Hi Carl
Me too. But when I try to visualize what you are
saying, I imagine that you are imagining (!!) a
triangle whose 30 degree angle is in the centre of the
circle, and the sides extend to the edge. It sounds
like you are saying because the side meets the
circumference of the circle at 90 degrees, that is
therefore the angle of that triangle. Is that what you
mean? But then, that would apply to both those 2 long
sides. So the triangle would be 30 90 90. Impossible.

Besides, there is no semicircle anyway. I think we
might be following a red herring here, perhaps?
Sorry if I am totally mistaken.
Best wishes
Justin

--- Carl Adkins <ca93901@...> wrote:

> Apparently I missed something. It's hard for me to
> grasp something without a picture of what you are
> attempting to do. All the geometric facts I gave
> you are true.
>
> Carl Adkins
> ----- Original Message -----
> From: Justin .
> To: geometry@yahoogroups.com
> Sent: Thursday, November 16, 2006 7:08 AM
> Subject: [Norton AntiSpam] Re: Re: [geometry] Re:
> How to find an angle in a triangle, and how to
> calculate an irregular area?
>
>
> If you think about that carefully Carl, that would
> give me a triangle with 2 angles being right
> angles,
> right? Don't think so. Or did I miss something?
>
> --- Carl Adkins <ca93901@...> wrote:
>
> > An angle inscribed in a semicircle is always a
> right
> > angle. You state the small angle is 30 degrees,
> so
> > you have a 30-60 right triangle. The side
> opposite
> > the 30 degree angle is half of the hypotenuse
> (aka
> > diameter), which would be equal to the radius.
> So,
> > you have a right triangle with sides of 2X, X,
> and
> > X*(sqrt of 3), with an area equal to
> (X/2)*[X(sqrt
> > of 3)].
> >
> > Carl Adkins
> > ----- Original Message -----
> > From: Justin .
> > To: geometry@yahoogroups.com
> > Sent: Wednesday, November 15, 2006 8:24 PM
> > Subject: [Norton AntiSpam] Re: [geometry] Re:
> How
> > to find an angle in a triangle, and how to
> calculate
> > an irregular area?
> >
> >
> > Hi Everyone
> > I have an idea on how to do this calculation of
> > irregular area. I will try to explain my idea to
> > see
> > if you all think it makes good sense and would
> be
> > accurate. As I am not a mathematician, it may be
> > wrong, and, sorry I do not know the conventional
> > way
> > to express it, so forgive me for my misuse/abuse
> > of
> > terms and symbols, but hopefully you can
> > understand
> > what I will try to express.
> >
> > By measuring 6 diameters, I have 12 triangles.
> The
> > small angle of all the triangles are all the
> same
> > - 30
> > degrees. I do not know the formula for areas of
> a
> > triangle, but, there must be a very simple one
> > when
> > the small angle and its 2 lengths are known
> > (Anyone
> > know the formula?)
> >
> > So from that it is easy to calculate the area of
> > the
> > shape.
> >
> > However, this shape has 12 straight sides,
> whereas
> > the
> > actual shape it represents (the real tube
> > crossection)
> > is more curvey. If this irregular curve were
> > approximately a straight line, I would expect
> the
> > curve to be going "over" AND "under" the outer
> > side of
> > the triangle equally, and thus the result to be
> > fair.
> > However, as the irregular curve is approximately
> a
> > circle, I would expect it to be going "over"
> more
> > than
> > "under", by which I mean, more of the irregular
> > curve
> > (ie more AREA) would lie OUTSIDE the triangles
> > than
> > INSIDE. Thus, I would expect the area resulting
> > from
> > the 12 sided shape to be an underestimate.
> >
> > I have this suggestion for correction:
> >
> > - Call the area of the 12 sided shape MUM.
> > - Generate a circle where
> > radius = length b of isosceles triangle (with
> > angle
> > inbetween its 2 b sides being 30 degrees and
> area
> > being MUM/12 )
> > Call the area of this circle DAD
> >
> > - DAD minus MUM = BABY
> >
> > Okay, so, I presume the real area of the
> irregular
> > shape must be greater than MUM, and less than
> DAD.
> > Do
> > you think? (It may help to visualize that the
> > shape
> > looks somwhat like the crossection of a
> > treetrunk.)
> >
> > So I guess that the formula would be
> > Area = MUM + (BABY * x)
> > where x is less than 1.
> >
> > Does that sound right? Then, I do not know what
> x
> > should be! Any suggestions?
> >
> > Also, as the complete formula includes the
> formula
> > of
> > the triangles and so on, and it takes all the
> > steps
> > for the different diameters, if anyone can write
> > this
> > more clearly, it would be great! Plus, this
> should
> > be
> > very simple (for those who know how!) to turn
> into
> > a
> > little program (or is it called a routine?). So
> > that
> > all I have to do is input the diameters, the
> > number of
> > diameters and the angle (between the diameters,
> > which
> > anyway is implicite in the number of diameters),
> > and
> > then it will do all of this calculation.
> >
> > Thank you all!
> > Justin
> >
> > --- Ben Saucer <bsaucer2@...> wrote:
> >
> > > At 11:48 PM 11/10/2006, you wrote:
> > >
> > > >Hi all
> > > >Just to further clarify on the second of my
> too
> > > >questions:
> > > >
> > > >I am not concerned with volume. I am trying
> to
> > work
> > > >out area. The area of a crossection (a number
> > of
> > > >crossections actually). And it seems to me
> also
> > > that
> > > >the average diameter will not have a direct
> > > >relationship to the area, surely? For
> example,
> > a
> > > line
> > > >may have its longest diameter at 20cm and its
> > > shortest
> > > >diameter at 0cm. Although its average
> diameter
> > is
> > > >therefore 10cm, its area is 0. That's why I
> > thought
> > > >there may be a more accurate way of working
> it
> > out.
> > > >For example, wouldn't an elipse use a
> different
> > > >formula for working out the area than a
> circle?
> > > What I
>
=== message truncated ===




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