378

MISCELLANEOUS

TRACT 38.

Note, that this formula is adapted to the mean tempe-rature of the air 55°. But, for every degree of temperaturedifferent from this, in the medium between the temperaturesat the top and bottom of the altitude a, that altitude willvary by its 435th part; which must be added, when thatmedium exceeds 55°, otherwise subtracted.—Note, also, thata column of 30 inches of mercury varies its length by aboutthe y-rs part of an inch for every degree of heat, or ratherT 5 -SS of the whole volume.

But the formula may be rendered much more convenientfor use, by reducing the factor 10592 to 10000, by changingthe temperature proportionally from 55°. Thus, as the diff.592 is the 18th part of the whole factor 10592 ; and as 18 isthe 24th part of 435 ; therefore the correspondent change oftemperature is 24°, which reduces the 55° to 31°. So that

the formula is, a = 10000 x log. — fathoms, when the tem-

perature is 31 degrees ; and for every degree above that, theresult is to be increased by so many times its 435th p$rt.

PROBLEM XXXIV.

To divide a Given Circle into any proposed number of EqualParts, by means of other Circles Concentric with the Givenone.

This problem is now added here in the appendix, havingbeen omitted in its proper place, Tract xiv. vol. 1, besideanother problem, allied to this, as well in their nature as intheir fate and consequences.

A particular case of the present problem was first of all,as far as I know, proposed in that useful and valuable littleannual work, the Ladies’ Diary, for the year 1709, in thisform, viz, Seven men bought a grinding-stone, of 5 feet or60 inches in diameter : and they agreed together, that eachshould grind off an equal share ; so that one, beginning first,should grind his 7th part off the stone ; then a second should