Calculators

Stadium Calculator

Area, perimeter, radius, straight-side length, overall length and width for any stadium shape — also called a discorectangle, obround, racetrack or pill. Nine solve modes with a live diagram and step-by-step working.

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What is a stadium?

A stadium is a two-dimensional shape made of a rectangle with a semicircle attached to each of the two shorter sides. Because the two semicircles share the same radius as the rectangle's half-width, the shape has no corners — it is the 2D equivalent of a running-track outline. It goes by several other names depending on the field: discorectangle in physics, obround in mechanical drafting, racetrack in athletics, and pill in pharmaceutical or UI-design contexts. All refer to the same shape and use the same formulas.

It's fully described by just two numbers: the straight-side length a (the rectangle's length) and the cap radius r (half of the overall width). Every other measurement — area, perimeter, overall length, arc length — falls out of those two.

Stadium, case by case

Area of a stadium

The middle rectangle has area 2ar. The two semicircular caps together make one full circle of radius r, area πr². Add them.

A = 2 a r + π r²
astraight side length (the rectangle's length)·rcap radius (= half the width)
ar
Example
Given
a = 8
r = 3
Substitute
A = 2·8·3 + π·3²
= 48 + 9π
Answer
A ≈ 76.274

Perimeter of a stadium

Two straight sides of length a plus the two semicircular arcs of radius r, which together form one full circle of circumference 2πr.

P = 2 a + 2 π r
aone straight side length·rcap radius
ar
Example
Given
a = 8
r = 3
Substitute
P = 2·8 + 2π·3
= 16 + 6π
Answer
P ≈ 34.850

Total length and width

The overall bounding box of a stadium is L long and W wide. Because the caps are semicircles, the width equals the cap diameter and the length equals the straight side plus one full diameter.

L = a + 2 r
W = 2 r
Ltotal outer length·Wtotal outer width
arLW
Example
Given
a = 8
r = 3
Substitute
L = 8 + 2·3
W = 2·3
Answer
L = 14
W = 6

Given a and Area, solve for r

With the straight side fixed, area is quadratic in the radius. Solving πr² + 2ar − A = 0 gives one positive root; the other is negative and unphysical.

r = (−a + √(a² + π A)) / π
astraight side (fixed)·Atarget area
ar
Example
Given
a = 8
A = 76.274
Substitute
r = (−8 + √(64 + π·76.274)) / π
Answer
r ≈ 3

Solve everything from L and W

The most common real measurement pair — the outer length and width of the shape (running track, capsule pill, slot cutout). Radius is half the width; the straight section is the length minus the width.

r = W / 2
a = L − W
Loverall outer length·Woverall outer width
arLW
Example
Given
L = 14
W = 6
Substitute
r = 6/2
= 3
a = 14 − 6
= 8
Answer
Then A = 2·8·3 + π·3² ≈ 76.274

Reverse-engineer from Area and Perimeter

If you know both the area and the perimeter (from a floor plan or a drawing spec) you can back out the shape uniquely — provided the pair is physically possible, i.e. P² ≥ 4πA.

π r² − P r + A = 0
a = P/2 − π r
Aknown area·Pknown perimeter
ar
Example
Given
A = 76.274
P = 34.850
Substitute
r = (34.850 − √(34.850² − 4π·76.274)) / (2π)
Answer
r ≈ 3
then a = 34.850/2 − 3π ≈ 8

Formulas at a glance

Area A = 2 a r + π r²
Perimeter P = 2 a + 2 π r
Overall length L = a + 2 r
Overall width W = 2 r
Cap arc length (both) ℓ = 2 π r
Isoperimetric bound P² ≥ 4 π A (equality only when a = 0)

All formulas — every calculation mode

The exact symbolic chain each mode of this calculator evaluates. Same notation as the widget: a is the straight side, r is the cap radius, L = a + 2r is the overall length, W = 2r is the overall width, A is the area and P is the perimeter.

1. Find Area & Perimeter — given a, r
A = 2 a r + π r²
P = 2 a + 2 π r
L = a + 2 r
W = 2 r
2. From overall L and W
r = W / 2
a = L − W
A = 2 a r + π r²
P = 2 a + 2 π r
3. Find a — given A and r
a = (A − π r²) / (2 r)
P = 2 a + 2 π r
4. Find r — given A and a (quadratic in r)
π r² + 2 a r − A = 0
r = (−a + √(a² + π A)) / π
P = 2 a + 2 π r
5. Find a — given P and r
a = (P − 2 π r) / 2
A = 2 a r + π r²
6. Find r — given P and a
r = (P − 2 a) / (2 π)
A = 2 a r + π r²
7. From L and P
r = (P − 2 L) / (2 (π − 2))
a = L − 2 r
A = 2 a r + π r²
8. From L and A (quadratic in r)
(4 − π) r² − 2 L r + A = 0
r = (L − √(L² − (4 − π) A)) / (4 − π)
a = L − 2 r
P = 2 a + 2 π r
9. From A and P (quadratic in r)
π r² − P r + A = 0 (requires P² ≥ 4 π A)
r = (P − √(P² − 4 π A)) / (2 π)
a = P/2 − π r

Where the formulas come from

Area. Slice the stadium into three pieces: the middle rectangle (dimensions 2r × a) and the two end semicircles. The rectangle contributes 2ar. The two semicircles of radius r, glued along their diameters, are geometrically identical to one full circle of radius r, contributing πr². Summing gives A = 2ar + πr².

