1

Georgios H. Vatistas August 2006

Mass-Spring-Damper Mechanical system and its Electrical Analogue.

There is large number of mechanical devices where MSD (Mass-Spring-Damper) system constitutes the fundamental building block for their mathematical modeling. The basic mechanical components of this system are the spring and the damper. There are complete courses in the engineering curriculum that deal systematically with this topic. Some of these courses are: Modeling Simulation and Analysis of Physical Systems (MECH 370), Fundamentals of Control Systems (MECH 373), Mechanical Vibrations (MECH 373), Fluid Power Control (MECH 448), Basic Circuit Analysis ELEC 273), Principles of Electrical Engineering (ELEC 275), and several others. Here we will only give a brief introduction to the subject, always in the context of the first engineering math course (ENGR 213). The Spring For a linear spring:

F s = k y The spring constant k is the slope of the Fs - y curve. It has units:

k ! F sy

= Nm

=

kg m

s 2

m =

kg

s 2

Square bracket [] means units of the parameter(s) inside the bracket.

If the spring is stiff then the value of k is relatively large, if it soft k is small.

0

y

F s

unstrechedposition

2

Fs (

N )

y ( m )

k is the slope

The Damper

A damper is a dissipative device that converts mechanical energy into heat. This device is mathematically modeled by:

F d = c V = c dy

dt, F d always opposes the motion

The units of c: c !

F dV

= Nm / s

= N sm

= kg m

s 2 s 1

m =

kgs

0

y

c

3

A relatively large value of c indicates hefty opposition to motion (and high

dissipation), while a small value indicates the opposite. The Spring-Mass System

0

W = m g

yo

m

no load (unstreched possition)

loaded(static equilibrium)

y

k

Fs (

N )

y ( m )

k is the slope

Fs o

y o 0

4

When the solid body with mass m is attached the spring will stretch until Fs = W. At this point the system is in static equilibrium, or,

forces acting on the body ! = 0 or W - F s o = W - k y o = 0 ! W = k y o

Starting from this static position let us begin to pull the mass down with a varying velocity (an acceleration dV dt).

0

W = m g

yo

m

loaded(static equilibrium)

y

k

Fso

Fso

Fs

Fs

no loadpossition

Newtons 3rd law

0

y

positionwithout

the mass

Equilibrium possition(with the

mass attached

________________________________________________________________________

When a solid material is cut for the sake of analysis then the internal forces that bind the solid together must be included in the free-body diagram, see for example the solid rod before and after the cut.

before after

F i F i

+ ve x - direction Note: that when we put the two parts (on the right) together to obtain the uncut rod, the total force is from Newtons 2nd law Fi - Fi = 0. ________________________________________________________________________

From Newtons 2nd law: ddt

m V = forces acting external to the body !

5

d

dt m V = m d V

dt + V d m

dt = W - F s o -F s = W - k y o - k y

Since we are dealing with a solid body, dm/dt = 0. Also, from above W = k yo. Then, Newtons second law applied to the mass (free-body-diagram)gives,

m d Vdt

= - k y or m d 2 y

dt 2 = - k y remember that V =

d y

dt

Rearranging:

d 2 y

dt 2 + k

m y = 0 with characteristic equation r 2 + k

m = 0

Alike to all equations, the differential equation must possess dimensional homogeneity i.e. all terms must have exactly the same units:

d 2 y

dt 2 !

y

t 2 = m

s 2 i.e. units of acceleration

d indicates an action; take the infinitesimal difference and as such it has no units. Similarly d 2 has no units.

km

y !

kg

s 2

kg = m

s 2 i.e. again units of acceleration

Hence, the equation it does possess dimensional homogeneity.

The above equation is a second order ordinary linear equation with constant coefficients, its general solution is:

y t = A cos km

t + B sin km

t

The argument of transcendental functions such as for example the trigonometric, exponential, logarithmic etc must be dimensionless (clear number),

km

t !

kg

s 2

kg s = 1 which is a clear number

Let us now consider the case where the mass is pulled down until y = yin, it is stopped for a while, and then at t = 0 is released. The initial conditions are then:

6

i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 Application of the second initial condition yields,

dy

dt

t = 0 = - A k

msin k

m 0 + B k

m cos k

m 0 = 0 or B = 0

The first initial condition gives that A = yin. = 0.2 m. Therefore, the position of the top of the mass is then given by:

y t = 0.2 cos ! t where ! (circular frequency) = k/m in rads / s

Since the cosine function appears in the solution, the mass will oscillate. Let us now plot amplitude y as a function of time for different values of k/m. The results are given in the following figure.

