**What is the formula for calculating the error?**

**How to determine the error of the measurement result?**

**1) Absolute error.**

The absolute error is usually denoted by the capital Greek letter delta (Δ).

To find the absolute error, you should use the formula:

**Δ = | x - x0 |**

Δ is the absolute error,

x is the approximate (practical) value of the measured quantity,

x0 is the exact (true / theoretical) value of the measured quantity.

The absolute error has the same unit of measurement as the measured quantity. For example: if the measured value is measured in meters, then the absolute error will be measured in meters, if meas. we measure the value in kilograms, then the absolute error is also in kilograms. And so on.

**2) Relative error.**

Relative error is usually indicated by the lowercase Greek letter delta (δ).

To find the relative error, you should use the formula:

**δ = | x - x0 | / x0**

δ is the relative error,

x is the approximate (practical) value of the measured quantity,

x0 is the exact (true / theoretical) value of the measured quantity.

Relative error is a dimensionless quantity. The relative error either has a unit of measure 1 (fractions of a unit) or is measured in percent.

To translate the relative error from fractions of a unit to percent, you need to multiply it by 100.

**δ (%) = δ * 100 = (| x - x0 | / x0) * 100**

For example, consider this problem.

The student measured the length of the pencil with a ruler. As a result of measurements, the student received a result equal to 152 mm. The true length of the pencil, measured with a caliper, is 151.7 mm. Question: what is the absolute and relative error of the student’s measurement result?

1) Find the absolute error.

Δ = | x - x0 | = | 152 mm - 151.7 mm | = | 0.3 mm | = 0.3 mm.

2) Find the relative error.

δ = | x - x0 | / x0 = (| 152 mm - 151.7 mm | / 151.7 mm) * 100% = (0.3 mm: 151.7 mm) * 100% = 0.198%.

Answer: **Δ = 0.3 mm, δ = approx. 0.198%** (approximation).

## 1.1. Errors in metrology

Not a single measurement is free of errors, or, more precisely, the probability of a measurement without errors approaches zero. The nature and causes of errors are very diverse and many factors influence them (Fig. 1.2).

The general characteristic of influencing factors can be systematized from various points of view, for example, according to the influence of the listed factors (Fig. 1.2).

According to the measurement results, the errors can be divided into three types: systematic, random and misses.

** Systematic errors**in turn, they are divided into groups because of their occurrence and the nature of the manifestation. They can be eliminated in various ways, for example, by introducing amendments.

**fig. 1.2**

** Random errors**caused by a complex set of changing factors, usually unknown and difficult to analyze. Their influence on the measurement result can be reduced, for example, by repeated measurements with further statistical processing of the obtained results by the probability theory method.

TO ** misses**gross errors that occur when sudden changes in the conditions of the experiment belong. These errors are also random in nature, and should be excluded after identification.

The measurement accuracy is estimated by measurement errors, which are divided by the nature of occurrence into instrumental and methodological and by calculation method into absolute, relative and reduced.

** Instrumental**the error is characterized by the accuracy class of the measuring device, which is given in its passport in the form of normalized main and additional errors.

** Methodical**the error is due to the imperfection of methods and means of measurement.

** Absolute**the error is the difference between the measured G

_{u}and true G values of the quantity, determined by the formula:

Note that the quantity has the dimension of the measured quantity.

** Relative**the error is found from equality

** Given**the error is calculated by the formula (accuracy class of the measuring device)

where g_{the norms} - the normalizing value of the measured value. It is taken equal to:

a) the final value of the scale of the device, if the zero mark is on the edge or outside the scale,

b) the sum of the final values of the scale, excluding signs, if the zero mark is located inside the scale,

c) the length of the scale, if the scale is uneven.

The accuracy class of the device is established during its verification and is the standardized error calculated by the formulas

where ΔG_{m} - the largest possible absolute error of the device,

G_{k} - the final value of the measurement limit of the device, c and d - coefficients that take into account the design parameters and properties of the measuring mechanism of the device.

