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draw the logarithmic curve, whose subtangent atevery point equals CV.

To find an intermediate ordinate SO,

Bisect the distance CL in S, and through S drawSO parallel to CV. Make ST in the axis on one sideS equal CV and SX in the axis on the other, equalLG and on TX as a diameter describe the semicircleTOX, cutting SO in O, SO is the mean to CV andLG extreme proportionals. In like manner, by ex-treme and mean proportionals, any point in the curvemay be found.

PutYW = w, YL=/(,CV=/', m =2.302585, LG= >w.Then Log. w = Log. nh = f m (Log. w — Log. n .)

TO DESCRIBE THE CATENARY EQUALLY STRONG IN

EVERY POINT, AND TO DETERMINE ITS INTRADOS.

Let the vertical line C A be the radius of curvature Fig. 2at the vertex. Through A draw AR at right anglesto CA, and with CA radius and C centre, draw thequadrant of a circle AB, and the right line CR,making any angle with CA, cutting the quadrant ABin m and AR in R. Continue the ordinate DE, takenequal to C A, and drawn to a logarithmic curve, whoseaxis or assimptote is in the line AR, and whose sub-tangent equals CA, and upon it make DL equal CR.

Draw LM parallel to the assimptote of the loga-rithmic curve, and cutting that curve in M. FromA, the vertex of the arch, take the absciss AI equalto LM, and draw the ordinate IH equal in length tothe circular arc Am. In the same manner find anyother point U. Through AHU, draw the catenary