My experience with motors is that they are usually specified as Ke = RMS voltage line-to-line / 1000RPM. The reason RMS voltage line-to-line is used, is because it is very easy to measure with a multimeter + doesn't need an oscilloscope.
Kv would be proportional to the reciprocal, and I'd expect it to be RPM / RMS line-to-line voltage.
In your case: 2193RPM / 0.738Vrms l-l = 2972RPM / Vrms,l-l, that's 15% higher than expected. There is bound to be some error in measurement (voltage in scopes is usually 1-2% or so off worstcase), and in part-to-part variation of the motor, but it sounds a little different.
You consult with the manufacturer. They may have information on short term overload performance, or performance at <100% duty cycles.
If they don't, or if they won't stand by their motor under specified overload conditions, then you can't rely on the motor.
However in non safety critical applications such as hobbyist or experimental conditions, you can estimate a range of conditions under which it'll probably work.
Loosely, you can use it for short enough bursts that it won't exceed its rated temperature - 1.1Nm at a 50% duty cycle gives the same mean torque load as 0.55Nm continuous operation, and therefore, for short enough bursts, should be safe - leaving the question, what does "short enough" mean?
That's where the thermal time constants come in. They have the same meaning as the time constant (= RC) in an RC network, allowing you to calculate the rate at which the voltage (or temperature here) rises to its final value.
One simple way of using this is to calculate the half-life or time taken to reach half the final value, which is 0.693* the time constant, or 32 seconds for the (46s) winding time constant. After 32 seconds at twice the rated power, it will reach half the final temperature, which should be within the temperature rating.
Of course it needs to cool before repeating the operation, or subsequent 32 second bursts may exceed the rated temperature.
Modeling that properly would require simulation, including heat transfer to the case (whose temperature rises more slowly, with a much longer time constant) and cooling terms according to airflow past the windings (if it's not a sealed motor) or over the case if it is. You can use R-C networks and an electrical simulator like the built-in one to approximate the thermal model.
Or experiment on a motor.
But (without having done the simulation), if your application allows running the motor less than 50% of the time, in bursts less than about 15 seconds with cooling periods of 30 seconds, I think it'll probably work.
Best Answer
Please keep in mind how each are defined
\$K_e\$ is defined as the PEAK line voltage per mechanical rotor velocity with a fundamental equation \$K_e = \frac{V_{pk,ll}}{\omega_m}\$
\$K_t\$ is defined as the PEAK torque per phase current with a fundamental equation \$K_t = \frac{T}{A}\$
The units of \$K_e = \frac{V \cdot s}{rad} = V\cdot s\$ (since radians are unitless
The units of \$K_t = \frac{N\cdot m}{A} = \frac{J}{\frac{C}{s}} = \frac{J\cdot s}{C} = V\cdot s \$
\$K_e\$ and \$K_t\$ have exactly the same units and in the ideal case ( no mechanical drag, no magnetic saturation) they are comparable. I say comparable not equal because there is the \$\sqrt{3}\$ factor due to one being line-line and the other being phase.
In practice... \$K_t\$ is defined at rated current and as such there is magnetic saturation resulting in \$ K_t < \sqrt(3) \cdot K_e \$ (how "less than" depends on saturation)
From your comment
\$ K_t < \sqrt(3) \cdot K_e \$
0.55 < \$\sqrt{3} \cdot\$ 0.318
0.55 < 0.5508
Thus your electrical machine datasheet is in alignment