra-operative images. The angle between a line connecting the ischial ring points and the reference element indicates any obliquely change between the pre-operative, preferably standing pelvic anterior/posterior x-ray image and the at least one intra-operative x-rays. The reference element may be a radiolucent bar extending parallel to the operating table surface, the radiolucent bar having two radiopaque markers thereon. Alternately, the radiolucent bar has three radiopaque markers, each marker located at an apex of a triangle. The pre-operative determination of the desired cup abduction and anteversion angles based on the standing x-ray are about 40 to 45 degrees and 15 to 20 degrees respectively. The calculations are performed by a computer using digital image analysis software receiving input from a digital x-ray machine and an operating room navigation tracking system. A pubic tilt may be is calculated by comparing the distance between the promontory points and the pubic symphysis on both the pre-operative standing x-rays and the at least one intra-operative x-ray. 
     As used herein when referring to bones or other parts of the body, the term “proximal” means close to the heart and the term “distal” means more distant from the heart. The term “inferior” means toward the feet and the term “superior” means toward the head. The term “anterior” means toward the front part or the face and the term “posterior” means toward the back of the body. The term “medial” means toward the midline of the body and the term “lateral” means away from the midline of the body 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIGS. 1A to 1C  are reference diagrams showing pelvic obliquity, pelvic rotation and pelvic tilt which three axial movements typically occur when moving from a standing position to a prone position; 
         FIG. 2  shows nine possible anatomic landmarks on a pelvic anterior/posterior x-ray; 
         FIG. 3  shows three points on a pelvic anterior/posterior x-ray including the right and left promontory points and the pubic symphysis; 
         FIG. 3A  shows a lateral view of a lateral x-ray showing a left promontory point and the pubic symphysis; 
         FIG. 4  shows an artist&#39;s rendition of an anterior/posterior standing x-ray showing the three points of  FIG. 3  and dimensions A, B and C; 
         FIG. 5  shows an artist&#39;s rendition of a lateral x-ray showing the anatomic points shown in  FIG. 3A  including the dimensions D, E and F therebetween; 
         FIG. 6  shows a reference system mounted on an operating table including a radiolucent guide bar and two radiopaque markers; 
         FIG. 7  shows a pelvis area of a patient located on an operating table showing the right and left promontory points, pubic symphysis and the base points of the right and left ischial rings; 
         FIG. 8  is a anterior/posterior x-ray showing the anatomic landmark points on the pelvis; 
         FIG. 8A  is an artist&#39;s rendition of a pelvis of a patient oriented on an operating table having a reference element mounted thereon with the pelvis in a first orientation having a pelvic tilt; 
         FIG. 8B  is an artist&#39;s rendition of the pelvis of  8 A showing the various components for a reaming operation including the chosen landmarks. 
         FIG. 8C  is a lateral x-ray of the patient&#39;s pelvis of  FIGS. 8 to 8B ; 
         FIG. 8D  is a picture of the anterior pelvic x-ray showing the promontory points and the symphysis; 
         FIG. 8E  shows three views of a pelvis, in an idealized position on an operating table on three standing image planes with an acetabular impactor at 45% version. 
         FIGS. 9 and 9A  are an artist&#39;s rendition of an intra-operative x-ray image taken in the anterior/posterior and lateral views of a pelvis on an operating table in a second position; 
         FIG. 10  shows an example of pelvis shift in the standing plan x-ray when a patient is placed on an operating table. 
         FIGS. 10A and 10B  are anterior and posterior intra-operative views of a pelvis on an operating table showing 45° inclination and 20° anteversion; 
         FIGS. 11 through 11D  show the reorientation of the acetabular cup impactor/reamer from a pre-operative plan based on a standing x-ray to an intra-operative plan accounting for movement of the pelvis with 10° tilt; 
         FIGS. 12 through 12L  show the repositioning of an acetabular cup impactor/reamer from an inter-operative plan of 45° inclination and 20° anteversion with 8° rotation of the pelvis when placed on an operating table as set forth in example 4 of the present application; 
         FIGS. 13 through 13G  shows the repositioning of an acetabular cup impactor or reamer from a pre-operative plan based on a standing x-ray with 45° inclination and 20° anteversion desired based on the standing x-ray with the pelvis being rotated in 10° of obliquity as set forth in example 5 of the present application; 
         FIGS. 14 through 14E  shows the repositioning of an acetabular cup impactor/reamer where the desired orientation is 45° inclination and 20° anteversion based on a standing x-ray with 10° of obliquity imparted to the pelvis when being placed on an operating table as set forth in example 6 of the present application; 
         FIGS. 15 through 15D  of the present application show the repositioning of an acetabular cup impactor from a desired 45° inclination and 20° anteversion based on a standing x-ray to a pelvis oriented on an operating table with 10° tilt, 8° rotation and 10° obliquity change from the standing x-ray as set forth in example 7, with  FIGS. 15 and 15A  showing the standing x-ray anterior/posterior plane, the central plane and a rotational plane of the pelvis along with the x, y and z vectors lying on each of the planes; and 
         FIG. 16  shows a preoperative image showing an individual with one leg shorter than the other (non-zero pelvic tilt in the x-ray). 
     
