Can we motivate mathematically why wind turbines almost always have 3 flappers and aeroplane propellers can...












3












$begingroup$


Firstly I know some might frown upon a question so very broad and applied as this one. It really may not be a well defined mathematical question as some people would prefer on the site. I am okay with moving this to physics or some other more suitable SE if enough people want to do that.





Supposedly both air propellers as well as wind turbines want to utilize the displacement of air in order to either turn it into motion or energy.



Is there some mathematical justification for the choice in difference of number of blades of an aircraft or a wind turbine?



How could we set up some kind of calculation for this? Which equations would be useful?





Own work What I suspect would be very useful is to use differential equations to model how flow of air effects acceleration of flappers (and other way around) and how this affects generation of electricity.










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$endgroup$








  • 3




    $begingroup$
    By "flappers", do you mean "blades"?
    $endgroup$
    – John Hughes
    Jan 16 at 16:42










  • $begingroup$
    @JohnHughes Ah yes! God, I knew there was a better word for it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:43






  • 1




    $begingroup$
    Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
    $endgroup$
    – Hagen von Eitzen
    Jan 16 at 16:46












  • $begingroup$
    @HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
    $endgroup$
    – mathreadler
    Jan 16 at 16:47






  • 1




    $begingroup$
    See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
    $endgroup$
    – cgiovanardi
    Jan 17 at 0:52
















3












$begingroup$


Firstly I know some might frown upon a question so very broad and applied as this one. It really may not be a well defined mathematical question as some people would prefer on the site. I am okay with moving this to physics or some other more suitable SE if enough people want to do that.





Supposedly both air propellers as well as wind turbines want to utilize the displacement of air in order to either turn it into motion or energy.



Is there some mathematical justification for the choice in difference of number of blades of an aircraft or a wind turbine?



How could we set up some kind of calculation for this? Which equations would be useful?





Own work What I suspect would be very useful is to use differential equations to model how flow of air effects acceleration of flappers (and other way around) and how this affects generation of electricity.










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    By "flappers", do you mean "blades"?
    $endgroup$
    – John Hughes
    Jan 16 at 16:42










  • $begingroup$
    @JohnHughes Ah yes! God, I knew there was a better word for it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:43






  • 1




    $begingroup$
    Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
    $endgroup$
    – Hagen von Eitzen
    Jan 16 at 16:46












  • $begingroup$
    @HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
    $endgroup$
    – mathreadler
    Jan 16 at 16:47






  • 1




    $begingroup$
    See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
    $endgroup$
    – cgiovanardi
    Jan 17 at 0:52














3












3








3


1



$begingroup$


Firstly I know some might frown upon a question so very broad and applied as this one. It really may not be a well defined mathematical question as some people would prefer on the site. I am okay with moving this to physics or some other more suitable SE if enough people want to do that.





Supposedly both air propellers as well as wind turbines want to utilize the displacement of air in order to either turn it into motion or energy.



Is there some mathematical justification for the choice in difference of number of blades of an aircraft or a wind turbine?



How could we set up some kind of calculation for this? Which equations would be useful?





Own work What I suspect would be very useful is to use differential equations to model how flow of air effects acceleration of flappers (and other way around) and how this affects generation of electricity.










share|cite|improve this question









$endgroup$




Firstly I know some might frown upon a question so very broad and applied as this one. It really may not be a well defined mathematical question as some people would prefer on the site. I am okay with moving this to physics or some other more suitable SE if enough people want to do that.





Supposedly both air propellers as well as wind turbines want to utilize the displacement of air in order to either turn it into motion or energy.



Is there some mathematical justification for the choice in difference of number of blades of an aircraft or a wind turbine?



How could we set up some kind of calculation for this? Which equations would be useful?





Own work What I suspect would be very useful is to use differential equations to model how flow of air effects acceleration of flappers (and other way around) and how this affects generation of electricity.







pde physics applications electromagnetism






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 16:38









mathreadlermathreadler

15.4k72263




15.4k72263








  • 3




    $begingroup$
    By "flappers", do you mean "blades"?
    $endgroup$
    – John Hughes
    Jan 16 at 16:42










  • $begingroup$
    @JohnHughes Ah yes! God, I knew there was a better word for it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:43






  • 1




    $begingroup$
    Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
    $endgroup$
    – Hagen von Eitzen
    Jan 16 at 16:46












  • $begingroup$
    @HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
    $endgroup$
    – mathreadler
    Jan 16 at 16:47






  • 1




    $begingroup$
    See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
    $endgroup$
    – cgiovanardi
    Jan 17 at 0:52














