Floating a ping pong ball in the air just using a pen!

Caution: Apparently this problem is harder than it seems!

There is a well known phenomena, which I first learnt about when I was a 10 years old kid. You can levitate a ping pong(or whatever not-so-heavy) ball in the air by blowing throw an empty pen body or using a hair dryer. An interesting demonstration can be seen here. While this demonstration is usually misleadingly explained using Bernoulli's principle (for instance look at this or other citation links from 62 to 66 here), it seems that it is explained using the Coandă effect. The Coandă effect is the tendency of a fluid jet to be attracted to a nearby surface. I am looking for a simple(and humbly realistic) model which is capable of answering the following questions quantitatively: Assume the ball's mass is $m$ and its radius is $r$, the diameter of our pen is $d$, the volumetric flow rate of air through the pen body is $Q$ and the gravitational constant is $g$

1. What is the minimum of $Q$ which makes this demonstration possible?
2. Perhaps using this minimum rate $Q$ and reasonable numbers, how can I estimate the longest time that a normal person can levitate a ping pong ball in the air using his breath? I have noticed that the smaller $d$ is the longer I can levitate the ball. For a normal pen body, my personal experience is about 3 seconds.
3. An estimation of the maximum angle that the pen can make with vertical direction?
4. If we deflect the ball by a tiny amount, what will be the frequencies of motion along the vertical and horizontal direction?
• Does anyone have a pen and a ping-pong ball around? – Ali Jul 4 '13 at 20:08
• 4 may be answerable if the pressure gradient at the surface of the ball can be taken to be $\rho v^2 /R$ for radius $R$ of curvature of the streamlines near the surface. The force could be calculated from integrating that over a volume that goes linearly with the displacement of the ball, but I'm not sure which volume is appropriate. I will try to calculate this explicitly tomorrow. – Chay Paterson Jul 7 '13 at 21:05
• @ChayPaterson I personally think, that is how 4 should be answered. For others reading your answer, please try to justify it. I am looking forward to your answer. Good Luck – Ali Jul 8 '13 at 1:49
• Event the slightest flow is going to levitate the ping pong ball. The question is really about the feedback mechanism the keeps the ball near the fluid stream. – ja72 Jul 8 '13 at 15:01
• @ja72 Not really. See BebopButUnsteady's answer. – Ali Jul 8 '13 at 15:50

Here's my quantitative attempt at $4.$ and $1.$:

The Coandă effect here is the tendency of the airflow to adhere to the surface of the ball. This means that near the surface of the ball, the streamlines are curved with a radius of curvature approximately equal to the radius of the ball $R$; this curvature results in a pressure gradient just as it does in the theory of lift which I assume to be constant over the surface:

$\vec{\nabla} P = - \frac{\rho v^2}{R} \vec{e}_r$

(Anecdotally, the "circulation" aerodynamicists use to calculate the lift per unit width is related to a certain integral of this quantity over an appropriate cross sectional area.)

This points radially inwards to some equilibrium point, at which the net forces acting on the ball cancel. In practice, the ball will hang slightly below this point due to gravity, Edit: see below.

When the ball is displaced from this position normally to the flow by some distance $z$, we can calculate the total restoring force acting on the ball by integrating $\vec{\nabla} P$ over the volume of the ball under two assumptions:

$1.$ the curvature and equilibrium position of the airflow is negligibly disturbed by moving the ball like this, and so independent of $z$

$2.$ the net volume of the displaced ball that's relevant to the integration (i.e. that contributes to an *un*balanced force) is approximately equal to $\pi R^2 z$, the cross sectional area of the ball times the displacement $z$

with these assumptions, the restoring force is

$\vec{F} = - \frac{\rho v^2}{R} \pi R^2 \vec{z} = -\pi \rho v^2 R \vec{z}$

which we can immediately recognise as a harmonic oscillator

$\vec{F} = -k \vec{z} = -m \omega^2 \vec{z}$

where $m$ is the ball's mass, and get for natural frequency

$\omega = v \sqrt{\frac{\pi\rho R}{m}}$.

or for an easier-to-measure frequency of oscillation $\nu$ in terms of $Q$, estimating $v = Q/A$,

$\nu = \frac{Q}{A} \sqrt{\frac{\rho R}{4 \pi m}}$

using the cross sectional area of the ball for $A = \pi R^2$, because $v$ is the velocity at which air travels over the ball's surface, and not the velocity at which it leaves the pen.

