Comparison of quantities by their numerical values. Comparison of numbers and values by length, volume, mass. Comparison of averages
Valery Galasyuk- Academician of the AES of Ukraine, General Director of the COWPERWOOD auditing firm (Dnepropetrovsk), Member of the Presidium of the Council of the Union of Auditors of Ukraine, Member of the Audit Chamber of Ukraine, Chairman of the Audit Commission of the Ukrainian Society of Appraisers, Deputy Chairman of the Board of the Association of Taxpayers of Ukraine, Deputy Chairman of the Commission for Evaluation of Efficiency investment activity of the Ukrainian Society of Financial Analysts, leading appraiser of the Ukrainian Society of Appraisers
Victor Galasyuk– Director of the Department of Credit Consulting of the information and consulting company “INCON-CENTER” (consulting group “COWPERWOOD”), Master of Economics of the Enterprise, laureate of competitions for young appraisers of the Ukrainian Society of Appraisers
Mathematics is the only perfect method
allowing himself to be led by the nose
Einstein
My job is to tell the truth, not to make you believe in it.
Rousseau
This article is devoted to the fundamental problem that arises in the process of numerical comparison of quantities. The essence of this problem lies in the fact that, under certain conditions, various methods of numerical comparison of the same quantities fix a different degree of their inequality. The uniqueness of this problem lies not so much in the fact that it has not yet been solved, although it would seem that the procedures for numerical comparison have been thoroughly studied and do not raise questions even among schoolchildren, but in the fact that it has not yet been adequately reflected in public consciousness and, more importantly, in practice.
As you know, you can compare two values numerically either by answering the question “How much is one value greater than the other?” Or by answering the question “How many times is one value greater than the other?”. That is, in order to numerically compare two quantities, you must either subtract one from the other (), or divide one by the other (). At the same time, as studies have shown, there are only two initial types of criteria for numerical comparison of quantities: and , and none of them has the exclusive right to exist.
Only 13 qualitatively different variants of the ratio on the numerical axis of the values of the two compared values X and Y are possible (see Fig. 1) .
When comparing two values X and Y based on the comparison criterion with any variant of their ratio on the number axis, there are no problems. Indeed, regardless of the values of X and Y, the comparison criterion uniquely characterizes the distance between points X and Y on the real axis.
However, the use of the comparison criterion to compare the values of X and Y in some cases of their relationship on the number axis can lead to problems, since in these cases the values of the X and Y values can have a significant impact on the results of the comparison. For example, when comparing the values of 0.0100000001 and 0.0000000001, corresponding to option 5 on the Galasyuk's beads, using the comparison criterion shows that the first number is greater than the second by 0.01, and using the comparison criterion shows that the first number is greater than the second by 100 000,001 times. Thus, with a certain ratio of compared values on the numerical axis, the comparison criterion indicates slight degree of inequality compared values X and Y, and the criterion of comparison points to a significant degree of their inequality.
Or, for example, when comparing the values \u200b\u200bof 1,000,000,000 100 and
1 000 000 000 000, corresponding to the same option 5 on the "rosary of Galasyuk", the use of the comparison criterion shows that the first number is greater than the second by 100, and the use of the comparison criterion shows that the first number is approximately equal to the second, since it is greater than the second number only in 1.0000000001 times. Thus, with a certain ratio of compared values on the numerical axis, the comparison criterion indicates significant degree of inequality compared values X and Y, and the criterion of comparison points to a slight degree of their inequality.
Since the problem discussed in this article arises only when using the comparison criterion , then to study it, we consider the comparison of two quantities m and n based on the comparison criterion. To compare these quantities, we divide m on the n: .
Analysis of the results of comparison of values m and n can be carried out in two stages: at the first stage, we take the denominator of the ratio unchanged - the value n, on the second numerator - the value m(see fig. 2).
To carry out the first stage of the analysis, we construct a graph of the dependence of the ratio on the value m(see Fig. 3), while it should be noted that when n=0 relation is not defined.
