+----------------------+-------------+-----------------------+ | Substances. | Expansions. | Authors. | +----------------------+-------------+-----------------------+ |White Deal, { | ·0000022685 | Captain Kater. | | { | ·0000028444 | Dr. Struve. | |English Flint Glass, | ·0000047887 | Dulong and Petit. | |Iron (cast), { | ·0000061700 | General Roy. | | { | ·0000065668 | Dulong and Petit. | |Iron (wire), | ·0000068613 | Lavoisier and L. | |Iron (bar), | ·0000069844 | Hasslar. | |Steel (rod), | ·0000063596 | General Roy. | | | | {Commissioners of | |Brass, | ·0000104400 | {Weights and Measures | | | | {--mean of several | | | | {experiments. | |Lead, | ·0000159259 | Smeaton. | |Zinc, | ·0000163426 | Ditto. | |Zinc (hammered), | ·0000172685 | Ditto. | |Mercury _in bulk_, | ·00010010 | Dulong and Petit. | +----------------------+-------------+-----------------------+ From this table it is easy to determine the length of a rod of any substance the expansion of which shall be equal to that of a rod of given length of any other substance. The lengths of such rods will be inversely proportionate to their expansions. If, therefore, we divide the lesser expansion by the greater (supposing the rod the length of which is given to be made of the lesser expansible material), and multiply the given length by this quotient, we shall have the required length of a rod, the expansion of which will be equal to that of the rod given. For example:--The expansion of a rod of steel being, from the above table, ·0000063596, and that of brass, ·0000104400; if it were required to determine the length of a rod of brass which should expand as much as a rod of steel of 39 inches in length, we have ·0000063596/·0000104400 = ·6091, which, multiplied by 39, gives 23·75 inches for the length of brass required. We shall here, in order to facilitate calculation, give the ratio of the lengths of such substances as may be employed in the construction of compensation pendulums. TABLE II.