Percentage drop. % drop = (static − residual) ÷ static × 100. It is a measure of how hard the distribution system has to work to deliver the flow you just took. The standard fireground reading of it: a drop of 0–10 % means three times the measured flow is available in addition; 11–15 % means two times; 16–25 % means the same amount again; more than 25 % means less than what is being delivered is available. The first-digit method is the same thing without the percentages — compare the drop against the first digit of the static pressure. A drop equal to that digit means three times available, twice that digit means two times, three times that digit means the same amount again. On a 72 psi static with a 24 psi drop: the first digit is 7, so 24 is more than three times 7 — less than the measured flow is available. Both methods agree, which is the point of the second one: you can do it in your head on the fireground. The bands are whole numbers, so a drop between two of them — 15.3 %, say — is read into the more conservative of the two, the one that says there is less water available.

Available fire flow at 20 psi residual. The Hazen-Williams-derived relationship used by NFPA 291 and the fire service: Q_R = Q × (S − 20)^0.54 ÷ (S − R)^0.54, where Q is the flow measured, S the static pressure, R the residual pressure. It gives the total flow the system could deliver while still holding 20 psi at the hydrant, which is the pressure a pumper generally needs at its intake to draft from the main without cavitating.

Pitot discharge. Q = 29.83 × c × d² × √p, where c is the outlet coefficient (0.90 for a smooth outlet, lower for a rough or corroded one), d the inside diameter of the outlet in inches and p the pitot pressure. Take the reading at the centre of the stream, with the pitot blade held square, at the point where the stream is at its widest.

Worked example. Static 72 psi, residual 48 psi, measured flow 800 gpm. Drop = 24 psi, % drop = 24 ÷ 72 × 100 = 33 % — more than 25 %, so less than 800 gpm more is available. Available at 20 psi residual: Q_R = 800 × (72 − 20)^0.54 ÷ (72 − 48)^0.54 = 800 × 8.445 ÷ 5.564 ≈ 1,210 gpm. Consistent with the other two methods: the system can push about 1,210 gpm in total, so only about 410 gpm is available on top of the 800 already flowing — less than the measured flow, exactly as the 33 % drop said.

The two failure modes. First, a drop of less than 10 % is not a usable test — the gauges and the timing are not accurate enough to resolve it, and NFPA 291 says so. Flow more, or flow from a second hydrant. Second, the whole calculation is only as good as the static reading: if the system was already under load when you took it — a neighbouring main flowing, a pump running, demand at that hour — the "available" figure is optimistic by whatever was already being drawn. Test at times of known demand, record the demand, and never treat one test as the permanent answer. Mains get throttled, valves get shut, and a hydrant test is a snapshot of a moving system.

Disclaimer. A calculation aid for training and pre-planning. Actual flow testing must follow NFPA 291 and your department's SOPs, with the residual pressure never allowed to fall below what the system or the pumper requires. Sources: NFPA 291, Recommended Practice for Fire Flow Testing and Marking of Hydrants · IFSTA Essentials of Fire Fighting · AWWA M17. Last reviewed: 2026-09-21.