220 CHAPTER XI 



components, namely A F and A G. Along C A produced lay off yl £ = np, 

 and resolve A E into vertical and horizontal components, namely A H 

 and A K. 



Along A H \2iy oH H L = A F, and along K A l&y oft K M = A G. 

 Complete the parallelogram A M N L. Then AN is the resultant of the 

 forces p and np, m magnitude and direction ; ^ L is the vertical com- 

 ponent of forces p and n p, and ^ M is the horizontal component. 



XT AT- L o- . li + cos a J A TT L " J. |i +cosa 

 Now AF = p cos - = p \'— -^ — and .4 H = np cos- = np^J- -- 



whence AL = A H + H L = A H + A F ^ {p + np) JL±_£2!^ 

 If the hydraulic pressure act on the top roll and is V, 



+ cos a _ ^ , -^^^ \/2 F 



and p = 



V=ip+ np)J^±^^^ or p +-np= /^ ^ 



^^ ^ \ 2 V I + COS a 



V2F 



(W + I) Vl + COS a 



whence, whatever the ratio of p to np, or whatever the settings adopted, 

 the sum total pressure exerted perpendicularly on the bagasse is constant 

 when V and a are constant. 



As a decreases, i.e., when the vertical angle becomes steeper, cos « 

 increases ; hence, with decrease of the vertical angle the hydraulic load or 

 value of V must be increased to keep the value p -{- np the same. 



The problem which presents itself in this connection is : — What is the 



a/2 V 

 best value of n ? For example, let . = = 500 tons. Then p may 



VI + cos a 



be 50 tons, when np will be 450 tons and n — g. With a different setting 

 p may be 100 tons when np — 400 tons and n = 4. So far as the writer 

 is aware, there is no very delinite information on this point ; in other words, 

 the problem resolves itself into the question whether the front roller is to be 

 regarded as a feeding roUer and the back roller as a crushing roller, or whether 

 the front roller is to be regarded as a crushing roller also. In the latter case 

 the values of p and np tend to approach each other, but the maximum single 

 pressure obtained decreases. This problem may also be expressed as the 

 question : — Will better results be obtained by two crushings at a lower 

 intensity, or by one very light one and a second very heavy one ? The ex- 

 periments of the writer quoted earlier point to the obtaining of better results 

 when p and np are equalized as far as possible. 



A • A r^ . • ™ ^ \ — cos a J . ^ . • '^ ^ /'I — cos a 



Agam A K =np sm — =np'\\ and^G — ^sm -—pyi ■ 



It ro<^ a 



whence AM = AK - AG ^p {n-z) \- ^^^^ = H 



and hence =r = p [n — i) and p = 



a/ I — cos a (^ ~ l) V I — cos a 



But H is the horizontal component of the forces p and np pressing the 

 top roller against the brasses on the feed side of the mill, often referred to 

 as the side thrust. When n = i or p = np there is no side thrust, and the 

 side thrust increases as n increases and as a increases; that is to say, as 

 the vertical angle becomes flatter. 



