Sod Calculator (Square Feet, Pallets and Topdressing)

Enter the size of the lawn and how you lay the sod. The gap, waste factor, topdressing and prices can be left blank (then there is no waste, no topdressing and no cost).

Units

What to find

How to enter the lawn size

Length (m)

Width (m)

Laying pattern

Size of one piece (cm, length × width)

Area per bundle or roll (m²)

Pieces per bundle

Waste factor (%, optional)

Bag size (liters)

Topdressing depth over the sod (mm, optional)

Sod price ($, optional)

Topdressing price ($ per bag, optional)

Change the area and pieces per bundle or roll and the size of one piece to match the product (they vary by farm and product). Enter the topdressing in two parts - the thin layer spread over the sod, and the fill for the gaps between pieces in the spaced and checkerboard patterns.

Result and figure

Enter the lawn area and the laying pattern on the left 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

Enter the size of the lawn (length × width, square feet, the diameter of a circle, or acres), and you get the number of sod pallets and pieces you need on the spot. The area per pallet (often 400 to 500 ft²) and the pieces per pallet can be changed to match the product
Choose how you lay it: solid (pieces butted tight), spaced (with gaps) or checkerboard. For the spaced pattern, the share of the area covered by sod is worked out from the gap width, so you can see how much sod the gaps save (a checkerboard uses half)
Enter a waste factor (%) for pieces cut at the edges and damaged sod. The formula "amount needed = net amount × (1 + waste factor)" shows why you buy a little extra
From the topdressing depth and the depth to fill the gaps, the page also works out how much topdressing you need (ft³ and yd³) and the number of bags (such as 0.75 ft³ bags; the bag size can be changed)
You can also switch to the reverse: the area you can cover with the sod you have. Enter a sod price (per pallet or per ft²) and a topdressing price (per bag) to get an estimated cost. A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are also on this page

The numbers of pallets and pieces and the topdressing amount are estimates. The size of a sod piece and the area per pallet vary by farm and product, so enter the values shown for the product you buy. This page covers the sod itself and the topdressing spread on it. For gravel around the lawn, use the "Gravel Calculator", and for the soil under the sod or in garden beds, use the "Soil Calculator".

What is this calculation used for?

Laying sod yourself (planning a DIY order)

For solid laying, the pallets you need are close to the lawn area divided by the area per pallet. For example, for a yard 80 × 50 ft (4,000 ft²) with a 5% waste factor, the sod area with waste is $4000 \times 1.05 = 4200$ ft², so you need $\lceil 4200 \div 400 \rceil = 11$ pallets (1,650 pieces) of 400 ft² each.
Sod is a living plant, so it should be laid the day it arrives or the next day. Working out the amount with waste ahead of time lets you order everything at once, instead of letting sod dry out while you go back for more.

Saving on sod with the spaced or checkerboard pattern

On a large lawn, the amount of sod is the main cost. Laying 16 × 24 in pieces with 1.5 in gaps gives a share covered of about 0.861, a saving of about 14%, and a checkerboard uses half. For 8,000 ft² with a 5% waste factor, solid laying needs 21 pallets, the spaced pattern $\lceil 8000 \times 0.861 \times 1.05 \div 400 \rceil = 19$ pallets, and a checkerboard $\lceil 8000 \times 0.5 \times 1.05 \div 400 \rceil = 11$ pallets.
However, the open ground takes one or two growing seasons to fill in, and only spreading grasses (such as zoysia, Bermuda and St. Augustine) fill it well. In the meantime there is more weeding and more topdressing for the gaps. You can compare the sod you save with the extra work in numbers before choosing the pattern.

Deciding how many bags of topdressing to buy

After laying sod, spread a thin layer of topdressing (about 1/8 to 1/4 in) over it, and fill the gaps if you left any. For 400 ft² laid with 1.5 in gaps (share covered 0.861), 1/8 in over the sod and 3/4 in in the gaps take about 7.65 ft³, or 11 bags of 0.75 ft³. Laid solid, only the layer over the sod is needed: about 4.17 ft³, or 6 bags.
Even a thin layer adds up on a large lawn: 1/8 in over 1,000 ft² is about 10.4 ft³. Working out the bags ahead of time saves a second trip, and the same formula works for yearly topdressing later.

Checking the quantities in a landscaper's quote

A landscaping quote lists items like "sod 4,200 sq ft" and "topdressing ○ yd". Work out lawn area × share covered × (1 + waste factor) yourself, and you can see whether the difference from the quote is the waste factor or something else.
Knowing where the quantities come from lets you look at materials, labor, the soil bed and grading separately, and compare quotes more easily. (In real projects, the site condition and the grass variety change the amounts and prices, so the quantities alone cannot tell you whether a price is fair.)

Finding how far leftover sod will go

When sod is left over, or a neighbor gives you some, the reverse calculation tells you how far it will go. With 2 pallets of 400 ft², you can cover about 800 ft² laid solid, $800 \div 0.861 \approx 930$ ft² with 1.5 in gaps, and 1,600 ft² in a checkerboard.
Questions like "is there enough for the bare patch by the driveway?" or "do I have to use a checkerboard to cover the whole yard?" can be answered just by counting pallets.

Formulas and figures

Lawn area (rectangle, circle, acres)

Standard notation (the usual math form)

$S$ $=$ $l$ $\times$ $w$

$S$ $=$ $\pi$ $\times$ $\left(\dfrac{d}{2}\right)$ $2$

$S$ $=$ $x$ $\times$ $43560$

In words (symbols replaced with words)

③ $S$: lawn area (rectangle) $=$ ① $l$: length $\times$ ② $w$: width

⑥ $S$: lawn area (circle) $=$ $\pi$ $\times$ ④ half the diameter $d$ (the radius) ⑤ squared

⑧ $S$: lawn area (from acres) $=$ ⑦ $x$: size in acres $\times$ $43560$

The formula in words

① Take the $l$: length

② multiply it by the $w$: width

③ and you get the $S$: lawn area of a rectangle

④ Take half the diameter $d$ (the radius) ,

⑤ square it and multiply by pi $\pi$ (about 3.14)