As seen in Figure 3, if n=const, n¹0, then for |m|→∞ the relation | |→∞, and for |m|→0 the relation | |→0.
To implement the second stage of the analysis, we construct a graph of the dependence of the ratio on the value n(see Fig. 4), while it should be noted that when n=0 relation is not defined.
As seen in Figure 4, if m=const, m¹0, n¹0, then for |n|→∞ the relation | |→0, and for |n|→0 the relation | |→∞. It should be noted that as the values of | n| equal changes | n| involve ever smaller changes in attitude | |. And when approaching zero values | n| equal changes | n| entail ever-greater changes in attitude | |.
Summarizing the results of stages I and II of the analysis, we present them in the form of the following table, including in it the results of the comparison analysis based on the initial type of criteria (see Table 1). Situations in which X=0 and Y=0 are not considered here. We hope to analyze them in the future.
Table 1
Generalized results of the analysis of comparison of valuesXandY
based on two original types of comparison criteria
(X¹ 0 andY¹ 0)
|
7. Galasyuk V.V. How many initial types of cost-effectiveness criteria should there be: one, two, three…?//Stock market.-2000.-№3.-p.39-42. 8. Galasyuk V.V. On two initial types of cost-effectiveness criteria//Questions of evaluation, Moscow.-2000.-№1.-p.37-40. 9. Poincaré Henri. About science: Per. from French-M.-Nauka. Main edition of physical and mathematical literature, 1983.-560 p. 20.10.2002 |
First, consider the problem of comparing the value measured in the experiment with the constant a. The value can only be determined approximately by calculating the average over the measurements. We need to find out if the relation holds. In this case, two tasks are posed, direct and inverse:
a) from a known value, find a constant a that is greater than c given probability
b) find the probability that , where a is a given constant.
Obviously, if then the probability that is less than 1/2. This case is of no interest, and further we will assume that
The problem is reduced to the problems discussed in Section 2. Let X and its standard be defined by measurements
The number of measurements will be considered not very small, so there is a random variable with a normal distribution. Then from the Student's criterion (9), taking into account the symmetry of the normal distribution, it follows that for an arbitrarily chosen probability, the condition
Let's rewrite this expression in the following form:
where are the Student's coefficients given in Table 23. Thus, the direct problem is solved: a constant a is found, which with probability exceeds
The inverse problem is solved using the direct one. Let us rewrite formulas (23) as follows:
This means that you need to calculate t from the known values of a, select the row with the data in table 23 - and find the corresponding value from the value of t. It determines the desired probability
Two random variables. It is often required to establish the influence of some factor on the quantity under study - for example, whether (and how much) a certain additive increases the strength of the metal. To do this, it is necessary to measure the strength of the original metal and the strength of the alloyed metal y and compare these two quantities, i.e. find
The compared values are random; Thus, the properties of a certain grade of metal vary from heat to heat, since the raw materials and the melting regime are not strictly the same. Let's denote these quantities by . The magnitude of the studied effect is equal and it is required to determine whether the condition is met
Thus, the problem is reduced to comparing random variable with the constant a discussed above. The direct and inverse comparison problems in this case are formulated as follows:
a) according to the measurement results, find the constant a, which exceeds with a given probability (i.e., estimate the magnitude of the effect under study);
b) determine the probability that where a is the desired effect size; this means that it is necessary to determine the probability with which
To solve these problems, it is necessary to calculate z and the variance of this quantity. Let's look at two ways to find them.
Independent measurements. Let us measure the value in experiments, and the value in experiments independent of the first experiments. We calculate the average values using the usual formulas:
These means are themselves random variables, and their standards (not to be confused with the standards of single measurements!) are approximately determined by unbiased estimates:
Since the experiments are independent, the random variables x and y are also independent, so when calculating them mathematical expectations are subtracted, and the variances are added:
A slightly more accurate estimate of the variance is:
Thus, its dispersion is also found, and further calculations are made using formulas (23) or (24).
Consistent measurements. A higher accuracy is obtained by another method of processing, when in each of the experiments simultaneously measure . For example, after the release of half of the melt, an additive is added to the metal remaining in the furnace, and then metal samples from each half of the melt are compared.
