l to 13.52 feet, which is the diameter of the
rolling circle. The diameter of the wheel is 19 ft. 4 in., so that the
diameter of the rolling circle is about 2/3ds of the diameter of the wheel,
and this is a frequent proportion. The depth of the paddle board is 2 feet,
and the difference between the diameters of the wheel and rolling circle
will be 5.8133, which will make the difference of their radii 2.9067; and
adding to this the depth of the paddle board, we have 4.9067, the fourth
power of which is 579.64, which, divided by four times the depth of the
paddle board, gives us 72.455, the cube root of which is 4.1689, which,
diminished by the difference of the radii of the wheel and rolling circle,
leaves 1.2622 feet for the distance of the centre of pressure from the
upper edge of the paddle board in the case of light immersions. The radius
of the wheel being 9.6667, the distance from the centre of the wheel to the
upper edge of the float is 7.6667, and adding to this 1.2622, we get 8.9299
feet as the radius, or 17.8598 feet as the diameter of the circle in which
the centre of pressure revolves. With 22 strokes per minute, the velocity
of the centre of pressure will be 20.573 feet per second, and with 10.62
miles per hour for the speed of the vessel, the velocity of the rolling
circle will be 15.576 feet per second. The effective velocity will be the
difference between these quantities, or 4.997 feet per second. Now the
height from which a body must fall by gravity, to acquire a velocity of
4.997 feet per second, is about .62 feet; and twice this height, or 1.24
feet, multiplied by 62-1/2, which is the number of Lbs. weight in a cubic
foot of water, gives 77-1/2 Lbs. as the pressure on each square foot of the
vertical paddle boards. As each board is of 20 square feet of area, and
there is a vertical board on each side of the ship, the total pressure on
the vertical paddle boards will be 2900 Lbs.

557. _Q._--What pressure is this equivalent to on each square inch of the
pistons?

_A._--A vessel of 200 horses power will have two cylinders, each 50 inches
diameter, and 5 feet stroke, or thereabout. The area of a piston of 50
inches diameter is 1963.5 square inches, so that the area of the two
pistons is 3927 square inches, and the piston will move through 10 feet
every revolution; and with 22 strokes per minute this will be 220 feet per
minute, or 3.66 feet per second. Now, if the effective velocity of the
centre of pres