Brick Calculator (Patios, Paths, Garden Edges and Low Walls)

Choose paving or a wall, then enter the area (or the length and height), the brick size, the joint width and the waste factor. The bricks per pallet, price, mortar bag and bricks you have can be left blank (then they are not calculated).

Units

What are you building? Brick size

Length (mm)

Width (mm)

Height (mm)

How the bricks lie (face up) How to enter the paving size

Length (m)

Width (m)

Joint width between bricks (mm)

Waste factor (%)

Bricks per pallet (optional)

Brick price ($, optional)

Joint mortar (when the joint width is more than 0)

Mixed mortar density (kg/L)

Yield per bag (L, optional)

Bricks you have (optional, reverse)

For sand joints (bricks laid tight with sand swept in), 0 to 3 mm is common; for mortar joints, about 10 mm. A waste factor of 5 to 10% is common, and 10 to 15% for patterns with angled cuts such as herringbone. With sand joints, the "joint mortar" volume works as a guide for the joint sand (for its weight, set the density to about 1.5 for dry sand).

Result and figure

On the left, choose paving or a wall, enter the area (or the length and height) and the brick size, and press "Calculate". The result will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formulas and figures
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

For paving (paths, patios, the floor of a garden bed), enter the area (length × width, square feet or a circle), the brick size, how the bricks lie and the joint width, and you get the number of bricks on the spot
For a wall (garden bed edges, low walls), the page finds the bricks per course, the number of courses and the total from the length and height (or the courses)
Choose a modular brick (7-5/8 × 3-5/8 × 2-1/4 in) or a firebrick (9 × 4-1/2 × 2-1/2 in) from the list, or type in the size of used or other bricks
At the same time you get the bricks needed with waste (breakage, cuts and spares), the number of pallets, the joint mortar (ft³, pounds and bags), and an estimated cost if you enter a price
Enter the bricks you have, and you also find the area you can pave or the length of wall you can build with them

The counts are estimates based on "area ÷ area per brick" and "bricks per course × courses". The real count depends on the pattern and on how many cut pieces the shape leaves, so allow extra with the waste factor. For thin tiles, use the "Tile Calculator". Concrete block walls (such as 8 × 8 × 16 in CMU) are not covered here, since the core fill and rebar work differently; use the "Concrete Block Calculator" for them.

What is this calculation used for?

Edging a garden bed with bricks (a DIY classic)

For example, the edge around a 5 × 3 ft garden bed is 16 ft (192 in) long. With modular bricks and 3/8 in joints, each course takes $\lceil 192 \div 8 \rceil = 24$ bricks, 3 courses (about 7-7/8 in high) take 72 net, and with a 5% waste factor you need $\lceil 72 \times 1.05 \rceil = 76$ bricks.
A bed has four corners where bricks are cut or offset, so buying a few extra keeps the work from stopping. You can work out the soil to fill it with the "Soil Calculator".

Paving a front walkway or garden path

A path 2 ft wide and 12 ft long (24 ft²) laid flat with 1/8 in sand joints gives an area per brick of $7.75 \times 3.75 \approx 29.06$ in², so the net count is $24 \times 144 \div 29.06 \approx 118.9$, and with a 5% waste factor, $\lceil 24 \times 144 \times 1.05 \div 29.06 \rceil = 125$ bricks.
A path is narrow, so the edges make up a large share and you cut bricks more often. For a curved path, a waste factor closer to 10% is safer.

A patio or barbecue area floor

A 10 × 6 ft (60 ft²) patio of modular bricks laid flat with 3/8 in joints takes 270 net, or 284 with a 5% waste factor, and about 1.53 ft³ of joint mortar. The same area with the bricks on end (the 3-5/8 × 2-1/4 in face up) gives an area per brick of $4 \times 2.625 = 10.5$ in², so the net count is $60 \times 144 \div 10.5 \approx 823$, more than three times as many.
The larger the area, the more the way the bricks lie affects the count and the cost, so compare both the look and the numbers before you decide.

Building a pizza oven or fire pit with firebrick

For a firebox wall 5 ft (60 in) long and 18 in high in firebrick (9 × 4-1/2 × 2-1/2 in) with 1/8 in joints, each course takes $\lceil 60 \div 9.125 \rceil = 7$ bricks and the courses are $\lceil 18 \div 2.625 \rceil = 7$, so 49 net, or 54 with a 10% waste factor.
Firebrick costs more per brick than common brick, so an accurate count goes straight into the cost. For the refractory mortar and how to use it, follow the product instructions.

Reading a contractor's estimate

A contractor's estimate lists quantities such as "brick paving 60 sq ft" or "brick edging 10 LF × 4 courses". Knowing these formulas, you can follow which sizes and which waste factor the brick count came from, and sort out what to ask in the meeting. (A real estimate also includes the gravel base, bedding sand, mortar and labor, so the brick count alone cannot tell you whether the price is fair.)

