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Powers of a Length A , Products and Divisions, Graphically.
Posted:
Aug 15, 2005 2:02 PM


Powers of a Length A , Products and Divisions,Graphically by Compass and Ruler. [The theory simply lies on the "similar orthogonal triangles"]
Draw a line horizontally of the Unit length of the Ruler. Let it be [ej] ,and A a vertical line [jh] ,at j corresponding to the Length A ,picked up by the compass. Then [eh] will be the hypotenuse of the orthogonal triangle [ejh]. Let THETA be the angle whose tangent is: [hj]/[ej]=A/1=A. We extend the line [ej] to [k],and [eh] to [q]. We use the compass and place the length A on [ek] , so we get a point on it [m] , that [em]=A. Then , we draw a vertical line on [em] at [m],which meets the line [eq] at [n]. Then [mn]=[A^2] , and tanTHETA=[ A^2] / [ A]= A So we have graphically obtained the square of the Length A , BY COMPASS and RULER . We pick up the length [A^2] lay it along line [ek] so we get a point on it [r] ,so that [er]=A^2 , the vertical on it ,meets [eq] at [s] ,so [rs]=A^3, and tanTHETA=A. By doing so we get the powers we need of a length graphically by compass and ruler. So any series that involves powers of a length x [ such as e^x=1+x+[1/2!]x^2+... , log x , trigs, etc ], may be drawn graphically. Of course a similar method is involved for the product of two lengths , inverses ,etc. The inverse 1/A is obtained from the original triangle taking as vertical the Unit Length of the ruler. Then the horizontal length is 1/A and tanTHETA=A. If we fit another Length B vertically on [ek] at t , so that it meets [eq] at [v] , so that B= [tv] , then the horizontal length [et] is the product [1/A]*[B] or the ratio B/A , and tanTHETA=[B]/[B/A]= A .
Copyright 2005 Eur Ing Panagiotis Stefanides
Regards from Athens,
Panagiotis Stefanides http://www.stefanides.gr/



