Math 2.0

Geophysics Definition

Unit hydrograph

Builds on Convolution

Definition

Let a drainage basin have area AA, and let D>0D > 0 be a time interval. Excess rainfall is the part of the rainfall that becomes direct runoff, and direct runoff is the discharge at the outlet minus the base flow. Suppose that a unit depth of excess rainfall falls at a uniform rate over the whole basin during the interval [0,D][0, D]. The DD-hour unit hydrograph of the basin is the function UD(t)U_D(t), t≥0t \ge 0, that gives the resulting direct runoff discharge at the outlet. The model assumes that the basin is linear and time invariant. Then, if excess rainfall depths P1,…,PMP_1, \dots, P_M (in units) fall during the successive intervals [(m−1)D,mD][(m-1)D, mD], the direct runoff is Q(t)=∑m=1MPm UD(t−(m−1)D),Q(t) = \sum_{m=1}^{M} P_m \, U_D\big(t - (m-1)D\big), where UD(s)=0U_D(s) = 0 for s<0s < 0.

Remarks

The total volume under a unit hydrograph equals the volume of one unit of excess rainfall over the basin: ∫0∞UD(t) dt=A⋅(unit depth)\int_0^\infty U_D(t)\,dt = A \cdot (\text{unit depth}). For example, take D=1D = 1 hour and sampled values UD=0,2,5,3,1,0U_D = 0, 2, 5, 3, 1, 0 at hours 0,…,50, \dots, 5. For depths P1=2P_1 = 2 and P2=1P_2 = 1, the runoff at hour 22 is 2⋅5+1⋅2=122 \cdot 5 + 1 \cdot 2 = 12, a discrete convolution. A non-example is the hydrograph of a storm of duration DD with two units of excess rainfall. It is not a unit hydrograph, since the rainfall depth is not one unit; under linearity it equals 2UD2 U_D. The limit D→0D \to 0 gives the instantaneous unit hydrograph u(t)u(t), the impulse response of the basin, and then Q(t)=∫0ti(τ) u(t−τ) dτQ(t) = \int_0^t i(\tau)\, u(t - \tau)\, d\tau for an excess rainfall intensity ii.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Hydrograph
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DefinesUnit hydrograph
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Changed3 November 2026

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