A Catechism of the Steam Engine

by C. E. John Bourne

553. Q.--All this relates to the action of the paddle when the vessel is at rest: will you explain its action when the vessel is in motion?

A.--When the wheel of a coach rolls along the ground, any point of its periphery describes in the air a curve which is termed a cycloid; any point within the periphery traces a prolate or protracted cycloid, and any point exterior to the periphery traces a curtate or contracted cycloid--the prolate cycloid partaking more of the nature of a straight line, and the curtate cycloid more of the nature of a circle. The action of a paddle wheel in the water resembles in this respect that of the wheel of a carriage running along the ground: that point in the radius of the paddle of which the rotative speed is just equal to the velocity of the vessel will describe a cycloid; points nearer the centre, prolate cycloids, and points further from the centre, curtate cycloids. The circle described by the point whose velocity equals the velocity of the ship, is called the rolling circle, and the resistance due to the difference of velocity of the rolling circle and centre of pressure is that which operates in the propulsion of the vessel. The resistance upon any part of the float, therefore, will vary as the square of its distance from the rolling circle, supposing the float to be totally immerged; but, taking into account the greater length of time during which the extremity of the paddle acts, whereby the resistance will be made greater, we shall not err far in estimating the resistance upon any point at the third power of its distance from the rolling circle in the case of light immersions, and the 2.5 power in the case of deep immersions.

554. Q.--How is the position of the centre of pressure to be determined?

A.--With the foregoing assumption, which accords sufficiently with experiment to justify its acceptation, the position of the centre of pressure may be found by the following rule:--from the radius of the wheel substract the radius of the rolling circle; to the remainder add the depth of the paddle board, and divide the fourth power of the sum by four times the depth; from the cube root of the quotient subtract the difference between the radii of the wheel and rolling circle, and the remainder will be the distance of the centre of pressure from the upper edge of the paddle.

555. Q.--How do you find the diameter of the rolling circle?

A.--The diameter of the rolling circle is very easily found, for we have only to divide 5,280 times the number of miles per hour, by 60 times the number of strokes per minute, to get an expression for the circumference of the rolling circle, or the following rule may be adopted:--divide 88 times the speed of the vessel in statute miles per hour, by 3.1416 times the number of strokes per minute; the quotient will be the diameter in feet of the rolling circle. The diameter of the circle in which the centre of pressure moves or the effective diameter of the wheel being known, and also the diameter of the rolling circle, we at once find the excess of the velocity of the wheel over the vessel.

556. Q.--Will you illustrate these rules by an example?

A.--A steam vessel of moderately good shape, and with engines of 200 horses power, realises, with 22 strokes per minute, a speed of 10.62 miles per hour. To find the diameter of the rolling circle, we have 88 times 10.62, equal to 934.66, and 22 times 3.1416, equal to 69.1152; then 934.66 divided by 69.1152 is equal 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 pressure and the velocity of the pistons had been the same, then a pressure of 2900 Lbs. upon the vertical paddles would have been balanced by an equal pressure on the pistons, which would have been in this case about .75 Lbs. per square inch; but as the effective velocity of the centre of pressure is 4.997 feet per second, while that of the pistons is only 3.66 feet per second, the pressure must be increased in the proportion of 4.997 to 3.66 to establish an equilibrium of pressure, or, in other words, it must be 1.02 Lbs. per square inch. It follows from this investigation, that, in radial wheels, the greater part of the engine power is distributed among the oblique floats.