This assumption is a bit questionable as none of the air actually passes through this area, but this is the only finite area relevant to the problem. At worst, the true area that this depends on, probably the "average" area that the flow passes through in the plane that passes through the ball's equator, will be $A$ times some numerical constant.

It's important to note that this scales linearly with $Q$, so blowing twice as hard will result in oscillations that are twice as fast.

The international standards are $R$ = 20 mm and $m$ = 2.7 g, so I would guess the following formula:

$\nu = 0.767$ m$^{-1} \frac{Q}{A}$

Using BebopButUnsteady's reasonable $Q$ of $6$ L/s, and the above $R$,

$\nu = 3.66$ Hz

Which sounds fairly reasonable to me. But again, the linear dependence on $Q$ is important, and for weaker flows it will be proportionately slower.

And as a final note, my earliest assumption that the pressure gradient was constant over the surface means this formula doesn't distinguish between oscillations parallel the stream and oscillations normal to it.

In practice, it's known that the inward pressure gradient will be lower at the points of the ball that are "in line with" the stream than those that lie further outside since the streamlines curve in the opposite direction, so oscillations parallel to the stream will have a lower frequency than the normal/"lateral" ones in general.

PS: The final formulas here will probably only hold up to some experimental "fudge factor", which may be interpreted as the numerical constant that appears in that approximation of $v \approx Q/A$: if $v = f Q/A$, then $f$ is the fudge factor. But this is fluid mechanics, and fudge factors are inevitable.

Edit: estimate for part $1.$

If we assume that the ball's displacement due to gravity is less than one radius, then we can say that

$$mg < \pi \rho v^2 R^2 = \rho Q^2 / \pi R^2$$

given the above definitions of $R$ and $Q$, or

$Q > R \sqrt{\frac{\pi mg}{\rho}} \approx 5.77$ mL/s

and

$v > \frac{1}{R} \sqrt{\frac{mg}{\pi \rho}} \approx 4.59$ m/s

which is far less than $6$ L/s, but may still be in line with what those toys that sometimes come in Christmas crackers. The velocity looks somewhat more intuitively reasonable. If these are too low to sustain experimentally, the reason for this may be that the flow is too unsteady when maintained by human breath.

In some videos, the amplitude of oscillations when suspended in a hair-drier flow can apparently be appreciably larger than one radius, so the assumption about $z<R$ may be out by some factor of up to 5; however, assuming this would make the final results for theoretical minimum $v$ or $Q$ smaller by a factor of $\approx 0.44$, so it is still an informed guess for an upper bound.

Since the assumptions involved in deriving this force only strictly apply to displacements normal to the stream, this is the minimum flow to suspend the ball horizontally; answering the question about maximum suspension angle (there may not be one) would require a more detailed description of the pressure distribution around the ball, which may well be available in the literature for laminar flow around a sphere.

• Can you add something for the other parts of the question? – Ali Jul 9 '13 at 5:48
• @Ali BebopButUnsteady dealt with 2. already with a very flexible method, so I've added something for 1. and some closing remarks about a strategy for 3. – Chay Paterson Jul 9 '13 at 17:16

Let me give the naivest possible estimate, so that people have something to criticize. Assuming that the most of the jet interacts with the ball and is deflected at a substantial angle then the force on the ball is roughly the momentum flow through the pen. In your units this is $\rho_{air} Q^2/(\pi d^2)$. Saying the force to levitate a ball is $1\times 10^{-2} N$, your pen's cross section is of order 10 $mm^2$ squared, we get $Q$ of order $1 L/s$.

