## Calculating Machine Principality of Liechtenstein (economic and customs union with Switzerland)

## Introduction

These computing examples complete the short instructions sheet "YOUR CURTA CALCULATOR"
which accompanies every CURTA machine, and it is assumed that the use of the machine for
effecting the four arithmetical rules, as described therein, is understood.

Every example in this collection relates to both CURTA models, with the exception of those which are
marked "only CURTA Model II".  The two machines differ only in the number of figures which they can
handle.  These are:

Setting        Counter        Result
Register      Register       Register

for the CURTA Model I            8             6              11
for the CURTA Model II          11             8              15

In every example the following abbreviations are used:

S.R.  =  Setting Register;
C.R.  =  Counter Register;
R.R.  =  Result Register.

The expression "Machine ready" signifies that:

1. 1.    S.R., C.R. and R.R. have been cleared ;
2. 2.    The operating handle is in its zero stop position;
3. 3.    The carriage is in position 1;
4. 4.    The reversing lever is in its normal (upper) position.
The following list of contents will give the reader a survey of the subjects considered.

## CONTENTS

#### General

Division by breaking down (subtractive method)
Division by multiplying by a reciprocal
The rule of three
The rule of three in a single calculation  (only CURTA Model II)
Extended rule of three Calculation of roots
Continued multiplication  a * b * c * d . . .  etc.
Cubes without intermediate notes

#### Commerce and Industry

Checking of invoices and goods
Percentage calculations        A)    Percentage increase
B)    Percentage decrease
C)    Profit margin
D)    Compound percentages
E)    Profit and loss
F)    Capital and interest
Costing
Costing with simultaneous control
(only CURTA Model II)
Calculations with nines transfer
Exchange calculations
Calculations with English currency

#### Statistics

Simultaneous accumulation of a sum and a sum of squares
(only CURTA Model II)
Compilation of arithmetic mean and standard deviation

#### Technical and Survey calculations

Division into a negative number (complementary number)
Calculation of co-ordinates
Determination of the amount of silver in an alloy
(only CURTA Model II)
Determination of the angles in an acute-angled triangle, given three sides
(only CURTA Model  II)
Determination of the side of an obtuse-angled triangle
(given the two other sides and their included angle)
Calculation of area from co-ordinates
Calculation of the distance between two points, given their coordinates
(using Pythagoras' theorem)
Calculation of distance and azimuth
(only CURTA Model II)
Linear interpolation

## General

### Division by breaking down (subtractive method) (5 or 6 figures are required in the answer.)

To begin with it is necessary to add together the three terms in the numerator.  In order to accumulate the
sum as far as possible in the R.R., it is necessary to move the carriage as far to the right as possible.  But
care must be taken to move the carriage only so far to the right that space is left to accommodate the
most significant figures.  Put the carriage in position 5.  Set decimal points between positions 2 and 3 in S.R.,
and positions 6 and 7 in R.R.

Step 1.  Set the first term of the sum on the right hand side of S.R. (00 . . . 9526.80).
One plus turn of the operating handle.
Step 2.  Set 00 . . . 37.51 in S.R.  One plus turn.
Step 3.  Third term in S.R.  One plus turn.

Sum in R.R. is now 10209.93. Clear only the C.R. and S.R.

Now follows the division by 3 by the breaking down method:

Carriage is in position 6.  Set the divisor 3 in S.R. as far to the left as possible, but not so far that a minus
turn would cause a negative number to be indicated in R.R.; i.e. set 3 in position 5.  Place a decimal point
between positions 4 and 5 of S.R.

Place the reversing lever in its lower position.

With the operating handle pulled upwards we continue to subtract the contents of S.R. from the contents
of R.R. until the latter indicates a negative number, i.e. until a row of 9's appears (or possibly a row of 9's
and an 8); thus turn the handle in its minus position and observe R.R.  After 4 minus turns R.R. indicates
9 ... 8209930.  Now make one plus turn; a row of 0's now appears in R.R., and the number in R.R. is in
fact 0 ... 1209.930 ... 0.  This number is the remainder after dividing the original dividend by the number
now in C.R., i.e. 3, which is the most significant figure in the quotient.

Set the carriage in the next lower position, i.e. position 5.  Again turn the handle in its minus position until
a row of 9's appears in R.R., which occurs in this case after 5 turns.  Make one plus turn.

Carriage in position 4.  Turn the handle in its minus position until a row of 9's appears in R.R., as described
previously, and again one plus turn.

Carriage in position 3, 4 minus turns, one plus turn.

