Proceedings of the Tomsk Technological Institute of Emperor Nicholas II

Vol. XXVI. 1912. No. 2.

On the Question of the Sharpening Angle of a Milling Cutter

From the field of investigation into the operation of an ordinary cutting tool it follows that, depending on the quality of the metal being machined, the angle of the working cutter must also change in order to preserve the principle of the most advantageous operation. Thus, according to Taylor’s experimental investigations, it follows:

A) For cutting cast iron and hard grades of steel with a carbon content of about 0.45%, the cutter should have a cutting angle of \(68^\circ\), with a back clearance angle of \(8^\circ\).

B) For cutting soft grades of steel with a carbon content of less than 0.45%, the cutting angle of the cutter should be \(61^\circ\), with the same back clearance angle of the cutter.

C) For cutting the chilled skin of cast iron, the cutting angle should be \(86\text{--}90^\circ\), with the same value of the back clearance angle of the cutter.

D) For steel equal in hardness to tire steel, the cutting angle is \(74^\circ\), under the same conditions with respect to the values of the back clearance angle of the cutter.

Thus, from the preceding it is clearly evident that for cutters of ordinary design there exists a quite definite practical dependence of the cutting angle of the cutter on the character of the metal being machined. It is entirely natural to suppose that a similar dependence also exists for more perfected types of cutters, such as milling cutters. But in reality, in practice, no such dependence is observed: the cutting angle of modern milling cutters, with very few exceptions, is \(90^\circ\). The principal reason for such a phenomenon in modern milling cutters lies in the fact that a milling cutter is a very valuable tool, and therefore the desire to design a milling cutter for the greatest possible cycle of operations is quite natural; but at the same time the milling cutter must be well preserved, and consequently must have the most durable cutting angle, equal to \(90^\circ\). It is clear that under such conditions, with a constant cutting angle of \(90^\circ\), it is sometimes necessary to expend a considerable quantity of energy, absorbed by the milling cutter, uselessly—energy that could have been avoided if there were a considerable set of

cutters with various cutting angles. But the benefit obtained in saving the expenditure of energy absorbed by the cutter during operation, in the majority of practical cases, does not repay in such cases the costs that must be incurred to assemble such an assortment of cutters with various cutting angles. Besides the economic reason just indicated, there is also a practical reason that prevents changing the cutting angle of cutters, namely: only in the case of end mills and of the simple cylindrical form is it practically possible, without great difficulty in their manufacture, to change the cutting angle, making it less than \(90^\circ\); but in this case the front face of the cutter tooth does not pass through the axis of the cutter. To reproduce the same front face of a cutter tooth with a deviation of the cutting angle from \(90^\circ\) is, for most form cutters, a practically impossible task. Indications that, with a decrease in the cutting angle of a cutter from the usual \(90^\circ\), the work expended by it will decrease were made as early as the end of the last century by the Frenchman Desgrandchamps¹, who proposed manufacturing cylindrical cutters in which the front face of the cutter tooth is directed not along the radius of the cutter, but along a certain tangent to a circle several times smaller in diameter than that of the cutter. Despite the obvious saving in the work of a cutter of this latter design, the cutter with undercutting of its teeth, owing to the economic and practical considerations indicated above, has not gained currency on the modern market. The best factories—Brown & Sharpe, Pratt & Whitney, Reinecker, Loewe, and others—currently manufacture cutters with a cutting angle of \(90^\circ\). In order to illustrate that, when the cutting angle of a cutter is reduced from \(90^\circ\) to the most advantageous dimensions for ordinary cutters, there will be a saving in the work absorbed by the cutter, a rather simple oil dynamometer was designed and made in the mechanical workshops of the Tomsk Technological Institute; by means of the diagrams of an ordinary indicator it made it possible to determine the torque of the cutter and, from it, to trace the operating conditions of a cylindrical cutter when the sharpening angle of the cutting teeth of the latter was changed.

The initial investigations on this question were made in the mechanical workshops of the Tomsk Technological Institute by students, and more detailed ones, under quite uniform conditions, were made later, independently of the first.

The construction of the above-mentioned dynamometer is as follows (Figs. 1 and 2 of the plate appended to the article, \(1/5\) natural—

ON THE ISSUE OF THE SHARPENING ANGLE OF A MILLING CUTTER.