-0.4

-0.2

0

0.2

0.4

0 2 4 6 8 10 12

1 4 9

y (

t )

in

m

t in s

k / m = 1 4 9

The period of oscillations T is,

T =

2 !"

= 2 ! m/k in s while the frequency is given by:

f = 1

T = 1

2 ! k/m in cycles / s or hz

Frequency f is known as the natural frequency of the system and will designated henceforth by fn.

7

It is clear that this system will undulate indefinitely. The reason behind this

unrealistic behavior is that in the mathematical modeling of this system the energy dissipation due to friction inside the spring and air have been neglected. The last paradox (contrary to what is expected) will be rectified in a subsequent section. Is also apparent that as the k/m value goes up, the number of oscillations within the same time interval (or fn) of the system increases. On one hand, if m is kept constant, increasing the stiffness of the spring (k) will result into a higher frequency of oscillations. On the other hand, increasing the mass m while keeping k constant will result into a lower frequency of the system.

A practical way to find fn of the system, by only one simple measurement, is as follows.

The static deflection of the spring-mass system was given previously by:

W = k y o ! m g = k y o ! g

y o = k

m Then

f n = 1

2 ! k/m = 1

2 ! g/y o in hz

Therefore, the natural frequency of the system cam be obtained by loading the

mass to the sprig and measure the static deflection yo. The Mass-Spring-Damper System

As in the previous section, starting from the static equilibrium position let us now begin to pull the mass down with a varying velocity (an acceleration dV/ dt). Remember that the damper opposes the motion. Application of Newtons second law yields,

m d 2 y

dt 2 = - F s - Fd = - k y - c

d y

dt ! m

d 2 y

dt 2 + c

d y

dt + k y = 0

or

d 2 y

dt 2 + c

m d y

dt + k

m y = 0

8

position

without

the mass0

W = m g

m

y

k

Fs

Fs

c

Fd

Fd

0

y

Equilibrium

position

(with the mass

attached

position

without the

mass

The above equation is a second order, ordinary, homogeneous, linear equation with constant coefficients. Its characteristic equation is:

r 2 + c

m r +

k

m = 0 ! r1, 2 =

- c

m c

m

2 - 4 k

m

2 Depending on the values of k, m, and c the roots of the characteristic equation could be (A) two real and distinct, (B) two real double, and (C) two complex conjugates. The three cases will be examined next. CASE A

Let k = 1 N/m, m = 1 kg, and c = 3 Ns / m. Then

r1, 2 = - 3

2 5

2

The general solution is therefore

y t = A exp - 3

2 + 5

2 t + B exp - 3

2 - 5

2 t

At first the arguments of the exponential function appears to have the units of s ([t]

s). This off course it is not true. In order to simplify the algebra we have not included the units of r 1,2 which are:

r1, 2 ! -

c

m c

m

2 - 4 k

m

2 !

- c

m c

m

2 - 4 k

m

2

9

= - 3

2

kg

s kg 3

2

kg

s kg

2 - 4

kg

s 2 kg = - 3

2 1s

32

1s

2 - 4 1

s 2 1s

= - 3

2 5

2 1

s Therefore, the arguments of the exponentials functions are dimensionless,

exp - 3

2 5

2 1s

t

Like before, let the initial conditions be:

i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 From the first and second initial conditions yield respectively

1

5 = A + B and A -

3

2 +

5

2 + B -

3

2 -

5

2 = 0

Solution of the previous system of equations gives

A = 3 + 5

10 5

and B = - 3 + 5

10 5

Then

y t = 1

10 5 3 + 5 exp - 3

2 + 5

2 t + - 3 + 5 exp - 3

2 - 5

2 t

Because the velocity V (t) is dy/dt, then

V t = - 1

5 5 exp - 3

2 + 5

2 t - exp - 3

2 - 5

2 t

CASE B

Let us now decrease the value of c to 2 Ns / m while keeping the rest of the parameters the same. In this case the roots of the characteristic equation are r1 = r2 = -1/2. The general solution is thus