Volunteering to build garden beds at a school or park

When volunteers build garden beds at a school or park, working out the bricks in advance helps with budget requests and cuts down on supply runs. For example, a bed 26 ft around with a 2-course edge takes $\lceil 312 \div 8 \rceil = 39$ bricks per course × 2 = 78 net, or 82 with a 5% waste factor.
If bricks are left from last time, following the formula backward tells you how long an edge they will make. For example, 40 bricks in 2 courses is 20 per course, and at an 8 in horizontal pitch that is 160 in, about 13.3 ft of edging.

Formulas and figures

Area per brick with joint

Figure

Standard notation (the usual math form)

$a$ $=$ $($ $p$ $+$ $g$ $)$ $\times$ $($ $q$ $+$ $g$ $)$

In words (symbols replaced with words)

④ $a$: area per brick with joint $=$ $($ ① $p$: long side of the face $+$ ② $g$: joint width $)$ $\times$ $($ ③ $q$: short side of the face $+$ $g$: joint width $)$

The formula in words

① Take the $p$: long side of the face (the side showing)

② add the $g$: joint width to get the space one brick takes along the long side,

③ add the same joint width to the $q$: short side of the face to get the space along the short side, and multiply the two

④ to get the $a$: area per brick with joint

Quick example

For modular bricks (7-5/8 × 3-5/8 × 2-1/4 in) laid flat (the 7-5/8 × 3-5/8 in face up) with 3/8 in joints, the area one brick takes is

$a$: area per brick $=$ $($ length (7.625 in) $+$ joint (0.375 in) $)$ $\times$ $($ width (3.625 in) $+$ joint (0.375 in) $)$

$(7.625 + 0.375) \times (3.625 + 0.375) = 8 \times 4 = 32\,\mathrm{in^2}$

$144 \div 32 = 4.5\ \text{bricks per ft}^2$

Key idea

The "face" here is the side of the brick that shows: on top when paving, or on the front of a wall. Which side shows depends on how the brick lies: laid flat it is length × width (7-5/8 × 3-5/8), on edge it is length × height (7-5/8 × 2-1/4), and on end it is width × height (3-5/8 × 2-1/4). The smaller the face showing, the smaller the area per brick, so the same area takes more bricks. Bricks are set with a joint (a gap) between them, so the space one brick really takes is "face size + joint width". The key is to add the joint width to both sides. For bricks laid tight with sand joints, use $g = 0$. Put all sizes in the same unit before calculating (square feet × 144 = square inches).

Bricks for paving (with waste)

Figure

Standard notation (the usual math form)

$N$ $=$ $\lceil$ $S$ $\times$ $($ $1 +$ $r$ $)$ $\div$ $a$ $\rceil$

In words (symbols replaced with words)

④ $N$: bricks needed with waste $=$ $\lceil$ ① $S$: paving area $\times$ $($ $1 +$ ② $r$: waste factor $)$ $\div$ ③ $a$: area per brick with joint $\rceil$

The formula in words

① Take the $S$: paving area

② multiply it by "1 + $r$: waste factor " to add extra for breakage, cuts and spares,

③ divide by the $a$: area per brick with joint to see how many bricks it is, and round up any decimal (the symbol $\lceil\ \rceil$ means "round up")

④ to get the $N$: bricks needed with waste

Quick example

For a 10 × 6 ft (60 ft²) patio with the bricks above (32 in² each) and a 5% waste factor, the bricks needed are

$N$: bricks needed $=$ $\lceil$ paving area (60 ft² = 8,640 in²) $\times$ $($ $1 +$ waste factor (0.05) $)$ $\div$ area per brick (32 in²) $\rceil$

$N_{0} = 60 \times 144 \div 32 = 8640 \div 32 = 270$

$8640 \times (1 + 0.05) \div 32 = 9072 \div 32 = 283.5$

$\lceil 283.5 \rceil = 284$

Key idea

"Paving area ÷ area per brick" before the waste factor is the net count $N_0$ (270 bricks in the example, which is 4.5 bricks per square foot), and the result shows it too. Bricks are sold one at a time, so always round up any decimal. Rounding down or to the nearest leaves you short. The waste factor $r$ covers cut-offs at the edges, bricks broken in delivery or during the work, and spares for relaying, and 5 to 10% is common. It depends on the pattern: running bond (each row offset by half a brick) wastes less, since a cut piece can start the next row, while herringbone, cut at 45° along the edges, leaves more waste, around 10 to 15%. The waste factor is applied to the area as "1 + r" (1.05 for 5%).

