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Trigonometry sq. FD + sq. DB = sq. BF Almagest Ch 10 Book I Ptolemy

moohah

New member
Jun 30, 2013
2
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%)
 

moohah

New member
Jun 30, 2013
2