In this case, in essence, in each experiment, the value of one random variable is measured immediately, which must be compared with the constant a. Measurements are then processed according to formulas (21)–(24), where z must be substituted everywhere.
The variance for consistent measurements will be smaller than for independent ones, since it is due only to a part of random factors: those factors that consistently change do not affect the spread of their difference. Therefore, this method allows to obtain more reliable conclusions.
Example. An interesting illustration of the comparison of values is the determination of the winner in those sports where judging is carried out "by eye" - gymnastics, figure skating, etc.
Table 24. Judging scores
Table 24 shows the protocol of dressage competitions at the 1972 Olympics. It can be seen that the spread of judges' marks is large, and not a single mark can be recognized as grossly erroneous and discarded. At first glance, it seems that the reliability of determining the winner is low.
Let's calculate how correctly the winner is determined, i.e. what is the probability of the event . Since both riders were scored by the same judges, the matched measurement method can be used. According to table 24, we calculate by substituting these values into formula (24) and get .
Choosing a row in table 23, we find that this value of t corresponds to From here, i.e., with a probability of 90% gold medal awarded correctly.
Comparison by independent measurement method will give a slightly worse score, since it does not use the information that the marks were given by the same judges.
Comparison of variances. Let it be required to compare two experimental methods. Obviously, the more accurate method is the one in which the variance of a single measurement is smaller (of course, if the systematic error does not increase). So, we need to establish whether the inequality is satisfied.
Average values
In clinical medicine and public health practice, we often encounter quantitative characteristics (height, number of days of incapacity for work, blood pressure levels, visits to the clinic, population on the site, etc.). Quantitative values can be discrete or continuous. An example of a discrete value is the number of children in a family, pulse; an example of a continuous value is blood pressure, height, weight (the number can be a fraction, turning into the next)
Each numerical value of the unit of observation is called option(x). If you build all options in ascending or descending order and indicate the frequency of each option (p), then you can get the so-called variation series.
A variational series having a normal distribution graphically represents a bell (histogram, polygon).
To characterize variation series, which has a normal distribution (or Gauss-Lyapunov distribution), two groups of parameters are always used:
1. Parameters characterizing the main trend of the series: average value (`x), mode (Mo), median (Me).
2. Parameters characterizing the dispersion of the series: standard deviation (d), coefficient of variation (V).
average value(`x) is a value that determines by one number the quantitative characteristic of a qualitatively homogeneous population.
Fashion (Mo)- the most common variant of the variation series.
Median (Me)- a variant that divides the variation series into equal halves.
Standard deviation(d) shows how, on average, each option deviates from the mean.
Coefficient of variation (V) determines the variability of the variation series in percent and makes it possible to judge the qualitative homogeneity of the studied population. It is advisable to use for comparison the variations of various traits (as well as the degree of variability of very different groups, groups of individuals different types, for example, the weight of newborns and seven-year-old children).
Limits or limits(lim) is the minimum and maximum value option. the simplest way to characterize a variational series, indicate its scope, the minimum and maximum values of the series, i.e. his limits. However, the limits do not indicate how individual members of the population are distributed according to the trait under study, therefore, the above two groups of parameters of the variation series are used.
There are different modifications of the calculation of the parameters of the variational series. Their choice depends on the variation series itself and technical means.
Depending on how the sign varies - discretely or continuously, in a wide or narrow range, simple unweighted, simple weighted (for discrete quantities) and an interval variational series (for continuous quantities).
Grouping of rows is carried out with large numbers observations following way:
1. Determine the range of the series by subtracting the minimum option from the maximum.
2. The resulting number is divided by the desired number of groups (the minimum number is 7, the maximum is 15). This is how the interval is defined.
3. Starting from the minimum option, build a variation series. The boundaries of the intervals should be clear, excluding the entry of the same option into different groups.