Bricks per course and number of courses for a wall

Figure

Standard notation (the usual math form)

$n$ $=$ $\lceil$ $L$ $\div$ $($ $l$ $+$ $g$ $)$ $\rceil$

$C$ $=$ $\lceil$ $H$ $\div$ $($ $h$ $+$ $g$ $)$ $\rceil$

In words (symbols replaced with words)

③ $n$: bricks per course $=$ $\lceil$ ① $L$: wall length $\div$ $($ ② $l$: brick length $+$ $g$: joint width $)$ $\rceil$

⑥ $C$: courses $=$ $\lceil$ ④ $H$: wall height $\div$ $($ ⑤ $h$: brick height $+$ $g$: joint width $)$ $\rceil$

The formula in words

① Take the $L$: wall length

② divide it by the $l$: brick length plus the joint width $g$ (the horizontal pitch of one brick), and round up any decimal

③ to get the $n$: bricks per course

④ Take the $H$: wall height

⑤ divide it by the $h$: brick height plus the joint width $g$ (the vertical pitch of one course), and round up any decimal

⑥ to get the $C$: courses

Quick example

For a garden bed edge 10 ft (120 in) long and 12 in high, built with modular bricks and 3/8 in joints, the bricks per course and the courses are

$n$: bricks per course $=$ $\lceil$ length (120 in) $\div$ $($ length (7.625 in) $+$ joint (0.375 in) $)$ $\rceil$

$C$: courses $=$ $\lceil$ height (12 in) $\div$ $($ height (2.25 in) $+$ joint (0.375 in) $)$ $\rceil$

$120 \div (7.625 + 0.375) = 120 \div 8 = 15 \quad \rightarrow \quad n = 15$

$12 \div (2.25 + 0.375) = 12 \div 2.625 \approx 4.57 \quad \rightarrow \quad \lceil 4.57 \rceil = 5$

Key idea

For a wall, count "how many bricks in a row" and "how many rows high" separately, not by area. The horizontal pitch is "brick length + joint width", and the vertical pitch is "brick height + joint width" (in a wall, the length × height face is on the front and the width is the wall thickness). Cutting bricks to fit at the ends, or offsetting each course by half a brick (running bond), puts half bricks at the ends of courses, but one brick cut in two makes two halves, so "bricks per course × courses" still counts right. If you want to set the courses first, enter them directly; "courses × vertical pitch" is then the built height (5 courses × 2.625 in = 13.125 in in the example). Estimating tables for modular brick often use about 6.75 bricks per square foot of wall (3 courses to 8 in); this calculator uses the actual pitch, which gives about 6.86.

Bricks for a wall (with waste)

Standard notation (the usual math form)

$N$ $=$ $\lceil$ $n$ $\times$ $C$ $\times$ $($ $1 +$ $r$ $)$ $\rceil$

In words (symbols replaced with words)

④ $N$: bricks needed with waste $=$ $\lceil$ ① $n$: bricks per course $\times$ ② $C$: courses $\times$ $($ $1 +$ ③ $r$: waste factor $)$ $\rceil$

The formula in words

① Take the $n$: bricks per course

② multiply it by the $C$: courses to get the net count $N_0$,

③ then multiply by "1 + $r$: waste factor " and round up any decimal

④ to get the $N$: bricks needed with waste

Quick example

For the example above (15 per course, 5 courses) with a 5% waste factor, the bricks needed are

$N$: bricks needed $=$ $\lceil$ per course (15) $\times$ courses (5) $\times$ $($ $1 +$ waste factor (0.05) $)$ $\rceil$

$15 \times 5 = 75$

$75 \times 1.05 = 78.75 \quad \rightarrow \quad \lceil 78.75 \rceil = 79$

Key idea

Add the waste factor for breakage and cuts to the net count $N_0 = n \times C$ (75 bricks in the example). A wall is only cut at the ends of courses, so it usually needs less waste than paving (around 5%), but a bed with many corners or a curved edge needs more cuts, so allowing extra is safer. When you buy by the pallet, divide the bricks needed $N$ by the bricks per pallet $k$ and round up: $B = \lceil N \div k \rceil$. The 79 bricks in the example, from pallets of 500, is $\lceil 79 \div 500 \rceil = 1$ pallet (for a small job, buying loose bricks is usually better).

Joint mortar

Standard notation (the usual math form)

$V$ $=$ $N_{0}$ $\times$ $($ $a$ $-$ $p\,q$ $)$ $\times$ $t$

In words (symbols replaced with words)

⑤ $V$: joint mortar $=$ ① $N_{0}$: net bricks $\times$ $($ ② $a$: area per brick with joint $-$ ③ $p\,q$: face area of one brick $)$ $\times$ ④ $t$: joint depth

The formula in words

① Take the $N_{0}$: net bricks

② multiply it by the $a$: area per brick with joint minus

③ $p\,q$: face area of one brick (the joint area around one brick),

④ then by the $t$: joint depth (the depth of the brick)

⑤ to get the $V$: joint mortar

Quick example

For the patio above (270 net bricks, 32 in² each, face 7.625 × 3.625 = 27.640625 in², and since they lie flat, the joint depth is the brick height, 2.25 in), the joint mortar is

$V$: mortar $=$ net (270) $\times$ $($ area per brick (32 in²) $-$ face area (27.64 in²) $)$ $\times$ depth (2.25 in)

