# Tag Info

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It all depends how tight your coiling is. If the wire doesn't touch itself, i.e. without any contact surface excluded with itself, then the area of the wire exposed is simply the area of the wire : $$A = 2\pi rL$$ which gives immediately the length of wire you wish (with $d$ the diameter of your wire) : $$L = \frac{A}{d\pi}$$ $$L = ... 0 The key idea here is that the volume of the wire before and after flattening it should be almost exactly the same, since you aren't removing or adding any metal, just smushing it around. If L is the length of the wire, r_{wire} is the radius of the wire, t is the thickness of the disk, and A is the desired area of the disk,$$V_{wire} = L ( \pi ...

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A circle is a degenerate ellipse, and you can also think of a circle as having two foci (on top of one another) as the eccentricity approaches 1. The foci of conic sections in general originate from the approach in which the curves are defined - using a focus (point) and directrix (line). This approach leads to rational parametric expressions for the conic ...

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The equation of an ellipse whose semi-major and semi-minor axes are parallel to the x and y axes is given by: $$(\frac{x-h}{a})^2+(\frac{y-k}{b})^2=1$$ (where $a$,$b$ are the lengths of its semi-major and semi-minor axes respectively.) A focus, $c$ is defined by $c^2=a^2-b^2$, and therefore there can be two foci at a distance of $(a^2-b^2)^{1/2}$ on both ...

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The foci are simply points that define the ellipse by the relation $c^2 = a^2 - b^2$, where $c$ equals the length of each one of the foci to the center and $a$ is the length of a focus to the end of the ellipse. For a circle, $a$ = $b$. Given any two foci, a point on the ellipse is a point that is equal I he sum of he lengths of the foci.

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If you're both going at a constant speed, lets say 35 MPH, then it would appear so. This is because the distance is shorter in the inner arc in which you are traveling compared to the outer arc in which the other vehicle is traveling.

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This is really just a comment on Dargscisyhp's post - please don't upvote this. The equation Dargscisyhp has given you is a vector equation i.e. the $\vec{x}_i$s are vectors. You do the calculation by adding vectors. If you're uncomfortable with this then you can find the $(x_c, y_c, z_c)$ coordinates for the centre of mass by using three separate ...

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$$\vec{X}_{cm} = \frac{1}{M} \sum_i m_i \vec{x}_i$$ where $m_i$ is your mass of particle $i$, $\vec{x}_i$ is the position of particle i, and $M$ is your total mass.

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