'Average Angle'

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Strilanc
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'Average Angle'

Postby Strilanc » Wed May 14, 2008 4:55 am UTC

I've often found myself wanting the average of a bunch of angles. But thinking about it more, angles are not so easy to average.

I used the definition "x is the average of a set of angles if the sum of all angles to the left equals the sum of all angles to the right" (exactly opposite and equal don't count). The first thing I noticed with this definition is there may be multiple averages: {0, pi} has averages 0, pi/2, pi, and 3pi/2. In fact every set of two distinct angles has at least two averages.

So I refined the definition to require that the sum of absolute differences be minimized at the average (vs other averages). This eliminates most multiple averages, but not all. Unfortunately, it also eliminates the pi/2 and 3pi/2 answers for {0,pi} (which are perhaps more intuitive than 0 and pi being averages for that set).

I think this all has to do with the fact that division over the reals (modulo some constant) is a multi-valued function.

So it comes down to this: what is a good definition of the average of a set of angles? How do you compute it? Given my above definition, can you have more than 2n averages for a set of n angles?
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Re: 'Average Angle'

Postby ++$_ » Wed May 14, 2008 5:09 am UTC

Strilanc wrote:So I refined the definition to require that the sum of absolute differences be minimized at the average (vs other averages). This eliminates most multiple averages, but not all. Unfortunately, it also eliminates the pi/2 and 3pi/2 answers for {0,pi} (which are perhaps more intuitive than 0 and pi being averages for that set).
I don't see how this eliminates [imath]\pi/2[/imath] or [imath]3\pi/2[/imath]. Nevertheless, you should probably minimize the sum of the squares of the differences instead.

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Re: 'Average Angle'

Postby csrjjsmp » Wed May 14, 2008 5:10 am UTC

To find the average of two angles ABC and XYZ, with vertices at B and Y respectively, move one to share a side with the other, so that BC coincides with XY, and B=Y. Bisect angle ABZ.
In general, you can't find the average of more than two angles.
Edit: If you use a system in which you assign measures to angles, you can average those in the normal way.
Last edited by csrjjsmp on Wed May 14, 2008 9:35 am UTC, edited 2 times in total.

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Cycle
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Re: 'Average Angle'

Postby Cycle » Wed May 14, 2008 5:57 am UTC

The best definition I can think of is to represent the angles as points on the unit circle (in C), sum them, and then take the argument.

This is undefined for samples that are perfectly balanced, but I think that's ok (there's no good way to define the average of {0, pi}). It gives you what you would expect (a unique answer) whenever it's defined (and it usually is). For example, the average of {0, pi/2} is pi/4.

As an interpretation: if you placed weights, corresponding to the given angles, on a CD balanced on a beer bottle, the average tells you which direction it will fall.

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Strilanc
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Re: 'Average Angle'

Postby Strilanc » Wed May 14, 2008 3:42 pm UTC

Cycle wrote:The best definition I can think of is to represent the angles as points on the unit circle (in C), sum them, and then take the argument.

This is undefined for samples that are perfectly balanced, but I think that's ok (there's no good way to define the average of {0, pi}). It gives you what you would expect (a unique answer) whenever it's defined (and it usually is). For example, the average of {0, pi/2} is pi/4.

As an interpretation: if you placed weights, corresponding to the given angles, on a CD balanced on a beer bottle, the average tells you which direction it will fall.


Good idea: just calling it undefined when there are multiple averages makes it much simpler (divide by zero, anyone?).
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Re: 'Average Angle'

Postby ventusignis » Wed May 14, 2008 9:38 pm UTC

There are two parts of an angle. Position (which shall be represented by using the angle bisector in standard form) and Width (How wide the angle is, like 90 degrees for perpendicular legs) (I totally made up these terms). If you average the position, and average the width, you can get an average angle with it's own position and width.

And by average, I mean add up all 'n' elements and dividing the sum by 'n'

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Strilanc
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Re: 'Average Angle'

Postby Strilanc » Wed May 14, 2008 11:27 pm UTC

ventusignis wrote:There are two parts of an angle. Position (which shall be represented by using the angle bisector in standard form) and Width (How wide the angle is, like 90 degrees for perpendicular legs) (I totally made up these terms). If you average the position, and average the width, you can get an average angle with it's own position and width.

