Suppose that an ellipse is rolling along a line. If we follow the path of one of the foci of the ellipse as it rolls, then this path formes a curve - namely an undulary. Now consider the following diagram

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The points $F$ and $F'$ are the forci of the ellipse, which is rolling along the line $KT$. The line $FT$ is the tangent to the curve traced out by $F$. Now the following folds:

$FT$ is is perpendicular to $FK$, so the normal to the locus of $F$ passes through $K$.

Why is this always the case? I read this in 3 several books, but every time without a proof. Is it that obvious? Can anyone explain it or give a proof/source?

Best regards!

  • $\begingroup$ In "Differential Geometry and It's Applications" from John Oprea I read the following sentence: "A general property of this sort of roulette is the fact that the direction of the normal to the curve traced by F passes through the point of contact with the line." That is exactly the property I want to proof. Any ideas? $\endgroup$ – Paul85 Nov 18 '14 at 9:22
  • $\begingroup$ $K$ is the center of the "infinitesimal rotation" of the point $F$. So the velocity of $F$ is perpendicular to $FK$. $\endgroup$ – Emanuele Paolini Nov 18 '14 at 9:25

Let $F(t)$ be the curve described by point $F$ and $K(t)$ the curve described by point $K$. When $K$ is on the line (ground) its velocity is zero by assumption (rolling on the line): $$ K'(t) = 0. $$ But we know that $|F(t)-K(t)|$ is constant (the ellipse is rigid) so: $$ 0 = \frac{d}{dt} (F(t)-K(t))^2 = (F(t)-K(t)) \cdot (F'(t) - K'(t)) = (F(t) -K(t)) \cdot F'(t) $$ which means that the velocity of $F$ (which is along the tangent line to the curve $F(t)$) is orthogonal to the segment $FK$.

  • $\begingroup$ First of all thanks for your answer. I've an additional question. You said that $|F(t) - K(t)|$ is constant. Is it? I think we only know that $|F_{1}(t) - K(t)| + |F_{2}(t) - K(t)|$ is constant and equal to $2a$ where $a$ is the length of the major axis, by definition of the ellipse (and $F_{1}$, $F_{2}$ are the forci of the ellipse). $\endgroup$ – Paul85 Nov 18 '14 at 13:06
  • $\begingroup$ $F$ and $K$ are fixed on the ellipse, so their distance is fixed. At time $t$ it happens that $K(t)$ is on the line, at other times the tangent point will be different. $\endgroup$ – Emanuele Paolini Nov 18 '14 at 13:18

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