Analyses of a new four-facet drill

Abstract

In this paper, a new four-facet drill was presented and analyzed. For evaluating the performance of the new and conventional drills, the geometrical analyses of the rake faces and flank surfaces as well as the prediction of cutting forces and torques were conducted. The results showed that the normal rake and clearance angles along the cutting lips of the new drill are always positive and a positive constant, respectively. In contrast, those of the conventional one are negative at certain points on cutting lips close to the drill axis and variable, respectively. Therefore, the cutting forces and torques of the new drill in drilling can be reduced as compared with the conventional one. The prediction of cutting forces and torques in drilling the workpiece made of plain-carbon steel (AISI1045) showed that the reductions of the total thrust force and torque along cutting lips for the new drill over the conventional drill are up to 50.12 and 26.54 %, respectively, and hence, the tool life of the new drill is expected to be longer than the conventional one’s.

Appendix

Assuming the rake face is represented by the polar coordinates (r, φ) and its cross-section at plane z = 0 is the bold cure as shown in Fig. 16. The point P of the rake face before grinding becomes the point Q. From Fig. 16, it is apparent that:

Fig. 16

Rake face

$$ \varphi =\beta +\psi $$

(A1)

The rake face is rotated by angle β from point P to point Q. The elevation is changed by t ^′ ′. The angle β is proportional to z and will equal 2π when t ^′ ′ = L, hence

$$ \beta =\frac{2\pi}{L}{t}^{\prime \prime } $$

(A2)

$$ {t}^{\prime \prime }={t}^{\prime } \cot \rho $$

(A3)

$$ {t}^{\prime }= r\cdot c\mathrm{os}\psi $$

(A4)

From Eqs. A1 to A4, we obtain the following:

$$ \begin{array}{l}\psi =\varphi +\frac{2\pi}{L} r \cos \psi \cdot \cot \rho, or\hfill \\ {} H\left( r,\varphi \right)=\varphi -\psi +\frac{2\pi}{L} r \cos \psi \cdot \cot \rho =0\hfill \end{array} $$

(A5)

Equation A5 is the equation of the rake face cross-section at plane z = 0. If z ≠ 0, the rake face is rotated by angle $\frac{2\pi}{L} z$. Hence, the cross-section of the rake face for z ≠ 0 is as follows:

$$ \psi =\varphi +\frac{2\pi}{L} r \cos \psi \cdot \cot \rho +\frac{2\pi}{L} z $$

(A6)

$$ \psi ={ \tan}^{-1}\left(\frac{y}{x}\right),\ r=\sqrt{x^2+{y}^2} $$

(A7)

From Eqs. A6 and A7, we obtain the following:

$$ {F}_5\left( x, y, z\right)= z-\frac{L}{2\pi}\left({ \tan}^{-1}\left(\frac{y}{x}\right)-\psi \right)+\sqrt{x^2+{y}^2} \cos \psi \cot \rho =0 $$

(A8)