# Brachistochrone Under Air Resistance by MikeJenny

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```									Brachistochrone Under Air
Resistance
Christine Lind
2/26/05
SPCVC
Source of Information:
Collaborator:
Brachistochrone Setup:
• Initial Point
P(x0,y0)
• Final Point
Q(x1,y1)
• Resistance Force
Fr
• Slope Angle

Geometric Constraints

• Parametric Approach:
– Start by using arclength
(s) as the parameter

• Parametrized by arclength:

x'(s)  cos( (s))  0
y'(s)  sin( (s))  0
(curves parametrized by arclength have unit speed)


Energy Constraint?
• Normally we use
conservation of energy to
solve for velocity in terms
of the other variables

• We have a Non-
Conservative system, so
what do we do?
Energy Constraint?

• Energy is lost to work
done by the resistance
force:
E  W
T  U  W


Energy Constraint
T(t)  1 mv 2
• Non-conservative system:               2

U(t)  mgy
dT dU dW
                              
t
dt   dt   dt             W (t)            Frvd
0

• Constraint parametrized by time:
                                 d  d 
 gsin  R(v)  
dv
0      v 
dt                       dt  ds 

• Constraint parametrized by arclength:
      vv'gsin  R(v)  0 
Problem Formulation:
• Boundary & Initial Conditions:
x(0)  x 0 ,   y(0)  y 0 ,   v(0)  v 0
x(l)  x1,     y(l)  y1,
• Minimize the time integral:
ds
T  0
l

               v
x'cos  0
• Other constraints:
y'sin   0
How do we incorporate them?
vv'gsin   R(v)  0

Lagrange Multipliers
• Introduce multipliers, vector:
(s), (s), (s) q(s)  (v, x, y,,, ,)
• Create modified functional:

l
F[q,l]     G(s,q,q')ds
                      0

where
1
G(s,q,q')    (x'cos )  (y'sin)  (vv'gsin  R(v))
         v
Euler-Lagrange Equations
 G(s,q,q')ds
l
F[q,l] 
0

• System of E-L equations:

G d G  
   0
q ds q' 

        G             G                   G 
          q0           q1  G  q'  l  0
q' s 0       q' s l             q' s l
7 Euler-Lagrange Equations:
1      dR 
v'  v ' 0
                             Note:
v 2
 dv 
 (s)  C1
 sin   cos  g cos  0
(s)  C2
 ' 0
' 0
x'cos  0                          Note:

y'sin   0                          Constraints Appear
vv'gsin   R(v)  0                 as E-L equations!
Natural Boundary Conditions:
1
 1 cos1  1 sin 1  1 gsin 1  R(v1 )  0
v1
1v1  0

Note:

v1 is not necessarily zero, so:
1  (l)  0
Lagrange Multipliers - Solved!
• Using:  sin   cos  g cos  0
s l

1
 1 cos1  1 sin 1  1gsin 1  R(v1)  0
v1
• Determine the Lagrange Multipliers:
cos1              sin 1
 
 (s)        , (s)             ,   Note:
v1                 v1
(s),(s) constants
sin(  1 )
(s)                            (s)=((s))
gv1 cos
First Integral
• Recall:
1
G(s,q,q')    (x'cos )  (y'sin)  (vv'gsin  R(v))
v
No explicit s-dependence!       (s) 
cos1
, (s) 
sin 1
,
v1                 v1

• First Integral:                               sin(  1 )
G                       (s) 
G(q,q')  q'      const.                     gv1 cos
q'

1
  cos   sin(gsin  R(v))  const.

v

Parametrize by Slope Angle

cos1              sin 1
 (s)       , (s)             ,
v1                 v1
sin(  1 )
(s) 
gv1 cos
Parametrize by Slope Angle
• Define f() to be the inverse function of
(s):
s  f (),      f (0 )  0

• f() continuously differentiable, monotonic
• Now we minimize:

Ý              Ý
ds  f d   
fd

1
T                   Still need constraints...
0    v
Modified Functional
• Transform modified problem in terms of :
x( 0 )  x 0 , y( 0 )  y 0 ,

1
L[p, 0 ,1]             H(,p, p)d
Ý
0                 x(1 )  x1,      y(1 )  y1,
v( 0 )  v 0 ,   f ( 0 )  0
p()  (v,s  f (),x, y,, ,)