⑥ and you get the $S$: lawn area of a circle

⑦ Multiply the $x$: size in acres by 43,560 (the ft² in 1 acre)

⑧ and you get the $S$: lawn area from acres

Quick example

The areas of a yard 80 ft long and 50 ft wide, a round lawn 30 ft across, and a quarter-acre lawn are

$S$: lawn area (rectangle) $=$ length (80 ft) $\times$ width (50 ft)

$80 \times 50 = 4000\,\mathrm{ft^2}$

$\pi \times \left(\dfrac{30}{2}\right)^{2} = \pi \times 225 \approx 706.86\,\mathrm{ft^2}$

$0.25 \times 43560 = 10890\,\mathrm{ft^2}$

Key idea

The lawn area is how much ground you will cover with sod, and it is the starting point for the number of pallets. For an L-shaped yard or one with curves, split it into rectangles and add their areas, or enter the square feet from a plan or quote directly under "Area". For a round lawn, measure the diameter; the area is half the diameter (the radius) squared, times pi. If you know the size in acres, multiply by 43,560 to get square feet (1 acre = 43,560 ft²).

Share covered by sod (by laying pattern)

Figure

Standard notation (the usual math form)

$c$ $=$ $1$

$c$ $=$ $\dfrac{a}{a + g}$ $\times$ $\dfrac{b}{b + g}$

$c$ $=$ $\dfrac{1}{2}$

In words (symbols replaced with words)

① $c$: share covered (solid) $=$ $1$

④ $c$: share covered (spaced) $=$ ② $\dfrac{\text{length } a}{\text{length } a + \text{gap } g}$ $\times$ ③ $\dfrac{\text{width } b}{\text{width } b + \text{gap } g}$

⑤ $c$: share covered (checkerboard) $=$ $\dfrac{1}{2}$

The formula in words

① Solid laying puts the pieces tight together, so the $c$: share covered is 1 (100%)

② For the spaced pattern, take length $a$ ÷ (length $a$ + gap $g$)

③ multiply it by width $b$ ÷ (width $b$ + gap $g$)

④ and you get the $c$: share covered for the spaced pattern

⑤ A checkerboard uses every other square, so the $c$: share covered is $\dfrac{1}{2}$ (50%)

Quick example

For 16 × 24 in pieces laid with 1.5 in gaps, the share covered by sod is

$c$: share covered $=$ 16 ÷ (16 + 1.5) $\times$ 24 ÷ (24 + 1.5)

$\dfrac{16}{17.5} \times \dfrac{24}{25.5} = \dfrac{384}{446.25} \approx 0.861\ \ (86.1\%)$

Key idea

Adding the gap $g$ to the length $a$ and width $b$ of one piece gives "$(a + g) \times (b + g)$", the area each piece takes up in the spaced pattern (the piece plus the gap to its right and below). The share of that area taken by the sod itself, $a \times b$, is $c$. Written in parts, it is the share along the length, $\dfrac{a}{a+g}$, times the share along the width, $\dfrac{b}{b+g}$. With 1.5 in gaps it is about 86%, so you need about 14% less sod. With a gap of 0, $c = 1$, the same as solid. A checkerboard uses every other square, so it is exactly half ($c = \dfrac{1}{2}$). In the US, sod is usually laid solid, with the pieces butted tight and the joints staggered like bricks. The spaced and checkerboard patterns save sod, but the open ground takes one or two growing seasons to fill in as the grass spreads by stolons (this works for spreading warm-season grasses such as zoysia, Bermuda and St. Augustine). Weeds come up easily in the meantime, so choose the pattern by weighing the sod you save against the extra work.

Sod area with waste

Standard notation (the usual math form)

$S'$ $=$ $S$ $\times$ $c$ $\times$ $\left(1 +\right.$ $\dfrac{r}{100}$ $\left.\right)$

In words (symbols replaced with words)

④ $S'$: sod area with waste $=$ ① $S$: lawn area $\times$ ② $c$: share covered $\times$ $\left(1 +\right.$ ③ $r$: waste factor (%) ÷ 100 $\left.\right)$

The formula in words

① Take the $S$: lawn area

② multiply it by the $c$: share covered to get the net sod area, then multiply by

③ "1 + waste factor $r$ ÷ 100" (1.05 for a 5% waste factor)

④ and you get the $S'$: sod area with waste

Quick example

For 4,000 ft² laid solid (share covered 1) with a 5% waste factor for the cuts at the edges

$S'$: sod area with waste $=$ lawn area (4,000 ft²) $\times$ share covered (1) $\times$ $\left(1 +\right.$ waste factor (5%) ÷ 100 $\left.\right)$

$4000 \times 1 \times \left(1 + \dfrac{5}{100}\right) = 4000 \times 1.05 = 4200\,\mathrm{ft^2}$

Key idea

The lawn area times the share covered is the net sod area, the area the sod takes up when it fits perfectly. In practice you cut off pieces to fit the edges, and a pallet may hold some pieces that were damaged or broke apart in delivery, so the exact net amount usually falls short. That is why the usual practice is to buy "net amount × (1 + waste factor)", and 5 to 10% is a common waste factor. Use less (about 5%) for a rectangular yard with few cuts, and more (toward 10%) for a yard with curved edges, flower beds or stepping stones. With a waste factor of 0, you get the net amount as it is.

Pallets needed (rounded up)

Figure

Standard notation (the usual math form)

$B$ $=$ $\lceil$ $S'$ $\div$ $A_b$ $\rceil$

In words (symbols replaced with words)

④ $B$: pallets needed $=$ ③ $\lceil$ ① $S'$: sod area with waste $\div$ ② $A_b$: area per pallet $\rceil$

The formula in words

① Take the $S'$: sod area with waste

② divide it by the $A_b$: area per pallet to see how many pallets it is

③ Round up any decimal (the symbol $\lceil\ \rceil$ means "round up")

④ and you get the $B$: pallets needed

Quick example

The number of pallets needed for 4,200 ft² of sod (with waste), sold in pallets of 400 ft², is

$B$: pallets needed $=$ $\lceil$ sod area (4,200 ft²) $\div$ 1 pallet (400 ft²) $\rceil$

$4200 \div 400 = 10.5$

$\lceil 10.5 \rceil = 11$

Key idea

Sod is sold by the whole pallet, so if the division leaves a decimal, always round up (10.5 pallets → 11). Rounding to the nearest or rounding down leaves you short. The symbol $\lceil\ \rceil$ (the ceiling function) stands for this rounding up. You buy 11 × 400 = 4,400 ft², so 4,400 − 4,200 = 200 ft² is left over. Pallet sizes vary: 400 to 500 ft² is common, and many sod farms also sell by the piece or by the square foot for small jobs. Change $A_b$ to the area shown by the sod farm or store.