… magnitude), a mandrel for the milling cutter under test was fastened in the hollow spindle of an ordinary universal milling machine; this mandrel was of a somewhat different construction in comparison with that for ordinary universal milling machines. In the spindle of the machine the right-hand part of this mandrel was fastened immovably and rotated together with the stepped pulley of the machine; the indicated right-hand part of this mandrel ended in a special fork, by means of which the force from the belt of the machine was transmitted to the pistons of the hydraulic cylinders (1, 2), which formed one whole with the cylindrical mandrel on which the milling cutter under test (3) was fastened. Oil was poured under the pistons of the hydraulic cylinders, and, after a number of necessary oscillations and other precautionary measures, the air was carefully removed from under the pistons of the cylinders (1, 2). The cutter mandrel, as is seen from the drawing, has inside it a channel (5), which is connected by special tubes with the oil cylinders of the apparatus (1, 2), and the indicated channel (5) is connected with the piston of an ordinary indicator. The cylindrical part of the mandrel on which the cutter sits, with respect to the right-hand part of the mandrel (the fork), which forms one whole with the spindle of the machine, can rotate; but the connection of the two halves of the mandrel into one whole takes place only through the pistons of the oil cylinders. Thus, the force overcome by the cutter during work produces a definite pressure on the oil under the pistons of the cylinders (1, 2), and the oil from the latter, through channel (5) of the cutter mandrel, acts on the piston of the indicator. The bar (12) from which chips were removed by the cutter was fastened on the table of the machine by means of simple stops (13, 13). To obtain the diagram on the indicator, the cord of the latter was connected, by means of guide rollers, with the table of the machine; consequently, depending on the amount of feed of the table during the cutter’s work, the drum of the indicator gradually rotated, and the pencil of the indicator, depending on the amount of oil pressure on the pistons of the apparatus, traced the pressure curve on the diagram. The diagram obtained in this way (Fig. 3, for a cutter with a cutting angle of \(80^\circ\) and a chip thickness of \(4\text{ mm}\)) represents a continuous shaded area and was not entirely convenient for measurements. To eliminate this last drawback and to obtain a clearer representation of the work of all the teeth of the cutter, for the subsequent tests a special drum (4) with a helical groove (11) was fastened next to the cutter on the latter’s mandrel. The length of that part (4) of the drum where the helical groove was located was less than the circumference of the indicator. In

...the pin (10), which forms one piece with the oblique flat lever (8), could, when pressed, move in the spiral groove of the drum (4). This latter lever (8), with its cylindrical guide, can move lengthwise along the milling-cutter arbor in the slots of special grips (7,7), by which the lever (8) is suspended from the oblique stake of the machine. To the end (9) of the cylindrical rod of the lever (8) is fastened the end of the cord of the indicator. By changing the length of the spiral on the drum (4) (interchangeable drums were installed), it was possible, when pressing the pin (10) into the hollow of the spiral, with the aid of the indicated indicator, to take diagrams of the oil pressure of the dynamometer pistons for \(1/2\), 1, and 2 revolutions of the milling cutter during operation. Fig. 4 gives a copy of one such diagram for two revolutions of the milling cutter with a cutting angle of \(80^\circ\) and a chip thickness of \(4^{m}/_{m}\). Fig. 5 is a copy of a pressure diagram for two revolutions of the milling cutter with a cutting angle of \(90^\circ\) and for a chip thickness of \(4^{m}/_{m}\). The work of each tooth of the milling cutter is clearly visible on the last two diagrams. The total number of all diagrams taken was very considerable, but in order to obtain general conclusions on the given question, no fewer than three characteristic diagrams for each case were taken for calculation.

For all the indicator diagrams obtained, the mean indicator pressures were calculated; then, from all the mean indicator pressures obtained and pertaining to each diagram separately, the mean indicator pressures were calculated for each layer thickness. The results of all calculations are placed in the following tables (scale of the indicator spring: 12 mm = 1 kg.)

Table of mean indicator pressures for diagrams taken with the first method of drum rotation.

The following table contains the values of the mean indicator pressures for the diagrams taken with the second method of rotating the indicator drum; in this table are given the mean indicator pressures for those diagrams which were taken while working with the bushing, the helical groove of which corresponded to two revolutions of the milling cutter.

