To find the horse-power of a wind-engine. HP = A X V2 / 1,100,000

When HP = effective horse-power;

A = area of sails in square feet; V = velocity of the wind in feet per second. To find the area of sails required for a given horse-power.

A = Hp X 1,100,000 / V2

The best effect is obtained when the total surface of the sails presented to the wind does not cover more than a quarter of the surface of the whole disk described by the radial arms or whips.

To find the force of wind. P = 0.002288 V2; P = 0.00422 V12; P = 0.0023 V2 X sin X. When P = pressure in lbs. per square foot; V = velocity in feet per second; V1 = velocity in miles per hour; X = angle of incidence of direction of the wind with the plane of the surface when it is oblique. To find the angle of the sails.

α = 23o - 18D2 / R2.

When α = angle of the sail with the plane of motion at any part of the sail;

D = distance of any part of the sail from the axis in feet; R = total radius of sail in feet.

To find angle of shaft with horizon. a = 8 degrees on level ground; = 15 degrees on high ground. To find breadth of whip. B = 1 /30 W; D =1/40 W; B1 = 1 /60 W;

When W = length of whip in feet; W1 = width of sail in feet; B = breadth of whip at axis in feet; D = depth of whip at axis in feet; B1 = breadth of whip at tip in feet; D1 = depth of whip at tip in feet; Divided by the whip in the proportion of 5 to 3, the narrow portion being nearest to the wind. W11 = 1/5 W;

D11 = 1/7 W. When W11 = width of sail at axis;

D11 = distance of sail from axis. Cross-bars from 16 to 18 inches apart. Velocity of tip of sails = 2.6 V, nearly. In examining the ratio between the velocity of the wind and the number of revolutions of the wheel-shaft Mr. Smeaton obtained the result in table below, for Dutch sails, in their common position, when the radius of the wheel was 30 feet:

Number of Revolutions of Wheel-shaft per Minute.

3 5

6

Velocity of Wind in an Hour.

2 miles

4 "

5 "

Ratio between Velocity of the Wind and Revolutions of Wheel-shaft. 0.666 0.800 0.833

Supposing the radius of the sail to be 30 feet, then the sail will commence at 1/6th, or 5 feet from the axis, where the angle of inclination will be 72°, at 2/6ths or 10 feet from the axis will be 71°, and so on.

In order to utilize the maximum effect of wind, therefore, it is necessary to load the wind-engine so that the number of revolutions of the wheel is proportional to the velocity of the wind.

To find proper number of revolutions of a wind-mill. N = 3.16 X V / L X sin U; if U = 16o,

N = 11.5 V / L ;

When N = number of revolutions of wheel per minute; V = velocity of the wind in feet per second:

= radius of center of percussion in feet;

R = extreme radius of wheel in feet; R1 = inner radius of wheel in feet; U = mean angle of sails to the plane of motion.