Carriage in position 2, 4 minus turns, one plus turn.

Carriage in position 1.  Make one minus turn and zero appears in R.R., which indicates that the division
is at an end.  (In the case of a division where there is a remainder, the remainder appears in R.R.) Check
that the number 3 (the divisor) is in R.R.

From the rule concerning decimal points in a division we have 6 (R.R.) - 4 (S.R.) = 2 decimal places in
C.R.  Thus we set a decimal point between positions 2 and 3 of C.R. and read the result (i.e. the quotient)
as 3403.31.  Return the reversing lever to its normal position. By using a CURTA Model II (which has at
its disposal 8 places in C.R.) a quotient of 8 significant figures may be obtained.

### Division by multiplying by a reciprocal

(If we require a number of dividends to be divided by the same divisor we can set the divisor as a constant
number in S.R. and build up the successive dividends in R.R., without resetting the divisor in S.R. and
clearing C.R. and R.R. each time.  The successive quotients appear in C.R. and can be noted down. See
the Costing example below.

If, however, a large number of dividends are to be divided by the same divisor, it is simplest to calculate
the reciprocal of the divisor once and for all, and place this quantity as a constant multiplier in S.R. The
divisions are then transformed into multiplications, e.g.

1633    /  11.7
341.5  /  11.7
67.8  /  11.7

The division 1 / 11.7 to 6 or 7 figures gives us the reciprocal of the divisor 11.7, namely 0.0854701.  We
place this number in the right hand end of S.R., clear C.R. and R.R., and set a decimal point between 7 and
8 of S.R.  The three divisions may now be carried out as three normal multiplication, namely,

0.0854701 * 1633    = 139.572
0.0854701 *  341.5  =  29.1880
0.0854701 *   67.8  =   5.79487

An experienced computor will dispense with the intermediate clearing of C.R. and R.R. after each
multiplication, and will build up the contents of C.R. to each successive multiplier by appropriate plus and
minus turns of the handle.

### The rule of three First method (subtractive division)

First we form the product 180 * 46 and then divide by 144.  The multiplication 180 * 46 is performed in
such a way that the product appears as far to the left as possible in R.R.

Set the number 46 in S.R. by means of knobs 2 and 1.  With the carriage in position 6 make two plus turns
and in position 5 2 minus turns (short cut multiplication).  The product 8280.000 is now in R.R. (with three
decimal places in accordance with the decimal rule).

Now follows the division 8280 / 144 by the subtractive method (see the previous example).  Clear
only C.R.  Set the number 144 in the right hand of S.R.  Place the reversing lever in the lower position.
The process of subtractive division now tales place with the carriage starting in position 5.  In this case
the division produces no remainder and the result (quotient) in C.R. is 57.5.

Return the reversing lever to its normal position.

Second method

In numerous calculations involving the rule of three the calculation of the intermediate quantity a / c is
also required; e.g.

If certain articles cost 180 francs/gross. how much does each article cost, and how much do 46 articles
cost?

First we must calculate the cost of one article (180 / 144) by additive division, so the number 144 is set
in S.R. by means of knobs 3-1, and with the carriage in position 6, the dividend 180 is built up figure by
figure in R.R. by positive turns of the operating handle.  The quotient appearing in C.R. is 1.25 = the cost
of one article.

Since we are about to multiply the cost of one article (1.25) by 46, and the number 1.25 is already in
C.R., clear only R.R.  Set the number 46 in the right hand end of S.R. and place the reversing lever in
the lower position.

Starting with the carriage in position 4. reduce every figure in C.R. to zero with plus turns of the handle.
This indicates a multiplication by 1.25, and the reduction of the contents of C.R. to zero is a check. (See
the examples  here , 34, 43.)  The result in R.R. is 57.50 francs, namely the price of 46 articles.
(Return the reversing lever to its normal position.)

The rule of three in a single calculation (only CURTA Model II)

In those cases in which the numbers involved are given only to a small number of figures, the rule of
three may be affected at one blow by using a CURTA Model II, the quotient being obtained with 4 or 5
figures. The division 1764 / 144 is carried out at the left hand end of R.R. by the additive method.  At the same
time the quotient of this division is multiplied by 375 in the right hand end of R.R. S.R. 14400000375.

Place the carriage in position 5 and by turning the operating handle in the appropriate manner build up
the number 1764 figure by figure in R.R.  R.R. now reads 176400004593750 and C.R. 00012250.  1764
is the dividend 12.25 is the quotient 1764 / 144 and 4593.75 is the final result (i.e. the product of
12.25 and 375). S.R. 34.40000087.2. Carriage in position 4.  (Since the first figure of the dividend is smaller than the first
figure of the divisor room for one more digit [i.e. position 15] in R.R., must be left free.)  Build up the
number 19.45 from the number 34.4.

