"mg=\\rho_wgV_\\text{submerged},\\\\\\space\\\\\nV_\\text{submerged}=\\frac{m}{\\rho_w}=\\frac{SG\u00b7\\rho_w\u00b7V}{\\rho_w},\\\\\\space\\\\\nV_\\text{submerged}=SG\u00b7V=SG\u00b7\\frac{\\pi d^2}{4}L." The volume above the water surface is
"\\Delta V=V-V_\\text{submerged}=(1-SG)\u00b7\\frac{\\pi d^2}{4}L." Divide by length to find the area of the segment above the water:
"A=(1-SG)\u00b7\\frac{\\pi d^2}{4}."
The area of the segment can be expressed in terms of areas of a triangle and sector:
"A=A_s-A_t=\\pi\\theta-x\u00b7(r-h),\\\\\n\\theta=2\\arctan\\frac xh.\\\\\\space\\\\\nA=2\\pi\\arctan\\frac xh-x\u00b7(r-h).\\\\\\space\\\\\nx^2=r^2-(r-h)^2=2rh-h^2.\\\\\\space\\\\\nh=0.26\\text{ m}." Therefore, the depth is
"D=d-h=0.54\\text{ m}."
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