Number of pieces

Standard notation (the usual math form)

$P$ $=$ $B$ $\times$ $n_b$

In words (symbols replaced with words)

③ $P$: number of pieces $=$ ① $B$: pallets needed $\times$ ② $n_b$: pieces per pallet

The formula in words

① Take the $B$: pallets needed

② multiply it by the $n_b$: pieces per pallet

③ and you get the $P$: number of pieces

Quick example

The number of pieces in 11 pallets of 150 pieces each is

$P$: number of pieces $=$ pallets (11) $\times$ 1 pallet (150 pieces)

$11 \times 150 = 1650$

Key idea

Once you know the number of pallets, the number of pieces is "pallets × pieces per pallet". 150 pieces of 16 × 24 in (2.67 ft² each) make a 400 ft² pallet, but the count depends on the piece size, so match the product. The number of pieces helps you estimate how many fit in a row and how many rows you will lay, and how long it will take to place them one by one with gaps.

Topdressing (over the sod + in the gaps)

Figure

Standard notation (the usual math form)

$V$ $=$ $S$ $\times$ $\dfrac{t_s}{12}$ $+$ $S$ $\times$ $\left(1 -\right.$ $c$ $\left.\right)$ $\times$ $\dfrac{t_j}{12}$

In words (symbols replaced with words)

⑤ $V$: topdressing (ft³) $=$ ① $S$: lawn area (ft²) $\times$ ② $t_s$: depth over the sod (in) ÷ 12 $+$ $S$: lawn area (ft²) $\times$ $\left(1 -\right.$ ③ $c$: share covered $\left.\right)$ $\times$ ④ $t_j$: gap fill depth (in) ÷ 12

The formula in words

① Take the $S$: lawn area (ft²)

② multiply it by the $t_s$: depth over the sod (in) ÷ 12 to get the volume of the layer over the sod.

③ Multiply the lawn area by "1 − $c$: share covered " to get the area of the gaps,

④ multiply that by the $t_j$: gap fill depth (in) ÷ 12 to get the gap fill, and add the two

⑤ to get the $V$: topdressing (ft³)

Quick example

For 400 ft² laid with gaps (share covered 0.861), with 1/8 in of topdressing over the sod and the gaps filled 3/4 in deep, the topdressing needed is

$V$: topdressing $=$ area (400 ft²) $\times$ 0.125 in ÷ 12 $+$ area (400 ft²) $\times$ $(1 -$ 0.861 $)$ $\times$ 0.75 in ÷ 12

$400 \times \dfrac{0.125}{12} \approx 4.17\,\mathrm{ft^3}$

$400 \times (1 - 0.861) \times \dfrac{0.75}{12} \approx 3.48\,\mathrm{ft^3}$

$4.17 + 3.48 \approx 7.65\,\mathrm{ft^3} \approx 0.28\,\mathrm{yd^3}$

Key idea

Topdressing is a volume: "area × depth". The area is in square feet and the depth in inches, so divide the depth by 12 to turn it into feet before multiplying (1/8 in → 0.0104 ft). Bags are labeled in cubic feet, so the answer in ft³ can be compared directly; for bulk delivery by the cubic yard, divide by 27 (1 yd³ = 27 ft³). Topdressing has two jobs. One is a thin layer (often 1/8 to 1/4 in) spread over the new sod to protect the roots and even out bumps. The other is filling the gaps between pieces in the spaced and checkerboard patterns up to the top of the sod (the soil layer of sod is usually 1/2 to 1 in thick), which helps the grass spread into them. Solid laying has no gaps ($c = 1$), so the second part is 0 and only the layer over the sod remains. With gaps, the formula shows that filling them takes a lot of topdressing for the area (in the example, the gaps need almost as much as the layer over the whole lawn). The empty squares of a checkerboard use the same formula, but because they cover half the lawn, some people fill them with top soil up to the sod and then finish with a thin topdressing. In that case, set the gap fill depth to 0 and work out the top soil with the "Soil Calculator". The gaps can also settle when people walk on them during the work, so buying a few extra bags is a safe choice. The soil you spread and level before laying the sod (the soil bed under it) is separate from this topdressing; work out its volume (area × depth) with the "Soil Calculator".

Bags of topdressing (rounded up)

Standard notation (the usual math form)

$K$ $=$ $\lceil$ $V$ $\div$ $k$ $\rceil$

In words (symbols replaced with words)

④ $K$: bags of topdressing $=$ ③ $\lceil$ ① $V$: topdressing (ft³) $\div$ ② $k$: bag size (ft³) $\rceil$

The formula in words

① Take the $V$: topdressing (ft³)

② divide it by the $k$: bag size (ft³) to see how many bags it is

③ Round up any decimal

④ and you get the $K$: bags of topdressing

Quick example

The number of 0.75 ft³ bags for 7.65 ft³ of topdressing is

$K$: bags of topdressing $=$ $\lceil$ topdressing (7.65 ft³) $\div$ 1 bag (0.75 ft³) $\rceil$

$7.65 \div 0.75 = 10.2$

$\lceil 10.2 \rceil = 11$

Key idea

Topdressing is also sold by the whole bag, so round up just as with pallets. Bagged topsoil and topdressing often come in 0.75 ft³ bags, and also in 1 ft³ and 1.5 ft³ bags; enter the size on the bag as $k$. For a large lawn, the bag count grows quickly, and buying in bulk by the cubic yard (V ÷ 27) is usually cheaper. If a bag shows only its weight, use the volume printed on it, or estimate from the density (sandy mixes are roughly 80 to 100 lb per ft³).