$32 - 27.640625 = 4.359375\,\mathrm{in^2}$

$270 \times 4.359375 \times 2.25 \div 1728 \approx 1.53\,\mathrm{ft^3}$

Key idea

The area per brick minus the face area is the joint area that belongs to one brick (the L-shaped gap on its right and top). Multiply by the joint depth (the depth of the brick, the dimension that does not show: the height when laid flat, or the width in a wall) to get the joint volume per brick, and multiply by the net count for the whole job. Divide cubic inches by 1,728 to get cubic feet. The result also shows this net amount with the waste factor (for squeeze-out and leftover mortar), its weight using the mixed density (about 125 lb per ft³), and the bags, found by dividing by the yield per bag and rounding up. This assumes the joints are completely filled, so if you fill the bottom of the joints with sand and only top them with mortar, you need less. To mix your own from cement and sand, use the "Mortar and Concrete Mix Calculator" for the amounts. A wall works the same way: think of mortar in the joints around the front face (length × height), as thick as the wall (the brick width). That means counting one bed joint (the mortar bed under the brick) and one head joint beside each brick, the same as for a concrete block wall. With sand joints (0 to 1/8 in), the volume from this formula works as a guide for the joint sand. Sand is lighter than mortar (dry sand is about 94 lb per ft³), so change the density to see its weight.

Estimated cost

Standard notation (the usual math form)

In words (symbols replaced with words)

$T$ $=$ $u$ $\times$ $Q$

③ $T$: estimated cost $=$ ① $u$: unit price $\times$ ② $Q$: quantity

The formula in words

① Take the $u$: unit price (per brick or per pallet)

② multiply it by the $Q$: quantity (bricks or pallets with waste, to match the price)

③ to get the $T$: estimated cost

Quick example

The cost of 284 bricks (with waste) at $0.75 each is

$T$: estimated cost $=$ unit price ($0.75) $\times$ quantity (284)

$0.75 \times 284 = 213$

Key idea

The key is to match the unit of the price and the quantity. For a price per brick, multiply by the brick count $N$; for a price per pallet, multiply by the pallets $B$. The cost on this page uses the bricks with waste. Besides the bricks, the joint mortar, the gravel and sand base, adhesive, tools and delivery cost extra.

For paving, the bricks needed are "paving area × (1 + waste factor) ÷ area per brick with joint", rounded up. For a wall, they are "bricks per course (length ÷ horizontal pitch) × courses (height ÷ vertical pitch) × (1 + waste factor)", rounded up. The joint mortar is "net bricks × joint area around one brick × joint depth".

Symbols and terms

Symbols

$l$	el	The brick length (the longest side), from the first letter of "length". 7-5/8 in (7.625 in) for a modular brick.
$w$	double-u	The brick width (the shorter side seen from above), from the first letter of "width". 3-5/8 in for a modular brick.
$h$	aitch	The brick height (thickness), from the first letter of "height". 2-1/4 in for a modular brick.
$p,\ q$	pee, cue	The long and short sides of the face (the side on top when paving, or on the front of a wall). Which two of $l, w, h$ they are depends on how the brick lies (laid flat, $p = l,\ q = w$).
$g$	gee	The joint width (the gap between bricks), from the first letter of "gap". 0 to 1/8 in for sand joints, 3/8 in for mortar joints.
$a$	ay	The area per brick with joint, found with $a = (p+g)(q+g)$. From the first letter of "area".
$S$	ess	The paving area (the size of the path, patio or bed floor), given as length × width, total area (ft²) or the diameter of a circle.
$r$	ar	The waste factor, the extra for breakage, cuts and spares. In the formulas it is a decimal (0.05 for 5%) applied as "1 + r". From the first letter of "rate".
$N_{0}$	N sub zero	The net bricks (without waste). $S \div a$ for paving and $n \times C$ for a wall.
$N$	en	The bricks needed with waste, found with $N = \lceil N_{0}(1+r) \rceil$. From the first letter of "number".
$L$	capital el	The wall length (for a garden bed edge, the total around the bed).
$H$	capital aitch	The wall height. It is not used when you enter the courses directly.
$n$	small en	The bricks per course, found with $n = \lceil L \div (l+g) \rceil$.
$C$	see	The number of courses, found with $C = \lceil H \div (h+g) \rceil$ or entered directly. From the first letter of "course".
$t$	tee	The joint depth, which is the depth of the brick (the dimension that does not show). The height $h$ when laid flat, the width $w$ in a wall. From the first letter of "thickness".
$V$	vee	The joint mortar (volume), found with $V = N_{0}(a - pq)t$. From the first letter of "volume".
$k$	kay	The bricks per pallet.
$B$	bee	The pallets needed, found with $B = \lceil N \div k \rceil$. From the first letter of "bundle".
$u$	you	The brick price (per brick or per pallet), from the first letter of "unit price".
$Q$	capital cue	The quantity that matches the price unit (bricks or pallets), from the first letter of "quantity".
$T$	capital tee	The estimated cost (bricks only), from the first letter of "total".
$\lceil x \rceil$	ceiling of x	The symbol for rounding up to a whole number, called the ceiling function. (Example - $\lceil 283.5 \rceil = 284$, $\lceil 4 \rceil = 4$)

