- #1
MrCrabs
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I have this crazy idea to make an air hockey table out of some junk laying around the basement.
Before I waste my time messing around with an old table and vacuum I figured I would brush up on my basic physics to figure some stuff out.
Lets say we want to lift up a lightweight hockey puck. How much are pressure do we need?
The surface has 1/16" holes spaced every 1 (in both x and y)
The puck has a 2.5" diameter.
The puck has a mass of 18 grams, 0.018 kg
The force on the puck due to gravity is 0.1764 N => 0.040466 lbf
The pressure needed to lift the puck is equal to the force of gravity in lbs / the area of air that hits the puck.
The puck will cover 4 holes at once.
Each 1/16" hole has an area of 0.003068 in^2.
So the puck covers 0.0123 in^2.
PSI = 0.040466 / 0.0123 = 3.29 PSI
Does that logic make sense?
Before I waste my time messing around with an old table and vacuum I figured I would brush up on my basic physics to figure some stuff out.
Lets say we want to lift up a lightweight hockey puck. How much are pressure do we need?
The surface has 1/16" holes spaced every 1 (in both x and y)
The puck has a 2.5" diameter.
The puck has a mass of 18 grams, 0.018 kg
The force on the puck due to gravity is 0.1764 N => 0.040466 lbf
The pressure needed to lift the puck is equal to the force of gravity in lbs / the area of air that hits the puck.
The puck will cover 4 holes at once.
Each 1/16" hole has an area of 0.003068 in^2.
So the puck covers 0.0123 in^2.
PSI = 0.040466 / 0.0123 = 3.29 PSI
Does that logic make sense?