Study of an Irregular Cross Vault

10 Diameter Semi-Circle intercepts 20 Diameter Circle Section
Plan Angles for all calculations are as per the diagram
 Vault Height = 5 For Height = 5 : 20 Diameter Circle Section Width 2 = 10 2 – 5 2 = 75 20 Diameter Circle Section Width = 8.66025 Semi-Circle Plan Angle = arctan (5 ÷ 8.66025) = 30° 20 Diameter Circle Section Plan Angle = arctan (8.66025 ÷ 5) = 60° Valley Ellipse Run = 5 ÷ sin 30° = 10 Equation of Valley Ellipse = x 2 + 4 y 2 = 100 Semi-Major Axis = 10    Semi-Minor Axis = 5 Major Axis = 20    Minor Axis = 10 Let the 10 Diameter Circle horizontal dimension = 4 For Width = 4 : Height of 10 Diameter Circle 2 = 5 2 – 4 2 = 9 Height of 10 Diameter Circle = 3 Corresponding horizontal dimension of the 20 × 10 Valley Ellipse = 4 ÷ sin 30° = 8 For Width = 8 : Height of 20 × 10 Valley Ellipse 2 = ( 100 – 8 2 ) ÷ 4 = 9 Height of 20 × 10 Valley Ellipse = 3 Projection of dimension to 20 Diameter Circle = 8 × sin 60° = 6.92820 For Width = 6.92820 : Height of 20 Diameter Circle 2 = 10 2 – 6.92820 2 = 52 Height of 20 Diameter Circle = 7.21111 Height of 20 Diameter Circle Section = 7.21111 – 5 = 2.21111 ... a fatal contradiction since the HEIGHTS at CORRESPONDING HORIZONTAL DIMENSIONS must be EQUAL Let the 20 Diameter Circle Section height = 3 Height of 20 Diameter Circle = 3 + 5 = 8 Width of 20 Diameter Circle 2 = 10 2 – 8 2 = 36 Width of 20 Diameter Circle = 6 ... corresponding run in plan = 6.92820 The POINT on the SURFACE of the VAULT does not align with the CORRESPONDING POINT in PLAN

 Solution #1: Make the 20 Diameter Circle Section a Semi-Ellipse Semi-Major Axis = 8.66025    Semi-Minor Axis = 5 Major Axis = 17.32051    Minor Axis = 10 Equation of Ellipse = x 2 + 3 y 2 = 75 For 20 × 10 Valley Ellipse horizontal dimension = 8 : Projection to 17.32051 × 10 Ellipse = 8 × sin 60° = 6.92820 Height of 17.32051 × 10 Ellipse 2 = ( 75 – 6.92820 2 ) ÷ 3 = 9 Height of 17.32051 × 10 Ellipse = 3 HEIGHTS at CORRESPONDING HORIZONTAL DIMENSIONS are EQUAL

 Solution #2: Make the 10 Diameter Circle a Semi-Ellipse Projection of 20 Diameter Circle to Valley Ellipse = 10 ÷ sin 60° = 11.54701 Semi-Major Axis = 11.54701    Semi-Minor Axis = 10 Major Axis = 23.09401    Minor Axis = 20 Equation of Ellipse = 100 x 2 + 133.33333 y 2 = 13333.33333 Projection of 23.09401 × 20 Valley Ellipse width through 30° Plan Angle : 11.54701 × sin 30° = 5.77350 NOTE : The Minor Axis is the Horizontal Dimension on the x-axis; the Major Axis is the Vertical Dimension on the y-axis. Semi-Major Axis = 10    Semi-Minor Axis = 5.77350 Major Axis = 20    Minor Axis = 11.54701 Equation of Ellipse = 100 x 2 + 33.33333 y 2 = 3333.33333 Let the horizontal dimension on the 11.54701 × 20 Ellipse = 4 Height of 11.54701 × 20 Ellipse 2 = ( 3333.33333 – 100 × 4 2 ) ÷ 33.33333 = 52 Height of 11.54701 × 20 Ellipse = 7.21111 The Plan Angles are the same, therefore the corresponding horizontal dimension on the 23.09401 × 20 Valley Ellipse remains as per the previous calculation at 4 ÷ sin 30° = 8 Height of 23.09401 × 20 Valley Ellipse 2 = ( 13333.33333 – 100 × 8 2 ) ÷ 133.33333 = 52 Height of 23.09401 × 20 Valley Ellipse = 7.21111 Corresponding horizontal dimension of the 20 Diameter Circle = 8 × sin 60° = 6.92820 The vertical dimension of the 20 Diameter Circle remains 7.21111 HEIGHTS at CORRESPONDING HORIZONTAL DIMENSIONS are EQUAL
 The full heights of the circles and ellipses need not be used in the construction of the vault. Any value ( for example, 5 ) may be subtracted from all the vertical dimensions to create a new level reference line.