For example, for a voltmeter with a constant relative error, the equality

Relative and reduced errors are related by the following dependencies:

a) for any value of the reduced error

b) for the greatest reduced error

From these relations it follows that when measuring, for example, with a voltmeter, in a circuit at the same voltage value, the relative error is greater, the smaller the measured voltage. And if this voltmeter is selected incorrectly, then the relative error can be commensurate with the value of G_{n} that is unacceptable. Note that in accordance with the terminology of the problems to be solved, for example, when measuring voltage G = U, when measuring current C = I, the letter designations in the formulas for calculating errors must be replaced with the corresponding symbols.

**Example 1.1.** Γ voltmeter_{m}= 1.0%, U_{n} = G_{the norms}, G_{k} = 450 V, measure the voltage U_{u} equal to 10 V. Let us estimate the measurement errors.

*Answer.* The measurement error is 45%. With such an error, the measured voltage cannot be considered reliable.

With limited options for choosing a device (voltmeter), the methodological error can be taken into account by a correction calculated by the formula

**Example 1.2** Calculate the absolute error of the V7-26 voltmeter when measuring voltage in a DC circuit. The accuracy class of the voltmeter is given by the maximum reduced error γ_{m}= ± 2.5%. The limit of the voltmeter scale U used in the work_{the norms}= 30 V.

*Decision.* The absolute error is calculated by the known formulas:

(since the given error, by definition, is expressed by the formula, from here you can find the absolute error:

Important steps in the measurement process are processing the results and rounding rules. The theory of approximate calculations allows, knowing the degree of accuracy of the data, to evaluate the degree of accuracy of the results even before the steps are taken: to select data with an appropriate degree of accuracy, sufficient to ensure the required accuracy of the result, but not too large to save the calculator from useless calculations, to streamline the calculation process, freeing him from those calculations that will not affect the exact numbers of the results.

When processing the results, rounding rules apply.

If the first of the discarded digits is more than five, then the last of the stored digits is incremented by one.*Rule 1*

If the first of the discarded digits is less than five, then no increase is made.*Rule 2*

If the discarded digit is five, and there are no significant digits behind it, then rounding is done to the nearest even number, i.e. the last stored digit remains unchanged if it is even, and increases if it is not even.*Rule 3*

If there are significant digits after five, then rounding is done according to rule 2.

Applying rule 3 to rounding a single number, we do not increase the rounding accuracy. But with multiple rounding, excess numbers will be found about as often as not enough. Mutual error compensation will provide the most accurate result.

A number obviously exceeding the absolute error (or in the worst case equal to it) is called *marginal absolute error.*

The margin of error is not well defined. For each approximate number, its marginal error (absolute or relative) must be known.

When it is not explicitly indicated, it is understood that the marginal absolute error is half the unit of the last discharge. So, if an approximate number of 4.78 is given without indicating the marginal error, then it is assumed that the limiting absolute error is 0.005. Due to this agreement, you can always do without specifying the marginal error of the number rounded according to rules 1-3, i.e., if the approximate number is denoted by the letter α, then

, where Δn is the limiting absolute error, and δ_{n} - marginal relative error.

In addition, when processing the results are used ** rules for finding errors** amount, difference, product and quotient.

The limiting absolute error of the sum is equal to the sum of the limiting absolute errors of the individual terms, but with a significant number of errors of the terms, the errors are usually mutually compensated, therefore, the true error of the sum only in exceptional cases coincides with the marginal error or is close to it.*Rule 1*

The marginal absolute error of the difference is equal to the sum of the marginal absolute errors of the reduced or subtracted.*Rule 2*

The limiting relative error is easy to find by calculating the limiting absolute error.

The limiting relative error of the sum (but not the difference) lies between the smallest and largest of the relative errors of the terms.*Rule 3*

If all terms have the same limiting relative error, then the sum has the same limiting relative error. In other words, in this case, the accuracy of the sum (in percentage terms) is not inferior to the accuracy of the terms.