And by average, I mean add up all 'n' elements and dividing the sum by 'n'


Ummm, what? Angles don't have a width. Adding up all the elements and dividing by n will either give multiple answers or sometimes an incorrect answer. For example: 3/2 pi and 0. The 'falling' average is 7/4 pi, but naive sum-and-divide will give 3/4 pi.
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ventusignis
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Re: 'Average Angle'

Postby ventusignis » Thu May 15, 2008 1:10 am UTC

What I was thinking was there were multiple angles in a unit circle, each with their own pairs of rays. what I was refering to as 'width' was the angle measure actually.
3/4 pi and 0 ? Are these angles in standard position?

Actually, I don't know what an average of an angle is anymore.

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Re: 'Average Angle'

Postby Teki-Teki » Thu May 15, 2008 2:16 am UTC

More generally, if we have some angles [imath]\theta_k[/imath], we can define the average simplistically as [math]\frac{1}{n}\sum_k \theta_k[/math]$. But the trouble is that actually any rotation of this angle through [imath]\frac{2\pi}{n}[/imath] radians will give an angle that could be interpreted as the "average".

Basically, you can think of it as the problem you get with square roots: [imath]\sqrt{4} = \pm 2[/imath], except that here you can have more than just 2 possible answers. The real reasons lie in the complex plane:

Consider [math]z_k = e^{i\theta_k}[/math] and [math]Z = \prod_k z_k .[/math] Then the average is really the argument of [math]Z^{\frac{1}{n}},[/math] which of course has n possible values unique up to a power of [math]\zeta = e^{\frac{2\pi i}{n}} .[/math]

Note that this definition still works if you allow your angles to be negative, which can happen if you're using directed angles. In general, though, directed angles are pretty useless except to make proofs airtight for different configurations.

Also, why is my math not working?

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Re: 'Average Angle'

Postby jestingrabbit » Thu May 15, 2008 2:31 am UTC

Teki-Teki wrote:Also, why is my math not working?

Probably not using the right skin. More here

viewtopic.php?f=17&t=21939

If I'm right you need to go to the user control panel and change to one that is supported, either prosilver or prosilver-left.

Totally agree with your analysis too.
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Re: 'Average Angle'

Postby skeptical scientist » Thu May 15, 2008 3:46 am UTC

Teki-Teki wrote:Also, why is my math not working?

Most of the time you are using [math] when you should be using [imath]. The [math] tag puts a formula on its own line, while [imath] puts it inline in the text. For the last one, you have a missing argument for \frac according to jsMath. The correct usage is \frac{numerator}{denominator}, and you seem to be using \frac{{num}{denom}}. In other words, you should have \zeta = e^\frac{2\pi i}{n} instead of what you have (although I would prefer [imath]\zeta = \exp(\frac{2\pi i}{n}[/imath]) since having a fraction in an exponent makes things hard to read).
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Re: 'Average Angle'

Postby Nimz » Thu May 15, 2008 7:35 am UTC

Teki-Teki wrote:why is my math not working?
Like skeptical scientist said, your bad \frac is due to where you put your brackets. e^{\frac{2\pi i}{n}} would have worked, but again, \exp is nice in the case of having a fraction within the exponent. Something else you may wish to try is putting your punctuation inside the [math] tags. That way they don't get placed on the next line. If you want to make sure they don't affect the mathematics, wrap them up in \mbox{}. E.g.
If you consider Euler's famous equation, [math]e^{i\pi} = -1,[/math] it is plain to see why logs of negative numbers should be complex.
If you consider Euler's famous equation, [math]e^{i\pi} = -1,[/math] it is plain to see why logs of negative numbers should be complex.

[imath]\frac{2}{3}\cdot 10^5 \mbox{. Some stuff in an mbox. } 2^{2^{2^{\dots}}}[/imath]
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Teki-Teki
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Re: 'Average Angle'

Postby Teki-Teki » Thu May 15, 2008 8:11 pm UTC

The tip worked - I didn't realize my skin had changed when I logged in. All good now, and darn that was a dumb LaTeX mistake for me to make after this long.

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Re: 'Average Angle'

Postby BeetlesBane » Fri May 16, 2008 7:43 am UTC

To return to the OP. srtilanc why are you wanting "the average of a bunch of angles"? This more than anyhing else will determine the answer to your question. It may meaningful to ignore the equivalence of two angles that differ by a integral number of rotations; in this case, the familiar real number procedures will work quite well.

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Strilanc
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Re: 'Average Angle'

Postby Strilanc » Fri May 16, 2008 2:30 pm UTC

It was just an in-general question I was thinking about.
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Re: 'Average Angle'

Postby 0SpinBoson » Fri May 16, 2008 10:04 pm UTC

Here's another way: Take each angle \alpha as a unit vector with direction \alpha as measured from the x axis. Resolve components, add them all, find the overall direction, divide by the number.


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