``` P R O C E D U R E S A N D T R I C K S O F C A L C U L A T I N G ======================================================================= Instruction Manuals (eg. CURTA & WALTHER) are sources of ancient wisdom (A) NEGATIVE RESULTS ******************** NEGATIVE RESULTS are displayed in the arithmetic unit as the COMPLEMENT of the next higher 10, 100, 1000, ... EXAMPLE: -12 = 99...9988 -------- To get the "TRUE VALUE", two mechanical procedures are available ... (a) with BackTransfer: ---------------------- Make the BACK TRANSFER 1 SUBTRACTIVE TURN shows on the right side the "true value" ( ignore the 99..99 on the left side !) (b) without BackTransfer: ------------------------- Set the COMPLEMENT RESULT into Input Unit Make a 1st subtractive turn ( result should be ZERO ) A 2nd SUBTRACTIVE TURN shows on the right side the "true value" ( ignore the 99..99 on the left side !) (B) SHORTENED METHOD OF MULTIPLICATION ************************************** Operations that work, when the machine has a COUNTER WITH 10s-CARRY EXAMPLE(1): 13,974 * 9 = 125,766 ----------- Instead of multiplying 13,974 by 9, let us calculate 13,974 * (10 - 1), or (13,974 * 10) - 13,974 ; thus accomplishing the calculation in two turns instead of nine: Set carriage to 2nd position. Enter 13,974 and make one positive turn. This additive turn produces the multiplication by 10. Move the carriage to the 1st position and make one negative turn. By means of these two turns, the calculation is finished. The counter shows 9 and the result = in the arithmetic unit shows the PRODUCT = 125,766 ======= EXAMPLE(2): 345,67 * 89 = 30764,63 ----------- Instead of 17, the calculation will be done in 3 steps: o Carriage to 3rd position o Enter 345,67 o One positive turn ( = Multiplication with 100 ), counter shows 100 o Carriage to 2nd position. One subtractive turn, counter shows 90 o Carriage to 1st position. One subtractive turn, counter shows 89 == The arithmetic unit shows the PRODUCT = 30764,63 ======== (C) SHORTENED METHOD OF CALCULATING DISCOUNT & NET SUM: ******************************************************* EXAMPLE: Item Price = 7,683.00 \$ -------- - 3 % = 230.49 \$ ----------------------- Net Sum = 7,452.51 \$ ========== Simultanious multiplication of 2 different numbers by the same factor: Set 3(%) at far left and 97(%) at far right of the input unit. Multiply these numbers with the item price (= shown in the counter). The arithmetic unit will show both(!), the DISCOUNT & NET SUM. (D) MULTIPLE DIVISIONS WITH THE SAME DIVISOR: MAKE MULTIPLICATIONS! ******************************************************************* At 1st calculate the reciprocal of the divisor 11.7 ==>> 0.0854701 At 2nd use it as a factor: 1633.0 : 11.7 0.0854701 * 1633.0 = 139.573 314.5 : 11.7 ===>>> 0.0854701 * 341.5 = 29.188 ... ... 67.8 : 11.7 0.0854701 * 67.8 = 5.795 (E) DIVISION (Additive Method; Division by Multiplication) ********************************************************** EXAMPLE: 123,456 : 789 = 156.47148 -------- In this 2nd method of division, the divisor is entered into input unit. The dividend will be build up (from left to right) in the arithmetic unit and the quotient comes in the counter; all 3 numbers are visible! Shift the carriage to the far right. Enter 789. Make 1 additive turn with the crank (a 2nd add.turn get 1578, too big! Make 1 sub.turn). Shift carriage left and make 5 add.turns to get 11835. Shift carriage left and make 6 add.turns to get 123084. Shift carriage left and make 4 add. turns to get 1233996. Shift carriage left and make 7 add. turns to get 12345483. Shift carriage left and make 1 add.turn to get 123455619. Shift carriage left and make 5 add.turns to get 12345601350 (too big?). Make 1 sub.turn to get 12345593460. Shift carriage left and make 8 add.turns to get 12345599772; an additional add.turn get 12345600561. The former number is clother to 123456, so a sub.turn