## Step 6: Release Mechanism Math - Part 2

So far, we have the dimensions that we need to ensure that each arm grabs the pendulum at the right location and releases it at the right location. Now, we need to make sure that all arms release all pendulums at the same time.

After doing a little sketching, I came up with a simple mechanism that works. Other than the arm, there are two components in this mechanism. To mess with the electrical engineers, I will call them the rotor and stator.

If the picture doesn't do it justice, here's a breakdown of how the thing works:

Each arm (red) has a slot cut into its side. The rotor (green) has two dowels, one on each side, which extend into the slots of both neighboring arms. It is pinned to the backs of two stators (blue). All rotors are attached to a dowel at a constant distance from the stator pin. Thus, when one rotor moves, all rotors move at the same angular velocity. With the right dimensions for the rotors and stators (the arm dimensions have already been calculated), all pendulums can be released at the same time.

The setup for this problem is easier than it might sound.

Solving for z and Q

We need to find values for z and Q such that at a given angle, Ï, the tip of the arm is a horizontal distance L*sin(θ) from the pendulum attachment point.

We already have S from the previous problem. J is an arbitrary value (it determines the width of the pendulum wave frame).
R_c is a value that we will solve for later. It is the distance from the pendulum release point to the line S. It runs parallel to z.

To solve this problem, we will use 3 equations:

z/h = R_c/H
Q*sin(Ï) = h
J = Q*cos(Ï) + SQRT(z^2-h^2)

We solve:

z/h = R_c/H => h = z*H/R_c;

Q*sin(Ï) = h => Q = z*H/(R_c*sin(Ï));

J = Q*cos(Ï) + SQRT(z^2-h^2) => J = z*H*cos(Ï)/(R_c*sin(Ï)) + SQRT(z^2-(z*H/R_c)^2)
=> J = z*H*cot(Ï)/R_c + z*SQRT(R_c^2-H^2)/R_c => J = (z/R_c)*[H*cot(Ï) + SQRT(R_c^2-H^2)]
=> z = J/(R_c*(SQRT(R_c^2-H^2) + H*cot(Ï)))

And

Q = z*H/(R_c*sin(Ï))

Solving for R_c

R_c is the sum of s (not big S from the previous problems) and the edge length of the arm (not shown) beginning at the pin. As mentioned on the previous step, the actual length of the arm is different than the value R we calculated previously. Additionally, the angle Ï is different than the angle Ï just calculated above. Sorry for any confusion.

The actual length, by Pythagoras, is:

R_actual= SQRT((R^2-t^2/4)

where t is the thickness of the arm.

So,

R_c = R_actual + s.

In order to find s, we need to find the value for α, the angle opposite of s.

From the diagram, we see that α = 180 - (Ï + 90 - β/2) => α = 90 - Ï + β/2

where Ï = asin(h/R), and β/2 = asin(t/(2*r)).

So, α = 90 - asin(h/R) + asin(t/(2*r)).

tan(α) = 2*s/t => s = t*tan(α)/2 => s = t*tan(90 - asin(h/R) + asin(t/(2*R)))/2

Thus,

R_c = SQRT(R^2-t^2/4) + t*tan(90 - asin(h/R) + asin(t/(2*R)))/2

For my pendulum wave, I chose the release thickness (t) to be 1/2".

We designed each arm in the release mechanism to let go of its respective pendulum when the rotor is at a constant angle (Ï). Thus, each rotor can be coupled to all other rotors. The implication of this is that since all rotors (when coupled) have the same angular velocity, and all pendulums will be released when the rotor is at angle Ï, all pendulums will be released at the same time.
<p><a href="http://fy.chalmers.se/%7Ef7xiz/TIF080/pendulum.pdf." rel="nofollow">http://fy.chalmers.se/~f7xiz/TIF080/pendulum.pdf.</a></p><p>it's not available anymore???</p>
<p>I love your top view design! How unique! Have you seen this giant version made by the STEAM Carnival &amp; Two Bit Circus? <iframe allowfullscreen="" frameborder="0" height="281" src="//www.youtube.com/embed/YbGev0--ryk" width="500"></iframe>I bet a life-size version of yours would be AMAZING!</p>
<p>How much would you charge to make one of these. my fianc&eacute; loves it. Otherwise I'll make a run to hobby lobby and break out the tools.</p>
Very cool release mechanism. I must say though, all that math is making that sponge thing in my head throb.
Thanks for the feedback! Yes, I was a little worried about posting the math... In hindsight, it might not have been a bad idea to have the last page an appendix of sorts, dedicated solely to math for those interested. I know it's a lot to wade through. <br> <br>Btw, you can always skip directly to the end for the bill of materials :)
And whos says science isn't fun!
Wow! That is amazing! You must be a genius or something with all of that math stuff, but I think I'll just follow your dimensions. This is a great instructable.
Ha, thanks! The dimensions should get you through. If you wait a couple months, I should have another instructable up detailing construction. However, go ahead and give it a try if you want. Feel free to ask if you have any questions!
Beautiful! That's a great science-on-the-beach exhibit :-)
Why thank you! <br> <br>Sand might result in a less-than-best viewing surface, however. ;)