    
    
     DETAILED DESCRIPTION 
     The method for aligning an insertion instrument for an acetabular cup of the present invention will now be described. Such an insertion instrument may be a reamer or impactor. For definitional purposes, pelvic obliquity and tilt rotation are shown respectively in  FIGS. 1A ,  1 B and  1 C. Initially a pre-operative digital standing A/P (anterior/posterior)( FIGS. 2 and 3 ) and lateral view x-ray ( FIG. 3A ) images of a pelvis  10  are taken. A magnification marker (not shown) may be included in these two x-rays. 
     The goal of the invention is that a pre-operative x-rays ( FIGS. 2 and 4 ) are taken and a plan established for the best acetabular cup position (inclination and version) relative to right and left acetabulum pelvis  10  for the individual patient. The pre-operative A/P plane is reestablished the calculations relative to the intra-op pelvis position ( FIG. 3 ). 
     The pre-operative cup position plan for inclination and version is then applied to the reestablished plane. In order to do this, at least some of the changes that took place in tilt, obliquity and rotation from their pre-operative position (angular changes) are determined and then used with the acetabular cup impactor. This can be done without placing any pins or other elements in contact with the body. 
     In the preferred embodiment, the x-ray images are taken with the patient standing. A standing image naturally orients the pelvis to the individual patients natural pelvic tilt, obliquity and rotation. 
     Referring to  FIG. 2  to  FIG. 5 , the surgeon identifies at least five or more specific points on the images (three on A/P x-ray and two on lateral x-ray). The three points on the A/P x-ray are preferably the left and right promontory  20 ,  22  and pubic symphysis  24 . The two points on the lateral x-ray are the lateral promontory point  22  and the pubic symphysis  24 . Other possible pelvic anatomy points are shown in  FIG. 2  as the bases of the left and right ischial rings  30 ,  32 , the left and right inside obturator foramen  34 ,  36  and the left and right acetabular teardrops  37  and  39 . It may not be necessary to compare every pre and intra-operative dimension to calculate tilt, rotation and obliquity changes. 
     Computer software may be used to calculate lengths and angles between these five points and retains the calculated dimensions to compare to a future correlated intra-operative dimension on intra-operative x-rays. For example,  FIG. 4  shows lengths A, B C and D based on points  20 ,  22  and  24  and  FIG. 5  shows lengths D, E and F based on points  22  and  24  and a point  25  at the origin of a right triangle formed by points  22 ,  24 . 
     Referring to  FIGS. 6 and 7 , the intra-operative routine starts with the patient being placed on an operating 46 table usually on the side opposite the hip being replaced. The surgeon performs the operation up to the point of reaming the acetabulum for acetabular cup. Prior to taking an intra-operative image, a reference plane is established. The reference plane is used to translate the angular changes to angular dimensions for a single measurement system, such as a navigation system, to use to direct the reamer and acetabular cup placement. Referring to  FIGS. 6 and 7 , preferably a radiolucent reference system  38  generally denotes as (system  38 ) with two or three system  38   1  of  FIG. 7 , radiopaque markers  44 ,  45  and  47 , is attached to the table. Reference system  38 ,  38   1  needs to be located so it is able to be included in the intra-operative x-ray image. Reference system  38 ,  38   1  have a horizontal bar  40  which and can be mounted on a post  42  vertically mounted on operating table. The reference element could also be just the post with radiopaque markers, and the post could be a patient positioning post for a peg board. The reference systems  38 ,  38   1  could conceivably be the actual operating room table  46 . Preferably a navigation system tracker  50  is then placed on the reference bar  40 . Magnification markers could be included in or on reference bar  40 . Bar  40  could be made out of radiolucent material with at least one magnification marker imbedded in the bar. The markers could be spherical in shape and the points taken could be the centers of the markers. 
     As shown in  FIGS. 8A to 8E , at least one intra-operative digital x-ray image (preferably an A/P image  FIG. 8 ) is taken making sure the reference bar  40  is within the image (see  FIGS. 8A to 8E ). The surgeon identifies at least 5 specific points on the intraoperative A/P image. The preferred embodiment has the surgeon identifying seven points. The promontory points  20 ,  22  and pubic symphysis point  24  are similar to the pre-operative image. The two ischial ring points  30 ,  32  are used to detect obliquity change. (Note: Other points that are meant be symmetrical about the pelvic anatomy can be chosen. Examples are: inside left and right obturator foramen  34 ,  36 , and left and right acetabular teardrops  37 ,  39 . The two ischial points or other symmetrical points are chosen to make calculating the obliquity easier. The line between the two promontory points could also be used, negating the need for two other symmetrical points.) 
     Software is used to calculate key lengths and angles between points on intraoperative image. Note that the specific points are at the center of the pubic symphysis  24 , the two promontory points  20 ,  22 , a point  35  at the center between the promontory points, and a point  49  that is 90° to the line between the promontory points. By determining the angular changes in tilt obliquity and rotation of the pelvis, the software can recreate a virtual standing image in reference to the pelvic position on the table ( FIG. 8E ).  FIG. 8E  shows the pelvis in a perfect prone position where there is no tilt rotation or obliquity and impactor is done at 45° inclination and 0° anteversion relative to table  46 . The standing image plane is shown as  37  and is a frontal or coronal plane. 
     The software identifies abduction and anteversion angles relative to the reference in order to achieve the pre-operative desired cup position as shown in  FIGS. 3 and 4 . Referring to  FIGS. 5 ,  6 ,  7 ,  8  and  9 , the preferred embodiment uses commercially available navigation system is used to determine the position of the cup impaction angle versus the reference bar  40 . A tracker  50  is placed on bar  40  and a tracker  52  is placed on an cup impactor  54 . Impactor  54  includes shaft  56  extending along axis  59  which engages the inside of an acetabular cup shell or reamer  5  located in acetabulum  11 ,  13 . The impactor is adjusted to the angles α and β calculated by a computer program while looking at a monitor. Preferably, the same monitor as the digital image monitor. Alternate embodiments could obtain the desired angles via manual goniometers or protractors, or via electronic means suh as with an inclinometer. Software for the required calculation is commercially available. 
     The surgeon then reams the acetabulum at indicated calculated angles with a standard acetabular reamer. If the surgeon suspects that the pelvis has moved prior to cup impaction, then an x-ray of the at least five points can be retaken. The surgeon impacts cup at the calculated angles α and β. 
     The following examples show calculations and workflow of various possible intraoperative pelvic orientations. 
     Example 1 
     This example shows how the cup impactor is positioned relative to reference bar  40  to obtain the desired inclination and anteversion. The pelvis is shown in a perfect orientation similar to  FIG. 8D . 
     Referring to  FIGS. 9 and 9A  there is shown a “perfect” pelvic orientation on table  46 . Also shown is the acetabular cup impactor  54  with shaft  56  shown at a 45° inclination 0° anteversion and zero tilt, rotation and obliquity with respect to reamer or impactor  57 . 
     As shown in  FIGS. 9 and 9A  pelvis  10  did not alter position from the A/P image (pelvis in a perfect position) when placed on table  46 . The basic pelvis orientation is exactly 90° to the standing x-ray. The plane is normal to the table. This is shown by the 0° alignment between the planes of the two x-rays (pre-standing and intraoperative prone). The plane that the standing x-ray was taken at is shown with the line  59  of  FIG. 9 . The plane through line  59  is normal to the table. The impactor is placed in the reamed acetabulum and rotated up to 45° in relation to the reference bar  40 . In general, the plane that the standing x-ray was taken at is first found, and then impactor  54  is angled for inclination along that plane, and then anteversion is placed normal to the plane by pivoting in the acetabulum (in example 1 this is 0°). The bar  40  defines the plane normal to the intra-operative x-ray view and parallel to the front edge of the table. It is perpendicular to the cross-table x-ray image taken in the operating room. The front edge of the operating room table could also be used as a reference, however, the size of the x-ray image may not be able to capture both the table and the anatomy required in the same shot. Having a reference bar  40  attached to the table allows it to be moved to a position that is within the image and not blocking the anatomic points needed for the calculations. The navigation tracker  50  is attached to whatever the reference is, whether it be the bar or the table itself. 
     Referring to  FIG. 10 , the navigation system first recreates the standing x-ray plane  37 , defines a central plane  39  then calculates the amount the pelvis has moved off that reference plane. A rotational plane  35  is also shown. Reference bar  40  defines plane normal to x-ray view and parallel to table. 
     Example 2 
     Referring to  FIGS. 10A and 10B , similar “perfect” pelvic orientation on the OR table as with Example 1. Standing x-ray plane is normal to the table shown with acetabular cup impactor at a now desired 45° inclination ( FIG. 10A ) and 20° anteversion ( FIG. 10B ). The cup impactor  54  would be oriented 45° from the reference bar  40  and 20° off the reference bar  40 . The two navigation trackers  50 ,  52 , one of the reference bar, and one on the impactor helps facilitate finding these angles for the impactor. The orientation is now 45 DEG INCLINATION, 20 ANTEVERSION, 0 TILT, 0 ROTATION and 0 OBLIQUITY. 
     Example 3 
     Here the cup impactor orientation is 45° INCLINATION, 20° ANTEVERSION, AND THE PELVIC ORIENTATION IS 10° TILT (to be confirmed below), 0° ROTATION and 0° OBLIQUITY. Dimensions for calculations for determining amount of tilt are pre-op images are shown in  FIGS. 11 ,  11 A and  11 B. Intra-op image: A/P image shown in  FIG. 11A : May have tilt. This is to be verified by the calculation outlined below. A Pre-op A/P, preferably standing image is taken and shown in  FIG. 11 . An Intra-operative A/P image is taken and shown in  FIG. 11A . The 3.179 dimension between points  24  and  35  is compared to the pre-op dimension of 2.599 between points  24  and  49 . Since the 3.179 is greater than 2.599, it indicates that the pelvis has tilted forward (positive tilt) by a certain amount. Dimensions for calculations for determining amount of tilt: A pre-operative lateral image is shown in  FIG. 11B  with lengths B, C, and E and angle D. 
     Table 1 refers to the dimensions of  FIGS. 11A and 11B . 
                                     TABLE 1                       Feature length/               Letter   Name Given   angle   Pre-op   Intra-op                   A   Promontory   Between   4.173   4.173           Line   Promontory               points       B   Normal Line   On A/P Image:   2.599   3.179               (90 deg)               between Prom               Line and Pub               Sym.               On Lateral               Image:               vertical               distance               between Prom               pts and Lat               Image               Horizontal               Line       C   Lateral Image   Promontory to   4.412   N/A as no           Hypotenuse   Pub Sym       lat image                       taken but                       4.412                       remains                       the same       D   Lateral Image   Angle between   36.094   To be           Tilt Angle   Hypotenuse and   deg   calculated               line parallel               to floor       E   Horizontal   Pre-op line   Not   N/A           Line   parallel to   important               the floor or   as no               OR table   rotation           Tilt Angle   Relative               change in               pelvic tilt               from the pre-               op angle D to               the intra-op               angle D                    
From  FIGS. 11A and 11B : Sin angle D=3.179/4.412
 