Terms

brick	A building unit of fired clay. The common US size is the modular brick, 7-5/8 × 3-5/8 × 2-1/4 in (8 × 4 × 2-2/3 in nominal with a 3/8 in joint), used for garden edges, paths, patios and walls. Paving bricks (pavers) are often a full 4 × 8 in, and used or imported bricks vary, so measure them and enter the size.
firebrick	A brick made to stand high heat, used for pizza ovens, barbecue pits, fire pits and around wood stoves. The standard size is 9 × 4-1/2 × 2-1/2 in (the "9-inch straight"), a little larger than a common brick, so choose the right size when counting.
joint	The gap between bricks. Mortar joints are about 3/8 in, and sand joints (bricks laid tight with sand swept in) are 0 to 1/8 in. The joint makes each brick take a little more area, so the count goes down slightly.
face	The side of the brick that shows, on top when paving or on the front of a wall. Which side shows depends on how the brick lies.
laid flat	Paving with the largest face (length × width, 7-5/8 × 3-5/8 in) on top. Each brick covers the most area, so the same area takes the fewest bricks. This is the most common way for paths and patios.
laid on edge	Paving with the brick on its long edge, showing the length × height face (7-5/8 × 2-1/4 in). The narrow faces are used for path borders and accent patterns, and a paving laid on edge is also very strong.
laid on end	Setting the brick upright with the smallest face (width × height, 3-5/8 × 2-1/4 in) on top. It is often used for a garden bed edge of bricks standing in a row (a soldier course edging), and takes the most bricks for the same area.
running bond	The most basic pattern, with each row offset by half a brick. The joints do not line up, which makes it strong, and the other half of a brick cut at the end can start the next row, so there is little waste. The figures on this page use this pattern.
herringbone	A pattern of bricks set at right angles to each other in a zigzag (like a fish skeleton). It looks striking, but the edges need many 45° cuts, so allow 10 to 15% waste.
basket weave	A pattern of pairs of bricks laid alternately across and along, like a woven basket. It works because two brick widths plus a joint equal about one brick length (for pavers, a full 4 × 8 in fits exactly). The waste is about the same as running bond.
sand joint	Laying bricks tight (or with a gap of up to about 1/8 in) and sweeping fine sand (such as polymeric sand) into the joints. With no mortar, it is easy to redo or replace bricks, but weeds and settling happen more easily.
mortar joint	Filling the gap between bricks (about 3/8 in) with mortar to set them. Walls are always built this way, with each course set on a bed of mortar and the head joints filled too. The joint mortar can be estimated with the formula on this page.
waste factor	The extra, in percent, for cuts, cut-offs, breakage and spares on top of the net amount. 5 to 10% is common, and 10 to 15% for angled patterns or shapes with many corners.
cut-off	The leftover piece when a brick is cut to fit at an edge or corner. How much there is depends on the pattern and shape, and not all of it can be used again, so it is covered by the waste factor.
net	The bare minimum amount, without waste or spares. On this page, "paving area ÷ area per brick" and "bricks per course × courses" are the net counts.
course	One horizontal row of bricks in a wall. A garden edge "4 courses high" is 4 rows stacked (about 10-1/2 in with modular bricks and joints).
mixed mortar	Mortar after cement, sand and water are mixed, ready to use. An 80 lb bag of mortar mix makes roughly 0.6 to 0.7 ft³ (follow the bag). Mixed mortar weighs about 125 lb per ft³.
pallet	A unit for buying bricks in bulk, often about 500 (called a cube). Buying by the pallet from a masonry supplier can lower the price per brick. Enter the bricks per pallet to get the pallets needed, rounded up.
rounding up	Changing a number with a decimal part to the next whole number. Bricks are sold one at a time, so bricks, courses, pallets and bags are always rounded up.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.

Area of rectangles and circles (Grades 3–7)	Knowing that the area of a rectangle is length × width Knowing that the area of a circle is radius × radius × pi (or, from the diameter, diameter × diameter × pi ÷ 4)
Converting units of length and area (Grades 4–6)	Knowing that 1 ft = 12 in Knowing that converting an area multiplies by the conversion number twice, as in $1\,\mathrm{ft^2} = 12 \times 12 = 144\,\mathrm{in^2}$
Multiplying and dividing decimals (Grade 5)	Understanding decimal calculations such as $7.625 + 0.375$ and $12 \div 2.625$ (a calculator can do the arithmetic)
Percents (Grade 6)	Knowing that "5% more" is the same as "× 1.05"
Rounding (Grades 3–4)	Knowing the difference between rounding up, rounding down and rounding to the nearest Being able to explain in your own words why material counts are rounded up
Volume of a box (Grade 5)	Knowing that volume is area × height (depth), and that $1\,\mathrm{ft^3} = 12 \times 12 \times 12 = 1728\,\mathrm{in^3}$

