Makes perfect sense. Although, I can’t see why you would use a square shape in comparison to a circular one when most penises are elliptically shaped. I think this topic has been covered in the various posts on measuring penis volume as well.
The approximate cross-sectional area of a penis should be calculated using the formula for the area of an ellipse. This is especially the case if a penis is noticeably elliptical, which I imagine most are to some degree.
The formula for the area of an ellipse is a slight variation on that for a circle. Area = pi*A*B where A is the long axis and B is the short axis. So, if your penis has a thickness of 2” when you measure from one side to the other across the top and a thickness of 1.5” when you measure from one side to the other from top to bottom, A = 2”/2 = 1” and B = 1.5”/2 = 0.75”. This particular penis would have a cross-sectional area of Area = (3.14)(1”)(0.75”) = 2.355 inches squared.
You can also see how the equation doesn’t stray far from Area = pi*r^2 by calculating the area of an ellipse with equal axes. Every circle is an ellipse, but not every ellipse is a circle.
Accurately calculating the perimeter of an ellipse requires more than just the values of the long and short axes. However, an approximate calculation can be made using the formula Pe = 2*pi*[1/2 (A^2 + B^2)]^1/2
What it all boils down to is the ratio of A to B for your penis. A penis with a 1:1 ratio is circular and a penis with a ratio exceeding about 1.15:1 is around the low end of being noticeably elliptical. So, how does this all relate to perimetral girth measurements? Does having an elliptical penis really put you at a disadvantage?
Let’s compare four penises. The first is 8” x 6” with a uniform circular cross-section of the shaft, the second is 8” x 6” with an A/B ratio of 1.15:1, the third is 8” x 6” with an A/B ratio of 1.25:1 and the fourth is 8” x 6” with an A/B ratio of 1.5:1. I could be wrong, but I don’t think too many penises have a long axis that exceeds the value of the short axis by more 50%.
#1
—-
Given:
P = 6”
Circular Shape
P = 2*pi*r
r = 6 / (2*pi)
r = 0.955”
Area = pi*r^2
Area = 2.86 inches squared
#2
—-
Given:
P = 6”
A/B = 1.15:1
==> A = 1.15B
P = 2*pi*[1/2 (A^2 + B^2)]^1/2
Sub A = 1.15B
6 = 2*pi*[1/2 (2.32B^2)]^1/2
B = 0.887”
A = 1.15B
A = 1.02”
Area = A*B*pi
Area = 2.84 inches squared
#3
—-
Given:
P = 6”
A/B = 1.25:1
==> A = 1.25B
P = 2*pi*[1/2 (A^2 + B^2)]^1/2
Sub A = 1.25B
6 = 2*pi*[1/2 (2.56B^2)]^1/2
B = 0.844”
A = 1.25B
A = 1.06”
Area = A*B*pi
Area = 2.81 inches squared
#4
—-
Given
P = 6”
A/B = 1.5:1
==> A = 1.5B
P = 2*pi*[1/2 (A^2 + B^2)]^1/2
Sub A = 1.5B
6 = 2*pi*[1/2 (3.25B^2)]^1/2
B = 0.749”
A = 1.5B
A = 1.12”
Area = A*B*pi
Area = 2.63 inches squared
===============================
What does all this say?
Comparing penises #2, #3 and #4 to penis #1 reveals:
Penis #1 is 0.7% bigger than penis #2
Penis #1 is 1.8% bigger than penis #3
Penis #1 is 8.7% bigger than penis #4
The differences in area between penises #1 and #2 and #1 and #3 are essentially negligible. And even if you have an elliptical penis with a axial ratio of 1.5:1, you are still 91.3% as big as someone that has a perfectly round penis with the same perimeter as yours. I really don’t know if ratios greater than 1.5:1 are common but if they aren’t, cross-sectional shape doesn’t seem to play a great role in filling your woman. However, a “flat” penis might have a harder time putting pressure on the G-spot area than a more rounded penis. I really don’t know though.. I don’t have a G-spot.
===============================
After wasting my time typing this out, I realized that a few Excel plots would have been far more efficient. Attached are three plots. The first plot shows cross-sectional area relative to A/B ratio for a 6” perimeter. The second and third plot size percentage relative to the baseline 6” perimeter/1:1 penis. Just to clarify, all values assume a 6” perimeter. The only variable is the A/B ratio.
Cheers,
G&G