b a c; 2 3; 1 3/5; 6 2/5; 5

n a m; 8

Here I have five against three, that is 5 at b and 3 at c, taking a to be the pole of a balance. And if you wished to know how much the 3 at c lifts of the 5 at b, you will say thus, by the rule of three, first exchanging the 5 into the place of the 3 and the 3 into the place of the 5, in order to level the balance. Then say: if the 5 truly proves at c to be 3, what in proportion will the 3 be that is found exchanged at b? You will say, by the said rule of three, that this 3 is one and 4/5; whence, one and 4/5 being taken from the 5 at b, there will remain to you 3 and 1/5. One may also say that the 3 at c lifts another 3 on the far side of the pole, so that the pole is burdened by 6.

Here one may clearly grasp how the centre of the beam n m rests upon the extremity of the beam that supports it. That being so, we must confess that the beam n m is held up entirely upon the extremity of the beam that supports it, that is beneath a, and gives none of its weight to any other part.

1 3/5 6 2/5; b a c

This one is tested, and is properly of the nature of a reversed balance: that is, whereas the upright balance has its pole support the attached weights, here the cords of the weights support the pole, which receives account of the weight of the whole beam.

The proportion of the weights that the cords b and c of the beam (or shaft) held up by them will feel will be the same as that which the spaces enclosed between b a and c a have among themselves; that is, a c being a subquadruple of b a, the weight felt at b will likewise be a subquadruple of that felt at c. That is: divide 8 by 5, which will be 40 fifths, and give 8 of those fifths to b, meaning one and 3/5, and then give 32 to c, which are 6 and 2/5, and so you will have perfectly apportioned it.

a b; c d; e t

Here it seems one may conclude that the line b c feels no weight, inasmuch as if the beam c d is loaded entirely upon the line a e, certainly all the rest of the support can be taken away, and the beam laid upon it will make no change.

If you consider well the weight that the beam gives of itself to its support over the line b t, by the reasons demonstrated above, you will find that the said line supports 32 fortieths of that beam c d, that is the 8 tenths, if the line a e did not contradict it, which, being beneath the centre of that beam, supports the whole.