Equation for the angular position of a slider crank mechanism

[Mech_LS24]

TL;DR Summary

I can't figure out how to setup a equation for the angle position of a slider crank mechanism. As the mechanism is driven by an angular velocity, the angle (theta) changes constantly. Theta is the angle which rotates the crank. I would like find the angle α (alpha) during the rotation of crank.

Hi!

I would like to know how I could define an equation for α with given the two lengths of the rods and angle theta (θ). I sketched the situation below, the problem arises when the rod Lab has pasts the 180 degrees.

Like to hear.

 

[anuttarasammyak]

It seems
\alpha+\gamma+2\pi-\theta_2=\pi.

 

[Mech_LS24]

It seems
α+γ+2π−θ2=π.

For θ2 you are correct (angles in radians). But using the same formula for θ1 doesn't work.

I would like to use θ as input in a worksheet, therefore one formula is demanded. So, the input will be angles between 0 and 360 degrees. When is larger as 180 there should be added a minus sign in the formula, this is what I want to avoid. But I can't figure out that formula because I get the errors with + and - signs which is also seen here with this formula.

 

[anuttarasammyak]

If you want common equation for $\theta_1$ and $\theta_2$ you should set up a systematic rule to measure all the angles like $\theta_1$ and $\theta_2$ are both measured counterclockwise from x axis. In your figure I see two $\alpha$ angles and two $\gamma$ angles. In which way of measuring rule these angles show these same values ?

 

[Mech_LS24]

What do you mean with systematic way? I got two constants, Lab and Lbc, the two rod which moves the 'piston'. That should be used for the input of the angles?

In your post I see red text with yellow background?

 

[anuttarasammyak]

I mean
How do you define $\angle ACB$ for B upside and B downside ?
How do you define $\angle ABC$ for B upside and B downside ?
How do you define $\angle CAB$ for B upside and B downside ?

 

[Mech_LS24]

Sorry, you're right, I am going to draw it. Please give me a moment

 

[Mech_LS24]

The sketches below is what I mean. In situation 2, I need to add an α' for getting the right γ-angle.

I changed the definitions of the angles, to correspond with the Law of sinus.

 

[anuttarasammyak]

Do you mean same equation holds for the same two angles and large / small $\alpha$ ?

 

[Mech_LS24]

Do you mean same equation holds for the same two angles and large / small $\alpha$ ?

Yes, I need angle β for calculating the velocity for rod Lbc.

 

[Mech_LS24]

Maybe this sketch gives a better indication. I need to know the angle between Vb (velocity of the rod) and Vb (tangent velocity of rod AB, so perpendicular with rod AB). This angle is depended on the input angle θ.

Could you help me with finding an equation for that?

 

[anuttarasammyak]

Though it might be not along your way, an analytic way is by introducing xy coordinate
A(0,0)
B(L_{AB}\cos\theta,L_{AB}\sin\theta)
C(L_{AC}(\theta),0)
and the equation is
(L_{AC}(\theta)-L_{AB}\cos\theta)^2+L_{AB}^2\sin^2\theta=L_{BC}^2
L_{AC}(\theta)=L_{AB}\cos\theta+\sqrt{L_{BC}^2-L_{AB}^2\sin^2\theta}
The velocity of rod is
\frac{d}{dt}L_{AC}(\theta)=\frac{d}{d\theta}L_{AC}(\theta)\ \ \dot{\theta}

 

[Mech_LS24]

Thanks,

What makes the x-coordinate of C the formula below? Shouldn't it not just be Lac?
C(L_{AC}(\theta),0)

And the equation should be without the squares under the root, like below?:
L_{AC}(\theta)=L_{AB}\cos\theta+\sqrt{L_{BC}-L_{AB}\sin\theta}

For example:
3^2 = 5^2 - 4^2
3 = sqrt((5^2) - (4^2)) doesn't make sense.

And, how to get that angle at point B in a single equation?

How did you come to that velocity differential equation? Could you give an example how to differentiate that velocity?

 

[anuttarasammyak]

Shouldn't it not just be Lac?

I wrote it to make it clear that it is a function of $\theta$.

Imagine what happens in your sketch when you choose minus sign.

$L_{AC}$ given, you can get $\angle ABC$ by sine law.

$\frac{df(x)}{dt}=\frac{df}{dx}\frac{dx}{dt}$ is the formula used. $L_{AC}$ is a function of $\theta$ and $\theta$ is a function of time.

 

[Mech_LS24]

I wrote it to make it clear that it is a function of θ.

, I just wanted to be sure.

LAC given, you can get ∠ABC by sine law.

df(x)dt=dfdxdxdt is the formula used. LAC is a function of θ and θ is a function of time.

Good point, I am going to draw and calculate on scale for different situations later today, thanks!

 

[Mech_LS24]

I have made the drawings on scale (with 4 different positions) and written the calculation for Lac as well. I ended up with the same equation as you for the length of Lac, so I do understand it. But I am issuing two problems now:

1. With the law of sines I end up with two solutions, where 99.7 degrees should be the right outcome. The only way I could find that angle is by first calculation the angle of γ and use the summation towards the 180 degrees.
2. An equation for the angle between Vb and Vb' is still hard to determine. As I have the issue that using 90 degrees as reference would result in minus values (see situation 2, on the page with the main sketches upper right).

Hopefully you get the message, I tried my best for the sketches and approaches towards the equations/calculations.

*Afterwards, I will look at the differential equation. I have some documentation about it now, so going to read that :)

 

[anuttarasammyak]

An equation for the angle between Vb and Vb' is still hard to determine. As I have the issue that using 90 degrees as reference would result in minus values (see situation 2, on the page with the main sketches upper right).

The angle you say seems $\angle ABC$ - $\pi/2$ where
\angle ABC = \sin^{-1} [\frac{L_{AB}\cos\theta+\sqrt{L_{BC}^2-L_{AB}^2\sin^2\theta}}{L_{BC}}\ sin\theta\ ]
Do you have a problem of its sign ?

 

[Mech_LS24]

You are right, I managed it to plot it in Excel, see result below.

Thanks for your help and active participation! :)

 

[Bence2002]

You are right, I managed it to plot it in Excel, see result below.

Thanks for your help and active participation! :)
View attachment 287140

I got the sam task as homework, but I am stuck in excel because of the negative values under the root..How did you manage to solve it?

 

[jack action]

I got the sam task as homework, but I am stuck in excel because of the negative values under the root..How did you manage to solve it?

The value under the root cannot be negative, it is a design constraint.

The only way it can be negative is if $L_{AB} \gt L_{BC}$. If that is the case, when $AB$ is at 90° with the cylinder $AC$, the link $BC$ wouldn't be long enough to reach the cylinder. The mechanism would therefore bind before reaching that point.

 

[Bence2002]

The value under the root cannot be negative, it is a design constraint.

The only way it can be negative is if $L_{AB} \gt L_{BC}$. If that is the case, when $AB$ is at 90° with the cylinder $AC$, the link $BC$ wouldn't be long enough to reach the cylinder. The mechanism would therefore bind before reaching that point.

Thank you!
I managed to solve it later, I made a stupid mistake
It's all good now :)