My question is if you have a prostate left after that many cores were taken? ;)
Valid question. Here's a back of napkin calculation of the percentage removed based on a few likely assumptions:
- cylindrical volume = cross sectional area X aggregate length
- cross sectional area for typical 18-gauge needle = pi X r-squared, or 3.14X(.5 mm)X(.5 mm) = 0.78 square mm
- length = (93 cores)X(15 mm/core, typical) = 1,395 mm aggregate length
- therefore, volume = (0.78 square mm)X(1,395 mm) = 1,088 cubic mm prostate material
A prostate 63g is same as 63 cubic cm, which is 63,000 cubic mm.
So therefore, (1,088 cubic mm) / (63,000 cubic mm) = 1.7% is a rough estimate of the material removed.