One division, run in both directions
Pressure is load over area. Area is load over pressure. Everything on this page is one of those two, and the only reason it is not a slide rule exercise is the unit juggling between square inches and square feet, which is where the arithmetic usually goes wrong by a factor of 144.
The defaults, worked through. Twelve thousand pounds at one leg, sitting on a bare foot of 100 square inches — that is 0.694 square feet, so 17,280 pounds per square foot, or 120 psi, or 8.64 tons per square foot. Divide the same 12,000 pounds by a bearing figure of 2,000 pounds per square foot and you get 6 square feet, which is 864 square inches: 29.4 inches square, or 33.2 inches across if it is round. The foot on the machine is 100 square inches. The division asks for 8.64 times that.
Why pads are always bigger than they look like they should be
Because area is a squared quantity and everybody estimates in linear ones. A 24 inch square pad is 576 square inches, four square feet, and under 12,000 pounds it produces 3,000 pounds per square foot. Bumping it to 30 inches gives 900 square inches — 56 percent more area for 25 percent more dimension — and the pressure drops to 1,920.
The same relationship bites in the other direction, which is the part worth remembering. When a pad will not fit where it has to go and somebody turns an 18 inch pad through 45 degrees or uses a half sheet instead of a full one, halving the area doubles the pressure and halving the dimension quadruples it. The physical change looks minor and the number does not.
It is also why the step from a bare steel foot to almost any pad is the biggest single improvement available, and why the step from a decent pad to a bigger one is a lot less dramatic. Going 100 to 576 square inches divides the pressure by 5.76. Going 576 to 900 divides it by 1.56.
The ground is not one thing
The arithmetic treats bearing pressure as a single number for the whole setup, and that is the assumption most likely to be wrong on a real site. Ground varies from leg to leg on the same machine. One outrigger over undisturbed material and the next over the backfill of a service trench dug last month are two different soils, and the pad size that came out of the division was worked out for whichever number you were given, not for both.
Things that change the ground and are invisible from a seat: a backfilled trench, a service run, a vault or basement lid, a soakaway, made ground, a kerb or slab edge that the pad is partly bridging, a slope, a recently thawed surface, and rain since the assessment. None of that is arithmetic and none of it is on this page.
What is deliberately absent
No verdict. This page does not compare the area it calculated to the pad you entered and does not tell you any pad is adequate, because adequacy involves the ground, the machine, the configuration, how load spreads through the pad, and the pad manufacturer own rating — which is a separate figure covering how much the pad will take and how it may be supported, and has nothing to do with area.
It also does not produce the outrigger load. That comes from the machine load chart for the configuration you are actually in, and it changes as the boom slews and extends: the loaded leg is not the same leg all day. And it does not offer a bearing pressure value, because that is a property of specific ground assessed by somebody qualified, not something to be taken off a table.
Machines tipping on ground that gave way is one of the ways this work kills people, and it happens too fast to react to. The arithmetic here is for planning what to put on the truck. The setup decision belongs to the machine instructions, a trained operator and whoever is responsible for the ground.
Neighbouring pages that stop at the same line: the boom lift reach calculator for the geometry above the ground, and the truss point load calculator for loads that go up instead of down.
Questions people ask
How do I calculate outrigger pad size?
Divide the load at the leg by the ground bearing pressure figure you have been given, then convert to the shape you are using. Twelve thousand pounds over 2,000 pounds per square foot is 6 square feet, which is 864 square inches — 29.4 inches square or 33.2 inches in diameter. Both of the numbers you divide with come from elsewhere: the load from the machine chart for your actual configuration, the pressure from someone qualified to assess that ground.
How much pressure does a bare outrigger foot put into the ground?
A great deal. A 10 by 10 inch float is 100 square inches, which is 0.694 square feet, so 12,000 pounds through it is 17,280 pounds per square foot — 120 psi, or 8.64 tons per square foot. That is the reason pads exist, and it is why the first pad under a bare foot does more than any later increase in pad size.
Why does a slightly bigger pad help so much?
Because area goes with the square of the dimension. Twenty-four inches square is 576 square inches; thirty inches square is 900. That is 25 percent more dimension for 56 percent more area, and the pressure falls in the same proportion. The relationship runs the other way too, which is why cutting a pad down to fit a tight spot costs far more than it appears to.
What ground bearing pressure should I use?
Not one from a web page. Allowable bearing pressure is a property of the specific ground and it is assessed by somebody qualified to assess it — a geotechnical report, the engineer for the project, or the party responsible for the site. It also varies within one setup: undisturbed material under one leg and trench backfill under the next are not the same ground, and the arithmetic assumes a single figure for all of them.
Does a big enough pad make the setup safe?
That is not a question arithmetic can answer, and this page does not attempt it. Area is one factor among the ground itself, the machine, the configuration, how the load spreads through the pad, and the pad own rating from its manufacturer, which is separate from its size. Machines tip when the ground under a leg gives way, quickly enough that nobody in the way moves. Setup belongs to the machine instructions, a trained operator, and whoever is responsible for the ground.