The calculation of the parameters of the variational series is carried out from the central variant. If the series is continuous, then the central variant is calculated as half the sum of the initial variant of the previous and subsequent groups. If this is a discontinuous series, then the central variant is calculated as half the sum of the initial and final variant in the group.
Calculation of the parameters of the variation series
Algorithm for calculating the parameters of a simple unweighted variational series:
1. Arrange options in ascending order
2. Sum all options (Sx);
3. By dividing the sum by the number of observations, an unweighted average is obtained;
4. Calculate the serial number of the median (Me);
5. Determine the median variant (Me)
6. Find the deviation (d) of each option from the average (d = x -`x)
7. Square the deviation (d 2);
8. Sum d 2 (Sd 2);
9. Calculate the standard deviation by the formula: ± ;
10. Determine the coefficient of variation by the formula: 
.
11. Make a conclusion about the results.
Note: in a homogeneous statistical population, the coefficient of variation is 5-10%, 11-20% - medium variation, more than 20% - high variation.
Example:
In the resuscitation and intensive care unit, 9 patients with vascular lesions of the brain were treated. Duration of treatment for each patient in days: 7, 8, 12, 6, 4, 10, 9, 5.11.
1. We build a variation series (x): 4,5,6,7,8,9,10,11,12
2. Calculate the sum option: Sx = 72
3. Calculate the average value of the variation series: =72/9=8 days;
4. 
;
5. Me n =5 =8 days;
| x | d | d2 |
| -4 | ||
| -3 | ||
| -2 | ||
| -1 | ||
| +1 | ||
| +2 | ||
| +3 | ||
| +4 | ||
| S=72 | S=0 | Sd2=60 |
9. 
(days);
10. The coefficient of variation is: ;
Algorithm for calculating the parameters of a simple weighted variation series:
1. Arrange the options in ascending order, indicating their frequency (p);
2. Multiply each option by its frequency (x * p);
3. Sum products xp (Sxp);
4. Calculate the average value by the formula (`x)= ;
5. Find the serial number of the median;
6. Determine the variant of the median (Me);
7. The most common variant is taken as fashion (Mo);
8. Find deviations d of each option from the average (d = x - `x);
9. Square the deviations (d 2);
10. Multiply d 2 by p (d 2 *p);
11. Sum d 2 *p (Sd 2 *p);
12. Calculate the standard deviation (s) by the formula: ± ;
13. Determine the coefficient of variation by the formula: .
Example.
Systolic blood pressure was measured in girls aged 16 years.
| Systolic blood pressure, mm Hg x | Number of examined, p | x*p | d | d2 | d2*p |
| -11.4 | 130.0 | 260.0 | |||
| -9.4 | 88.4 | 265.2 | |||
| -7.4 | 54.8 | 219.2 | |||
| -5.4 | 29.2 | 175.2 | |||
| -1.4 | 2.0 | 20.0 | |||
| +0.6 | 0.4 | 9.6 | |||
| 2.6 | 6.8 | 40.8 | |||
| 4.6 | 21.2 | 84.8 | |||
| 6.6 | 43.6 | 130.8 | |||
| 10.6 | 112.4 | 337.2 | |||
| 12.6 | 158.8 | 317.6 | |||
| n=67 | Sxp=7194 | Sd 2 p=1860.4 |
mmHg.;
MmHg.
