The direction of wind is horizontal, or nearly so, when unobstructed. Precipitous mountains or tall buildings deflect the wind considerably from its usual horizontal direction. Its power usually does not exceed 30 pounds per superficial foot except on elevated places, where it sometimes reaches 50 pounds or more. This is the pressure upon a vertical surface; roofs, however, generally present to the wind an inclined surface. The effect of a horizontal force on an inclined surface is in proportion to the sine of the angle of inclination; the direction of this effect being at right angles to the inclined surface. The force thus acting may be resolved into forces acting in two directions - the one horizontal, the other vertical; the former tending, in the case of a roof, to thrust aside the walls on which the roof rests, and the latter acting directly on the materials of which the roof is constructed - this latter force being in proportion to the sine of the angle of inclination multiplied by the cosine. This will be made clear by the following explanation. Referring to Fig. 83, let D KE be the angle of inclination of the roof, D E being equal to one foot. Bisect D K at A; draw A L parallel with EK; make A L equal to the horizontal pressure of the wind upon one foot superficial of a vertical plane. Draw A C perpendicular to DK, and LF parallel with A C from .Fdraw FC parallel with EK; draw A B parallel with BE. The sides of the triangle L A F represent the three several forces in equilibrium: LA is the force of the wind; L F is the pressure upon the roof; and A F is the force with which the wind moves on up the roof towards D. Now, to find the relation of the force of the wind to the strain produced by it in the direction A C, we have -

rad.: sin.:: FC: A C; F C = LA; therefore - rad.: sin.:: LA: A C - LA sin.; A C = F sin.;

or, the strain perpendicular to the surface of the roof equals the force of the wind multiplied by the sine of the angle of inclination. When A C represents this strain, then, of the two forces referred to above, B C represents the horizontal force, and A B the vertical force. To obtain this last force, we have -

rad.: cos.:: A C: A B. Putting for A C its value as above, we have -

rad.: cos.:: .F sin.: A B = F sin. cos.; V = F sin. cos.;

Fig. 83.

or, the vertical effect is equal to the product of the force of the wind upon a superficial foot into the sine and the cosine of the angle of inclination. This result is that which is due to the pressure of the wind upon so much of the inclined surface as is covered by one square foot of a vertical surface. The wind, acting horizontally through one foot superficial of vertical section, acts on an area of inclined surface equal to the reciprocal of the sine of inclination, and the horizontal measurement of this inclined surface is equal to the cosine of the angle of inclination divided by the sine. This may be illustrated from Fig. 83, thus -

sin.: rad.:: D E: D K.

D E equals 1 foot; therefore -

sin.: rad.:: 1: D K = 1/sin;

or, the surface acted upon by one square foot of sectional area equals the reciprocal of the sine of the angle of inclination. Again, the horizontal measure of this inclined surface may be obtained thus -

sin.: cos.:: D E: KE = cos/sin;

or, KE, the horizontal measurement, equals the cosine of the angle of inclination divided by the sine.

In the figure, make K G equal to one foot; then we have -

Ke: Kg:: V: W;

in which V, as above, represents the vertical pressure due to the wind acting upon the surface KB, and W the vertical pressure due to the wind acting upon the surface KH, or so much as covers KG, one foot horizontal.

Now we have, as above, K E equal to cos/sin K G = 1, and

V = F sin. cos. Substituting these values, we have, instead of the above proportion -

cos/sin:I:: F sin. cos.: W; from which -

W = Fsin. cos/.cos./sin. = Fsin. (94.)

or, the vertical effect of the wind upon so much of the roof as covers each square foot horizontal, is equal to the product of the force of the wind per square foot into the square of the sine of the angle of inclination.

Example. - When the force of the wind upon a square foot of vertical surface is 30 pounds, what will be the vertical effect per square foot horizontal upon a roof the inclination of which is 26o 33', or 6 inches to the foot?

Here we have F = 30, and the sine of 26o 33' is 0.44698; therefore -

W= 30x 0.446982 = 5.9937. This is conveniently solved by logarithms; thus -

 log. 30 = 1.4771213 0.44698 = 9.6502868 0.44698 = 9 ..6502868 5..9937 = 0.7776949

or, the vertical effect is (5 .9937, or) 6 pounds.

The form of equation (94.) may be changed; for, in a right-angled triangle, the sine of the angle at the base is equal to the perpendicular divided by the hypothenuse; which, in the case of a roof, is the height divided by the length of the rafter; or -

Sine = height/rafter = h/2.

Therefore, equation (94.) may be changed to -

W = F h2/12; (95.)

or, the vertical effect upon each square foot of a roof is equal to the product of the force of the wind per foot into the square of the height of the roof at the ridge, divided by the square of the length of the rafter (the height and length both in feet.)

Example. - When the force of the wind is 30 pounds, the height of the roof 10 feet, and the length of the rafter 22.36 feet, what will be the vertical effect of the wind? Here we have F = 30, h - 10, and l = 22.36; and -

W = 30 x 102 /22.362 = 6.