Positioning three weights on a suspended rod
Multiplying weight by arm-length to fix counterweights and spacing
A ruled horizontal bar drawn at the top hangs three weights along the line a n m o, tagged with the numbers 3 6 1 and 15 30 12 5 3. The note works through the balancing arithmetic: with the middle weight's arm divided into 5 parts, it multiplies arm length by weight (2 times 6 makes 12) to assign 12 of the 15 parts of a to the weight 3 as counterweight, leaving 3 parts for the arm of the least weight. Setting 3 weights against 5, it gives 5 spaces against 3 on the middle weight's arm, so the three weights end up in their proper places.
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Suspended rod carrying three weights (a n m o)
Three unequal weights hang from a graduated bar along the line a n m o. Each is positioned so that its arm and load balance the others, the middle weight serving as the reference whose arm is divided into equal parts.
Multiplying arm by weight to balance (2 x 6 = 12)
With the middle weight's arm split into 5 parts, multiply the arm of the greater weight by its weight: the arms being 2, '2 times 6 makes 12', so give 12 of the 15 parts of a to the 3 as counterweight. The remaining 3 parts form the arm of the least weight, and against 5 one sets 5 spaces to 3, placing all three weights in their proper sites.