John Hughes’s Garden Problem

John Hughes’s Garden Problem
Solution by Chris Petrie

In the drawing the line GD is parallel to the line C_iE. The area GDB is to be found interms of y. The half-width w, and dimensions x and h are given.

      R_o² = w² + (R_o – x)² = w² + R_o² – 2 R_ox + x²
      R_o = (w² + x²) / 2 x
      R_i² – w² + (R_i – (x–h))² = w² + R_i² – 2R_i(x–h) + (x–h)²
      R_i = [w² + (x–h)²] / 2 (x–h)
      Ð GC_iE = α = sin^-1(y/R_i)
      Ð DC_oE = β = sin^-1(y/R_o)
      Ð BC_iE = θ = sin^-1(w/R_i)
      Ð BC_oE = φ = sin^-1(w/R_o)

By inspection, the area GDB is equal to:
      §BC_oD + ΔBC_oC_i – §BGC_i – ΔGC_iC_o – ΔC_oDG

      §BC_oD = π R_o² (φ – β) / 2 π = R_o² (φ – β) / 2

      ΔBC_oC_i = (R_i + h – R_o) w / 2

      §BGC_i = R_i²(θ – α) / 2

      ΔGC_iC_o = (R_i + h – R_o) y / 2

      ΔC_oDG = (R_i + h – R_o + R_o cos β – R_i cos α) y / 2

Calculate the area GDB for a series of y between 0 and w.
Plot the area against y.
Lift off values of y for any particular area.

John Hughes’s Garden Problem Solution by Chris Petrie

John Hughes’s Garden Problem
Solution by Chris Petrie