R.R.  19.44976  0  49.30.  C.R.  000.5654.

### Extended rule of three  Firstly, the division 325 / 12 is carried out by the additive method to give 5 figures in the quotient; the
division is commenced with the carriage in position 5.  The quotient 27.083 appears in the first 5
figures of C.R.  Clear only R.R.  Set 677 in the right hand end of S.R.  Place the reversing lever
to its lower position.  Starting with the carriage in position 1 reduce the number in C.R. to zero by plus
tuns of the handle (see  here ).  The product 18355.191 appears in R.R.

Carriage in position 6.  Set the new divisor 119 at the right hand end of S.R.  Carry out the division by
means of the subtractive method as far as position 1 of the carriage.

R.R. now indicates a remainder of 0.028, C.R. indicates the quotient 154.077.   (Replace the
reversing lever in its normal position.)  By using this method it is possible that a small rounding off
error can occur which, however, is insignificant in most practical cases.  If it is required to eliminate
this error completely, it is recommended that first the factors in the denominator be multiplied
together and noted down, and secondly, the factors in the numerator be multiplied together with the
product being accumulated as far to the left as is possible n R.R.  The divisor may then be set in S.R.
and the division carried out by the subtractive method.

### Calculation of roots

Square Roots (Hermann's Method)

In the method to be described it is supposed, that by means of a slide rule or auxiliary tables or by
judicious guessing, an approximate square root has been found.  We wish to obtain a better
approximation.

Let N be the approximate value of R, the square root of R**2, and denote the error in the approximation
by E, so that R = N + E.  The method proceeds by setting N in S.R., multiplying by N (which appears
in C.R.) to produce N**2 in R.R.  The quantity 2N is then set in S.R. (without clearing R.R. or C.R.), and
R**2 is built up from N**2 in R.R.  Since it follows that (if we neglect  E**2) E is added to C.R.  Since C.R. already contained N, C.R. now reads
N + E the new approximation.

The square root of  150 = ?

Initial approximation = 12.2.

Carriage position 6.  Set 12.2 in S.R. with knobs 3 to 1.  Multiply by 12.2 starting from position 6.

Carriage position 6:  1 plus turn.
Carriage position 5:  2 plus turns.
Carriage position 4:  2 plus turns.

Set a decimal point between position 6 and 5 of R.R.: 148.84000.

S.R.  Set twice the initial approximation (i.e. 24.4) in the same positions as were used for the initial
approximation before.  By turning the handle in the appropriate manner in consecutive positions of the
carriage, increase the number in R.R. to 150, thus

Carriage position 4:  1 plus turn.
Carriage position 3:  5 minus turns.
Carriage position 2:  3 minus turns.
Carriage position 1:  5 plus turns.

C.R. now gives to 6 correct figures the root 12.2475.

This method determines as many additional correct figures as there were correct figures in the
approximation.  In this case the initial approximation is correct to three figures, and the desired
approximation to 6 figures.  From this rule it is unnecessary to proceed further; furthermore we have
in this particular case come very near to the required number (150) with 149.999.

If one requires 8 correct figures in the root and a CURTA Model II is available, one can repeat the
previous example using the initial approximation 12.25 (i.e. with 4 correct figures).  One then obtains
the root  12.247449  in C.R.

Remark: The rule that one gains as many correct figures in the root as one has to start with is capable
of a limited number of exceptions.  (See page 49 same method by subtraction.)

Cube roots
CURTA users who have at their disposal our CURTA Tables may, by use of these tables, obtain a
cube root correct to 5 figures by means of an addition and one subsequent multiplication.  These
tables will be forwarded free of charge on application to our headquarters.

Use of tables can be avoided by using an extension of the method for obtaining square roots as
described in the last section.

Let N be the approximate value of R, the cube root of , and denote the error in the approximation
by E, so that The computation proceeds by arriving at a situation in which we have in R.R., in S.R.,
and  in C.R.  We then build up the contents of R.R. to , i.e. from the above equation, adding
approximately  to the contents of C.R.  C.R. thus contains a quantity which approximates to  N + E.

It will be appreciated that (and this remark applies equally well to the method for deriving square
roots) if the derived approximation is insufficiently accurate, it may be improved by repeating the
process.

The cube root of  132.651  =  ?