  • 3




    $begingroup$
    By "flappers", do you mean "blades"?
    $endgroup$
    – John Hughes
    Jan 16 at 16:42










  • $begingroup$
    @JohnHughes Ah yes! God, I knew there was a better word for it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:43






  • 1




    $begingroup$
    Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
    $endgroup$
    – Hagen von Eitzen
    Jan 16 at 16:46












  • $begingroup$
    @HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
    $endgroup$
    – mathreadler
    Jan 16 at 16:47






  • 1




    $begingroup$
    See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
    $endgroup$
    – cgiovanardi
    Jan 17 at 0:52








3




3




$begingroup$
By "flappers", do you mean "blades"?
$endgroup$
– John Hughes
Jan 16 at 16:42




$begingroup$
By "flappers", do you mean "blades"?
$endgroup$
– John Hughes
Jan 16 at 16:42












$begingroup$
@JohnHughes Ah yes! God, I knew there was a better word for it.
$endgroup$
– mathreadler
Jan 16 at 16:43




$begingroup$
@JohnHughes Ah yes! God, I knew there was a better word for it.
$endgroup$
– mathreadler
Jan 16 at 16:43




1




1




$begingroup$
Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
$endgroup$
– Hagen von Eitzen
Jan 16 at 16:46






$begingroup$
Note that there are wind turbines with only one(!) blade, but probably no such propellers. Also, there are those wind-mills with very high number of blades that we know from wild west movies (for pumping water in dry land)
$endgroup$
– Hagen von Eitzen
Jan 16 at 16:46














$begingroup$
@HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
$endgroup$
– mathreadler
Jan 16 at 16:47




$begingroup$
@HagenvonEitzen okay, I only based my premise on all commercial wind turbines I have ever seen, which are rather big, white and three-bladed, like at least 99% of the ones which I have seen.
$endgroup$
– mathreadler
Jan 16 at 16:47




1




1




$begingroup$
See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
$endgroup$
– cgiovanardi
Jan 17 at 0:52




$begingroup$
See physics.stackexchange.com/questions/11194/… and citeseerx.ist.psu.edu/viewdoc/…
$endgroup$
– cgiovanardi
Jan 17 at 0:52










2 Answers
2






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oldest

votes


















2












$begingroup$

Let's take for granted that for wind turbines, fluid dynamics and economics dictate that we want as few blades as possible, for efficiency. Let's also take for granted that for any rotating apparatus, mechanics dictates that we want the center of mass and the moment of inertia to be invariant under rotations, for stability. The "why" of these dictates probably belongs on other sites.



For the mathematics: We want to place the blades at the roots of unity, $z^n-1=0$. The center of mass is the sum of the solutions, which is the coefficient of $z^1$, which is $0$ if and only if $n>1$.



The moment of inertia requirement means that we also want a sum of the form $sum_kcos^2(kalpha+theta)$ to be constant. Using the double angle formula, this is equivalent to something like $sum_kcos(2kalpha+theta)=0$. If $n$ is odd, this reduces to the same constraint in the previous paragraph, so we just need $n>1$, which means $ngeq 3$. If $n$ is even, this reduces to the previous constraint on $n/2$, so we need $n/2>1$, which means $ngeq 4$. Combining the even and odd cases, we have a constant moment of inertia iff $ngeq 3$.



So for both wind turbines and aircraft propellers, we need $ngeq 3$.



For wind turbines in particular, the least $n$ that solves $ngeq 3$ is, of course, $n=3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
    $endgroup$
    – mathreadler
    Jan 17 at 9:45










  • $begingroup$
    I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
    $endgroup$
    – John Hughes
    Jan 17 at 12:15












  • $begingroup$
    (1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
    $endgroup$
    – Chris Culter
    Jan 17 at 18:35



















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$begingroup$

I conjecture that this is not a mathematical problem at all, but rather an engineering and legal one.



You want to extract lots of energy from the wind with your turbine. If you put tons of blades close together, each will interfere with the airflow over the next one, so there's clearly an upper limit. The assembly needs to be balanced to avoid huge non-axial torques, so you need more than 1 blade (although Hagen suggests that in some cases this may not be true...or maybe his "one blade" is a diameter rather than a radius, i.e., what I'd call a 2-blade setup).



The towers (often) have maximum height restrictions, and you can only fit so many of them in a given space, so cramming in more blades makes sense, hence 3-blade rather than 2-blade. But why not 4 or 6 or 12 blade ones?



I'm guessing that each additional blade gives a bump in output, but perhaps not in proportion to the number of blades: the 5th blade increases the output less than the 4th one did, and so on. And blades are expensive. And more blades requires a more substantial axle, gearbox, etc., not to mention a stronger tower. So there's a point where the additional cost for the $n$th blade exceeds the value of the incremental power generated.



