Peak Discharge Rational Method Calculator
Calculate peak discharge rational method with our free science calculator. Uses standard scientific formulas with unit conversions and explanations.
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer
Peak Discharge Rational Method Calculator
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Formula: Q = C x i x A / 360
Worked example — Peak discharge: 1.65 m3/s (10-year event)
Formula
Q = C x i x A / 360
The Rational Method formula Q = C x i x A / 360 estimates peak stormwater runoff from a catchment. C is the dimensionless runoff coefficient (0 to 1) reflecting how much rainfall becomes runoff based on land cover and soil type — higher values for impervious surfaces, lower for vegetated areas. i is the design rainfall intensity (mm/hr) at the time of concentration for the chosen return period. A is the catchment area in hectares (ha). The divisor 360 converts mm/hr and hectares to m³/s. The result Q gives the peak discharge used to size culverts, stormwater pipes, and detention basins.
Worked Examples
Example 1: Suburban Residential Subdivision
Problem:A = 12 ha of medium-density housing, C = 0.55, Tc = 15 min. The 10-year IDF curve gives i = 90 mm/hr at a 15-minute duration.
Solution:Q = C x i x A / 360 Q = 0.55 x 90 mm/hr x 12 ha / 360 i x A = 90 x 12 = 1080 1080 / 360 = 3.0 Q = 0.55 x 3.0 = 1.65 m3/s
Result:Peak discharge: 1.65 m3/s (10-year event)
Example 2: Asphalt Car Park
Problem:A = 1.5 ha of asphalt paving and roof, C = 0.90, Tc = 10 min. The 5-year IDF curve gives i = 120 mm/hr at a 10-minute duration.
Solution:Q = C x i x A / 360 Q = 0.90 x 120 mm/hr x 1.5 ha / 360 i x A = 120 x 1.5 = 180 180 / 360 = 0.5 Q = 0.90 x 0.5 = 0.45 m3/s
Result:Peak discharge: 0.45 m3/s (5-year event)
Example 3: Mixed-Cover Catchment (Composite C)
Problem:A = 60 ha made up of 20 ha paved (C = 0.90), 30 ha lawn on clay (C = 0.25) and 10 ha woodland (C = 0.15), Tc = 45 min. The 25-year IDF curve gives i = 60 mm/hr at a 45-minute duration.
Solution:Area-weighted C = (20 x 0.90 + 30 x 0.25 + 10 x 0.15) / 60 = (18 + 7.5 + 1.5) / 60 = 27 / 60 = 0.45 Q = C x i x A / 360 Q = 0.45 x 60 mm/hr x 60 ha / 360 i x A = 60 x 60 = 3600 3600 / 360 = 10.0 Q = 0.45 x 10.0 = 4.5 m3/s
Result:Composite C: 0.45 | Peak discharge: 4.5 m3/s (25-year event)
Frequently Asked Questions
What is Peak Discharge (Rational Method)?
Peak discharge is the maximum instantaneous rate of stormwater runoff leaving a catchment during a design storm, reported here in m3/s. The Rational Method is the procedure used to estimate it: Q = C x i x A, in which a dimensionless runoff coefficient C scales the design rainfall intensity i falling on catchment area A. It appears above as Q = C x i x A / 360 only because that divisor converts mm/hr and hectares into m3/s. It returns that single peak value only — no hydrograph and no runoff volume — which is enough to size an inlet, culvert, roadside channel or storm sewer for conveyance, but not enough on its own to size a detention basin. FHWA HDS-2 and ASCE/WEF MOP 77 both set it out as the standard procedure for small urban and highway drainage areas.
How is Peak Discharge (Rational Method) calculated?
Q = C x i x A / 360 is evaluated directly, but the 360 fixes the units rather than merely requiring them to be consistent: i must be in mm/hr and A in hectares, and Q then comes out in m3/s. Entering m/hr and m2 gives a wrong answer even though those units are self-consistent with each other. The working order is to fix a return period, estimate the time of concentration Tc, read i from the local IDF curve at a duration equal to Tc, take C from a land-use table (area-weighted where cover is mixed), then multiply. The result assumes rainfall of uniform intensity over the whole catchment for the full duration of Tc.
Why does the Rational Method peak discharge formula divide by 360 in metric units?