Angle=46.098 deg (intra-op angle D)
 
Tilt angle=angle of intra-op image minus angle of standing image
 
Tilt angle=46.098-36.094
 
Tilt angle˜10 degrees
 
 FIGS. 11C and 11D  show what angles the navigation would set the impaction at.
 
Inclination=45 deg (remains unchanged)
 
Version=20 deg anteversion minus 10 deg positive tilt=10 deg as shown
 
     Example 4 
     The cue orientation pelvic orientation in this example is 45 DEG INCLINATION, 20° ANTEVERSION, and the intra-operative pelvic orientation is 0° TILT, 8° ROTATION (to be confirmed below) and 0° OBLIQUITY.  FIGS. 12 and 12A  show an inter-operative top view ( FIG. 12 ) and front view ( FIG. 12A ) of what a pelvis at 8 deg of Rotation looks like.  FIG. 12  shows standing x-ray plane  37  and rotational plane  140 . Calculations for determining amount of rotation: Previously (similar to  FIGS. 11 and 11C ) taken pre-operative images are shown in  FIGS. 12B and 12C .  FIGS. 12D and 12E  are A/P Intra-operative images which show the pelvis may have rotated. This is to be verified. The following table  2  refers to the dimensions in  FIGS. 12B to 12E . 
                                                           TABLE 2                       Feature length/               Letter   Name Given   angle   Pre-op   Intra-op                                A   Promontory   Between   4.173   4.133           Line   Promontory               points       B   Normal Line   On A/P Image:   2.599   2.599               (90 deg)               between Prom               Line and Pub               Sym.               On Lateral               Image: vertical               distance               between Prom               pts and Lat               Image               Horizontal Line       C   Lateral Image   Promontory to   4.412   N/A as no           Hypotenuse   Pub Sym       lat image                       taken but                       remains                       the same       D   Lateral Image   Angle between   36.094 deg   N/A as no           Tilt Angle   Hypotenuse and       lat image               line parallel       taken but               to floor       remains                       the same       E   Horizontal   Pre-op line   Not   To be           Line   parallel to the   important   calculated               floor or OR               table       F   Midpoint Line   Line between   Not   Not               Prom Line   important   important               Midpoint and               Pub Sym       G   Rotational   Distance   0   0.496           Offset   between Normal           Distance   Line and               Midpoint Line               on Promontory               Line       H   Rotational   Angle of Pelvic   0   To be           Angle   Rotation of       calculated               Intra-op               relative to               Pre-op                    
Rotational Angle
 
