14 mph winds?

Question

would 14mph winds be able to push my wood and glass door open a little if not completely closed already? well closed but not completely clicked. i also have a metal screen in front of the door, it would have to be able to push through the screen to push the door open ( screen closes in the opposite direction) so screen itself would not move

Answer 1

yes

Answer 2

You can use the drag equation to get an idea of the force available from the wind: http://en.wikipedia.org/wiki/Drag_equati...

Answer 3

More than likely yes.

It would have probably occurred with much lower wind speed also.

Something to keep in mind however is that unless you were actually measuring the wind speed outside the door at that exact moment that it happened, you will never know what the actual wind speed was which caused it to happen.

Any wind is not a steady state condition.

A 14 mph wind is an average with gusts exceeding that and lulls below that average.

If you wanted to find out what total weight of pressure was required to open the door, you could take a kitchen scale and slowly press it against the door while reading scale until the door opens.

You could then extrapolate form the other known information by calculating the total exposed surface area, which in your case would be the total area of the metal screen mesh, less the total amount of mesh surface area and go from there.

That is the really difficult part since the wire mesh screen will have more or less resistance to the air depending on it's mesh.

There is where any further assistance becomes difficult since I can not think of an easy way of doing that calculation.

They only way I can think of would be if you knew the diameter of the wire used and counted the number of strands per inch to get a percentage of area per inch that the material occupied, otherwise it would only be guesswork.

Even if you went to all the trouble of doing these calculations and measurements, I suspect you may always wonder, 'Did I actually latch that door completely, or did someone try to enter the house?'

I am hoping that someone a bit more clever than me comes up with an easy way of doing the calculation since I am going to let this 'bug me' now for another couple of days!