Basic Math
© by Ed Seykota, 1999
Energy of a
Ball:
A ball of mass m has velocity v with components x, y and z.
The kinetic energy, Ek = mv²/2 where v² = x² + y² + z².
Pressure and Force of a Ball in a Cube:
A ball is in a cube of side s.
The pressure in the cube, P, is Energy/Volume or Ek/s³.
As the ball bounces back and forth between the walls
it changes momentum, p = mv, with a frequency n = x/s.
The force on side x, Fx, is momentum change times frequency.
Fx = 2mx * x/2s = mx²/s.
Some Notes and Derivations
Two Types of Energy
Balls bouncing around, bumping into each other,
with no net velocity in any one direction
have a static energy proportional to temperature.
Balls with a net velocity in one direction
have a dynamic energy, Ek = mv²/2.
The Great Magic Disk invites a study
of the conversions between these two forms of energy. See Units
of Measure.
Volume Flux Through an Aperture:
The following derives, using fluid dynamics, equations for the escape velocity
of the air through the aperture and the height of the gap.
For Reynolds numbers above about 3 (I estimate the Reynolds number for a
levitator much higher), Q, the volume flux through a circular aperture, in terms
of C, the orifice coefficient (about .6), A, the area of the aperture, dP, the
pressure differential across the aperture and p, the density of the fluid, is,
per Vogel, page 187:
Q = C*Pi*A²*sqrt(2*dP/p)
Note that for high Reynolds numbers, the volume flux depends on fluid density,
the square of area and the suare root of pressure drop; at low Reynolds numbers
the volume flux depends on the cube of area, the first power of the pressure
drop and m, the dynamic viscosity:
Q = A³*dP/3m
Assuming that at high, and not too high, Reynolds numbers, the shape of the
orifice is irrelevant, we then have A, the area of the aperature:
A = 2*Pi*Rh*G, and
Q = C*Pi*(2*Pi*Rh*G)²*sqrt(2*dP/p)
Q = k * G²*sqrt(dP)
Also:
Solving for the gap:
Volume Flux = f(Gap, dPressure)
Gap = f(Volume Flux, Aperture)
Second Order Effects:
Velocity(r) = f(Escape Velocity, Pressure(r))
Total Head = f(Pressure(r), Velocity(r))
Virtual Rim = f(Total Head(r), Ambient Pressure)
Pressure Above Disk = f(Hole Pressure, Pressure(r), Virtual Rim)
The Math
The following derives, using standard physics and the 1/r theory, an equation
for the forces on the disk as a function of pressures and radii. Note: no such
derivation exists based on the Bernoulli principle.
Fw = Fu (1  (Ph/Pa)(Rh/Rd)²  2*(Rh/Rd)(1  Rh/Rd))
Terms:
Pa  ambient pressure (pascals)
Ph  pressure in hole (pascals)
Px  pressure just past aperture (pascals)
Rd  radius of disk (meters)
Rh  radius of hole (meters)
Fw  down force on disk from weight of disk (newtons)
Fh  down force on disk under the hole (newtons)
Fd  down force on disk under gap hole (nestons)
Fu  up force on underside of disk from ambient pressure (newtons)
G  height of the gap between the top of the disk and the ceiling
Pi  3.14159
Assumptions:
Air is noncompressible.
Static pressure drops to ambient pressure across the aperture (Pa = Px).
This drop accelerates the molecules to their maximum "escape"
velocity.
The molecules, having momentum, keep moving and fan out radially.
