CIRCLE INTERSECTION

In order to draw arch and elliptical pipe correctly you must calculate the radius intersections given the ASTM measurements.  The measurements given in ASTM C506 for Arch Pipe and C507 for Elliptical pipe do not give intersection points or t Start and t Stop to draw the shapes perfectly and I’ve also found the arcs do not always intersect.  I slightly modify the R3 dimension to force intersection with R1 and R2 so we can evaluate the function.  To properly draw arch and elliptical pipe we need to calculate intersection points of all curves and calculate the t start and t stop for the parametric equation of each curve.

ArchPipe-ASTMC506

Arch Pipe DimensionsModified to Intersect
EquivalentWall ThicknessRiseSpanABCR1R2R3
152.2511180.3754.68754.9687522.87510.6254.05
182.513.522-0.2565.7527.513.755.26
2431828.53.43755.906259.6562540.687514.56254.596
303.522.536.253.757.687512.093755118.756.032
36426.62543.3754.1258.562515.56222.56.38
424.531.312551.1255.062510.0625187326.257.57
4853658.5611.5937520.584308.76
545.540656.6251322.687592.533.3759.83
60645737.514.687525.2812510537.511.21875
727548891731.43751264512.5625

 

This example will calculate the intersection of R1 and R3 for 24″ equivalent arch pipe.

First step is to calculate the variables a,b,c,d.  In the arch pipe diagram above you can see the centroid of the pipe in line with the center of R3 is given point (0,0) and we will calculate the center points of both circles from that point.  Please not capital letters are different variables than lower case variables and are pulled directly from ASTM.

Center R1 =(a,b)= (0,R1-B) = (0,40.6875-5.90625) = (0,34.78125)

Center R3 =(c,d)= (C,0) = (9.65625,0)

Two circles with Radius R1 and R3 can be represented by the equations below:

\[R1^2= (x-a)^2+(y-b)^2    \\\   and  \\\    R3^2=(x-c)^2+(y-d)^2\]

\[40.6875^2= (x-0)^2+(y-34.78125)^2    \\\   and  \\\    4.596^2=(x-9.65625)^2+(y-0)^2\]

Now we need to calculate the intersection point of the two circles above.

\[x1,3=\frac{a+c}{2}+\frac{(c-a)(R1^2-R3^2)}{2D^2}\pm2\frac{b-d}{D^2}\partial\]

\[y1,3=\frac{b+d}{2}+\frac{(d-b)(R1^2-R3^2)}{2D^2}\pm2\frac{a-c}{D^2}\partial\]

\[D=\sqrt{(c-a)^2+(d-b)^2}\]

\[\partial=\frac{1}{4}\sqrt{(D+R1+R3)(D+R1-R3)(D-R1+R3)(-D+R1+R3)}\]

We will start with calculating D.

\[D=\sqrt{(c-a)^2+(d-b)^2}\]

\[D=\sqrt{(9.65625-0)^2+(0-34.78125)^2}\]

\[D=36.09679\]

Next we need to calculate:

\[\partial=\frac{1}{4}\sqrt{(D+R1+R3)(D+R1-R3)(D-R1+R3)(-D+R1+R3)}\]

\[\partial=\frac{1}{4}\sqrt{(36.09679+40.6875+4.596)(36.09679+40.6875-4.596)(36.09679-40.6875+4.596)(-36.09679+40.6875+4.596)}\]

\[\partial=4.22564\]

Now we can calculate the actual intersection points.

\[x1,3=\frac{a+c}{2}+\frac{(c-a)(r1^2-r3^2)}{2D^2}\pm2\frac{b-d}{D^2}\partial\]

\[x1,3=\frac{0+9.65625}{2}+\frac{(9.65625-0)(40.6875^2-4.596^2)}{2*36.09679^2}+\frac{34.78125-0}{36.09679^2}*4.22564\]

\[x1,3=11.10972\]

\[y1,3=\frac{b+d}{2}+\frac{(d-b)(r1^2-r3^2)}{2D^2}\pm2\frac{a-c}{D^2}\partial\]

\[y1,3=\frac{34.78125+0}{2}+\frac{(0-34.78125)(40.6875^2-4.596^2)}{2*36.09679^2}+\frac{0-9.65625}{36.09679^2}*4.22564\]

\[y1,3=-4.48538\]

R1 and R3 intersect at point (11.10972,-4.48538).

Next we need to calculate the intersection point of R2 and R3.  If we go through the same process we get (14.0918,1.0729).

To draw Radius R3 on the right side of the arch pipe wet Calculation for R3 need to calculate the range of t from the intersection of R1,R3 to R2,R3.

 

 

 

 

We first calculate tStart using the intersection point of R1 and R3 (11.10972,-4.48538).

\[tStart= 2\pi-\arcsin(\frac{-y}{R3})\]

\[tStart= 2\pi-\arcsin(\frac{-(-4.48538)}{4.596})\]

\[tStart=4.9322\]

We next calculate tStop using the intersection point of R2 and R3 (14.0918,1.0729).

\[tStop= \arcsin(\frac{y}{R3})\]

\[tStop= \arcsin(\frac{1.0729}{4.596})\]

\[tStop= 0.235608\]

With these calculations complete we can draw R3 with our parametric equations derived on the intersecting cylinders page by substituting R3 for the hole radius a as shown below and evaluating over the tStart to tStop+2*pi interval.

\[4.9322<=t<=(2*\pi+0.235608)\]

\[\begin{pmatrix} x(t) \\y(t) \\z(t) \end{pmatrix}=\begin{pmatrix} ((R3 * \cos(t) + Offset)*(cos(360-Angle))) -(\sqrt{b^2-[(R3 * cos(t))+Offset]^2} *sin(360-Angle))\\(R3* \cos(t) + Offset )*sin(360-Angle)+(\sqrt{b^2-[(R3 * cos(t))+Offset]^2}*cos(360-Angle)) \\R3 * \sin(t)+Elevation+\frac{R3}{2} \end{pmatrix}\]

After completing this same process for R1 and R2 you have a complete function to draw arch pipe in three dimensions.