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Quarter-Tank Problem -- from Wolfram MathWorld

Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...



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Quarter-Tank Problem -- from Wolfram MathWorld

https://mathworld.wolfram.com/Quarter-TankProblem.html

Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...



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https://mathworld.wolfram.com/Quarter-TankProblem.html

Quarter-Tank Problem -- from Wolfram MathWorld

Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...

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      Quarter-Tank Problem -- from Wolfram MathWorld
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      Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...
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      Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...

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      Quarter-Tank Problem -- from Wolfram MathWorld
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      Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...

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      Finding the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be one quarter full amounts to plugging A=piR^2/4 (one quarter of the area of a full circle) into the equation for the area of a circular segment of radius R, A=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2) This gives equality between the two shaded areas in the above figure, resulting in 1/4pi=cos^(-1)(1-x)-(1-x)sqrt(2x-x^2), where x=h/R and R is the...
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