Calculating Windspeed after Brick Launch from Dumpster Lid

[matt621]

Hello,

We had a wind storm. We have a dumpster with a lid. On that lid we put a heavy brick to keep the lid closed during wind storms. It's worked great for years. But we had a very strong wind storm and it launched that brick quite a distance. I'm trying to figure out the wind speed that would cause that.

Brick is 4.14#s
Distance traveled 55 feet.

The lid is pivoted on one side so the wind has to lift first, then once it catches the wind, it'll flip that lid nearly instantly and off the brick goes. So the launch height above the ground would probably be about 5'. (The height of the dumpster is about 3' and the lids are 3' so if the brick is w/in a foot of the edge of the lid that would put a launch height about 5'.)

The angle of launch? I would guess 30 to 40 degrees, but that's just a guess but don't have any clue as to the actual trajectory. Just imagine a big flat plate, secured at one end with a pivot and it opens up. The brick is usually about a foot from the end. So as the wind picks up the leading edge it'll go nearly vertical nearly instantly once the wind gets under it.

The lid is 3' x 3'. So about 9 square feet.

Can anyone help me figure this out?

Thanks

 

[jrmichler]

There are four main steps to solve this.

#1: How much wind to catch the lid and flip it up. A similar problem is the loading on a roof overhang during a hurricane. The loading is the area of the overhang times the wind velocity pressure times 2.0. The uplift coefficient of an overhang is 2.0. You can calculate the force to lift the lid from the weight and dimensions of the lid plus the weight and location of the brick, or you can measure it directly with a spring scale. The result will be the minimum wind to lift the lid. The actual wind was likely greater than that.

#2: The lid is up, and the wind force is accelerating it. This is an F = ma problem where the wind force is a function of lid position, and the wind velocity is unknown. Search lift and drag on a flat plate to learn about wind forces as a function of lid position. This problem can be solved by a numerical simulation with different wind velocities. The goal is to calculate lid velocity vs time and lid velocity vs lid position for different wind speeds.

#3: Now you have some curves of lid velocity vs time, and lid velocity vs position, you can calculate the brick velocity vs lid position. The brick initially moves with the lid. At some lid velocity and position, the brick starts to slide. The free body diagram of the brick includes the force of lid acceleration, gravity, and centrifugal force. At some point, the brick slides off the lid. Do this calculation for each wind velocity. This is a second numerical simulation problem. The goal is calculate the velocity, direction, and height at which the brick leaves the lid for different wind velocities.

#4: When the brick left the lid, it had a velocity, direction, and height. All of those variables are a function of the wind velocity. It came to a stop at a distance and height. In between, it was subject the force of gravity plus the wind. This is a third numerical simulation problem.

Or you can just say "It was blowing so hard it threw that brick 55 feet".

 

[matt621]

I think it's safe to assume the brick did not slide down much once the lid started to rise up. Even just a little opening would cause the lid to flip nearly instantly. So the starting height above ground would be about 5'. So what I did was use a cannon ball trajectory and if I select a launch angle of about 0 degree (I think this is a safe assumption) it shows 68mph for a distance of 55', which the official weather station showed wind speed of 69mph. (They don't show gusts.)

Thanks