Estimated cost

Standard notation (the usual math form)

$T$ $=$ $u_t$ $\times$ $B$ $+$ $u_s$ $\times$ $K$

In words (symbols replaced with words)

⑤ $T$: estimated cost $=$ ① $u_t$: sod price $\times$ ② $B$: pallets needed $+$ ③ $u_s$: topdressing price $\times$ ④ $K$: bags of topdressing

The formula in words

① Multiply the $u_t$: sod price (per pallet)

② by the $B$: pallets needed to get the sod cost.

③ Multiply the $u_s$: topdressing price (per bag)

④ by the $K$: bags of topdressing to get the topdressing cost, and add the two

⑤ to get the $T$: estimated cost

Quick example

If you buy 11 pallets of sod at $250 each and 11 bags of topdressing at $5 each, the materials cost

$T$: estimated cost $=$ price ($250 per pallet) $\times$ pallets (11) $+$ price ($5 per bag) $\times$ bags (11)

$250 \times 11 + 5 \times 11 = 2750 + 55 = 2805$

Key idea

The key is to match the unit of the price and the amount. If the sod price is per pallet, multiply by the rounded-up pallets $B$; if it is per ft², multiply by the sod area you buy ($B \times A_b$). Buying by the pallet means you pay for the rounded-up part (the leftover) too, so when the area falls just over a pallet, buying the last part by the piece can cost less. The soil bed, fertilizer, a lawn roller rental and delivery fees are extra.

Reverse (the area you can cover with the pallets you have)

Standard notation (the usual math form)

$S_{\mathrm{cov}}$ $=$ $n$ $\times$ $A_b$ $\div$ $\left(1 +\right.$ $\dfrac{r}{100}$ $\left.\right)$ $\div$ $c$

In words (symbols replaced with words)

⑤ $S_{\mathrm{cov}}$: area you can cover (ft²) $=$ ① $n$: pallets you have $\times$ ② $A_b$: area per pallet $\div$ $\left(1 +\right.$ ③ $r$: waste factor ÷ 100 $\left.\right)$ $\div$ ④ $c$: share covered

The formula in words

① Take the $n$: pallets you have

② multiply it by the $A_b$: area per pallet to get the area of the sod you have,

③ divide by 1 + waste factor $r$ ÷ 100 to take out the waste, then divide by the

④ $c$: share covered (gaps let the sod cover more ground)

⑤ and you get the $S_{\mathrm{cov}}$: area you can cover

Quick example

With 2 pallets of 400 ft² each, a waste factor of 0 and the spaced pattern (share covered 0.861), the area you can cover is

$2 \times 400 = 800\,\mathrm{ft^2}$

$800 \div \left(1 + \dfrac{0}{100}\right) = 800\,\mathrm{ft^2}$

$800 \div 0.861 \approx 929.7\,\mathrm{ft^2}$

Key idea

The reverse just follows the formula for the amount needed backward. Pallets × area per pallet gives the area of the sod you have; taking out the waste gives the usable area; and dividing by the share covered gives the lawn area you can cover. Laid solid ($c = 1$), the usable area is the area you can cover. With 1.5 in gaps (share covered 0.861) you can cover about 1.16 times as much, and with a checkerboard twice as much. Use it to answer questions like "how far will the leftover sod go?" or "is this enough if I lay it in a checkerboard?". The topdressing for the area you can cover uses the same formula as before.

To find how much sod you need, multiply the lawn area by the share covered for your laying pattern to get the net sod area, add the waste factor, then divide by the area per pallet and round up. Topdressing is the volume "area × depth" (the layer over the sod plus the gap fill) in ft³, divided by the bag size and rounded up.

Symbols and terms

Symbols

$S$	ess	The lawn area (ft²): the ground you will cover with sod, found from length × width, entered directly, from the diameter of a circle, or from acres. S is often used for area and is said to come from "square" or "surface".
$l$, $w$	el, double-u	The length and width of a rectangle (ft), from the first letters of "length" and "width".
$d$	dee	The diameter of a round area (ft), from the first letter of "diameter". The radius is $d \div 2$.
$\pi$	pi	Pi, about 3.14. The area of a circle is radius × radius × pi.
$x$	ex	The size in acres. Multiply by 43,560 (the ft² in 1 acre) to get square feet.
$c$	see	The share covered by sod: the part of the lawn area taken up by the sod itself. It is 1 for solid, $\dfrac{a}{a+g} \times \dfrac{b}{b+g}$ for spaced, and $\dfrac{1}{2}$ for a checkerboard. From the first letter of "coverage".
$a$, $b$	ay, bee	The length and width of one piece of sod (in). A common piece is about 16 × 24 in, but it varies by product. Used for the share covered in the spaced pattern and for the layout figure.
$g$	gee	The gap width (in): the space left between pieces in the spaced pattern, from the first letter of "gap". 1 to 2 in is typical.
$r$	ar	The waste factor (%), from the first letter of "rate". It is the percent added to the net amount for cut edges and damaged sod; 5 means 5% more (× 1.05).
$S'$	S prime	The sod area with waste (ft²), found with $S' = S \times c \times (1 + r \div 100)$. The mark $\prime$ at the upper right means "a changed version of $S$" and is read "prime".
$A_b$	A sub b	The area per pallet (ft²), often 400 to 500 ft². A is for "area", and the subscript b is for "bundle" (a pallet or bundle of sod).
$n_b$	n sub b	The pieces per pallet (for example, 150 pieces of 16 × 24 in make 400 ft²). n is for "number", and the subscript b is for "bundle".
$B$	bee	The number of pallets needed, from the first letter of "bundle". It is found with $B = \lceil S' \div A_b \rceil$.
$P$	pee	The number of pieces, from the first letter of "piece". It is found with $P = B \times n_b$.
$V$	vee	The amount of topdressing (ft³), from the first letter of "volume". It is the layer over the sod plus the gap fill; 1 yd³ = 27 ft³.
$t_s$	t sub s	The depth of topdressing over the sod (in). t is for "thickness", and the subscript s is for "surface". Divide by 12 to turn it into feet (1/8 in → 0.0104 ft).
$t_j$	t sub j	The depth of topdressing in the gaps (in). The subscript j is for "joint". Fill up to the top of the sod (the soil layer of sod is usually 1/2 to 1 in thick).
$k$	kay	The bag size (ft³). 0.75 ft³ bags are common, and 1 ft³ and 1.5 ft³ bags are also sold.
$K$	capital kay	The number of bags of topdressing, found with $K = \lceil V \div k \rceil$.
$u_t$, $u_s$	u sub t, u sub s	The sod price (per pallet, or per ft²) and the topdressing price (per bag). u is for "unit price", and the subscripts t and s are for "turf" and "soil".
$T$	tee	The estimated cost (materials only; the soil bed, fertilizer and delivery are not included), from the first letter of "total".
$n$	en	In the reverse calculation, the number of pallets you have, from the first letter of "number".
$S_{\mathrm{cov}}$	S sub cov	The lawn area you can cover with the sod you have (ft²). The subscript cov is short for "cover".
$\lceil x \rceil$	ceiling of x	The symbol for rounding up to a whole number, called the ceiling function. (Example - $\lceil 10.5 \rceil = 11$, $\lceil 4 \rceil = 4$)