Referring to  FIG. 12E  to find the Rotation Angle (H), first find length E.
 
B 2 +E 2 =C 2  
 
E 2 =C 2 −B 2  
 
E 2 =(4.412) 2 −(2.599) 2  
 
E=3.565
 
     The Lateral Image Horizontal Line dimension E found above, 3.565″, is now used to help calculate the amount of intra-operative pelvic rotation. The Rotational Offset Distance (0.496″),  FIG. 12D , was found on the intra-op digital image device. Using these two numbers, the Rotational Angle can be found. 
     Sin H=0.496/3.565 
     Rotational Angle H=7.997 deg 
     The pelvis has rotated nominally 8 degrees 
     Note: 
     A less optimal way of determining the degree of rotation would be to compare the pre-op Promontory Line distance (4.173) to the intra-op distance (4.133). The 4.133 distance is a projection of the 4.173 distance at the rotation angle. 
     Cos rotation angle=4.133/4.173 
     Rotation angle=7.939 degrees 
       FIG. 12F  shows what 8 degrees of rotation looks like from a side view with sanding x-ray plane  37  and rotation plane  137 . 
     Calculations for finding the impactor angles to be used by Navigation are shown in  FIGS. 12G to 12J . 
     Step 1: Project the impactor angle position (45/20) onto the plane of angle change. In this case it is the Rotation Plane shown in  FIG. 12G . 
               TAN   ⁢           ⁢   45   ⁢   °     =     h   R                   SIN   ⁢           ⁢   45   ⁢   °     =     h   L                   COS   ⁢           ⁢   45   ⁢   °     =     R   L                   SIN   ⁢           ⁢   20   ⁢   °     =     B   R                 F   =     SIN   ⁢           ⁢   20   ⁢   °   ⁢           ⁢   R                 F   =       (     SIN   ⁢           ⁢   20   ⁢   °     )     ⁢     (     COS   ⁢           ⁢   45   ⁢   °     )     ⁢   L                   TAN   ⁢           ⁢   α     =     h   F                   TAN   ⁢           ⁢   α     =       sin   ⁢           ⁢   45   ⁢   °L         (     sin   ⁢           ⁢   20   ⁢   °     )     ⁢     (     cos   ⁢           ⁢   45   ⁢   °     )     ⁢   L                   α   =     71.118   ⁢   °           
Step 2: Rotate 8° as shown in  FIG. 12H  (71.118-8°.
 
               cos   ⁢           ⁢   71.118   ⁢   °     =     F     L   R                     cos   ⁢           ⁢   63.118   ⁢   °     =       (     F   +   f     )       L   R                     F     cos   ⁢           ⁢   71.118   ⁢   °       =       (     F   +   f     )       cos   ⁢           ⁢   63.118   ⁢   °                     F   .3236     =       F   .4521     +     f   .4521                   F   =       .7157   ⁢   F     +     .7157   ⁢   f                     .2843   ⁢   F     =     .7157   ⁢   f                 f   =     .3973   ⁢   F           
Step 3: As shown in  FIG. 12I .
 
               From   ⁢           ⁢   Before   ⁢     :     ⁢           ⁢   F     =       (     cos   ⁢           ⁢   70   ⁢   °     )     ⁢     (     cos   ⁢           ⁢   45   ⁢   °     )     ⁢   L                 f   =     .3973   ⁢   F                   F   +   f     =         (     cos   ⁢           ⁢   70   ⁢   °     )     ⁢     (     cos   ⁢           ⁢   45   ⁢   °     )     ⁢   L     +     .3973   ⁢     (     cos   ⁢           ⁢   70   ⁢   °     )     ⁢     (     cos   ⁢           ⁢   45   ⁢   °     )     ⁢   L                     F   +   f     =     .3378   ⁢   L                   TAN   ⁢           ⁢   70   ⁢   °     =       D   R     F                   TAN   ⁢           ⁢   ω     =       D   R       F   +   f                     TAN   ⁢           ⁢   ω     =       TAN   ⁢           ⁢   70   ⁢   °   ⁢           ⁢     F   .         .3378   ⁢   L                     TAN   ⁢           ⁢   ω     =         (     TAN   ⁢           ⁢   70   ⁢   °     )     ⁢     (     COS   ⁢           ⁢   70   ⁢   °     )     ⁢     L   ⁡     (     COS   ⁢           ⁢   45   ⁢   °     )       ⁢   L       .3378   ⁢   L                     TAN   ⁢           ⁢   ω     =       1.9670   ⁢           ⁢   ω     =       63.052   ⁢   °     =         90   ⁢   °     -     63.052   ⁢   °       =     26.947   ⁢   °                 
Step 4: Find cup impactor inclination angle as shown in  FIG. 12J .
 
               TAN   ⁢           ⁢   63.118   ⁢   °     =       h   R       (     F   +   f     )                       From   ⁢           ⁢   Before   ⁢     :     ⁢           ⁢   F     +   f     =     .3378   ⁢   L                   SIN   ⁢           ⁢   γ     =       h   R     L                   SIN   ⁢           ⁢   γ     =         (     TAN   ⁢           ⁢   63.118   ⁢   °     )     ⁢     (   .3378   )     ⁢   L     L                   SIN   ⁢           ⁢   γ     =   .6663               γ   =     41.786   ⁢   °           
The following angular values are used for the navigation system to effectively impact a cup at a desired standing position of 45°/20° (inclination/anteversion) and are shown in  FIGS. 12K and 12L .  FIG. 12L  shows standing plane  37  and rotated plane  140 .
 