Initial approximation  =  5.0.
Carriage in position 6:  Set 5 in S.R. with knob 1.
Multiply by 5 in position 6.
R.R.:  25.  Set a decimal point in C.R. between positions 5 and 6.
Clear C.R. and R.R.  Set 25 in S.R. with knobs 2 and 1.
Multiply by 5 in position 6.  Replace 25 in S.R. by 75 (= 3 * 25).
Increase the contents of R.R. to 132.651, thus

Carriage in position 5:  1 plus turn.
Carriage in position 3:  2 plus turns.
Carriage in position 1:  1 plus turn.

R.R. now contains  132.65075.  C.R. contains 5.10201.  The cube root of  132.651  is 5.1.  If the
process is repeated with the first approximation  5.10201, C.R. finally contains 5.1.

Further roots may sometimes be computed by application of the preceding methods, e.g.

The sixth root of  14  =  ?

Firstly, we compute the square root 3.74166 and then the cube root of this number, namely
the sixth root of  14  =  cube root of  1.5524.

Continued multiplication

a * b * c * d * . . . etc.

Introductory remark: The limit of a computation of this type is reached when the number of figures
in the product is equal to the number of figures in the Result Register, i.e. 11 figures for the CURTA
Model I and  15 figures for the CURTA Model II.  Before the computation is begun it is possible to
estimate the number of figures in the different factors.  One can, if the need arises, omit the last figures
of the partial products in order to avoid an excessively long end product.

Obviously, one can compute a function of the form  a * b * c * d * . . .  etc.  by multiplying the first two
factors together, setting the product in S.R., multiplying by the third factor, and so on.  Use of this
method, however, means that each partial product must be set at each stage into S.R.  We now
describe two methods which may readily be applied in many cases.

First Method

38 * 24 * 57 * 63.44  =  ?

I.    S.R.:    Set the number  38  with knobs 2 and 1.
C.R.:    Develop  24  (normal multiplication).
R.R.:    Result  912  =  partial product I.

II.  S.R.:    Set the next factor diminished by a tenth, i.e.  56.9  =  (57 - 1/10th)  with knobs 3 to 1.

Carriage:  Place the last figure on the right of S.R. immediately under the first figure on the left of the
partial product in R.R., thus move the carriage to position 3.

Handle:  Continue to turn the operating handle in its plus position until the number immediately above
the 9 in S.R. goes to 0.  One thus observes this number in R.R. and continues to turn the handle until it
reads 0 (optical control).  513 012 now appears in R.R.

Carriage:  Position 2.  One observes the second place in R.R.; after one plus turn this indicates a 0.

Carriage:  Position 1.  One observes the first figure of R.R.; this indicates  0  after  2  plus turns.  R.R.
then reads  519 840.  The partial product II is  51 984.0  (because 56.9 was set instead of  57  which gives
one decimal place in R.R.).  The partial product II was derive as follows:
(56.9 * partial product I)  +  1/10th partial product I  =  57  *  partial product I

III.  S.R.:    Set the number  63.439  =  (63.44 - 1/10th) with knobs 5 to 1. (minus 1/10th in last digit)

Carriage:  Position 6.  Turn the handle in its plus position and at the same time observe the 6th position
of R.R.  When this reaches  0  R.R. reads  31 720 019 840.

Carriage:  Position 5.  When the 5th number in R.R. has ben reduced to  0,  R.R. reads  32 354 409 840.
Proceed in this manner from position to position up to and including position 1 of the carriage.  Finally,
R.R. indicates  32 978 649 600.  The final product, after bearing in mind the three decimal places of the
number, is thus  3 297 864.96.

Remark:  It can occur that the number of figures in the partial product is greater than the number of
positions through which the carriage may be moved.  The next example shows nevertheless, that the
same method can still be used.  It is, however, advisable to arrange the computation in such a manner,
where possible, that the factor with the largest number of figures is set last in S.R. (see the first
example).

63.44 * 38 * 24 * 57  =  ?  (the result is already known).  The multiplication  63.44  *  38  =  2410.72  is
normal.  Next set  23.9  in S.R., with knobs 3 to 1.

Carriage:  Position 6.  By turning the handle in its plus position in positions  6, 5, 4, 3, 2 and 1  of the
carriage, reduce every digit in R.R. to  0.  R.R. now reads  57 857 280.  The partial product is  57 857.280.

Continuation of the calculation with a CURTA Model I.

Carriage:  Position 6.

