In laparoscopic ventral hernia surgery, the ideal overlap in order to secure in-growth is regarded by many authors as 5 cm in every direction from the abdominal wall defect, and many authors have the impression that larger hernias need a larger overlap. Exact placement of the mesh is mandatory for exact overlap, but calculating an overlap by 5 cm, i.e. a mesh 10 cm larger than the defect in all directions, will probably allow for some variation in placement and shrinkage—and still allow secure in-growth. Attempts to calculate mesh-defect ratios have been founded on area calculations or defect-size-mesh-size ratios. In a search for a simpler parameter to express overlap, I found no area-based or ratio-based method that could be applied easily to statistical analysis. In order to achieve a simply expressed parameter for statistical analysis, I hereby propose the following overlap coefficient: The proposed coefficient is based on the assumption of an ideal overlap as a ratio expressed by a single number. The coefficient can be modified easily if different assumptions of ideal overlap are made. The coefficient is calculated from the least overlap in two directions—length and width of mesh versus defect, respectively—therefore yielding a logical argument: the least overlap is decisive. When dividing the difference between mesh size and defect size by the desired overlap one gets a number, the coefficient. If the overlap is 4 cm, where one wants 5 cm, the number becomes k = 0.8—i.e. 20% ‘‘off target’’. Conversely, if the overlap is 6 cm the number becomes k = 1.2; also 20% ‘‘off target’’ but to the ‘‘good side’’. The same deviation is expressed if one defines another target for overlap, e.g. a 3-cm overlap when 4 cm would be desirable: k = 0.75. Thus, ideal and exact overlap should give a coefficient, k = 1; larger overlaps are[1 and smaller overlaps\1. The smallest value for length versus width is decisive, as the defect size is measured by two simple perpendicular estimates and corresponding mesh placement. In the example with 5 cm as the ideal overlap, a constant can be established: 1/(5 cm ? 5 cm) = 0.1 cm. Formula (logical argument):
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