2 3 ° Mr. Atwood’s Disquisition on 
Let the sides of a vessel be plane surfaces, inclined to each 
other at an angle of 30° ; the vessel’s inclination from the up- 
right — 15°; BA =t = 100 ; GE = d = 13 ; the angle SAH = 
105 ; AHS = 6 o° : BCS == go 0 . By calculating the value of 
the line GZ, according to the solution just given, it is found 
that SA3 xsa 
6Vb 
X 
1 SB 3 xsn 
and_ 6v7~ x 
v/ 
, s z xi—b z , 4? x v' i 
4 H — v- H - 
b 
4 + 
S*X !• 
4s x I — a 2 
3 - 1*55 
3- 10 75 
Sum of these values = 6.2230 
d s - - = 3-365 
Finally, the measure of the vessel’s stability, — 
or 
+ 
SA 3 x s a 
6Vb 
SB 3 x 5 
6Vc 
x v/ 
- 
-XV 
4 + 
s z x 1 —b z 
+ 
4s x v' i — a 1 
4 + 
s x 1 — c 2 
4$ X v' 1 • 
ds=z GZ = 2.838 
If the weight of the vessel should be 1000 tons, the force of 
stability will be equivalent to that weight of pressure, acting at 
the distance of 2.83 from the axis ; or the weight of 37.0 tons, 
acting at the distance of 30 from the axis. 
If the sides should be inclined at an angle of 6o°, instead of 
30°, the measure of stability will be 2.92 ; and the effort to turn 
the vessel equal to 1000 tons, acting at the distance 2.92, or 
38.4 tons acting at the distance of 30 from the axis. 
The sides of vessels are not unfrequently formed so as to 
coincide with the sides of an isosceles wedge, or are so little 
curved as to approximate nearly to that figure, at least so far 
as that portion of the sides extends which may be immersed in, 
or may emerge from, the water, by the vessel’s inclination. The 
preceding solution being expressed in terms which are rather 
