Hi
I ask for a cartesian equation of 3-BARS CURVE...
Or at least the parametric equation...
Or at least the polar equation...
Help me, please
Hi Ackbeet!
We consider an articulate quadrilateral ABCD, with a fixed bar AB, a bar AC (first crank), a bar BD (second crank), and a bar CD (connecting rod).
The 3-bars curved is the curve traced from a certain point on the connecting rod to rotating of the first crank...
Thanks very much for the picture. That's very helpful.
So, let me get this straight: AC is rotating about A, and BD is rotating about B. You have a variable length connecting rod from C to D. You are interested in obtaining the path that a particular point (call it E) traces out as the two cranks rotate. Is this correct? If so, I have a number of questions for you:
1. At what rate is AB rotating?
2. At what rate is BD rotating?
3. Exactly what point E are you interested in?
4. I note that you've drawn AC < BD. Are you merely indicating that the two cranks are not necessarily the same length?
5. Are AB and BD rotating "in phase"?
Let's suppose the following: $\displaystyle A$ is the origin, $\displaystyle E$ is the midpoint of $\displaystyle CD$, $\displaystyle AC$ rotates with constant angular velocity $\displaystyle \omega_{1}$, and $\displaystyle BD$ rotates with constant angular velocity $\displaystyle \omega_{2}$. Then point $\displaystyle C$ has the coordinates
$\displaystyle \vec{AC}=AC\langle\cos(\omega_{1}t),\sin(\omega_{1 }t)\rangle$ and the point $\displaystyle D$, has the coordinates
$\displaystyle \vec{AD}=\vec{AB}+BD\langle\cos(\omega_{2}t+\delta ),\sin(\omega_{2}t+\delta)\rangle.$
You can see that the angle $\displaystyle \delta$ I've included in order to allow any phase difference.
The midpoint $\displaystyle E$ would then have coordinates
$\displaystyle \dfrac{1}{2}(\vec{AC}+\vec{AD}).$
HI Akbeet and Opalg
I thank you for your interest to my problem…
I wanted to clarify to Ackbeet that the rotations of two cranks AC and BD are not independent, but that of BD depends from that of AC.
Just, if I make to rotate crank AC, it moves the connecting rod CD that, in its turn, can make to rotate or oscillate other crank BD.
Moreover AB, BD, CD, CE are not prefixed, but they are parameters.
And the curve is only traced for details values of these parameters (the connecting rod must have fixed length).
Therefore placing AC=a, CD= b, BD=c, CE=d, the curve has an equation (cartesian implicit) of the type
f (x, y, a, b, c, d) =0.
And this is the equation that I look for!
Opelg then points out a generalization of my problem, in which E it is not on the connecting rod, but, loyal with it, to a sure distance, that it increases of an other parameter the curve…
I think that also in this case the equation, even if complicated, exists,
You're also going to need to call the length $\displaystyle AB$ something. I'll call it $\displaystyle f$. Ok, with your updated information, here are my thoughts:
Choose the coordinate system so that point $\displaystyle A$ is the origin, and vector $\displaystyle \vec{AB}=\langle f,0\rangle.$ Now the only real differences between what I did in my last post (post # 8) is that point $\displaystyle E$ is not the midpoint of $\displaystyle CD$, and I'm also thinking that the angle of length $\displaystyle c$ may not be uniform angular motion. Instead, point $\displaystyle E$ is a distance $\displaystyle d$ from $\displaystyle C$. Therefore, point $\displaystyle E$ is a percentage $\displaystyle \tfrac{d}{b}$ of the distance along $\displaystyle CD$, right? Therefore, we can write
$\displaystyle \vec{AE}=\dfrac{d}{b}\,\vec{AC}+\left(1-\dfrac{d}{b}\right)\vec{AD}.$
This checks out. If $\displaystyle d=b$, then this equation reduces to $\displaystyle \vec{AE}=\vec{AC}.$ On the other hand, if $\displaystyle d=0,$ then this equation reduces to $\displaystyle \vec{AE}=\vec{AD}.$
It seems to me that there could be some mechanical limitations on this system. Let's assume, as per your drawing, that $\displaystyle a < c.$ What if $\displaystyle c+f-a>b?$ Then this says that if length $\displaystyle a$ is in the positive $\displaystyle x$ direction, its most extreme position, you're not going to be able to get length $\displaystyle c$ to the positive x direction. In order for this mechanical system to work properly, you'd probably need to have $\displaystyle a+b=c+f.$
So let's assume the mechanical aspects all work out. We will have the following equations:
$\displaystyle \vec{AC}=a\,\langle\cos(\omega_{1}t),\sin(\omega_{ 1}t)\rangle$
$\displaystyle \vec{AD}=\langle f,0\rangle+c\,\langle\cos(\theta),\sin(\theta)\ran gle,$ and
$\displaystyle \vec{AE}=
\dfrac{d}{b}\,\vec{AC}+\left(1-\dfrac{d}{b}\right)\vec{AD}=
\dfrac{ad}{b}\,\langle\cos(\omega_{1}t),\sin(\omeg a_{1}t)\rangle+\left(1-\dfrac{d}{b}\right)\left(\langle f,0\rangle+c\,\langle\cos(\theta),\sin(\theta)\ran gle\right).$
The one piece missing from this set of equations is an expression for $\displaystyle \theta.$ You could think of $\displaystyle \theta$ as one of two solutions the intersection of two circles: one centered at $\displaystyle C$ with radius $\displaystyle b$, and one centered at $\displaystyle B$ with radius $\displaystyle c$.
So these are some ideas. They are essentially equivalent, I think, to the link Opalg provided.
I thank you very for your information, than me they will be useful for my voice on the sextic curves...
I wanted to wonder to you now, if you can generalize the problem, like suggested from Opalg, putting the point E not in the straight PQ, ma on triangle CED, like indicated in attached one..
May you?
Sure you can. See here for an example of that. Your triangle might not be an equilateral triangle, but use whatever angles you have in order to do the rotations required.I wanted to wonder to you now, if you can generalize the problem, like suggested from Opalg, putting the point E not in the straight PQ, ma on triangle CED, like indicated in attached one..
Thanks!
With your information I have created a 3bars animation (see attachment)
3barre parametrica LINES.zip