Dyne's LD-F1 (D-O concept) build

[Dyne]

Does anyone understand levers well enough to double-check my logic here?


First, my belief is that LD-F1's lift acts as a third class lever (The Effort E is applied *between* the fulcrum F and the load L).  Because I didn't sufficiently consider this before buying the servo, I now need to find the Mechanical Advantage (M) of this lever for any given placement of the servo, to make sure that I position it where it is most likely to work successfully.  


I know the formula for M.  If A is the distance from Fulcrum to Effort, and B is the distance from Fulcrum to Load, then M = A / B.

This is also the inverse ratio for the forces acting at points A and B.  That is, M = L / E.  If we solve for E (the force needed to lift the weight at point L), then E = L / M.

Since I'm concerned with forces, a higher M is better.  The more that the load gets divided by, the better.  This also means that the closer A is to B, the better.

With a third class lever, however, M is always less than 1 (A is always less than B).  If M was greater than 1, then this wouldn't be a third class lever anymore.  If M equaled 1, it would mean that A and B are the same length, so the servo would be directly lifting the load, it could use its full power, and the lever would be irrelevant.  Obviously neither is the situation we have here.

E (the weight of L divided by M) must be below the maximum thrust of the servo.  Alternately, the maximum thrust multiplied by M must be above the weight of L.  Probably with a decent safety factor.


Figuring out length B is simple, since those are two known pivot points in my lift mechanism.  They are separated by 274.181 mm

The problem is that I found it a bit hard to understand what the value of A should be in my situation.  The "lever" is not a real physical part, and the effort isn't actually applied directly to it but rather to some point above it.  So I needed a generalized method to find A.  

I've came up with (and dismissed) a couple of low confidence ideas before I arrived at what I have below, so I hope it's correct.



My current thinking is based on the fact that the lever only cares where along its length the effort is applied.  Any given lever with the effort applied at an arbitrary distance from one end is exactly equivalent to a parallel lever of the same length with effort applied at the same relative distance.

This image may make the situation clearer:




The (non-physical) lever is the purple line FL, which has length B.  

The servo is the line between points E and P.  It has length 104 mm because the servo actually has minimum 102 and I applied a 2 mm buffer to avoid the endstop.  Neither point E nor P must necessarily be where they are shown, as that's what I'm trying to determine.  But E *must* fall somewhere along the black dashed line (which is just an offset from the real underside of the body shell -- the purple dashed line -- to account for mounting hardware).

As you can see, I've drawn a line parallel to line FL and with the same length.  That's the parallel lever.  This new line is the one that passes through point E, wherever E happens to be along the black dashed line in the setup I'm checking.  I've named its endpoints "vF" for virtual fulcrum, and "vL" for virtual load.  Now L, F, vF and vL form the vertices of a rectangle.  I don't suppose that it must be a rectangle, rather than an arbitrary parallelogram, but for drawing it out, this is the easiest way.

My thinking is that the line between E and vF is the effective value of A.  You can imagine projecting this point onto the lever (line FL) using a line drawn parallel to the small sides of the rectangle, if that helps.

Does this make any sense?  



In this diagram, this gives me A = 166.30.  With B fixed at 274.181, I get M = A/B = ~0.61.


The weight of the parts that I have currently assembled is around 1.25 kg or 2.67 lbs.  There are more parts to add, so let's estimate the final weight (L) as 3.5 lbs or 1.59 kg.  

Using E = L / M, as described above, we have E = 3.5 / 0.61 = 5.83 lbs.  The servo needs to apply 5.83 pounds of force to lift 3.5 lbs with this particular setup.  Since the most we can have is 11.5, that's well within spec, so the servo would work.

Specifically, it would work IF this servo has a long enough stroke in this arrangement to lift as far as I need it to, and it doesn't collide with any other parts or occupy any impossible positions.  I know already that it's problematic to put point P concentric with the center of the lower stalk axles, as shown here, because the hole size in the servo is about half the diameter of the axle.  Also, to reach the full 45 degrees of lift, the servo would need a longer than 50 mm stroke.  Not by much, but still.

That's why I'm checking different solutions.  Gotta find the optimal combination of stroke length, lift range, and mechanical advantage that doesn't cause other problems.


Edit: I did think of another possible way to find A, which is to draw an arc centered at F with the radius FE, down from point E until it intersects the line FL at a point I'll call R (it'd be near the letter B in the diagram). Then A is the length FR. This would give me a slightly longer A than above, but not by much since the rectangle isn't very wide.

Or I could project the SERVO through an arc centered at P to figure out where it intersects FL (where a servo of this length would attach, if that lever was a real physical part). No idea if either of these make any mathematical sense as far as calculating the lever is concerned, though.