220 CHAPTER XI 



components, namely A F and A G. Along C A produced lay off A E = n p, 

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

 and A K. 



Along A H lay off H L = A F, and along K A lay off K M = A G. 

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

 forces p and n p, in magnitude and direction ; A L is the vertical com- 

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



Now A F = p cos = p +GOSa andJff = n 



whence 4 L = ,4#+#L=,4#+.4F=(^+^) J 

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



VI + COS a 

 -V/2 V 



r (n + I) A/I + cos a 

 whence, whatever the ratio of p to w^>, 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 a 

 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 -f COS a 



be 50 tons, when np will be 450 tons and n = 9. 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 definite 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 roller 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. 



Again AK=npsin =np-\l - smdAG psm -= 



|I COS a 

 /> \ 



2 f ' "2 



whence AM = A K - A G =p (n- i)^^- = H 



and hence , = p (n i) and p 



V I COS a (n i) 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. 



