There are exactly 3 left! So we move them up. To add the three tens we need to look at the lower deck of the second column from the right. But before we should change the 2 value5 beads into one of value 10, that is one bead of the lower deck from the second column from the right. Now we still need to add the 3 tens to our abacus. Then we change the five value1 beads for one value5 bead, that is we move one value5 bead down and the five value1 beads down.īut we still have 2 value1 beads to add so we immediately move 2 of them up again. Three of the 5 beads in that column are already up so we need to split the 4 into 2+2 in our head. Now we just need to add 4 more value1 beads. For the 5 we move one bead from the upper deck from the column farthest to the right down.
We can immediately split it into 5 and 4. When you have the first number in place we start with the second number.
In the second column we move one bead from the upper deck down and one bead from the lower deck up. The 6 is bigger than 5 so we split it into 5 + 1. For the three ones we simply move the three beads in the first column of the lower deck upwards. Then we first display the 63 which contains three ones and 6 tens. We are done and you can see that 4 + 3 makes 7.Īgain we put all the beads in their starting position.
Multiply with chinese abacus free#
Now we have enough free beads again to move the remaining 2 beads up. So we split the 3 in 1+2 in our head and change the 5 moved-up beads with one bead from the upper deck from the same column. After moving one more bead up there are no more beads left to move up for the two that still need to be moved upward. For this you move 4 beads from the column of the lower deck from the farthest right up. Let's look at a very simple addition to see how it works on the abacus: Let's add 4 +3.Īfter bringing the abacus in its starting position you begin with displaying the first number, the 4 in our case. The 5 beads in the lower deck of the second column from the right each have the potential value 10, the 2 beads in the upper deck of the second column from the right each have the potential value 50, the 5 beads in the lower deck of the third column from the right each have the potential value 100 and so on.Īnd as soon as you move a bead you actually give it its value. The 2 beads in the upper deck of that column each have the potential value of 5. Here each bead has the potential value 1. We start with the 5 beads in the lower deck of the column the farthest to the right. The five beads of each colum of the lower deck together have the same potential value as one of the 2 beads of each colum of the corresponding upper deck. In its start position or when it shows 0, the beads in the lower deck are all moved down while the beads in the upper deck are all moved up. It has several colums with 5 beads on each column of the lower deck and 2 beads on each colum of the upper deck. It is a fully working abacus so you can try out all the examples we will be looking at! Look at the abacus and play around with it.
Multiply with chinese abacus how to#
The value of the beads and how to read numbers from the abacus
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