Tsaacs—Faughn—Schick-Sargent: Deep-Sea Moorings 309 
Step 3. Calculate drag frontal area of wire 
Step 4. Calculate drag of various components 
Step 5. Calculate weight in water Fy of various components 
Step 6. Sum weight, Fy and drag Fy; (individually) 
Step 7. Calculate wire angle =are tan SF'y/2Fy 
The wire shape may now be derived graphically by constructing successive 
foree-vector diagrams starting at the anchor and proceeding upward as shown in 
table 6 and figure 13. Note that figure 13 shows the graphic steps of only the lower 
three segments into which the mooring has been divided. Force diagrams and 
configuration are simultaneously constructed. 
TABLE 6 
SAMPLE CALCULATION OF FoRCES 
(Sample calculations for a mooring using 2,500 fathoms of .120-inch piano wire which weighs 
500 pounds and has a breaking strength of 3,190 pounds. As the anchor for this mooring must 
weigh a minimum of 536 pounds or a maximum of 1,414 pounds, an anchor weighing from 700 
to 900 pounds would be a reasonable choice.) 
Depth v Vv y2 Area Fu = Fa Fy = Fv 
(fathoms) (knots) (ft/sec) (ft?/sec?) (£t?) (Ibs.) (Ibs.) (Ibs.) (Ibs.) 
SKS is eee EMG: cee aston ae 90.0 ee Sse ees 
BUOY Ae esiact nals alae oer shee ne 10.0 100.0 —780 —780 
NOOS200 Ne faccteyeveisvers 1.00 1.69 2.86 6 18.9 118.9 +40 —740 
200=300. 220.52... 0.750 1.27 1.62 6 10.7 129.6 20 —720 
300-500........... 0.486 0.818 0.669 12 8.9 138.5 40 —680 
500-1000.......... 0.439 0.742 0.550 30 18.2 156.7 100 — 580 
1000-1500......... 0.370 0.626 0.392 30 13.0 169.7 100 —480 
1500-2000......... 0.302 0.512 0.262 30 8.6 178.3 100 —380 
2000-2500 
(bottom)....... 0.234 0.397 0.158 30 5.0 183.3 100 — 280 
CALCULATION OF RESTORING FORCES ON A SURFACE FLOAT 
Imagine a pennant having n floats equally spaced along its length. Let the slope 
of the pennant at the submerged float of the taut mooring be 6,, and the slope at 
the skiff be zero. Then as an approximation we may consider that at each line float 
the slope of the pennant changes by the same amount A@, and that it is perfectly 
straight between floats. Then 
A6= (1/n) 6 
and the slope above the mth float (fig. 14) is expressed by 
On = On SO) 
If Fy, is the buoyancy of a float, then the force it exerts in a direction at equal 
angles to the two pennant segments above and below it is expressed by 
Fy, = ¢08 (Am-1—- Y2A0) Fg (20) 
Therefore the force in the direction of the pennant segment above the float and in 
the downward sense is expressed by 
F, = YF, /sin Y2A8 (21) 
Although it is clear that the figure assumed for the array is imaginary and never, 
