224 Tables and Rules for Suspension Bridges. [April, 



1.0352 x secant of 15<>. 



9.8602 = length of upper link. 

 .2588 = X sine of deflexion 15°. 



2.5418 = deflection of upper link, 

 ft. in. ft. in. 



5 X 16 + 9.8602 X 2 == 99.7204 length of chain. 

 The sum of column No. 5 = 7.5068 deflection of ditto. 

 Ditto No. 6 = 1.5014 rise of roadway. 



7.5068 -f- 1.5214 + 3.5 = 12.5082 height of the point of s-us- 



pension at standard. 

 N. B. Column 5 is found by multiplying column 4 by 5 feet. 

 Column 6 is one-fifth of column No. 5. 

 Column 7 is equal to columns 5th -f- 6th -j- 3 .5 feet. 



The geometrical construction of this problem will answer as a 

 proof to the foregoing rule, and will be of assistance in making plans 

 of suspension bridges. 



C^J> 



In the right-angled triangle ABC make the angle A = 15* = 

 angle of suspension, and the side AB = 5 feet = length of one link 

 of the chain. Divide the side CB into as many spaces, commencing 

 at C, as there are drop -bars in \ the space — 8| spaces, and join Aw 

 A 2 n, &c. From the centre A with the radius AB describe the arc BD, 

 and complete the lines shewing the sines and cosines of the angles 

 formed by the line AB and the radii Aw, A 2 w, A 3 w, &c. Then as 

 these radii are parallel to the links of the chain, the sines of the 

 angles E 1, E 2, E 3, &c. are the differences between the lengths of 

 the drop-bars 1, 2, 3, 4, &c. and the cosines of these angles are the 

 spaces which the links of the chain occupy in the space of the bridge. 

 Supposing n == length of the centre drop-bar, the other drop-bars will 

 be as follows : 

 Centre bar w. 



8th, w + E 8. 



7th, n + E 8 + E 7. 



6th, n -f- E 8 + E 7 + E 6, and so on. This does not in- 

 clude the rise of the road, however, which is an arbitrary quantity. 



