Question 1121277
{{{(L+1.25*sqrt(S)-9.8*3sqrt(D))/.686}}} < 24 
 L is the boats length ( in meters ),
 S is the sail area ( in square meters , and
 D is the displacement ( in cubic meters ).
:
 Displacement is the volume of the boat that is underwater.
 Yacht designers try to make the value of the left side of the inequality as close to 24 as possible. 
The table below lists data on the AC45 Wingsail Catamaran, a yacht that was built to complete in the 2013 America’s cup.
 You are going to design a new yacht that will satisfy the America’s Cup rule and have twice the length and twice the sail area of the AC45z 
Length : 13.45 meters 
Sail area : 93.7 square meters 
Displacement: 1.3 cubic meters
: 
What are the requirements of your yacht? 
L = 26.9 m
S = 187.4 sq/m
D = ?
:
{{{(26.9+1.25*sqrt(187.4)-9.8*3sqrt(D))/.686}}} =24 
{{{(26.9+17.11-9.8*3sqrt(D))/.686}}} =24
{{{(44-9.8*3sqrt(D))/.686}}} =24
multiply both sides by .686
{{{44-9.8*3sqrt(D)}}} = 16.464
subtract 44 from both sides
{{{-9.8*3sqrt(D)}}} = 16.464 - 44
{{{-9.8*3sqrt(D)}}} = -27.536
{{{3sqrt(D)}}} = {{{-27.536/(-9.8)}}}
{{{3sqrt(D)}}} = +2.81
cube both sides
D = {{{2.81^3}}}
D = 22.2 cu/m is the displacement of the larger boat that still meets the America cup requirements
However:
Check
{{{(26.9+1.25*sqrt(187.4)-9.8*3sqrt(22.2))/.686}}} = 24.001
that's a little over 24, make the displacement = 22.3 cu/m and we get
{{{(26.9+1.25*sqrt(187.4)-9.8*3sqrt(22.3))/.686}}} = 23.9
;
that would be a lot of boat below the waterline!