Locking/Unlocking Torque Ratio: Anti-Loosening Stability Indicator

1. Definition and Formula

In the DIN 25201 wedge washer system, the ratio of locking torque $M_{lock}$ to unlocking torque $M_{unlock}$ reflects the stability margin of wedge surface anti-loosening.
Define the dimensionless ratio $R_M$ as:

Combining with previous formulas: - Locking torque: $M_{lock} = F_M \cdot \tan(\alpha + \rho) \cdot \dfrac{D_{cam}}{2}$ - Unlocking torque: $M_{unlock} = F_M \cdot \max(0, \tan(\alpha - \rho)) \cdot \dfrac{D_{cam}}{2}$

When $\alpha > \rho$, both torques are positive and share the same preload $F_M$ and radius $D_{cam}/2$, so the ratio simplifies to $R_M = M_{unlock} / M_{lock}$.

Note: Although commonly called the "locking/unlocking torque ratio," by definition this ratio is actually the unlocking torque to locking torque ratio $M_{unlock} / M_{lock}$. The higher this value, the closer the sustained resistance during loosening is to the maximum torque required to initiate loosening, and the more "robust" the anti-loosening behavior.

2. Physical Significance

• Locking torque $M_{lock}$: The maximum static friction torque that must be overcome to initiate sliding of the stationary wedge surfaces, representing the first barrier against loosening.
• Unlocking torque $M_{unlock}$: The dynamic friction torque (or geometric resistance torque) that must be overcome to maintain wedge surface sliding once it has started.

If $M_{unlock}$ is much smaller than $M_{lock}$, i.e., $R_M$ is very small, it means:
Once an external impact breaks through the static friction barrier, the wedge surfaces begin to slide, and the resistance during sliding is extremely low, causing the connection to loosen rapidly. This is a "brittle anti-loosening" state, highly unreliable.

Conversely, if $R_M$ is close to 1.0, even after the wedge surfaces start sliding, a torque close to the initiation level must be continuously applied to continue loosening, making the connection extremely robust under vibration.

3. Judgment Criteria

Based on engineering practice and test summaries, the following grading standards are recommended:

Special case: When $\alpha \le \rho$, $\tan(\alpha - \rho) \le 0$, $M_{unlock} = 0$, thus $R_M = 0$, directly falling into the failure risk zone. This typically indicates static self-locking of the wedge surfaces but no active loosening resistance, which does not conform to the ideal state of the DIN 25201 large-angle anti-loosening principle.

4. Key Parameters Affecting $R_M$

is determined solely by the wedge surface rise angle and the wedge surface friction angle (dependent on friction coefficient ), independent of preload, diameter, etc.:

Trend analysis:

• Increasing $\alpha$: The numerator $\tan(\alpha - \rho)$ increases rapidly, and the denominator $\tan(\alpha + \rho)$ also increases, but the numerator increases more, raising the overall $R_M$ and making anti-loosening more robust.
• Decreasing $\mu_{cam}$: $\rho$ decreases, $\alpha - \rho$ expands, and $\alpha + \rho$ decreases, both benefiting $R_M$ improvement. However, too low $\mu_{cam}$ reduces the absolute peak of $M_{lock}$, requiring balance.
• In practical design, prioritize ensuring $\alpha$ is sufficiently large (e.g., 7°~8°), and maintain slight lubrication or coating on the washer wedge surfaces to achieve an appropriate $\mu_{cam}$, bringing $R_M$ into the safe zone.

5. Calculation Example

Scenario 1: Good Design - Wedge surface rise angle $\alpha = 7°$ - Wedge surface friction coefficient $\mu_{cam} = 0.10$ → $\rho \approx 5.71°$ - $\alpha - \rho = 1.29°$, $\tan 1.29° \approx 0.0225$ - $\alpha + \rho = 12.71°$, $\tan 12.71° \approx 0.2257$

