Quite some time ago I handed y'all a quandary, to wit, proving that sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a circle and there are an odd number of them. (It's even if there's an even number of them -- you've got this nice pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) So, I ran across this page: <https://en.wikipedia.org/wiki/ List_of_trigonometric_identities#Linear_combinations> It gives a rule upon which to base a proof! There's still not a direct proof, because doing so ends up being redundant. However, somewhere in there is a fairly easy proof that with some restrictions (N > 1, ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're into 2CVs), 7, 9, etc. It has to do with DFTs of vectors of odd number... -- Tim Wescott Wescott Design Services http://www.wescottdesign.com I'm looking for work -- see my website!
Hah! Why 5-lug wheels balance
Started by ●January 20, 2017
Reply by ●January 20, 20172017-01-20
On 1/20/2017 6:31 PM, Tim Wescott wrote:> Quite some time ago I handed y'all a quandary, to wit, proving that > sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a > circle and there are an odd number of them. > > (It's even if there's an even number of them -- you've got this nice > pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) > > So, I ran across this page: > <https://en.wikipedia.org/wiki/ > List_of_trigonometric_identities#Linear_combinations> > > It gives a rule upon which to base a proof! There's still not a direct > proof, because doing so ends up being redundant. However, somewhere in > there is a fairly easy proof that with some restrictions (N > 1, > ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're > into 2CVs), 7, 9, etc. > > It has to do with DFTs of vectors of odd number...So why do some cars use 4 lugs per wheel? -- Rick C
Reply by ●January 20, 20172017-01-20
It is fairly obvious. The trick is with radiator fans, balancing them when angles are unequal. Simple if even number of blades, e.g. for 6 blades: make angles 50,60,70 degrees. But how when 5 blades?
Reply by ●January 20, 20172017-01-20
On Friday, January 20, 2017 at 7:18:27 PM UTC-6, pedr...@lycos.com wrote:> But how when 5 blades?Add any balanced configuration to any other balanced configuration and the result will still be balanced, no matter what the number of blades nor the relative rotation between the two configurations. Same goes for subtracting a smaller balanced configuration (if one is present) from a larger balanced configuration.
Reply by ●January 20, 20172017-01-20
On 01/20/17 18:31, Tim Wescott wrote:> Quite some time ago I handed y'all a quandary, to wit, proving that > sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a > circle and there are an odd number of them. > > (It's even if there's an even number of them -- you've got this nice > pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) > > So, I ran across this page: > <https://en.wikipedia.org/wiki/ > List_of_trigonometric_identities#Linear_combinations> > > It gives a rule upon which to base a proof! There's still not a direct > proof, because doing so ends up being redundant. However, somewhere in > there is a fairly easy proof that with some restrictions (N > 1, > ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're > into 2CVs), 7, 9, etc. > > It has to do with DFTs of vectors of odd number... >This seems pretty straightforward: If there is only one angle theta in your set then the sum isn't zero. If there are two or more, consider the sum of the complex exponentials exp(i.theta) for all the evenly-spaced values of theta in your set; this sum will be unchanged by multiplication by exp(i.eta) where eta is the difference between two successive values of theta (since this multiplication permutes the summands). But the only complex number fixed by a non-trivial rotation about 0 is 0. So your cosine sum is the real part of 0. Hope that helps. REB
Reply by ●January 20, 20172017-01-20
rickman wrote:> On 1/20/2017 6:31 PM, Tim Wescott wrote: >> Quite some time ago I handed y'all a quandary, to wit, proving that >> sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a >> circle and there are an odd number of them. >> >> (It's even if there's an even number of them -- you've got this nice >> pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) >> >> So, I ran across this page: >> <https://en.wikipedia.org/wiki/ >> List_of_trigonometric_identities#Linear_combinations> >> >> It gives a rule upon which to base a proof! There's still not a direct >> proof, because doing so ends up being redundant. However, somewhere in >> there is a fairly easy proof that with some restrictions (N > 1, >> ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're >> into 2CVs), 7, 9, etc. >> >> It has to do with DFTs of vectors of odd number... > > So why do some cars use 4 lugs per wheel? >Smaller wheels, I presume. -- Les Cargill
Reply by ●January 21, 20172017-01-21
On 1/20/2017 10:48 PM, Les Cargill wrote:> rickman wrote: >> On 1/20/2017 6:31 PM, Tim Wescott wrote: >>> Quite some time ago I handed y'all a quandary, to wit, proving that >>> sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a >>> circle and there are an odd number of them. >>> >>> (It's even if there's an even number of them -- you've got this nice >>> pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) >>> >>> So, I ran across this page: >>> <https://en.wikipedia.org/wiki/ >>> List_of_trigonometric_identities#Linear_combinations> >>> >>> It gives a rule upon which to base a proof! There's still not a direct >>> proof, because doing so ends up being redundant. However, somewhere in >>> there is a fairly easy proof that with some restrictions (N > 1, >>> ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're >>> into 2CVs), 7, 9, etc. >>> >>> It has to do with DFTs of vectors of odd number... >> >> So why do some cars use 4 lugs per wheel? >> > > Smaller wheels, I presume.Unbalanced smaller wheels? -- Rick C
