Force, momentum and umbrellas

By Marcus Wilson 21/06/2011

Perhaps the most difficult of the ‘big’ challenges facing modern engineering is designing a lightweight umbrella that doesn’t turn inside out with the slightest breath of wind. There you are, walking down a Portsmouth Street with your umbrella up, when a bus comes along and its wake is sufficient to rip the umbrella inside out. And you get very wet.

A brief physics-based estimate reveals the scale of the problem facing the umbrella-technologists. The force exerted by the wind on an object is equal to the rate at which it intercepts momentum from the air. It’s roughly equal to the cross-sectional area of the object, times the density of the air, times the velocity of the air squared.  The velocity gets squared because while  the momentum of a volume of gas is equal to its mass times its velocity,  the mass intercepted per second is also proportional to velocity. That gives a factor of velocity squared.

So, for my mother’s ex-umbrella, with a cross-sectional area of maybe 3/4 of a metre squared (circle of radius 50 cm), air of density 1 kg per metre cubed, and a wind speed of 10 metres per second, which isn’t terribly large, we get an estimate of force of 75 newtons. That’s the force that a 7.5 kg mass has under earth’s gravity.  With a six-spoked umbrella, it’s the equivalent of hanging about bag of sugar on each spoke.  That’s a lot for a lightweight umbrella to cope with. So no surprises that it failed on me.

Just a couple of days now before I return back to NZ, which is probably when the weather here will improve.