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TB 55190023210
G3.4 TOTAL BARGE RESISTANCE . Add the results of frictional wave forming and wind resistance from the previous
three sections.
G3.5 EXAMPLE. Assume a berthing barge (YRBM) is to be towed Dimensions are:
Length
265 feet
Width
65 feet
Draft
7 feet
Hull Depth
12 feet
Deck house
220 ft x 55 ft x 25 ft
Desired tow speed is 10 knots, maximum head wind IS 20 knots Bottom conditions are average. The hull has rake
shaped ends.
G3.5.1 Frictional Resistance. Estimate the average length below the waterline as 250 feet, with the bottom being 245
feet.
S = 7 x 2 x (250 + 65) + (245 x 65)
= 20,335 ft2
R = 65 x 20,335 x (10/6)2
= 36,716 lbs
G3.5.2 Wave Forming Resistance.
B = 65 x 7 = 455
G = 2.85 x 455 x 0 2 x 102 x 1 2
= 31,122 lbs.
In this case a 20knot wind could have generated an average wave height of 5 feet. Curve No. 1, Figure G2 predicts
an added resistance of 2,500 pounds. This is well within the 5,187 pounds catchall correction included in the formula's 1
2 factor, so do not add the additional 2,500 pounds.
If the tow encountered 10foot average wave height, and was structurally adequate to continue, the added resistance of
12,000 pounds would be added to the total resistance.
G3.5.3 Wind Resistance.
C = 65 x (127) + (25 x 55)
= 1,700 ft2
Where:
C = frontal area
W = 1,700 x .004 (20 + 10)2 x .6
= 3,672 lbs.
G3.5.4 Total Resistance.
R + G + W = 36,716 + 31,122 + 3,672
= 71,510 lbs
Inspecting Figure 61 shows that this tow is within the capability of the TATF and ATS 1 Classes, at the assumed towing
speed of 10 knots Clearly, other towing ships, ag., the ARS 50 and ATF 76 Classes, would be adequate for this tow at
lower speeds
G20


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