At 10:51 AM 11/10/2006, you wrote:

>Hi guys!
>I hope you can help me out. It's for my work! Here are 2 questions:
>
>1) On a lathe, I need to know an angle for the taper. It boils down to
>this:
>Length = 2cm
>Taper = .5mm (and some others too).
>
>So what is the formula? It seems to be a right angled triangle, with
>one length 20mm and one length .5mm, so what does that make the small
>angle? (Also the actual formula would be useful as I need it for other
>tapers too.)

It's a simple trigonometry problem.

I assume those are the legs of the right triangle. In that case, the
tangent of the angle is equal to the ratio of the opposite leg over
the adjacent leg. If, on the other hand, the measurements given are
those of the short leg and the hypotenuse, then the sine of the angle
is equal to the opposite leg over the hypotenuse.

Use the arcsine or arctangent function to find the angle.

>2) I have to measure the internal dimentions of a tube with an
>iregular surface. So, I want to know the crossectional areas. At
>regular intervals down the length of the tube, I will take a number of
>measurements of diameter. For example, at 5 cm down the tube, I will
>take, let's say north south diameter, east west, north-east
>south-west, etc. Perhaps 6 or 8 such diameters. If it were a circle of
>course they'd all be the same measurement. But it is irregular. What I
>want to do is actually work out the area. What formula can I use to
>work it out? I am no mathematician, but, I figure if I merely average
>out the diameters it will give me an erroneous result. But I feel sure
>there must be a way to work this out.


This is a calculus problem. Express the figure in polar coordinates
where the radius r is a function of the angle theta. For each point
along the figure, construct a thin isosceles triangle with the apex
at the center (origin), and the center of the base on the perimeter
of the figure. Use some small number for the thin apex angle and call
it delta-theta. The base of the triangle is r times delta-theta. The
area of the triangle is 1/2 of r^2 times delta theta. Continue
constructing thin triangles all the way around the figure using the
same delta-theta as the angle, measure the radius, and add up the areas.

Since the figure is irregular, you can guess where the center is, and
do all calculations based on that.

As an actual integral, it's written as integral ( theta = 0 to 2pi )
r(theta)^2 * d-theta / 2


Ben Saucer
e-mail: bsaucer1@...
web page: www.saucersdomain.com
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