Before dealing with Callendar’s formula, the term “degrees on the platinum scale” will be explained. Such degrees are obtained by assuming that the increase of resistance of platinum is uniform at all temperatures; that is, that the temperature-resistance curve is a straight line, and not a parabola. For example, a piece of platinum wire of 2·6 ohms resistance at 0° C. will show an increase to 3·6 ohms at 100° C.—an addition of 1 ohm for 100°. We now assume that a further augmentation of 1 ohm, bringing the total to 4·6 ohms, will represent an increase of 100°, or a temperature of 200°. Similarly, a total resistance of 5·6 ohms would indicate 300°, and 12·6 ohms 1000°. The temperature scale obtained by this process of extrapolation is called the “platinum scale,” and differs considerably from the true or gas scale, the difference becoming greater as the temperature rises. This is indicated in fig. 32, in which A represents the true parabolic relation between resistance and temperature, and B the assumed straight-line relation. Reading from curve A, the temperature corresponding to 8 ohms resistance is 600° C.; but from B the same resistance is seen to represent only 545° C., which is the “temperature on the platinum scale” to which this resistance refers. An inspection of fig. 32 shows that at all temperatures, except between 0° and 100°, the platinum-scale readings for given resistances are less than those indicated on the gas scale. Callendar’s formula is expressed in terms of the difference between the gas-scale and platinum-scale readings, and takes the form _t_ - _p_ = δ{(_t_/100)^2 - (_t_/100)}, where _t_ = temperature on the gas scale, _p_ = temperature on the platinum scale. δ = a constant, depending upon the purity of the wire. [Illustration: FIG. 32.—CONNECTION BETWEEN RESISTANCE OF PLATINUM AND TEMPERATURE: A, ON GAS SCALE; B, ON PLATINUM SCALE.] In order to determine the value of δ, it is necessary to measure the resistance of the wire at 0°, 100°, and a third temperature, which should be considerably above 100°. The readings at 0° and 100° are requisite to establish the platinum scale of temperatures; the third reading is required to calculate the value of δ, as _p_ and _t_ are equal at 0° and 100°, these points forming the basis of both scales. An example is appended to make this matter clear. _Example._—A platinum wire has a resistance in ice of 2·6 ohms; in steam, 3·6 ohms; in boiling sulphur, 6·815 ohms. To find the value of δ, the boiling point of sulphur being 444·5 on the gas scale. Since an increase of (3·6 - 2·6) = 1 ohm is produced by 100°, the increase observed in boiling sulphur, (6·815 - 2·6) = 4·215 ohms, will represent a temperature, on the platinum scale, of (4·215 × 100)/1 = 421·5° _p_. Applying Callendar’s formula, (444·5 - 421·5) = δ{(444·5/100)^2 - (444·5/100)}