Terms

sod	Grass grown on a sod farm and cut out with a thin layer of soil, laid on prepared ground to make a lawn. Unlike a lawn started from seed, it is green the day you lay it. In the US, sod of grasses such as Kentucky bluegrass, tall fescue, Bermuda, zoysia and St. Augustine is sold.
sod piece	One rectangle of sod (also called a slab), cut with its soil. A common size is about 16 × 24 in (2.67 ft²) with a soil layer 1/2 to 1 in thick. Sod is also sold in rolls, such as 2 × 5 ft. Sizes vary by farm and product.
pallet	The usual unit for buying sod in the US. One pallet often covers 400 to 500 ft² (for example, 150 pieces of 16 × 24 in make 400 ft²). Many sod farms also sell single pieces or by the square foot for small jobs. (In the UK and other metric countries, turf is often sold in rolls of about 1 m².)
solid laying	Laying the pieces tight against each other with no gaps, usually with the joints staggered like bricks. It uses the most sod (share covered 1), but gives a full lawn right away and leaves little room for weeds. This is the usual way sod is laid in the US.
spaced pattern	Laying the pieces with a gap between them. The gaps save sod (about 14% less with 1.5 in gaps on 16 × 24 in pieces), and if you fill them with topdressing, spreading grasses fill them in over time.
checkerboard	Laying a piece in every other square, like a checkerboard. It uses half the sod, but the empty squares take one or two growing seasons to fill in, and weeds need attention in the meantime.
gap	The space between pieces of sod, like the joints between tiles or bricks. In the spaced and checkerboard patterns, the gaps are filled with topdressing and the grass stems grow into them.
topdressing	Sand or soil spread thinly over the lawn and into the gaps. It protects the roots and stems, levels bumps and helps the grass spread. It is sold in bags (often 0.75 ft³) or in bulk by the cubic yard.
soil bed	The soil you mix in and level before laying sod. It goes under the sod, unlike topdressing, which goes on top. This page does not cover it; work out its volume (area × depth) with the "Soil Calculator".
waste factor	The percent added to the net amount for pieces cut off at the edges and sod damaged in delivery. 5 to 10% is common.
zoysia	A warm-season grass with fine, dense blades. It turns brown in winter and green again in spring, and spreads by stolons, so it fills gaps well. It is grown in the southern and transition zones of the US, and it is also the grass most used in Japanese gardens.
acre	A US unit of land area. 1 acre = 43,560 ft² (about 4,047 m²), so a quarter acre is 10,890 ft². Lot sizes are often given in acres.
rounding up	Changing a number with a decimal part to the next whole number. Pallets and bags are sold one at a time, so their counts are always rounded up.
pi	The number of times the distance around a circle is longer than its diameter, about 3.14. The area of a circle is radius × radius × pi.

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 (about 3.14). If you only know the diameter, halve it to get the radius
Ratios, percents and multiplying fractions (Grades 5–6)	Showing "what part of the whole" as a fraction such as $\dfrac{16}{17.5}$ or a decimal such as $0.9$ Knowing that multiplying the share along the length by the share along the width gives the share of the area ($\dfrac{3}{4} \times \dfrac{4}{5} = \dfrac{3}{5}$) Knowing that "5% more" is the same as "× 1.05"
Volume and converting units (Grades 5–6)	Knowing that volume is base area × height. On this page, that is lawn area × topdressing depth Knowing that 1 ft = 12 in, and turning a depth of 1/8 in into feet (0.125 ÷ 12 ≈ 0.0104 ft) Knowing that 1 yd³ = 27 ft³ (a cube 1 yd, or 3 ft, on each side holds 3 × 3 × 3 = 27 cubes of 1 ft)
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 pallets and bags are rounded up
Working backward with division (Grades 6–7)	Turning a multiplication into a division to find the original amount, as in "area ÷ share = original area"

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 lawn area (rectangle)

Length (ft)	80
Width (ft)	50
Lawn area (ft²)	=B1*B2

Table to find the lawn area (circle, acres)

Circle diameter (ft)	30
Lawn area (ft², circle)	=PI()*(B1/2)^2
Size (acres)	0.25
Lawn area (ft², from acres)	=B3*43560

Table to find the share covered (spaced pattern)

Piece length (in)	16
Piece width (in)	24
Gap width (in)	1.5
Share covered	=B1/(B1+B3)*B2/(B2+B3)

Table to find the sod area with waste

Lawn area (ft²)	4000
Share covered (solid 1, checkerboard 0.5)	1
Waste factor (%)	5
Sod area with waste (ft²)	=B1B2(1+B3/100)

Table to find the pallets needed

Sod area with waste (ft²)	4200
Area per pallet (ft²)	400
Pallets needed	=ROUNDUP(B1/B2,0)

Table to find the number of pieces

Pallets needed	11
Pieces per pallet	150
Number of pieces	=B1*B2

Table to find the topdressing (over the sod + in the gaps)

Lawn area (ft²)	400
Depth over the sod (in)	0.125
Share covered	0.861
Gap fill depth (in)	0.75
Topdressing (ft³)	=B1B2/12+B1(1-B3)*B4/12
Topdressing (yd³)	=B5/27