Navigation version angle=26.947 deg
 
Navigation inclination angle=41.786 deg
 
     Example 5 
     Here the cup impactor orientation is 
     45° INCLINATION, 20° ANTEVERSION, AND THE PELVIC ORIENTATION IS 0° TILT, 0° ROTATION, and 10° OBLIQUITY (to be confirmed below).  FIGS. 13 and 13A  show a top view and front view of what a pelvis at 10 deg of Obliquity looks like. To find obliquity changes, two points are identified on the intra-op pelvis (refer to  FIG. 9 ) to determine the angle of obliquity. The pelvis obliquity has changed by 10 degrees from the perfect position of 90 degrees. 
     Calculations for finding the impactor angles to be used by Navigation: 
     Step 1: As shown in  FIG. 13B , project pre angle change impactor inclination angle onto plane of angle change. In this case it is the Standing Image Plane  37  shown in  FIG. 13 . Calculation angles are shown in plane of standing x-ray in  FIG. 13C . 
     Referring to  FIG. 13C , project onto plane  37  (standing x-ray) that the change in angle will take place. 
     
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               45 
               ⁢ 
               ° 
             
             = 
             
               R 
               L 
             
           
         
       
       
         
           
             R 
             = 
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               45 
               ⁢ 
               ° 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               20 
               ⁢ 
               ° 
             
             = 
             
               A 
               R 
             
           
         
       
       
         
           
             A 
             = 
             
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   20 
                   ⁢ 
                   ° 
                 
                 ) 
               
               ⁢ 
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   45 
                   ⁢ 
                   ° 
                 
                 ) 
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               45 
               ⁢ 
               ° 
             
             = 
             
               H 
               L 
             
           
         
       
       
         
           
             H 
             = 
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               45 
               ⁢ 
               ° 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               γ 
             
             = 
             
               
                 H 
                 A 
               
               = 
               
                 
                   SIN 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   45 
                   ⁢ 
                   ° 
                 
                 
                   
                     ( 
                     
                       COS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       20 
                       ⁢ 
                       ° 
                     
                     ) 
                   
                   ⁢ 
                   
                     ( 
                     
                       COS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       45 
                       ⁢ 
                       ° 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       L 
                     
                     ) 
                   
                 
               
             
           
         
       
       
         
           
             γ 
             = 
             
               46.7808 
               ⁢ 
               ° 
             
           
         
       
     
     Step 2: Rotate 10 deg of Obliquity in plane of rotation (Standing Image Plane). 
     Step 3: Calculate impactor version angle as shown in  FIG. 13D . 
     
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               46.7808 
               ⁢ 
               ° 
             
             = 
             
               
                 
                   A 
                   
                     L 
                     1 
                   
                 
                 ⁢ 
                 
                   L 
                   1 
                 
               
               = 
               
                 A 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   46.7808 
                   ⁢ 
                   ° 
                 
               
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               36.7808 
               ⁢ 
               ° 
             
             = 
             
               
                 
                   
                     ( 
                     
                       A 
                       + 
                       a 
                     
                     ) 
                   
                   
                     L 
                     1 
                   
                 
                 ⁢ 
                 
                   L 
                   1 
                 
               
               = 
               
                 
                   ( 
                   
                     A 
                     + 
                     a 
                   
                   ) 
                 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   36.7808 
                   ⁢ 
                   ° 
                 
               
             
           
         
       
       
         
           
             
               A 
               .6848 
             
             = 
             
               
                 A 
                 .8009 
               
               + 
               
                 a 
                 .8009 
               
             
           
         
       
       
         
           
             A 
             = 
             
               
                 .8550 
                 ⁢ 
                 A 
               
               + 
               
                 .8550 
                 ⁢ 
                 a 
               
             
           
         
       
       
         
           
             
               .1450 
               ⁢ 
               A 
             
             = 
             
               
                 .8550 
                 ⁢ 
                 
                   a 
                   ⁢ 
                   
                     
 
                   
                   . 
                   a 
                 
               
               = 
               
                 .1695 
                 ⁢ 
                 A 
               
             
           
         
       
       
         
           
             
               From 
               ⁢ 
               
                   
               
               ⁢ 
               Before 
               ⁢ 
               
                 : 
               
               ⁢ 
               
                   
               
               ⁢ 
               a 
             
             = 
             
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   20 
                   ⁢ 
                   ° 
                 
                 ) 
               
               ⁢ 
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   45 
                   ⁢ 
                   ° 
                 
                 ) 
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             a 
             = 
             
               .1126 
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               s 
               ⁢ 
               
                   
               
               ⁢ 
               o 
               ⁢ 
               
                 : 
               
               ⁢ 
               
                   
               
               ⁢ 
               A 
             
             = 
             
               .6644 
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               A 
               + 
               a 
             
             = 
             
               .7771 
               ⁢ 
               L 
             
           
         
       
       
         
           
             Solve 
             ⁢ 
             
                 
             
             ⁢ 
             for 
             ⁢ 
             
                 
             
             ⁢ 
             σ 
           
         
       
       
         
           
             
               Tan 
               ⁢ 
               
                   
               
               ⁢ 
               σ 
             
             = 
             
               d 
               
                 .7771 
                 ⁢ 
                 L 
               
             
           
         
       
       
         
           
             
               Tan 
               ⁢ 
               
                   
               
               ⁢ 
               20 
               ⁢ 
               ° 
             
             = 
             
               d 
               
                 .6644 
                 ⁢ 
                 L 
               
             
           
         
       
       
         
           
             d 
             = 
             
               .2418 
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               Tan 
               ⁢ 
               
                   
               
               ⁢ 
               σ 
             
             = 
             
               
                 
                   
                     .2418 
                     ⁢ 
                     L 
                   
                   
                     .7771 
                     ⁢ 
                     L 
                   
                 
                 ⁢ 
                 σ 
               
               = 
               
                 17.283 
                 ⁢ 
                 ° 
               
             
           
         
       
     
     The new position of the impactor handle is 17.286 degrees off the reference. This is one of the input angles needed for navigation and is shown in  FIG. 13E . 
     Now calculate the inclination angle of the impactor as shown in  FIG. 13F . 
     
