Weighing a Parked Jet With a Tyre Gauge, and Why the Answer Is a Bracket

A jet is parked on the apron. You have a pressure gauge, a tape measure and no scale anywhere on the airfield. How heavy is it?

The usual first answer is that the question has no answer, because weighing needs a scale and there is not one. That reply is wrong in an interesting way. The aircraft is already standing on four scales, one under each main wheel, and each of them has a dial. What follows is the reading, the interval it sits in, and the exact place where the reading stops being trustworthy.

A pressure gauge is a force gauge divided by an area

Pressure is defined as force per unit area. Rearranged, that definition says a known pressure acting over a known area is a force, and nothing else is needed to get one from the other:

The air inside a tyre carries the load above it. Where the rubber has flattened against the tarmac, the ground pushes back with a force equal to the internal pressure times the flattened area. Sum that over the wheels and you have the weight of everything resting on them.

The chain, with every guess declared

Two quantities have to be supplied. The first is the pressure, and a main gear tyre runs near 200 pounds per square inch, far above anything on a road vehicle. The second is the contact patch, which you kneel down and measure: call it 10 inches by 20 inches, so 200 square inches per tyre and 800 square inches over four main tyres. The product is

A short ton is 2000 pounds, so 160,000 pounds is 80 of them. That number wants a footnote before it travels: in metric units the same weight is 72.575 tonnes, since a pound is 0.45359237 kilograms exactly. Anyone reading “eighty tons” as eighty tonnes has gained ten percent for free. Pinning the conversion separately is cheap insurance.

The answer is an interval, and here is the interval

Neither input was measured to three digits, so quoting the output to three digits would be a lie about the method. The honest deliverable is the range the answer occupies once each guess is pushed to the edge of what is plausible. Take the pressure anywhere in 180 to 220 pounds per square inch and each patch anywhere in 160 to 240 square inches:

Because the chain is a product of factors bounded independently, those two corners are the true minimum and maximum rather than a quantile: sweeping the whole integer grid inside those ranges, all 3321 combinations of it, produces nothing outside. The upper end is 1.32 times the round answer and the lower end is 0.72 times it, so the entire interval fits inside two thirds and three halves of 160,000. End to end the bracket spans a factor of 1.83, which is under a factor of two.

Corner arithmetic is pessimistic on purpose, since it assumes both guesses fail in the same direction at once. Drawing the two inputs log-uniformly across their ranges instead, 200,000 times, puts the median at 155,937 pounds with a 5th to 95th percentile band of 126,041 to 193,037, and no single draw lands further than a factor of 1.39 from the headline. Two different notions of “how wrong could this be” agree that the answer is good to roughly a factor of two, which is what an order-of-magnitude question asks for.

Calibrate on something you can check

A method that only ever produces unfalsifiable numbers is not a method. So run the identical procedure on an object whose weight is public. A passenger car tyre sits near 32 pounds per square inch and its patch is about 30 square inches, four of them:

That is squarely where a mid-size saloon actually sits. The same two-line calculation, with inputs a factor of six apart in pressure, lands on a checkable answer. This is the second leg of the argument and it shares no step with the first, since it tests the physics rather than re-testing the arithmetic.

Where the identity stops being exact

Equation (1) is exact. The step from equation (1) to an aircraft is not, and there are three distinct places it leaks. Naming them is more useful than adding digits.

The first is structural. A pneumatic tyre is not an ideal membrane. Its sidewall and belts are stiff and carry part of the load themselves, so the air does less work than the model assumes and the patch flattens less than $W/p$ would predict:

The inequality has a definite direction, which is the useful part. The estimate is a floor, not a two-sided guess. For carcass shares up to about fifteen percent, the chain recovers between 85 and 100 percent of the true weight, and it never overshoots for that reason.

The second is bookkeeping. Four tyres is a narrowbody main gear, and the nose gear was left out. It carries something like five percent of the weight, so the answer is low by about that much again, on top of the sidewall effect. A widebody standing on sixteen main tyres needs a proportionally larger patch total before the same arithmetic means anything, and the failure mode there is mixing a tyre count from one aircraft with a patch size from another.

The third is the most interesting, because it is about the shape of the reasoning rather than the physics. Choosing 200 square inches per patch is suspiciously convenient: it is close to what equation (1) would hand you if you already knew the aircraft's weight and divided by 200. A guess informed by the answer is circular, and a Fermi chain assembled from such guesses will always look successful. The defence is procedural. The patch is not a guess, it is a measurement, and it takes a tape measure and thirty seconds. Anything you can walk over and measure is an input; anything you look up because the answer is known is not.

The general statement

Nothing above is specific to aircraft. For any object resting on $n$ pneumatic tyres at pressures $p_i$ with patches $A_i$, the load supported is

where $C$ is whatever the carcass carries. Setting $C = 0$ gives the floor. The method fails outright whenever $C$ stops being small: a solid tyre, a flat tyre, a tracked vehicle, or a tyre so overloaded that the sidewall is folded onto the rim. In those cases there is no air pressure doing the supporting and equation (1) is being applied to a quantity that is not the load path at all.

Sources and further reading

• Pressure as force per unit area, which is the whole of the argument: Pressure
• Why the units cancel without a conversion factor: Dimensional analysis and Pounds per square inch
• The style of reasoning, and why the deliverable is a decade rather than a digit: Fermi problem
• Enrico Fermi's own order-of-magnitude estimate of the Trinity yield from the displacement of scraps of paper is the canonical worked example of the genre, reported in his 1945 observation notes and reproduced in Trinity (nuclear test).
• The exact pound: 1 lb = 0.45359237 kg by international agreement, 1959. See Pound (mass).

Every figure quoted here was recomputed in exact rational arithmetic, and the bracket in equation (3) was confirmed by sweeping the full integer grid rather than by sampling it. None of it asserts the weight of any particular aircraft, which is the point: the claim is the interval and the direction of the bias, not the digits.