Table to find the bags of topdressing

Topdressing (ft³)	7.65
Bag size (ft³)	0.75
Bags of topdressing	=ROUNDUP(B1/B2,0)

Table to find the estimated cost

Sod price ($ per pallet)	250
Pallets needed	11
Topdressing price ($ per bag)	5
Bags of topdressing	11
Estimated cost ($)	=B1B2+B3B4

Table to find the area you can cover (reverse)

Pallets you have	2
Area per pallet (ft²)	400
Waste factor (%)	0
Share covered	0.861
Area you can cover (ft²)	=B1*B2/(1+B3/100)/B4

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). "PI()" is pi.
B4 in the third table is about 0.861 (with 1.5 in gaps, sod covers about 86%), B4 in the fourth table is 4,200, B3 in the fifth table is 11 pallets, and B3 in the sixth table is 1,650 pieces.
The seventh table divides the depths in inches by 12 to turn them into feet before multiplying by the area. B5 is about 7.64 ft³ and B6 about 0.28 yd³. For solid laying, enter 1 in B3 and the gap fill becomes 0. B3 in the eighth table is 11 bags, and B5 in the ninth table is $2,805.
In the last table, B5 is the area you can cover (about 929.1 ft² with the rounded share 0.861).

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 lawn area (rectangle)

Length (ft)	80
Width (ft)	50
Lawn area (ft²)	=B1*B2

Table to find the lawn area (circle, acres)

Circle diameter (ft)	30
Lawn area (ft², circle)	=PI()*(B1/2)^2
Size (acres)	0.25
Lawn area (ft², from acres)	=B3*43560

Table to find the share covered (spaced pattern)

Piece length (in)	16
Piece width (in)	24
Gap width (in)	1.5
Share covered	=B1/(B1+B3)*B2/(B2+B3)

Table to find the sod area with waste

Lawn area (ft²)	4000
Share covered (solid 1, checkerboard 0.5)	1
Waste factor (%)	5
Sod area with waste (ft²)	=B1B2(1+B3/100)

Table to find the pallets needed

Sod area with waste (ft²)	4200
Area per pallet (ft²)	400
Pallets needed	=ROUNDUP(B1/B2,0)

Table to find the number of pieces

Pallets needed	11
Pieces per pallet	150
Number of pieces	=B1*B2

Table to find the topdressing (over the sod + in the gaps)

Lawn area (ft²)	400
Depth over the sod (in)	0.125
Share covered	0.861
Gap fill depth (in)	0.75
Topdressing (ft³)	=B1B2/12+B1(1-B3)*B4/12
Topdressing (yd³)	=B5/27

Table to find the bags of topdressing

Topdressing (ft³)	7.65
Bag size (ft³)	0.75
Bags of topdressing	=ROUNDUP(B1/B2,0)

Table to find the estimated cost

Sod price ($ per pallet)	250
Pallets needed	11
Topdressing price ($ per bag)	5
Bags of topdressing	11
Estimated cost ($)	=B1B2+B3B4

Table to find the area you can cover (reverse)

Pallets you have	2
Area per pallet (ft²)	400
Waste factor (%)	0
Share covered	0.861
Area you can cover (ft²)	=B1*B2/(1+B3/100)/B4

The same formulas as in Excel work as is (ROUNDUP and PI have the same names). 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

length_ft = 80            # length (ft)
width_ft = 50             # width (ft)
pattern = "joint"         # laying pattern: "solid" / "joint" (spaced, with gaps) / "checker" (checkerboard)
piece_a_in = 16           # length of one sod piece (in)
piece_b_in = 24           # width of one sod piece (in)
gap_in = 1.5              # gap width (in). Only used for the spaced pattern
pallet_area_ft2 = 400     # area per pallet (ft2). Often 400 to 500
pallet_pcs = 150          # pieces per pallet
loss_rate = 5             # waste factor (%). Cut edges and damaged sod. 0 for none
soil_thickness_in = 0.125 # topdressing depth over the sod (in)
joint_depth_in = 0.75     # gap fill depth (in). Used for the spaced and checkerboard patterns
bag_ft3 = 0.75            # bag size (ft3)
price_per_pallet = 250    # price per pallet of sod ($)
price_per_bag = 5         # price per bag of topdressing ($)

area_ft2 = length_ft * width_ft                                     # lawn area (ft2)
if pattern == "joint":
    coverage = piece_a_in / (piece_a_in + gap_in) * piece_b_in / (piece_b_in + gap_in)  # share covered by sod
elif pattern == "checker":
    coverage = 0.5
else:
    coverage = 1.0
turf_area_ft2 = area_ft2 * coverage                                 # net sod area (ft2)
required_ft2 = turf_area_ft2 * (1 + loss_rate / 100)                # sod area with waste (ft2)
pallets_needed = math.ceil(required_ft2 / pallet_area_ft2)          # pallets needed (rounded up)
pieces = pallets_needed * pallet_pcs                                # number of pieces
joint_depth = joint_depth_in if pattern != "solid" else 0           # solid laying has no gaps
soil_ft3 = area_ft2 * soil_thickness_in / 12 + area_ft2 * (1 - coverage) * joint_depth / 12  # topdressing (ft3)
soil_bags = math.ceil(soil_ft3 / bag_ft3)                           # bags of topdressing (rounded up)
cost = pallets_needed * price_per_pallet + soil_bags * price_per_bag  # estimated cost ($)

print(f"Lawn area: {area_ft2} ft2")
print(f"Share covered: {coverage:.3f} ({coverage * 100:.1f} %)")
print(f"Sod area with waste: {required_ft2:.2f} ft2")
print(f"Pallets needed: {pallets_needed} ({pieces} pieces)")
print(f"Topdressing: {soil_ft3:.1f} ft3 ({soil_ft3 / 27:.2f} yd3) -> {soil_bags} bags")
print(f"Estimated cost: ${cost:,}")

# Reverse: the area you can cover with the pallets you have (2 pallets, no waste, spaced)
have_pallets = 2
coverable_ft2 = have_pallets * pallet_area_ft2 / (1 + 0 / 100) / coverage
print(f"Area you can cover with {have_pallets} pallets laid with gaps: {coverable_ft2:.2f} ft2")

Runs with the standard library only. math.ceil() rounds up (the ⌈ ⌉ in the formulas). Replace the sizes, pattern and pallet details at the top with your own numbers and run it. If the sod price is per ft², use pallets_needed * pallet_area_ft2 * price instead of pallets_needed * price_per_pallet.