If now from the obtained mean indicator pressures for each layer thickness we subtract the pressure caused by friction in the apparatus, we obtain the absolute pressure, dependent on the resistance to cutting of the milling cutter. For the diagrams taken with the first method of rotating the drum during work with a milling cutter whose cutters have a cutting angle \(=90^\circ\), one should subtract from the mean indicator pressures \(0,225\ \mathrm{kg}/\mathrm{cm}^2\); for all the remaining diagrams one should subtract from the mean indicator pressures \(0,1375\ \mathrm{kg}\), since in the latter case the work was carried out with an apparatus which, after preliminary disassembly, had been cleaned and lubricated, and consequently the friction in it was considerably less, and during operation it was incomparably more sensitive. The results of the calculations obtained are placed in the following table.

Table of mean (corrected) indicator pressures for diagrams taken with the first method of rotating the drum.

Table of average (corrected) indicator pressures for diagrams taken with the second method of rotating the drum.

With the aid of the average indicator pressures obtained in corrected form, it is necessary to calculate the moments acting on the pistons of the oil apparatus; for this there is the following equation:

\[ M = l . S . p_i, \]

where

\(M\)—the moment acting on the pistons

\(l\)—the arm between the pistons, equal to the distance between the axes of the pistons, \(= 200\) mm.

\(S\)—the area of the piston; the area of the piston is determined by the formula

\[ S = \frac{\pi d^2}{4}, \]

where \(d\)—the diameter of the piston \(= 4,5\) cm; consequently the area of the piston is

\[ S = \frac{\pi . 4,5^2}{4} = 15,806\ \mathrm{cm}^2. \]

\(p_i\)—the average indicator pressure for each layer thickness, in \(\mathrm{kg}/\mathrm{cm}^2\).

The moments calculated by the above equation are placed in the following table.

On the Issue of the Sharpening Angle of a Milling Cutter

With the aid of the moments obtained, it is possible to determine the circumferential forces on the milling-cutter teeth that are in operation at each given moment;

for this there is the equation

\[ M = W \cdot r \quad \text{or} \quad W = \frac{M}{r}, \quad \text{where} \]

\(W\) — the circumferential force acting on the milling-cutter teeth.
\(M\) — the moment on the rods of the instrument.
\(r\) — the outside radius of the milling cutter \(= 50.5\ \mathrm{mm}\).

The results of the calculations made according to the above equation are placed in the following table.

Now it is necessary to determine how many teeth of the milling cutter simultaneously take part in removing each layer thickness. We have a drawing (Fig. 6), where \(D\) is the outside diameter of the milling cutter, \(d\) is the thickness of the layer being removed, \(\psi\) is the angle corresponding to that part of the circumference of the milling cutter with the teeth simultaneously removing a layer of thickness \(d\).

The angle \(a\), corresponding to one tooth of the milling cutter (with the total number of cutter teeth \(= 22\)), is determined as follows:

\[ a = \frac{360}{22} = 16^\circ 21' 49'' \]

The angle \(\psi_1\), corresponding to the thickness of the layer being removed \(d = 1\ \mathrm{mm}\), is determined from the equation

\[ \frac{D}{2} - d = \cos \psi \cdot \frac{D}{2} \quad \text{or} \quad \cos \psi = \frac{\frac{D}{2} - d}{\frac{D}{2}}, \]

where

\[ \frac{D}{2} = \text{radius of the cutter} = 50.5\ \mathrm{mm}, \quad d = 1\ \mathrm{mm}, \]

therefore

\[ \cos \psi_1 = \frac{50.5 - 1}{50.5} = \frac{49.5}{50.5} = 0.9802, \quad \text{whence} \quad \psi_1 = 11^\circ 25'. \]

From comparing the two angles \(\alpha\) and \(\psi_1\), it is evident that, for a thickness of the layer being removed \(d = 1\ \mathrm{mm}\), only one tooth of the cutter participates in the work at each given moment. For a thickness of the layer being removed \(d = 2\ \mathrm{mm}\), the angle \(\psi_2\) is determined from the equation

\[ \cos \psi_2 = \frac{\frac{D}{2} - d_2}{\frac{D}{2}} = \frac{50.5 - 2}{50.5} = \frac{48.5}{50.5} = 0.9604. \]

\[ \psi_2 = 16^\circ 11'. \]

From the second comparison of the two angles \(\alpha\) and \(\psi_2\), it is evident that, for a thickness of the layer being removed \(d = 2\ \mathrm{mm}\), one tooth participates in the work at each given moment.

