in which _l_ = the distance in English miles L′ _d_ at which the light would strike the ocean’s surface. We then reduce this value of H′ by the correction for mean refraction, which permits the light to be seen at a greater distance, and which = 2 _l_² ------, (2.) 21 2 _l_² 2 _l_² 4 _l_² So as to get, H = ------ - ------ = ------ (3.) 3 21 7 an expression which at once gives the height of the tower required, if the eye of the mariner were just on the surface of the water at _d_, where the tangent between his eye at S and the light at L would touch the sea. We must, therefore, in the first instance, find the distance _d_ S = _l′_, which is the radius of the visible horizon due to the height SS′ = _h_ of his eye above the water, and is, of course, at once obtained conversely by the expression:-- √(7 _h_) _l′_ = -------- (4.) 2 Deducting this distance from SL, the whole effective range of the light, we have L _d_ = _l_, and operating with this value in the former equation, 4_l_² H = ----- 7 we find the height of the tower which answers the conditions of the case.[81] From the above data the following Table has been computed. [81] In the above expressions _l_ and _l′_ are given in English miles, which in Scotland may be considered as bearing to nautical miles the ratio of 5280 to 6088. In order to convert a distance given in nautical miles to English miles, all that is needful is to add the log of the number of nautical miles to log 5280, and subtract log 6088.