Six-speed machine tool gearbox design

(a) Progression ratio and speeds

For $z$ speeds in GP: $n_{max}/n_{min} = \varphi^{\,z-1}$.

$$\varphi = \left(\frac{400}{125}\right)^{1/5} = 3.2^{0.2} = 1.262 \approx 1.26 \ (\text{R10 series})$$

Spindle speeds ($\times 1.26$ each step): 125, 160, 200, 250, 315, 400 rpm.

(b) Structure and ray diagram

Six speeds $= 3 \times 2$. Structural formula: 3(1) x 2(3), where the number in brackets is the characteristic $x$ (the step is $\varphi^{x}$).

• Group 1 (3 speeds, $x=1$): ratios $1,\ 1/\varphi,\ 1/\varphi^{2} = 1,\ 0.794,\ 0.630$.
• Group 2 (2 speeds, $x=3$): ratios $1,\ 1/\varphi^{3} = 1,\ 0.5$.

Check of range: group 1 $\varphi^{2}=1.6$, group 2 $\varphi^{3}=2$, both below the limit of 8, and no gear ratio is smaller than 1/4.

Belt drive: motor 1440 rpm to shaft I at 400 rpm, ratio $1440/400 = 3.6$.

 Shaft I     Shaft II      Shaft III (rpm)
  400 ----> 400 ---------> 400  (x1)
      \               \---> 200  (x0.5)
       \--> 315 --------> 315
        \            \---> 160
         \-> 250 -------> 250
                      \---> 125

Shaft II runs at 400, 315, 250 rpm. Each speed of shaft II gives two speeds on shaft III: the same speed (ratio 1) or three steps lower (ratio 0.5), giving 400, 315, 250 and 200, 160, 125 rpm.

(c) Gear tooth numbers

Stage 1 (sum 70): ratio $= z_1/z_2$.

Stage 2 (sum 60):

All pinions have more than 17 teeth (no undercutting for 20 degree teeth).

Check of actual spindle speeds: 400, 317.9, 251.2, 200.0, 159.0, 125.6 rpm. Errors against the standard speeds are below 1%, within the permitted $\pm 10(\varphi-1)\% = 2.6\%$.

Answer: $\varphi = 1.26$; speeds 125, 160, 200, 250, 315, 400 rpm; structure $3(1)\times2(3)$; teeth 35/35, 31/39, 27/43 and 30/30, 20/40.