The head needed at a bifurcation in order to get the supply into a branch or distributary is always small unless the velocity is high. For a velocity of 3 feet per second the head required is only about ·16 ft., for 2 ft. per second ·1 ft. On an Inundation Canal which has no weir across the river, the mean supply downstream of the regulator (which is built a few miles down the canal lest it should be damaged by the shifting of the river) is, as has been mentioned, about half the full supply. The command in such canals is not generally very good. A distributary can often obtain only mean supply and it should be designed so as to command the country when it is carrying mean supply. A detailed description of Inundation Canals in Northern India, is given in _Punjab Rivers and Works_. Let M, F, m, f, be the mean and full supply discharges at the heads of a canal and of an average distributary on it and let the number of distributaries be n. It has been seen (Chap. I. Art. 6.) that M = ·8F about. Let k be the proportion of the supply lost by absorption in canal and branches. Then n m = (1 - k) M = ·8 (1 - k) F. If the distributaries all run with full supplies--at the time of greatest demand--for 4 days out of 5, then, nf = 1·25 (1 - k)F f 1·25 - = ---- = 1·56 m ·8 Since k depends on the wetted area, it is not likely to be so great for F as for M, but the above gives a general idea of the ratio of the full kharif discharge to the mean kharif discharge. On a large canal the circumstances of the distributaries will not all be similar. Some will run full for a greater proportion of their time than others. They can be divided into groups and the ratio of the full to the mean supply calculated for each group. The mean supply is, as above stated, obtained from the area to be irrigated, and the duty as estimated at the distributary head.