In contrast to the sum, the difference in the approximate numbers may be less accurate than the decrement and subtract. The loss of accuracy is especially great when the reduced and subtracted little differ from each other.

The limiting relative error of the product is approximately equal to the sum of the limiting relative errors of the factors: δ = δ*Rule 4*_{1}+δ_{2}or, more precisely, δ = δ_{1}+δ_{2}+δ_{1}δ_{2}where δ is the relative error of the product, δ_{1}δ_{2}- the relative errors of the factors.

1. If approximate numbers with the same number of significant digits are multiplied, then the same significant digits should be stored in the work. The last saved number will not be completely reliable.

2. If some factors have more significant digits than others, then before multiplying the first ones should be rounded, keeping in them as many digits as the least accurate factor or another one (as a spare), it is useless to keep further numbers.

3. If it is required that the product of two numbers have a predetermined number that is completely reliable, then in each of the factors the number of exact numbers (obtained by measurement or calculation) should be one more. If the number of factors is more than two and less than ten, then in each of the factors the number of exact digits for a full guarantee should be two units more than the required number of exact digits. In practice, it is quite enough to take only one extra figure.

The limiting relative error of the quotient is approximately equal to the sum of the limiting relative errors of the dividend and divisor. The exact value of the marginal relative error always exceeds the approximate. The percentage of excess is approximately equal to the maximum relative error of the divider.*Rule 5*

**Example 1.3** Find the limiting absolute error of the particular 2.81: 0.571.

*Decision.* The limiting relative error of the dividend is 0.005: 2.81 = 0.2%, the divisor is 0.005: 0.571 = 0.1%, and the quotient is 0.2% + 0.1% = 0.3%. The limiting absolute error of the quotient will approximately be 2.81: 0.571 · 0.0030 = 0.015

So, in the particular 2.81: 0.571 = 4.92, the third significant figure is not reliable.

**Example 1.4** Calculate the relative error of the readings of the voltmeter included in the circuit (Fig. 1.3), which is obtained if we assume that the voltmeter has infinitely high resistance and does not introduce distortion into the measured circuit. Classify the measurement error for this task.

**fig. 1.3**

*Decision.* We denote the readings of the real voltmeter by AND, and the voltmeter with infinitely large resistance by AND ∞. The desired relative error

,

Since R_{AND} >> R and R> r, then the fraction in the denominator of the last equality is much less than unity. Therefore, we can use the approximate formula valid for λ≤1 for any α. Assuming that in this formula α = -1 and λ = rR (r + R) -1 R_{AND} -1, we obtain δ ≈ rR / (r + R) R_{AND} .

The greater the resistance of the voltmeter compared to the external resistance of the circuit, the smaller the error. But condition R is a sufficient but not necessary condition for the smallness of δ. The error will be small also in the case when the condition r≤R_{AND} , i.e. the resistance of the voltmeter is much greater than the internal resistance of the current source. In this case, the external resistance can be arbitrarily large.

*Answer.* The error is systematic methodological.

**Example 1.5** The following devices are included in the direct current circuit (Fig. 1.4): A - type 330 ammeter of accuracy class K_{BUT} = 1.5 with a limit of measurement I_{k} = 20 A, A_{1} - ammeter type M 366 accuracy class K_{A1} = 1.0 with a limit of measurement I_{k1} = 7.5 A. Find the largest possible relative error in measuring current I_{2} and the possible limits of its actual value, if the devices showed that I = 8.0A. and I_{1} = 6.0A. Classify a dimension.

**fig. 1.4**

*Decision.* We determine the current I_{2} according to the testimony of the device (excluding their errors): I_{2}= I-I_{1}= 8.0-6.0 = 2.0 A.

We find the absolute error moduli of ammeters A and A_{1}

For A, we have equality for the ammeter

Find the sum of absolute error modules:

Therefore, the greatest possible of the same value, expressed in fractions of this value, is 1. 10 3 - for one device, 2 · 10 3 - for another device. Which of these devices will be the most accurate?