Perimeter. The boundary consists of the two straight edges of length a on the top and bottom, plus the two circular arcs on the left and right. Each arc is exactly half of a circle of radius r, so together they contribute 2πr. Summing gives P = 2a + 2πr.

Overall length and width. The width is the diameter of the caps: W = 2r. The length is the rectangle's length plus one full diameter (half a diameter from each cap): L = a + 2r.

Isoperimetric bound. Among all closed 2D curves with a fixed perimeter, the circle maximises area — this is the classical isoperimetric inequality, P² ≥ 4πA. For a stadium, eliminating a between the area and perimeter formulas produces the quadratic πr² − Pr + A = 0. Its discriminant P² − 4πA must be non-negative, giving the same bound (equality holds only when the straight side collapses, a = 0, and the stadium becomes a pure circle).

Worked example — a 400 m running track

A standard IAAF athletics track measures 400 m along the inside lane and is a stadium shape. If the inner semicircular curves have radius r = 36.5 m, what straight-side length gives a 400 m inner perimeter?

P = 2 a + 2 π r ⇒ a
= (P − 2 π r) / 2
a = (400 − 2 π × 36.5) / 2 ≈ (400 − 229.336) / 2 ≈ 85.332 m

The overall inside length is L = a + 2r ≈ 85.332 + 73 ≈ 158.33 m and the width is W = 2r = 73 m. The enclosed infield area is A = 2ar + πr² ≈ 2 × 85.332 × 36.5 + π × 36.5² ≈ 6229 + 4185 ≈ 10 414 m².

What this tool does for you

  • Nine calculation modes — solve from any pair among (a, r, L, W, A, P), including the two quadratic-in-r modes competitor calculators skip
  • Live SVG diagram labeled with a, r, L, W and the computed values
  • Reports area, perimeter, straight side, radius, overall length and width, diameter and total arc length
  • Unit selector (mm, cm, m, km, in, ft, yd) with automatic m² conversion for the cost estimator
  • Show/hide step-by-step working for every mode — including the quadratic-formula substitution
  • Handles the a = 0 degenerate case (full circle) and rejects (A, P) pairs that violate P² ≥ 4πA
  • Material or cost estimator — enter a price per m² and get the total

Frequently asked questions

+What is a stadium shape in geometry?
A stadium is a rectangle capped on both ends by semicircles of the same radius. Its width equals the cap diameter, and its perimeter is two straight sides plus one full circle's worth of arc.
+Is a stadium the same as a discorectangle, obround, racetrack or pill?
Yes — all of those names refer to the same 2D shape. Stadium is the standard mathematical term; discorectangle is common in physics; obround is common in mechanical drafting; racetrack is common in athletics; and pill is common in pharmaceutical or icon-design contexts.
+What are the formulas for area and perimeter?
With straight-side length a and cap radius r:
A = 2 a r + π r²
P = 2 a + 2 π r
The overall bounding-box length is L = a + 2r and the width is W = 2r.
+Which values do I actually need to enter?
Any two of a, r, L, W, A, P are enough to determine the shape. This calculator supports every common pairing: (a, r), (L, W), (A, r), (A, a), (P, r), (P, a), (L, P), (L, A) and (A, P).
+What happens when the straight side a is zero?
The stadium collapses to a full circle of radius r: area = πr², perimeter = 2πr, length = width = 2r. The calculator accepts a = 0 as a valid input in every mode that takes a directly.
+Why do some (Area, Perimeter) inputs return no solution?
For any real stadium, P² ≥ 4πA (the classical isoperimetric bound; equality holds only for the pure circle). If your entered A and P violate this inequality, no positive radius satisfies both equations simultaneously — the tool reports the conflict instead of returning a spurious answer.
+How is a stadium different from a rounded rectangle?
A rounded rectangle has its corners softened by an arc whose radius is independent of the rectangle's width, so it keeps some flat top and side edges. A stadium is the special case where the corner radius equals half the width — the two arcs on each end meet to form a complete semicircle, and no flat side remains.
+How is this different from the 3D capsule?
A capsule is what you get when you rotate a stadium around its long axis of symmetry — a cylinder with a hemisphere glued onto each end. For 3D volume and surface area, use the Volume Calculator (capsule mode) and the Surface Area Calculator; the 2D formulas on this page don't carry over directly.
+What is this shape used for in real life?
Standard 400 m running tracks, hockey rink centre-face-off area outlines, oval racetracks, pill-shaped tablets, obround slots and washers in machine design, stadium-shaped buttons in UI design, and any bounding shape where sharp corners need to be replaced by circular caps.

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