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

Lawn area (rectangle, circle, acres)

S = l × w,  S = π × (d ÷ 2)²,  S = x × 43560

S = l \times w,\quad S = \pi \left(\frac{d}{2}\right)^{2},\quad S = x \times 43560

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>S</mi><mo>=</mo><mi>l</mi><mo>&#xD7;</mo><mi>w</mi>
    <mo>,</mo>
    <mi>S</mi><mo>=</mo><mi>&#x3C0;</mi>
    <msup>
      <mrow><mo>(</mo><mfrac><mi>d</mi><mn>2</mn></mfrac><mo>)</mo></mrow>
      <mn>2</mn>
    </msup>
    <mo>,</mo>
    <mi>S</mi><mo>=</mo><mi>x</mi><mo>&#xD7;</mo><mn>43560</mn>
  </mrow>
</math>

S = l * w,  S = pi * (d/2)^2,  S = x * 43560

{l*w, Pi*(d/2)^2, x*43560}

S := l*w;  S := Pi*(d/2)^2;  S := x*43560;

S = l*w; S = pi*(d/2)^2; S = x*43560;

S = l × w, S = π (d/2)^2, S = x × 43560

Share covered by sod (by laying pattern)

c = 1 (solid),  c = a ÷ (a + g) × b ÷ (b + g) (spaced),  c = 1/2 (checkerboard)

c = 1,\quad c = \frac{a}{a + g} \times \frac{b}{b + g},\quad c = \frac{1}{2}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>c</mi><mo>=</mo><mn>1</mn>
    <mo>,</mo>
    <mi>c</mi><mo>=</mo>
    <mfrac><mi>a</mi><mrow><mi>a</mi><mo>+</mo><mi>g</mi></mrow></mfrac>
    <mo>&#xD7;</mo>
    <mfrac><mi>b</mi><mrow><mi>b</mi><mo>+</mo><mi>g</mi></mrow></mfrac>
    <mo>,</mo>
    <mi>c</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac>
  </mrow>
</math>

c = 1,  c = a/(a + g) * b/(b + g),  c = 1/2

{1, a/(a + g)*b/(b + g), 1/2}

c := 1;  c := a/(a + g)*b/(b + g);  c := 1/2;

c = 1; c = a/(a + g)*b/(b + g); c = 1/2;

c = 1, c = a/(a + g) × b/(b + g), c = 1/2

Sod area with waste

S' = S × c × (1 + r ÷ 100)

S' = S \times c \left(1 + \frac{r}{100}\right)

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msup><mi>S</mi><mo>&#x2032;</mo></msup><mo>=</mo><mi>S</mi><mo>&#xD7;</mo><mi>c</mi>
    <mrow><mo>(</mo><mn>1</mn><mo>+</mo><mfrac><mi>r</mi><mn>100</mn></mfrac><mo>)</mo></mrow>
  </mrow>
</math>

S' = S * c * (1 + r/100)

s*c*(1 + r/100)

Sp := S*c*(1 + r/100);

Sp = S*c*(1 + r/100);

S' = S c (1 + r/100)

Pallets needed (rounded up)

B = ⌈S' ÷ A_b⌉

B = \left\lceil \frac{S'}{A_b} \right\rceil

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>B</mi><mo>=</mo>
    <mo>&#x2308;</mo><mfrac><msup><mi>S</mi><mo>&#x2032;</mo></msup><msub><mi>A</mi><mi>b</mi></msub></mfrac><mo>&#x2309;</mo>
  </mrow>
</math>

B = |~ S' / A_b ~|

Ceiling[sp/ab]

B := ceil(Sp/Ab);

B = ceil(Sp/Ab);

B = ⌈S'/A_b⌉

Number of pieces

P = B × n_b

P = B \times n_b

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi><mo>=</mo><mi>B</mi><mo>&#xD7;</mo><msub><mi>n</mi><mi>b</mi></msub>
  </mrow>
</math>

P = B * n_b

b*nb

P := B*nb;

P = B*nb;

P = B × n_b

Topdressing (over the sod + in the gaps)

V = S × t_s ÷ 12 + S × (1 − c) × t_j ÷ 12

V = S \times \frac{t_s}{12} + S \left(1 - c\right) \frac{t_j}{12}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>V</mi><mo>=</mo><mi>S</mi><mo>&#xD7;</mo>
    <mfrac><msub><mi>t</mi><mi>s</mi></msub><mn>12</mn></mfrac>
    <mo>+</mo><mi>S</mi>
    <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>c</mi><mo>)</mo></mrow>
    <mfrac><msub><mi>t</mi><mi>j</mi></msub><mn>12</mn></mfrac>
  </mrow>
</math>

V = S * t_s/12 + S * (1 - c) * t_j/12

s*ts/12 + s*(1 - c)*tj/12

V := S*ts/12 + S*(1 - c)*tj/12;

V = S*ts/12 + S*(1 - c)*tj/12;

V = S × t_s/12 + S (1 − c) t_j/12

Bags of topdressing (rounded up)

K = ⌈V ÷ k⌉

K = \left\lceil \frac{V}{k} \right\rceil

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>K</mi><mo>=</mo>
    <mo>&#x2308;</mo>
    <mfrac><mi>V</mi><mi>k</mi></mfrac>
    <mo>&#x2309;</mo>
  </mrow>
</math>

K = |~ V / k ~|

Ceiling[v/k]

K := ceil(V/k);

K = ceil(V/k);

K = ⌈V/k⌉

Estimated cost

T = u_t × B + u_s × K

T = u_t \times B + u_s \times K

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>T</mi><mo>=</mo><msub><mi>u</mi><mi>t</mi></msub><mo>&#xD7;</mo><mi>B</mi>
    <mo>+</mo><msub><mi>u</mi><mi>s</mi></msub><mo>&#xD7;</mo><mi>K</mi>
  </mrow>
</math>

