Reading this paper for a project. In section 2.1 it says an approximate formula for the smooth curve described by the edge of the ski
is y ≈ $x^2/2R_{SC} − d$. Why is the $x^2$ value divided by $2R_{SC}$? I get that they are trying to fit a parabola to a circle of radius $R_{SC}$ and chord length $C$, but I don't understand what geometry they did to divide by $2R_{SC}$. Thanks for any help.
[Math] Fitting a Circle Arc to a Parabola
algebra-precalculuscalculusgeometry
Related Solutions
(I had been meaning to blog about roulettes a while back, but since this question came up, I'll write about this topic here.)
I'll use the parametric representation
$$\begin{pmatrix}2at\\at^2\end{pmatrix}$$
for a parabola opening upwards, where $a$ is the focal length, or the length of the segment joining the parabola's vertex and focus. The arclength function corresponding to this parametrization is $s(t)=a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))$.
user8268 gave a derivation for the "cycloidal" case, and Willie used unit-speed machinery, so I'll handle the generalization to the "trochoidal case", where the tracing point is not necessarily on the rolling circle's circumference.
Willie's comment shows how you should consider the notion of "rolling" in deriving the parametric equations: a rotation (about the wheel's center) followed by a rotation/translation. The first key is to consider that the amount of rotation needed for your "wheel" to roll should be equivalent to the arclength along the "base curve" (in your case, the parabola).
I'll start with a parametrization of a circle of radius $r$ tangent to the horizontal axis at the origin:
$$\begin{pmatrix}-r\sin\;u\\r-r\cos\;u\end{pmatrix}$$
This parametrization of the circle was designed such that a positive value of the parameter $u$ corresponds to a clockwise rotation of the wheel, and the origin corresponds to the parameter value $u=0$.
The arclength function for this circle is $ru$; for rolling this circle, we obtain the equivalence
$$ru=s(t)-s(c)$$
where $c$ is the parameter value corresponding to the point on the base curve where the rolling starts. Solving for $u$ and substituting the resulting expression into the circle equations yields
$$\begin{pmatrix}-r\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-r\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$
So far, this is for the "cycloidal" case, where the tracing point is on the circumference. To obtain the "trochoidal" case, what is needed is to replace the $r$ multiplying the trigonometric functions with the quantity $hr$, the distance of the tracing point from the center of the rolling circle:
$$\begin{pmatrix}-hr\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-hr\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$
At this point, I note that $r$ here can be a positive or a negative quantity. For your "parabolic trochoid", negative $r$ corresponds to the circle rolling outside the parabola and positive $r$ corresponds to rolling inside the parabola. $h=1$ is the "cycloidal" case; $h > 1$ is the "prolate" case (tracing point outside the rolling circle), and $0 < h < 1$ is the "curtate" case (tracing point within the rolling circle).
That only takes care of the rotation corresponding to "rolling"; to get the circle into the proper position, a further rotation and a translation has to be done. The further rotation needed is a rotation by the tangential angle $\phi$, where for a parametrically-represented curve $(f(t)\quad g(t))^T$, $\tan\;\phi=\frac{g^\prime(t)}{f^\prime(t)}$. (In words: $\phi$ is the angle the tangent of the curve at a given $t$ value makes with the horizontal axis.)
We then substitute the expression for $\phi$ into the anticlockwise rotation matrix
$$\begin{pmatrix}\cos\;\phi&-\sin\;\phi\\\sin\;\phi&\cos\;\phi\end{pmatrix}$$
which yields
$$\begin{pmatrix}\frac{f^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}&-\frac{g^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\\\frac{g^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}&\frac{f^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\end{pmatrix}$$
For the parabola as I had parametrized it, the tangential angle rotation matrix is
$$\begin{pmatrix}\frac1{\sqrt{1+t^2}}&-\frac{t}{\sqrt{1+t^2}}\\\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}$$
This rotation matrix can be multiplied with the "transformed circle" and then translated by the vector $(f(t)\quad g(t))^T$, finally resulting in the expression
$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}+\frac1{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\begin{pmatrix}f^\prime(t)&-g^\prime(t)\\g^\prime(t)&f^\prime(t)\end{pmatrix}\begin{pmatrix}-hr\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-hr\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$
for a trochoidal curve. (What those last two transformations do, in words, is to rotate and shift the rolling circle appropriately such that the rolling circle touches an appropriate point on the base curve.)
Using this formula, the parametric equations for the "parabolic trochoid" (with starting point at the vertex, $c=0$) are
$$\begin{align*}x&=2at+\frac{r}{\sqrt{1+t^2}}\left(ht\cos\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)-t-h\sin\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)\right)\\y&=at^2-\frac{r}{\sqrt{1+t^2}}\left(h\cos\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)+ht\sin\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)-1\right)\end{align*}$$
A further generalization to a space curve can be made if the rolling circle is not coplanar to the parabola; I'll leave the derivation to the interested reader (hint: rotate the "transformed" rolling circle equation about the x-axis before applying the other transformations).
Now, for some plots:
For this picture, I used a focal length $a=1$ and a radius $r=\frac34$ (negative for the "outer" ones and positive for the "inner" ones). The curtate, cycloidal, and prolate cases correspond to $h=\frac12,1,\frac32$.
(added 5/2/2011)
I did promise to include animations and code, so here's a bunch of GIFs I had previously made in Mathematica 5.2:
Inner parabolic cycloid, $a=1,\;r=\frac34\;h=1$
Curtate inner parabolic trochoid, $a=1,\;r=\frac34\;h=\frac12$
Prolate inner parabolic trochoid, $a=1,\;r=\frac34\;h=\frac32$
Outer parabolic cycloid, $a=1,\;r=-\frac34\;h=1$
Curtate outer parabolic trochoid, $a=1,\;r=-\frac34\;h=\frac12$
Prolate outer parabolic trochoid, $a=1,\;r=-\frac34\;h=\frac32$
The Mathematica code (unoptimized, sorry) is a bit too long to reproduce; those who want to experiment with parabolic trochoids can obtain a notebook from me upon request.
As a final bonus, here is an animation of a three-dimensional generalization of the prolate parabolic trochoid:
Another proof: Since both CI and BD are perpendicular to AB, CI is parallel to BD. The fact that C is the midpoint of AD implies that I is the midpoint of AB.
Best Answer
Write the equation of the circle as $x^2 + (y - (R - d))^2 = R^2$. Divide through by $R^2$ and rewrite everything in terms of the dimensionless quantities $\xi = x/R$, $\eta = y/R$, and $\epsilon = d/R$.
Consider $\eta$ and $\epsilon$ to be small quantities. Expand everything out and drop all second-order terms (like $\epsilon^2$, $\eta\epsilon$, etc.). You should get $y = {x^2\over 2R} - d$.
To spell it out: divide through by $R^2$ in $x^2 + (y - (R - d))^2 = R^2$ to get $$\left({x\over R}\right)^2 + \left({y\over R} - \left(1 - {d\over R}\right)\right)^2 = 1\,.$$ Define the variables I mentioned above: $$\xi^2 + (\eta - (1-\epsilon))^2 = 1\,.$$ Expand everything out: $$\xi^2 + \eta^2 - 2\eta + 2\eta\epsilon + 1 - 2\epsilon + \epsilon^2 = 1\,.$$ Drop $\eta^2$, $\eta\epsilon$, and $\epsilon^2$: $$\xi^2 - 2\eta - 2\epsilon = 0\,.$$ Rearrange as $$\eta = {\xi^2\over 2} - \epsilon\,.$$ Multiply through by $R$ to get $y = {x^2\over 2R} - d$.