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- Thread starter pechan
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- #2

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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?

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These are the calculations I came up with. I think I am missing something as this seems like a large number at 150k lbs.

Your Thoughts?

- #4

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The formula I found is

Fe being the required effort.

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

make that my load?

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

- #5

Baluncore

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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.

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 lets say the retracted length is (s + k) and the extended length is (2*s + k), with k determined by the standard used.

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.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.

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 lets say the retracted length is (s + k) and the extended length is (2*s + k), with k determined by the standard used.

Last edited:

- #6

Baluncore

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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.

- #7

Bystander

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

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