parametric equations of a trochoid

[iDoof]

i have in my book the parametric equations for a trochoid (trochoid is the shape made by taking a point at distance d from the center of a circle with radius r, then tracing the path created by that point as the circle rolls along a straight line)

here are the equations:

x = rt - d*sin t;
y = r - d*cos t;

can someone help me figure out how this was derived (qualitatively)? i'm having trouble figuring out what each part of the equation means...does that make sense? i.e. what exactly does 't' mean here? some angle, i presume?...



thanks for your time and input everyone, i hope i make sense.

 

[iDoof]

ok i'll simplify the question:

what are the parametric equations for a trochoid?

 

[soroban]

Hello, iDoof!

i have in my book the parametric equations for a trochoid
(the shape made by taking a point at distance d from the center of a circle with radius r,
then tracing the path created by that point as the circle rolls along a straight line)

Here are the equations: .\(\displaystyle \begin{array}{cc}x\:=\:r\theta\,-\,d\cdot\sin\theta \\ y\:=\:r\,-\,d\cdot\cos\theta\end{array}\)

Can someone help me figure out how this was derived (qualitatively)?
i'm having trouble figuring out what each part of the equation means
Does that make sense? i.e. what exactly does 't' mean here? some angle, i presume? . yes!

        |   *             *
        |  *       C   r   *
        |  *       *-------*E
        |  *    d/θ|       *
        |   * P*---+B     *
        |    *     |     *
        |   D  *   |   *
        +---------***-----
        |O         A

The circle has center \(\displaystyle C\) and radius \(\displaystyle CD\,=\,CA\,=\,CE\,=\,r\).
Point \(\displaystyle P\) is a fixed distance \(\displaystyle d\) from \(\displaystyle C\).

At the beginning (when \(\displaystyle \theta\,=\,0\)), \(\displaystyle C\) is on the y-axis.
. . Point \(\displaystyle P \\)is \(\displaystyle d\) units below \(\displaystyle C\).

As the circle rolls to the right, radius \(\displaystyle CD\) turns through an angle of \(\displaystyle \theta\).
. . And point \(\displaystyle P\) is in the position seen in the diagram above.


Let us determine the \(\displaystyle x\)-coordinate of point \(\displaystyle P\).

We can see that: .\(\displaystyle x\;=\;OA\,-\,PB\)
. . \(\displaystyle OA\) is the distance the circle rolled to the right.
. . . . This is the length of arc \(\displaystyle DA\,=\,OA\,=\,r\theta\)
. . \(\displaystyle PB\) is in the right triangle \(\displaystyle PBC:\;PB\,=\,d\cdot\sin\theta\)

Hence: .\(\displaystyle x\;=\;r\theta\,-\,d\cdot\sin\theta\) . <--- half the answer!


Now determine the \(\displaystyle y\)-coordinate of point \(\displaystyle P\).
. . The "height" of point \(\displaystyle P\) is \(\displaystyle y\;=\;BA\;=\;CA\,-\,CB\)
. . We know that \(\displaystyle CA\,=\,r\)
. . From right triangle \(\displaystyle PCB:\;CB\,=\,d\cdot\cos\theta\)

Hence: .\(\displaystyle y\;=\;r\,-\,d\cdot\cos\theta\) . <--- the other half!