Re: Average chord

On Fri, 09 Dec 2005 21:10:22 -0700, Virgil
<ITSnetNOTcom#virgil@xxxxxxxxxxx> wrote:

>In article <aoikp11rb4u6bfn6i163b1htm3rv3gq6k5@xxxxxxx>,
> quasi <quasi@xxxxxxxx> wrote:
>
>> On 9 Dec 2005 19:17:13 -0800, "Narasimham" <mathma18@xxxxxxxxxxx>
>> wrote:
>>
>> >>If two pairs of points are almost diametrically
>> >>opposed their average distance will be almost 2.
>> >>How is the above possible?
>> >
>> >Maximum 2 , Minimum 0.
>>
>> Average 1.
>
>Depends on how the averaging is weighted. If it is weighted by arc
>lengths then it averages out to more.
>(1/pi)*Integral[ t = 0..pi, sqrt( 2 * ( 1 - cos(t) ) ]

You may have missed the context.

Earlier in this thread we discussed averages over the whole circle,
but here we are just talking about the average of just 2 distances:

2 and 0,

so the average is just (2+0)/2 = 1.

quasi
.

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