;
Me=108 mm Hg; Mo=108 mmHg
Algorithm for calculating the parameters of a grouped variational series by the method of moments:
1. Arrange the options in ascending order, indicating their frequency (p)
2. Hold grouping option
3. Calculate the central variant
4. The variant with the highest frequency is taken as the conditional average (A)
5. Calculate the conditional deviation (a) of each central option from the conditional average (A)
6. Multiply a by p (a * p)
7. Summarize the products of ar
8. Determine the value of the interval y by subtracting the central option from the previous
9. Calculate the average value according to the formula:
;
10. To calculate the conditional square deviation, the conditional deviations are squared (a 2)
11. Multiply a 2 * p
12. Sum up the products a * p 2
13. Calculate the standard deviation by the formula
Example
Data available for men aged 30-39 years
| mass, kg x | Number of surveyed p | Middle option x s | a | a 2 | a 2 *p | a*r | Accumulated Frequencies |
| 45-49 | 47,5 | -4 | -4 | ||||
| 50-54 | 52,5 | -3 | -9 | ||||
| 55-59 | 57,5 | -2 | -14 | ||||
| 60-64 | 62,5 | -1 | -10 | ||||
| 65-69 | 67,5 | ||||||
| 70-74 | 72,5 | ||||||
| 75-79 | 77,5 | ||||||
| 80-84 | 82,5 | ||||||
| 85-89 | 87,5 | ||||||
| sum |
- arithmetic mean
; - standard deviation; 
- mean error
Reliability assessment
Statistical assessment of the reliability of the results of a medical statistical study consists of a number of stages - the accuracy of the results depends on the individual stages.
In this case, there are two categories of errors: 1) errors that cannot be taken into account in advance mathematical methods(mistakes in accuracy, attention, typicality, methodological errors, etc.); 2) representativeness errors associated with sample research.
The magnitude of the error of representativeness is determined by both the sample size and the diversity of the trait and is expressed as the mean error. The average error of the indicator is calculated by the formula:
where m is the average error of the indicator;
p is a statistical indicator;
q is the reciprocal of p (1-p, 100-p, 1000-p, etc.)
n is the number of observations.
When the number of observations is less than 30, an amendment is introduced into the formula:
The error of the mean value is calculated by the formulas:
; 
;
where s is the standard deviation;
n is the number of observations.
Example 1
289 people left the hospital, 12 died.
Lethality will be:
; ;
When conducting repeated studies, the average (M) in 68% of cases will fluctuate within ±m, i.e. the degree of probability (p) with which we obtain such confidence limits for the mean is 0.68. However, this degree of probability usually does not satisfy researchers. The smallest degree of probability with which they want to obtain certain boundaries for the fluctuation of the mean (confidence limits) is 0.95 (95%). In this case, the confidence limits of the mean must be expanded by multiplying the error (m) by the confidence factor (t).
Confidence coefficient (t) - a number showing how many times the error of the mean value must be increased in order to assert with a given number of observations with the desired degree of probability (p) that the mean value will not go beyond the limits obtained in this way.
At p=0.95 (95%) t=2, i.e. M±tm=M+2m;
At p=0.99 (99%) t=3, i.e. M±tm=M+3m;
Comparison of averages
When comparing two arithmetic averages (or two indicators) calculated for different periods of time or under slightly different conditions, the significance of the differences between them is determined. In this case, the following rule applies: the difference between the averages (or indicators) is considered significant if the arithmetic difference between the compared averages (or indicators) is greater than two square root from the sum of the squares of the errors of these means (or indicators), i.e.
(for compared averages);
(for comparable indicators).
Subject: maths
The name of the educational and methodical kit (EMC): "RITM»
Lesson topic: Comparison of numbers and quantities by length, volume, mass.
Lesson type: Systematization and generalization of knowledge.
The purpose of the lesson: to teach first-graders to establish connections "scheme-feature"; restore in their memory ways of comparing objects according to the characteristics studied; generalize and consolidate the material about quantities (for example, the quantities length, volume, mass).
Lesson objectives:
To form the ability to describe the results of observations of the properties of objects (color, shape, size, material, volume, area, mass);
To form the ability to single out sets of objects or figures that have a common feature;
Train mental operations, develop motor skills of small muscles, the ability to self-control, develop communication skills;
To educate students in a relationship of business cooperation (goodwill towards each other, respect the opinions of others, be able to listen to comrades);
Instill interest in the subject.
Planned results:
Personal :
To form educational and cognitive interest in the material;
The ability to evaluate their own work and the work of their comrades;
Take responsibility for your work;
Develop motivation for learning and learning;
readiness and ability for self-development, development of tolerance.