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               36.7808 
               ⁢ 
               ° 
             
             = 
             
               H 
               
                 ( 
                 
                   A 
                   + 
                   a 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               From 
               ⁢ 
               
                   
               
               ⁢ 
               Before 
               ⁢ 
               
                 : 
               
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   A 
                   + 
                   a 
                 
                 ) 
               
             
             = 
             
               .7771 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               H 
               L 
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               
                 
                   
                     ( 
                     
                       TAN 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       36.7808 
                       ⁢ 
                       ° 
                     
                     ) 
                   
                   ⁢ 
                   .7771 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   L 
                 
                 ) 
               
               L 
             
           
         
       
       
         
           
             δ 
             = 
             
               35.5166 
               ⁢ 
               ° 
               ⁢ 
               
                   
               
               ⁢ 
               shown 
               ⁢ 
               
                   
               
               ⁢ 
               in 
               ⁢ 
               
                   
               
               ⁢ 
               FIG 
               ⁢ 
               
                   
               
               ⁢ 
               13 
               ⁢ 
               
                 G 
                 . 
               
             
           
         
       
     
     The following angular values are used for the navigation system to effectively impact a cup at 45/20. Navigation version angle=17.283 deg; Navigation inclination angle=35.520 deg. 
     Example 6 
     Here the cup impactor orientation is 45 DEG INCLINATION, 20° ANTEVERSION, AND THE PELVIC ORIENTATION IS 0° TILT, 8° ROTATION (to be confirmed below) and 10° OBLIQUITY (to be confirmed below). Calculations for determining amount of Obliquity and Rotation: Use the method in Example 4 to calculate the amount of Rotation, and use the method in Example 5 to calculate the amount of Obliquity. Calculations for finding the impactor angles to be used by Navigation: 
     Step 1: 
     Use the method in Example 5 to find the following angles for the cup impactor for 10 deg Obliquity: 
     Cup impactor version=17.283 deg 
     Cup impactor inclination=35.516 deg 
     Step 2: 
     Now project these onto the plane for rotation, rotate 8 deg, and project back to plane that impactor is on (See  FIG. 14 ). 
       FIG. 14  shows the cup impactor position prior to rotating 8 degrees 
     
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               35.516 
             
             = 
             
               A 
               L 
             
           
         
       
       
         
           
             A 
             = 
             
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   35.516 
                 
                 ) 
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               35.516 
             
             = 
             
               h 
               L 
             
           
         
       
       
         
           
             h 
             = 
             
               
                 ( 
                 
                   SIN 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   35.516 
                 
                 ) 
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   
                     90 
                     ⁢ 
                     ° 
                   
                   - 
                   
                     17.283 
                     ⁢ 
                     ° 
                   
                 
                 ) 
               
             
             = 
             
               B 
               A 
             
           
         
       
       
         
           
             
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   72.717 
                 
                 ) 
               
               ⁢ 
               
                 ( 
                 
                   COS 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   35.516 
                 
                 ) 
               
               ⁢ 
               L 
             
             = 
             B 
           
         
       
       
         
           
             
               .2418 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
             = 
             B 
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               ω 
             
             = 
             
               
                 h 
                 B 
               
               = 
               
                 
                   
                     ( 
                     
                       SIN 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       35.516 
                     
                     ) 
                   
                   ⁢ 
                   L 
                 
                 
                   
                     ( 
                     
                       COS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       72.717 
                     
                     ) 
                   
                   ⁢ 
                   
                     ( 
                     
                       COS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       35.516 
                     
                     ) 
                   
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   L 
                 
               
             
           
         
       
       
         
           
             ω 
             = 
             
               67.400 
               ⁢ 
               ° 
             
           
         
       
     
     Now rotate 8 degrees as shown in  FIG. 14A 
 
σ=ω−8°
 
σ=67.400−8°
 
σ=59.400°
 
     Now project back to plane of impactor as shown in  FIGS. 14A ,  14 B and  14 C. 
     
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               72.717 
             
             = 
             
               m 
               B 
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               m 
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 72.717 
                 ) 
               
               ⁢ 
               B 
             
             = 
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               67.400 
             
             = 
             
               B 
               
                 L 
                 1 
               
             
           
         
       
       
         
           
             
               L 
               1 
             
             = 
             
               B 
               
                 COS 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 67.400 
               
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               59.400 
             
             = 
             
               
                 B 
                 + 
                 b 
               
               
                 L 
                 1 
               
             
           
         
       
       
         
           
             
               L 
               1 
             
             = 
             
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
               
                 COS 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 59.400 
               
             
           
         
       
       
         
           
             
               B 
               
                 COS 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 67.400 
               
             
             = 
             
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
               
                 COS 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 59.400 
               
             
           
         
       
       
         
           
             
               B 
               .3843 
             
             = 
             
               
                 B 
                 .5090 
               
               + 
               
                 b 
                 .5090 
               
             
           
         
       
       
         
           
             B 
             = 
             
               .7549 
               + 
               
                 .7549 
                 ⁢ 
                 b 
               
             
           
         
       
       
         
           
             
               .2451 
               ⁢ 
               B 
             
             = 
             
               .7549 
               ⁢ 
               b 
             
           
         
       
       
         
           
             b 
             = 
             
               .3246 
               ⁢ 
               B 
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               m 
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               72.717 
             
             = 
             
               m 
               B 
             
           
         
       
       
         
           
             
               
                 TAN 
                 ⁡ 
                 
                   ( 
                   72.717 
                   ) 
                 