will bring it back in the arithmetic unit. Estimating the decimal marker, the QUOTIENT = 156.47148 is visible in the counter. ========= (F) CHAIN MULTIPLICATIONS WITH BACK-TRANSFER ******************************************** BRUNSVIGA-13RM or WALTHER-WSR160 have build-in "Back-Transfer", means that an intermediate result in the arithmetic unit can be brought back into input unit (ref. to the manuals) without new entering! EXAMPLE(1): 123 * 45 * 67 = 370,845 (1st Product 123 * 45 = 5,535) ----------- ======= EXAMPLE(2): 345 * 6.78 \$ = 2,339.10 \$ ----------- - 35 % = 818.68 \$ ------------------------- Net Sum = 1,520.42 \$ ========== Multiply as usual 345 * 6.78 and back-transfer the intermediate product 2339.10 into input unit. Multiply with 35(%). The intermediate result 818.69 is in arithmetic unit and the multiplier 35 is in the counter. Calculate the NET SUM in transfering 35 into 65 WITH ONLY 3 ADD.TURNS on 2nd position. This means: Build the difference 35% to 100% (G) CHAIN MULTIPLICATIONS WITHOUT BACK-TRANSFER *********************************************** EXAMPLE: 123 * 45 * 67 = 370,845 (1st Product 123 * 45 = 5,535) -------- The multiplication 123 * 45 is performed in the normal way. 5,535 is shown in the arithmetic unit. Clear the input register. Set it to 5,535 and make a backward turn, causing the arithmetic unit to zero (=Proof!). Set counter to zero. Multiply with 67 and get 370,845. ======= (H) MULTIPLICATION INVOLVING ADDITION OF PRODUCTS ************************************************* EXAMPLE: 123 * 45 + 67 * 89 = 11,498 (1st Product 123 * 45 = 5,535) -------- The multiplication 123 * 45 is performed in the normal way. 5,535 is shown in the arithmetic unit. Clear the input register and the counter. Multiply 67 * 89 (short-cut!) and the arithmetic unit shows 11,498. ====== (I) CUBES WITHOUT BACK-TRANSFER ******************************* EXAMPLE: 3273 = 34,965,783 -------- Multiply the square 327 * 327 = 106,929 as usual. The counter shows 327. Building up the counter to 106,929 starting from the left-hand digits, so as not to interfere, during multiplication, with remaining figures on the arithmetic unit. In this way there is no need to write down the square, nor to memorize it: Carriage on the 6th position. The 1st figure of the arithmetic unit beeing 1, the 1st figure of the counter needs to become 1, so make one additive turn. The following figure of the result being zero, the carriage can skip a place. Carriage to the 4th position; the next digit of the result dial is 6, so turn the crank until the counter shows 6 in the corresponding place. Carriage to 3rd position. The corresponding digit of the result dial beeing 9, turn the handle until 9 appears next in the counter. The next digit on both result dial and counter is 2. The carriage can therefore skip a place. Carriage to 1st position. On the result dial is 9 and on the counter 7. Two turns of the crank changes the latter digit to 9. The calculation is completed. The input unit shows 327, the counter shows the square 106,929 and the arithmetic unit therefore shows the required CUBE = 34,965,783. ========== (K) CALCULATING SQUARE ROOTS **************************** 1st METHOD ( Toepler ): ----------------------- This method is based on the astonishing fact, that the Arithmetic Serie of n odd numbers is the nth square root! Examples: 1 + 3 = 4 = 22 1 + 3 + 5 = 9 = 32 1 + 3 + 5 + 7 = 16 = 42 1 + 3 + 5 + 7 + 9 = 25 = 52 ... EXAMPLE: sqrt( 3029.4 ) = 55.040 -------- The number is to be split up into groups of two digits, starting in both directions from the decimal point. Eg. 30 | 29 . 