Arch Calculator helps builders find arc length, radius, sweep angle, segment area, formwork area, and masonry unit count from span, rise, depth, unit width, and joint size on site.
How It’s Calculated
An arch built as a single circular segment is defined by two dimensions: the span (the straight-line width between the two springing points) and the rise (the vertical height from that springing line up to the crown). From just those two numbers, the entire circle the arch is cut from can be reconstructed.
The radius of that circle comes from the chord-and-sagitta relationship:
$$R = \frac{W^2}{8H} + \frac{H}{2}$$
where $W$ is the span and $H$ is the rise. Once the radius is known, the central angle the arc sweeps through follows from the inverse sine of half the span over the radius:
$$\theta = 2\arcsin\left(\frac{W}{2R}\right)$$
Arc length is radius times angle in radians, $$L = R\theta$$, and the area enclosed between the arc and its chord (the segment area) is $$A_{seg} = \frac{R^2}{2}(\theta – \sin\theta)$$. Multiplying arc length by the arch’s depth, how far the curved intrados surface extends into the wall, gives the exposed curved surface area: $$A_{intrados} = L \times D$$.
Laying masonry units along that arc length is a packing problem. With unit width $B$ and mortar joint thickness $J$, $N$ units create $N-1$ internal joints, so the number of units needed is the smallest whole number satisfying $N(B+J) \geq L + J$:
$$N = \left\lceil \frac{L + J}{B + J} \right\rceil$$
Because $N$ is always rounded up to a whole unit, the units and joints laid end to end run slightly longer than the arc itself. That leftover, the fit allowance, equals $(N \times B) + ((N-1) \times J) – L$, and in practice gets absorbed by trimming the end units or spreading it across the joints.
Worked Example
Take a segmental arch with a 48 in span, a 12 in rise, an 8 in depth, 2.5 in wide masonry units, and 0.375 in mortar joints.
First the radius: $$R = \frac{48^2}{8 \times 12} + \frac{12}{2} = \frac{2304}{96} + 6 = 24 + 6 = 30 \text{ in}$$
The half-span-to-radius ratio is $48 / (2 \times 30) = 0.8$, so the central angle is $\theta = 2\arcsin(0.8) = 106.26°$ (1.8546 rad). Arc length follows directly: $L = 30 \times 1.8546 = 55.64$ in, about 7.64 in longer than the straight 48 in span.
The segment area works out to $\frac{30^2}{2}(1.8546 – 0.96) = 402.57$ sq in, and multiplying the 55.64 in arc length by the 8 in depth gives an intrados surface area of 445.10 sq in.
For the masonry layout, each unit-plus-joint module is $2.5 + 0.375 = 2.875$ in. Dividing $(55.64 + 0.375) / 2.875 = 19.48$ and rounding up means 20 units and 19 joints are needed, which together run 57.125 in, 1.49 in longer than the arc, the fit allowance to trim or distribute.
| Result | Value |
|---|---|
| Arc length | 55.64 in |
| Radius | 30.00 in |
| Center depth below springline | 18.00 in |
| Central angle | 106.26° |
| Masonry units needed | 20 |
| Joints | 19 |
| Fit allowance | 1.49 in |
| Intrados surface area | 445.10 sq in |
| Segment area | 402.57 sq in |
| Arc length over span | 7.64 in |
| Rise:span ratio | 1:4.00 |
| Tangent (springing) angle | 53.13° |
| Radius:rise ratio | 2.50:1 |
What the Result Means
The radius and central angle describe the circle the arch is struck from. A 30 in radius with a 106° sweep, as in the example above, is a moderate segmental arch: curved, but well short of a semicircle.
The center of that circle sits 18 in below the springline, which is normal for a flatter arch; it just means the point you’d swing a radius from during layout is below the springline, not above it.
The unit count is always rounded up, so the fit allowance is never zero unless the arc length happens to divide evenly into whole unit-plus-joint modules.
A small allowance, under a couple of inches for typical brick sizes, is normal and gets taken up in the mortar joints or by trimming the last piece. A large allowance usually means the unit width doesn’t suit that particular arc length well.
The rise:span ratio, expressed as 1:X, tells you how flat the arch is. A 1:2 ratio is a full semicircle, the roundest a single-radius arch gets, and the ratio climbs as the arch gets shallower, so 1:4 is noticeably flatter than 1:2 and 1:8 flatter still. The tangent angle is half the central angle, roughly the angle the arc departs from horizontal at each springing point.
What Changes the Result
Rise has an outsized effect on radius because it sits in the denominator of the dominant term, $W^2/(8H)$. Halving the rise on a fixed span roughly doubles the radius, which is why very flat arches need centering forms struck from a radius many times the span, while a full semicircular arch needs a radius equal to just half the span.
The calculator won’t accept a rise greater than half the span. Past that point there’s no single circular arc with a level springline that produces it, so the calculation halts rather than return a distorted result.
Unit width and joint thickness only enter through the rounding-up in the unit count formula, so small changes in arc length can push the count to an extra whole unit even when the change looks minor. Joint thickness can be set to zero for dry-stacked units, but unit width can’t, since it’s the denominator of that division.
Depth only affects the intrados surface area. It has no effect on radius, arc length, or unit count, since it’s multiplied in at the end as the exposed-face dimension.
Frequently Asked Questions
How do you find the radius of an arch from its span and rise?
Use $R = W^2/(8H) + H/2$, where $W$ is the span and $H$ is the rise, in the same units. A 48 in span with a 12 in rise gives a 30 in radius; the same span with a 6 in rise, a flatter arch, jumps to a 78 in radius, since rise sits in the denominator of the dominant term.
How long is the arc of an arch compared to its span?
Always longer, never shorter, since a curve between two points is always longer than the straight line between them. For a 48 in span with a 12 in rise the arc runs 55.64 in, about 7.64 in more than the span itself; flatter arches stay closer to the span length, steeper ones diverge more.
How many bricks or blocks does a segmental arch need?
Divide the arc length plus one joint thickness by the unit width plus joint thickness, then round up: $N = \lceil (L+J)/(B+J) \rceil$. For a 55.64 in arc with 2.5 in units and 0.375 in joints, that comes out to 20 units and 19 joints.
What’s the maximum rise for a given arch span?
Half the span. At exactly rise = span/2 the arch is a full semicircle with its radius equal to the rise; push the rise any higher and there’s no single circular arc with a level springline that fits, so the geometry breaks down.
Why doesn’t the fit allowance come out to zero?
Because the unit count is always rounded up to a whole number, laying that many units and joints end to end almost always runs a little longer than the actual arc length. That small excess, 1.49 in in the 48 in span example, is normal and gets absorbed in the mortar joints or trimmed from the end pieces.
What does a rise-to-span ratio like 1:4 mean?
It means for every 1 unit of rise there are 4 units of span, a moderately flat segmental arch. A 1:2 ratio is a full semicircle, the roundest a single-radius arch gets; ratios above that, 1:6, 1:8, and beyond, describe progressively flatter arches.
Does the arch depth change how many masonry units I need?
No. Depth only multiplies into the intrados surface area calculation; it has no effect on the radius, arc length, or unit count, which are driven entirely by span, rise, unit width, and joint thickness.
Can mortar joints be zero width?
Yes, joint thickness can be set to zero for dry-stacked or butted units. Unit width can’t be zero, though, since it’s the denominator that determines how many units fit along the arc.