Ý
f
Ý Ý            ( Ý Ý              Ý Ý           Ý
H(,p, p)    ( x  f cos )  y  f sin )  (vv  f gsin  f R(v))
Ý
            v
          1                                                   
G(s,q,q')    (x'cos )  (y'sin )  (vv'gsin   R(v))

          v                                                   
7 Euler-Lagrange Equations
ÝdR  d v   0
Ý 
f
 v  f
Ý        
v 2
      dv  d
d 1                                         
   cos   sin   (gsin   R(v)) 0 First Integral!
d v                                         
  0
Ý                              1      dR 
v'  v ' 0
 
Ý
 0                         v 2
 dv 
Ýcos  0                 sin   cos  g cos  0
x f
Ý
 ' 0
Ý Ý
y  f sin   0                          (Old Equations)
' 0
Ý Ý            Ý
vv  f gsin   f R(v)  0 x'cos  0
y'sin   0
vv'gsin   R(v)  0
Same Natural B.C.’s
& Lagrange Multipliers
1
 1 cos1  1 sin 1  1 gsin 1  R(v1 )  0
v1
1v1  0

cos1              sin 1
 ( )        , ( )            ,
                       v1                 v1
sin(  1 )
( ) 
gv1 cos
Solve for v()
• Using Lagrange Multipliers and First
Integral:
d 1                                     
   cos   sin  (gsin   R(v)) 0
d v                                     

• Obtain:
               1 cos1 sin(  1 )
                  R(v)  0
v v1 cos   gv1 cos
Solve for Initial Angle 0
• Evaluate at 0:
1 cos1 sin(  1 )
                  R(v)  0
v v1 cos   gv1 cos

• Obtain Implicit Equation for
initial slope angle:

v0             sin( 0  1)
cos 0   v 0 R(v 0 )               0
v1                 gv1
Solving for f()
• Rearrange E-L equation:
ÝdR  d v   0
Ý 
f
 Ý f
v        
v 2
     dv  d

• Obtain ODE:
Ý            v 3
        f ( ) 
dR
v2
1
dv
( Recall that we already have v(), 0, & initial
condition f(0) = 0 )

Solving for x() and y()
• Integrate the E-L equations
Ý Ý
x  f cos  0
Ý Ý
y  f sin  0
• Obtain


x( )  x 0             Ý
f ( )cosd
                        0



y( )  y 0             Ý
f ( )sin d
0
Seems like we are done...
• What about parameters 1 & v1?
Appear everywhere, due to:

cos1             sin 1             sin(  1)
 ( )        ,   ( )         ,   ( ) 
v1                 v1                gv1 cos

• How can we solve for them?

Newton’s Method...
• Use the equations for x() and y() and the
corresponding boundary conditions:


1
M(1,v1 )  x 0              Ý
f ( ,1,v1 )cosd  x1  0
0



N(1,v1 )  y 0              Ý
f ( ,1,v1 )sin d  y1  0
0

• Now we really are done!

Example: Air Resistance
Example: Air Resistance
• Take R(v) = k v
(k - coefficient
of viscous friction)
– Newtonian fluid
– first order approx. for air resistance
• Let x0 = 0, y0 = 0, v0 = 0,
v0               sin( 0  1 )
cos 0     v 0 R(v 0 )                0
v1                   gv1
 0 = /2                      v        2 sin( 0  1 )
cos 0  0  kv0                   0
v1               gv1
Solve for v()
1 cos1 sin(  1 )
                      R(v)  0
v v1 cos     gv1 cos
cos1     sin(  1 ) 2
1         v             kv  0
v1 cos     gv1 cos


g cos1               kv1 sin(  1 )cos 
v                 
1 1 4                      
2k sin(  1 )         g       cos 1
2


( take the negative root to satisfy v(0) = 0 )

Many Calculations...
Results:

(Straight Line)

(Cycloid)
Conclusions
• Different approach to the Brachistochrone
– parametrization by the slope angle 
– use of Lagrange Multipliers
• Gained:
– analytical solution for non-conservative
velocity-dependent frictional force
• Lost ( due to definition s = f() ):
– ability to descibe free-fall and cyclic motion
Questions ?

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