84 



Combining the load on the shaft due to the l)elt pull with that due to its 

 weight, the resultant load will be 



V x'^ -L,Ws--y)2> (14) 



hence the friction load will be 



If W. = 3 (ild^ X 3.36 L ) we have 



F = i 9 V x2 f [y + 3 (ii d2 X ?>.36 L )]2 ( ^'^ ^ 



Taking a specific case in which the cross belts are assumed to drive horizon- 

 tally on each side of the line shaft, and the main belt to make an angle of 30° 

 with the horizontal, we have 



The velocity of intermediate belting is so variable that any assumption of 

 speed must be regarded as applying to a particular case or representative of a cer- 

 tain type of factory, and can not be taken as general. In many machine shops 

 the average speed of intermediate belts is not more than 500 feet per minute; in 

 others the average speed is more than twice as great, and in wood-working shops 

 it is still greater. 



For our present purpose we shall assume an average speed of 660 feet per 

 minute for belts running from the main shaft to a secondary or countershaft, and 

 four times this speed for the velocity of belt from engine .to main shaft, that is 



y =r = 4. 



Substituting these values in ( 13) we have 

 ^ 1 , lOOOd^^N 1 . , . x-, 



therefore F = ^-o] ^ [ ^ ( H ' N cos =c ] ^ + [ i (| d « N sin oc ) + 7.9 d ^ L ] ^ 



= V [ 0.025 d •' N ] ^ + [ 0.014 d ^« N + 0.6 d -' L 1 ^ ( 16 ) 



From the formula for the power absorbed by friction we have 



F^ = H-P"- = 3l5^fl^'°^^-P''-=^-^^«^^^^' ^''^ 

 hence the ratio of power absorbed by friction to the horse-power which the shaft 

 is capable of safely transmitting will be 



H. P„. 0.0^ 8dN F 0.08 F 



ITPT = 0.01 d ^ N = "d^ P^"" '''''• 