And looking at common windmills suggests that this point is at about $n = 3$. :)






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$endgroup$













  • $begingroup$
    I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:57










  • $begingroup$
    I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
    $endgroup$
    – John Hughes
    Jan 16 at 17:26












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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let's take for granted that for wind turbines, fluid dynamics and economics dictate that we want as few blades as possible, for efficiency. Let's also take for granted that for any rotating apparatus, mechanics dictates that we want the center of mass and the moment of inertia to be invariant under rotations, for stability. The "why" of these dictates probably belongs on other sites.



For the mathematics: We want to place the blades at the roots of unity, $z^n-1=0$. The center of mass is the sum of the solutions, which is the coefficient of $z^1$, which is $0$ if and only if $n>1$.



The moment of inertia requirement means that we also want a sum of the form $sum_kcos^2(kalpha+theta)$ to be constant. Using the double angle formula, this is equivalent to something like $sum_kcos(2kalpha+theta)=0$. If $n$ is odd, this reduces to the same constraint in the previous paragraph, so we just need $n>1$, which means $ngeq 3$. If $n$ is even, this reduces to the previous constraint on $n/2$, so we need $n/2>1$, which means $ngeq 4$. Combining the even and odd cases, we have a constant moment of inertia iff $ngeq 3$.



So for both wind turbines and aircraft propellers, we need $ngeq 3$.



For wind turbines in particular, the least $n$ that solves $ngeq 3$ is, of course, $n=3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
    $endgroup$
    – mathreadler
    Jan 17 at 9:45










  • $begingroup$
    I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
    $endgroup$
    – John Hughes
    Jan 17 at 12:15












  • $begingroup$
    (1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
    $endgroup$
    – Chris Culter
    Jan 17 at 18:35
















2












$begingroup$

Let's take for granted that for wind turbines, fluid dynamics and economics dictate that we want as few blades as possible, for efficiency. Let's also take for granted that for any rotating apparatus, mechanics dictates that we want the center of mass and the moment of inertia to be invariant under rotations, for stability. The "why" of these dictates probably belongs on other sites.



For the mathematics: We want to place the blades at the roots of unity, $z^n-1=0$. The center of mass is the sum of the solutions, which is the coefficient of $z^1$, which is $0$ if and only if $n>1$.



The moment of inertia requirement means that we also want a sum of the form $sum_kcos^2(kalpha+theta)$ to be constant. Using the double angle formula, this is equivalent to something like $sum_kcos(2kalpha+theta)=0$. If $n$ is odd, this reduces to the same constraint in the previous paragraph, so we just need $n>1$, which means $ngeq 3$. If $n$ is even, this reduces to the previous constraint on $n/2$, so we need $n/2>1$, which means $ngeq 4$. Combining the even and odd cases, we have a constant moment of inertia iff $ngeq 3$.



So for both wind turbines and aircraft propellers, we need $ngeq 3$.



For wind turbines in particular, the least $n$ that solves $ngeq 3$ is, of course, $n=3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
    $endgroup$
    – mathreadler
    Jan 17 at 9:45










  • $begingroup$
    I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
    $endgroup$
    – John Hughes
    Jan 17 at 12:15












  • $begingroup$
    (1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
    $endgroup$
    – Chris Culter
    Jan 17 at 18:35














2












2








2





$begingroup$

Let's take for granted that for wind turbines, fluid dynamics and economics dictate that we want as few blades as possible, for efficiency. Let's also take for granted that for any rotating apparatus, mechanics dictates that we want the center of mass and the moment of inertia to be invariant under rotations, for stability. The "why" of these dictates probably belongs on other sites.



For the mathematics: We want to place the blades at the roots of unity, $z^n-1=0$. The center of mass is the sum of the solutions, which is the coefficient of $z^1$, which is $0$ if and only if $n>1$.



The moment of inertia requirement means that we also want a sum of the form $sum_kcos^2(kalpha+theta)$ to be constant. Using the double angle formula, this is equivalent to something like $sum_kcos(2kalpha+theta)=0$. If $n$ is odd, this reduces to the same constraint in the previous paragraph, so we just need $n>1$, which means $ngeq 3$. If $n$ is even, this reduces to the previous constraint on $n/2$, so we need $n/2>1$, which means $ngeq 4$. Combining the even and odd cases, we have a constant moment of inertia iff $ngeq 3$.



So for both wind turbines and aircraft propellers, we need $ngeq 3$.