The total head opposes the ambient pressure at the periphery of the gap and
establishes the virtual radius of the disk. 1/r Law: The static pressure at a
point in the gap = Ph*(Rh/r)
Analysis of forces as a function of pressures and radii:
Up force on underside: Fu = Pa*Pi*Rd²
Down force under hole: Fh = Ph*Pi*Rh²
Down force in a thin ring in gap: dFr = Px*(Rh/r) * 2*Pi*r*dr
Rearranging: Fr = 2*Pi*Px*Rh
Down force under entire gap: 2*Pi*Px*Rh(Rd  Rh)
Now from force balance: Fw + Fh + Fd = Fu
So: Fw = Fu  Fh  Fd
Substituting:
Fw = Pa*Pi*Rd#178;  Ph*Pi*Rh²  2*Pi*Px*Rh(Rd  Rh)
Fw = 2*Pi* (Pa*Rd#178;  Ph*Rh²  2*Px*Rh(Rd  Rh))
Fw = Fu (1  (Ph/Pa)(Rh/Rd)²  2*(Px/Pa)(Rh/Rd)(1  Rh/Rd))
Fw = Fu (1  (Ph/Pa)(Rh/Rd)²  2*(Rh/Rd)(1  Rh/Rd))
Reasonablility Tests:
For a levitator with a very, very small hole:
Rh << Rd
Fw = Fu (1  0  0)) = Fu
The levitator lifts a weight equal to the ambient force on the bottom of the
disk.
For a disk with no overhang and a vacuum in the hole:
Rh = Rd and Ph = 0
Fw = Fu (1  0  0)
The levitator lifts a weight equal to the ambient force on the bottom of the
disk.
For a disk with no overhang and a half vacuum in the hole:
Rh = Rd and Ph = Pa/2
Fw = Fu (1  1/2  0)
The levitator lifts a weight equal to half the ambient force on the bottom of
the disk.
For a disk with no overhang:
Rh = Rd
Ph > Pa
Fw = Fu (1  Ph/Pa  0)
Fw = Fu (1  Ph/Pa) < 0
The levitator lifts a negative weight. Thatis, it only balances by pushing down
on a lighterthatair disk.
For a disk with the motor off:
Ph = Pa
Fw = Fu (1  (Rh/Rd)²  2*(Rh/Rd(1  Rh/Rd))
Fw = Fu (1  (Rh/Rd)²  2*(Rh/Rd) + 2*(Rh/Rd))²)
Fw = Fu (1  2*(Rh/Rd) + (Rh/Rd))²)
Fw = Fu (1  (Rh/Rd))²
The levitator lifts a weight somewhere between zero and the ambient force on the
bottom of the disk, depending on the ratio of the radii of the hole and disk. At
this zero air velocity limit, the equilibrium gap is also zero, so the disk acts
like a suction cup.
For a typical disk:
Rh = Rd/4 and Ph = 2*Pa
Fw = Fu (1  (2)(1/4)² 2*(1/4)(1  1/4))
Fw = Fu (1  1/8  3/8) = Fu/2
The levitator lifts a weight equal to half the ambient force on the bottom of
the disk.
If you don't agree with these derivations, and/or have suggestions and comments,
I'd like to hear from you. Good debates make for good science.
Units of Measure
Quantities 
Dimensions 
SI Units 
Length, Distance 
L 
Meter (m) 
Area, Surface 
L² 
Square Meter (m²) 
Volume 
L³ 
Cubic Meter (m³) 
Time 
T 
Second (s) 
Velocity, Speed 
L/T 
Meters per Second (m/s) 
Acceleration 
L/T² 
Meters per Second Squared (m/s²) 
Mass 
M 
Kilogram (kg) 
Force 
ML/T² 
Newton (N) or kgm/s² 
Density 
M/L³ 
Kilogram per Cubic Meter (kg/L³) 
Work 
ML² /T² 
Joule (J) or NewtonMeter (Nm) 
Power 
ML²/T³ 
Watt (W) or Joule per Second (J/s) 
Pressure, Shear Stress 
M/LT² 
Pascal (Pa or Ns/m²) 
Dynamic Viscosity 
M/LT 
Pascal Second (Pas or Ns/m²) 
Kinematic Viscosity 
L²/T 
Squate Meter per Second (m²/s) 
Dear Professor Brad Snyder (UNR) 7846939: For our conversation,
scheduled for Monday May 10 at 11:00 AM, you asked for clear diagrams
defining the problems. I hope the following is clear. If anything is
unclear or ambiguous, please let me know at 8327377. Non disclosure
request: please keep these matters confidential until the end of 1999.
