# Volume of a Square to Round Transition

A 'square to round transition' or 'round to square transition' is a 3-D geometric shape that resembles a cross between a conical frustum and a square pyramidal frustum. It has a square base that gradually morphs into a circular face at the top. Usually the round face is smaller than the square face, but it can also be the opposite.

Square to round transitions are often created by tradesmen to connect pieces of ductwork of different sizes and shapes. In some industrial applications it is useful to know the interior volume of a square to round transition. If you know the height of the object, the side length of the square base, and the diameter of the circular hole at the top, then you can plug these variables into a formula to find the volume.

The volume formula is derived using integral calculus. Here we give the formula, how the formula is developed, and some examples of calculating the volume.

## Various Views of a Square to Round Transition

## Top View of a Square to Round Transition

## Volume Formula for Square to Round Transition

If the side length of the square base is S, the diameter of the circle at the top D, and the height of the figure H, then the volume is give by the equation

**Volume = (H/3)[S^2 + SD + (π/4)D^2]**

This formula works regardless of whether D is smaller than S or D is greater than S. This formula does not include the volumes of short cylindrical and rectangular pyramidal pieces connected to the ends of the square to round transition.

## How the Square to Round Transition Volume Formula is Derived Using Geometry and Calculus

Between the square base and the round hole at the top, the horizontal cross-sections of the square to round transition are squares with rounded corners, a shape sometimes called a squircle. If the radius of the quarter circular arc at the corners is R and the length of the straight line segment along the sides is L, then the area of the figure is given by the equation

Squircle Area = L^2 + 4RL + πR^2

See diagram below for a visual explanation of this formula. This area equation is derived by partitioning the shape into a square, four rectangles, and four quarter circular arcs, and then adding the areas of all these shapes.

Imagine slicing the shape from top to bottom into thin squircle-shaped sheets. Now imagine making an infinite number of slices of infinitesimally small thickness. The volume of the square to round transition is the infinite sum of these infinitesimally thin squircle sheets. This is formulated as an integral and solved using calculus.

As you move along the z-axis of the figure, the lengths of the straight line segments of the squircle cross-sections decrease linearly from S to 0. At the same time, the radii of the arcs at the corners increase linearly from 0 to D/2.

Now consider a cross-sectional slice taken at an arbitrary height of x. The variable x ranges from 0 to H. The straight slide length of a cross-section taken at height x is L = S - (S/H)x. The corner arc radius of a cross-section taken at height x is R = (D/(2H))x.

Substituting these expressions in the formula L^2 + 4RL + πR^2 gives us the area of the cross section at height x. Call this area function f(x).

f(x) = [S - (S/H)x]^2 + 4(D/(2H))x[S - (S/H)x] + π[(D/(2H))x]^2

The function f(x) is a quadratic (2nd degree) polynomial in x. The volume of the square to round transition is given by the integral

∫ f(x) dx, {0 ≤ x ≤ H}

In this definite integral, x is the variable and S, D, and H are constants. The limits of integration are 0 to H. After finding the antiderivative and plugging in the limits of integration, final expression works out to

**∫ f(x) dx, {0 ≤ x ≤ H} = (H/3)(S^2 + SD + (π/4)D^2)**

## Example 1

A square to round transition has a base length of 1 foot, a top diameter of 6 inches, and a height of 14 inches. Attached to the top is a piece of a circular duct with a diameter of 6 inches and a height of 7 inches. Attached to the base is a piece of a square duct with a height of 8 inches and a a cross-sectional side length of 1 foot. What is the total volume of this component in cubic inches?

The total volume of this shape is the volume of the square to round transition plus the volume of the cylinder at the top plus the volume of the rectangular prism at the base. The volumes of the cylinder and rectangular base are, respectively

Cyl.Vol = π*7*(6/2)^2 ≈ 197.9203 in^3

Rect.Prsm.Vol = 12*12*8 = 1152 in^3

The volume of the square to round transition is found using the volume formula derived above with S = 12, D = 6, and H = 14. This gives us

(14/3)(12^2 + 12*6 + (π/4)6^2) ≈ 1139.9469 in^3

Therefore, the total volume of the object is 197.9203 + 1152 + 1139.9469 ≈ 2489.8672 cubic inches. In cubic feet this is about 1.4409 cubic feet.

## Example 2

A square to round transition has a volume of 41.89 liters. If the length of the base is 35 cm and the height is 34 cm, what is the diameter of the circular hole at the top?

For this problem, we need to convert 41.89 liters to 41890 cubic cm. Using S = 35, we have the equation

41890 = (34/3)(35^2 + 35D + (π/4)D^2)

17πD^2 + 2380D - 84020 = 0

D ≈ 23.2 cm and -67.8 cm

Since the only positive solution to the quadratic equation is 23.2, that must be the diameter of the circular top of the square to round transition.

## Comments

Will this formula work if the circle part is wider than the square part? I do not see that type of round2square transition discussed.

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