Re: Near Zone Calculation Dear Jeff, the existence of a near zone with ultrasonic transducers has always the same physical reason: diffraction by the boundaries. Therefore the extension and the shape of this zone are determined by the shape of the transducer and the wavelength in the material, resp. frequency and sound velocity. To normalize the equations for sound pressure this influence of shape, frequency and velocity is eliminated by using the relative distance Z = (z v)/(aa f), where z is the distance, v sound velocity, a half the size of the transducer and f is frequency. The condition to be fulfilled when using this normalization is: size a must be much greater than the wavelength, which is always met in nondestructive testing in the megahertz range. The nearfield distance is defined as the distance to the last pressure maximum on the acoustical axis. Expressed in relative distances this means N = k (aa f)/ v, with a dimensionless factor k. For a circular, plane transducer this factor is k = 1.0, which together with the radius a of the disk transducer  gives the equation you know: Ncirc = (aa f)/ v = (DD f)/ (4v). For a rectangular transducer having a large side 2a and a small side 2b the same notation can be used: Nrect = k (aa f)/ v, where the factor k is a function of the side ratio (b/a). This ratio is always between 0 (line source) and 1 (square source). There is no simple equation to calculate the factor k depending on (b/a), but I can offer you a diagram which I made long time ago, showing this factor k over the ratio (b/a). The values of k were determined by me on the following way: 1. by numerical calculation the sound pressure on the acoustical axis was calculated for rectangular transducers with varying ratio (b/a). 2. from these values the location of the last pressure maximum was determined to a precison of 5 decimals expressed in relative distances (aa f)/ v. 3. these values for N depending on (b/a) gives the curve shown in the attached diagram. To give some examples: a) The nearfield distance of a line source of length 2a is N( 0) = 0.988 (aa f)/ v. b) the nearfield length of a transducer with (b/a) = 0.5 is N(0.5) = 1.014 (aa f/ v). c) the nearfield length of a transducer with (b/a) = 1 is N(1.0) = 1.368 (aa f) v). Diagram by separate mail kind regards Udo Schlengermann