The 360 is a unit conversion, not an empirical constant. One mm/hr on one hectare delivers 0.001 m x 10,000 m2 = 10 m3/hr, and 10/3600 = 1/360 m3/s, so C x i x A / 360 returns m3/s directly from mm/hr and hectares. With area in km2 the divisor becomes 3.6. The US customary form needs no constant: Q in cfs = C x i (in/hr) x A (acres), because 1 in/hr over 1 acre is 1.008 cfs and the 0.8 percent difference is conventionally ignored.
How do I choose the runoff coefficient C for a Rational Method peak discharge estimate?
C comes from published land-use tables such as those in FHWA HDS-2 and the ASCE/WEF manuals of practice: asphalt or concrete pavement 0.70–0.95, roofs 0.75–0.95, downtown business districts 0.70–0.95, single-family residential 0.30–0.50, suburban residential 0.25–0.40, parks and cemeteries 0.10–0.25. Mixed catchments take an area-weighted composite. Because Q is directly proportional to C, redevelopment that lifts a composite C from 0.30 to 0.75 multiplies peak discharge by 2.5 for the same storm.
Why is peak discharge highest when storm duration equals the time of concentration in the Rational Method?
A storm shorter than Tc leaves part of the catchment not yet contributing at the outlet. A storm longer than Tc covers the whole area, but IDF curves fall steeply with duration, so i drops. The product peaks at a duration of Tc, which is where the design intensity is read. Because IDF curves are steepest at short durations, an error in Tc moves Q as much as an error in C does, and most manuals impose a 5-minute minimum Tc since published IDF data rarely extends below it.
How is the time of concentration estimated for a Rational Method peak discharge calculation?
Kirpich's 1940 formula is the usual choice: Tc in minutes = 0.0195 x L^0.77 x S^-0.385, with L the longest flow path in metres and S the average slope in m/m; the US customary form uses 0.0078 with L in feet. Kirpich calibrated it on seven small, steep, well-channelled agricultural catchments in Tennessee of roughly 0.5–45 ha, so practice multiplies Tc by 0.4 for overland flow on asphalt or concrete and by 0.2 for concrete-lined channels. NRCS TR-55 segmental sheet, shallow-concentrated and channel flow suits mixed urban surfaces better.
What catchment size limit applies before a Rational Method peak discharge estimate becomes unreliable?
FHWA HDS-2 recommends the method only for drainage areas up to about 80 ha (200 acres) and only where flood storage is not significant; many state DOT and municipal manuals cut that to 20–40 ha. Above that size rainfall is no longer uniform over the whole catchment for a full Tc, and channel routing and storage attenuate the peak. Larger or storage-influenced catchments call for NRCS TR-55 curve-number procedures or an event model such as HEC-HMS or EPA SWMM.
Can a small impervious sub-area produce a higher peak discharge than the whole catchment in the Rational Method?
It can, and most drainage manuals require the check. If a small paved sub-area drains straight to the outlet while the rest of the catchment is pervious with a long Tc, the sub-area's short Tc selects a far higher intensity from the IDF curve. Q computed for that sub-area alone can exceed Q for the full catchment, so the inlet, pipe or culvert is sized on whichever value is larger.
How does the peak discharge from the Rational Method feed into culvert and storm sewer sizing?
Q becomes the design flow in Manning's equation, Q = (1/n) x Af x R^(2/3) x S^(1/2) in SI units, where Af is the flow cross-section and R the hydraulic radius, solved for the size that carries it. Roughness matters: Chow (1959) lists n = 0.010–0.013 for straight concrete culvert pipe, 0.021–0.030 (0.024 typical) for corrugated-metal storm drain, and 0.025–0.033 for clean straight natural channels. Culverts must also be checked for inlet control and allowable headwater under FHWA HDS-5.
Does the design return period change the peak discharge given by the Rational Method?
Only through i, the rainfall intensity — the return period and Tc fields steer which IDF value you enter, they are not in the equation itself. A 100-year IDF curve gives a larger i than the 10-year curve at the same duration, so Q scales with that ratio; in the bare equation C and A do not change. One refinement does adjust C, however. Because tabulated C values were derived from frequent storms, ASCE and FHWA HDS-2 practice multiplies C by a frequency factor Cf of 1.1 for the 25-year, 1.2 for the 50-year and 1.25 for the 100-year event, capping C x Cf at 1.0. This page applies no Cf, so enter the already-adjusted C if your design manual requires one.
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Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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