y t = A t + B exp - t

2

10

Imposition of the initial conditions (i) and (ii) yields

A = t

10 and B =

t

5

Therefore the position y ( t ) and the velocity V ( t ) are given by

y t = 1

5 1 + t

2 exp - t

2 and V t = - t

20 exp - t

2

CASE C

Decrease now c further, say to 1 Ns / m. while keeping the rest of the parameters constant. In this case the roots of the characteristic equation are complex conjugate:

r1, 2 = - 1

2 i 3

2

Then

y t = exp - 1

2 t A cos 3

2 t + B sin 3

2 t

The initial conditions yield

A = t

5 and B =

t

5 3

Therefore the position and velocity are given respectively by

y t = 1

5 exp - t

2 cos 3

2 t + 1

3

sin 3

2 t

and

V t = 2

5 3 exp - t

2 sin 3

2 t

The discriminant of the characteristic equation is,

! = cm

2 - 4 k

m

11

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 2 4 6 8 10

A y

B y

C y

y (

t )

in

m

t in s

A. OVER DAMPED

B. CRITICALLY DAMPED

C. UNDER DAMPED

-0.15

-0.1

-0.05

0

0.05

0 2 4 6 8 10

A V

B V

C V

V (

t )

in

m /

s

t in s

C. UNDER DAMPED

B. CRITICALLY DAMPED

A. OVER DAMPED

For CASE A. greater than zero (c is relatively large) the system is OVER DAMPED

12

CASE B. equal to zero the system is CTITICALLY DAMPED

CASE C. less than zero (c is relatively small) the system is UNDER DAMPED

The position and velocity for the three above case are plotted in the figures given

above.

It is evident from thee graphs that as t increases ( t ) the system tends to static equilibrium state i.e. y ( t ) 0 and V ( t ) 0. The reason behind this behavior is that in all the cases considered here we have accounted in the mathematical model for dissipation of mechanical energy (damper). The system will only oscillate (with a diminishing amplitude) when it is under damped. The analysis considered here constitutes the basic methodology in the design of car, truck, locomotive, and aircraft suspension systems. In this case however, the spring instead of being stretched it is compressed. Nevertheless, the analysis is the same.

k c

m

0

y

positionwithout

the mass

Equilibrium possition(with the

mass attached

In the case of an aircraft, in addition to the damping of any bumps on the runway

(providing thus comfort to passengers) it must also absorb the touchdown impact during landing.

When the car mechanic wishes to find out if the car suspension needs repairs (in addition to other diagnostic techniques) it pushes the vehicle down and then lets it go. If the car oscillates excessively, the mechanic then knows that the damping provided is not enough (under damped system), the hydraulic damper has lost liquid (due to leaks) and it must be replaced. Resonance of Spring-Mass-Damper System

Let us now return to the last problem and add an externally applied force F ( t ). The equation in this case is,

13

m d 2 y

dt 2 + c

d y

dt + k y = F t

or

d 2 y

dt 2 + c

m d y

dt + k

m y = 1

m F t = f t

position

without

the mass0

W = m g

m

y

k

Fs

Fs

c

Fd

Fd

0

y

Equilibrium

position

(with the mass

attached

position

without the

mass

Suppose ` f t = A sin ! f t Then

d 2 y

dt 2 + c

m d y

dt + k

m y = A sin ! f t

First simplify the system by removing the damper

d 2 y

dt 2 + k

m y = A sin ! f t

The homogeneous solution is

y h t = B cos km

t + D sin km

t Assume a particular solution of the form:

y p t = Q cos ! f t + M sin ! f t

14

Then

dy p

dt = - Q ! f sin ! f t + M ! f cos ! f t

and

dy p

2

dt 2 = - Q !f

2 cos ! f t - M !f 2 sin ! f t

Inserting yp and yp into the differential equation we have:

