It looks like Heisenberg's uncertainty principle surprisingly predicts the same values for maximum velocity for elementary particles as I got here earlier from different approach and actually different formula. I assume minimum uncertainty in position is the Planck length, the rest comes straight out from the Heisenberg's uncertainty principle
\begin{eqnarray}
\sigma_x \sigma_p \geq \hbar \nonumber \\
l_p\sigma_p \geq \hbar \nonumber \\
l_p\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\geq \hbar \nonumber \\
l_p\frac{\frac{\hbar}{\bar{\lambda}}\frac{1}{c}v}{\sqrt{1-\frac{v^2}{c^2}}}\geq \hbar \nonumber \\
\frac{\frac{1}{c}v}{\sqrt{1-\frac{v^2}{c^2}}}\geq \frac{\bar{\lambda}}{l_p} \nonumber \\
\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\geq \frac{\bar{\lambda}}{l_p}c \nonumber \\
\frac{v^2}{1-\frac{v^2}{c^2}}\geq \frac{\bar{\lambda}^2}{l_p^2}c^2 \nonumber \\
v^2\leq \frac{\bar{\lambda}^2}{l_p^2}c^2 \left(1-\frac{v^2}{c^2}\right) \nonumber \\
v^2\left(1+\frac{\bar{\lambda}^2}{l_p^2}\right)\leq \frac{\bar{\lambda}^2}{l_p^2}c^2\nonumber \\
v^2\leq \frac{\frac{\bar{\lambda}^2}{l_p^2}c^2}{\left(1+\frac{\bar{\lambda}^2}{l_p^2}\right)}\nonumber \\
v\leq \frac{c}{\sqrt{1+\frac{l_p^2}{\bar{\lambda}^2}}}
\end{eqnarray}
Taylor expansion shows that this for any practical significance (for hypothetical high energy physics experiments) gives the same value for any observed subatomic particle as my max velocity formula, not so strange as they both are limits on the velocity. For the Planck mass particle the formula above cannot be used, as the momentum of the Planck mass particle likely always is \(m_pc\) that leads to another and exactly the same prediction as my max velocity formula, early working paper (my max velocity formula now in 3 published papers, soon in 4 and 5)
Early working paper, comments and in particular harsh critics welcome