Almost the size, and also almost the weight which is only 640 kg.
With my experiences on tires and torus, I was seeing if we could do something simpler by carrying a turbine shrouded by the torus and suspended with a kite or a Flettner balloon.
As wind turbines quickly become too heavy, we should limit ourselves to small wind turbines of 5 or 10 kW maximum, perhaps of type rope-drive Kiwee @Kitewinder , for individual use, in bumper car mode allowing collisions without consequences between neighbors’ AWES.
If, as the tests have shown, the 42 cm diameter tire makes it possible to increase the wind speed by a quarter, this would almost double the power if we do not take into account the negative parameters: the power would have to be measured with a turbine and not just the increase in wind speed.
Cons: the torus shape adds a lot of drag, especially because the outer half torus area adds a large area leading to useless drag.
I evaluated the total drag with the torus, as between 1.5 and 2 times the drag of a non-shrouded wind turbine sweeping a doubled area.
Indeed, if the power is almost doubled, the same goes for the drag or thrust for the turbine (“normal” thrust coefficient Cd of 0.9 before being doubled when the turbine is augmented). We must add the drag by the torus itself, and whose coefficient Cd would be 0.8, as shown on Figure 4:
So a drag calculation was sketched. I am not sure if it is correct.
0.42 m diameter tire; diameter throat = 0.24 m. Length Lt = 0.09 m (= thickness of the tire). Lt/D throat = 0.375. D useful shroud = 0.33 m. Lt/D shroud = 0.272. Full torus area = 0.093274 m²; outer half torus area = 0.053 m²; inner half torus area = 0.08548 m² - 0.0452 m² = 0.04 m²; diameter 0.33 m disc including inner half torus area = 0.08548 m²; inside 0.24 m diameter D throat area = 0.0452 m²; 0.08548/0.0452 = 1.89, so 1.9; complete area of 0.42 m diameter disc = 0.138474 m², so about 3 times diameter throat area; outer half torus (useless drag) area = 0.053 m² with 0.8 Cd with 90° angle of attack 0.42 m diameter tire; diameter throat = 0.24 m. Length Lt = 0.09 m (= thickness of the tire). Lt/D throat = 0.375. D useful shroud = 0.33 m. Lt/D shroud = 0.272. Full torus area = 0.093274 m²; outer half torus area = 0.053 m²; inner half torus area = 0.08548 m² - 0.0452 m² = 0.04 m²; diameter 0.33 m disc including inner half torus area = 0.08548 m²; inside 0.24 m diameter D throat area = 0.0452 m²; 0.08548/0.0452 = 1.89, so 1.9; complete area of 0.42 m diameter disc = 0.138474 m², so about 3 times diameter throat area; outer half torus (useless drag) area = 0.053 m² with 0.8 Cd with 90° angle of attack according to the Figure 4, so slightly higher than diameter throat area, and slightly lower (0.0424) by taking account of the 0.8 Cd. Power is 1.9 times the turbine area, while drag (thrust) is 0.9 Cd x 1.9 x 00.45 = 0.07695 for the turbine, plus 0.8 Cd x 0.093274 m² (full torus area) = 0.0746192, so 0.1515692. A turbine sweeping 1.9 times more would have a drag of 0.07695, so almost 2 times less for the same power. But perhaps it is not correct to count both augmented drag of the turbine and drag of the inner half torus area (0.04 m² x 0.8 = 0.032), which leads to 0.1195692, so only 1.5538 times more drag. The useless part of drag of the outer torus area represents 0.053 m² x 0.8 Cd = 0.0424.