Q km

- !f 2 cos ! f t + M

km

- !f 2 sin ! f t = 0 cos ! f t + A sin ! f t

From the above equation

Q km

- !f 2 = 0 and M k

m - !f

2 = A " Q = 0 and M = A

km

- !f 2

The total solution is then

y t = B cos k

m t + D sin k

m t + A

km

- !f 2

sin ! f t

Application of the following initial conditions: i. t = 0, y = 0 ii. t = 0, Vin = dy/dt = 0 gives:

y t = A

1 - !f 2

sin ! f t - ! f sin t

Remember k = 1 N/m and m = 1 kg. Suppose that the amplitude of the externally applied force F ( t ) divided by the mass i.e. f(t) is equal to 0.1 (m/s2), then

y t = 1 10 1 - !f

2 sin ! f t - ! f sin t

The natural circular frequency n is

! n = k

m = 1 rads

s

15

It is not difficult to see that as f n = 1, y ( t ) .

y t lim! f " ! n = 1

= sin ! n t - ! n sin t 1

10 1 - !f 2

lim! f " ! n = 1

= #

This condition is known as resonance. In reality however the amplitude of y ( t ) will increase, as f n, but because of friction (damper) it will never (even theoretically) go to infinity. The presence of singularities (like in the present situation) is an indication that an important physical feature (friction in this case) has not been incorporated into the mathematical representation of reality.

-1.5

-1

-0.5

0

0.5

1

1.5

0 10 20 30 40 50 60

0.9

0.6

0.3

y (

t )

in

m

t in s

! f

We are now ready to include the damper. The equation is then,

d 2 y

dt 2 + c

m d y

dt + k

m y = A sin ! f t

Let, like before, k = 1 N/m, m = 1 kg, and c = 1 Ns / m, or

d 2 y

dt 2 +

d y

dt + y = A sin ! f t

The homogeneous solution is

y h t = exp - 1

2 t B cos 3

2 t + D sin 3

2 t

16

Assume a particular solution of the form:

y p t = Q cos ! f t + M sin ! f t Then

dy p

dt = - Q ! f sin ! f t + M ! f cos ! f t

and

dy p

2

dt 2 = - Q !f

2 cos ! f t - M !f 2 sin ! f t

Inserting yp , yp, and yp into the differential equation we have: - Q !f

2 cos ! f t - M !f 2 sin ! f t - Q ! f sin ! f t + M ! f cos ! f t

Q cos ! f t + M sin ! f t = 0 cos ! f t + A sin ! f t From the above equation

M = A 1 - !f

2

1 - !f 2 2 + ! f

2 and Q = -

A ! f

1 - !f 2 2 + ! f

2

The particular solution is then

y p t =

A

1 - !f 2 2 + ! f

2 1 - !f

2 sin ! f t - ! f cos ! f t

The total solution is then

y t = exp - 1

2 t B cos 3

2 t + D sin 3

2 t

+ A

1 - !f 2 2 + ! f

2 1 - !f

2 sin ! f t - ! f cos ! f t

Application of the initial conditions: i. t = 0, y = 0 ii. t = 0, Vin = dy/dt = 0 yields y t = y h t + y p t

17

where

y h t = A = 0.1 ! f exp -

1

2 t

1 - !f 2 2 + ! f

2 cos 3

2 t -

1 - 2 !f 2

3 sin 3

2 t

and

y p t =

A = 0.1

1 - !f 2 2 + ! f

2 1 - !f

2 sin ! f t - ! f cos ! f t

Making use of the identity

C 1 cos x + C 2 sin x = C 1 2

+ C 2 2

sin x + arctan C 1

C 2

Since in our case

C 1 = - ! f and C 2 = 1 - !f 2

then the particular solution transforms into:

y p t = ! p sin " f t - arctan " f

1 - "f 2

Similarly

y h t = ! h sin 3

2 t - arctan 3

1 - 2 "f 2

where

! p = A = 0.1

1 - "f 2 2 + " f 2

and ! h = A = 0.1 " f

1

3 1 - 2 "f

2 2+ 1 exp - 1

2 t

1 - "f 2 2 + " f

2

In the following figure the homogeneous and particular solutions are plotted separately.

18

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 5 10 15 20 25 30

B

B

yh (

t )

, y

p (

t )

in (

m )

t in s

yh ( t )

yp ( t )

From the above graphs we conclude the following: (a) Due to the presence of exp( - 0.5 t ) term, the homogeneous solution will rapidly

diminish with time to zero*. In this problem, after 20 s the percentage of the homogeneous solution amplitude of y is less than 0.004 % of the total ( h / x100, = h + p ). Therefore, yh is only important at the start up phase.