S.R.:  Set  56.9  with knobs 5 to 3 (knobs 1 and 2 to 0).  By turning the handle in its plus position in
positions 6 to 1 of the carriage reduce each digit in turn to  0  (first the 8th digit, then the 7th digit, and
so on).  The second and first digits of R.R. are not reduced to  0.  R.R. now reads  3 297 860.4080.  We
can identify this result with that produced in the previous example.  If complete accuracy is required
for the last two figures the computation may be conducted as follows:

S.R.:  Set  56.9  with knobs 3 to 1 (knobs 4 and 5 to 0).

Carriage:  Position 2.  By turning the handle in its plus position reduce the digits in positions 8 to 2 to 0.
Carriage:  Position 1.  This digit is already  0, the calculation is finished.  R.R. reads  3 297 864.9600.

Continuation of the calculation with a CURTA Model II.

The carriage of the CURTA Model II may be moved through 8 positions, and in this instance the
calculation in hand may be completed in the normal manner as described in I.

However, as an exercise for the case in which the number of figures in the partial product exceeds the
number of positions through which the carriage may be moved, and in order to fully appreciate the
principle described, it is advisable to carry out the exercise given for the CURTA Model I.  We
therefore request you to perform the previous method on your CURTA Model II.  The end result in
R.R. is the same as before.

Second Method

38 * 24 * 57 * 63.44  =  ?

Proceed as for section I of the previous example.  This method differs from the previous one in that
the next factor is reduced by a unit in the last figure instead of by a tenth.  Thus set  57 - 1 = 56  in
S.R. with knobs 2 and 1.

Clear only C.R.

Carriage:  Position 3.
Handle:  The arrow points to 9 in R.R.  Memorize this number and carry out  9
plus turns of the handle.   Naturally, a 9 appears in C.R.

Carriage:  Position 2.  The arrow now points to a 1 in R.R., thus one plus turn of the handle.

Carriage:  Position 1.  The arrow points to a 2 in R.R., thus two turns of the handle.  R.R. shows partial
product II, namely  51 984,  and C.R. indicates partial product I, namely  912.  Set in R.R. the third factor
minus a unit, namely  56.  Partial product II was arrived at as follows:
(56 * Partial product I) + (1 * Partial product I),  (which was already in R.R.)  =  57 * Partial product I.

Clear only C.R., and set in the right hand end of S.R. the factor 63.44 with the last figure diminished by
a unit, i.e.  63.43.  Beginning with the carriage in position 5 turn the handle in its plus position so as to
build up in C.R. the number now in R.R.  The final result indicated in R.R. is  3 297 864.96.

The advantage of this second method is that no zeros appear at the right hand end of R.R., and thus
the full capacity of the machine can be utilized.  It is, however, necessary to check the calculations by
noting that the partial products appear in C.R., whereas in the first method this was automatically
assured by the appearance of a  0  in the successive positions of R.R.

#### Cubes with intermediate notes

327**3  =  ?

We first obtain  327**2, i.e.  327 * 327  by normal multiplication.  The square  106 929  is built up in R.R.
S.R. indicating 327 and C.R. also indicating 327 are not cleared.

C.R. will now be built up to  106 929  starting with the carriage in position 6 in order that the least
significant figures in R.R. will not be modified in the ensuing multiplications.

Carriage:  Position 6.  The arrow points to a  1  in R.R.  In the corresponding position of C.R. there is a  0;
thus one plus turn.

Carriage:  Position 5.  The arrow points to a  0.  In position 5 of C.R. there is similarly a  0.  Thus this
position can be skipped over.

Carriage:  Position 4.  The arrow points to a  6  in R.R., thus turn the handle in its plus position until C.R.
also indicates a  6  in this position.

Carriage:  Position 3.  The arrow points to a  9  in R.R.  Thus turn the handle in its plus position until the
3rd position in C.R. also indicates  9.

Carriage:  Position 2.  In this position both R.R. and C.R. indicate a  2.  Thus this position can be skipped
over.

Carriage:  Position 1.  In this position R.R. indicates a  9,  and C.R. a  7;  thus two plus turns.

Now the basic number  327  is in S.R., in C.R. the square  106 929,  and in R.R. the 3rd power
34 965 783.

Higher Powers

Clearly the capacity of the machine used limits the order of the power to be computed. In the previous
case for example the fourth power may be computed with a Model I by setting the cube and multiplying
by 327.  With the result  11433811041  the limit of the capacity or R.R. is reached.  Higher powers may
be computed only by discarding the last 3 or 4 figures in S.R.