For wind turbines in particular, the least $n$ that solves $ngeq 3$ is, of course, $n=3$.






share|cite|improve this answer









$endgroup$



Let's take for granted that for wind turbines, fluid dynamics and economics dictate that we want as few blades as possible, for efficiency. Let's also take for granted that for any rotating apparatus, mechanics dictates that we want the center of mass and the moment of inertia to be invariant under rotations, for stability. The "why" of these dictates probably belongs on other sites.



For the mathematics: We want to place the blades at the roots of unity, $z^n-1=0$. The center of mass is the sum of the solutions, which is the coefficient of $z^1$, which is $0$ if and only if $n>1$.



The moment of inertia requirement means that we also want a sum of the form $sum_kcos^2(kalpha+theta)$ to be constant. Using the double angle formula, this is equivalent to something like $sum_kcos(2kalpha+theta)=0$. If $n$ is odd, this reduces to the same constraint in the previous paragraph, so we just need $n>1$, which means $ngeq 3$. If $n$ is even, this reduces to the previous constraint on $n/2$, so we need $n/2>1$, which means $ngeq 4$. Combining the even and odd cases, we have a constant moment of inertia iff $ngeq 3$.



So for both wind turbines and aircraft propellers, we need $ngeq 3$.



For wind turbines in particular, the least $n$ that solves $ngeq 3$ is, of course, $n=3$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 19:46









Chris CulterChris Culter

21.5k43888




21.5k43888












  • $begingroup$
    Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
    $endgroup$
    – mathreadler
    Jan 17 at 9:45










  • $begingroup$
    I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
    $endgroup$
    – John Hughes
    Jan 17 at 12:15












  • $begingroup$
    (1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
    $endgroup$
    – Chris Culter
    Jan 17 at 18:35


















  • $begingroup$
    Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
    $endgroup$
    – mathreadler
    Jan 17 at 9:45










  • $begingroup$
    I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
    $endgroup$
    – John Hughes
    Jan 17 at 12:15












  • $begingroup$
    (1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
    $endgroup$
    – Chris Culter
    Jan 17 at 18:35
















$begingroup$
Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
$endgroup$
– mathreadler
Jan 17 at 9:45




$begingroup$
Indeed very good points about inertia and stability of construction. Maybe cost of construction and maintenance comes into play pressing n downwards to 3.
$endgroup$
– mathreadler
Jan 17 at 9:45












$begingroup$
I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
$endgroup$
– John Hughes
Jan 17 at 12:15






$begingroup$
I'm puzzled by the moment of inertia stuff. Do you mean invariant under ALL rotations, or just rotations in the plane of the blades? Whichever it is, when the sum of squared cosines is constant, differentiating leads to a sum of terms of the form $2 sin cos$, which ends up being a sum of sines: $sum_k sin(2kalpha + 2theta) = 0$. Ignoring all of that, though, the sum of squared cosines confuses me: the moment of inertia of a (thin) rod in the $xy$-plane, about the rod's center, is $frac{1}{12}ml^2$ about the $z$-axis. That's independent of $theta$. I must be missing something.
$endgroup$
– John Hughes
Jan 17 at 12:15














$begingroup$
(1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
$endgroup$
– Chris Culter
Jan 17 at 18:35




$begingroup$
(1) Invariant under rotations in the plane of the blades. (2) Sure, we can just re-parameterize $theta$ until the expression looks the way we want. (3) We need to worry about the other components of the moment of inertia.
$endgroup$
– Chris Culter
Jan 17 at 18:35











2












$begingroup$

I conjecture that this is not a mathematical problem at all, but rather an engineering and legal one.



You want to extract lots of energy from the wind with your turbine. If you put tons of blades close together, each will interfere with the airflow over the next one, so there's clearly an upper limit. The assembly needs to be balanced to avoid huge non-axial torques, so you need more than 1 blade (although Hagen suggests that in some cases this may not be true...or maybe his "one blade" is a diameter rather than a radius, i.e., what I'd call a 2-blade setup).



The towers (often) have maximum height restrictions, and you can only fit so many of them in a given space, so cramming in more blades makes sense, hence 3-blade rather than 2-blade. But why not 4 or 6 or 12 blade ones?



I'm guessing that each additional blade gives a bump in output, but perhaps not in proportion to the number of blades: the 5th blade increases the output less than the 4th one did, and so on. And blades are expensive. And more blades requires a more substantial axle, gearbox, etc., not to mention a stronger tower. So there's a point where the additional cost for the $n$th blade exceeds the value of the incremental power generated.



