The formula to calculate the Efficiency is:
\[ \eta = \frac{W \cdot \tan(\psi) \cdot d}{W_{\text{load}} \cdot \tan(\psi + \Phi) \cdot d + \mu_{\text{collar}} \cdot W_{\text{load}} \cdot R_{\text{collar}}} \]
The Efficiency of an electric motor is defined as the ratio of usable shaft power to electric input power.
Let's assume the following values:
Using the formula:
\[ \eta = \frac{60 \cdot \tan(0.4363323129985) \cdot 0.06}{53 \cdot \tan(0.4363323129985 + 0.21816615649925) \cdot 0.06 + 0.16 \cdot 53 \cdot 0.02} = 0.643256958272496 \]
| Weight (W) | Helix Angle (ψ) | Mean Diameter of Screw (d) | Load (Wload) | Limiting Angle of Friction (Φ) | Coefficient of Friction for Collar (μcollar) | Mean Radius of Collar (Rcollar) | Efficiency (η) |
|---|---|---|---|---|---|---|---|
| 55 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.589652211750 |
| 56 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.600373161054 |
| 57 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.611094110359 |
| 58 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.621815059663 |
| 59 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.632536008968 |
| 60 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.643256958272 |
| 61 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.653977907577 |
| 62 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.664698856882 |
| 63 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.675419806186 |
| 64 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.686140755491 |
| 65 kg | 0.4363323129985 rad | 0.06 m | 53 N | 0.21816615649925 rad | 0.16 | 0.02 m | 0.696861704795 |