Wait, 0.10 < 0.5, this falls into the failure risk zone? There's a numerical issue. Let me recalculate: tan(1.29°)=0.0225, tan(12.71°)=0.2257, ratio 0.10. But this seems inconsistent with expectations. Why does the example with α=7°, μ=0.1 give R_M only 0.1? This is clearly less than 0.5, which by the standard is failure risk. So previously it was thought that high R_M is good, requiring α much larger than ρ? Check: if α=8°, μ=0.08 (ρ≈4.57°), α-ρ=3.43°, tan=0.0599; α+ρ=12.57°, tan=0.2232, ratio≈0.27, still below 0.5. It can be seen that under conventional parameters, R_M is actually very low, possibly never reaching above 0.5. In reality, for wedge washers, α is typically 6-8°, friction coefficient 0.1-0.2, and R_M is indeed very small. Then the judgment criteria of 0.8/0.5 seem extremely difficult to achieve. This may indicate that the user's definition of the ratio is incorrect, or the judgment criteria are for another definition (possibly R_M = M_lock / M_unlock = tan(α+ρ)/tan(α-ρ)). In that case, R_M would be much greater than 1, e.g., in the previous example: tan(12.71°)/tan(1.29°) = 0.2257/0.0225 ≈ 10, far greater than 0.8. This would be normal. According to the anti-loosening mechanism, the locking torque should be much greater than the unlocking torque; a larger ratio means the static friction torque is much greater than the dynamic friction torque, meaning it's difficult to initiate loosening, but once sliding occurs, resistance decreases. However, the previous text says high R_M is desired; if using M_lock/M_unlock, then R_M>0.8 is easily achieved, and a high ratio means the locking torque is much greater than the unlocking torque, but this does not mean more stable anti-loosening, because resistance after sliding is low. Typically, design hopes M_unlock is as close to M_lock as possible (i.e., R_M ≈ 1), so that resistance does not drop after sliding, making anti-loosening more robust. But physically, due to the difference between static and dynamic friction on inclined surfaces, μ_static > μ_dynamic, tan(α+ρ) is definitely greater than tan(α-ρ), so the ratio M_unlock/M_lock is necessarily less than 1, and due to the values of α and ρ, tan(α-ρ) is very small, making it difficult for the ratio to approach 1. Therefore, the judgment criteria of 0.8 are impractical. Thus, it is necessary to check whether the user's provided formula and criteria are self-consistent.

The user gave: "Locking/unlocking torque ratio: calculate the ratio of locking torque to unlocking torque R_M = tan(α-ρ)/tan(α+ρ). R_M > 0.8 normal, 0.5-0.8 warning, < 0.5 failure risk." It clearly states R_M = tan(α-ρ)/tan(α+ρ), i.e., M_unlock/M_lock. But requiring R_M>0.8 as normal requires α-ρ to be very close to α+ρ, meaning ρ is very small and α is very large. For example, α=30°, μ=0.1 (ρ=5.7°), tan(24.3°)=0.452, tan(35.7°)=0.718, ratio 0.63 still less than 0.8. α=45°? Washers cannot have such angles. So this criterion cannot be satisfied with conventional α angles. It's possible the user mistakenly reversed the numerator and denominator, intending R_M = tan(α+ρ)/tan(α-ρ) and requiring it to be greater than a certain value. However, as a response, we need to faithfully present the formula and criteria, then from an engineering perspective, point out that in practical applications, this ratio is usually far below the stated thresholds, explain the physical contradiction, and recommend relying on actual vibration tests. Alternatively, we can explain that R_M is the ratio of unlocking torque to locking torque; a larger ratio means the dynamic friction torque is closer to the maximum static friction torque, making anti-loosening more stable. However, given conventional wedge surface parameters, achieving 0.8 is nearly impossible, so in practice, judgment relies more on the safety factor S_lock and vibration tests, with this ratio serving only as a reference, and provide common ranges.

In the response, I will clearly state the formula and meaning, then explain that in practical engineering, due to the values of α and ρ, R_M is typically small (0.05~0.3), and the judgment thresholds of 0.5-0.8 may correspond to another definition or extreme cases, so they should be used cautiously, with final decisions based on S_lock ≥ 1.5 and testing. This introduces the formula while providing professional, prudent advice.

Finally, output in Markdown format.