Archimedes Estimate of Pi

[Home]   [Puzzles & Projects]    [Delphi Techniques]   [Math Topics]   [Library]   [Utilities]

 

Search

 

Search DelphiForFun.org only

Support DFF

 If you shop at Amazon anyway,  consider using this link. We receive a few cents from each purchase.   Thanks.

In Association with Amazon.com

 

Support DFF

 If you benefit from the website,  in terms of knowledge, entertainment value, or something otherwise useful, consider making a donation via PayPal  to help defray the costs.  (No PayPal account necessary to donate via credit card.)  Transaction is secure.

 

 

Contact

Feedback:  Send an e-mail with your comments about this program (or anything else).

 

Search DelphiForFun.org only

 

 

 

 

Problem Description

Archimedes made a significant contribution to mathematics when he developed the first iterative method for calculating the value of  Pi with greater precision that then existed.  This program satisfies my curiosity about how he did it. 

Background & Techniques

Here is an illustrated description of Archimedes' method for estimating Pi developed a couple of hundred years BC. Starting with circumscribed and inscribed hexagons for a given circle, he calculated the perimeters and used the ratio of each perimeter to the diameter of the given circle as an estimate of Pi ( = Circumference/Diameter). Since the circumscribed perimeter will be larger than the circumference of the circle, the derived estimate of Pi will be too large. Similarly the inscribed perimeter will be too small and produce a low estimate. As the number of sides increases, the two estimates converge towards Pi.

 

Pi is between 3.0 and 3.464
(but it gets better)!

Archimedes started with what was probably the largest number of sides for which he knew the actual ratio of the side lengths to the diameter, a hexagon. He then cleverly derived the ratios for 12 sided regular polygons and repeated the same logic for 24, 48, and 96 sided polygons. Most online explanations of Archimedes procedure  cheat and use trigonometry but here we'll do it the way Archimedes did it using only algebra and geometry. The English translation of Proposition 3 in "The Measure of a Circle" used as my source and  is from "The Works of Archimedes" translated from Greek by T. L. Heath and published in 1897. A free download is available here.

After 4 iterations, Archimedes determined that the true value of Pi lies between 310/71 and 31/7 . Decimal terminology had not yet been developed so he used rational fraction estimates for irrational values, starting with estimates for the square root of 3, but making sure that the error introduced only made his estimate worse than if he could have used the true value. Pretty smart fellow!

In the text below, I'll describe the "clever" part in getting from the known perimeter of the hexagon to the perimeter of the 12 sided polygon. Actually, he only needed to concentrate on the length of one side and what happens to side lengths if we bisect the N sides to create a polygon with 2N sides. The technique for interior and exterior side lengths both use Euclid's Angle Bisector Theorem: "The angle bisector any angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle." It is from Euclid's Book VI, Proposition 3 and, as Heath did, we'll refer to it as EUCL VI.3 when it is used  in the following explanations. Archimedes tended to omit intermediate steps and explanations in his proofs. Heath added some explanatory text in his translation and  I've added more explanation for the parts that weren't as obvious to me as they must have been to Archimedes and Heath.
 


Circumscribed Detail

Here is Archimedes' derivation of the Pi estimate for circumscribed polygons.  Heath placed his explanations in [ ] brackets which I have tried to retain.  For the exterior hexagon, he considers the triangle formed by connecting point O, the circle's center to point A, the point of tangency of a side, to point C, the end of that side. He knows that angle AOC is 1/3 of a right angle (30 degrees to us). He also knows that, for such a triangle (which we commonly call a 30-60-90 triangle):

Archimedes then says:

(1) the ratio of OA:AC [= sqrt(3):1] >265:153 and
(2) the OC:AC ratio is [= 2:1] = 306:153.

Me: From this we can derive a 1st estimate of Pi as the perimeter divided by the diameter (twice the length of OA ) or 12*AC/(2*OA) = 6*(AC/OA)=6*1/sqrt(3)=3.464 or, for Archimedes since decimal notation did not yet exist, it would be ratios of the semi-perimeter to the radius = 6/(265/153) = 918/265 which also happens to be 3.464.  We'll use this same concept to track the values for the rest of the exercise.   