Using the Model II the fourth power also may be computed without clearing S.R. and C.R. by
developing the approximate number in C.R. in a manner similar to that which has been described in the
third power, the cube being fully contained in C.R.  For higher powers it is suggested that some
convenient number (square, 3rd power, etc.) be set in S.R., in which case, when larger numbers are
being dealt with, the necessity for curtailing the number is indicated in advance.

## STATISTICS

Simultaneous accumulation of a sum and a sum of squares
(only with CURTA Model II) + 6925    6925**2
+ 3289    3289**2
- 1721    1721**2
+ 2987    2987**2
s = ?     S = ?

Move the clearing lever to the left hand end of R.R.  Set decimal point between positions 11 and 10
of R.R.  Set a  1  with knob  11  and the number 6925 with knobs  4  to  1  in S.R.  Develop 6925 with the
handle.

In C.R. there now appears the number 6925, on the right of R.R. the square 47 955 625 and
on the left of R.R. as a check the multiplier 6925.  Clear only the left hand end of R.R.;  i.e.
move the clearing lever only over the number  6925  as far as the  0  and return it to its original
position.

Set the next number, i.e.  3289, at the right hand end of S.R. (retaining the  1  in position  11).  Develop
3289  with the handle so that the digit in S.R. generates the multiplier at the left of R.R. and acts as a
check.

We have now accumulated the sum of the first two numbers in C.R., and the sum of the first two squares
at the right hand end of R.R.  As a check the multiplier  3289  is at the left hand end of R.R.  Again we
clear only multiplier  3289  at the left hand end of R.R.

Set the next number in the right hand end of R.R.  Since the sum is being accumulated in C.R. we move the
reversing lever to its lower position.  Develop the number  1721  normally with the handle.  Again
clear only the left hand end of R.R., (i.e. the multiplier  1721).  Move the reversing lever to its normal
position.  Set the next number at the right of S.R. and multiply it by itself.

The final result at the right hand end of R.R. is 70 657 156 = S, and in C.R. we have 11 480 = s.

At the left of R.R. the last number  2987  stands as a check.  You may again partially clear so that the sum
S  remains by itself for further calculations.

#### Computation of the arithmetic mean and standard deviation

Given  N  observations .  The arithmetic mean is given by and the standard deviation by In order to facilitate the calculation we reduce each observation by a known constant  x0  and in this
manner reduce the number of figures in the numbers used in the calculation.  We have and The calculation can be carried out by the CURTA in the following manner, where the capacity of the
Model II is always sufficient.  If the number of figures in, and the dispersion of, the observations are not
too large, the CURTA Model I may also be used, as the following example indicates: S.R.:  005.30001;  Carriage in position 3.
Multiply by  5.3  and clear only C.R.
To mark the decimal points always use the white movable markers.

S.R.:  006.40001;  Carriage in position 3.  Multiply by  6.4.  Clear only C.R.  After these two steps we
have in R.R.:
069 ,  05 / 0 / 11 , 700
i.e. Proceed in this manner, and after  7  such steps we have in R.R.:    252 , 84 / 0 / 41 , 200
or Clear only C.R.
Set in S.R. and N, thus:  041.20007.
Carry out the division  41.2 / 7  in the right hand part of R.R. by the subtractive method.  Thus move the
reversing lever to its lower position and commence the division with the carriage in position 4.  When the
division is finished C.R. indicates  005.885,  that is: The mean value of the observations is thus R.R. contains  0.1037800005.  Up to the last  5  (which is the remainder of the division) that is: Clear only C.R.  Carriage in position  6.
S.R.:  Set  N(N-1),  i.e.  0.42.00000,  in S.R.  Carry out the division by the subtractive method.
Result in C.R.:  0.247.

By means of a slide rule or a CURTA the square root of this number is found to be  0.497.  The observation
measurement reads finally: (Return the reversing lever to its normal position.)

## TECHNICAL AND SURVEY CALCULATIONS

### Division into a negative number (complementary number)  S.R.:  Set  3.15  at the extreme right.  Set a decimal point in front of knob  2  of S.R.
Carriage in position  4:  Normal multiplication by  17.5  in positions  4,  5  and  6  of the carriage.
R.R.:  . . . 55.125 . . .  Set decimal points in front of position  4  in C.R. and position  6  of R.R.

Clear only C.R.

S.R.:  Set  9.6  with knobs  3  and  2.  Set knob  1  to zero.  Place the reversing lever in its lower position.

Develop  23.3  in C.R. by turning the handle in its minus position in positions  4,  5  and  6 of the carriage
(negative multiplication).
R.R.:  . . . 99831.445 . . .  =  Complement of the dividend.  Do not clear R.R.

