We’ll use an average 5,000 lb vehicle as our example recovering vehicle. The two vehicles are attached with the strap, loose at first. The recovering vehicle proceeds forward with a bit of gas, reaching only 5 mph when reaching the end of the strap. At this point the vehicle has gathered kinetic energy equal to 1/2*mass*velocity^2.
KE = 0.5 * 5000 lb * (5 mi/hr)^2 = 5.7 kJ
We will assume for this instance that the stuck vehicle will remain stuck and will not budge (worst case)… so all of the recovering vehicle’s energy transfers into the strap and is turned into elastic potential energy. This stored energy will be equal to the kinetic energy that the truck had. This stored energy relates to the force exerted on each end by the following: energy = average force * distance. The distance is how far the strap stretches. The average force is assuming the rope exerts constant force, which ours does not. Because it’s force exerted most closely resembles a linear relationship to the stretch, the average force should be multiplied by 2 to get the maximum exerted force (which is all we are interested in here)… assuming the system reaches equilibrium without failure.
In instance 1, we will use a dynamic strap, which can stretch about 6 feet.
5.7 kJ / (6 ft) * 2 = 2,089 lbf (well within the safe range of most straps)
For instance 2, we used a static strap, which we will assume stretches only 4 inches before reaching equilibrium.
5.7 kJ / (4 in) * 2 = 25,072 lbf (enough to snap a strap or possibly rack your frame)
For the last instance, what if we used a chain, which has extremely minimal stretch. So we will say 0.5″…
5.7 kJ / (0.5 in) * 2 = 200,576 lbf (you will certainly break something!!)
So, I hope this gives you a real world, numerical understanding of why dynamic straps should ALWAYS be used in dynamic vehicle-to-vehicle recoveries.
* I did not show unit conversions for the sake of simplicity (there were a lot)