*Decision.* The accuracy of the device is characterized by a value opposite to the error (the more accurate the device, the smaller the error), i.e. for the first device this will be 1 / (1. 10 3) = 1000, for the second - 1 / (2. 10 3) = 500. Note that 1000> 500. Therefore, the first device is more accurate than the second twice.

You can come to a similar conclusion by checking the correspondence of the errors: 2. 10 3/1. 10 3 = 2.

*Answer.* The first device is two times more accurate than the second.

**Example 1.6** Find the sum of the approximate measurements of the device. Find the number of valid characters: 0.0909 + 0.0833 + 0.0769 + 0.0714 + 0.0667 + 0.0625 + 0.0588+ 0.0556 + 0.0526.

*Decision.* Adding all the measurement results, we get 0.6187. The maximum maximum error of the sum is 0.00005 · 9 = 0.00045. So, in the last fourth digit of the sum, an error of up to 5 units is possible. Therefore, we round the amount to the third digit, i.e. thousandths, we get 0.619 - the result in which all the signs are correct.

*Answer.* 0.619. The number of valid characters is three decimal places.

## Absolute error

**The absolute error of the number** call the difference between this number and its exact value. __ Let's look at an example.__: 374 students study at school. If you round this number to 400, then the absolute measurement error is 400-374 = 26.

To calculate the absolute error, it is necessary to subtract the smaller from a larger number.

There is a formula for absolute error. Denote the exact number by the letter A, and the letter a - approximation to the exact number. An approximate number is a number that differs slightly from the exact one and usually replaces it in calculations. Then the formula will look like this:

Δa = Aa. How to find the absolute error by the formula, we examined above.

In practice, absolute error is not enough to accurately measure. It is rare when you can accurately know the value of a measured quantity in order to calculate the absolute error. Measuring a book 20 cm long and making an error of 1 cm, we can consider the measurement with a big error. But if an error of 1 cm was made when measuring a wall of 20 meters, this measurement can be considered as accurate as possible. Therefore, in practice, the determination of the relative measurement error is more important.

The absolute error of the number is recorded using the ± sign. __ for example__, the length of the roll of wallpaper is 30 m ± 3 cm. The boundary of the absolute error is called the marginal absolute error.

## Relative error

**Relative error** they call the ratio of the absolute error of a number to that number itself. To calculate the relative error in the example with students, we divide 26 by 374. We get the number 0.0695, translate it into percentages and get 6%. Relative error is indicated by percent, because it is a dimensionless quantity. Relative error is an accurate estimate of the measurement error. If we take an absolute error of 1 cm when measuring the lengths of segments of 10 cm and 10 m, then the relative errors will be respectively 10% and 0.1%. For a length of 10 cm, the error of 1 cm is very large, this is an error of 10%. And for a ten-meter length, 1 cm does not matter, only 0.1%.

Distinguish between systematic and random errors. The error is called systematic, which remains unchanged during repeated measurements. Random error arises as a result of exposure to the process of measuring external factors and can change its value.

## Error Counting Rules

Для номинальной оценки погрешностей существует несколько правил:

- when adding and subtracting numbers, it is necessary to add their absolute errors,
- when dividing and multiplying numbers, you need to add the relative errors,
- when exponentiation, the relative error is multiplied by the exponent.

Approximate and exact numbers are written using decimal fractions. Only the average value is taken, since the exact value can be infinitely long. To understand how to write these numbers, you need to learn about the correct and doubtful numbers.

The numbers are called true, the discharge of which exceeds the absolute error of the number. If the digit category is less than the absolute error, it is called doubtful. __ for example__, for the fraction 3.6714 with an error of 0.002, the numbers 3.6.7 will be true, and the dubious ones will be 1 and 4. In the approximate number record, only the correct numbers are left. Fraction in this case will look like this - 3.67.

## What did we learn?

Absolute and relative errors are used to evaluate the accuracy of measurements. Absolute error is the difference between an exact and an approximate number. Relative error is the ratio of the absolute error of a number to the number itself. In practice, relative error is used, since it is more accurate.