Metasubject:
regulatory:
Be able to determine and formulate the goal in the lesson with the help of a teacher;
Speak the sequence of actions in the lesson;
Understand the learning objective of the lesson; carry out the solution of the educational problem under the guidance of the teacher;
Evaluate the correctness of the action at the level of an adequate retrospective assessment;
Express your guess;
cognitive:
Be able to navigate in your system of knowledge;
Find answers to questions using the textbook, your life experience and information obtained in the lesson;
communicative:
Formulate your own opinion and position;
Listen and understand the opinions of others;
Follow the rules of working in pairs;
Subject:
The ability to distinguish the properties of objects that are quantities from those properties that are not quantities;
Knowledge of what can be done with quantities: compare, measure;
Ability to compare quantities and their numerical values;
Ability to compare results;
Ability to work in a group.
Lesson equipment: demonstration cards with the names of features (length, volume, color, area, shape, perimeter, width, material, mass), cards (individual), scales, 4 cubes (outwardly the same, but different in mass - 2 cubes of the same mass), demonstration boat, presentation for the lesson.
Equipment: multimedia projector, computer, handout for group work (balloons, balls, boxes of different materials, different in size, balloons, wire), mathematical fan, cards for individual work.
TECHNOLOGICAL CARD OF THE LESSON
| Teacher activity | Student activities |
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| Hello. I am glad to welcome you. Let's create for successful work good mood. Look at each other with kind eyes. Smile at each other with a kind smile. Give each other a good look. Tell each other, quietly, a kind word. The mood is excellent. Let's get started Are you ready to start the lesson? Check your workplace. | Check readiness for the lesson. Listen to teachers. Share your mood, say kind words. Prepare for upcoming class work |
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| Setting goals and objectives for the lesson. Motivation learning activities students. |
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| Over what learning task we will work? How do we compare values? (by length, volume, mass) Can you compare quantities? Show with our special icons how you compare by length, perimeter, area, material, color, shape, mass, volume. | Children answer questions. They show with icons how to compare objects on some basis. |
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| Knowledge update |
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| 1. Concretely - practical work comparing objects on a given basis What does the diagram show? (objects are not the same for some reason) Tasks by groups: Find items that are not the same length Find objects that are not the same in mass Find items that are not the same size - Name the values. LENGTH COLOR WEIGHT VALUES VOLUME SMELL THE FORM Why are they big? | Mass, length, volume. They can be measured. We will talk about magnitudes. |
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| I give the group wires of the same length. Make broken lines from 2, 3, 4, 5 links. On what basis are broken lines the same? (material, length) | |||||||||||||||||||
| Comparison of values On what basis can two objects be compared? Which scheme is suitable? 1. Mouse and elephant. Compare by weight, size 2. Triangle and square. Compare by size or shape 3. Two vessels with water. Compare by volume. Krosh and Hedgehog decided to help Nyusha water the flowers. | View slides and compare |
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| Verbal counting Compare values by number of items. We put a sign more or less. How much more or less? Penguins 2 and 4 Fish 8 and 4 Keys 3 and 1 Alarm clocks and table lamps 3 and 4 What number is 2 more than 3, 4 What number is 1 less than 8, 3 less than 6, 1 more than 10 | Compare the number of pictures We show the answer using a mathematical fan |
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| Generalization and systematization of knowledge |
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| Guys, what are we going to talk about in class today? To learn more about them, I suggest you work in groups. Each group will be given a task to complete. practical work. The group must be responsible. One speaks, the other listens. Express your disagreement politely. If you don't understand, ask again. Everyone should work for the result. Task 1 group Insert the number that corresponds to the weight of the animal 8, 5 and 2
2) Put signs greater than, less than or equal to. Task 2 group 1) Measure the length of the fish with a ruler and write it down. swordsman
2). Restore record: Rate the work. Task 4 group Do practical work. 1) Measure the volume of a mug, glass, jar, using a measure - a mug and write down the data. 2) Restore the record: Rate the work. IV. Generalization. Group conclusions. 1 group - What was the value in your assignment? Which animal has the smallest mass? Kitty? Why should a person know this? 2 group
3group - How much did you work with? The conclusion is made according to the table that turned out in the course of the answers of the student.