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the area per brick with joint

Long side of the face p (in)	7.625
Short side of the face q (in)	3.625
Joint width g (in)	0.375
Area per brick a (in²)	=(B1+B3)*(B2+B3)

Table to find the bricks for paving (with waste)

Paving area S (ft²)	60
Waste factor r (%)	5
Area per brick with joint a (in²)	32
Net bricks (before rounding up)	=B1*144/B3
Bricks needed with waste N	=ROUNDUP(B1144(1+B2/100)/B3,0)

Table to find the bricks per course and courses for a wall

Wall length L (ft)	10
Wall height H (in)	12
Brick length l (in)	7.625
Brick height h (in)	2.25
Joint width g (in)	0.375
Bricks per course n	=ROUNDUP(B1*12/(B3+B5),0)
Courses C	=ROUNDUP(B2/(B4+B5),0)

Table to find the bricks for a wall (with waste)

Bricks per course n	15
Courses C	5
Waste factor r (%)	5
Net bricks	=B1*B2
Bricks needed with waste N	=ROUNDUP(B1B2(1+B3/100),0)

Table to find the joint mortar

Net bricks N0	270
Area per brick with joint a (in²)	32
Face area of one brick pq (in²)	27.640625
Joint depth t (in)	2.25
Joint mortar V (ft³)	=B1(B2-B3)B4/1728

Table to find the estimated cost

Unit price ($)	0.75
Quantity (bricks or pallets)	284
Estimated cost ($)	=B1*B2

After pasting, the upper rows of column B are your inputs and the last rows are calculated automatically.
"ROUNDUP(value, 0)" rounds up to a whole number (the ⌈ ⌉ in the formulas).
Square feet are turned into square inches with "*144", feet into inches with "*12", and cubic inches into cubic feet with "/1728". B4 in the first table shows 32.
In the second table, B4 is 270 and B5 is 284 bricks. In the third, B6 is 15 bricks and B7 is 5 courses. In the fourth, B4 is 75 and B5 is 79 bricks. In the fifth, B5 is about 1.53 ft³. In the sixth, B3 is $213. Just change column B to your own numbers.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the area per brick with joint

Long side of the face p (in)	7.625
Short side of the face q (in)	3.625
Joint width g (in)	0.375
Area per brick a (in²)	=(B1+B3)*(B2+B3)

Table to find the bricks for paving (with waste)

Paving area S (ft²)	60
Waste factor r (%)	5
Area per brick with joint a (in²)	32
Net bricks (before rounding up)	=B1*144/B3
Bricks needed with waste N	=ROUNDUP(B1144(1+B2/100)/B3,0)

Table to find the bricks per course and courses for a wall

Wall length L (ft)	10
Wall height H (in)	12
Brick length l (in)	7.625
Brick height h (in)	2.25
Joint width g (in)	0.375
Bricks per course n	=ROUNDUP(B1*12/(B3+B5),0)
Courses C	=ROUNDUP(B2/(B4+B5),0)

Table to find the bricks for a wall (with waste)

Bricks per course n	15
Courses C	5
Waste factor r (%)	5
Net bricks	=B1*B2
Bricks needed with waste N	=ROUNDUP(B1B2(1+B3/100),0)

Table to find the joint mortar

Net bricks N0	270
Area per brick with joint a (in²)	32
Face area of one brick pq (in²)	27.640625
Joint depth t (in)	2.25
Joint mortar V (ft³)	=B1(B2-B3)B4/1728

Table to find the estimated cost

Unit price ($)	0.75
Quantity (bricks or pallets)	284
Estimated cost ($)	=B1*B2

The same formulas as in Excel work as is (ROUNDUP has the same name). Copy the whole table, paste it into cell A1, and replace the numbers in column B with your own.

How to calculate it in Python

import math

# ===== Brick and joint (in) =====
brick_length_in = 7.625  # brick length (7-5/8)
brick_width_in = 3.625   # brick width (3-5/8)
brick_height_in = 2.25   # brick height, thickness (2-1/4)
gap_in = 0.375           # joint width (0 for sand joints)
loss_rate = 0.05         # waste factor (5% -> 0.05)

# ===== Paving: laid flat (length x width face up) =====
area_ft2 = 10 * 6                                  # paving area (10 ft x 6 ft)
face_p = brick_length_in                           # long side of the face (in)
face_q = brick_width_in                            # short side of the face (in)
joint_depth = brick_height_in                      # joint depth = brick height when laid flat (in)
brick_area = (face_p + gap_in) * (face_q + gap_in) # area per brick with joint (in2)
net_count = area_ft2 * 144 / brick_area            # net bricks (before rounding up)
bricks_needed = math.ceil(area_ft2 * 144 * (1 + loss_rate) / brick_area)   # bricks needed with waste
mortar_ft3 = net_count * (brick_area - face_p * face_q) * joint_depth / 1728   # joint mortar (ft3)

print(f"[Paving] Area per brick: {brick_area:.4f} in2")
print(f"[Paving] Net bricks: {net_count:.2f} -> rounded up {math.ceil(net_count)}")
print(f"[Paving] Bricks needed with waste: {bricks_needed}")
print(f"[Paving] Joint mortar: {mortar_ft3:.2f} ft3")

