And now, with the 42 cm tire within the 85 cm buoy, will the increase be even greater?
A windier day. Anemometer alone, average 6 m/s, peaking at 7.3 m/s, anemometer within 42 cm tire, average 7.2 m/s, peaking at 8.8 m/s (always 20% faster, at several wind speeds).
Then the whole in the buoy was not better, peaking at 8.3 m/s.
That said tests were performed again because (compared to the anemometer in the same 42 cm tire but with not taut enough wires):
Previous measurements were underestimated due to lack of tension in the wires holding the anemometer. This is the reason why other tests were carried out with a higher tension of all the wires holding the anemometer which was able to position itself facing the wind in all circumstances. The measured increase of efficiency was significant, being approximately 25 percent. Measurements in a windy day gave the following values at three different wind speeds: an anemometer alone and another anemometer in the 42 cm tire (as previously) in the same time: respectively 6 m/s and 7.5 m/s; 7.4 m/s and 10 m/s (unusual increase); 10.2 and 12.6 m/s. In all tests, the proportion of increase does not appear to depend on wind speed.
Anemometer in the tire well positioned with taut wires
A more compact option was added in these experiments.
Experiment with an anemometer inside the 85 cm diameter buoy (photo below). This was far less efficient than with the 42 cm diameter tire (see above). Wind speed inside the buoy was slightly higher than that in free air, approximately 5 to 10 percent more, instead of 25 percent increase with the tire. Perhaps the relatively greater thickness of the buoy further slows down the entry of air.
Today other measures were performed with the efficient 42 cm diameter tire.
Then other measures were performed by reversing the two anemometers, the one inside the tire on a tube, and the other one in free air. Said tube was moved through three positions, so that the anemometer rotor was placed in the center, then between the center and the edge, then at the edge, as shown in the next three photos. Wind speed in free air was 4 to 5 m/s. As previously, the anemometer inside the tire indicated a value 25 percent higher, so respectively 5 m/s for 4 m/s in free air, and 6.25 m/s for 5 m/s in free air,regardless of its position at the center or at the edge or between the two.The effectiveness of the 42 cm diameter tire as a shroud was therefore confirmed in the different experimental configurations.
(PDF) Flexible kite carrying a turbine within a torus-shaped balloon. Available from: https://www.researchgate.net/publication/388800040_Flexible_kite_carrying_a_turbine_within_a_torus-shaped_balloon [accessed Mar 02 2025].
Anemometer rotor in the center of the 42 cm tire
Anemometer rotor between the center and the edge of the 42 cm tire
Anemometer rotor on the edge of the 42 cm tire
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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 neighborsâ 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 (ânormalâ 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.
Another experiments with a thinner inner tube.
Anemometer in the 36 cm diameter inner tube
Anemometer inside the 36 cm diameter inner tube
It was used an inner tube with a thinner but irregular section of 4.5 cm on average instead of 9 cm for the 42 cm diameter tire. The tests with two anemometers including one in the free air showed higher values ââinside the said inner tube. These values were slightly higher than that of the 85 cm buoy, but lesser than that of the 42 cm tire. These values ââwere slightly higher than those of the 85 cm buoy, but lower than those of the 42 cm tire, and tended to come closer together as the wind speed increased, unlike the 42 cm tire: thus for a wind speed of 3 m/s, we had 3.7 m/s in the 36 cm inner tube, and for a wind speed of 4.5 m/s, we had 5.3 m/s, while a wind peak of 6.5 m/s corresponded to a peak of 7 m/s in the 36 cm inner tube.
Hi Pierre: You are very thorough in your research. I am surprised you get so much enhancement in speed from such a narrow inner-tube profile compared to its diameter! All very interesting. 
Hi Doug, at low wind speed (3 m/s vs 3.7 m/s), but not above (6.5 m/s vs only 7 m/s). The 42 cm tire is better, but the fatter buoy is worse.