Synchronous Field Shaping in Rotational Megavolt Therapy

The advantage of rotation in the treat-ment of deep tumors by supervoltage radiation is well known. On a transverse cross section the ionization is maximal and quite uniform in a circular area which includes the tumor, while it diminishes for points outside of that circle according to distance from the center. Sometimes the fact that the shape of this area is circular, and that its center lies on the axis of rotation, may be an undesirable limitation. Under the presently used rotational method of fixed beam cross section, the full dosage is delivered to a considerable mass of healthy tissue if the region to be treated is: (a) very unsym-metrical with respect to the axis of rotation (“unsymmetrical” indicates that the center of the tumor does not lie close to the axis of rotation on all transverse cross sections), or (b) of elongated (noncircular) cross section. Two technics for improving the distribution of ionization in rotational therapy will be presented in this paper. Technic Giving Circular Radiation Area of Predetermined Center and Radius If the treatment area is very unsymmetrical with respect to the axis of rotation (a, above), a minimum amount of ionization will be delivered to the healthy tissue if the beam is just wide enough to cover only the desired region, at all positions of rotation. This occurs if the x-ray beam follows the rotating tumor, as a searchlight beam follows the course of a circling airplane. Such a technic is presented for one cross section in Figure 1. Suppose that C1 is the center of rotation of the patient cross section. O1 and O2 are the centers of rotation of two absorbing disks (O1C2 = C2O2) and are symmetrical with respect to TC1. The centers A1 and A2 of the disks are positioned so that the segments O1A1 and O2A2 are parallel and equal. The midpoint, D2, of A1A2 coincides with that of EF, since A1E = A2F = r. Then C2D2 is parallel and equal to O1A1 and to O2A2.?? Extend the line TD2 until it intersects a line from C1 parallel to C2D2. The similar triangles TC1D1 and TC2D2 give:?? Now suppose that both disks and the patient cross section rotate around the points O1, O2, and C1 respectively with the same angular velocity.