(b) For relatively long time levels, the solution tends to the particular solution.

Therefore, when the transient effects have died down the system vibrates harmonically following closely the particular solution. At this level, the system oscillates with a frequency equal to the frequency of the forcing function (f).

We have seen previously that when the damping was neglected as f n the

amplitude of oscillations of the system was tending to infinity. This however is not true in real situations (when friction is present). The maximum amplitude at steady state (when the transient effects are over) for different values of c are given in the following figure. NOTE: Only the cases c = 0 Ns / m and c = 1 Ns / m (k = 1 N/m, m = 1 kg) were treated thoroughly here. One however can produce the rest of the curves following exactly the same procedure as before but with different c values.

The maximum amplitude of the particular case is obtained by:

dd ! f

A = 0.1

1 - !f 2 2 + ! f 2

= ! f 2 ! f

2 - 1

1 - !f 2 2 + ! f 2

3/2

= 0

or

* Theoretically when t .In reality however, infinity may be closer than we think. If a property cannot be sensed (detected, measured) its existence is uncertain. In engineering, t representing infinity may be taken as value of time, at which the magnitude of y reaches the limit of certainty of the best available measuring instrument.

19

! f 2 ! f 2 - 1 = 0

The root f = 1/2 gives the frequency where the amplitude is maximum (at 0.115 m).

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.5 1 1.5 2 2.5 3

!f ( in rads / s )

" (

in m

)

c = 1

0.3

0.5

0.7

00

" = A = 0.1

1 - !f 2 2 + c 2 ! f

2

Electrical Analog the RLC (resistance-inductance-capacitance) System (Electrical filter)

The study of physical phenomena by analogy is a common and useful method of scientific inquiry and technical application that has been used in many physical areas. Two systems, that may not necessarily resemble physically each other, are considered to be analogous, if and only if, both are described by the same set of equations (including the boundary and initial conditions). Such systems can be found in dynamics of bodies, electricity, magnetism, thermodynamics, hydraulics, etc. Recently advances in optics were triggered by accumulated knowledge in fluid mechanics. Therefore, this powerful technique can be used to cross-fertilize different classifications of scientific and technical knowledge.

In order to demonstrate the method we consider here the electrical circuit, shown bellow, which consists from an electrical source Vs, and inductor L, a capacitor C, and a resistor R.

20

L

C

R V RV S i or Vout

Application of Kirchhofs second law: the sum of voltage drop in each of the

components is equal to the electrical voltage source, we have: Voltage drop across Governing Equation components ____________________

L: L d i t

dt

R: i t R L

d i t

dt + i t R + 1

C q t = V s t

C: i t R ____________________ Where: C the capacitance (in F (s A / V , in fundamental units: s4 A2 m-2 Kg -1) i is the current (in A) L the inductance (in H = V s A-1 , in fundamental units: m2 kg s-2 A-2) q the electrical charge in the capacitor (Q = A s) R the resistance (in = V / , in fundamental units: m2 kg s3 A2) t the time (in s) VS the voltage source (V, in fundamental units: m2 kg s3 A1) VR The voltage drop across the resistor (V, in fundamental units: m2 kg s3 A1 )

The current i in the circuit and voltage Vout are given by

i t = d q t

dt and V out = i t R = R

d q t

dt

respectively. Due to the first of the above two identities, the governing equation transforms into:

L d 2 q t

dt 2 + R

dq t

dt + 1

C q t = V s t in V

Dimensional homogeneity of the equation:

21

Dividing through the equation by L yields

d 2 q t

dt 2 + R

L

dq t

dt + 1

C L q t = 1

L V

s t

Assuming a harmonic variation for the source we have:

d 2 q t

dt 2 + R

L dq t

dt + 1

C L q t = Aampl sin ! f t in

Vs

" Aampl = Vs

Comparing the mechanical and electrical systems Mechanical system

d 2 y

dt 2 + c

m d y

dt + k

m y = A sin ! f t

initial conditions:

i. t = 0, y = 0 ii. t = 0, Vin = dy/dt = 0 Electrical system

d 2 q t

dt 2 + R

L dq t

dt + 1

C L q t = A ampl sin ! f t

initial conditions:

i. t = 0, q = 0 ii. t = 0, iin = dq/dt = 0 We see that the two differ by scaling factors (constants). If now we let:

y ! q

c

m ! R

L

km

! 1 L C

(R/L = 1, 1/LC = 1, and Aampl = 0.1 as in the last example), then the expression for q(t) is:

q t = ! h sin 3

2 t - arctan 3

1 - 2 "f 2

+ ! p sin " f t - arctan " f

1 - "f 2

22

where

! p = Aampl = 0.1

1 - "f 2 2 + " f 2

and ! h = Aampl = 0.1 " f

1

3 1 - 2 "f

2 2+ 1 exp - 1

2 t

1 - "f 2 2 + " f

2

Because i(t) = dq(t)/dt, the current through the circuit can be obtained by

differentiating the above equation with respect to time. i t =

d q t

dt = ! h

3

2 cos 3

2 t - arctan 3

1 - 2"f 2

+! p " f cos " f t - arctan " f

1 - "f 2

The voltage across the resistor (VR or Vout), can then be calculated multiplying the resulting equation by R. Vout t = R i t = ! h

3

2 R cos 3

2 t - arctan 3

1 - 2"f 2

+! p " f R cos " f t - arctan " f

1 - "f 2

The capacitors charge and current through the circuit versus time are plotted in the subsequent figure. Note that this graph represents also the velocity of the mechanical system. It is evident that the magnitude of the current through the circuit depends on the frequency of the source. As the circular frequency approaches the value of 1/2, the current amplitude increases. For high frequencies the amplitude is relatively insignificantly small. Therefore this circuit is used to filter out frequencies other than the frequency that will resonate the system. This is the reason that the RLC circuit is also known as a filter. Filters can be found in many electrical and electronic systems.

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 5 10 15 20 25 30 35 40

q i

t , s

q ( t ) i ( t )

q (

t )

, i (

t )

Q

, A

Because of the similarity between the two systems, any of the developments given before for the mechanical system is also applicable to the electrical analog.

23

For example consider the LC circuit shown schematically below.

L

C

V s

A

B

The governing equation is:

d 2 q t

dt 2 + 1

C L q t = 0

Initially, switch A is closed and B is open. When the capacitor is charged with 0.2 Q, switch A is opened and then B is closed. Initial conditions i. t = 0, y = qin = 0.2 Q ii. t = 0, iin = dq/dt = 0 A Using the similarity transformation relationships

y ! q

c

m ! R

L

km

! 1 L C

The solution is: q t = 0.2 cos ! t In the equivalent mechanical system is given by:

! (circular frequency) = k/m In the electrical analogue (k/m 1/(LC)

24

! (circular frequency) = 1/LC in rads / s The natural frequency for the mechanical system is:

f n = 1

2 ! k/m

The natural frequency for the electrical LC circuit analogue is,

f n = 1

2 ! 1

L C 1

L C = 1

V s A

s AV

= 1s

Other equivalents systems can be constructed. If the solution is known in one then the solution in the analogous system can be obtained through a simple variable transformation.

25

The Power of Analogies (optional)

Two systems, that may not necessarily resemble physically each other, are analogous if both are described by similar set of equations, boundary, and initial conditions. A well-known paradigm of such systems is the shallow water hydraulics and the 2-dimensional compressible gas flows. This renowned likeness between the two problems suggests that the free surface profile in hydraulics corresponds to the density variation in gas dynamics. The refraction angle depends on the free surface rise and the gas density variation in liquid and compressible vortices respectively. Since the radial profiles of liquid elevation and density are similar, when illuminated, they must project analogous refracted patterns on the image plane.

Hydraulic Vortices

Kiehn, R. M. THE FALACO SOLITON Cosmic Strings in a Swimming Pool , Physics Department, University of Houston

Berry & Hajanal 1983,Optica Acta, 30 Sterling et al. Phys. Fluids, 30 (11), November 1987

26

Gas Dynamics Vortices

Theoretical

Vatistas, 2006, Transactions of CSME, 30 (1)

Experimental helicopter blade tip vortices

Bagai & Leishman 1993, Exp. Fluids, Vol. 15,, pp. 431 - 442.

Mathematical modeling of these two problems shows that the differential equations and boundary conditions describing the two phenomena are similar. Therefore, the physical problems must be analogous. The argument is confirmed by the experimental results shown above.