Vesica Piscis Related Proportions

Figure 1: Vesica Piscis Related Proportions

The Vesica Piscis, sometimes called the “Vesica Pisces,” is another one of the more important shapes in mathematics, which is formed by the intersection of two circles with the same radius. These are oriented in such a way that you have the center of each circle positioned so its perimeter intersects the center of the opposite circle, as shown in Figure 1.

The Vesica Piscis also appeared in the first proposition of Euclid’s Elements, where it shows how to use this configuration to construct an equilateral triangle using a compass and straight edge.

In this same diagram, we can also derive other important numbers often seen in sacred geometry, design, and architecture, including the square roots of numbers 1 through 5, in addition to the golden ratio.

Figure 2: Vesica Piscis Related Proportions Model Enlarged

Equilateral Triangle Construction

Referring to our enlarged Vesica Piscis model in Figure 2, this is a very useful diagram we can create by using only two circles, and it gives us a nice way to visualize these other related proportions. We can also construct the golden rectangle from these same two circles. The radius $r$ of both circles is determined to be of length one. So we can construct a triangle from this as follows,

\begin{align}
r = OA = OB = AB = 1
\end{align}

By definition, a triangle with all three sides being equal is equilateral. When the three sides are equal, we also know the three angles opposite to the three sides are equal. Since the sum of all three angles in a triangle add up to $180^{\circ}$, each angle in an equilateral triangle is given by,

\begin{align}
180^{\circ} \div 3 = 60^{\circ}
\end{align}

Unit Square and Root 2

The ubiquitous square is another shape in mathematics that can’t be overlooked, and we can construct the simplest and most elegant version of a square from our model. Since our circles are both defined with radius $r = 1$, we can extend a perpendicular line from origin $O$ to point $D$ on the perimeter of the yellow circle. Doing the same thing in the green circle gives us line $BE$. Now we have all four corners of our unit length square, $ODBE$.

It’s easy to calculate the diagonal $OE$ of this square by the Pythagoream Theorem,

\begin{align}
OE^2 = 1^2 + 1^2 \\\\
OE = \sqrt{2}
\end{align}

For the sake of completion, we can also notice that we have our first root value in this model as well, since,

\begin{align}
r = 1 = \sqrt{1}
\end{align}

Vesica Piscis and Root 3

Figure 3: Vesica Piscis Height and Area

ENLARGE IMAGE

The Vesica Piscis deserves a further field of study all on its own. If you’ve ever wondered about Gothic architecture and how they designed it, one of the things they used was the Vesica Piscis as their proportioning system. So we should know as much as we can about this mysterious shape and what makes it tick.

We can learn quite a bit about it by going back to our equilateral triangle in Figure 3. By calculating the height $h$ of this triangle, we can see that a line $AC$ from tip to tip of the vesica piscis is given by $2h$. Here again, this is easy to get from our model.

\begin{align}
1^2 = h^2 + \left(\frac{1}{2}\right)^2 \\\\
h^2 = 1 \; – \; \frac{1}{4} \\\\
h = \frac{\sqrt{3}}{2}
\end{align}

So since line $AC$ is given by $2h$, we can see the vertical height of a vesica piscis when $r = 1$.

\begin{align}
AC = \sqrt{3}
\end{align}

And more generally, for any vesica piscis constructed with two circles having a different value for $r$,

\begin{align}
r^2 = h^2 + \left(\frac{r}{2}\right)^2 \\\\
h = \frac{\sqrt{3}}{2}r \\\\
AC = 2h = \sqrt{3} \; r \\\\
\end{align}

Figure 4: Vesica Piscis Area

ENLARGE IMAGE

Calculating the Area of a Vesica Piscis
At first glance, calculating the area of a Vesica Piscis might not sound so easy. But once again, our equilateral triangle saves the day. Since we can form an arc from $OA$ through to $OB$, this creates a $60^{\circ}$ sector which is one sixth of our yellow $360^{\circ}$ circle.

