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Forum Index : Windmills : Need advise on a tower footing
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Gizmo Admin Group  Joined: 05/06/2004 Location: AustraliaPosts: 5119 |
Hi All. I want to build a new tower for a large Lenz2 design windmill. The tower will be a free standing latice type, with a drive shaft from the windmill at top to the alternator at ground level. I had planned to dig and pour some concrete footings, but thats not easy in the rocky ground here. Then I had a brain wave. I have a dozen concrete railway sleepers. These are bloody heavy, two people cant move one at all, I have to use an chain block just to drag one across the ground, and even then tractor at the other end of the chain was been dragged more than the sleeper. I was thinking of using these as the footings for my tower, the big advantage is I can move the tower if needed. Now the help I need. As I figure it, I have a lever, with one sleeper as the fulcrum, another sleeper as the weight, and the windmill as the force. 
The tower height ( A ) will be 8 meters tall. The windmill has a frontal area of 5 meters square. The weight of the sleepers is no known, but I can find out. The length of the base ( B ) is what I need to work out. So I need some maths. A formula to work out the length of B ( Once I know the sleeper weight ) for a given max wind speed, so the tower wont fall over. Or if I have a length for B, what wind speed would it take to push the tower over? This is simple stuff, and the answer could be found with a little research in old school books or the internet, but I thought it would be a fun quiz for you guys, and helpfull for anyone building a similar tower. ( And my brain is too tired to work it out ) Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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windstuffnow Newbie  Joined: 30/06/2006 Location: United StatesPosts: 31 |
I never learned meters and Kg so its kind of awkward for me to convert the numbers... I know it awkward the other way around as well... Basically it a simple lever formula once you find the pressure against the turbine. Force on the turbine would be .00492 x area x windspeed^2 = force area is in square feet windspeed in mph force is in pounds Then its a matter of force x height / base width. To come out right you would also need to consider the area of the tower that is exposed to the wind as well and use 1/2 the height for that calculation. Can't wait to see it !!! WindstuffEd |
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| RossW Guru  Joined: 25/02/2006 Location: AustraliaPosts: 495 |
If I remember correctly, the fact that the tower is vertical and the base is horizontal is largely immaterial. You can "straighten" it out like a standard lever+pivot. It then becomes ratiometric. Do you REALLY mean the windmill is 5 metres, square? (Ie, 25 square metres of area?). I'll assume it's bad phrasing and you mean 5 square metres :) Converting to the next contributors formula, 5m^2 = 54 square feet. Lets assume a wind gust of 100KMH (62MPH), Force on the turbine would be .00492 x 54 x 62^2 = 1021 lbs = 463Kg force. Round it off to 500Kg :) If your mast is 8m high, and the base was 8m wide and it pivoted on the corner, you would have that 500Kg acting at the "other" side. 8m high and 4m wide, you'd need 1000Kg to hold the bugger down. 8m high and only 2m wide (more like the proportion you've drawn) you'd need 2 tonnes to hold that puppy down! (Or have I made a fatal flaw in my assumptions somewhere?) |
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| Gizmo Admin Group Joined: 05/06/2004 Location: AustraliaPosts: 5119 |
Yeah your right Ross, its 5m^2 area ( 2.4 meters high and just over 2 meters wide ). The actual wind loading would be less, Ed might know the exact figures. The example above also doesn't take into account the other two sleepers, out to the sides. This would also contribute to the weight, and it also depends on which direction the wind is comming from, its actually a lot more complicated. I found out the weight of concrete is 2.4 tons per meter^3, so I'll measure up the sleepers this afternoon to work out their volume, and therefore the weight. This is fun, loved geometry as a kid. The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| windstuffnow Newbie Joined: 30/06/2006 Location: United StatesPosts: 31 |
With a 10ft square base ( 3.48m) you would need 2600 lbs per side or 1327 lbs ( about 600kg ) per leg to hold it in place and of course a structure that would hold it. A 55 gallon drum filled with dirt would weigh 1052 lbs (447 kg ). It sounds like those concrete structures would most likely work fine, one on each leg. Dirt weighs about 125 lbs per cu ft, concrete weighs about 150lbs per cubic ft so you could measure it and come up with a reasonable guess as to what they weigh. Fun stuff... |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Will the other weights have a varying contribution? Assume the base is square, and has the 4 faces aligned with the compass so we have a N, E, S and W side. If the wind is comming from due north, the tower will want to "tip over" to the south, thus the weight on the N side has to "hold it down". Assume your concrete sleepers weigh 1000Kg and are exactly the same length as the base, and lets assume the base is 2m a side. You have 1000Kg at the "opposite" side to the pivot, plus a further 2000Kg but it's