# Unique Pendulum Wave and Release Mechanism

10 Steps

## Step 5: Release Mechanism Math - Part 1

In order for the release mechanism to work, each release has to do two things: grab the pendulum, then release at the same time when all the pendulums are the same distance (Lsinθ) away from the support beam.

Before dealing with the timing for each release, let's figure out the geometry first. We need to find the length of the release arm and the horizontal distance between the release arm and the pendulum attachment point.

The pendulum scenario is depicted in the diagram below:

In the figure,

L is the distance from the pendulum support beam to the center of the pendulum bob. This value is known for each pendulum.

W is the horizontal distance from the pendulum attachment point to the center of the pendulum bob at release (L*sin(θ), anyone?).

R is the distance from the arm attachment point to the underside of the pendulum bob. This value is very unknown.

D is the distance that I want the arm to extend past the bob when the arm collects it. For my pendulum wave, I chose 0.5".

S is the horizontal distance from the pendulum attachment point to the arm attachment point. This value is also very unknown.

C is the vertical distance from the arm attachment point to the bottom of the pendulum bob. This value is selected arbitrarily (I chose 4").

H is the vertical distance the arm attachment to the bottom of the pendulum bob at release. From inspection, this value is L-L*cos(θ)+C.

Solving for R and S

We define the arm by two values: the length of the arm (R+D) and its required offset (S). Lets solve for these values in terms of the driving (known) variables.

For starters, we know that the distance W is equal the the sum of S and the horizontal distance between S and the pendulum release point. By using Pythagorean's theorem, we can say that

W=SQRT(R^2-C^2)+SQRT((R+D)^2-H^2)

Squaring both sides of this equation and isolating zero gives the quadratic equation:

((W^2-R^2+C^2-(R+D)^2+H^2)^2)/4-(R^2-C^2)((R+D)^2-H^2)=0

This is a painful calculation to solve by hand, so I suggest using a program like Matlab. Here is how I solved this equation in Matlab for R:

>> syms R C H W D                                                                                               % Define variables
>> solve([W^2-R^2+C^2-(R+D)^2+H^2]^2/4-(R^2-C^2)*((R+D)^2-H^2),R)                         % Solve for R

ans =

(W*(C^4 - 2*C^2*D^2 - 2*C^2*H^2 + 2*C^2*W^2 + D^4 - 2*D^2*H^2 - 2*D^2*W^2 + H^4 + 2*H^2*W^2 + W^4)^(1/2) - C^2*D + D*H^2 + D*W^2 - D^3)/(2*D^2 - 2*W^2)

-(W*(C^4 - 2*C^2*D^2 - 2*C^2*H^2 + 2*C^2*W^2 + D^4 - 2*D^2*H^2 - 2*D^2*W^2 + H^4 + 2*H^2*W^2 + W^4)^(1/2) + C^2*D - D*H^2 - D*W^2 + D^3)/(2*D^2 - 2*W^2)

Note that Matlab gives us two answers because we solved a quadratic. Which solution do we choose? To decide, plug in reasonable, positive values into both expressions and see which solution gives us a reasonable, positive answer (the second equation).

Thus,

R = -(W*(C^4 - 2*C^2*D^2 - 2*C^2*H^2 + 2*C^2*W^2 + D^4 - 2*D^2*H^2 - 2*D^2*W^2 + H^4 + 2*H^2*W^2 + W^4)^(1/2) + C^2*D - D*H^2 - D*W^2 + D^3)/(2*D^2 - 2*W^2).

Adding D to this value (as shown in the diagram) gives us the diagonal length of the arm from the arm attachment point to the bottom of the pendulum. This is because the arm will be a rectangle rather than a line.

With R solved, we can now use Pythagorean Theorem to solve for S:

S = SQRT(R^2-C^2).

Simple enough.

Note that the diagonal distance from the center of the arm pin location to the pendulum at release is R+D. The actual length of the arm from the pin is

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

where t is the width of the arm.
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