IMCA mechanical

IMCA mechanical calculator

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An italian calculator designed around 1958.

Inside

This calculator can perform the four basic operations :

Operands with 10 digits.
Accumulator with 13 digits.
Quotient with 8 digits.
Second factor, when multiplying, with 8 digits.
Copy back from accumulator to Operand.
Bell ring when overflow.

Cleaning the machine.

Operation

To input an operand, use the cursors B, the number appears in A. The lever L sets the operand to zero.
The lever K adds or substracts (according to the rotation) the operand in A to the accumulator in C.
Each rotation is counted in D, the counter. Its digits are white for additions and red for substractions.
The lever E set the counter back to zero and, if F is in the low position, it also resets the accumulator C.
The accumulator C can be switched left or right compared with A to sum directly on tens, hundreds, etc. The lever I/J switches the carrier one position left or right. For a free and fast move, use the lever H.
To copy C back to A, lower G (and F) and use the lever E to clear the accumulator as you copy it to A. Then, don't forget to raise G with a small pin on the right side to free every other movement.
The cursors M are free and you can use them to set the comma.

Add and Substract

Let's calculate 1524 + 97

Note	Actions	Operand A/B	Counter D	Accumulator C
Clear all	F- E L	0000000000	00000000	0000000000000
Set first number and add it to the accumulator	B=1524 K+	0000001524	00000001	0000000001524
Set second number and add it!	L B=97 K+	0000000097	00000002	0000000001621

Then 1524 + 97 = 1621

Let's calculate 20.14-19.66

Note	Actions	Operand A/B	Counter D	Accumulator C
Clear all, comma to second digit	F- E L M2	0000000000	00000000	0000000000000
Set first number and add it to the accumulator	B=2014 K+	0000002014	00000001	0000000002014
Set second number and substract it!	L B=1966 K-	0000001966	00000000	0000000000048

Then 20.14 - 1966 = 0.48

Multiply and Divide

How many hours in a year? 365 * 24

Note	Actions	Operand A/B	Counter D	Accumulator C
Clear all	F- E L	0000000000	00000000	0000000000000
Set first number and add it 4 times to the accumulator	B=365 K+ K+ K+ K+	0000000365	00000004	0000000001460
Shift right to the tens, and add the number twice to the accumulator	I K+ K+	0000000365	00000024	0000000008760

Then, there are 8760 hours in one year.

Sums and differences of products (314 * 12) + (15 * 24) - (47 * 31)

It's easy, each product is computed with the previous method, but, between them, you just clear the counter and not the accumulator! (Position F+ when using the lever E). The effect is that they are simply added!
If you want to substract, just rotate the other way with K-.

Approximate PI with the quotient 22 / 7 with three decimals.

The result will appear in Counter D, it's the number of turns that gives the quotient!

Note	Actions	Operand A/B	Counter D	Accumulator C
Clear all, comma to third digit for C and D	F- E L M3	0000000000	00000000	0000000000000
Set first number, shift carrier 3 positions to the right, add it to the accumulator	B=22 I I I K+	0000000022	00001000	0000000022000
Reset counter and set the divisor	F+ E B=7	0000000007	00000000	0000000022000
Substract 7 from 22 as many times as possible (), shift left one digit*	K- K- K- J	0000000007	00003000	0000000001000
Substract 7 from 10 as many times as possible, shift left one digit	K- J	0000000007	00003100	0000000000300
Substract 7 from 30 as many times as possible, shift left one digit	K- K- K- K- J	0000000007	00003140	0000000000020
Substract 7 from 20 as many times as possible, shift left one digit	K- K- J	0000000007	00003142	0000000000006

Then PI is close to 3.142

(*) Note : if you substract too many times, then you hear a bell ring indicating an overflow. Then, just turn it back with K+, you'll hear another bell ring!

Square root

Let's calculate the square root of 215.4

You have to cut this number in blocks of two digits starting from the comma (because 100 = 10²):

02 15.40

We'll start working on the first block, the 2. On my IMCA, to get a good precision, place the carrier to have the arrow pointing at the 7th digit in the counter D. The, we'll get 6 or 7 digits for the square root acording to the number of digits in the first block.

This method uses the fact that the sum of the odd numbers is a square (1+3+5=9=3² for example). More details and theory here.

Note	Actions	Operand A/B	Counter D	Accumulator C
Clear all, columns B7-B4 we program the value and send it to the accumulatorr	F- E L M3 B=2154 K+	0002154000	01000000	2154000000000
Clear the counter, column of the first block (B7) start with 1	F+ E L B=1	0001000000	00000000	2154000000000
Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left!	B=1 K- B=3 K- ♪ K+ ♪ B=2 J	0002000000	01000000	1154000000000
Columns B7-B6 program the next odd number, 21, Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left!	B=21 K- B=23 K- B=25 K- B=27 K- B=29 K- ♪ K+ ♪ B=28 J	0002800000	01400000	0194000000000
Columns B7-B5 program the next odd number, 281, Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left!	B=281 K- B=283 K- B=285 K- B=287 K- B=289 K- B=291 K- B=293 K- ♪ K+ ♪ B=292 J	0002920000	01460000	0022400000000

Then we would go on, columns B7-B4, with the odd value B=2921...

Then, the square root of 215.4 is close to 14.6