combined action of these two causes is, that a body weighing 195 lbs. at either pole will weigh but 194 over the equator. _The line of a falling body, called also the line of direction_, is interesting as being that direction in space at any point of the earth’s surface with reference to which all other directions are named, and by which they are to be determined. A few points remain to be named. _The flow of water is the result of the force of gravity_; the importance of this fact and its wide influence cannot be over stated; the gently falling dew, the mighty currents in the unfathomable depths of the ocean, as well as the rivulet merrily falling over the rocks to a lower level are all subject to the laws of terrestrial gravity. The upper surface of a liquid in a vessel exposed to the atmosphere is called _the free surface_ and is pressed downwards by the air under about 15 lbs. pressure per square inch. The free surface of a small body of a perfect liquid, at rest, is horizontal and perpendicular to the action of gravity although in large bodies of liquid, as lakes and ponds, the free surface is spherical, assuming the curvature of the earth’s surface. RULES RELATING TO THE VELOCITY OF FALLING BODIES. 1.—_To find the Velocity a falling Body will acquire in any given time._ Multiply the time, in seconds, by 32-1/6, and it will give the velocity acquired in feet, per second. _Example._ Required the velocity in seven seconds. 32-1/6 × 7 = 225-1/6 feet. Ans. 2.—_To find the Velocity a Body will acquire by falling from any given height._ Multiply the space, in feet, by 64-1/3, the square root of the product will be the velocity acquired, in feet, per second. _Example._ Required the velocity which a ball has acquired in descending through 201 feet. 64-1/3 × 201 = 12931; √12931 = 113·7 feet. Ans. 3.—_To find the Space through which a Body will fall in any given time._ Multiply the square of the time, in seconds, by 16-1/12, and it will give the space in feet. _Example._ Required the space fallen through in seven seconds. 16-1/12 × 7² = 788-1/12 feet. Ans. 4.—_To find the Time which a Body will occupy in falling through a given space._ Divide the _square root_ of the space fallen through by 4, and the quotient will be the time in which it was falling. _Example._ Required the time a body will take in falling through 402.08 feet of space. √402·08 = 20·049, and 20·049 ÷ 4 = 5·012. Ans. 5.—_The Velocity being given, to find the Space fallen through._ Divide the velocity by 8, and the square of the quotient will be the distance fallen through to acquire that velocity. _Example._ If the velocity of a cannon ball be 660 feet per second, from what height must a body fall to acquire the same velocity? 660 ÷ 8 = 82·5² = 6806·25 feet. Ans. 6.—_To find the Time, the Velocity per second being given._ Divide the given velocity by 8, and one-fourth part of the quotient will be the answer.