For a thickness of the layer being removed \(d = 3\ \mathrm{mm}\), the angle \(\psi_3\) is determined from the equation

\[ \cos \psi_3 = \frac{\frac{D}{2} - d_3}{\frac{D}{2}} = \frac{50.5 - 3}{50.5} = \frac{47.5}{50.5} = 0.9406 \]

\[ \psi_3 = 19^\circ 51' \]

From the last comparison of the angles \(\alpha\) and \(\psi_3\), it is evident that, for a thickness of the layer being removed \(d = 3\ \mathrm{mm}\), two teeth of the cutter participate at each given moment; the angle corresponding to the second tooth of the cutter is

\[ \psi_3' = \psi_3 - \alpha = 19^\circ 51' - 16^\circ 21'49'' = 3^\circ 29'11''; \quad \text{accordingly, } \cos 3^\circ 29'11'' = 0.9982. \]

The thickness of the layer \(d_3'\), removed by the 2nd tooth, is determined from the equation

\[ \cos \psi_3' = \frac{\frac{D}{2} - d_3'}{\frac{D}{2}} \quad \text{or} \quad d_3' = \frac{D}{2} - \frac{D}{2}\cos \psi_3' = 50.5 - 50.5 \cos 3^\circ 29'11'' \]

\[ = 50.5 - 50.5 \cdot 0.9982, \quad \text{i.e.} \]

\[ d_3' = 0.091\ \mathrm{mm}. \]

For the thickness of the layer being removed, \(d = 4\) mm, the angle \(\psi_4\) is determined from the equation

\[ \cos \psi_4 = \frac{\frac{D}{2} - d_4}{\frac{D}{2}} = \frac{50.5 - 4}{50.5} = \frac{46.5}{50.5} = 0.9208. \]

\[ \psi_4 = 22^\circ 57' 30''. \]

It is evident from this that, with a layer thickness \(d = 4\) mm, two teeth take part in the work simultaneously. The angle corresponding to the second tooth is

\[ \psi'_4 = \psi_4 - \alpha = 22^\circ 57' 30'' - 16^\circ 21' 49'' = 6^\circ 35' 41''; \]

the corresponding

\[ \cos 6^\circ 35' 41'' = 0.9934; \]

the thickness of the layer removed by the second tooth is

\[ = d'_4 = \frac{D}{2} - \frac{D}{2}\cos \psi'_4 = 50.5 - 50.5 \cdot \cos 6^\circ 35' 41'' = 50.5 \cdot 0.9934 = \]

\[ = 0.33\ \text{mm}. \]

To determine the thickness of the chip removed by one tooth of the milling cutter, let us take the drawing (Fig. 7), in which \(ebfiae\) represents the chip removed by one tooth of the cutter, \(\frac{D}{2}\) is the radius of the cutter, and \(d\) is the thickness of the layer being removed.

For the calculation it is necessary to assume that, during the time in which the cutter tooth removes the chip \(ebfiae\), the cutter itself moves by the amount \(mm_1\), equal to the feed per one tooth of the cutter, and that the curve \(ebf\) of the outer contour is an arc of radius \(= \frac{D}{2}\); in this case the chip thickness for some position corresponding to the angle \(y\) will be expressed as follows:

\[ X = ac \sin y, \]

where \(ac\) is equal to \(mm_1\), i.e., to the amount of feed of the cutter per one tooth. If the cutter has \(Z\) working teeth, \(n\) is the number of revolutions per 1 min., and \(V\) is the feed speed per second, then

\[ ac = \frac{60}{n}\cdot\frac{V}{Z}, \]

therefore

\[ X = \frac{60}{n}\cdot\frac{V}{Z}\cdot \sin y; \]

but the greatest chip thickness \(\delta\) will occur when \(y = \psi\), i.e.,

\[ \delta = \frac{60}{n}\cdot\frac{V}{Z}\sin \psi; \]

for the thickness of the layer being removed \(d\),

\[ \cos\psi=\frac{\frac{D}{2}-d}{\frac{D}{2}}, \]

and

\[ \sin\psi=\sqrt{\,1-\cos^{2}\psi\,}=\sqrt{\,4\left(\frac{d}{D}-\frac{d^{2}}{D^{2}}\right)}. \]

