1. Synthetic Unit Hydrographs
All the methods of deriving a UH discussed previously require observations of rainfall and runoff. However, for some drainage basins rain gages and/ or streams flow gages are not, therefore, rainfall – runoff data are not available. For those basins, some kind of techniques to generate UH (without using rainfall – runoff records) is needed. The UH so derived is called synthetic UH.
Three types of synthetic UHs:
(1) those relate hydrograph characteristics (peak discharge, time base,
time-to-peak, etc.) to basin characteristics;
(2) those based on dimensionless UH; and those based on models of
watershed storage. UH

2. Snyder’s Method - 1
Study area: US Appalachian highlands. 10 ~ 10,000 mi2 (30 ~30,000 km2)
Properties characterizes the response of watershed under various rainfall inputs:
(a)   Duration of rainfall excess;
(b)   Lag time;
(c)   Time base of UH;
(d)   Time to peak;
(e)   Peak discharge of UH;
(f) Shape of UH. UH

3. Snyder’s Method - 2
Snyder’s method allows the computations of (a) lag time (tL); (b) UH duration (tr); (c) UH peak discharge (qp); (d) Hydrograph time width at 50% and 75% (W50, W75) of peak flow UH

4. Snyder’s Method - 3
1. Lag time (tL): time from the center of rainfall – excess to the UH peak 
tL = C1Ct (LLc)0.3
where tL = Time [hrs]; C1 = 0.75 for SI unit; 1.0 for English unit; Ct = Coefficient which is a function of watershed slope and shape, 1.8~2.2 (for steeper slope, Ct is smaller); L = length of the main channel [mi, km]; Lc = length along the main channel to the point nearest to the watershed centroid UH

5. Snyder’s Method - 4
2. UH Duration (tr): tr = tL / 5.5 where tr and tL are in [hrs]. If the duration of UH is other than tr, then the lag time needs to be adjusted as tpL = tL (tR - tr) where tLR = adjusted lag time; tR = desired UH duration. 3. UH Peak Discharge (qp): or where C2 = 2.75 for SI unit; 640 for English unit; Cp = coefficient accounting for flood wave and storage condition, 0.4 ~ 0.8; qp = specific discharge, [m3/s/km2] or [ft3/s/mi2] To compute actual discharge, Qp = Aqp UH

6. Snyder’s Method - 5
4. Time Base (tb): Assuming triangular UH, tb = C3 / qp where tb – [hrs]; C3 = 5.56 for SI unit, 1290 for English unit. 5. UH Widths: or where CW, 75 = 1.22 for SI unit; 440 for English unit; CW, 50 = 2.14 for SI unit; 770 for English unit;. W50, W75 are in hours; Usually, 1/3 of the width is distributed before UH peak and 2/3 after the peak Remember to check that the volume of UH is close to 1 cm or 1 inch UH

7. Snyder’s Method - ExampleUH

8. SCS Dimensionless UH UH

9. SCS Dimensionless UH UH

10. Instantaneous Unit Hydrograph (IUH)

11. HK WSD 15-min Dimensionless UH
Dimensionless discharge: Q’ = Q(Lag+tr/2)/V
Dimensionless time: t’= 100  t / (Lag+tr/2)
where V = surface runoff from 1” of rainfall excess (in ft3);
Lag = basin lag time (min)
= time between the centroids of rainfall excess
and runoff hydrograph
=1.47  A with A = basin area (acres);
tr = 15min
15 min 1”
Lag t u UH

12. HK WSD 15-min Dimensionless UH