486 Proceedings of Royal Society of Edinburgh. [sess. 
some of the threads will come against the narrow parts of others 
in such a way that they will average up and fill the same space as 
an equal number of threads of a hypothetical yarn uniform 
throughout its length and with a diameter equal to the mean 
projection width of the real yarn. To find this mean value we 
may proceed in one or other of two ways. (1) The most ex- 
peditious method is to employ the integral calculus. We may 
consider, for purposes of calculation, that the twist is generated by 
keeping one thread stationary and rotating the other about the 
axis of the first as centre. Proceeding from section A to section 
B, the angle of twist grows from 0° to 90°, i.e. through J turn of 
twist. 
If we call the angle of twist 0 and consider any intermediate 
section C, the horizontal projection is AD or AB + CD + BC, 
but AB + CD = d, the diameter of the single thread, 
and BO = d 
BC = d cos 6 
. •. AD = d (1 + cos 0) 
And the sum of all the sections = d I (1 + cos 0)d6 between the 
limits 0 = 0 and 0 = 90° or ^ radians, 
or 
d J 2 ( 1 + cos 0)d0 = d 0 + sin 0 J 2 = d(^ 
+ 1 
Integral J + 1 
.*. the mean width of projection = ^ =- d = 
9 T 
-0 + i> 
= l*637fZ 
A graphical method of solving the problem . — The following 
graphical method will be intelligible to those who are not familiar 