If we let $S_a =$ Sector Area, and $r =$ radius, the area of this sector would be,

\begin{align}
S_a = \frac{1}{6}\pi r^2
\end{align}

If our equilateral triangle has side lengths $r$ = the radius of the circle, where $r = 1$ in our model, we can let $T_a$ equal the triangle area which is given by,

\begin{align}
T_a = \frac{1}{2}rh = \frac{\sqrt{3}}{4}
\end{align}

Referring to Figure 4, if we look at the outer segment of the $60^{\circ}$ circular sector shown in red, this can be calculated by subtracting the triangle area $T_a$ shown in orange, from the full sector area $S_a$ we calculated earlier. We can let $A_s$ equal the area of one sector segment and perform that calculation.

\begin{align}
A_s = S_a \; – \; T_a = \frac{1}{6}\pi \; – \; \frac{\sqrt{3}}{4} \\\\
A_s = \frac{4\pi \; – \; 6\sqrt{3}}{24} = \frac{2\pi \; – \; 3\sqrt{3}}{12}
\end{align}

So to get the area of the Vesica Piscis in our model, we can sum the area of 2 triangles and 4 sector segments. Let $VP_a$ equal the area of our Vesica Piscis, and here is our formula,

\begin{align}
VP_a = 2T_a + 4A_s = \frac{\sqrt{3}}{2} + \frac{2\pi \; – \; 3\sqrt{3}}{3} \\\\
VP_a = \frac{3\sqrt{3} + 4\pi \; – \; 6\sqrt{3}}{6} = \frac{4\pi \; – \; 3\sqrt{3}}{6}
\end{align}

For a more general formula relating a Vesica Piscis to the radius $r$ of any given circles, it can be shown that the area of an equilateral triangle can also be calculated by this formula,

\begin{align}
T_a = \frac{\sqrt{3}}{4}r^2 \\\\
\end{align}

In that case, we would have $r^2$ relating to both the area of the triangle $T_a$ and the area of the sector $S_a$. Performing the same basic calculations with both values relating to $r^2$ will give us the more general formula as follows,

\begin{align}
VP_a = \frac{1}{6}\left(4\pi \; – \; 3\sqrt{3}\right)r^2 \\\\
\end{align}

Golden Rectangle and Root 5

Figure 5: Golden Rectangle and Root 5

ENLARGE IMAGE

So far we have the values for root 1 through root 3, but we can also see root 4 from line $DF$ in Figure 5. This is the diameter of the circle, which is $2r$. When $r = 1$, then $DF = 2$. So from that we can see,

\begin{align}
DF = 2 = \sqrt{4}
\end{align}

No matter what you’re doing with the golden ratio, root 5 shows up everywhere. So we should look for that value in our model as well. To construct a golden rectangle, we can define a point $M$ which is midway between line $OB$ as shown in Figure 5. Then we can create a diagonal from the center base of the square $ODEB$ to the upper right corner $E$. Let’s define this diagonal to be line $ME$ and find its value,

\begin{align}
ME^2 = 1^2 + \left(\frac{1}{2}\right)^2 = \frac{5}{4} \\\\
ME = \frac{\sqrt{5}}{2}
\end{align}

We can also see from Figure 5 that triangles $\bigtriangleup MEB$ and $\bigtriangleup MFO$ are identical, so our root 5 is given by line $EF = 2ME$,

\begin{align}
EF = 2ME = \sqrt{5}
\end{align}

Furthermore, now that we have the diagonal $ME$ established, constructing a golden rectangle is straight forward. If we use $M$ as a center point, we can form an arc from $E$ through $G$. Then we can notice the distance from $O$ to $G$ is actually the golden ratio $\varphi$.

\begin{align}
OG = \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} = \varphi \\\\
\end{align}

In addition, we have rectangle $ODHG$ which is a golden rectangle!

Conclusion

How you can get all of that from two little circles is quite miraculous, and this hints at why the Vesica Piscis has made such a wonderful proportioning system throughout history! The Vesica Piscis also ties in very closely with our research on Gabriel’s Horn, so check out our latest article with more information about that relationship below,

GABRIEL’S HORN INTERSECTING PLANE APP

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