 # WInd Speed to Force Knots to Tons / Pounds per Square Foot

Wind Force
Force per square foot experienced when wind is blowing perpendicular to a surface is calculated using this formula:

``````                    F = 0.004V(squared)

F = wind force measured in pounds per square foot
``````

V = wind velocity in knots

``````                                 Wind Speed (kts)---------Force (lbs/square foot)
``````

10 ---------------------------0.4
15 ---------------------------0.9
20 ---------------------------1.6
25 ---------------------------2.5
30 ---------------------------3.6
34 ---------------------------4.6

``                                                     [I]Source: U.S. Navy Sailors Handbook [/I]``

This might be somewhat counterintuitive, when wind speeds increase from 10 to 15 knots the force doubles.

When wind speed doubles the force exerted increases by a factor of 4. If the force at 30 kts is 3.6 lbs/square foot the force at 60 knots calculates out to 14.4 lbs/ft(3.6 X 4 = 14.4)

If an operation is effected by winds it might be more useful to look at the problem the other way. That is start with a wind speed at which operations become marginal.

For example at 30 kts the wind force is 3.6. lbs/ft^2. The force is double (7.2 lbs/ft^2) at a wind speed of 42.5 kts.

D=(1/2)c_drhoV^2A

Where,

D = Drag Force
Rho = Density of fluid, combination of air, seawater, and maybe a bird
A = Area of affected area
c_d = Coefficient of drag of affected object, a flat plate would be 1.28 assuming perpendicular flow

Just make sure you check your units!

Don’t forget about the huge moment that will be placed on object.

[QUOTE=lm1883;168103]This is the one I use, which is from Rowe’s book.

Force in tons = (area/1000) (velocity squared / 18 )

Velocity is in meters per second (approx half of the value in knots)

Something to think about is the density of the Air. A 20 knot wind at say 25 degrees F is much stronger than a 20 knot wind at 70 degrees F. I seem to have misplaced the formula for that but a French Pilot had that equation and others posted somewhere on the web.[/QUOTE]

I came across The Hughes Formula at work. It takes into account the angle of the wind as well. It seems to be in widespread use generally in slightly different forms. It can be found here.

The version I use is simpler, it assumes the wind direction perpendicular. It doesn’t take into account the angle.

The advantage of Navy formula of course is it’s simple and is an improvement over a seat-of-the-pants guesstimate.

I had this explained to me by a ferry skipper once, he went on to say that there was also a formula for working out whether the cost of using x number of tugs to assist docking was greater than the cost of replacing y amount of fendering they were likely to damage if it went slightly wrong.

[QUOTE=Day Sailor;168190]I had this explained to me by a ferry skipper once, he went on to say that there was also a formula for working out whether the cost of using x number of tugs to assist docking was greater than the cost of replacing y amount of fendering they were likely to damage if it went slightly wrong.[/QUOTE]

In most ports the number of tugs used seems pretty standardized based on past practice etc. In my experience 99% of the time the best bet is the pilot’s advice. It is sometimes nice to know the minimum required bollard pull required based on wind speed. From what I’ve seen ports that have guidance use either a graph or a table like this one from the Columbia River Pilots:

](https://flic.kr/p/xo9nmf)image by kennebeccaptain, on Flickr[/IMG]

I’m pretty sure those calcs were intended to be used for sail-area calculations, they do not take into account any lateral resistance from the underwater portion of the hull.

[QUOTE=captjamied;168366]I’m pretty sure those calcs were intended to be used for sail-area calculations, they do not take into account any lateral resistance from the underwater portion of the hull.[/QUOTE]

That’s right, the graph is going to give approximate the amount of bollard pull to equal the force created by wind. It’s to get in the right ballpark.

The wind force at 48 kts is double what it is at 34 kts.

48 kts is the lower limit of Beaufort F-10 or Storm Force winds, 34 kts is the lower limit of Gale force winds or F-8.

The formula in post #1 comes from the Aug, 1999 Mariners Weather Log article in which the 34 kt Radius and the 1-2-3 rule for Hurricane avoidance was published.

One of the authors of that article is Michael Carr who was at the [I]El Faro[/I] hearings.