Validating the Slopes of the Ellipses
Geometric solution of Slope at any Point on the Ellipse

 The slopes of the ellipses at the Framing Points will be evaluated directly from known values according to the formulae:

 The slopes will also be tested using the equation:

 The slopes to be evaluted have been assigned negative values. Due to symmetry the slopes in question may actually be positive or negative (see sketches Solution #1 and Solution #2). 10 Diameter Semi-Circle intercepts 20 Diameter Circle Section: At ( 4 , 3 ) on the 10 Diameter Semi-Circle 10 Diameter Semi-Circle Slope = – ( 5 2 × 4 ) ÷ ( 5 2 × 3 ) = – 4/3 At ( 8 , 3 ) on the 20 × 10 Valley Ellipse 20 × 10 Valley Ellipse Slope = – ( 5 2 × 8 ) ÷ ( 10 2 × 3 ) = – 2/3 Projecting the 10 Diameter Semi-Circle Slope to the 20 × 10 Valley Ellipse: 20 × 10 Valley Ellipse Slope = – 4/3 × sin 30° = – 2/3 At ( 6.92820 , 7.21111 ) on the 20 Diameter Circle 20 Diameter Circle Slope = – ( 10 2 × 6.92820 ) ÷ ( 10 2 × 7.21111 ) = – .96077 Projecting the 20 × 10 Valley Ellipse Slope to the 20 Diameter Circle: – 2/3 ÷ sin 60° = – .76980 ... another inconsistent result Solution #1: Make the 20 Diameter Circle Section a Semi-Ellipse The horizontal dimension on the 17.32051 × 10 Ellipse = 6.92820 At ( 6.92820 , 3 ) on the 17.32051 × 10 Ellipse Slope of 17.32051 × 10 Ellipse = – ( 5 2 × 6.92820 ) ÷ ( 8.66025 2 × 3 ) = – .76980 In accord with the value returned by projecting the Valley Ellipse Slope through the Plan Angle Solution #2: Make the 10 Diameter Circle a Semi-Ellipse At ( 6.92820 , 7.2111 ) on the 20 Diameter Circle 20 Diameter Circle Slope = – ( 10 2 × 6.92820 ) ÷ ( 10 2 × 7.21111 ) = – .96077 Projecting the 20 Diameter Circle Slope to the 23.09401 × 20 Valley Ellipse: 23.09401 × 20 Valley Ellipse Slope = – .96077 × sin 60° = – .83205 At ( 8 , 7.2111 ) on the 23.09401 × 20 Valley Ellipse 23.09401 × 20 Valley Ellipse Slope = – ( 10 2 × 8 ) ÷ ( 11.54701 2 × 7.21111 ) = – .83205 Projecting the 23.09401 × 20 Valley Ellipse Slope to the 11.54701 × 20 Ellipse: 11.54701 × 20 Ellipse Slope = – .83205 ÷ sin 30° = – 1.66410 NOTE : The Minor Axis lies on the x-axis and the Major Axis lies on the y-axis Therefore the Slope = – [ Semi-Major Axis 2 × x ] ÷ [ Semi-Minor Axis 2 × y ] At ( 4 , 7.2111 ) on the 11.54701 × 20 Ellipse 11.54701 × 20 Ellipse Slope = – ( 10 2 × 4 ) ÷ ( 5.77350 2 × 7.21111 ) = – 1.66410