Calculating load for tidal movement

AI Thread Summary
Calculating tidal movement load involves using wind load formulas, specifically wind speed squared multiplied by 0.00431 and the area in square feet. For example, a 5 mph wind on a 100 sq. ft. surface results in a force of 10.77 lbs. The mass of salt water being approximately 854 times that of air raises questions about adapting wind load calculations for water. Different atmospheric conditions may also influence the weight of water, necessitating adjustments in calculations. Various standards and coefficients, such as those in API 2SK, are essential for accurate force assessments on ocean vessels.
jamesd442001
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Calculating wind load is wind speed squared x .00431 x area in sq. ft. A 5 mph wind with a vertical 100 sq. ft. wing is 10.77 lbs. The mass of salt water is around 854 times greater than the mass of air. Can you use the wind formula and multiply by 854? Does the weight of water at different atmospheres change this?
 
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jamesd442001 said:
Calculating wind load is wind speed squared x .00431 x area in sq. ft. A 5 mph wind with a vertical 100 sq. ft. wing is 10.77 lbs. The mass of salt water is around 854 times greater than the mass of air. Can you use the wind formula and multiply by 854? Does the weight of water at different atmospheres change this?

What is the context of this question? Various standards would apply and have things such as shape coefficients that will change the resultant force from environmental loadings.

The type of hull (presumably this is for an ocean vessel) may also introduce other factors to consider.

That being said, the final calculation will be some derivative of the basic equation:

F = PA

API 2SK is a good reference for this. For a ship hull the current force on the bow or stern (in lbf) is:

F_{cy} = C_{cy}SV_c^2

where,

S = wetted surface area of hull including appendages (ft^2)
V_c = current speed (knots)
C_cy = current force coefficient = 0.40 lbf/(ft^2 - knots)

Hope this helps.

CS
 
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