In English we have:


"beef" for "cow", "cattle"
"veal" for "calf"
"pork" for "pig"
"mutton" for "sheep"


I'm not aware of this separation for "fish", "goat" or "chicken" (Spanish has "pollo" and "gallina") and other poultry. Are these words used simply to distinguish the meat from the animal (i.e. to avoid saying "cow meat") or is there a psychological separation to avoid the association? I doubt the latter since these words developed when people were likely less squeamish than some are today. 

Why are there not meat words for some animals? 

What are some others I didn't list?


Question: A helical cycloid?
Answer: I'm adjusting my answer, which as TonyK points out was based on an incorrect assumption.  Hopefully it's correct now.

The center of the circle is not directly above the point of contact with the helix as my incorrect answer had asserted.  To see where it is, first picture the starting point at $(a,0,0)$, where the circle is parallel to the $y$-$z$ plane.  The tangent vector to the helix is in the direction of $(0,a,c)$, so the radius is perpendicular to this in the direction of $(0,-c,a)$.  So you have to add $\frac{r}{\sqrt{a^2+c^2}}(0,-c,a)$ to get to the center.  As the circle rolls along, the vector you add to get to the center will rotate around the $z$ axis to $\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.  So the center of the circle is parametrized by $(a\cos(t),a\sin(t),ct)+\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.

The arclength formula allows you to write down the rotation angle of the point about the center of the circle in terms of $t$ as $\frac{\sqrt{a^2+c^2}}{r}t$.  Let $b=\frac{\sqrt{a^2+c^2}}{r}$ from now on. 

If we pretend for a minute that the point was just going to rotate about the center of the circle at this rate while the circle remains fixed, and simplify the problem for now by assuming that the point starts above the center of the circle, then the vector we would add to the center of the circle to parametrize the point would be $(0,hr\sin(bt),hr\cos(bt))$.  When looking toward the $y$-$z$ plane from the positive $x$ axis, this motion would appear clockwise, as it should because the circle will be starting off to the right from this perspective.

But the vector we actually add to the center will also be rotated by $t$ about the $z$-axis, so putting it all together (with minimal simplification) we get 

$\begin{align*}
x&amp;=a\cos(t)+\frac{c}{b}\sin(t)-hr\sin(t)\sin(bt)\\
y&amp;=a\sin(t)-\frac{c}{b}\cos(t)+hr\cos(t)\sin(bt)\\
z&amp;=ct+\frac{a}{b}+hr\cos(bt).
\end{align*}$

Now to adjust the starting point, you can change $bt$ to $b(t-t_0)$ for suitable $t_0$.