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Brachistochrone Under Air Resistance

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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' 

 • Additional boundary conditions:
        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:
                                
                                      Additional
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

       Quadratic Formula:
     
          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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Description: Brachistochrone Under Air Resistance