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imsmooth Senior Member Joined: 07/02/2008 Location: United StatesPosts: 214
Posted: 10:42pm 14 Mar 2008
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The rotor creates a thrust, that when applied to the offset from the axis creates a moment. When this thrust exceeds the moment of the tail due to gravity the windmill will start to furl. I have no problems with this part.
This all assumes that the tail is fixed parallel with the wind. Nowhere did I see any consideration to the area of the tail. For an extreme case where the tail has zero surface area the tail will move with any rotor thrust because nothing is keeping the tail fixed in position. The rotor thrust does not have to exceed the weight of the tail because the wind can't keep it in position.
So, should the area of the tail at least equal the area of the blades? Does anyone have any answers or thoughts?
GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127
Posted: 12:43am 15 Mar 2008
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Hi imsmooth,
The tail area is not as critical as the blade. On my mill, the blade area is approx 0.3m^2. The tail is made from 5ply with and area 0.25m^2. The tail boom is 1.2m long. The boom components were made strong and as light as possible. I found my mill furled too easily. I just added some weight to the end of the boom to load the mill more to get the furling at my selected windspeed. My mill starts to furl at 7.5m/s windspeed and is fully furled at 11m/s.
I have seen mills with a blade area of 0.8m^2 on a yaw box with the same geometry as mine track well with a longer tail boom and 0.25m^2 fin area.
At some point with an individual design, fine tuning with regard to fin area and boom length and mass may be required to give good tracking and high wind protection.
cheers, Gordon. become more energy aware
imsmooth Senior Member Joined: 07/02/2008 Location: United StatesPosts: 214
Posted: 03:53am 15 Mar 2008
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Thanks Gordon, but it still doesn't answer how one determines the critical area needed. If the tail had not surface area the rotor would just rotate around the pole as it pushed the tail through the wind. The tail's surface area keeps it in-line with the wind. As the rotor turns the tail stays parallel with the wind and moves up the slanted axis. The vertical weight vector eventually balances the rotor force. Should the tail be at least equal to the surface area of the blades? Has anyone thought about this?
GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127
Posted: 04:50am 15 Mar 2008
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I was trying to give you an indication of the tail area not being too critical. The weight and boom length are more critical.
If you were to give some indication of rotor dia then this might help others to share some advice. Try a couple of square feet as a triangle attached to the end of the boom.
You could try and be technical by calculating the thrust developed by a certain fin area at a certain windspeed, at various angles to the wind. As long as the thrust was greater than the thrust from the blades at the boom length, it should keep the boom aligned to the wind at that windspeed.
I would try a couple of square feet first. If too small, just add some more. If it is too big it won't matter too much. The area does not need to be the same as the blades.
cheers, Gordon.become more energy aware
Gill Senior Member Joined: 11/11/2006 Location: AustraliaPosts: 669
Posted: 07:26am 15 Mar 2008
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G'day imsmooth,
I think I know what you are asking.
I have done some consideration in this area.
Firstly, The tails wind seeking ability and furling are two separate issues. And you are correct, furling will not occur satisfactorily, if at all, if the tails wind seeking ability is not correct first off.
So of these two issues, tail size and length is the first consideration and with prop offset for furling comes tail angle and offset. Note we have not started considering the needs of furling yet where the additional factors of tail weight and pivot angles come into play.
I have played with formula to calculate all these parameters but have continually run aground with the initial wind forces on the blades. With DIY blades, one cannot work on swept area, nor number of blades, or blade area, or profile, or blade angle, or angle of attack, or blade efficiency as all these factors come into play and I have yet to see a formula for this one figure that can be applied to PVC, wood or glass blades that give any workable figure to commence tail calculations with. By all means research it yourself and I for one eagerly await and will appreciate a generalised formula that makes sense and works. Then it's worthwhile to discuss tail calculations.
Until then we use known workable tail proportions and do OK.
Sorry, am in a rush to go out but that's the nitty gritty of it for me.was working fine... til the smoke got out.
