Branching Spring Arm discussion from Team 1107 reveal post 34
While I’m primarily reponsding to Murdomeek, anyone should feel free to chime in with corrections to my math, or questions.
Starting with basic physics:
Spring Potential Energy (SPE) kxx/2; where k is spring constant, x is displacement
The force from a spring is kx;
Gravity Potential Energy (GPE) for a mass (arm) at h (height) is mgh
The force from gravity is (also known as weight), is mg.
Torque around a pivot = (length of lever arm) * force; eg inch*pounds
SPE can be traded for GPE with the correct mechanical connection.
The arm motor is only needed to accellerate the arm,
and make up for friction and imperfect elastic losses,
and (of course) to lift the additional mass of the cargo.
Any disagreement so far in principle and equations for SPE and GPE?
Next step (and last for today),
calculate (or build example of) the correct mechanical connection for a single arm location,
where the net torque on the arm axle is 0, when the arm is horizontal.
The arm is modeled as a point-mass m, at the end of a massless arm length L, with the shoulder pivot at Y above the ground, with arm angle theta, where theta =
-90 degrees as straight down, cos(-90) = 0;
00 degrees as straight out parallel to the ground, cos(0) = 1
+90 degrees as straight up, cos(90) = 0;
The usual arm range is something like -45 to +45, for a total swing of ~ 90 degrees.
This means the horizontal distance from mass to pivot is cos(theta) * L
The spring is connected to a chain, which wraps around a sprocket.
The radius R of the sprocket is the length of the lever arm for the spring.
At the 0 degree position, the arm torque on the pivot is mg*L*cos(0) = mgL
This torque needs to be balanced by (spring force) * (spring lever arm).
Like so: mgL = kxR
So it should be obvious that you can change number of rubber bands (to change k),
or change how far they are stretched (x, the red bar by the pic of rubber band)
or the Radius of the sprocket ®,
until this makes the arm balanced at the horizontal position.
After all these words, I hacked up a picture to represent all the terms:
The rubber band is circled at the left, but doesn’t show up well in the picture.
The blue bar represents the unstretched default length of the rubber band.
The red bar shows the length of x, the amount of stretch.
Now you can bounce it upward with your finger, but it will settle back to the same position.
It will balance with 20 rubber bands (big k) with very small x (stretch),
or 1 rubber band (small k) with a bigger x.
The full form: mgL*cos(theta) = kxR, has 2 variables theta and x;
Since x is a function of theta (more theta rotation winds up the chain and increases x stretch)
there is only one solution, which is the balance point.