The pressure under the puck forces air to flow out through this curtain area. The air flow has to come out from under the puck through this curtain area. Curtain area is perimeter (diam x pi) times the height. Knowing the flow through the holes under the puck, you might then compare that to the 'curtain area' around the puck. Next, you have to determine the amount of puck lift. Note the flow will be different than the others since the pressure under the puck will be greater. Given this pressure under the puck, calculate the air flow through the holes that the puck is covering. You've calculated what pressure is needed to lift the puck. That blower pressure will be on the underside of the holes in your air hocky table.
That's your total flow which corresponds to a pressure the blower can create. There will be a pressure the fan can create which depends on flow. Blowers have a curve, such as shown on the web page about farming fans you noted about half way down the page. That's the basic idea, though it gets a bit complicated.
#AIR HOCKEY PUCKS HOW TO#
When you say "The calculation of how to determine the size of the holes has to do with the flow curve of your blower."ĭo you mean I should get a blower first, and figure out how many cfm it can put out at my desired psi, then go ahead and calculate what size holes are needed to match the air output of the blower? Using my total flow figure, and the pressure needed in the air chamber under the table, I could figure out if my fan motor is big enough. It gives flows vs pressure for various sized motors. It is about using a fan for crop drying, but Table 7 looks interesting.
I guess my question is if those formulas are valid for a large surface with multiple orifices. Then multiplied by 3196 (total number of holes on a 3' x 8' table with 1" hole spacing) to get the total flow. I used those equations for figure out how much air would flow thru 1 hole. I found this site about determining air flow through an orifice. however, you are correct about the skirt around the puck, and I agree with the rest of your analysis. When I first did the pressure calculations I used the area of the puck, like you did, and also arrived at the 0.00824 psi figure.īut then I started thinking too much, and thought it would depend on the area of the holes. The calculation of how to determine the size of the holes has to do with the flow curve of your blower. The only thing the holes are doing is allowing air to come in. But as long as there is little pressure drop from where the air comes through the holes to the place it exits at the skirt, the pressure will be uniform beneath the puck. If there's sufficient air, the flow out from under the puck along this perimeter skirt will equal the flow in.
If there is no additional air coming it, the air pressure will quickly equalize and the puck will drop back to the table. So the force upwards is equal to the pressure times the area. The pressure is uniform as long as the skirt is in contact with the table and can't get out. The pressure increases, creating a uniform pressure under the puck and inside the perimeter of this skirt. Pucks have a very small skirt around the perimeter just like a hovercraft. The puck is sitting on the table with no pressure under it. The pressure under the puck can be assumed to be uniform. 040466 lb puck ( oz) that is 2.5" in diameter, the pressure needed to lift the puck is: Pressure needed to lift the puck is independant of the size of the holes in the table.
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