$$T_m-J\frac{d²\Theta}{dt²}+F\frac{d\Theta}{dt}-T_L=0$$
$$T_m=k_Ti$$
Where \$T_m\$ is motor torque, \$T_L\$ is the load torque, \$J\$ is the moment of inertia, \$F\$ is the friction coefficient, \$\Theta\$ is the rotor angle. \$k_t\$ is torque constant.
There is no max. current vs. speed. It's just a nominal current which may not be exceeded if you look at the current as mean value. The peaks can be very high with respect to the nominal current, but they may not persist for long time - \$I^2t\$ limit.
If we ommit the transients of acceleraton/decelration and friction, then the formula becomes: \$T_m=T_L\$ or \$i=\dfrac{T_L}{k_T}\$
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
The maximum current in a permanent magnet motor is governed by two things, which are in order of importance ...
a) demagnetisation of the permanent magnets
b) heating
Demagnetisation will kill your motor in an instant, if the current goes over this limit at all, it's game over. Fortunately, it takes a rather large current to do it. Unless the manufacturer specifies it directly, it can be rather difficult / impossible to infer from the other specifications. You can assume it's above the peak rated current, but by what safety factor, 2x, 5x, 10x ? If it's not in the data sheet, one way is to contact the manufacturer, another is to test a motor, with the expectation of destroying it. After demagnetisation, the Kt will be lower, zero or even reversed, a quick measurement of output torque will detect this.
Heating is a bit more tractable, but still needs some guesswork. The limitation is the temperature of the windings, as the insulation is good only up to a certain temperature. Fortunately it's easy to estimate the temperature of the windings by measuring their resistance, copper has a relatively large 10% change in resistance for every 25C increase in temperature. Unfortunately, the specs probably do not tell you what the limiting temperature on the windings is. Best is to play safe and assume (say) 80C for the max temperature, \$t_{max}\$
The method is to measure the approximate \$I^2t_{max}\$ of the windings. Measure the starting temperature \$t_{initial}\$. Apply a 10 second pulse (short enough to be roughly adiabatic, long enough to be useful) of rated current, and estimate the temperature rise from the change in resistance. The \$I^2t\$ you have just applied, \$I^2t_{test}\$, is 10*Irated*Irated. The maximum \$I^2t\$ you can apply safely, \$I^2t_{max}\$, is $$I^2t_{max}=\frac{t_{max}-t_{initial}}{ temperature rise}I^2t_{test}$$
The maximum you can apply on thermal grounds is any current I, as long as the application time is less than $$\frac{I^2t_{max}}{I^2}$$ in a single pulse, starting with a cold motor. If you apply multiple pulses, then you must use less current, to keep the temperature down.