40 | 00 | 00 Each pair corresponds to one digit of the root. The root will have as many decimals as there are groups after the decimal point; therefore, in our example, three decimals. Set the carriage in the 5th position (corresponding to the number of digits to be obtained in the root). Build up the Arithmetic Series, given above, starting from the 5th input slot (likewise corresponding to the number of places in the root): o Set 1, one additive turn o Set 3, one additive turn o Set 5, one additive turn o Set 7, one additive turn o Set 9, one additive turn The intermediate result shown in the arithmetic unit is 2500.000000; the 5th digit of the root has been found. It is easyly seen that, if 11 were added, the required value would be surpassed. Increase the figure 9 now in the 5th input slot by 1. In this case change it to 10 by setting 1 in the 6th slot and 0 in the 5th. The input dial ( 100000 ) should now show double the value shown on the counter ( 50000 ), and it is as well to check that at this point. Set the carriage to the 4th position. Once more build up the series given, by setting 1, 3, 5, 7, 9, successively in the 4th slot, with an additive turn after each one. The intermediate result shown in the arithmetic unit is now 3025.000000; the 4th digit of the root is found. Increase by 1 the 9 in the 4th slot; this means set the 4th slot to 0 and add 1 in the 5th slot. The input dial then shows 110.000 and the counter 55.000. Set the carriage to the 3rd position. It is not possible to build up the series at all in this column, as the required value would be sur- passed by the addition of 1 (eg. 3036.010000). Pass directly to the following 2nd position. Set the carriage to the 2nd position. Build up the series from the 2nd slot by setting 1, 3, 5, successively, with an additive turn after each one. The intermediate result is 3028.300900. Increase by 1 the 5 set in the 2nd slot, giving 6. Set the carriage to the 1st position. Set 1, 3, 5, 7, 9, 11, 13, 15, 17, one after the other in the 1st slot, with an additive turn after each. To set 11, set 1 in the 1st slot, and increase the figure in the 2nd slot by 1, from 6 to 7; for 13, 15, 17, it is then only necessary to set 3, 5, 7 in the 1st slot. The intermediate result shown in arithmetic unit is now 3029.291521; setting and transfering 9 from the 1st slot into arithmetic unit gives 3029.401600 which is a better approximation to the required value. The counter shows the SQUARE ROOT = 55.040, which is correct to 5 figures. ====== 2nd METHOD ( Herrmann ): ------------------------ If an approximate root is known (or good guessed), the result can be obtained more quickly and, if necessary for more significant figures, the procedure can be repeated! EXAMPLE: sqrt( 150 ) = 12.24744871 ( with 10 significant figures ) -------- (1) Starting with 12.2 as a 3 FIGURES APPROXIMATION, it is always possible to find 6 SIGNIFICANT FIGURES of the SQUARE ROOT: Calculate 12.22 with the carriage in 6th position: Enter 12.2 in the far right input slots. 1 additive turn in the 6th position, 2 add.turns in the 5th position and 2 add.turns in 4th position of the carrier give the approximate SQUARE = 148.84000 in the arithmetic unit. Without clearing, enter the DOUBLE value (= 24.4) of the former input (12.2) into the same input slots. With the appropriate (add./sub.) turns in the ACTUAL and FOLLOWING positions of the carriage, make the best approximation: 1 additive turn in the 4th position (= 151.28000); 5 sub.turns in the 3rd position (= 150.06000); 3 sub.turns in the 2nd position (= 149.98680); 5 add. turns in the 1st position. The best approximation 149.99900 is in the arithmetic unit and in the counter is the SQUARE ROOT = 12.2475 =======(1) (2) If it is necessary to get 8 SIGNIFICANT FIGURES of the square root, REPEAT the procedure with 12.25 as the better approximation and start with the carriage in the 8th (most right) position... The best approximation 150.00000050 is in the arithmetic unit and in the counter is the SQUARE ROOT = 12.247449 =========(2) impressum: *********************************************************************** © C.HAMANN http://public.beuth-hochschule.de/~hamann 11/10/09 ```