Now he goes on to work on the  on the 12 sided polygon:

Archimedes:  First, draw OD bisecting the angle AOC and meeting AC in D.

Me:  So that now AD represents half the length of one side of a 12 sided polygon.

Archimedes:
Now CO:OA=CD:DA [By EUCL VI.3]
So that [(CO+OA):OA = CA:DA or]
(CO+OA):CA = OA:DA

Me: I had to step through it this way:

CO/OA=CD:DA
CO/OA+1 = CD/DA +1
==> CO/OA+OA/OA = CD/DA + DA/DA
==> (CO+OA)/OA = (CD+DA)/DA
==> [(CO+OA)/OA = CA/DA]
Multiply both sides of the above by OA/CA to get
==> (CO+OA):CA = OA:DA

Archmedes: Therefore [by (1) and (2)] OD/DA > 571/153             (3)

Me: The estimate for Pi, for the 12 sided polygon, if A. had bothered to record it, is
Pi < 12*AD/OA=12*153/571 = 3.215

Archimedes:
Hence OD2 : AD2) [= (OA2 + AD2) : AD2] > (5712 + 1532) : 1532
> 349350 : 23409,

So that OD/DA > 5911/8 : 153               (4)

Me: Pythagoras allows us to replace the square of hypotenuse OD with the sum of the squares of sides OA and AD.  We now have the ratios for the triangle AOD which we can bisect again, giving Archimedes' point E for the 24 sided polygon and giving an OA/AE ratio of 1162.125/153 and a Pi estimate of 24*153/1162.125=3.160.

Similarly Archimedes uses AOE to calculate AOF for the 48 sided case, he gets 2334 1/4 : 153 and we can calculate Pi < 153*48/2334.25 = 3.1462.

An finally, from AOG, the 96 side case, he gets 4673 1/2 : 153 and Pi < 153*96/4673.5 = 3.1428 which he expresses as 3 1/7.  His completed figure includes one additional line, AH, merely to show the full extent of a side for the final polygon

 



Inscribed Detail

Calculating Pi estimates for the inscribed polygons seemed trickier than for circumscribed. Again, I'll expand the Archimedes proof to add my explanations of things which may not be obvious to us mere mortals.

Archimedes: Let AB be the diameter of a circle and let AC meeting the circle at C make the angle CAB equal to one third of a right angle. Join BC.

Then AC:CB [sqrt(3):1)] < 1351 :780.

Me: This time the estimate of sqrt(3) is greater than the true value so we can be sure that it does not make the estimate, which will always be low, closer than the rest of the mathematics might  indicate.

We can make our first estimate of Pi from the known ratios From the given triangle; we know that the ratio of the diameter (AB) to one side of the hexagon (CB) is 1/2  so our initial (low) estimate of the value of Pi is 6/2 or 3.00.

Archimedes: First let AD bisect the angle BAC and meet BC in d and the circle in D. Join BD.

Then angle BAD = angle dAC = angle dBD and the angles at D and C are both right angles.

It follows that the triangles ADB [ACd], and BDd are similar.

Me: BAD = dAC by construction . Angles at D and C are right angles because they are angles inscribed in a semicircle which are always right angles. (Search the Web for "angle inscribed semicircle" for many proofs). The vertical angles at d are equal. So angle dAC = 180-90-Cda = 180-90-DdB = DBd, The three triangles thus all have the same three angle values and so are similar by definition.

Archimedes:

Therefore AD : DB = BD : Dd [=AC : Cd] =AB : Bd [Eucl. VI.3] = AB + AC : Bd + Cd = AB + AC : BC or BA + AC : BC = AD : DB

Me: It might be obvious to geometers, but it wasn't clear to me that if AC:Cd=AB:Bd then it is also true that AC:Cd = (AC+AB):(Cd+Bd). Here's my proof:

AC/Cd=AB/Bd = some constant, k. So AC= k*Cd and AB = k*Bd.

