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

[Dyne]

Ack, you're mixing lbs and mm!   Which I suppose could be more confusing because 5lb mass is only 5lbf if the interesting direction happens to include gravity.

If you're pointing out that I'm mixing metric and US units ... that's mostly because the servo rating was given in lbs of thrust rather than kg or newtons or whatever, and I typically weigh things in that as well, while I prefer using metric as the default length unit in CAD.

I'm confused.  It's not rotating around P?  In the earlier animation, I thought the bar rotated around it's endpoint?

In the earlier animation, I was showing the parallelogram.  I skipping calling out most of that in the more recent diagram, for simplicity's sake.  It just seemed to confuse the issue when I tried to describe the problem to friends.

To answer your question, the green arm does indeed rotate around P (and in the arrangement shown, the servo does as well, but this arrangement isn't final).  

The body shell (or rather the imaginary line between the pivots on each end of it) forms the arm parallel to that green bar, and this arm rotates around F.  Those are the two "ground" pivots of the four bar mechanism.  The other two pivots are L and the unlabeled green circle a bit forward of L.

I guess you're saying that the "real" lever is BC in the diagram you'd posted earlier?  Since that's the hinged point of the cover and the place it rotates about?

So, the lever is kinda like BC, except that it's even longer because the weight is all over at A?

Yeah, mostly I was using the old diagram's BC (FL in the new one) as the "real" lever because that's the closest part of the parallelogram to where the servo force is actually being applied, so it seemed best to use that bit, omitting the rest of the parallelogram (on the grounds that if I can't calculate the values when there's just one arm in the lever, adding an entire four-bar linkage into the mix is just going to make it worse).

It wasn't as clear in the old diagram, but in the real build, A and B are just two pivots on the same single part that forms the upper stalks of the neck (the "light green stuff"). So all of the weight is bearing on the entire length AB. I guess you could say SOME of the weight is all the way over at A, and presumably the rest at B, but your guess is as good as mine as to what percentage goes where. Or maybe all of it is *effectively* at A.

Two thoughts:

First is that the diagram of forces is going to change as the model moves from it's closed to fully open position.  You'll want to position your servo in a place that is highly effective in the worst positions.

Yes.  In one of my earlier attempts I was trying to calculate the effective distance of the effort from F by another method, and I realized that it always gave a range of results between the two positions because the equivalent of point E was shifting, sometimes quite substantially depending on the servo placement.  

I think it's sufficient to consider two cases: when the head is lifted, and when it is lowered but trying to rise.  The latter is, I assume, typically going to be the worst position, since it's fighting gravity and the angle is generally the furthest from perpendicular.  

When the head is already raised, the servo is just handling the static weight (which it is pretty good at, even unpowered), and when it's lowering, gravity is helping it, so the servo is mostly acting to keep it from just dropping.

Second is that you can simplify a rigid structure to a line for the calculations if it's rigid.  If the loads are at the ends, and if the effort is along the line, then it's easy enough to so the math.  if the effort isn't along the line, then it gets more complicated.  You'd have to have a second line where the servo attached to the fulcrum.  But the math's the same, presuming the servo's effort is perpendicular to it's line.

(Putting this next quote here, out of order, because it's related...)

Anyhoo. If you're applying the effort from your servo to E, then it's lever arm around F is going to be the line EF, not A.  AND that's presuming the effort being applied perpendicularly to the lever.  (The obvious other extreme would be the servo being mounted parallel to the EF lever.  In which case it doesn't matter how hard it pushes, all it can do is break the lever, it can't make it pivot.)

I think I reached a similar conclusion (as suggested in the edit that added the last paragraph or two) by extrapolating from thinking of the Effort from the fulcrum's point of view as a torque.  If I draw a circle centered at F that touches E, that's where the servo's force is applied anywhere that it can possibly move (in other words, E is a constant distance from the Fulcrum, being connected to it, albeit in a roundabout fashion).  

The math is basically 'just' saying "how much does this mass (or effort) want this lever to rotate around the endpoint?"  Which is really easy, it's just the force times the distance.  If you have 12 points with mass and 15 efforts in all sorts of weird places on a strange rigid structure all rotating around the same point, it's still the same math.  + or - forces multiplied by the length of their lever arms.

In other words, if, in your diagram, the lever rotates around F, and your mass load is on L (It's not, there's all that light green stuff to the left of that, so the mass has even more leverage).

That's true.  I was treating it all as being at L, mostly because, like I said, if I couldn't get a handle on calculating the leverage for ONE arm in a simplified arrangement, I certainly wasn't going to be able to do it for four arms and a weight distributed in some unknown ratio along the entire length of the line AB (in my old diagram).

And not just that, the arms of the mechanism also add some of the weight that has to be lifted by the servo.  I did include this weight in my numbers, at least...

To be clear, using the old diagram, some unknown portion of the weight is bearing down at A, some at B, some is the part connecting AB, some is the part AD (the green arm), and some is the body shell and accompanying stuff.  Maybe I should've treated all the weight as being at A in the old diagram.

But, assuming that it was on L, then the load would be the mass on L times the length LF (B) - if the B was parallel to the ground.  It's not, so you'd have to figure out how much the load was perpendicular to the LF (B) line.  Or just go with it because it's close to parallel.



Here's a video I found from a quick search.  I didn't check for accuracy or whatever, but it shows what happens if a force isn't perpendicular to the lever:

 [/url][url=https://www.youtube.com/watch?v=_HuxmF_1Z90]Ch 8 - Torque - Calculating Lever Arm and Torque - YouTube

I will have a look. Thanks for the thoughts.


For now, while I'll still be able to raise the head manually, I'm making the executive decision to punt -- I'll set aside the lift servo, at least until after Dragon Con.

It's not so much because of the lever stuff, but mostly because I started playing around with how to fit the various external components in, and it looks like the only space big enough for the battery is directly between the two green arms.



That will put severe constraints on where point P can be and may end up requiring me to move to a dual servo setup, which I am unlikely to be able to do until after the con.  I will probably take some time at the con to run the issue past other builders like Matt Hobbs and David Ferreira and see what they think (assuming they are coming this year).