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Math Help - sq. FD + sq. DB = sq. BF Almagest Ptolemy

  1. #1
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    sq. FD + sq. DB = sq. BF Almagest Ptolemy

    Hello, everyone I am reading through the Almagest by Ptolemy and I have a question I hope someone can resolve for me. In Book I Chapter 10 of the Almagest, Ptolemy sets up a semicircle ABC with two right triangles on the diameter AC. To view the text just click the link below:

    Claudius Ptolemy: The Almagest

    In this text Ptolemy proves sq. FD + sq. BD = sq. BF but when I try to do the math my calculations are off by 7/60^2 + 2/60^3 + 16/60^4 (or about 0.2%). If anyone can help me to improve the precision of my calculations it would be very much appreciated!

    Thanks in advance for any answers!

    Below you can find my calculations of the squares on FD, DB and BF using the sexagesimal (base-60) system employed by Ptolemy:

    We shall attempt to prove sq. FD + sq. DB = sq. BF with the values given in Ch 10 Book I of the Almagest. The line FD subtends an arc of 36 degrees and is equal to 37 parts 4 minutes and 55 seconds of the 120 parts of the diameter of the circle ABC. (Found by using the Table of Chords in Ch 11 Book I of the Almagest).


    60^0 60^-1 60^-2
    FD = 37 04 55
    a b c

    = (a + b+ c) (a + b + c)
    = a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2
    = a^2 + 2ab + 2ac + b^2 + 2bc + c^2
    = 37^2 + 2(37 * 4/60) + 2(37 * 55/60^2) + (4/60)^2 + 2(4/60 * 55/60^2) + (55/60^2)^2
    = 1369 + 296/60 + 4070/60^2 + 16/60^2 + 440/60^3 + 3025/60^4
    = 1369 + 296/60 + 4086/60^2 + 440/60^3 + 3025/60^4

    Now, since the above number is an improper fraction, we shall use a table to obtain the proper sexagesimal number.

    In the following table
    a = Initial Units of 60
    b = Previous Multiple of 60
    c = Units of 60 + Previous Multiple of 60
    d = Multiples of 60 (The integer obtained from c/60)
    e = Units of 60 (The remainder obtained from c/60)
    f = Base of Sixty

    a b c d e f
    3025 00 3025 50 25 60^-4
    0440 50 0490 08 10 60^-3
    4086 08 4094 68 14 60^-2
    0296 68 0364 06 04 60^-1
    1369 06 1375 22 55 60^ 0
    0000 22 0022 00 22 60^ 1


    Therefore sq. FD = 00, 22, 55; 04, 14, 10, 25
    or = 1375 + 4/60 + 14/60^2 + 10/60^3 + 25/60^4

    Now we shall find the square of DB. The line DB which subtends an arc of 60 degrees is equal to 60 parts of the 120 parts of the diameter of the circle ABC.

    DB = 60
    sq. DB = 3600

    Therefore the sq. DB is equal to 3600 or 01, 00, 00; 00, 00, 00, 00 in sexagesimal notation

    Now we shall find the square of BF. The line BF which subtends an arc of 72 degrees is equal to 70 parts 32 minutes and 3 seconds of the 120 parts of the diameter of the circle ABC.

    BF= 70^2 + 2(70 * 32/60) + 2(70 * 3/60^2) + (32/60)^2 + 2(32/60 * 3/60^2) + (3/60^2)^2
    = 4900 + 4480/60 + 420/60^2 + 1024/10^2 + 192/60^3 + 9/60^4
    = 4900 + 4480/60 + 1444/60^2 + 192/60^3 + 9/60^4

    Now, since the above number is an improper fraction, we shall use a table to obtain the proper sexagesimal number.

    In the following table
    a = Initial Units of 60
    b = Previous Multiple of 60
    c = Units of 60 + Previous Multiple of 60
    d = Multiples of 60 (The integer obtained from c/60)
    e = Units of 60 (The remainder obtained from c/60)
    f = Base of Sixty

    a b c d e f
    0009 00 0009 00 09 60^-4
    0192 00 0192 03 12 60^-3
    1444 03 1447 24 07 60^-2
    4480 24 4504 75 04 60^-1
    4900 75 4975 82 55 60^ 0
    0000 82 0082 01 22 60^ 1
    0000 01 0001 00 01 60^ 2

    Therefore sq. BF = 01, 22, 55; 04, 07, 12, 09
    or = 4975 + 4/60 + 7/60^2 + 12/60^3 + 9/ 60^4

    Now sq. FD + sq. DB = sq. BF

    And we shall construct another table similar to our previous tables

    In the following table

    Bases of 60: 60^2 60^1 60^0 60^-1 60^-2 60^-3 60^-4
    sq. FD: 00 22 55 04 14 10 25
    sq. DB: 01 00 00 00 00 00 00
    sq. BF: 01 22 55 04 07 12 09
    Error: 00 00 00 00 07 02 16

    Thus, our calculations are off by 7/60^2 + 2/60^3 + 16/60^4

    07/60^2 = 0.00194444444
    02/60^3 = 0.00000925925
    16/60^4 = 0.00001234567
    Total = 0.00196604937

    Or, our calculations are off by 0.196604937% (about 0.2%)
    Last edited by moohah; June 30th 2013 at 02:22 PM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: sq. FD + sq. DB = sq. BF Almagest Ptolemy

    It seems pretty clear that |FD|^2 + |BD|^2 = |BF|^2, from Pythagoras. My guess is that your error is due to rounding, since you only go as far as seconds in converting degrees to "parts of the diameter of the circle." If line BD is 60 units, and hence DE is 30 units, then the line FD is about 37.08204 units, which converts using your technique to 37, 4', 55.3416''. By truncating it at 55" you have introduced a small error - about 0.00026%. By my calculations the square of FD is 1375.078, whereas the square of your value is 1375.071. Put into you notation, the more accurate result is 1375, 4, 39, 30, 20.88 versus yours of 1375, 4, 14, 10, 25. This same type of error affects BF as well.

    However, even though your calculations have these rounding errors, the magnitude of the error is really not as bad as you state - the error of 0.00197 out of 70.5 is only about 0.0028%, so ignoring everything smaller than 1/60^2 is truly of minor consequence.

    Now, I have a question for you - since I am not familiar with this technique, can you please explain in a bit more detail what you mean by: "The line FD subtends an arc of 36 degrees ... " Where is this angle of 36 degrees?
    Last edited by ebaines; July 1st 2013 at 01:15 PM.
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