Now the thickness of the chip removed by one tooth of the cutter for a layer of thickness \(d\) will be expressed as follows:

\[ \delta=\frac{2.60\,V}{n.Z}\cdot \sqrt{\frac{d}{D}-\frac{d^{2}}{D^{2}}}. \]

This formula, for our case, can be considerably simplified if we substitute into it the numerical quantities common to all cases of work with the cutters under observation; thus, instead of \(V\), substitute the feed speed of the cutter per second,

\[ \frac{2.4}{60}=0.04\ \text{mm}; \]

instead of \(n\), the number of revolutions of the cutter per minute, \(=12\); instead of \(Z\), the number of cutter teeth working during one second,

\[ \frac{12}{60}\cdot 22=4.4; \]

instead of \(D\), the outside diameter of the cutter, \(=101\) mm; then the formula for \(\delta\) will take the following form:

\[ \delta=\frac{2.60.0.44}{12.4.4}\sqrt{\frac{d}{101}-\frac{d^{2}}{101^{2}}} =\frac{1}{11}\sqrt{\frac{101.d-d^{2}}{101^{2}}} =\frac{1}{1011}\sqrt{101d-d^{2}} \]

\[ \delta=\frac{1}{1011}\sqrt{d(101-d)}. \]

For a thickness of the layer being removed \(d=1\) mm, as is already known, one tooth of the cutter participates in the work at each given moment. The thickness of the chip removed by one tooth of the cutter is determined by the formula

\[ \delta_{1}=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\cdot \sqrt{1(101-1)} =\frac{10}{1011}, \]

i.e.,

\[ \delta_{1}=0.009891\ \text{mm}. \]

For a thickness of the layer being removed \(d=2\) mm, one tooth of the cutter participates in the work at each given moment. The thickness of the chip removed by one tooth of the cutter is determined by the same formula.

On the Issue of the Sharpening Angle of a Milling Cutter.

\[ \delta_2=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\sqrt{2(101-2)} =\frac{1}{1011}\sqrt{198}, \]

whence

\[ \delta_2=0.01392\ \mathrm{mm}. \]

Further, with the thickness of the layer being removed \(d=3\ \mathrm{mm}\), two teeth of the milling cutter take part in the work at each given moment. The thickness of the chip removed by one tooth of the milling cutter, for this case, is determined from the relation

\[ \delta_3'=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\sqrt{3(101-3)} =\frac{1}{1011}\sqrt{294}. \]

\[ \delta_3'=0.01696\ \mathrm{mm}. \]

The thickness of the chip removed by the other tooth of the milling cutter, for the same case, is determined from the equation

\[ \delta_3''=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\sqrt{0.091(101-0.091)} = \]

\[ =\frac{1}{1011}\sqrt{9.1827} \]

\[ \delta_3''=0.002997\ \mathrm{mm}. \]

With the thickness of the layer being removed \(d=4\ \mathrm{mm}\), as was indicated earlier, two teeth of the milling cutter take part in the work at each moment. The thickness of the chip removed by one tooth of the milling cutter, for the latter case, is determined from the relation

\[ \delta_4'=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\sqrt{4(101-4)} =\frac{1}{1011}\sqrt{388} \]

\[ \delta_4'=0.0195\ \mathrm{mm}. \]

The thickness of the chip removed by the second tooth of the milling cutter is determined from the equation

\[ \delta_4''=\frac{1}{1011}\sqrt{d(101-d)} =\frac{1}{1011}\sqrt{0.33(101-0.33)} =\frac{1}{1011}\sqrt{33.22}. \]

\[ \delta_4''=0.005701\ \mathrm{mm}. \]

Finally, to determine the cutting coefficient there is the general formula:

\[ W = k \delta b \quad \text{or} \quad R = \frac{W}{\delta b}, \]

where \(R\) is the cutting coefficient in kg/mm\(^2\);

\(W\) is the circumferential force on the cutters of the milling cutter working simultaneously. The value for \(W\) is taken from the table given above, according to each case.

\(\delta\) is the chip thickness. When removing a layer of thickness 1 and 2 mm, one cutter tooth takes part in the work at a time; consequently, for these cases, instead of \(\delta\) one must substitute the value of the thickness of the chip removed by one cutter tooth. When removing a layer of thickness 3 and 4 mm, two cutter teeth take part in the work at the same time; consequently, for these cases, instead of \(\delta\) one must substitute the value of the sum of the thicknesses of the chips removed by two teeth.