working linearly from "nothing" to 100% as you go from the pivot end (south) to the northern end. Together, it'd be the average of the total weight over half the length, either side, or another 1000Kg effective weight. Same if the wind came square from any side (e, s, w). If the wind comes from a "worst-case", exactly between two (NE say), then the turning moment would make the tower want to fall SW. It's going to be "tipping" on the SW corner, not along an edge. In this case, it's trying to lift the weights (all 4) but they're all along the base which is now the square root of the sum of the squares of two sides (ie, the diagonal, not just a 2m side). So we've got 4000Kg acting uniformly over a 2.8m base, or the equivalent of 1400Kg holding it down. (Not quite as bad as initially thought). I thought it was 2600Kg/m^3, but then that's probably wet! (I needed to work out how much the trucks were carrying when I built this place, and how much the roof was going to weigh! There was 133 m^3 of concrete in the roof slab alone, and 154 in the floor!) According to my sources, the standard size of a railway sleeper in Qld is 7ft x 9 inches x 4.5 inches. or about 2130 x 230 x 110 = 0.054m^3 = 130Kg Working back from all that, neglecting the weight of the tower itself (which you should consider, as it'll act pretty much uniformly down until its centre of gravity falls outside the area of its base), if you had a 2m square base and 8m high tower, you should only need about 50Kg of sideways force at the top of the tower to topple it. Hope you've got a few more of those sleepers, Giz! |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Hmm... just checked. The "Low-profile" concrete sleepers are the same size as the timber ones, but the "full profile" are about twice the size (but curved, so not sure their exact mass). Lets give it the benefit of the doubt, and roughly double my calculations above. I still feel it's a little light, but that'll depend a *LOT* on your anticipated winds! (Christmas, we got 107KMH gusts through here, but that was exceptional, we rarely get more than about 60) |
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| Gizmo Admin Group Joined: 05/06/2004 Location: AustraliaPosts: 5119 |
Ok they measure 2100 long by 230 wide by 260 high. Actually the bottom is 280 wide, top is 180 wide, works out as 230 wide overall for our calculations. Gives us a volume of 0.12558 meters, at 2.4 tons per meter, total weight is approx 300 kg each. So class, what does the length of my base need to be to survive 100kmh winds?  The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Well, we worked out 500 KgF with 5m^2 area and 100KMH, although that neglected the area of the tower itself. If your sleepers are 300Kg ea, neglecting any weight of the tower and any wind-load presented by it, and assuming the wind from the worst-case vector, you have 500Kg force at the top, we need 4000 Kg/m to hold it. You could stack a dozen of your sleepers on a 1m base, or by calculation you could use one of your 300Kg sleepers along each side and make the base 5.6m square. Are you sure it's 2.4x2 ? And is 100KMH reasonable up your way? |
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| Gizmo Admin Group Joined: 05/06/2004 Location: AustraliaPosts: 5119 |
Yeah the windmill will be based on a Lenz2 design, each wing will use one sheet of 2400 * 1200 ( 8 * 4 ) aluminium. Works out as a windmill 2.4 meters high, and just a tad over 2 meter diameter. I hope I dont get winds that fast, working on a worse case situation like a big storm. Remember the verticals dont furl out of high winds. In reality the wind area is a lot less than 5m^2 as lot of wind goes though the windmill. I actually have 8 sleepers I can use, 2 at each corner, so a 3 meter base will work, and not take up too much space. The sleepers have cast iron eyelets poking out where the rail line is attached, I can use these to bolt the tower down. The weight of the tower will help, at least untill it tilts past 10 or so degrees, in which case I would be hiding under the kitchen table. Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| Curbie Newbie Joined: 25/09/2009 Location: Posts: 1 |
I know this is an old post, I followed to here from a discussion on fieldlines, what does the .00492 represent in this formula??? Thanks, Curbie |
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GWatPE Senior Member  Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Hi Gizmo, all my windspeed data shows wind gusts at close to double the average windspeeds. A drag type machine presents the same area, as there is no furling mechanism with changing windspeed. Use a 4-10 safety factor to allow for those unexpected natural wind events. Probably need more sleepers, or bigger tower footprint. Gordon. become more energy aware |
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AMUN-RA Senior Member Joined: 10/03/2007 Location: AustraliaPosts: 144 |
Cyclone Larry, March 2006. After wreaking havoc in Innisfail, Larry has now weakened into a low pressure system and is not expected to be any further threat. At its peak, the cyclone registered winds of up to 290km/h and was graded a category five, the highest possible grading for a cyclone. just something to consider in your calcs remember your only down the road living on an exposed hill from your photos Mick Every day the sun shines & gravity sucks= free energy. |
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