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.
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.
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