Adopting this as the smallest angle which the two lights should subtend at the observer’s eye, we may find the least vertical distance between them which will cause them to appear as separate objects by the following formula: H = 2 Δ . tan θ in which Δ is the observer’s distance in feet; θ, half the subtense, = 1′ 39″; and H the required height of the tower between the two lights in feet. The following Table gives the height in feet corresponding to the distance in nautic miles, from 1 to 20 inclusive: the heights, which are the bases of _similar_ isosceles triangles, increase, of course, in an arithmetical series: +---------------+-----------------++---------------+-----------------+ |Distance of the|Vertical distance||Distance of the|Vertical distance| | observer in | in feet between || observer in | in feet between | | Nautic Miles. | the Lights. || Nautic Miles. | the Lights. | +---------------+-----------------++---------------+-----------------+ | 1 | 6·02 || 11 | 66·22 | | 2 | 12·04 || 12 | 72·24 | | 3 | 18·06 || 13 | 78·26 | | 4 | 24·08 || 14 | 84·28 | | 5 | 30·10 || 15 | 90·30 | | 6 | 36·12 || 16 | 96·32 | | 7 | 42·14 || 17 | 102·34 | | 8 | 48·16 || 18 | 108·36 | | 9 | 54·18 || 19 | 114·38 | | 10 | 60·20 || 20 | 120·40 | +---------------+-----------------++---------------+-----------------+ ~Leading Lights.~