– What actions did we perform with the values? | About quantities The guys work in groups. Each group has its own task. Showcase their work. They answer questions. |
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| Fizminutka |
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| Control of assimilation, discussion of the mistakes made and their correction |
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| calligraphic minute What number follows the number 6? precedes the number 7? Independent work. | Children answer questions Perform independent work on cards Checking group work We check the work together by reading the answers |
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| On what basis were the values compared? Page 103, no. 7 Tutorial and slides On what basis did Ira compare the objects? (by volume) Dasha? (by height) Tanya? (by weight) Compare items with a diagram. | Children look at drawings, diagrams, compare drawings with diagrams and draw conclusions. |
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| Drawing up tasks based on a drawing and a diagram Page 111, no. 18 Let's make a problem about father and son. According to the scheme, we determine what the values \u200b\u200bare equal to. What's this? (Package with vegetables and potatoes) On what basis can these objects be compared? (by weight) What's this? (buckets of water) On what basis can these objects be compared? (by volume) What's this? (2 fish) How can these objects be compared? | In groups, children collectively create tasks. Speak the text of the problem based on the drawing and diagram. Orally answer the question. Show on a diagram how much one value is greater or less than another |
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| Reflection (summarizing the lesson) |
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| What topic are we working on today? I got it all I've been having trouble I did not understand anything What were you having trouble with? What did you do easily, without difficulty? | Show with the help of a sign, their attitude to the lesson. Express their attitude. |
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| Additional task Task on logical thinking |
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| 1. Standing on one leg, the chicken weighs 1kg. How much does this chicken weigh when it stands on two legs? 2 identical watermelons weigh the same as 3 identical melons. Which is heavier: watermelon or melon? | Conclusion: A chicken, standing on two legs, weighs 1 kg. |
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Since the earliest times, people have been seriously interested in the question of how it is most convenient to compare quantities expressed in different values. And it's not just natural curiosity. The man of the most ancient terrestrial civilizations attached purely applied significance to this rather difficult matter. Correctly measuring the land, determining the weight of the product on the market, calculating the required ratio of goods in barter, determining the correct rate of grapes when harvesting wine - these are just a few of the tasks that often surfaced in the already difficult life of our ancestors. Therefore, poorly educated and illiterate people, if necessary, to compare the values, went for advice to their more experienced comrades, and they often took an appropriate bribe for such a service, and quite a good one, by the way.
What can be compared
Nowadays, this lesson also plays a significant role in the process of studying the exact sciences. Of course, everyone knows that it is necessary to compare homogeneous values, that is, apples with apples, and beets with beets. It would never occur to anyone to try to express degrees Celsius in kilometers or kilograms in decibels, but we have known the length of the boa constrictor in parrots since childhood (for those who do not remember: there are 38 parrots in one boa constrictor). Although parrots are also different, and in fact the length of the boa constrictor will vary depending on the subspecies of the parrot, but these are the details that we will try to figure out.
Dimensions
When the task says: "Compare the values of the quantities", it is necessary to bring these same quantities to the same denominator, that is, to express them in the same values for ease of comparison. It is clear that it will not be difficult for many of us to compare the value expressed in kilograms with the value expressed in centners or in tons. However, there are homogeneous quantities that can be expressed in different dimensions and, moreover, in different systems measurements. Try, for example, to compare kinematic viscosity values and determine which of the fluids is more viscous in centistokes and square meters per second. Does not work? And it won't work. To do this, you need to reflect both values in the same values, and already by the numerical value to determine which of them is superior to the opponent.
Measurement system
In order to understand what quantities can be compared, let's try to recall the existing measurement systems. To optimize and speed up settlement processes in 1875, seventeen countries (including Russia, the USA, Germany, etc.) signed a metric convention and defined metric system measures. To develop and consolidate the standards of the meter and kilogram, the International Committee for Weights and Measures was founded, and the International Bureau of Weights and Measures was set up in Paris. This system eventually evolved into the International System of Units, SI. At present, this system is adopted by most countries in the field of technical calculations, including those countries where traditionally in Everyday life national ones are used (for example, the USA and England).