Reply by ●January 21, 20172017-01-21
Tim Wescott <seemywebsite@myfooter.really> writes:> Quite some time ago I handed y'all a quandary, to wit, proving that > sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a > circle and there are an odd number of them. > > (It's even if there's an even number of them -- you've got this nice > pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) > > So, I ran across this page: > <https://en.wikipedia.org/wiki/ > List_of_trigonometric_identities#Linear_combinations> > > It gives a rule upon which to base a proof! There's still not a direct > proof, because doing so ends up being redundant. However, somewhere in > there is a fairly easy proof that with some restrictions (N > 1, > ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're > into 2CVs), 7, 9, etc. > > It has to do with DFTs of vectors of odd number...You can do it that way, but it's a little bit of overkill. All you need is the sum of a geometric series. Write all the thetas as \theta_k = 2\pi k/n + \theta_c I threw the "\theta_c" in there because the thetas may not line up with our coordinate system. Anyway, we want to prove that \sum_{k=0}^{n-1} \cos\theta_k = 0 So, for good measure, we'll also prove that \sum_{k=0}^{n-1} \sin\theta_k = 0 So multiply the second equation through by "i" and add them together: \sum_{k=0}^{n-1} \cos\theta_k + i\sin\theta_k = \sum_{k=0}^{n-1} e^{i\theta_k} = \sum_{k=0}^{n-1} e^{i(2\pi k/n+\theta_c)} = e^{i\theta_c}\sum_{k=0}^{n-1} e^{i(2\pi k/n} Now the sum is just 1 + r + r^2 + ... + r^{n-1}, where r = e^{i 2\pi/n} The sum of the geometric series is (1-r^n)/(1-r). The numerator of this expression is (1-e^{i 2\pi}). Since e^{i 2\pi} is 1, the numerator is zero, and therefore the whole sum is zero. Then you have both the real part and the imaginary part of the sum is zero, which gives the required result. Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear
Reply by ●January 21, 20172017-01-21
On Fri, 20 Jan 2017 21:48:20 -0600, Les Cargill <lcargill99@comcast.com> wrote:>rickman wrote: >> On 1/20/2017 6:31 PM, Tim Wescott wrote: >>> Quite some time ago I handed y'all a quandary, to wit, proving that >>> sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a >>> circle and there are an odd number of them. >>> >>> (It's even if there's an even number of them -- you've got this nice >>> pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) >>> >>> So, I ran across this page: >>> <https://en.wikipedia.org/wiki/ >>> List_of_trigonometric_identities#Linear_combinations> >>> >>> It gives a rule upon which to base a proof! There's still not a direct >>> proof, because doing so ends up being redundant. However, somewhere in >>> there is a fairly easy proof that with some restrictions (N > 1, >>> ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're >>> into 2CVs), 7, 9, etc. >>> >>> It has to do with DFTs of vectors of odd number... >> >> So why do some cars use 4 lugs per wheel? >> > >Smaller wheels, I presume.And cost. Many larger trucks have even numbers of studs/lugs, e.g., 8. http://www.discountedwheelwarehouse.com/images/Pacer-164P-LT-Mod-Polished-8lug-P.jpg My street car has 6 lugs/studs per wheel, like this: http://i40.tinypic.com/iycs4m.jpg Some older cars and many race cars have only one. --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus
Reply by ●January 22, 20172017-01-22
On 1/21/2017 5:06 PM, eric.jacobsen@ieee.org wrote:> On Fri, 20 Jan 2017 21:48:20 -0600, Les Cargill > <lcargill99@comcast.com> wrote: > >> rickman wrote: >>> On 1/20/2017 6:31 PM, Tim Wescott wrote: >>>> Quite some time ago I handed y'all a quandary, to wit, proving that >>>> sum_{\theta} cos(\theta) = 0, when \theta is evenly distributed on a >>>> circle and there are an odd number of them. >>>> >>>> (It's even if there's an even number of them -- you've got this nice >>>> pairing of cos(this) + cos(-this) = 0, and the proof is a few lines.) >>>> >>>> So, I ran across this page: >>>> <https://en.wikipedia.org/wiki/ >>>> List_of_trigonometric_identities#Linear_combinations> >>>> >>>> It gives a rule upon which to base a proof! There's still not a direct >>>> proof, because doing so ends up being redundant. However, somewhere in >>>> there is a fairly easy proof that with some restrictions (N > 1, >>>> ferinstance) a five-lug wheel balances, as does a 3-lug wheel (if you're >>>> into 2CVs), 7, 9, etc. >>>> >>>> It has to do with DFTs of vectors of odd number... >>> >>> So why do some cars use 4 lugs per wheel? >>> >> >> Smaller wheels, I presume. > > And cost. > > Many larger trucks have even numbers of studs/lugs, e.g., 8. > > http://www.discountedwheelwarehouse.com/images/Pacer-164P-LT-Mod-Polished-8lug-P.jpg > > My street car has 6 lugs/studs per wheel, like this: > > http://i40.tinypic.com/iycs4m.jpg > > Some older cars and many race cars have only one.Given the issues of symmetry which seem to be at play here, I would expect one lug to be a special case combining properties of both even and odd numbered lugs. -- Rick C