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