Force, momentum and umbrellas

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.

21 thoughts on “Force, momentum and umbrellas”

  • I am trying to develop a simple umbrella quality test to assess the chance of an umbrella turning inside out. I was planning on attaching a string to each of the spokes and applying a set force to see if the umbrella will flip inside out. Can you recommned what this force should be?

  • Marcus Wilson says:

    Fascinating comment Toni. Glad to see that someone is taking the problem seriously. I reckon any umbrella that can’t cope with a 40 km/h (11 metres per second) wind is as about as valuable as a Greek government bond. If you do the calculations (force = density times velocity squared times cross-sectional area) you’ll get about 120 newtons for each metre squared of area of the umbrella.
    So…if your umbrella radius is ‘r’, say, (measure it in metres) then your force is 120 N times pi times r squared. If you have M spokes on the umbrella, this means that each spoke has to take 120 pi r^2 / M newtons of force. Larger umbrellas are going to have a more difficult problem, since they have larger area.
    So, if the radius of your umbrella were 70 cm (=0.7 m) and it had 8 spokes, then your force per spoke would be about
    120 x 3.14 x 0.7 x 0.7 / 8 newtons = 23 newtons.
    If you hang weights under gravity to make your force, then a 1 kg mass gives you 9.8 newtons of force (call it about 10 newtons), so you’d hand 2.3 kg of mass on each spoke. Of course it depends on what you set as the wind speed where failing is tolerated. In windy places (like Portsmouth) you might want it higher than 40 km/h.
    You could I guess slowly increase the masses until the umbrella failed, then work backwards to work out the overall wind speed at which the umbrella would fail.
    Hope this helps.

  • Hi Marcus,
    Thank you so much for your comprehensive reply. If we look at the data provided by the UK Met office:
    it would be reasonable to say that you wouldn’t expect to be able to keep your umbrella up in a near gale – do you agree. In which case a strong breeze of 24 knots (44.448 km/h)would be the next category down . If we take the maximum value of 27 knowt (50km/h) then I guess this would be an acceptable limit to set for pass/fail criteria for a standard umbrella.
    Can you explain to me how you calculated the force per spoke?
    I was planning on joining the strings and pulling the umbrella inside out by applying the force to all of the spokes at the same time. Would this work?
    Thank you very much for your help. I have struggled to find anything in my school library to help with this.

  • Marcus Wilson says:

    Here’s how the equation works again.
    The force (in newtons) on the umbrella is given by
    Density of air (in kg/m3) x Area of umbrella (in m2) x velocity of air (in m/s) squared.
    Then divide this by the number of spokes to get force per spoke. Or, if you want to tie the strings together and just hang one weight on it, then you obviously don’t divide by the number of spokes. The equation comes from fluid mechanics, considering the rate of change of momentum of the air.
    The density of air is conveniently about 1.0 kg/m3, but you should be able to look up a value. The area of the umbrella obviously depends on how big the umbrella is – it will be approximately pi times r squared, where r is the radius (again, use m as the unit).
    So for any given velocity, you can work out a force.
    Of course, in practice, a 50 km/h average wind contains gusts that are much stronger. Also, the force on the umbrella won’t always be evenly distributed among the spokes depending on the relative orientation of the umbrella and wind. So you might find that your test will over-estimate how good an umbrella is.

  • Hi Marcus,
    Thank you once again for taking the time to explain this in more detail.
    I will let you know how I get on.
    Really appreciate your support and the quick responses I have received.
    Thank you

  • Hi
    It is probably to much to ask that this thread is alive after seven years…
    Does the shape and design of the umbrella canopy (Dome/pyramid e.g) have any effect on the force that applies? I figure a pointy or dropshaped have less drag than a flat surface…


    • Hello Par

      The thread is most certainly alive if you wish it to be!

      There will be shapes that help – different shapes do have different drag forces, as you say. But there’s no escaping the fact that if the wind gets under the umbrella, it’s hard to keep it umbrella-shaped.

  • Hello,

    I’m hoping you can provide some insight, & please let me know if I’m barking up the wrong tree here – But I’m looking at building an umbrella canopy to span the length of 100m down a road. It is a temporary installation, to be rigged once a year at an event. I’m concerned about wind inverting my umbrellas.
    My current structure is to rig the umbrellas with paracord 500 to the buildings on either side of the road, which is strong enough, as that can hold 250kgs of weight. It will be installed piece by piece, so that not all the weight is on one solid cord. (better weight-wise but timeous to install…)
    Anyway – Do you think that if I have weight (IE – a netting?) over the top of the umbrellas, will this help against the inversion issue? Is it necessary? The last thing I need is for them to all be flipped upside down & looking like a disaster. You have clearly put thought & effort into figuring out the dynamics around this & are clearly better with the numbers game as I am.

    Could you help ?

    • Hi Shannon. It will depend on how big the umbrellas are and how they are designed. But certainly I think you need to do some wind-loading calculations to work out what a safe wind speed would be.

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