I. S. Chizhov.

On the Question of the Degree of Reliability of Interference Fits

As is known, all fits, regardless of accuracy class, may be divided into three large groups.

The first group includes fits with a guaranteed minimum clearance; these are the so-called running fits.

The third group includes fits with a guaranteed minimum interference; these are the so-called press fits.

An intermediate position is occupied by the second group of fits. These are the so-called interference fits. Usually these fits serve for fixed joints of various machine parts that are subject to more or less frequent disassembly. Consequently, by the nature of the assemblies for which these fits are intended, they must provide between the mating parts a certain interference, greater or smaller depending on the specific conditions of the given assembly. But in view of the deviations permitted in the dimensions of the shafts and holes being manufactured, instead of the expected interferences, clearances are often found during assembly. Thus, for example, for an assembly having a nominal diameter greater than 30 and up to 50 mm with fit \(M_8/h\,7\) according to JSA (the so-called interference fit of \(2\frac{1}{2}\) accuracy class in the shaft system; see OST 1026), if the shaft diameter is made to the maximum and the hole diameter to the minimum, we obtain an interference of \(34\,\mu\), while if the shaft diameter is made to the minimum and the hole diameter to the maximum, we obtain a clearance of \(30\,\mu\).

Thus in this group of fits assemblies are possible both with interferences and with clearances, the latter, in the majority of cases, being undesirable for us.

When assigning tolerances for a given assembly, in order to evaluate one or another fit, it is very important for the designer, in addition to information on the possible clearances and interferences, also to know the degree of probability of obtaining clearances or interferences during assembly. The probability of obtaining precisely the expected interferences during assembly serves, as it were, as a measure of the reliability of the given fit. Unfortunately, most textbooks on tolerances and fits do not contain such data on the probability of obtaining clearances or interferences during assembly in the group of interference fits. The attempts found in the literature to approach the solution of this question cannot be considered successful.

Thus, Lesokhin in the book Tolerances in Mechanical Engineering, 4th ed., pp. 82–84, gives a calculation of the percentage of assemblies with clearances for blind and interference fits of the 2nd accuracy class (\(G\) and \(T\) in the hole system for the interval 50–80 mm). For the other interference fits of the 2nd class, and also for interference fits of the 1st and \(2\frac{1}{2}\) classes, data on the percentage of fits with clearances are not given¹. With regard to the indicated calculation, the following remarks must be made. The first remark, of a particular nature, concerns the calculation of the percentage of fits with clearances for fit \(T\).

In this calculation the author made an arithmetic error: in the book

\[ \sum_{20}^{0} m = 105, \]

whereas in fact it should be 210. In view of this the percentage

for assemblies with clearances should have been obtained by the author as 32.2%, instead of the indicated 16.1%.

The author's second error lies in the fundamental approach to solving the stated problem. Lesokhin's reasoning is approximately as follows (Fig. 1). Let a hole \(A\) be given with deviations \(BO_a\) and \(HO_a\), and a shaft \(B\) with deviations \(BO_b\) and \(HO_b\) (in microns). For each shaft with deviation \(n\), all holes with deviations from \(HO_a\) to \(n\) will give interferences, while holes with deviations from \(n\) to \(BO_a\) will give clearances. If the parts have deviations only in whole microns, then the total number of fits with this shaft will be \((BO_a - HO_a + 1)\). The number of fits with clearances is \(BO_a - n\). Assigning to the shaft deviations \(n\) different values from \(HO_b\) to \(BO_a\), we obtain the number of possible combinations with clearances equal to \(S\).

Fig. 2.

\[ S = \sum_{n=HO_b}^{n=BO_a} (BO_a - n) = \sum_{m=BO_a-HO_b}^{m=0} m. \]

The total number of all fits for the given shaft and hole will be
\((BO_a - HO_a + 1)\cdot(BO_b - HO_b + 1)\), and therefore the required percentage of fits with clearances will be

\[ (100 - W_0) = \frac{ 100 \displaystyle\sum_{BO_a-HO_b}^{0} m }{ (BO_a - HO_a + 1)(BO_b - HO_b + 1) } = \frac{ 100 \displaystyle\sum_{BO_a-HO_b}^{0} m }{ (\delta_a + 1)(\delta_b + 1) } \tag{1} \]

Here and below, \(W_0\) is understood to mean the percentage of fits with interferences.

