This "air glider" works as a hovercraft, using air pressure to lift itself and its load.
From a commercial model we get the following specifications:
- 8-1/2 in. x 36 in. (22 cm x 91 cm) dual pads (0.4 m2 total area).
- 750 pounds (~3500 N) of lifting capacity.
- 1.75 HP blower.
The required (relative) pressure to lift 3500 N of weight using a 0.4 m2 platform is
$p_{\rm cushion} = \frac{3500\,\mathrm{N}}{0.4\,\mathrm{m^2}} \approx 9000\,\mathrm{Pa}$
We can take the air inside the "cushion" as stagnated to calculate the velocity of the exit flow. Then the dynamic pressure at the exit of the gap must be about 9000 Pa, giving an air velocity of
$\frac{1}{2}\rho_{\rm air}\,v_{\rm gap}^2 = 9000\,\mathrm{Pa}$
$\frac{1}{2}\cdot 1.205\,\mathrm{kg\cdot m^{-3}}\,\cdot v_{\rm gap}^2 = 9000\,\mathrm{Pa}$
$v_{\rm gap} \approx 135\,\mathrm{m\cdot s^{-1}}$
If we assume the gap is about 0.2 mm (just a guess), the flow rate will be about
$Q = 4\cdot(0.22\,\mathrm{m} + 0.91\,\mathrm{m})\cdot 0.0002\,\mathrm{m}\cdot 135\,\mathrm{m\cdot s^{-1}} \approx 0.12\,\mathrm{m^3\cdot s^{-1}} $
and the required power
$P = Q\cdot p_{\rm cushion} \approx 0.12\,\mathrm{m^3\cdot s^{-1}}\cdot 9000\,\mathrm{Pa} \approx 1100\,\mathrm{W}$,
below the specified 1.75 HP (1305 W).