Following some rather convoluted Wikipedia articles, http://en.wikipedia.org/wiki/Spirometry and http://en.wikipedia.org/wiki/Lung_capacity, it seems that 1L is a normal breath, and we can do about 6 L/s at a peak or 1L for 6 seconds if we try. This seems to square with common experience.

• That is the most rough estimation I could make(like a Fermi question), more than that it is unable to provide equilibrium points and frequencies and the angle as well. But $\color{green}{+1}$ for your Fermi type answer. – Ali Jul 2 '13 at 6:56
• Also, my personal record is 3 seconds :) – Ali Jul 3 '13 at 6:44
• "So that people have something to criticize" -- haha, you know the physics stack exchange very well. – Nick Jul 8 '13 at 20:31

I'm a first year physics student, so my answer might not be satisfactory - but I hope it will give some insight to the problem.

1) from what I know we need to consider:

1. Drag - which I will address
2. Turbulence - which I know next to nothing about, and therefore I will ignore with hope someone will be able to expand.

we need the drag force to be equal to or greater than the weight of the ball.

using the formula $F_d= \frac{1}{2} \rho v^2 C_d A$ (can be found in wikipedia)

we get $mg = \frac{1}{2}\rho v^2 C_d \pi r^2$

$\rho$ is given here , and $C_d$ is given here (though it is velocity depended and should be measured).

the velocity is: $v=\frac{Q}{\pi \frac{d^2}{4} }=\frac{4Q}{\pi d^2}$

so we get $Q= \frac{d^2}{r}\sqrt{\frac{mg\pi}{8\rho C_d}}$

because not all air will contribute to lifting the ball, and because the flow of air is not steady, this Q is not the minimum but it should be somewhere near that. I'll try to experiment with this - in a month.

2) That will take some measurments: how much air do you blow in 3 seconds? what it's velocity (can be tested with a balloon, like the way you measure lung capacity)

4) Not sure, but I think that horizontal displacement will result in an overdamped "oscillation" - due to Bernoulli's effect.

• Thank you, I am really interested in fluid dynamics; and your question is really interesting. I'll be sure to experiment with this when I'm finish with my exams - if I'll find a better answer I'll be sure to keep you posted. – Bitx Jul 8 '13 at 10:17

Let me try to answer the first question.

Suppose that in a spacecraft, water flows along a circular channel, the distance from the bottom of the channel to the axis of the circular channel is $$r$$, the width of the channel is $$b$$, the depth of the channel is $$h$$ (water depth is also $$h$$), the density of the water is $$ρ$$, the angular velocity of the water is $$ω$$, no viscosity, no friction, no water tension, and the air pressure of the spacecraft is $$p$$(lower case character). What is the water pressure at the bottom of the channel? How to calculate?

Thanks to Chet Miller for the following answers:

$$P=p-ρrhω^2(1+h/2r)$$

The pressure at the bottom of the channel decreases with the increase of $$ω$$. Lower than atmospheric pressure (Note that the air pressure in the spacecraft is one atmosphere). Assuming that the water flows only through the upper half of the channel, there will be an upward force acting on the cylinder because of the atmospheric pressure below the cylinder.

For simplification, suppose that the table tennis ball is a cylinder A, the radius of A is $$r$$, the weight of A is $$w$$, the cross section of pen body is rectangular, the width of rectangle is $$b$$, and the height is $$h$$. Suppose that the gas flows through the upper half circle of A with rectangular cross section. Then the upward force $$F$$ can be calculated by integral of formula.

When $$F = w$$, we can calculate $$ω$$.

Velocity $$v$$ can be calculated from $$ω$$ and $$r$$

Finally, it can be calculated that:

$$Q=vbh$$

The lifting force of table tennis is not because of Bernoulli's theorem, but because of the centrifugal force of fluid curve movement.

The centrifugal force of the fluid makes the table tennis surface produce low pressure, and the low pressure of the sphere makes the table tennis ball suspended in the air.

• @ChetMiller What do you think of my answer,? – enbin zheng Apr 6 at 11:32
• @Ali What do you think of my answer,? – enbin zheng Apr 6 at 11:35