Division by e

This division is carried out by building up the dividend to zero by means of the divisor.  Return the
reversing lever to its normal position.  Clear only C.R.  Carriage in position  6.

Set  137.4  in S.R. with knobs  4  to  1.  Set decimal points in front of knob  1  in S.R. and in front of position
5  of C.R.  By turning the handle force the contents of R.R. as near zero as possible,  i.e. to  9999  etc.
Thus observe R.R. and in:

Carriage position  6:  1  plus turn
Carriage position  5:  2  plus turns
Carriage position  4:  2  plus turns
Carriage position  3:  7  plus turns
Carriage position  2:  3  minus turns
Carriage position  1:  5  plus turns

There is now in R.R. the nearest possible approximation to zero  (0.000450).
The answer  -1.22675  stands in C.R.

### Determination of the angles in an acute-angled triangle given three sides only

(only CURTA Model II) Given: a  =  16.78
b  =  13.63
c  =  20.33

To find:   We first compute the divisor and note it down.  The dividend is then formed as far to the left as possible in
R.R. and finally, by setting the divisor in S.R., subtractively divided.

Set the quantity  2 * 16.78,  i.e.  33.56,  in S.R. and multiply by  13.63.  The result  457.4228 is noted down.

Set the number  16.78  in S.R. by means of knobs  8  to  5:  00016780000.  Develop  16.78  in C.R. as
follows:
Carriage in position 5:  2 minus turns
Carriage in position 6:  2 minus turns
Carriage in position 7:  2 minus turns
Carriage in position 8:  2 plus turns

Clear C.R. only.  Set a decimal point between positions  12  and  13  of R.R.  Set the number  13.63  in
S.R. with knobs  8  to  5:  00013630000, and develop the number  13.63  in C.R. in positions  5  to  6.
R.R. now contains the sum of the first two squares:  467.3543.

Clear C.R. only.  Set the number  20.33  in S.R. with knobs  8  to  5:  0020330000.  Place the reversing
lever in its lower position.  Develop  20.33  in C.R. by turning the handle in its minus position in positions
5  to  8  of the carriage (negative multiplication).  In R.R. the third square has been subtracted from the sum
of the first two and the result is the dividend   54.0364.

Clear C.R. only.  Place the carriage in position 8.  Set the divisor  457.4228  (previously noted down) in S.R.
with knobs  7  to  1:  00004574228, and set a decimal point between knobs  5  and  4.  Now follows the
division by the subtractive method (see here).

The final result in C.R. after applying the decimal point rule is (Return the reversing lever to its normal position.)

### Determination of a side of an obtuse-angled triangle

(given the two other sides and their included angle) The distance A-B is unknown, or it is perhaps inconvenient to measure it.

Given: find  c  =  ?
Direct use of the classical formula is computationally inconvenient, due to
the large size of the numbers involved.  The best method of procedure is as follows:

1.  Obtain the values .
2.  Compute 3.  Then by Pythagoras' theorem To mark the decimal points always use the white movable markers.

S.R.:  21.47.  Develop  0.532712  in C.R. with the handle.

R.R.: .  Note this number down and do not clear.

C.R.:  By turning the handle change the contents to  0.846296.

R.R.: . . ., not cleared.

Carriage position 6.
S.R.:  Set b = . . . 32.14  to correspond with the number in R.R.

Handle:  1  minus turn.  If a complement appears in R.R., which is the case with the present example.

S.R.:  Set the modulus of the number in R.R.  (If , then set the number in R.R.)  In the present
example the number (rounded off) is . . . 13.970.  As a check carry out one plus turn when R.R. reads
(1)0000 . . .  or  9999 . . .  Clear R.R. and C.R.  Finally, multiply the number in S.R. by itself, thus
13.970 * 13.970.

R.R.: Clear C.R. only.
S.R.: .  Multiply by  11.437.

R.R.: Now compute the square root by Hermann's method (see here) or any other appropriate method.

The result is  18.054 in. = c.

### Calculation of area from co-ordinates

(by Elling's method)

The area to be computed is defined by the following co-ordinates, which are juxtaposed for the purpose
of calculation.

The calculation follows from the general formulae First we develop in C.R. by means of the handle.  Now set in S.R., must be multiplied by , thus

C.R.:  Develop 12.
S.R.:  Set 64 at the extreme right.
C.R.:  By means of the handle change to 68 (difference between and ).

The arrows in the following scheme indicate the further course of the calculation. S.R.:  72;
C.R. by means of the handle change to  100.

