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

[kresty]

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.  

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

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?  

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.

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.

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).  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.


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.)

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:

 Ch 8 - Torque - Calculating Lever Arm and Torque - YouTube