               
               ⁢ 
               B 
             
             = 
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               
                 δ 
                 ⁡ 
                 
                   ( 
                   
                     B 
                     + 
                     b 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 TAN 
                 ⁡ 
                 
                   ( 
                   72.717 
                   ) 
                 
               
               ⁢ 
               B 
             
             = 
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               
                 δ 
                 ⁡ 
                 
                   ( 
                   
                     B 
                     + 
                     
                       .3246 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       B 
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               3.2139 
               ⁢ 
               
                   
               
               ⁢ 
               B 
             
             = 
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               
                 δ 
                 ⁡ 
                 
                   ( 
                   
                     1.3246 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     B 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             2.4263 
             = 
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
           
         
       
       
         
           
             δ 
             = 
             
               67.601 
               ⁢ 
               ° 
             
           
         
       
       
         
           
             
               
                 Navigated 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 impactor 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 version 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 90 
                 ⁢ 
                 ° 
               
               - 
               
                 64.601 
                 ⁢ 
                 ° 
               
             
             = 
             
               22.399 
               ⁢ 
               ° 
             
           
         
       
     
     Find impactor inclination (shown in  FIG. 14D ). 
     
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               59.400 
             
             = 
             
               
                 h 
                 1 
               
               
                 ( 
                 
                   B 
                   + 
                   b 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               α 
             
             = 
             
               
                 TAN 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 59.400 
                 ⁢ 
                 
                   ( 
                   
                     B 
                     + 
                     b 
                   
                   ) 
                 
               
               L 
             
           
         
       
       
         
           
             
               From 
               ⁢ 
               
                   
               
               ⁢ 
               Before 
               ⁢ 
               
                 : 
               
               ⁢ 
               
                   
               
               ⁢ 
               .2418 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
             = 
             
               
                 B 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 and 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 b 
               
               = 
               
                 .3246 
                 ⁢ 
                 B 
               
             
           
         
       
       
         
           
             b 
             = 
             
               .3216 
               ⁢ 
               
                 ( 
                 .2418 
                 ) 
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             b 
             = 
             
               .0785 
               ⁢ 
               
                   
               
               ⁢ 
               L 
             
           
         
       
       
         
           
             
               
                 
                   
                     B 
                     + 
                     b 
                   
                   = 
                     
                   ⁢ 
                   
                     
                       .2418 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       L 
                     
                     + 
                     
                       .0785 
                       ⁢ 
                       L 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                   ⁢ 
                   
                     .3203 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     L 
                   
                 
               
             
           
         
       
       
         
           
             
               SIN 
               ⁢ 
               
                   
               
               ⁢ 
               α 
             
             = 
             
               
                 TAN 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 59.400 
                 ⁢ 
                 
                   ( 
                   .323 
                   ) 
                 
                 ⁢ 
                 L 
               
               L 
             
           
         
       
       
         
           
             α 
             = 
             
               32.798 
               ⁢ 
               ° 
             
           
         
       
     
     The following angular values are used for the navigation system to effectively impact a cup at  45 / 20 . Navigation version angle=22.399 deg ( FIG. 14C ) Navigation inclination angle=32.798 deg ( FIG. 14E ) 
     Example 7 
     Here the cup impactor orientation is 45° INCLINATION, 20° ANTEVERSION, AND THE PELVIC ORIENTATION IS 10° TILT (to be confirmed below), 8° ROTATION (to be confirmed below) and 10° OBLIQUITY (to be confirmed below). Calculations for determining amount of Obliquity and Rotation: Use the method in Example 3 to calculate the amount of Tilt Use the method in Example 4 to calculate the amount of Rotation 
     Use the method in Example 5 to calculate the amount of Obliquity 
     Calculations for Finding the Impactor Angles to be Used by Navigation: 
     Step 1: 
     Use the methods in Example 6 to find the impactor angles to be used for Navigation after applying 8 deg of Rotation and 10 deg of Obliquity. 
     Cup impactor version=22.399 deg 
     Cup impactor inclination=32.798 deg 
     Step 2: 
     Apply 10 deg of Tilt. 
       FIG. 15  shows the three basic x, y and z directions that angle changes can take place: Tilt, Obliquity, and Rotation. The three directions are along the three basic vector directions of a coordinate system. In Example 6, Rotation and Obliquity were applied 90 degrees to each other. Following this approach, Tilt would be rotated around the third vector Y (perpendicular to the page) of the coordinate system. In  FIG. 15 , Tilt changes would take place in the Central Plane  39  by rotating on the axis perpendicular (normal) to the Central Plane. The practical issue with this is that the pelvis would be lifting off the OR table in doing this.  FIG. 15A  illustrates the extreme example of the pelvis with the same Rotation and Obliquity as  FIG. 15 , but with 180 deg of Tilt applied. 
     The pelvis is no longer laying on the operating room table with the legs resting along the plane of the table. In this position, the portion of the legs near the pelvis would be lifted off the table. This is not a realistic situation as the patient, no matter what the pelvic tilt, is laying flat on the table. Although this example is extreme, it demonstrates that any pelvic tilt angle applied to the coordinate system would lift the legs up off the table. Calculations could be performed to project the cup inclination and version angles found in Example 6 along the Central Plane  39 , rotate the Tilt amount in this plane, and reproject as per the above examples. However, a more practical method would be to perform the calculations as per Example 6, and simply rotate these results by the Tilt angle amount found as per Example 3. 
     Method to apply 10 deg of Tilt: 
     Step 1: 
     Find the amount of Tilt as per Example 3. 
     Step 2: 
     Perform calculations as per Example 6 for Rotation and Obliquity. Navigation angles found in Example 6 are shown in  FIGS. 15B and 15C . 
     To apply 10 degrees of positive Tilt, simply subtract 10 degrees from 22.402 deg to yield 12.402 degrees. The cup impactor inclination angle of 32.798 deg remains the same. 
     This example uses the apparent obliquity change when looking across the reference from the front of the OR table. In reality, the pelvis has obliquely changed in the direction of the pelvic tilt. In this example, the obliquity would be 10 degrees off the front of table plane. The calculations below show what the actual obliquity would be (see  FIG. 15D ). 
     
