My main concern was not the lattice tower but the large amount of used material for the given swept area. In the photo it can be seen that the diameter D is about half the height H. So the swept area is 1/2 H^2. A HAWT with a radius R = H has a swept area pi * H^2 so more than a factor 6 higher.
The centrifugal force in the blades of a H-Darrieus rotor works perpendicular to the blade and it works always to the outside of the rotor. It gives a large bending moment in the blade root for a rotor with only one spoke assembly. The thrust force workes to the inside of the rotor if the blade is at the front side but to the outside if the blade is at the back side. So this is a fatigue load which has to be added to the load of the centrifugal force. So the total bending moment is maximal for the back blade. As there is only one spoke assembly half way the rotor height, the blade must be very strong where it is connected to the end of a spoke.
The centrifugal force in the blades of a HAWT works in the direction of the blade if the blade is turning in the rotor plane. So then it gives no bending moment in the blade root. The thrust force works perpendicular to the blade and it gives a bending moment in the blade root. However, if the blade bends backwards due to this thrust moment, the centrifugal force gives a forwards moment which partly compensates the moment of the thrust.
The design tip speed ratio lambda of a HAWT can be chosen in between 1 and about 10. Most big 3-bladed rotors have a lambda of about 8 resulting in a low solidity. The lambda of a H-Darrieus rotor has only one optimal value for a certain airfoil. In report KD 601, I have found that this lambda is about 4.2 for the NACA 0015 airfoil. If the lambda is chosen smaller than 4.2, the angle of attack alpha at the front and back blade becomes that large that the airfoil stalls. If the lambda is chosen larger than 4.2, alpha become smaller than the angle for which the Cd/Cl ratio is minimal. Because of the rather low lambda, a H-Darrieus rotor needs a rather large solidity and so the blades are rather heavy. Apart from the blades, one also needs a spoke assembly which connects the blades to the hub. To minimise drag losses of these spokes, they must have an aerodynamic shape. The weigh of a spokes has to be added to the weight of the blades. I think that therefore the weight of this H-Darrieus rotor will be about the same as that of a HAWT with a radius R = H.
In the photo it can be seen that the blades can turn around the blade axis. I expect that there is a mechanism in the spokes with which the three blades are connected to each other. Blade pitch control can have two good reasons for a H-Darrieus rotor. One is to make the rotor self starting and one is to limit the rotational speed at high wind speeds. To make the rotor self starting, the blade angle beta must be positive for the front blade and negative for the back blade. This requires a cyclic movement of the blade. To limit the rotational speed at high wind speeds by increasing the drag on the airfoil by stalling, beta must be negative at the front blade and positive at the back blade. So this also requires cyclic movement of the blade. Technically this is possible but it is rather complicated as it must also take the wind direction into account. Instead of a mechanical coupling in between the three blades, one also can use an individual motor which steeres each blade but then the motors have to be synchronised with each other and with the wind direction. So the pitch control mechanism of this H-Darrieus rotor will be more complicated and so more expensive than that of a normal big HAWT.
