# The Phyiscs of Air Hockey - Lifting up the puck

• MrCrabs
In summary, the conversation discusses the idea of making an air hockey table out of junk in the basement, and how to calculate the pressure needed to lift a lightweight hockey puck. The surface of the table has 1/16" holes spaced every 1" in both x and y directions, while the puck has a 2.5" diameter and a mass of 18 grams. The pressure needed to lift the puck is equal to the force of gravity divided by the area of air that hits the puck, which is found to be 3.29 PSI. However, the pressure under the puck can be assumed to be uniform due to the small skirt around the perimeter of the puck. The conversation also touches on the calculation of air flow through
MrCrabs
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?

You got it. [didn't check the math, but the logic is sound]

You might get a greater effective area then the area of all the holes since the puck is not flat. That said does an air hockey table even lift the puck anyway? As long as it reduces the weight it will reduce the friction.

Pressure needed to lift the puck is independant of the size of the holes in the table. For a .040466 lb puck ( oz) that is 2.5" in diameter, the pressure needed to lift the puck is:
P = .040466 lb / (pi/4*2.5^2) = .00824 psi
not 3.29 psi.

The pressure under the puck can be assumed to be uniform. It is not just the area of the holes.

Think of it this way. The puck is sitting on the table with no pressure under it. Pucks have a very small skirt around the perimeter just like a hovercraft. The pressure increases, creating a uniform pressure under the puck and inside the perimeter of this skirt. The pressure is uniform as long as the skirt is in contact with the table and can't get out. So the force upwards is equal to the pressure times the area. At some point the puck begins to lift. If there is no additional air coming it, the air pressure will quickly equalize and the puck will drop back to the table. If there's sufficient air, the flow out from under the puck along this perimeter skirt will equal the flow 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. The only thing the holes are doing is allowing air to come in. The calculation of how to determine the size of the holes has to do with the flow curve of your blower.

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.
But then I started thinking too much, and thought it would depend on the area of the holes... however, you are correct about the skirt around the puck, and I agree with the rest of your analysis. (See http://us.st11.yimg.com/us.st.yimg.com/I/jimcds_1986_24883136 for an image of an air hockey puck)

Now comes the flow questions...
I found this site about determining air flow through an orifice.
http://www.jegasho.net/rotron pages/orifice flow calculation.htm

I used those equations for figure out how much air would flow thru 1 hole. Then multiplied by 3196 (total number of holes on a 3' x 8' table with 1" hole spacing) to get the total flow.
I guess my question is if those formulas are valid for a large surface with multiple orifices.

Then I came across this site. http://www.extension.umn.edu/distribution/cropsystems/DC5716.html
It is about using a fan for crop drying, but Table 7 looks interesting. It gives flows vs pressure for various sized motors.

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.

When you say "The calculation of how to determine the size of the holes has to do with the flow curve of your blower."
Do 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?

Last edited by a moderator:
When you say "The calculation of how to determine the size of the holes has to do with the flow curve of your blower."
Do 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?

That's the basic idea, though it gets a bit complicated. Blowers have a curve, such as shown on the web page about farming fans you noted about half way down the page. There will be a pressure the fan can create which depends on flow. Calculate flow through all holes. That's your total flow which corresponds to a pressure the blower can create. That blower pressure will be on the underside of the holes in your air hocky table.

You've calculated what pressure is needed to lift the puck. Given this pressure under the puck, calculate the air flow through the holes that the puck is covering. Note the flow will be different than the others since the pressure under the puck will be greater.

Next, you have to determine the amount of puck lift.

Knowing the flow through the holes under the puck, you might then compare that to the 'curtain area' around the puck. Curtain area is perimeter (diam x pi) times the height. The air flow has to come out from under the puck through this curtain area. The pressure under the puck forces air to flow out through this curtain area. Given the pressure under the puck and total curtain area gives you a flow rate. The amount of flow out this curtain area has to equal the flow through the holes under the puck.

Knowing the flow through the curtain area = flow through the holes under the puck, you can calculate the lift of the puck (curtain area is a function of lift).

How much lift is acceptable? If I had to guess, I'd say about .010".

Hope that helps.

"Knowing the flow through the curtain area = flow through the holes under the puck, you can calculate the lift of the puck (curtain area is a function of lift). "

Q_Goest, do you know of an equation to calculate the flow through the curtain area? I assume it is Q=VA, but I need a formula to get V based off the curtain area shape I think. The problem I am working on involve the air flowing out from underneath a sheet of paper.

whtrojan2005 said:
Q_Goest, do you know of an equation to calculate the flow through the curtain area? I assume it is Q=VA, but I need a formula to get V based off the curtain area shape I think. The problem I am working on involve the air flowing out from underneath a sheet of paper.
Do you know how to calculate velocity from Bernoulli's equation? You have a dP so you should be able to determine velocity. You have a curtain area, so you can find flow rate as you've already shown (Q=VA). The only issue with using Bernoulli's is it won't take into account the typical restrictions due to geometry which are often lumped into one variable we call the discharge coefficient. For an area like this, there is no good resource I'm aware of that will give that to you. From experience, I'd suggest a Cd of between 0.6 to 0.8.

If you're doing this for a sheet of paper, that's different though because you don't have a single restriction (ie: the curtain area doesn't exist). I'd model that as two flat sheets separated by a constant air gap. Here, the velocity changes radically from relatively high velocity where the air is being inserted to a very low velocity where it meets atmosphere. It's more like the flow of air through a pipe, but a pipe that spreads out such that velocity is a function of the distance it's moved away from the hole squared. I don't know how you'd do that one. Maybe someone else here can help with that. I'd suggest you start a new thread.

can anyone help out here?

no, it will not be a rocket - just imagine a regualr air pump - NO hot air at all.
The air used, will be the same temperature as the surounding one.

spark727 said:
no, it will not be a rocket - just imagine a regualr air pump - NO hot air at all.
The air used, will be the same temperature as the surounding one.

Have a look at the site.

I'd start be working out what force you need to achieve the desired acceleration. You can break it down into the force needed to hold it stationary against gravity (W=mg) and the force needed to accelerate in the absence of gravity (F=ma). Then W+F is the total force required.

## 1. How does the air cushion affect the movement of the puck in air hockey?

The air cushion created by the air hockey table reduces the friction between the puck and the surface, allowing the puck to move more easily and quickly. This results in a faster and smoother game.

## 2. What role does air pressure play in lifting up the puck in air hockey?

The air pressure in the table creates a cushion of air that lifts the puck slightly off the surface. This allows the puck to glide smoothly and quickly across the table, making it easier for players to control and manipulate.

## 3. Can the angle of the air jets affect the lift of the puck in air hockey?

Yes, the angle of the air jets can affect the lift of the puck. If the jets are angled too high, the puck may lift off the table completely, making it difficult to control. If the jets are angled too low, there may not be enough lift to keep the puck in motion.

## 4. How does the weight of the puck affect its movement in air hockey?

The weight of the puck can affect its speed and trajectory. A heavier puck may be more difficult to lift and control, but it may also be harder for the opponent to defend. A lighter puck may be easier to lift and maneuver, but it may also be more susceptible to outside forces, such as air currents.

## 5. Is there a limit to how high the puck can be lifted in air hockey?

Yes, there is a limit to how high the puck can be lifted. The angle and force of the air jets, as well as the weight and shape of the puck, all play a role in determining the maximum height the puck can reach. Additionally, the design of the table may also have an impact on the maximum lift of the puck.

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