tunnel job: finding 'radius' from center of ellipse to point

DaveWW00

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Aug 6, 2006
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:?:
I am trying to figure out some weird geometry for a tunnel job at work and need one thing to help me figure out a small part of it.

I need to know the formula to calculate the radius from the center of an ellipse to a point on the ellipse. Anybody know this?
 
Re: Urgent- Need help fast- "Radius" of Ellipse

DaveWW00 said:
:?: I need to know the formula to calculate the radius from the center of an ellipse to a point on the ellipse. Anybody know this?
First, you said “radius from the center of an ellipse to a point on the ellipse.”, but that is not a constant! The distance from the center of an ellipse to a point on the ellipse varies as the point varies.
\(\displaystyle \begin{array}{l}
\frac{{\left( {x - h} \right)^2 }}{{a^2 }} + \frac{{\left( {y - k} \right)^2 }}{{b^2 }} = 1 \\
\left\{ \begin{array}{l}
x = a\cos (t) + h \\
y = b\sin (t) + k \\
\end{array} \right.\quad ,0 \le t \le 2\pi \\
\\
d(t) = \sqrt {a^2 \cos ^2 (t) + b^2 \sin ^2 (t)} \\
\end{array}.\)
The first is the standard equation of an ellipse with center (h,k).
The second is the same rewritten in parametric form.
Using parametric form the last equation gives the distance from the center of the ellipse to a point on the ellipse corresponding to any value of t.

This may not be what you had in mind. If please do reply with clarification.
 
I think I may know what you mean. If so, just sub in your values for what I have in my example.

Suppose you have a road going through a semi-elliptical tunnel(like in the diagram).

The road is 30 feet wide and the highest part of the arch is 10 feet.

Say, we wanted to find the height of the arch 6 feet from the center of the road.

tunneltc9.gif


Use the general ellipse equation formula \(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

In this example, a=15(major axis, the road) and b=10(minor axis, the height at center).

We could write this as \(\displaystyle V(\pm{15},0)\;\ M(0, \pm{10})\)

Substitute x=6: \(\displaystyle \L\\\frac{6^{2}}{15^{2}}+\frac{y^{2}}{10^{2}}=1\)

=\(\displaystyle \L\\\frac{36}{225}+\frac{y^{2}}{100}=1\)

Now, solve for y:

\(\displaystyle \L\\\frac{y^{2}}{100}=1-\frac{36}{225}\)

\(\displaystyle \L\\=\frac{y^{2}}{100}=\frac{21}{25}\)

\(\displaystyle \L\\y^{2}=\frac{2100}{25}=84\)

\(\displaystyle \L\\y=\sqrt{84}=2\sqrt{21}\approx{9.165}\;\ ft\)

There, the height of your tunnel 6 feet on either side of the center of the road is about 9 feet 2 inches.

Now, if you wanted to pull a tape from the center of the road to that point on the arch we just calculated, you could do the Pythagoras thing.

\(\displaystyle \L\\\sqrt{6^{2}+84}=2\sqrt{30}=\approx{10.95}\;\ ft\)

About 10' 11-3/8"

I hope this is on the lines of your query.
 
first reply was more like what i needed and i figured out what i needed from that. Thanks for both your help though.
 
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