GHS
However, in parallel with the generally accepted standard of standards, another, less convenient CGS system (centimeter-gram-second) developed. It was proposed in 1832 by the German physicist Gauss, and in 1874 modernized by Maxwell and Thompson, mainly in the field of electrodynamics. In 1889, a more convenient ISS (meter-kilogram-second) system was proposed. Comparing objects by the size of the reference values of the meter and kilogram is much more convenient for engineers than using their derivatives (centi-, milli-, deci-, etc.). However, this concept also did not find a mass response in the hearts of those for whom it was intended. All over the world, it was actively developed and used, therefore, calculations in the CGS were carried out less and less, and after 1960, with the introduction of the SI system, the CGS practically fell into disuse. At present, the GHS is actually used in practice only in calculations in theoretical mechanics and astrophysics, and then because of the simpler form of recording the laws of electromagnetism.
Step-by-step instruction
Let's analyze an example in detail. Suppose the problem is: "Compare the values of 25 tons and 19570 kg. Which of the values is greater?" The first thing to do is to determine in what quantities we have given values. So, the first value is given in tons, and the second - in kilograms. At the second step, we check whether the compilers of the problem are trying to mislead us by trying to force us to compare heterogeneous quantities. There are also such trap tasks, especially in quick tests, where 20-30 seconds are given to answer each question. As we can see, the values are homogeneous: both in kilograms and in tons, we measure the mass and weight of the body, so the second test was passed with a positive result. The third step, we translate kilograms into tons or, conversely, tons into kilograms for ease of comparison. In the first version, 25 and 19.57 tons are obtained, and in the second: 25,000 and 19,570 kilograms. And now you can compare the magnitudes of these values with peace of mind. As can be clearly seen, the first value (25 tons) in both cases is greater than the second (19,570 kg).
Traps
As mentioned above, modern tests contain a lot of deception tasks. These are not necessarily tasks that we have analyzed, a rather harmless-looking question can turn out to be a trap, especially one where a completely logical answer suggests itself. However, the deceit, as a rule, lies in the details or in a small nuance that the compilers of the task are trying to disguise in every possible way. For example, instead of the question already familiar to you from the analyzed problems with the formulation of the question: "Compare values where possible" - the compilers of the test can simply ask you to compare the indicated values, and choose the values themselves strikingly similar to each other. For example, kg * m / s 2 and m / s 2. In the first case, this is the force acting on the object (newtons), and in the second - the acceleration of the body, or m / s 2 and m / s, where you are asked to compare the acceleration with the speed of the body, that is, absolutely heterogeneous quantities.
Complex comparisons
However, very often two values are given in tasks, expressed not only in different units of measurement and in different systems of calculation, but also different from each other in specifics. physical sense. For example, the statement of the problem says: "Compare the values of the dynamic and kinematic viscosities and determine which liquid is more viscous." In this case, the values are indicated in SI units, that is, in m 2 / s, and dynamic - in CGS, that is, in poise. How to proceed in this case?
To solve such problems, you can use the instructions presented above with a small addition to it. We decide in which of the systems we will work: let it be generally accepted among engineers. In the second step, we also check if this is a trap? But in this example, too, everything is clean. We compare two fluids in terms of internal friction (viscosity), so both values are homogeneous. The third step is to convert from poise to pascal second, that is, to the generally accepted units of the SI system. Next, we translate the kinematic viscosity into dynamic, multiplying it by the corresponding value of the density of the liquid (table value), and compare the results obtained.
Out of the system
There are also non-systemic units of measurement, that is, units that are not included in the SI, but according to the results of the decisions of the convening of the General Conference on Weights and Measures (GCVM), acceptable for sharing with the SI. It is possible to compare such quantities with each other only when they are reduced to a general form in the SI standard. Non-systemic units include such units as minute, hour, day, liter, electron volt, knot, hectare, bar, angstrom and many others.