T = u_t * B + u_s * K

ut*b + us*k

T := ut*B + us*K;

T = ut*B + us*K;

T = u_t × B + u_s × K

Reverse (the area you can cover with the pallets you have)

S_cov = n × A_b ÷ (1 + r ÷ 100) ÷ c

S_{\mathrm{cov}} = \frac{n \times A_b}{\left(1 + \frac{r}{100}\right) c}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>S</mi><mi>cov</mi></msub><mo>=</mo>
    <mfrac>
      <mrow><mi>n</mi><mo>&#xD7;</mo><msub><mi>A</mi><mi>b</mi></msub></mrow>
      <mrow><mrow><mo>(</mo><mn>1</mn><mo>+</mo><mfrac><mi>r</mi><mn>100</mn></mfrac><mo>)</mo></mrow><mi>c</mi></mrow>
    </mfrac>
  </mrow>
</math>

S_cov = (n * A_b) / ((1 + r/100) * c)

n*ab/((1 + r/100)*c)

Scov := n*Ab/((1 + r/100)*c);

Scov = n*Ab/((1 + r/100)*c);

S_cov = (n × A_b)/((1 + r/100) c)

How to have ChatGPT do the calculation

You are a quantity calculation assistant for landscaping. 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 laying sod on a yard 80 ft long and 50 ft wide. The sod pieces are 16 × 24 in, and one pallet covers 400 ft² (150 pieces).
Find each of the following:
1. The lawn area (ft²)
2. For solid laying with a 5% waste factor, the sod area with waste (ft²), the pallets needed (rounded up) and the number of pieces
3. For a spaced pattern with 1.5 in gaps, the share covered by sod (16 ÷ (16 + 1.5) × 24 ÷ (24 + 1.5)) and the pallets needed with a 5% waste factor
4. For the spaced pattern, the topdressing needed (ft³ and yd³) for 1/8 in over the sod and the gaps filled 3/4 in deep, and the number of 0.75 ft³ bags (rounded up)
5. With only 2 pallets, the area I can cover with no waste and the spaced pattern (ft²)

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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\(S\)	ess	The lawn area (ft²): the ground you will cover with sod, found from length × width, entered directly, from the diameter of a circle, or from acres. S is often used for area and is said to come from "square" or "surface".
\(l\), \(w\)	el, double-u	The length and width of a rectangle (ft), from the first letters of "length" and "width".
\(d\)	dee	The diameter of a round area (ft), from the first letter of "diameter". The radius is \(d \div 2\).
\(\pi\)	pi	Pi, about 3.14. The area of a circle is radius × radius × pi.
\(x\)	ex	The size in acres. Multiply by 43,560 (the ft² in 1 acre) to get square feet.
\(c\)	see	The share covered by sod: the part of the lawn area taken up by the sod itself. It is 1 for solid, \(\dfrac{a}{a+g} \times \dfrac{b}{b+g}\) for spaced, and \(\dfrac{1}{2}\) for a checkerboard. From the first letter of "coverage".
\(a\), \(b\)	ay, bee	The length and width of one piece of sod (in). A common piece is about 16 × 24 in, but it varies by product. Used for the share covered in the spaced pattern and for the layout figure.
\(g\)	gee	The gap width (in): the space left between pieces in the spaced pattern, from the first letter of "gap". 1 to 2 in is typical.
\(r\)	ar	The waste factor (%), from the first letter of "rate". It is the percent added to the net amount for cut edges and damaged sod; 5 means 5% more (× 1.05).
\(S'\)	S prime	The sod area with waste (ft²), found with \(S' = S \times c \times (1 + r \div 100)\). The mark \(\prime\) at the upper right means "a changed version of \(S\)" and is read "prime".
\(A_b\)	A sub b	The area per pallet (ft²), often 400 to 500 ft². A is for "area", and the subscript b is for "bundle" (a pallet or bundle of sod).
\(n_b\)	n sub b	The pieces per pallet (for example, 150 pieces of 16 × 24 in make 400 ft²). n is for "number", and the subscript b is for "bundle".
\(B\)	bee	The number of pallets needed, from the first letter of "bundle". It is found with \(B = \lceil S' \div A_b \rceil\).
\(P\)	pee	The number of pieces, from the first letter of "piece". It is found with \(P = B \times n_b\).
\(V\)	vee	The amount of topdressing (ft³), from the first letter of "volume". It is the layer over the sod plus the gap fill; 1 yd³ = 27 ft³.
\(t_s\)	t sub s	The depth of topdressing over the sod (in). t is for "thickness", and the subscript s is for "surface". Divide by 12 to turn it into feet (1/8 in → 0.0104 ft).
\(t_j\)	t sub j	The depth of topdressing in the gaps (in). The subscript j is for "joint". Fill up to the top of the sod (the soil layer of sod is usually 1/2 to 1 in thick).
\(k\)	kay	The bag size (ft³). 0.75 ft³ bags are common, and 1 ft³ and 1.5 ft³ bags are also sold.
\(K\)	capital kay	The number of bags of topdressing, found with \(K = \lceil V \div k \rceil\).
\(u_t\), \(u_s\)	u sub t, u sub s	The sod price (per pallet, or per ft²) and the topdressing price (per bag). u is for "unit price", and the subscripts t and s are for "turf" and "soil".
\(T\)	tee	The estimated cost (materials only; the soil bed, fertilizer and delivery are not included), from the first letter of "total".
\(n\)	en	In the reverse calculation, the number of pallets you have, from the first letter of "number".
\(S_{\mathrm{cov}}\)	S sub cov	The lawn area you can cover with the sod you have (ft²). The subscript cov is short for "cover".
\(\lceil x \rceil\)	ceiling of x	The symbol for rounding up to a whole number, called the ceiling function. (Example - \(\lceil 10.5 \rceil = 11\), \(\lceil 4 \rceil = 4\))

Sod Calculator (Square Feet, Pallets and Topdressing)

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

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Also for the yard. Find the number of deck boards, joists and posts from the deck size

Fence Calculator (Posts, Panels, Rails and Pickets)

Find the number of posts, panels and pickets for a fence around the yard

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