Cheers Gill _Cairns, FNQ
imsmooth Senior Member Joined: 07/02/2008 Location: United StatesPosts: 214
Posted: 03:20pm 15 Mar 2008
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I'm going to take a crack at this:
P = Power/sec = F*distance/sec = F*V, or P/V = F (thrust force)
Also, P per second = (0.5)*M*V^2, where the mass of air/second = M = r*A*V, where r = 1.225 kg/m^3
Therefore, F = P/V = 0.5 * r*A*V * V^2 / V
F = 0.5*(1.225)*A*V^2
However, the wind initially is parallel to the tail so there is no component striking it. If the tail moves 1 degree the horizontal wind component is V*sin(1). Once the tail moves about 5 degrees the component is V*sin(5)
The rotor thrust, which we must balance, is D^2 * V1^2 /24. As the rotor moves out of the wind, the thrust vector is cos(x). Rotorthrust = K*D^2 * V1^2 * cos(x)/24. K represents the percentage of the Betz limit. The standard rotor thrust equations assumes the ideal turbine with 59% power extraction. If our rotor has a 30% efficiency then K = 0.5 (30/59).
Thus, 0.61*A * V^2 * sin(x) = K*cos(x)*D^2 * V1^2 / 24. The velocity drops out giving:
A = K*cos(x)*D^2 / 14.7*sin(x)
For an 8 foot rotor, 30% efficency, D = 2.64m and when the tail is 15 degrees into the wind we need a tail area of about 0.7m^2 to not get pushed by the rotor thrust. This is not furling. It is simply the area needed so the tail tracks the wind. THe tail will stay 15 degrees into the wind and as the rotor thrust increases beyond this point, the rotor moves on the center axis and the tail *now* begins to furl.
I have to comment also that this equation is assuming ideal rotor thrust at the Betz limit. Rotors develop less thrust so the actual area will be less.
How does this math sound?Edited by imsmooth 2008-03-17
Gill Senior Member Joined: 11/11/2006 Location: AustraliaPosts: 669
Posted: 07:30pm 15 Mar 2008
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By all means play with those calculations from Fieldlines.
Unfortunately I believe you will find the tail size will be grossly oversized, the same as our furling calculations on this site give a furling tail weight that is too heavy.
As I said above, the problem is with the calculation of prop thrust. The calculation commonly used is one for wind loading when constructing the mast and guy wires. Here the prop thrust is taken using prop swept area as if it were a solid disc. This is fine for designing a mast but the over calculation is sh*t for our tail design needs.
Swept area is too large and blade surface area is too small. The actual figure needed to calculate thrust lies in between these two. You can recognise the architects thrust calcs by the fact that the number of blades(solidity) is not included.
A detailed calc would take into account the factors I mentioned previously. Due to the variety of props we make, only a fool would come up with an accurate formula and it would be so complex none of us non aerodynamic engineer types could understand it. However I think a rule of thumb relationship could be worked out to give an approximation.
With a workable prop thrust the other calcs follow somewhat easier.
I wish you every success.was working fine... til the smoke got out.
Cheers Gill _Cairns, FNQ
imsmooth Senior Member Joined: 07/02/2008 Location: United StatesPosts: 214
Posted: 01:36am 16 Mar 2008
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As a more practical point, I have a question as I am getting close to designing my tail. I originally was going to weld a piece of flat steel to a rod, but now I think I'm going to use flat plywood that is painted. Any thoughts? I am concerned that the elements will eventually destroy the wood, even if painted. On the other hand, I am afraid that if the steel sheet comes loose I will have a nasty object falling from 30-40 feet.
KiwiJohn Guru Joined: 01/12/2005 Location: New ZealandPosts: 691
Posted: 02:16am 16 Mar 2008
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Regarding ply versus steel I would go for ply everytime if it is going to be left out in the weather, especially if you are near the sea. In those conditions exterior grade ply will outlast thin steel everytime!
Gill Senior Member Joined: 11/11/2006 Location: AustraliaPosts: 669
Posted: 10:25am 16 Mar 2008
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As another alternative try aluminium.
I acquired some large pieces of aluminium after the last big cyclone hit our region. The sheets used for official roadside signage are of a particularly hard and rigid alloy. I don't know if the same stuff is available in the US or where where best to (legally) source some from but it works well.
was working fine... til the smoke got out.
Cheers Gill _Cairns, FNQ
Ben Newbie Joined: 10/08/2007 Location: AustraliaPosts: 18
Posted: 09:41am 17 Mar 2008
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I had an original tail from coringated iron which did not furl as it was to heavy so I cut it down to reduce weight ,then it was to small and it was hunting in strong winds.My latest tail is 1200mm long on a 3 blade timber fan 2400mm diam and is made from laserlite roofing with two light steel braces to keep its shape it is working very well so far.
Notice. New forum software under development. It's going to miss a few functions and look a bit ugly for a while, but I'm working on it full time now as the old forum was too unstable. Couple days, all good. If you notice any issues, please contact me.
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