Yours truly, Ed Seykota 
Definitions, Symbols and Units:
States 
Symbol 
Units 
Notes 
Mass 
M 
Kg 
Air: d ~ 1.2 kg/m3 
Momentum 
Mo 
Kgm/s 
Mo = M * U 
Thermal Energy 
Te 
Kgm^{2}/s^{2} 
Te = PV = NRT 
Rates and Flows 



Mass Flux 
Mdot 
Kg/s 
Mdot = dM/dt 
Momentum Flux 
Modot 
Kgm/s^{2} 
Modot = dMo/dt 
Thermal Energy Flux 
Tedot 
Kgm^{2}/s^{3} 
Tedot = dTe/dt 
Problems:
Problem #1: Air escapes through a round orifice in a solid plate. 
Given:
0 < P < 5 atm (gauge); P_{0} = 0 atm.; T = 70 F; T_{0}
= 50 F; D = 1/16".
Find the flows at the exit:
Find Mdot, Modot, Tedot 
Problem #2: A circular plate over the orifice creates a ringshaped
"valve" at the center and changes the flows. 
Given:
0 < P < 5 atm; P_{0} = 0 atm.; T = 70 F; T_{0} =
50 F; D = 1/16", h ~ .001 m.
Find the flows at the exit:
Find Mdot; Modot; and Tedot. 
Problem #3: Air leaving the valve area continues to experience
skinfriction drag. 
Given:
0 < P < 5 atm; P_{0} = 0 atm.; T = 70 F; T_{0} =
50 F; D = 1/16", E_{0} ~ .0000003 m.
Find drag from skin friction:
Derive U (velocity), p (pressure) and d (density) from the satates as
functions of M, Mo, and Te.
Find drag as a function of U, p and d. 
Problem #4: Difference Equations:
To do: Create control volumes from concentric rings of thickness dR,
radiating out from the orifice between the plates, like nested bicycle
tires. (The rings in this panel are shown on their sides.) During the
traverse across dR, within a control volume, Mo and Te change slightly as
a function of drag and pressure gradient across dR. For control volume[r],
write equations for these small changes. 
Continued … 
… continuation.
There is (1) a pressure gradient from the inside to the outside and (2)
some skin friction across the top and the bottom. These act to change the
momentum and thermal energy of the air passing through. 
Very small section of a ring showing direction of mass flux. Width (w)
is actually curved slightly and continues around as the circumference of
the ring. 
Mathematics
Fluid velocity  Fluid
flux  Total energy  Radial
Action Theory  Thermal
Energy and Pressure  Kinetic
Energy and Pressure  Units
of Measure
Fluid velocity
The air at the entrance to the gap sees a clear way ahead and expands
into it such that it loses thermal (pressure) energy and gains radial
(velocity) energy. To simplify the math, I assume that right at the
entrance to the gap: (1) the pressure instantly drops right to ambient
pressure and (2) the air instantly achieves its maximum "escape
velocity." These assumptions are best for Reynolds numbers above 3
and I feel the Levitator qualifies. A more rigorous examination would
show these assumptions are not exactly accurate although the
discrepancies would not seem to effect the overall Theory of Radial
Action.
1/2 mv_{r}^{2} = dP * V_{0}; kinetic
energy gain from thermal energy; dP = P_{0}  P_{A}. v_{r}^{2 }= 2 * dP / D; velocity squared equals
twice pressure drop over density; v_{r}^{ }= sqrt(2 * dP / D); by taking square
root.

For a pressure drop of 25 newtons/cm^{2} and a density of .01
gram/cm^{3}, (seatofpants estimate)
v_{r}^{ }= sqrt(2 * (25 newtons/cm^{2} )
/ (.01 gram/ cm^{3}) = sqrt(5000 m^{2}/sec^{2}); v_{r}^{ }= 70 m/sec = 200 ft/sec = 120 miles per
hour, approx
