# Calculating hydraulic cylinder size wind turbine tower

• pechan
In summary, a standard low cost hydraulic cylinder will have a retracted length, (between the eyes), of it's stroke plus about 12.25 inches. The extended length must then be 2 * stroke plus 12.25”. So let's say the retracted length is (s + k) and the extended length is (2*s + k), with k determined by the standard used.When I separate the top tower element into a 10 ft length centred on 65 ft, weighing 500 lbs, and a generator at 70 ft weighing 500 lbs, I get moments 1 through 5 of …m1; 10 * 2500 = 25,000.m2; 30 * 1700 = 51,000.m3;f

#### pechan

Hello, I could use some help determining what size hydraulic cylinder I need to be able to tilt my turbine tower up. What I have is a 70 ft tower where the first 20 feet weigh 2500 lbs, the 2nd 20 weigh 1700 lbs and the 3rd 20 ft weigh 1000 lbs and the last 10 feet weigh 1000 lbs. For a total weight of 6200 lbs. The Tower is hinged at the 2 ft point and the cylinder base will be attached 1 ft away from the base of tower and attached to the tower at the 6.5 ft point. How would I determine the force required to get this vertical?

Would I need to split the tower into 70 weights for each foot and multiply these by their distance from the cylinder attachment point?
ie. For the first 20 feet 2500/20 =125 lbs, so I run 20 calculations starting with 1 ft X 125 = 125 ... 20 ft X 125 lbs = 2500 lbs.
Adding up all calculations and that would give me the force required to lift my tower? These are the calculations I came up with. I think I am missing something as this seems like a large number at 150k lbs.

Obviously this is a class 3 lever.
The formula I found is Fe = Fl dl / de
Fe being the required effort.

Fl being the load.(tower weighs 8000 lbs)

dl being the distance from the load to the point of effort (cylinder attachment point on tower)

de being the distance from the fulcrum (tower hinge) to the point of effort ( cylinder attachment point on tower) ( this is 4.5 feet)

Since this is a distributed load can I just find the center of gravity of the tower and

So if the center of gravity ends up being at 35 ft could I solve this with the following equation?

Fe = 8000 * 35 / 4.5
Fe = 280k /4.5
Fe=62,222 lbs

Thanks

You will need to model the tower as 5 separate centres of mass. One for each of the four tower sections and one for the generator assembly. The torque needed to lift or lower the tower will be the sum of those five components.
The Tower is hinged at the 2 ft point and the cylinder base will be attached 1 ft away from the base of tower and attached to the tower at the 6.5 ft point.
The forces on the tower near the cylinder attachment point and the fulcrum will be extreme. The 2 ft hinge and 1 ft offset give a base of less than 2.25 ft, working against the 6.5 ft point.It would be better to move the cylinder base away from the tower base so that the cylinder starts in a vertical position and ends diagonally when the tower is up. The torque needed to stand the tower will be a maximum when the tower is horizontal.

A standard low cost cylinder will have a retracted length, (between the eyes), of it's stroke plus about 12.25 inches. The extended length must then be 2 * stroke plus 12.25”. So let's say the retracted length is (s + k) and the extended length is (2*s + k), with k determined by the standard used.

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When I separate the top tower element into a 10 ft length centred on 65 ft, weighing 500 lbs, and a generator at 70 ft weighing 500 lbs, I get moments 1 through 5 of …
m1; 10 * 2500 = 25,000.
m2; 30 * 1700 = 51,000.
m3; 50 * 1000 = 50,000.
m4; 65 * 500 = 32,500.
m5; 70 * 500 = 35,000
Total moment = 193,500. foot pounds.

For a cylinder with stroke s and a constant k, the height of the tower hinge point will be y = s + k. The distance of the cylinder base from the tower base will be x. That is found by solving a right angle triangle. The base of the triangle is x, while the height is y + x. The triangle to solve has a hypotenuse of z = 2 * s + k, which is the extended length of the cylinder. (x is one root of the quadratic equation).

For a ram with a stroke of s = 4 ft, and with k = 1.02 ft, I get x = 3.353 ft.
The cylinder force in pounds will then need to be (193,500. / 3.353) = 57,710. lbs, (about 25.75 tons).

Typical agricultural cylinders will safely handle 2000 psi. So the piston area will need to be about (57,710. / 2000.) = 28.86 square inches. That is a radius of 3.03” or a bore of about 6 inches diameter. Using a shorter stroke will increase the cylinder internal diameter requirement. That will significantly increase the cost. Using a longer cylinder will require a higher hinge point with a more slender cylinder rod. The cylinder eye pins must be lubricated and quite free to turn or the cylinder rod will bend under load and the tower will crash.

Summary:
The cylinder will have a bore of 6", a stroke of 4 ft, a retracted length of 5.02 ft and extended length of 9.02 ft.
The tower hinge will be 5.02 ft above the ground.
The cylinder base pin will be 3.353 ft from the base of the tower.
The cylinder rod pin will attach 3.353 ft along the tower from the hinge.

I think you should consider a lighter lattice structure tower.

How often were you planning on raising and lowering your tower?