Considering this thermometer as a vessel consisting of a glass bulb and a long glass stem of fine and uniform bore, hermetically sealed and containing only mercury and mercury vapour, he explained the numerical relation between the temperature as shown by the instrument and the volumes of the mercury and vessel. The scale is really defined by the method of graduation adopted. Two points of reference are marked on the stem at which the top of the mercury stands when the vessel is immersed (1) in melting ice, (2) in saturated steam under standard atmospheric pressure. The stem is divided into parts of equal volume of bore between these two points and beyond each of them. For a centigrade thermometer the bore-space between the two points is divided into 100 equal parts, and the lower point of reference is marked 0 and the upper 100, and the other dividing marks are numbered in accordance with this along the stem. Each of these parts of the bore may be called a degree-space. Now let the instrument contain in its bulb and stem, up to the mark 0, N degree-spaces, and let v be the volume of a degree-space at that temperature. The volume up to the mark 0 will be Nv, at that temperature; and if the substance of the vessel be quite uniform in quality and free from stress, N will be the same for all temperatures. If v₀ be the volume of a degree-space at the temperature of melting ice the volume of the mercury at that temperature will be Nv₀. If G be the expansion of the glass when the volume of a degree-space is increased from v₀ to v by the rise of temperature, then v = v₀(1 + G). The volume of the mercury has been increased therefore to (N + n)v₀(1 + G) by the same rise of temperature, if the top of the column is thereby made to rise from the mark 0 so as to occupy n degree-spaces more than before. But if E be the expansion of the mercury between the temperature of melting ice and that which has now been attained, the volume of the mercury is also Nv₀(1 + E). Hence N(1 + E) = (N + n)(1 + G). This gives n = N(E － G)⧸(1 + G). If we take, as is usual, n as measuring the temperature, and substitute for it the symbol t, we have, since N = 100(1 + G₁₀₀)⧸(E₁₀₀ － G₁₀₀), t = 100 {(1 + G₁₀₀)⧸(1 + G)} {(E － G)⧸(E₁₀₀ － G₁₀₀)} ... (D)