----- 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@comcast.net> 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
> >am measuring is closer to an elipse. However it is
> >more irregular. It may aid you if I give you the
> image
> >of a lake. How would we work out the area of a
> lake,
> >when our available data is a number of diameter
> >measurements of the lake?
> >By the way I figured if I actually used the
> diameters
> >to plot out a map on graph paper, I could count the
> >little squares to find the area!! However, as I
> have
> >to do a large number of such calculations, that is
> not
> >practical. I feel sure there must be a suitable
> >formula.
> >
> >Thank you!
> >Best wishes
> >Justin
>
> My previous answer was based on knowing the radii
> from some center.
> However, knowing diameters is more difficult, since
> they may not all
> meet at a common center.
>
> There might be a way to use a tree as a "center",
> but instead of
> using a common angle, you must measure the distance
> of each point of
> the lake from the tree, and the distance between
> consecutive chosen
> points on the lake. The area of the triangles (no
> longer isosceles)
> require a different formula. Also, triangles formed
> from points on
> the near side of the lake must be subtracted from
> the triangles from
> the far side of the lake. Some triangles have no
> area at all.
>
>
> Ben Saucer
> e-mail: bsaucer1@comcast.net
> web page: www.saucersdomain.com
> ICQ: 20610314
>
>

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