The error of Eng. Lesokhin in the reasoning set out above consists in the fact that he regards the occurrence in assembly of each of the shafts (as well as of the holes) with a given deviation as equally probable. Meanwhile, experience and theoretical considerations show that the distribution of parts over the tolerance field can in no case be regarded as uniform. With respect to the law of distribution of parts over the tolerance field, according to the existing investigations carried out in this direction, the following conclusion may be drawn:

1. Where the process of manufacturing parts, whether a shaft or a hole, is organized so that the final size of the part is determined by the corresponding preliminary setting of the machine tool, tool, or fixture, there we have the right to expect a symmetric distribution of parts over the tolerance field according to the so-called normal Gaussian distribution curve. This includes machining on automatic machines, machining on set-up turret and revolving machines with stops, reaming and drilling of holes, centerless grinding, etc.

2. Where the process of manufacturing the part is constructed in such a way that bringing it to the final size is left to the skill of the operator, the distribution curve is asymmetrical. Owing to the worker's natural desire to keep closer to the pass side

of the gauge, the vertex of the distribution curve is displaced by a certain distance toward the beginning of the tolerance field.

For mass and large-scale production, where work with limiting gauges on the basis of the principles of interchangeability is exceptionally widespread, the first type of construction of technological processes is precisely the most characteristic, and, consequently, the corresponding normal distribution of parts in the tolerance field according to the Gaussian curve is encountered most often.

The equation of the Gaussian curve may be represented in the following form

\[ W(v)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(v-M)^2}{2\sigma^2}} \tag{2} \]

Here \(\pi=3.1415\);

\[ e=\text{the base of the system of natural logarithms }—2.71828; \]

\(\sigma=\) the root-mean-square deviation of the individual variants of the statistical population from the arithmetic mean;

\(M=\) the arithmetic mean of the given statistical population;

\(v=\) the current values of the variant, in the present case the current values of the actual deviation of the part from the nominal size;

\(W(v)=\) the frequency function of occurrence of each \(v\).

The general form of the curve is shown in Fig. 2.

Theoretically, both branches of the curve have the value \(W=0\) at \(-\infty\) and \(+\infty\), i.e., both branches asymptotically approach the \(v\)-axis. As is seen in the normal distribution curve, the greatest frequency is possessed by parts with deviations equal to the arithmetic mean.

If, in the above diagram, the area enclosed between the curve and the \(v\)-axis is taken as unity, then the area between the curve, the \(v\)-axis, and two given ordinates \(bb'\) and \(cc'\) determines the probability that a randomly selected part will fall within the limits marked by the ordinates \(bb'\) and \(cc'\).

In equation (2) there enters the unknown quantity of the root-mean-square deviation \(\sigma\). The quantity \(\sigma\) characterizes the degree of dispersion of the given statistical population. It is most convenient, instead of \(v\), to lay off on the abscissa axis the deviation \(v\) from the arithmetic mean, expressed in fractions of \(\sigma\), i.e. the quantity

\[ \frac{v-M}{\sigma}=t. \]

Calculation shows that within the limits from \(t=-1\) to \(t=+1\) are contained \(68.3\%\) of all values of \(v\), within the limits from \(t=-2\) to \(t=+2\) are contained \(95.5\%\) of all values of \(v\), and within the limits from \(t=-3\) to \(t=+3\) are contained \(99.74\%\) of all values of \(v\). This means that only \(0.26\%\) of parts have deviations greater than \(\pm 3\sigma\), and since half of these parts (i.e., \(\sim 0.13\%\)) have deviation \(+3\sigma\) for shafts and with devia-

Fig. 2.

—minus \(3\sigma\) for holes can be corrected by reworking, then we may assume that, within \(\pm 3\sigma\), all parts manufactured to the given tolerance must be contained. Not knowing the actual value of \(\sigma\) for the law of distribution of parts over the tolerance zone, we may suppose that the magnitude of the tolerance is entirely contained within \(\pm 3\sigma\), i.e., if the magnitude of the tolerance is denoted by \(\delta\), then \(\delta = 6\sigma\), and therefore

\[ \sigma=\frac{\delta}{6}. \]

Indeed, if the distribution has a large scatter and \(\sigma > \dfrac{\delta}{6}\), then a considerable number of parts will fall outside the tolerance zone (Fig. 3). Such a large amount of scrap must lead to an immediate review of the technological process and its readjustment. If, on the other hand, \(\sigma < \dfrac{\delta}{6}\), this indicates that the tolerance is not being fully used and the work is being performed with greater accuracy than is necessary. We also cannot regard such conditions as normal. Therefore the distribution with

\[ \sigma=\frac{\delta}{6} \]

must be recognized as the most acceptable. Having adopted this assumption, we have completely determined quantitatively the law of distribution of parts within the tolerance zone, for each fit of any class. We may now proceed to solving the problem that interests us.

