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abver · May 10, 2011, 01:58 PM

explain formula : (minor arc (AB) - chord (AB)) * (diameter / 8) = segment of a circle (minor arc (AB) / 2)

mmresd · May 11, 2011, 11:32 AM

Is this for an ellipsis?

abver · May 11, 2011, 12:45 PM

no, if diameter is 10 ; (15,7 - 10) * (10 / 8) = 7,125 cm 2

abver · May 11, 2011, 05:36 PM

or (7,85 - 7,071.. ) * (10 / 8) = 0,97375

Unknown008 · May 12, 2011, 10:12 AM

Again one of your crazy formulae? :rolleyes:

abver · May 12, 2011, 02:57 PM

Now l realize that you are the expert not only in math but also psychiatry (lucky for you). This formula is correct (easily verifiable) so can not be crazy, to which you do not understand, l will survive (do not worry)

ebaines · May 13, 2011, 08:31 AM

What Abver is trying to describe (in his usual obtuse way) is that the area of the segment of a circle formed by the half-angle AB is equal to his formula. The area of the yellow shaded part in the figure below is what he's talking about.

You can derive it like this:

The yellow area = area of sector AC minus area of the internal triangle AOC:

$ \frac 1 2 R^2 \frac {\theta} 2 - \frac 1 2 R^2 \sin(\frac {\theta} 2 ) $

This can be manipulated into Abver's formula using the identity

$ \sin (\frac {\theta} 2) = \frac {\sqrt { 2 - 2 \cos \theta}} 2 $

$ \frac 1 2 R^2 \frac {\theta} 2 - \frac 1 2 R^2 \sin(\frac {\theta} 2 ) = (R \theta - R \sqrt{2 - 2 \cos \theta} (\frac {2R} 8 )\\ = (Arc\ length\ AB - chord\ length\ AB) * \frac D 8 $

Unknown008 · May 13, 2011, 08:46 AM

Originally Posted by abver

now l realize that you are the expert not only in math but also psychiatry (lucky for you). this formula is correct (easily verifiable) so can not be crazy, to which you do not understand, l will survive (do not worry)

Yes, but what is the point (use) of this question if it's 'easily verifiable'? Are you testing the people who answer here?

The first poster did a query, and you immediately turned a mere 'no' and simply putting in numbers that you don't even explain where you got. The diameter is 10, where did you get 15,7 and 10, which are respectively the minor arc and chord of AB? Nothing of that, no no. What you are doing is like doing a wish, and pray that the solution appears on the screen of your computer.

abver · May 14, 2011, 10:24 PM

That image is correct, an explanation should be fine

Unknown008 · May 15, 2011, 02:03 AM

ebaines already gave you an explanation -.-