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701a19.jpg
Spice Candy
701a19
>>138609
Every inch of water above a point increases the pressure on that point by 0.037 pounds.
I would estimate that Sheep stands about 5' tall, which makes the room about 8' tall.
The top of the window is about the same height, and it looks to be about 1.5' tall. Adjusting for parallax, it looks to be about 2' wide.
When the room is full, there will be 3' of water above the top, and 4.5' of water above the bottom. Since it's a linear change we can say that the average pressure on the window will be the same as at the midpoint, 3.75'.
45" of water means 1.665PSI. The surface area is 432"^2, so the total water pressure on the window will be about 720lbs. We can ignore the ambient pressure since that's being applied to both sides, meaning 720 is the force difference.
Kicking the window would be useless, but hitting it with something like a wrench could work.
However, we shouldn't ignore the pressure on the door itself. The door goes about 0.5' above the window, and about 3.5' below it. It's about 3' wide. The total area of the door, then, is about 5.5'x3', so 16.5'^2. The water level difference will be 2.5' to 8' for an average of 5.25' thus an average pressure of 2.331PSI acting on an area of 2376'^2 for a total force imbalance of 5,540lbs.
The clincher is that all that force is going to be restrained by the hinges and latch, so depending on the construction of the door there's a good chance it'll give way under that much pressure - which is about equal to the weight of two cars.
I could have done all that in metric, but I didn't feel like doing things the easy way - can somebody check my math?
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