Fig. 3.

First let us solve the problem in general form. A certain fit is given to us by tolerance zones arranged in a definite manner with respect to the zero line (Fig. 4). The distribution curves of the corresponding parts in the tolerance zone are shown on the right. We shall denote the mean deviations of the parts by \(M\) with the corresponding subscript. Let us denote the mean interference between part \(A\) (hole) and part \(B\) (shaft).

\[ M_b - M_a = S_0 \]

Take an arbitrary hole \(A_1\) and an arbitrary shaft \(B_1\). Let the interference between them (Fig. 4) be denoted by \(S_1\). What is the probability of obtaining an interference equal to \(S_1\) for various combinations of shaft and hole?

Fig. 4.

Let the distribution curve of the aggregate \(A\) (holes) be

\[ f_a(v-M_a)=\frac{1}{\sigma_a\sqrt{2\pi}}e^{-\frac{(v-M_a)^2}{2\sigma_a^2}} \tag{3} \]

and the distribution curve of the aggregate \(B\) (shafts)

\[ f_b(v-M_b)=\frac{1}{\sigma_b\sqrt{2\pi}}e^{-\frac{(v-M_b)^2}{2\sigma_b^2}} . \tag{4} \]

Let each of the aggregates contain \(n\) parts. Each pair of parts gives some interference \(S\). Consequently, the number of possible combinations giving different magnitudes of interference will be \(n^2\). The question is, how many of these pairs give an interference lying in the interval from \((S_1)\) to \((S_1+dS_1)\).

To answer this, choose a very limited region of hole values \(dv\) (shaded in the drawing). It contains \(n f_a(v-M_a)\,dv\) holes. Here by \(f_a(v-M_a)\) is meant the value of the distribution function for the argument \(=(v-M_a)\). It is clear that only those shafts which lie between the ordinates from \((v+S_1)\) to \((v+S_1+dS_1)\), together with the holes mentioned, give interferences from \(S_1\) to \(S_1+ds_1\). But the number of these shafts will be \(n f_b(v-M_b+S_1)\,ds_1\). The combinations of \(n f_a(v-M_a)\,dv\) holes with \(n f_b(v-M_b+S_1)\,ds_1\) shafts will give combinations with the desired interferences from \(S_1\) to \(S_1+ds_1\).

The number of these combinations is

\[ n f_a(v-M_a)\,dv\, n f_b(v-M_b+S_1)\,ds_1 = n^2 f_a(v-M_a) f_b(v-M_b+S_1)\,dv\,ds_1 . \]

Integrating with respect to \(v\) from \(-\infty\) to \(+\infty\), we obtain the total number of the required combinations

\[ = n^2 \int_{-\infty}^{+\infty} f_a(v-M_a) f_b(v-M_b+s_1)\,dv\,ds_1 . \]

We have taken the limits of integration from \(-\infty\) to \(+\infty\), although the values of \(v\) extend from \(-\dfrac{\delta_a}{2}\) to \(+\dfrac{\delta_a}{2}\). As indicated above, within the limits from \(-\dfrac{\delta_a}{2}\) to \(+\dfrac{\delta_a}{2}\) there are contained \(99.74\%\) of all values of \(v\), so that, by integrating from \(-\infty\) to \(+\infty\), we introduce into the calculation a number of values exceeding the actual one by \(0.26\%\). Meanwhile, the simplicity of the calculation and of the derivation gains greatly when the limits from \(-\infty\) to \(+\infty\) are adopted.

Dividing this number by the total number of combinations of shafts and holes, \(n^2\), we obtain the probability \(W(S_1)\,ds_1\) of obtaining interference \(=s_1\) after substituting the values \(f_a^{\,2}_a(v-M_a)\) and \(f_b^{\,2}_b(v-M_b+S_1)\)

\[ W(s_1)\,ds_1 = \int_{-\infty}^{+\infty} \frac{1}{\sigma_a\sqrt{2\pi}} e^{-\frac{(v-M_a)^2}{2\sigma_a^2}} \cdot \frac{1}{\sigma_b\sqrt{2\pi}} e^{-\frac{[v-(M_b-S_1)]^2}{2\sigma_b^2}} \,dv\,ds_1 . \]