# ===== Wall: 10 ft long, 12 in high =====
wall_length_in = 10 * 12
wall_height_in = 12
per_course = math.ceil(wall_length_in / (brick_length_in + gap_in))    # bricks per course
courses = math.ceil(wall_height_in / (brick_height_in + gap_in))       # courses
stack_net = per_course * courses                                       # net bricks
stack_needed = math.ceil(stack_net * (1 + loss_rate))                  # bricks needed with waste

print(f"[Wall] {per_course} per course x {courses} courses = {stack_net} net bricks")
print(f"[Wall] Bricks needed with waste: {stack_needed}")

Runs with the standard library only. math.ceil() rounds up (the ⌈ ⌉ in the formulas). Replace the sizes, area and waste factor at the top with your own numbers and run it (the example gives 284 bricks and about 1.53 ft³ of mortar for paving, and 15 × 5 = 75 → 79 bricks for the wall).

How to write it in LaTeX and other math languages (copy and paste)

Area per brick with joint

a = (p + g) × (q + g)

a = (p + g)(q + g)

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>a</mi>
    <mo>=</mo>
    <mrow><mo>(</mo><mi>p</mi><mo>+</mo><mi>g</mi><mo>)</mo></mrow>
    <mo>&#x2062;</mo>
    <mrow><mo>(</mo><mi>q</mi><mo>+</mo><mi>g</mi><mo>)</mo></mrow>
  </mrow>
</math>

a = (p + g)(q + g)

(p + g)*(q + g)

a := (p + g)*(q + g);

a = (p + g)*(q + g);

a = (p + g)(q + g)

Bricks for paving (with waste)

N = ⌈S × (1 + r) ÷ a⌉

N = \left\lceil \frac{S (1 + r)}{a} \right\rceil

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>N</mi>
    <mo>=</mo>
    <mo>&#x2308;</mo>
    <mfrac>
      <mrow><mi>S</mi><mo>&#x2062;</mo><mo>(</mo><mn>1</mn><mo>+</mo><mi>r</mi><mo>)</mo></mrow>
      <mi>a</mi>
    </mfrac>
    <mo>&#x2309;</mo>
  </mrow>
</math>

N = |~ (S (1 + r)) / a ~|

Ceiling[s (1 + r)/a]

N := ceil(S*(1 + r)/a);

N = ceil(S*(1 + r)/a);

N = ⌈S(1 + r)/a⌉

Bricks per course and number of courses for a wall

n = ⌈L ÷ (l + g)⌉,  C = ⌈H ÷ (h + g)⌉

n = \left\lceil \frac{L}{l + g} \right\rceil,\quad C = \left\lceil \frac{H}{h + g} \right\rceil

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>n</mi>
    <mo>=</mo>
    <mo>&#x2308;</mo>
    <mfrac><mi>L</mi><mrow><mi>l</mi><mo>+</mo><mi>g</mi></mrow></mfrac>
    <mo>&#x2309;</mo>
    <mo>,</mo>
    <mi>C</mi>
    <mo>=</mo>
    <mo>&#x2308;</mo>
    <mfrac><mi>H</mi><mrow><mi>h</mi><mo>+</mo><mi>g</mi></mrow></mfrac>
    <mo>&#x2309;</mo>
  </mrow>
</math>

n = |~ L / (l + g) ~|,  C = |~ H / (h + g) ~|

{Ceiling[L/(l + g)], Ceiling[H/(h + g)]}

n := ceil(L/(l + g));  C := ceil(H/(h + g));

n = ceil(L/(l + g)); C = ceil(H/(h + g));

n = ⌈L/(l + g)⌉, C = ⌈H/(h + g)⌉

Bricks for a wall (with waste)

N = ⌈n × C × (1 + r)⌉

N = \lceil n C (1 + r) \rceil

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>N</mi>
    <mo>=</mo>
    <mo>&#x2308;</mo>
    <mi>n</mi><mo>&#x2062;</mo><mi>C</mi><mo>&#x2062;</mo>
    <mo>(</mo><mn>1</mn><mo>+</mo><mi>r</mi><mo>)</mo>
    <mo>&#x2309;</mo>
  </mrow>
</math>

N = |~ n C (1 + r) ~|

Ceiling[n c (1 + r)]

N := ceil(n*C*(1 + r));

N = ceil(n*C*(1 + r));

