A spacing is not a bar count — this page turns the spacing into rows, columns, a cut list and an order
← Back to the calculatorSpacing is an interval; what you buy is bars. Clear width 238 in ÷ 12 in = 19 spaces + 1 = 20 bars; clear length 286 in ÷ 12 in = 23 spaces + 1 = 24 bars. Total 20 × 24 ft + 24 × 20 ft = 960 ft of #4 bar, 641.3 lb, crossing at 480 points.
Open the calculator for your own dimensions, or read the four steps below.
Search for a rebar spacing calculator and most of what comes back hands you one number: the spacing you already typed in, turned into a weight. That is the easy direction. The one that matters on site is the other way round — you have the spacing from the drawing and you need to know how many bars to cut, how many pieces, how many intersections there are to tie, and how many whole bars to buy.
The gap is that a spacing is not a count. Spacing is a fixed interval; the clear distance you are spanning almost never divides into it evenly, so the bar count is a step function and every row of bars finishes with a short space at one end. Three of the numbers people ask for — bars, intersections, chairs — all fall out of the same two counts: rows and columns.
This page works a 24 ft × 20 ft slab with #4 bar at 12 in on centre both ways and shows every figure, so you can run your own drawing through it.
A two-way mat is two sets of bars. Work each set separately, and be clear about which direction each set runs: the bars you count across the width are the ones that run the length, and the bars you count across the length are the ones that run the width.
Measure the clear distance between the first and last bar positions, not the outside of the slab. The first bar sits one cover in from the edge, and the number that matters is to the bar centre, which is the cover plus half a bar diameter. For a #4 bar at 3/4 in cover that is 3/4 + 1/4 = 1 in, and we use 1 in at each edge:
Here is the part the thin calculators skip. The division above is not exact, and the leftover is not a rounding error to be waved away — it is a real, physical short space at the end of every row, and where it sits is a layout decision you have to make and then write down.
Read the arithmetic plainly: 238 in of clear contains 19 whole 12 in spaces, which uses 228 in. That leaves 10 in. The layout rule is floor(clear ÷ spacing) + 1, so you place 20 bars and the surplus of 10 in falls into the last space. Every row of that mat has one 12 in bay fewer and one 10 in bay, and the two end distances from bar centre to slab edge are not equal.
| Spacing | Bars across 20 ft | Whole spaces | Last space | Bars across 24 ft | Last space |
|---|---|---|---|---|---|
| 6 in | 40 | 39 | 4 in | 48 | 4 in |
| 8 in | 30 | 29 | 6 in | 36 | 6 in |
| 10 in | 24 | 23 | 10 in | 29 | 8 in |
| 12 in | 20 | 19 | 10 in | 24 | 10 in |
| 16 in | 15 | 14 | 14 in | 18 | 14 in |
| 18 in | 14 | 13 | 4 in | 16 | 14 in |
| 24 in | 10 | 9 | 22 in | 12 | 22 in |
Swipe the table sideways for more columns →
Two things fall out of that table. First, the last space swings wildly with spacing — 4 in at 6 in o.c., 22 in at 24 in o.c. — and on a small job that short bay at 24 in o.c. is nearly another full bay, so the row looks wrong on the ground even though the arithmetic is right. Second, the bar count does not move in proportion to the spacing: dropping from 12 in to 10 in adds four bars across the width, not two, because you cross a whole extra interval.
floor(clear ÷ spacing) + 1 gives you. Where the drawing shows bars centred in the panel, or the mat ties into a wall or footing at both ends, split the surplus instead: subtract the remainder from the clear distance first, divide that by the spacing, and set the first and last bays equal. Either is workable. What is not workable is not knowing which one you built.Once the count is right the weight is a multiplication, and it is worth seeing how hard the spacing bites. Run the same 24 ft × 20 ft slab at eight spacings with a #4 bar and the steel doubles every time you halve the spacing — because the length of bar in a mat is proportional to 1 ÷ spacing, not to anything more forgiving.
| Spacing | Bars 24 ft | Bars 20 ft | Total length | Weight (#4) | Intersections | lb per 1,000 sq ft |
|---|---|---|---|---|---|---|
| 6 in | 40 | 48 | 1,920 ft | 1,282.6 lb | 1,920 | 2,672 lb |
| 8 in | 30 | 36 | 1,440 ft | 961.9 lb | 1,080 | 2,004 lb |
| 10 in | 24 | 29 | 1,156 ft | 772.2 lb | 696 | 1,609 lb |
| 12 in | 20 | 24 | 960 ft | 641.3 lb | 480 | 1,336 lb |
| 16 in | 15 | 18 | 720 ft | 481.0 lb | 270 | 1,002 lb |
| 18 in | 14 | 16 | 656 ft | 438.2 lb | 224 | 913 lb |
| 24 in | 10 | 12 | 480 ft | 320.6 lb | 120 | 668 lb |
| 32 in | 8 | 9 | 372 ft | 248.5 lb | 72 | 518 lb |
Swipe the table sideways for more columns →
Read the last column as the transferable number: 1,336 lb of #4 per 1,000 sq ft at 12 in o.c. both ways, and 668 lb at 24 in — exactly half, as it must be. The intersections column is what you tie and, in a two-layer mat, what you support:
On the 24 ft × 20 ft slab at 12 in o.c. there are 480 intersections; tying 50% of them with 8 in of 16 ga wire is 1.7 lb. On a 3 ft chair grid the same slab takes 8 × 7 = 56 chairs. Tighten the spacing to 6 in and the intersections quadruple to 1,920, so the wire goes to 6.7 lb and the number of places that need supporting more than triples. Every one of those numbers traces back to the two counts in Step 1.