Denoting \(v-M_a=x\) and replacing \(M_b=M_a+S_0\), we obtain

\[ W(s_1)= \frac{1}{\sigma_a\sigma_b\,2\pi} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{2\sigma_a^2}+\frac{(x-S_0+1)^2}{2\sigma_b^2}} \,dx . \]

We transform the exponent as follows:

\[ -\frac{1}{2} \left[ x^2\left(\frac{1}{\sigma_a^2}+\frac{1}{\sigma_b^2}\right) -2x\frac{S_0-S_1}{\sigma_b^2} +\frac{(S_0-S_1)^2}{\sigma_b^2} \right] \]

Let us denote

\[ \frac{1}{\sigma_a^2}+\frac{1}{\sigma_b^2}=\alpha, \]

\[ \frac{S_0-S_1}{\sigma_b^2}=\beta,\qquad \frac{(S_0-S_1)}{\sigma_b^2}=\gamma . \]

Then the exponent at \(e\) will take the form

\[ -\frac{1}{2}\left(x^2\alpha-2x\beta+\gamma\right) \]

Taking \(\alpha\) outside the brackets, we obtain:

\[ -\frac{\alpha}{2}\left[x^2-2x\frac{\beta}{\alpha} +\left(\frac{\beta}{\alpha}\right)^2 -\left(\frac{\beta}{\alpha}\right)^2 +\frac{\gamma}{\alpha}\right] = \frac{\alpha}{2}\left(x-\frac{\beta}{2}\right)^2 -\frac{\gamma\alpha-\beta^2}{2\alpha} \]

Substituting the values of \(\alpha\), \(\beta\), and \(\gamma\) into the last term, after reduction and transformations we obtain

\[ \frac{\gamma\alpha-\beta^2}{2\alpha} = \frac{(S_0-S_1)^2}{2(\sigma_a^2+\sigma_b^2)} \]

As we see, this expression does not depend on \(x\). Substituting the transformed expression for the exponent into the probability equation, we obtain

\[ W(S_1)=\frac{1}{\sigma_a\sigma_b 2\pi} \int_{-\infty}^{+\infty} e^{-\frac{\alpha}{2}\left(x-\frac{\beta}{\alpha}\right)^2} e^{-\frac{(S_0-S_1)^2}{2(\sigma_a^2+\sigma_b^2)}}\,dx \]

The last factor, as independent of \(x\), may be taken outside the integral sign,

Furthermore,

\[ \int_{-\infty}^{+\infty} e^{-\frac{\alpha}{2}\left(x-\frac{\beta}{\alpha}\right)^2}\,dx = \sqrt{\frac{2\pi}{\alpha}}. \]

Substituting all this into the equation for \(W(S_1)\), we obtain

\[ W(S_1)= \frac{1}{\sqrt{\sigma_a^2+\sigma_b^2}\sqrt{2\pi}}\, e^{-\frac{(S_0-S_1)^2}{2(\sigma_a^2+\sigma_b^2)}}. \]

Introducing the notation

\[ \xi^2=\sigma_a^2+\sigma_b^2, \]

we finally obtain

\[ W'(S_1)=\frac{1}{\xi\sqrt{2\pi}}\, e^{-\frac{(S_0-S_1)^2}{2\xi^2}} \tag{5} \]

i.e., the frequency function of the interference \(S_1\) is also a Gaussian curve with arithmetic mean \(S_0\) and standard deviation equal to the hypotenuse of a right triangle whose legs are the standard deviations of the shaft and the hole.

If, again, the area between the curve and the abscissa axis is taken as unity, then the area enclosed between the ordinates \(S_a\), \(S_b\), the abscissa axis, and the curve gives the probability that an arbitrary shaft and hole will give an interference within the limits from \(S_a\) to \(S_b\) (Fig. 5).

Having the general law of distribution of interferences, it will not be difficult to compute also the probability that, for a given positive \(S_0\), two arbitrary parts (shaft and hole) will likewise give a positive interference. The required probability \(W_0\) will be obtained by integrating the function \(W(S_1)\) with respect to \(S_1\) over the limits from \(0\) to \(-\infty\).

