Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Topic: Powers of a Length A , Products and Divisions, Graphically.
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
Eur Ing Panagiotis Stefanides

Posts: 567
Registered: 12/3/04
Powers of a Length A , Products and Divisions, Graphically.
Posted: Aug 15, 2005 2:02 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2016. All Rights Reserved.