And looking at common windmills suggests that this point is at about $n = 3$. :)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:57










  • $begingroup$
    I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
    $endgroup$
    – John Hughes
    Jan 16 at 17:26
















2












$begingroup$

I conjecture that this is not a mathematical problem at all, but rather an engineering and legal one.



You want to extract lots of energy from the wind with your turbine. If you put tons of blades close together, each will interfere with the airflow over the next one, so there's clearly an upper limit. The assembly needs to be balanced to avoid huge non-axial torques, so you need more than 1 blade (although Hagen suggests that in some cases this may not be true...or maybe his "one blade" is a diameter rather than a radius, i.e., what I'd call a 2-blade setup).



The towers (often) have maximum height restrictions, and you can only fit so many of them in a given space, so cramming in more blades makes sense, hence 3-blade rather than 2-blade. But why not 4 or 6 or 12 blade ones?



I'm guessing that each additional blade gives a bump in output, but perhaps not in proportion to the number of blades: the 5th blade increases the output less than the 4th one did, and so on. And blades are expensive. And more blades requires a more substantial axle, gearbox, etc., not to mention a stronger tower. So there's a point where the additional cost for the $n$th blade exceeds the value of the incremental power generated.



And looking at common windmills suggests that this point is at about $n = 3$. :)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:57










  • $begingroup$
    I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
    $endgroup$
    – John Hughes
    Jan 16 at 17:26














2












2








2





$begingroup$

I conjecture that this is not a mathematical problem at all, but rather an engineering and legal one.



You want to extract lots of energy from the wind with your turbine. If you put tons of blades close together, each will interfere with the airflow over the next one, so there's clearly an upper limit. The assembly needs to be balanced to avoid huge non-axial torques, so you need more than 1 blade (although Hagen suggests that in some cases this may not be true...or maybe his "one blade" is a diameter rather than a radius, i.e., what I'd call a 2-blade setup).



The towers (often) have maximum height restrictions, and you can only fit so many of them in a given space, so cramming in more blades makes sense, hence 3-blade rather than 2-blade. But why not 4 or 6 or 12 blade ones?



I'm guessing that each additional blade gives a bump in output, but perhaps not in proportion to the number of blades: the 5th blade increases the output less than the 4th one did, and so on. And blades are expensive. And more blades requires a more substantial axle, gearbox, etc., not to mention a stronger tower. So there's a point where the additional cost for the $n$th blade exceeds the value of the incremental power generated.



And looking at common windmills suggests that this point is at about $n = 3$. :)






share|cite|improve this answer









$endgroup$



I conjecture that this is not a mathematical problem at all, but rather an engineering and legal one.



You want to extract lots of energy from the wind with your turbine. If you put tons of blades close together, each will interfere with the airflow over the next one, so there's clearly an upper limit. The assembly needs to be balanced to avoid huge non-axial torques, so you need more than 1 blade (although Hagen suggests that in some cases this may not be true...or maybe his "one blade" is a diameter rather than a radius, i.e., what I'd call a 2-blade setup).



The towers (often) have maximum height restrictions, and you can only fit so many of them in a given space, so cramming in more blades makes sense, hence 3-blade rather than 2-blade. But why not 4 or 6 or 12 blade ones?



I'm guessing that each additional blade gives a bump in output, but perhaps not in proportion to the number of blades: the 5th blade increases the output less than the 4th one did, and so on. And blades are expensive. And more blades requires a more substantial axle, gearbox, etc., not to mention a stronger tower. So there's a point where the additional cost for the $n$th blade exceeds the value of the incremental power generated.



And looking at common windmills suggests that this point is at about $n = 3$. :)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 16:52









John HughesJohn Hughes

65.2k24293




65.2k24293












  • $begingroup$
    I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:57










  • $begingroup$
    I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
    $endgroup$
    – John Hughes
    Jan 16 at 17:26


















  • $begingroup$
    I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
    $endgroup$
    – mathreadler
    Jan 16 at 16:57










  • $begingroup$
    I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
    $endgroup$
    – John Hughes
    Jan 16 at 17:26
















$begingroup$
I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
$endgroup$
– mathreadler
Jan 16 at 16:57




$begingroup$
I think your points are reasonable. I would also be surprised if exactly three blades would be optimal. But nevertheless, I am still interested in how to calculate it.
$endgroup$
– mathreadler
Jan 16 at 16:57












$begingroup$
I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
$endgroup$
– John Hughes
Jan 16 at 17:26




$begingroup$
I'd be surprised if exactly three were optimal. Probably if you could write down the equations, it'd turn out that $3.35$ blades were best...if only the problem were on a continuous domain. :(
$endgroup$
– John Hughes
Jan 16 at 17:26


















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