Fig. 5.

\[ W_0=\frac{1}{\xi\sqrt{2\pi}}\int_0^\infty e^{-\frac{(S_0-S_1)}{2\xi^2}}\,dS. \]

Denoting

\[ \frac{S_0-S_1}{\xi}=x, \]

we have \(-dS_1=\xi\,dx\) and

\[ W_0=-\int_{\frac{S_0}{\xi}}^{-\infty}\frac{\xi}{\xi\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx = \frac{1}{\sqrt{2\pi}}\int_{-\frac{S_0}{\xi}}^\infty e^{-\frac{x^2}{2}}\,dx = \frac{1}{2}\left[1+\Phi\left(\frac{S_0}{\xi}\right)\right]. \]

\[ W_0=\frac{1}{2}\left[1+\Phi\left(\frac{S_0}{\xi}\right)\right] \tag{6} \]

Here the quantity \(\Phi\left(\frac{S_0}{\xi}\right)\) is the value of the probability integral, which is best taken from existing tables giving the value of this integral for the given ratio \(\left(\frac{S_0}{\xi}\right)\). Tables of the probability integral may be found in any course on mathematical or variational statistics or in courses on probability theory. The value of the function \(\Phi\left(\frac{S_0}{\xi}\right)\), as \(\left(\frac{S_0}{\xi}\right)\) changes, at first increases rather rapidly, and then gradually slows down.

(see the diagram in Fig. 6). It is obvious that the curve \(\Phi\!\left(\dfrac{S_0}{\xi}\right)\) is the integral curve for the distribution function. From formula (6) and the form of the function (see Fig. 6) it is clear that, for small values of \(\dfrac{S_0}{\xi}\), the probability of a positive interference will be close to \(\dfrac{1}{2}\). Indeed, for an average interference equal to

Fig. 6.

or close to zero, the probability of the occurrence of a positive interference or a positive clearance is close to one half.

Moreover, from the same formula we see that the value \(W_0\) will also approach \(\dfrac{1}{2}\) for very large values of \(\xi\), and, consequently, \(\sigma_a\) and \(\sigma_b\).

We shall illustrate the application of the conclusions and arguments given above to specific fits by several examples.

Example 1. Compute the percentage of fits with clearances for a tight fit of accuracy class 2 (\(T\)) in the range \(50\text{--}80\ \mathrm{mm}\), shaft system. From OST 1022 we find the deviations:

for the shaft—the upper deviation \(BO_b = 0\),

lower deviation \(HO_b = -20\mu\);

for the hole—the upper deviation \(BO_a = 0\),

lower deviation \(HO_a = -30\mu\).

Therefore, the shaft tolerance is

\[ \sigma_b = BO_b - HO_b = 20\mu, \]

the hole tolerance is

\[ \sigma_a = BO_a - HO_a = 30\mu. \]

Minimum interference

\[ S_{\min} = HO_b - BO_a = -20 - 0 = -20\mu, \]

i.e. (a clearance of \(20\mu\)).

Maximum interference

\[ S_{\max} = BO_b - HO_a = 0 - (-30) = 30\mu. \]

Mean interference

\[ S_0 = S_{\mathrm{av}} = \frac{S_{\max}+S_{\min}}{2}=\frac{30-20}{2}=5\mu \]

Mean square deviation for shafts

\[ \sigma_b=\frac{\delta_b}{6}=\frac{20}{6}=\frac{10}{3}\mu \]

The same for the hole

\[ \sigma_a=\frac{\delta_a}{6}=\frac{30}{6}=5\mu. \]

Then the mean square deviation for the distribution curve of interferences from the mean interference is

\[ \xi=\sqrt{\sigma_a^2+\sigma_b^2}=\sqrt{5^2+\left(\frac{10}{3}\right)^2}\cong 6\mu. \]

The probability of obtaining positive interferences is

\[ W_0=\frac{1}{2}\left[1+\Phi\left(\frac{S_{\mathrm{av}}}{\xi}\right)\right] =\frac{1}{2}\left[1+\Phi\left(\frac{5}{6}\right)\right]. \]

We take the value of the probability integral function \(\Phi\left(\frac{5}{6}\right)\) from the table on p. 126 of Levinsky’s book A Short Course in Variation Statistics:

\[ \Phi\left(\frac{5}{5}\right)=0.5962. \]

Consequently,

\[ W_0=\frac{1}{2}(1+0.5962)=79.8\%. \]

Thus, \(79.8\%\) of assemblies will have positive interferences and \(20.2\%\) will have clearances. According to Lesokhin, by formula (1) we would obtain

\[ (100-W_0)= \frac{ 100\displaystyle\sum_{BO_a-HO_b}^{0} m }{ (\delta_a+1)(\delta_b+1) } = \frac{ 100\displaystyle\sum_{0+20}^{0} m }{ (20+1)(30+1) } = \frac{100\cdot210}{651} =32.2\%. \]

The calculated percentage of assemblies with clearances by the method we propose turns out to be lower than Lesokhin’s. This is quite understandable, since Lesokhin proceeds from an essentially incorrect “distribution curve” in the form of a rectangle (equal probabilities for any variant values of the shaft and the hole). In fact, the probabilities of the occurrence of shafts and holes that give clearances in the mating will be smaller than for shafts and holes that give the mean interference.