\(b\) is the chip width—38.5 mm.

First we shall determine the cutting coefficient for the first method of setting the indicator drum into rotation.

For a milling cutter with a cutting angle of the teeth \(=90^\circ\), with the thickness of the layer being removed \(d = 1\) mm, the cutting coefficient \(R\) is determined by the formula

\[ R = \frac{W}{\delta_1 \cdot b} = \frac{71.8}{0.009891 \cdot 38.5} = 188.55 \]

With layer thickness \(d = 2\) mm, the coefficient is

\[ R = \frac{W}{\delta_2 \cdot b} = \frac{82.5}{0.01392 \cdot 38.5} = 153.94 \]

With layer thickness \(d = 3\) mm, the coefficient \(R\) is determined from the equation

\[ R = \frac{W}{(\delta_3' + \delta_3'')\, b} = \frac{126.9}{(0.01696 + 0.002977)\, 38.5} = 165.13 \]

With layer thickness \(d = 4\) mm, the coefficient \(R\) is determined from the relation

\[ R = \frac{W}{(\delta_4' + \delta_4'')\, b} = \frac{173.8}{(0.0195 + 0.005701)\cdot 38.5} = \frac{173.8}{0.025201 \cdot 38.5} = 179.14. \]

For a milling cutter with a cutting angle of the teeth \(=80^\circ\), with layer thickness \(d = 1\) mm, the coefficient \(R\) is determined by the formula

\[ R = \frac{W}{\delta_1 \cdot b} = \frac{39}{0.009891 \cdot 38.5} = \frac{39}{0.3808} = 102.4. \]

For a layer thickness \(d = 2\ \mathrm{mm}\), the coefficient

\[ R=\frac{W}{\delta_2\cdot b} =\frac{56}{0{,}01392\cdot 38{,}5} =\frac{56}{0{,}5359} =104{,}5. \]

For a layer thickness \(d = 3\ \mathrm{mm}\), the coefficient

\[ R=\frac{W}{(\delta_3' + \delta_3'')b} =\frac{83}{(0{,}01696+0{,}002997)\,38{,}5} =\frac{83}{0{,}76846} =108{,}1. \]

For a layer thickness \(d = 4\ \mathrm{mm}\), the coefficient

\[ R=\frac{W}{(\delta_4' + \delta_4'')b} =\frac{101}{(0{,}0195+0{,}005701)\,38{,}5} =\frac{101}{0{,}9702} =104{,}1. \]

Then let us determine the values of the cutting coefficient by the second method of setting the indicator drum in rotation.

For a cutter with cutting-edge angle \(=90^\circ\), at a layer thickness \(d=1\ \mathrm{mm}\), the coefficient \(R\) is determined by the formula

\[ R=\frac{W}{\delta_1\cdot b} =\frac{51{,}6}{0{,}009891\cdot 38{,}5} =\frac{51{,}6}{0{,}3808} =135{,}5. \]

For a layer thickness \(d = 2\ \mathrm{mm}\), the cutting coefficient

\[ R=\frac{W}{\delta_2\cdot b} =\frac{99{,}4}{0{,}01392\cdot 38{,}5} =\frac{99{,}4}{0{,}5359} =185{,}5. \]

For a layer thickness \(d = 3\ \mathrm{mm}\), the cutting coefficient

\[ R=\frac{W}{(\delta_3' + \delta_3'')b} =\frac{138}{(0{,}01696+0{,}002997)\,38{,}5} =\frac{138}{0{,}76846} =179{,}6 \]

For a layer thickness \(d = 4\ \mathrm{mm}\), for the same cutting coefficient we have

\[ R=\frac{W}{(\delta_4' + \delta_4'')b} =\frac{170}{(0{,}0195+0{,}05701)\,38{,}5} =\frac{170}{0{,}9702} =175{,}2. \]

For a cutter with cutting-edge angle \(=80^\circ\), at a layer thickness \(d=1\ \mathrm{mm}\), the coefficient

\[ R=\frac{W}{\delta_1\cdot b} =\frac{31{,}5}{0{,}009891\cdot 38{,}5} =\frac{31{,}5}{0{,}3808} =82{,}72 \]