Adding these equations, we get AC+AB=k*Cd+k*Bd, so AC+AB= k*(Cd+Bd) and (AC+AB)/(Cd+Bd)=k. ===> (AC+AB) : (Cd+Bd) = k = AC:Cd

Archimedes: But AC:CB < 1351:780 from above while BA:BC = 2:1 = 1550:780

Therefore AD:DB < 2911 : 780     (1)

[Hence AB2:BD2 < (29112 + 7802) : 7802 < 9082321:608400]

Thus AB:BD < 3013 3/4: 780        (2)

Me: This is enough to give us our estimate for the 12 sided inscribed polygons. Because AB is the hypotenuse of right triangle ABD, AB:DB less than the calculated value 3013 3/4 :780 and Pi > 12* (780/3013.75) = 3.1057

Now the hard thinking is done and with the help of a calculator or computer, even most of the work is done. Archimedes bisects DAB with AE to get the 24 sided ratios. AB:BE < 1838 9/11: 240                  (4)

giving the 24 sided case estimate: Pi > 24*(240/1838.818) = 3.1325. Similarly 48 side case: AB:BF < 1009 1/6: 66 and Pi > 48*66/1009.167 = 3.1392 and 96 side case: AB:BG < 2017 1/6: 66 and Pi > 96*66/2017.167 - 3.1410

A few final notes: 

bulletArchimedes use of rational fractions raises the question: "When did the positional notation decimal system develop?"  The answer from http://en.wikipedia.org/wiki/Hindu-Arabic_numeral_system is:  "The Hindu–Arabic numeral system or Hindu numeral system is a positional decimal numeral system developed by the 9th century by Indian mathematicians, adopted by Persian (Al-Khwarizmi's circa 825 book On the Calculation with Hindu Numerals) and Arabic mathematicians (Al-Kindi's circa 830 volumes On the Use of the Indian Numerals), and spread to the western world by the High Middle Ages."
bulletThe statements about the values Archimedes used for  sqrt(3) seemed to need verification.  Here's what I found:
bulletFor the circumscribed case, 265/153 is indeed less than the sqrt(3).  This value represents the length of the radius which appears the denominator of the Pi estimate, making that estimate larger that the true value.
bulletFor inscribed, the assumed sqrt(3) value, 1351/780  is larger than the which, thanks to Pythagoras, forces the calculated polygon side value to be too small which in turn forces the Pi estimate to be too small.    
bulletIf you download the PDF version of the Heath text, he has an extensive introduction which is interesting but the the text used here is from page numbers 93-98 of "Measurement of a Circle"  and  starts around PDF page number 281.
bulletI entered many letter combinations describing lines, angles, and triangles as I transcribed and expanded on Archimedes' text.  In the process I'm sure that there is more than one typo.  If you work through the text and find any of these, please let me know.

Programmer's Notes:

There is nothing particularly complex about the code, just a lot of fiddling with trig functions to draw the figures simulating Archimedes'.  The results displayed in the string grid are based on the rational number approximations used by Archimedes, but for the graphics there was no reason not to use sine and cosine functions.  For readability, the detail images posted above show only the first iteration moving from 6 to 12 sides but the original figures and my simulation has additional lines for a side of the 24, 48 and 96 side cases.   This meant using little tricks like offsetting the letters labeling the lines representing shorter sides so they do not overlap.  I spent more hours this week working on the mathematics and text than on the code, so I'll keep this short.  Use the feedback links to inquire if you have questions about the code (or the math). 

The source code references  some routines from our DFF Library so, if you have not already done so, you will need to download the library before recompiling.        

Running/Exploring the Program 

bulletDownload source
bulletDownload  executable

Suggestions for Further Explorations

By abandoning the graphics and the rational number approximations, it would be interesting to add a button to carry the iteration process further, say 100 iterations to see how accurate the estimates would be with 1030 sided polygons.  J
  .
   

 

Original:  March 28, 2010

Modified:  March 29, 2010

 

  [Feedback]   [Newsletters (subscribe/view)] [About me]
Copyright © 2000-2014, Gary Darby    All rights reserved.