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 TB 55-1900-232-10
G-3.4 TOTAL BARGE RESISTANCE . Add the results of frictional wave forming and wind resistance from the previous
three sections.
G-3.5 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.
G-3.5.1 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
G-3.5.2 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 20-knot wind could have generated an average wave height of 5 feet. Curve No. 1, Figure G-2 predicts
an added resistance of 2,500 pounds. This is well within the 5,187 pounds catch-all correction included in the formula's 1
2 factor, so do not add the additional 2,500 pounds.
If the tow encountered 10-foot average wave height, and was structurally adequate to continue, the added resistance of
12,000 pounds would be added to the total resistance.
G-3.5.3 Wind Resistance.
C = 65 x (12-7) + (25 x 55)
= 1,700 ft2
Where:
C = frontal area
W = 1,700 x .004 (20 + 10)2 x .6
= 3,672 lbs.
G-3.5.4 Total Resistance.
R + G + W = 36,716 + 31,122 + 3,672
= 71,510 lbs
Inspecting Figure 6-1 shows that this tow is within the capability of the T-ATF 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
G-20
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