For a layer thickness \(d=2\ \mathrm{mm}\),

\[ R=\frac{W}{\delta_2\cdot b} =\frac{36{,}5}{0{,}01392\cdot 38{,}5} =\frac{36{,}5}{0{,}5359} =68{,}11. \]

For a layer thickness \(d = 3\) mm.

\[ R=\frac{W}{(\delta_3' + \delta_3'')b} =\frac{57.9}{(0.01696+0.002997)38.5} =\frac{57.9}{0.76846} =75.35. \]

For a layer thickness \(d = 4\) mm.

\[ R=\frac{W}{(\delta_4' + \delta_4'')b} =\frac{81.6}{(0.0195+0.005701)38.5} =\frac{81.6}{0.9702} =84.11. \]

The mean value of the cutting coefficient for a milling cutter with a cutting angle of the cutters \(=90^\circ\), for both directions of rotation of the indicator drum, will be

\[ R=\frac{188.55+153.94+165.13+179.14+135.5+185.5+179.6+175.2}{8}=170.32. \]

The mean value of the cutting coefficient for a milling cutter with a cutting angle of the cutters \(=80^\circ\), for both directions of rotation of the indicator drum, will be

\[ R=\frac{102.4+104.5+108.1+104.1+82.72+68.11+75.35+84.11}{8}=91.17. \]

The results of all calculations of the cutting coefficient may be presented in the following table.

Thus, from the data of the last table it follows quite clearly that, when the cutting angle is reduced from 90° to 80° in an ordinary cylindrical milling cutter, while all other working conditions are kept the same, there is a saving in the expenditure of the work absorbed by it of about 58%. The cutter with an angle of 90° was ordered from the L. Leve works; the one with an angle of 80° was made in the mechanical workshops of the Institute. Assuming that the considerable gain in the expenditure of the energy absorbed by the cutter with a cutting angle of 80°, as compared with the cutter with a cutting angle of 90°, is explained simply by the difference in the manufacture and grinding of the cutter, in the mechanical workshops of the Institute, from the same steel from which the cutter with a cutting angle of 80° had been made, a cutter with a cutting angle of 90° was made, and observations were carried out with it; in this case the cutter worked very little and was undoubtedly very sharp, whereas in the previously mentioned experiments the cutters worked for several days without sharpening, but under identical conditions—with lubrication by soapy water¹. The diagrams were taken for one revolution of the cutter; the working conditions were the same as before for the cutter with an angle of 90°, and on the basis of such diagrams, by the method of calculation given earlier, the mean cutting coefficient for the same material (open-hearth iron) under the same working conditions was determined.

For a thickness of the layer removed of 1 mm — 82.2 kg/mm
 ”  ”  ”  ”  2 ” — 113.2 ”
 ”  ”  ”  ”  3 ” — 101 ”
 ”  ”  ”  ”  4 ” — 103.2 ”

The mean value of the cutting coefficient from the last observations is 99.9 kg/mm. Thus, even in the latter case, when the working conditions of the cutter were very favorable for it—the cutter was very sharp—the mean value of the cutting coefficient, 99.9 kg/mm, is greater than the mean value of the cutting coefficient obtained for the cutter with a cutting angle of 80°—91.17 kg/mm. The change of the cutting angle from 90° to 80° was achieved only by changing the direction of the front face of the tooth; at 80° it did not pass through the center of the cutter, as it did for 90°, but along a tangent to a certain circle of smaller diameter (than that of the cutter), while preserving all the other dimensions of the tooth.

It was not possible to increase the width and thickness of the chip removed, owing to the weakness of the machine.

From all that has been set forth above, it follows quite clearly that the fundamental rule which applies to cutters of ordinary design, and which states that, as the cutting angle is reduced from \(90^\circ\) to the most advantageous limits, the energy absorbed by it also decreases, is fully applicable to cutters of more complex design—milling cutters; but only the practical and economic reasons indicated earlier prevent this principle from being put into practice.

T. I. Tikhonov.

The arrangement of the dynamometer on the universal milling machine.

Fig. 7.

Fig. 1.

Fig. 5.

To the article by P. I. Til:

“On the question of the angle
of milling cutters.”

Fig. 5.

Fig. 2.

For the article by T. I. Tikhonov:

“On the issue of the sharpening angle of a milling cutter.”

1. Observations on the operation of the cutter being tested were made for 3–4 hours a day, and only during this period were the cutters in operation. ↩↩