Chis Petrie’s “grandad” problem

Many yeas ago, his grandad suggested to the young Chris Petrie that he calculate the angle of the dangling rod in the diagram. It is rumoured that the grandad added that it would keep Chris quiet on a rainy day. Was this a case of “Spare the rod and spoil the child”?

In order to make things easy, we are to assume that the two strings are weightless, and the magic knots allow complete freedom or movement.

In order to make things difficult, all four lengths in the diagram are in general different.
The aim is to find the maximum value of h.
      h   =   ½ (h1 + h2)
      h12   =   s12 –  x2
      h22   =   s22 –  y2
      4 l2   =   (L + x + y)2 + (h1 – h2)2

Differentiate with respect to x:
           dh
dx 		  =   0   =   1
2 		  ( dh1
dx 		+ dh2
dx 		)
           h1 dh1
dx 		  =   – x
           h2 dh2
dx 		  =   – y dy
dx
           0   =   (L + x + y) ( 1 + dy
dx 		)   +   (h1 – h2) ( dh1
dx 		– dh2
dx 		)

Use the first equation to eliminate dh2/dx in the other 3 equations :
           h1 dh1
dx 		  =   – x
           h2 dh1
dx 		  =   y dy
dx
           0   =   (L + x + y) ( 1 + dy
dx 		)   +  2 (h1 – h2) dh1
dx

Substitute to first equation into the other 2 :
           h1 h2 dh1
dx 		  =   – h2 x   =   h1 y dy
dx
           0   =   h1 (L + x + y) ( 1 + dy
dx 		)   +  2 h1 (h1 – h2) dh1
dx 		  =   (L + x + y) ( h1 + h1 dy
dx 		)   –  2 x (h1 – h2)

Multiply the second equation through by y and substitute the first equation :
           0   =   (L + x + y) ( h1 y + h1 y dy
dx 		)   –  2 x y (h1 – h2)   =   (L + x + y) ( h1 y – h2 x )   –  2 x y (h1 – h2)

We now have four equations in the four unknowns h1, h2, x and y :
      h12   =   s12 –  x2
      h22   =   s22 –  y2
      4 l2   =   (L + x + y)2 + (h1 – h2)2
      (L + x + y) ( h1 y – h2 x )     =    2 x y (h1 – h2)


      16 x2 y2 l2   =   4 x2 y2 (L + x + y)2 + (L + x + y)2 ( h1 y – h2 x )2
                          =   4 x2 y2 (L + x + y)2 + (L + x + y)2 ( h12 y2 – 2 h1 h2 x y + h22 x2 )
                          =   (L + x + y)2 ( h12 y2 – 2 h1 h2 x y + h22 x2 + 4 x2 y2 )
If s1 = s2 = s
      h22 h12   =   h22 s2 –  h22 x2
      h12 h22   =   h12 s2 –  h12 y2
      0   =   h12 s2 –  h12 y2 – h22 s2 +  h22 x2
      0   =   s2 (h12 – h22) –  (h1 y +  h2 x)(h1 y –  h2 x)


      h22 h12   =   h22 s12 –  h22 x2
      h12 h22   =   h12 s22 –  h12 y2
      0   =   h12 s22 –  h12 y2 – h22 s12 +  h22 x2
      0   =   (h1 s2 – h2 s1) (h1 s2 + h2 s1) –  (h1 y +  h2 x)(h1 y –  h2 x)