297. But though the error introduced, by considering all the arch-stones above the joint that makes an angle of 60 degrees with the horizon as pressing wholly on the centre, is not a very considerable one, it may be desirable to use a method that approaches nearer to the truth when the arch is circular. However inclined a joint may be, a certain portion of the force of the arch-stone will be lost in friction, except when the joint becomes vertical. And in any case the pressure perpendicular to the curve of the centre will be expressed by the equation W (sin. a - f cos. a) = perpendicular pressure = P. But it will be more convenient to measure the angle a from the vertical line passing through the crown; then W (cos. a - f sin. a) = P.

If the angle included between the joints of one arch-stone be denoted by a, and the stones be alike in weight, and in the portion of the arch they occupy, then the pressure of any number n of arch-stones upon the centre will be expressed by the equation -

W (cos. n/2 a x sin.n+1/2 a - f sin. n/2 a x sin. n+1/2 a )* = pressure = P. sin. 1/2 a.

The magnitude of the arc a being ascertained, the sines and cosines to a radius of unity may be taken from a table

*This expression is equal to W (sum of cosines of n a - f x sura of Bines of na), and for the sura of the sines and cosines see Hutton's 'Course of Mathematics.' of natural sines. But the calculation will be more simple under the following form: -

W x sin. n+1/2 a /sin. 1/2 a x (cos. 1/2 na -fsin. 1/2 na). [A]

The computation becomes easier by using a table of logarithms.

"When the arch-stones are small, the pressure upon the centre is greater than when they are large, weight for weight; and as an arch-stone will seldom be smaller than would be embraced by one degree of the arch, the result obtained as above may be assumed as sufficiently accurate, the error being always in excess until each of the arch-stones becomes less than one degree. Now the number of degrees between the middle or crown and the angle of repose is 58; therefore by equation [A] the whole pressure upon the semi-centre may be determined. The equation being arranged thus:

W x ( cos. n/2 a x sin. n+1/2 a - f xsin. n/2 a x sin. n+1/2 a) [B] sin. 1/2 a. sin. 1/2 a.

The calculation can be made by logarithms as follows, the usual logarithmic radius being assumed.

Example. - If a = 1 degree, n = 58, and (by Art. 292) f = 0.625, we have

Log. cos. n/2 a = log. cos. 29°

=

9.941819

Log. sin. n+1/2 a a = log. sin. 29° 30'

=

9.692339

19.634158

Log. sin. 1/2 a = log. sin. 0° 30' ..

=

7.940842

11.693316

Deduct Log. Radius ......

=

10.000000

Log. 49.36 ..........

=

1.693316

Log.f =log. 0.625 ......

=

1 .795880

Log. sin. n/2 = log. sin. 29o ..

=

9.685571

Log. sin. n+1/2 a........

=

9.692339

19.173790

Log. sin. 1/2 a..........

=

7.940842

11.232948

Deduct Log. Radius...........................

=

10.000000

Log. 17.1 ..........

=

1.232948

And 49.36 - 17.1 = 32.26; consequently, 32.26 W is equal to the pressure of the semi-arch upon the centre, where W is the weight of a part of the ring of arch-stones embraced by one degree.