Example 2. Calculate the percentage of assemblies with clearances for an interference fit of the 2nd accuracy class in the interval \(50\)—\(80\) mm. Shaft basis system. According to OST 1022 and OST 1030 we have:

\[ \delta_a=31\mu,\quad \delta_b=20\mu.\quad \text{Hence, }\xi= \sqrt{\left(\frac{\delta_a}{6}\right)^2+\left(\frac{\delta_b}{6}\right)^2} = \sqrt{\left(\frac{31}{6}\right)^2+\left(\frac{20}{6}\right)^2} =6.15 \]

\[ S_{\min}=-28\mu\;(\text{clearance}) \qquad \Phi\left(\frac{S_0}{\xi}\right)=\Phi\left(\frac{5}{6.15}\right)=0.5837 \]

\[ S_{\max}=+23\mu. \]

\[ S_{\mathrm{av}}=S_0=-5\mu \quad \text{(clearance).} \]

Consequently, the probability of the occurrence of clearances will be

\[ W_0=\frac{1}{2}(1+0{,}5837)\simeq 0{,}792, \]

i.e., 79.2% will be with clearances and only 20.8% with interferences.

By this method one can calculate the percentage of assemblies with interferences also for other fits. Table No. 1 presents the results of calculations made by the author for interference fits of all accuracy classes in various diameter intervals.

Table No. 1.

Percentage of assemblies with interferences in interference fits of the 1st, 2nd, and 2½ classes.

From consideration of this table it follows that the fluctuations in the percentage of fully reliable fits (with positive interference) in different intervals of a given accuracy class are very insignificant and are explained by the roundings in the specified deviations of the various fits. In exactly the same way, the table shows that the percentage reliability of each fit decreases with transition to coarser accuracy classes, which is quite natural and requires no explanation. A certain anomaly in this respect occurs only for the close fit.

Incidentally, it should be emphasized that the latter fit gives an overwhelming number of assemblies with clearances, and only as an exception are there encountered from 0.5—3.3% of assemblies with interferences.

The table given is, as a whole, of practical interest for the designer in assigning and selecting interference fits, in the case of a normal distribution of parts over the tolerance field.

In the case of an asymmetric distribution of parts over the tolerance field, the percentage of fits with interferences will, naturally, be higher. This percentage can be calculated by the same method; for this it is necessary to know the exact mathematical expression of the form of the distribution functions \(f_a(v-M_a)\) and \(f_b(v-M_b)\).

In conclusion it should be pointed out that, by means of formula (5), one can calculate the percentage of assemblies with interferences or clearances from \(S_1\) to \(S_2\); for this it is necessary to integrate \(W(S)\) with respect to \(S\) and substitute the corresponding limits in the integration.

References Used

1. Zykov. Application of the law of large numbers to the system of tolerances. Vestnik Metallopromyshlennosti, 1928, No. 5/6, and 1929, No. 5.

2. G. Schlesinger. Die Bewährung der Dinpassungen in Lichte der Statistik. Maschinenbau, 1929, Heft 9.

3. Buxbaum, Ing. Kontrolle des Zustandes und der Ausnützung von Maschinen und maschinellen Einrichtungen (see the book by Dr. Ing. G. Kienzie, Kontrollen der Betriebwirtschaft, Verlag v. Julias Springer, 1931).

4. Dr. Ing. Fischer. Feste Grenzlehren in der Feinmechanik. Maschinenbau, 1933, No. 23–24.

5. Ing. Obeltshauser. Arbeitsgenauigkeit von Automaten. Maschinenbau, 1928, Heft 11.

6. Pivovar V. A. Distribution of parts over the tolerance field. Vestnik Metallopromyshlennosti, 1935, No. 9.

7. Levinskii V. P. A Short Course in Variational Statistics. Moscow, 1935.

8. Lesokhin A., Eng. Tolerances in Machine Building. 4th ed.

1. Fits of \(2\frac{1}{2}\) accuracy class, OST 1016 and 1026, were approved after the publication of Lesokhin’s book. ↩