The Square-Cube Law (of Foursomes)
May. 13th, 2011 01:55 amLet's say an average sexual dyad has 2 sets of 4 pointy joints and a (full) bed surface is 53'x 80' or 8 joints/29 square feet or a quarter joint per square foot. If the number of partners is doubled that's over half a joint per square foot.
If range of motion of any given limb is a foot and a half at the pointy joint, then eight joints to the power of 1.5 feet divided among 29 square feet is 22.627417 over 29 = 0.780255759 or less than one joint per foot of bedspace. Double the number of limbs and you get 16^1.5 = 64 & 64 over 29 = 2.20689655 or more than two joints per foot of bedspace and that is really bad odds for not getting an elbow to the head.
I think threesomes would be the most I am willing to risk, mathematically speaking.
If range of motion of any given limb is a foot and a half at the pointy joint, then eight joints to the power of 1.5 feet divided among 29 square feet is 22.627417 over 29 = 0.780255759 or less than one joint per foot of bedspace. Double the number of limbs and you get 16^1.5 = 64 & 64 over 29 = 2.20689655 or more than two joints per foot of bedspace and that is really bad odds for not getting an elbow to the head.
I think threesomes would be the most I am willing to risk, mathematically speaking.