Motor and required torque.

So I need a motor that is able to provide [itex]\begin{equation}0.412Nm\end{equation}[/itex] of continuous torque (to overcome rolling resistance).

A BLDC I have is rated at [itex]\begin{equation}285W\end{equation}[/itex] and [itex]4700Kv (rpm/V)[/itex].

So I was wondering if my calculations are correct.

Using the relationship between Torque and Power and angular speed
[itex]
\begin{equation}
\begin{split}
\omega &\approx 700rads^{-1} \approx 6700rpm\\
T &= 0.412Nm \\
P &= 285W
\end{split}
\end{equation}
[/itex]

From this get the voltage from the motor specs
[itex]
\begin{equation}
\begin{split}
V &= \frac{6700}{4700}\\
&= 1.42V
\end{split}
\end{equation}
[/itex]

So now using the fact that
[itex]
\begin{equation}
\begin{split}
K_t &= \frac{1}{K_v}\\
T &= K_t i
\end{split}
\end{equation}
[/itex]

Therefore

[itex]
\begin{equation}
\begin{split}
i &= K_v \times T\\
&= 4700 \times 0.412\\
&= 1936A
\end{split}
\end{equation}
[/itex]

This result makes no sense because that is a tonne of current.

Assuming 100% efficiency (P = IV)
[itex]
\begin{equation}
\begin{split}
285 &= i \times V\\
i &= 196A
\end{split}
\end{equation}
[/itex]

Which again seems really really high.

Am I missing something somewhere? Will the motor not draw that amount of current at the given loads/speed? I am confused as I think the equations are correct....?