S.R.:  32;
C.R. by means of the handle change to  20.

S.R.:  68;
C.R. by means of the handle change to  44.

S.R.:  56;
C.R. by means of the handle change to  88.

S.R.:  56;
C.R. by means of the handle change to  60.

S.R.:  44;
C.R. by means of the handle change to  12.

R.R. contains  3856 = 2F.  F = 1928.

If the number of points given is even, then in practice we enter the last point twice.  The area contributed
by the existence of the added point is zero, and the validity of the scheme adopted is unaffected.
Depending on the order chosen in which to traverse the points, it may occur that the final result in R.R.
is negative.  The correct result in this case is the complement of this number.  The given example is
sufficient to establish the principle employed in calculating areas with the machine.  The same principle
also applies when the points lie in differing quadrants, but the sign convention must very strictly be
obeyed.

### Calculation of the distance between two points given their co-ordinates (Using Pythagoras' theorem)

The co-ordinates of the points  A  and  B  are known.  We wish to find the distance . Set a decimal point between knobs  3  and  2  of S.R.  Set  8.38  at the right of S.R.  With the carriage in
positions  4  to  6  develop the number  8.38  in C.R. by means of the handle.  Set a decimal point between
positions  8  and  7  of R.R.  The product in R.R.,  70.2244. will be accumulated with the next product
(7.86 * 7.86), and is therefore not cleared.

Clear C.R. only.

Set  7.86  at the right of S.R.  With the carriage in positions  4  to  6  develop the number  7.86  in C.R. by
means of the handle.  R.R. now contains the sum of the two squares:  132.004.  This number is not
cleared.

Clear C.R. and S.R. only.

The square root may now directly be computed by Hermann's method (see here).  The calculation here,
however, involves a subtraction.  Thus move the reversing lever to its lower position.  First
approximation  11.5.

Carriage in position  6.  Set the first approximation  11.5  in S.R. with knobs  5  to  5.  Develop the number
11.5  in C.R. by turning the handle in its minus position with the carriage in positions  6,  5,  and  4.

S.R.:  Change knobs  5  to  3  to  230  and in positions  3,  2,  and  1  of the carriage use the handle to bring
the number in R.R. as close to zero as possible.  C.R. then indicates the square root  11.4893  (R.R.
contains  0.0001).  The length = 11.49 in.

(Return the reversing lever to its normal position).

### Calculation of Distance and Azimuth between two points of known Co-ordinates

(Only CURTA Model II)   (To mark the decimal points always use the white movable markers.)

Carriage:  Position  7.
S.R.:  00094892.791     1  plus turn
S.R.:  00095141.420     1  minus turn
R.R.:      999751.371 . . .  =  complement of Do not clear R.R. Carriage:  Position  1.
S.R.:  00011779.323     1  minus turn
S.R.:  00011517.150     1  plus turn
R.R.:      999751.371 (see NOTE) / 737827 /                NOTE:  Use two contiguous markers here.

737827  =  complement of .
S.R. . . . 262.173 =  absolute value of .   1  plus turn.
R.R.:  999751.371 . . .  =  complement of .
Clear C.R.  Note down the absolute value of  248.629. (Division by using the divisor to bring the contents of R.R. up to zero (see here)).  Reversing lever in its
normal position.

Carriage in position  7:  1  plus turn
Carriage in position  6:  1  minus turn
Carriage in position  5:  5  plus turns
Carriage in position  4:  2  minus turns
Carriage in position  3:  4  plus turns
Carriage in position  2:  7 minus turns
Carriage in position  1:  9  plus turns Do not clear S.R. Clear C.R. and R.R. only.
Carriage in position  7.  1  plus turn.
R.R.:  000262.173 . . . = Dividend .
S.R.:  . . . 0.725601.  Carriage in position  6.  Clear C.R.
Move the reversing lever to its lower position.
Carry out subtractive division.
C.R.:  000361.318  =  D. Machine ready. Carry out the division  248.629  /  0.688116.
C.R.:  361.31844.  D  =  361.318.

### Linear Interpolation

One can obtain two values and of a function by looking in tables.  Suppose that we are interested in
an intermediate value .  This may be determined by the interpolation formula: For using the CURTA this is conveniently rearranged as: Example:  to find From tables: Since the argument is in steps of , we must
interpolate using as argument 14/60  =  0.233.  (An experienced computer will carry out the division by 60
Set at the right hand end of S.R., and in C.R. a decimal point between positions  3
Set in S.R., and complete the number in C.C. to  1.000 (multiplication by  1-0.233). 