How much ballast do I need?

April 17, 2016

As we enter the silly season, one thing can be certain wherever you are in the world, the weather is playing with us. Tragically as we have all seen it must be respected, it could be interpreted as grossly negligent to call any weather incident a ‘freak of nature’ especially when you consider  the level of monitoring technology, 24/hr news and forecasts, advances in mobile technology coupled with peoples raised awareness of climate related issues.


One issue we are often faced with is that fact that you cannot use ground anchors or stakes of any type, either due to the location of the item/structure, type of soil or the fact that the soil is already saturated or indeed the type of surface is such that staking is not an option e.g block paving, ornamental surfaces etc. In these cases you have to be able to provide the correct amount of surface ballast or kentledge if you come from the sea faring community. Lets keep things simple and call it ballast for now.


For demonstrative purposes, lets assume that we are going to have a structure that is 10mx5mx3m and will be clad on all faces with 100% wind resistant material ie not shade netting or scrim but more like branding vinyls or boards. In the UK we take as a rule of thumb that all structures must be able to withstand 25m/s wind pressure. Manufacturers of inflatable play equipment generally advise a maximum of 12m/s before needing to deflate but each case has to be considered individually. The greater the surface area the lower the maximum tolerance which can be as low as 6-9m/s. In order to prevent the structure from falling over due to wind load alone, there a few things we need to know and calculate. For the sake of our example the wind is only going to blow on the larger face of the structure and will be held down by 4 anchor points A,B,C,D. (fig.1)






The pivot at which the structure will fall around shall be referenced as the overturning moment and this depends on a couple of primary factors. 1) The direction of the wind and 2) the moment about which the structure can turn, marked as OT1 in the diagram above because the wind force being applied to the structure is coming from front to back, marked as ‘P’


As we know wind has a mass and a weight, therefore there is a figure known as a constant by which we can start our calculations, this is 0.613. The velocity of the wind that we need to maintain stability is known as V2 and in this case is 25m/s.



Therefore the wind pressure can be calculated as :-


P= 0.613 x V2

P=0.613 x (25)2



Looking at the graph below, we can see that wind load has already been calculated and at 25m/s it creates a load of 375N/m2. See fig 2. Table available at








The first step is to calculate the wind pressure at 25m/s (V2) (gust)


P=0.613 x 25 x 25 / 1000 = 0.383 kPa


The Pressure (P) is measured against the full area of the face of the structure:-


Area =  x * y, therefore

Area = 10 x5 = 50 m2


To establish the resultant force (R) on the face of the structure in kN


kN = P x A

therefore 0.383 x 50 = 19.15kN


The axis around the overturning moment shall be half of the height of the y axis (height) of the structure which in this case is 5m. Therefore the overturning moment is measured in kNm and is derived as


OTM = R x 5m

OTM = 19.15 x 5 = 95.75 kN



To work out the required quantity of ballast the next step is to determine the lever arm on the width of the structure which in this case is 3m.


The amount of individual ballast blocks acting against the pressure in our example can be more than 2, however the total should not be less than the total required and should be evenly spaced across the width of the structure. Note that the ballast at the rear of the structure would serve no purpose to the wind coming from the direction as shown in the drawing. Ballast (b) shall be known as (n)b where n is the number of ballast acting against the wind load. Therefore from above -


95.75 kNm = 2b x 3m

so : 2b = 95.75 /6 =15.95 kN

Therefore ballast A,B should equal 1.595 T


However…. In order to prevent the structure falling over when the wind is blowing from front or back ; up to a wind speed of 25m/s (gust speed)…


The weight of each ballast must not be less than 800Kg to ensure that each side is equal to the total required 1.6T


In this example we have not considered or incorporated the self weight of the structure, this would also counter any overturning moment, however, this is not a mistake, this allows for a common sense factor of safety (FOS - normally 1,4) to account for the many other variables such as , design, materials, design safety factor of material failure and so on. Additionally it should be noted that the above is demonstrative of the calculation process and that you must also have regard of other factors including but not limited to, dynamic (horizontal and vertical) and static loads


It is important to remember that ballast needs to be secured to the structure by a method of restraint capable of resisting the imposed load. . If the ballast is separate to the structure, then the type of ballast selected must be suitablke for the ground conditions upon which it sits, after all wouldn’t  want you ballast sliding along the floor! The co-efficient of materials and surfaces are a known factor but this subject presents a topic in its own right.


Things to avoid however would be to use rubber matting between your concrete ballast and a concrete surface as this would slip more easily than if the two concrete surfaces were in direct contact. Water kentledge in polypropylene tanks should also be avoided wherever possible, leaks often occur and we find many that have not been filled correctly in the first instance. Additionally, the friction coefficiency between polypropylene and grass, tarmac or concrete is low, meaning that once it starts to move, it will probably keep moving!!


Another point of failure, as alluded to earlier is the attachments of ballast to the structure, there is no point in using 3 or 8 strand polyester rope (10 mm) that has a minimum breaking strength of 1.2T and owing to the safety factor, a SWL of only 107Kg where  the ballast weighs approx 800kg. If you are using ratchet straps, they must be rated, have the SWL stamped somewhere on the label or strap and be subject to thorough examination and inspection. These cannot afford to fail.


By far the best type of attachment for ballast has to come from steel wire rope that has been rated, is well maintained, stored and inspected. It is acknowledge however that for the smaller weights involved SWR would not be needed considering that even a small diameter rope of 6.4mm has a minimum breaking strength of 2.4T and has a SWL of 489 Kg


The attachment of ballasts often require shackles, in the same way as the straps and ropes need to be robust, shackles must also be rated to the correct load handling characteristics. We often see in many structural failures that there is not just a single point of failure but actually there are multiples. Which gives way first always depends on the cause but if the strap holds, the shackle may not, if they both hold, will the ballast stay in its designed location. If all these pass, the structure should be safe to withstand the intended wind load.


The calculation of ballast requirements on this simple example is by no means simple, especially considering that many structures are not as symmetrical as our example. If you need to calculate the amount of ballast for any structure or inflatable play equipment and are unable to do so yourself, always take professional advice from a legally competent person such as a civil or structural engineer. In any case where ever you have a complex structure you should always get an independent design review which will indicate exactly how much ballast needed and where it should be placed.


Steve Riches (CMIOSH)



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