[satterly] study of A CATENARY 45 



A Straight edge was held up to just touch the chain and its line 

 marked. The tangent through the medial line of the chain was then 

 estimated and the distance of the point of contact from the lowest 

 point of the chain measured by simply pulling the chain taut along 

 a scale, one end of which rested on a nail driven through the link at A. 



25 



(a) At a point, P, the tangent of the slope was — ; 



25 ^^ 



therefore, s=lllx — = 95 cms. 

 29 



Direct measurement yielded the same result. 



24 



(b) At another point, Q, the tangent of the slope was — ; 



24 ^^ 



therefore, s=lllx — = 148 cms. 

 18 



Direct measurement gave 146 cms. 



4. The slope of the normal being known, the length of the 



normal PG could be calculated. For thepoint P mentioned in 3 (a) 



above, x was 863^ cms. and Y 1463^ cms. Knowing the slope of 



25 

 the tangent, we get by geometry MG (Fig. 3) = — x 1463^, and 



PG = U6}iJ 1 + ( — V= 194 cms. The length of PG could have 



■^^-(^ï- 



been measured directly, but this is the more convenient way of 



Y2 

 getting it. The formula— gives 193 cms. — a fair agreement, 

 c 



X X 



5. From s = c sinh - and y + c = c cosh -we get 

 c c 



(y + c2) - s2 = c^ 



. • . y2 + 2cy - s2 = O 



2cy = s- — y- 



c = i . (s + y) (s - y) 

 2 y 



This affords another way of giving c. Applying it to the point 

 B, where y= 105 -3 -21 -0 = 84 -3 cms., we get 



1 (161 + 84-3) (161 -84-3) ,,, 



c = - • ^ ■ —-^ =111 cms. 



2 84-3 



This method does not require anything more than the hanging 

 chain itself, the forces not being required, hence it is perhaps the 

 simplest method of getting c. 



We thus see how by experiment the formulae of the catenary can 

 be verified and the student brought to realise the mutual helpfulness 

 of mathematics and physics. 