N = ⌈nC(1 + r)⌉

Joint mortar

V = N₀ × (a − p q) × t

V = N_{0} (a - p q) t

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>V</mi>
    <mo>=</mo>
    <msub><mi>N</mi><mn>0</mn></msub>
    <mo>&#x2062;</mo>
    <mrow><mo>(</mo><mi>a</mi><mo>-</mo><mi>p</mi><mo>&#x2062;</mo><mi>q</mi><mo>)</mo></mrow>
    <mo>&#x2062;</mo>
    <mi>t</mi>
  </mrow>
</math>

V = N_0 (a - p q) t

n0 (a - p q) t

V := N0*(a - p*q)*t;

V = N0*(a - p*q)*t;

V = N_0 (a − pq) t

Estimated cost

T = u × Q

T = u \times Q

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>T</mi>
    <mo>=</mo>
    <mi>u</mi>
    <mo>&#xD7;</mo>
    <mi>Q</mi>
  </mrow>
</math>

T = u * Q

u*q

T := u*Q;

T = u*Q;

T = u × Q

How to have ChatGPT do the calculation

You are a quantity calculation assistant for brick work. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

I am using modular bricks (7-5/8 in × 3-5/8 in × 2-1/4 in). The joint width is 3/8 in and the waste factor is 5%.
1. For a 10 ft × 6 ft patio with the bricks laid flat (7-5/8 × 3-5/8 in face up), find the area per brick with joint (in²), the net bricks (area ÷ area per brick, with 1 ft² = 144 in²), and the bricks needed with waste (net × 1.05, rounded up).
2. For a garden bed edge 10 ft long and 12 in high, find the bricks per course (length ÷ (7-5/8 + 3/8) in, rounded up), the courses (height ÷ (2-1/4 + 3/8) in, rounded up), and the bricks needed with waste.
3. For item 1, find the joint mortar (net bricks × (area per brick − 7.625 × 3.625) × 2.25 in) in cubic feet (1 ft³ = 1,728 in³).

Show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula

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\(l\)	el	The brick length (the longest side), from the first letter of "length". 7-5/8 in (7.625 in) for a modular brick.
\(w\)	double-u	The brick width (the shorter side seen from above), from the first letter of "width". 3-5/8 in for a modular brick.
\(h\)	aitch	The brick height (thickness), from the first letter of "height". 2-1/4 in for a modular brick.
\(p,\ q\)	pee, cue	The long and short sides of the face (the side on top when paving, or on the front of a wall). Which two of \(l, w, h\) they are depends on how the brick lies (laid flat, \(p = l,\ q = w\)).
\(g\)	gee	The joint width (the gap between bricks), from the first letter of "gap". 0 to 1/8 in for sand joints, 3/8 in for mortar joints.
\(a\)	ay	The area per brick with joint, found with \(a = (p+g)(q+g)\). From the first letter of "area".
\(S\)	ess	The paving area (the size of the path, patio or bed floor), given as length × width, total area (ft²) or the diameter of a circle.
\(r\)	ar	The waste factor, the extra for breakage, cuts and spares. In the formulas it is a decimal (0.05 for 5%) applied as "1 + r". From the first letter of "rate".
\(N_{0}\)	N sub zero	The net bricks (without waste). \(S \div a\) for paving and \(n \times C\) for a wall.
\(N\)	en	The bricks needed with waste, found with \(N = \lceil N_{0}(1+r) \rceil\). From the first letter of "number".
\(L\)	capital el	The wall length (for a garden bed edge, the total around the bed).
\(H\)	capital aitch	The wall height. It is not used when you enter the courses directly.
\(n\)	small en	The bricks per course, found with \(n = \lceil L \div (l+g) \rceil\).
\(C\)	see	The number of courses, found with \(C = \lceil H \div (h+g) \rceil\) or entered directly. From the first letter of "course".
\(t\)	tee	The joint depth, which is the depth of the brick (the dimension that does not show). The height \(h\) when laid flat, the width \(w\) in a wall. From the first letter of "thickness".
\(V\)	vee	The joint mortar (volume), found with \(V = N_{0}(a - pq)t\). From the first letter of "volume".
\(k\)	kay	The bricks per pallet.
\(B\)	bee	The pallets needed, found with \(B = \lceil N \div k \rceil\). From the first letter of "bundle".
\(u\)	you	The brick price (per brick or per pallet), from the first letter of "unit price".
\(Q\)	capital cue	The quantity that matches the price unit (bricks or pallets), from the first letter of "quantity".
\(T\)	capital tee	The estimated cost (bricks only), from the first letter of "total".
\(\lceil x \rceil\)	ceiling of x	The symbol for rounding up to a whole number, called the ceiling function. (Example - \(\lceil 283.5 \rceil = 284\), \(\lceil 4 \rceil = 4\))

Brick Calculator (Patios, Paths, Garden Edges and Low Walls)

Table of Contents

What you can do on this page

What is this calculation used for?

Formulas and figures

Symbols and terms

Symbols

Terms

Good to know before you start

How to calculate it in Excel

How to calculate it in Google Sheets

How to calculate it in Python

How to write it in LaTeX and other math languages (copy and paste)

How to have ChatGPT do the calculation

How to Use

DataChef Features

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