As a fellow scientist, I hypothesize that the maximum size a man can balloon to and still have a rock hard 12 - 2 o'clock erection is under between 6.75-9.50 inches in length and 6-7.50 inches in girth, which includes the glans circumference as well. For example, at maximum ballooning stage, a man can reach 8.5 inches in length, 6.50 inches girth midshaft and 6.75 inches around in the glans or head. Thus, I would add these variables: 8.5 + 6.50 + 6.75 = 21.75 >From my calculations, it is nearly impossible for a man to have at least a 2 o'clock erection or higher angle with an index OF 22.00 OR GREATER. THE INDEX MUST BE BELOW 22.00. SO THE EXAMPLE SHOWN HERE IS VERY CLOSE TO THE MAXIMUM SIZE OF LENGTH, MIDSHAFT GIRTH AND HEAD SIZE A MAN CAN ATTAIN AND STILL HAVE A 2 O'CLOCK OR HIGHER ANGLE ERECTION. OF COURSE, THERE ARE OTHER PERMUTATIONS TO ACHIEVE THIS INDEX, BUT THEY MUST FOLLOW THE LENGTH AND GIRTH LIMITATIONS POSED ABOVE. I PROPOSE THAT THE MAXIMUM DIMENSIONS TO ACHIEVE THIS 2 O'CLOCK OR HIGHER ANGLE ERECTION ARE AS FOLLOWS: LENGTH: NO MORE THAN 9.5 INCHES IN LENGTH GIRTH: MIDSHAFT AND HEAD SIZE NO MORE THAN 7.5 INCHES IN CIRCUMFERENCE. HERE ARE THE POSSIBLE COMBINATIONS: Note that head size is either equal or slightly larger than the shaft to accentuate the drumstick effect you refer to. All the following numbers are in inches. Length midshaft glans INDEX TOTAL 6.75 + 7.50 + 7.50 = 21.75, 7.00 + 7.25 + 7.50 = 21.75, 7.25 + 7.25 + 7.25 = 21.75, 7.50 + 7.00 + 7.25 = 21.75, 7.75 + 7.00 + 7.00 = 21.75, 8.00 + 6.75 + 7.00 = 21.75, 8.25 + 6.75 + 6.75 = 21.75, 8.50 + 6.50 + 6.75 = 21.75, 8.75 + 6.50 + 6.50 = 21.75, 9.00 + 6.25 + 6.50 = 21.75, 9.25 + 6.25 + 6.25 = 21.75, 9.50 + 6.00 + 6.25 = 21.75 I WOULD APPRECIATE YOUR EXPERTISE AND VALUABLE INSIGHT REGARDING MY HYPOTHESIS AND RESEARCH. THANKS YOU FOR YOUR ATTENTION AND TIRELESS SUPPORT DR. LIN. PLEASE NOTE THAT WE ARE DISCUSSING THE MAXIMUM CAPACITY OF YOUR BALLOONING METHOD BECAUSE I KNOW THAT YOU BELIEVE THAT A 12 - 2 O'CLOCK 6-7 INCH LENGTH ERECTION IS SUFFICIENT FOR MOST WOMEN. THESE EXTREME SIZES ARE FOR THE WOMEN THAT CAN ACCOMODATE SUCH A STRONG YANG IMPLEMENT.
Another important class of problems involves cantilever beams. The bending moments (M) , shear forces (Q), and deflections (w) for a cantilever beam subjected to a point load at the free end and a uniformly distributed load are given in the table below[5].
Distribution | Max. value | |
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Cantilever beam with end load |
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M(x) = P(x − L) | MA = PL | |
Q(x) = P | Qmax = P | |
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Cantilever beam with uniformly distributed load |
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Q(x) = q(L − x), | QA = qL | |
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