A bar count is not yet something you can send to a supplier. Two more things happen between the count and the order: the bars get cut to length, and the long ones get spliced because they are longer than the stock you can buy.
Write the cut list first. One line per distinct bar length, with a mark, a size, a piece count and a cut length:
| Mark | Size | Pieces | Cut length | Runs in | Notes |
|---|---|---|---|---|---|
| A | #4 | 20 | 24 ft | length | 1 lap per bar on 20 ft stock |
| B | #4 | 24 | 20 ft | width | exactly one stock bar, no lap |
| Total | #4 | 44 lines | — | — | 960 ft net, 641.3 lb |
Swipe the table sideways for more columns →
Then apply the stock length. Mark B is already 20 ft and needs no thinking. Mark A is 24 ft, which does not fit a 20 ft bar, so each of those twenty bars becomes two pieces — a 20 ft piece and a 4 ft piece — joined by one lap of 40 × db:
| Item | 20 ft stock | 40 ft stock |
|---|---|---|
| Mark A, 24 ft lines | 2 pieces each, 1 lap | 1 piece each, no lap |
| Mark B, 20 ft lines | 1 piece each | 2 lines per 40 ft bar |
| Lap splices (#4, 20 in) | 20 | 0 |
| Bars bought | 64 of 20 ft | 32 of 40 ft |
| Length bought | 1,280 ft | 1,280 ft |
| Order weight | 855.0 lb | 855.0 lb |
Swipe the table sideways for more columns →
The two stock lengths come out level on material on this particular slab — the same 1,280 ft, because the 4 ft offcuts off mark A are nearly the length the two 40 ft bars waste on mark B. What is not level is the number of pieces and laps: 20 ft stock needs 64 bars and 20 laps, 40 ft stock needs 32 bars and none. That is the same trade the footing page works the other way round, and the estimator page explains why the offcut is a remainder rather than a percentage.
The floor(clear ÷ spacing) + 1 rule is not a convention invented here; it is the arithmetic of fitting a fixed interval into a fixed distance with a bar at every position, including both ends. The + 1 is the bar at the start; the floor is the whole number of intervals that fit before you run out of clear distance. The rule appears wherever bar layout is taught, from supplier handbooks to the worked examples on the calculator sites.
What most of those pages leave out is the third term of the same sentence: the leftover. Spacing calculators generally assume the spacing divides the panel, or they round the bar count and move on. Both hide the short bay — and the short bay is where a mat looks wrong on the ground, where the last bar falls short of the edge, and where the count changes by one if you decide to split the surplus instead. Our slab page gives the same rule from the other direction, starting from the area rather than the spacing.
Twenty bars running the 24 ft direction and twenty-four running the 20 ft direction, with 1 in from the slab edge to the first bar centre. Across the 20 ft width the clear distance is 240 − 2 = 238 in; divide by 12 in and you get 19 whole spaces, so 19 + 1 = 20 bars, each 24 ft long. Across the 24 ft length the clear is 288 − 2 = 286 in, which is 23 spaces and 24 bars of 20 ft. Total 20 × 24 + 24 × 20 = 960 ft of #4, or 641.3 lb, crossing at 480 points.
Because a fixed interval almost never divides a clear distance exactly, and the rule keeps a whole bar at the far end rather than a fractional one. On the 20 ft width, 238 in of clear holds 19 full 12 in bays — 228 in — and leaves 10 in over. That 10 in becomes the last space, so the row ends 10 in before the far edge instead of 12 in. If you want the row centred, split the 10 in between the two ends before you divide.
Split it where the drawing shows the mat centred in the panel, or where the mat ties into a wall or footing and the end bars need to be at a fixed position. Take it at one end where it is a slab crack-control mat and the crew is setting out from a single reference edge — that is faster and easier to check. Either is fine; what matters is that the cut list says which one you did, because the bar count can move by one if the layout changes.
Multiply rows by columns. Every bar running one way crosses every bar running the other way exactly once, so the count is simply the two bar counts multiplied. On the 24 ft × 20 ft slab at 12 in o.c. that is 20 × 24 = 480 intersections. That single number gives you the tie points, the wire, and — for a two-layer mat — the number of places that need a chair or a bar support.
It changes the weight in direct proportion to 1 ÷ spacing, so halving the spacing doubles the steel. On the same 24 ft × 20 ft slab with #4 bar: 24 in o.c. both ways is 480 ft and 320.6 lb, 12 in o.c. is 960 ft and 641.3 lb, and 6 in o.c. is 1,920 ft and 1,282.6 lb. Per 1,000 sq ft that is 668 lb, 1,336 lb and 2,672 lb. Tightening a spacing is the most expensive change you can make to a mat, which is why it is the engineer's call and not a field decision.
The drawing or the local code sets it, and this page does not override that. What other sources commonly quote for a residential slab on grade is 12 in to 18 in on centre in both directions for a #3 or #4 crack-control mat, and 18 in to 24 in for a heavier mat, with 18 in to 24 in drawn from the guidance published by the construction-software sites. Treat those as the range you will see on drawings, not as an engineering determination.
One line per distinct bar length, four columns: bar mark, bar size, number of pieces, and cut length. Group bars of the same length under one mark so the yard can batch them. Then add a lap wherever a bar line is longer than the stock length you intend to buy. On the 24 ft × 20 ft slab with 20 ft stock: mark A is the 24 ft lines, which become 40 pieces of 20 ft with one 20 in lap each; mark B is 24 pieces of 20 ft with no lap.
Work each zone as its own grid and add them. Where a slab is thickened or carries a point load, the drawing may tighten the spacing over part of the area: measure that zone's clear distance, apply floor(clear ÷ spacing) + 1 across it, then repeat for the rest of the slab. Do not average the two spacings — the bar count is a step function, so a partial tightening still adds whole bars at whole intervals.