For design class (grade $10$), we have to make a Learning Product/Tool. I decided to make an abacus with 2 horizontal rows (it s easier) like the one in this picture:
The abacus isn't for very complex big numbers. Addition and subtraction of numbers between $0$-$20$ ex. $5+10$ the highest possible sum limit probably $20+20=40$
Let the lower (horizontal) rod represent the place value $1$ and let the upper rod represent the place value $10$. Also, let's agree that on a rod if all the beads are on the left side then the meaning is zero. If one shifts $k$ beads from the left side to the right side the resulting configuration represents the number $10\times k$ on the upper rod and it represents simply $k$ on the lower rod.
For example, if one has $5$ beads on the right side of the lower rod and $9$ beads on the right side of the upper rod then this configuration represents the number
$$95=5+9\times 10.$$
$$56+36=92.$$
Then we first set the configuration representing $56$, that is, we shift $6$ beads to the right side of the lower rod and $5$ beads to the right side of the upper rod. Then we start shifting further beads from the left side to the right side of the rods as if we were to set the configuration representing $36$. We would have to shift $6$ beads from the left side of the lower rod. The problem is that we have only $4$ beads on the left side. So we shift them to the right side and shout ten. Then we shift back to the right side all the ten beads and shift one further bead to the right on the upper rod. We shouldn't forget about the remaining $2$ sticks that we had to move to the right side of the lower stick. (We've moved only four out of $6$ so far!) Now, we have two sticks on the right side of the lower rod and $5+1=6$ beads on the right side of the upper rod. Since we have to represent $36$ (when adding $36$ to $56$ we still have to move $3$ beads from the left side of the upper rod to its right side. Finally we are going to have $3+6=9$ beads on the right side of the upper rod and $2$ beads on the right side of the lower rod; this configuration, indeed, represents $92$, the correct result.
The largest number we can represent on the two rod abacus is $99$. If we moved one more bead to the right side on any of the rods then we would have to move the whole set of beads back to left side and move one rod to the right side on the rod one up. The problem is that we do not have a third rod above the firs two.
If the diminuend is larger than the substrahend the the result will be positive. If the substrahend is greater the the result will be negative. The abacus cannot handle negative numbers, so we have to memorize what the sign of the result will be and subtract always the the smaller number from the larger one.
Here is how to do that. The first step is that we set up the configuration belonging to the larger number. The we will have to try to move back beads from the left hand side as if the substrahend was set up. For example, let's see the following subtraction
$$36-54=-18.$$
Of course, we will do the subtraction
$$54-36=18$$
keeping in mind that the resultant ($18$) will be a negative number. The first step is that we set up the configuration representing $54$: $5$ beads on the right hand side of the upper rod and $4$ beads on the right hand side of the lower rod. Then we picture (imagine) that we have $3$ beads on the right side of the upper rod and $6$ beads on the right side of the lower rod. So, we will have to move $6$ beads from the left hand side of the lower rod to its left side. We will manage to move all the $4$ beads to the left side of the lower rod but the we will shout no more beads, let's borrow . Borrowing is easy. If we move one bead form the right side of the upper rod to its left side then we deserve that we shift all the $10$ beads of the lower rod from the left to the right. Then we can accomplish our task -- moving $2$ more beads to the left side of the lower rod. The remaining number of beads on the right side of the lower rod is $8$. Now, we move $3$ beads from the right side of the upper rod to its left side. Remember we've moved already one bead, after moving $3$ more the number of remaining beads will be $1$. The resulting configuration will represent the number $18$ and we remember that the sign is negative.
How to multiply numbers? Multiplication of integer numbers is repeated addition. We, however, will have to be careful with the signs.
Division? Division is repeated subtraction. Remainder is possible. Whatch the sign!
Surprise your classmates!
We saw that with the base $10$ two rod abacus the largest number one can represent on it is $99$. I will show now how to increase the largest number represented with $10$ beads on rods.
We have to use the number system base $11$. Agree that the value of the lower beads on the left is still $1$ but it is allowed to leave $10$ beads on the right hand side. The meaning will be $10$. The secret is that we will change the meaning of the upper beads. Let the meaning be $$k\times 11$$ in the case of the upper beads if there are $k$ of them on the right hand side.
The rules of calculations are all the same as it was discussed above. The only question is: How to rewrite numbers given in the base $10$ system to the base $11$ system? The rule is very simple. If the number is less or equal then $10$ then it will not change. If it is greater than $10$ then we will have to divide it by eleven and take the remainder. This remainder will be the leftmost digit in the base $11$ system. It is a good idea to denote $10$ by $A$. If the result of the division is less or equal then $A=10$ then we are done. If not, then divide by eleven again and write down the remainder.
Let's see an example. Let $115$ be given in the base $10$ system. Firs, divide this number by $11$. The remainder is $5$ and the result is $10=A$. This number will be represented on our revolutionary abacus as follows: $5$ beads on the right side of the lower rod and $A=10$ beads on the right side of the upper rod.
Build a $4$ rod abacus! In that case the largest number one can represent in the base $10$ system is $9999$. However in the base $11$ system the largest number will be $14640$.