# 3-bar equation

• Sep 23rd 2010, 09:40 PM
scifo
3-bar equation
Hi(Hi)
I ask for a cartesian equation of 3-BARS CURVE...(Wait)
Or at least the parametric equation...(Wait)
Or at least the polar equation...(Wait)
• Sep 24th 2010, 05:35 AM
Ackbeet
What in tarnation is a 3-bars curve? Can you describe in more detail?
• Sep 24th 2010, 07:38 AM
scifo
3-bars definition
(Hi)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...
• Sep 24th 2010, 07:40 AM
Ackbeet
Hmm. A picture would be very helpful here. Could you please include one?
• Sep 24th 2010, 08:13 AM
scifo
A 3-bar image
(Hi)
Voila!
• Sep 24th 2010, 08:23 AM
Ackbeet
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"?
• Sep 24th 2010, 08:51 AM
Opalg
This is a complicated problem. I doubt whether an analytic solution is feasible in general, whether in cartesian, parametric or polar form. There is an interesting web page here with an interactive program for numerical computations.
• Sep 24th 2010, 09:25 AM
Ackbeet
Let's suppose the following: $A$ is the origin, $E$ is the midpoint of $CD$, $AC$ rotates with constant angular velocity $\omega_{1}$, and $BD$ rotates with constant angular velocity $\omega_{2}$. Then point $C$ has the coordinates

$\vec{AC}=AC\langle\cos(\omega_{1}t),\sin(\omega_{1 }t)\rangle$ and the point $D$, has the coordinates

$\vec{AD}=\vec{AB}+BD\langle\cos(\omega_{2}t+\delta ),\sin(\omega_{2}t+\delta)\rangle.$

You can see that the angle $\delta$ I've included in order to allow any phase difference.

The midpoint $E$ would then have coordinates

$\dfrac{1}{2}(\vec{AC}+\vec{AD}).$
• Sep 25th 2010, 05:03 AM
scifo
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,
• Sep 27th 2010, 05:55 AM
Ackbeet
You're also going to need to call the length $AB$ something. I'll call it $f$. Ok, with your updated information, here are my thoughts:

Choose the coordinate system so that point $A$ is the origin, and vector $\vec{AB}=\langle f,0\rangle.$ Now the only real differences between what I did in my last post (post # 8) is that point $E$ is not the midpoint of $CD$, and I'm also thinking that the angle of length $c$ may not be uniform angular motion. Instead, point $E$ is a distance $d$ from $C$. Therefore, point $E$ is a percentage $\tfrac{d}{b}$ of the distance along $CD$, right? Therefore, we can write

$\vec{AE}=\dfrac{d}{b}\,\vec{AC}+\left(1-\dfrac{d}{b}\right)\vec{AD}.$

This checks out. If $d=b$, then this equation reduces to $\vec{AE}=\vec{AC}.$ On the other hand, if $d=0,$ then this equation reduces to $\vec{AE}=\vec{AD}.$

It seems to me that there could be some mechanical limitations on this system. Let's assume, as per your drawing, that $a < c.$ What if $c+f-a>b?$ Then this says that if length $a$ is in the positive $x$ direction, its most extreme position, you're not going to be able to get length $c$ to the positive x direction. In order for this mechanical system to work properly, you'd probably need to have $a+b=c+f.$

So let's assume the mechanical aspects all work out. We will have the following equations:

$\vec{AC}=a\,\langle\cos(\omega_{1}t),\sin(\omega_{ 1}t)\rangle$

$\vec{AD}=\langle f,0\rangle+c\,\langle\cos(\theta),\sin(\theta)\ran gle,$ and

$\vec{AE}=

The one piece missing from this set of equations is an expression for $\theta.$ You could think of $\theta$ as one of two solutions the intersection of two circles: one centered at $C$ with radius $b$, and one centered at $B$ with radius $c$.

So these are some ideas. They are essentially equivalent, I think, to the link Opalg provided.
• Sep 28th 2010, 11:22 PM
scifo
3-bars generalized
(Hi)I thank you very for your information, than me they will be useful for my voice on the sextic curves...(Clapping)
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?(Worried)
• Sep 29th 2010, 05:51 AM
Ackbeet
Quote:

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..
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.
• Oct 11th 2010, 12:31 AM
scifo
(Hi)Thanks!(Clapping)
With your information I have created a 3bars animation (see attachment)

Attachment 19270
• Oct 11th 2010, 12:40 AM
scifo
In the attachment :
click to OK for a step by step motion...
press to ENTER for the continuous motion...
click to CANCEL for exit...
• Oct 11th 2010, 05:54 AM
Ackbeet
Nice animation. I assume the slight inaccuracy in the left circle is insignificant? The tip of the arm does not seem to track with the circle that is already drawn there.