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               10 
               ⁢ 
               ° 
             
             = 
             
               d 
               h 
             
           
         
       
       
         
           
             
               COS 
               ⁢ 
               
                   
               
               ⁢ 
               10 
               ⁢ 
               ° 
             
             = 
             
               d 
               L 
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               L 
               h 
             
           
         
       
       
         
           
             
               TAN 
               ⁢ 
               
                   
               
               ⁢ 
               δ 
             
             = 
             
               
                 TAN 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 10 
                 ⁢ 
                 ° 
               
               
                 COS 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 10 
                 ⁢ 
                 ° 
               
             
           
         
       
       
         
           
             δ 
             = 
             10.15 
           
         
       
     
     The plane defined by the front of the table would be a plane thru d and h of  FIG. 15D . The plane defined by the tilt plane in the direction of the standing xray image plane would be defined by L and h. The actual obliquity would be 10.15 degrees not 10 degrees. However, the difference of 0.15 degrees is inconsequential in practice for cup positioning in hip surgery and that using 10 degrees for the calculations would be acceptable. Either method could be used. 
     The pre-operative image may show a leg shortening that needs correction as shown in  FIG. 16 . Note that the pelvis is in the pre-operative x-ray shown in  FIG. 16  is not parallel to the floor (the line joining points  30  and  32  is not parallel). A line through points  30  and  32  is angled at 5° from horizontal. The surgeon may want to correct for this. If a check for obliquity is taken in this image, it would show that the pelvis is 5 degrees off. The method of measurements set forth above allows for correcting this. To correct for the 5 degrees, a measurement for obliquity as shown in  FIG. 20  would be taken. But suppose the surgeon wants to keep the 5 degrees of obliquity, then the extra 5 degrees is added/subtracted from the intra-operative obliquity measurement and used to calculate the impactor inclination angle. 
     It is assumed that the intra-operative image is taken perpendicular (normal) to reference bar  40 . Normally, this is a good assumption, but in practice, it may be a few degrees off from being perpendicular. This could be accounted for in two ways: 
     The two or more radiopaque points in the reference bar could be at precise distance apart from each other. Any angular differences in the image would result in a difference in measurement between the points which can be calculated in the computer. 
     The preferred reference bar is radiolucent, and has two radiopague spheres ( 44 ,  45 ) embedded in reference bar  40  3 inches apart. Any angular differences in the image would result in a difference in measurement between spheres  44  and  45 . An image that is taken perpendicular to bar  40  would show a 3 inch measurement between the radiopaque spheres. However, an image at a 10 degree angle to the bar would show a distance of 2.954 inches long. So in practice, an image would be taken and the actual distance on the image compared against the known distance of 3 inches. It would be discovered that the image is being taken at an angle to the reference at that time. The x-ray direction could then be changed to account for it, or the calculated 10 degree angle could be used as a modifier to the follow-up dimensions to be used for obliquity, tilt, and rotation. 
     If for some reason it is suspected that the x-ray emitter is also taken at an angle to the floor, as in pointing more to the floor or away, another option would be to include a third radiopaque marker in the reference bar that could be used for angular corrections. 
     Using the above methodology to determine the pelvis location intra-operatively can have benefits beyond cup placement. Another use would be for drilling along a certain direction in times where it is desirable to place a screw in the area having the most bone. An example of this would be in a severe revision situation whereas bone erosion can comprise desirable areas to have fixation. 
     Alternately, a navigation tracker could be placed on the x-ray emitter itself and used to make automatic corrections. If for some reason it is suspected that the x-ray emitter is also taken at an angle to the floor, as in pointing more to the floor or away, another option would be to include a third radiopaque marker in the reference that could be used for angular corrections. The navigation tracker could be used for this as well. 
     This invention describes the benefits of taking a pre-operative standing image vs. a pre-operative supine image. A standing image takes into account the normal pelvic position for each individual patient as opposed to a laying down/supine image which can alter the pelvis position similar to laying down on the operating table. 
     However, the invention does not exclude supine images, both pre-operative and intra-operative. For example, for the direct anterior surgical approach, the patient is supine on the table. It could be argued that the pelvis may have similar angular changes for the supine pre-operative x-ray image to the intra-operative supine position on the table, and therefore, in a way, recreating a natural standing pelvic position due to the similar angular changes. 
     At a minimum, the two A/P promontory points and the public symphysis point needs to be taken. If only these three points are taken, the line between he promontory points could help dictate the obliquity angle, and an assumption could be made that the x-ray emitter is parallel to the operating table and floor, and therefore, any line parallel to the floor on the image could be the reference. This also assumes that the image detector is parallel to the operating table and/or floor. 
     A common operating room table can be adjusted in two ways: trendelenburg (about the long axis of the table) and lateral tilt (about the short axis of the table). It is conceivable that the table could be adjusted to account for pelvic obliquity and rotation prior to an intro-operative image to account for any angular changes. This would be an attempt to adjust the table in order to place the pelvis at the “perfect” position 90° to the floor as described above, and therefore a surgeon could simply use the impactor at the 45°/20° position to the floor. There are two major issues with this that this invention addresses. First, there are only two adjustments with the table, not three. There is not an adjustment for tilt and therefore the surgeon would need to somehow adjust for this. Second, multiple intra-operative images would have to be taken in order to get the exact position needed for the pelvis. For instance, the table would have to be tilted to remove any intra-operative pelvic rotation. An image would be taken, and a guess as to how much the table would be adjusted to remove rotation. Another image taken, and further adjustment until the pubic symphysis is directly positioned over the sacrum. This would be a visual positioning, as well as adjusting for obliquity with trendelenburg table adjustments. 
     Although the invention herein has been described with reference to particular embodiments, it is to be understood that these embodiments are merely illustrative of the principles and applications of the present invention. It is therefore to be understood that numerous modifications may be made to the illustrative embodiments and that other arrangements may be devised without departing from the spirit and scope of the present invention as defined by the appended claims.