Scissors Geometry

[kgwhipp]

I've been staring at these calculations for a while now and I've hit a stumbling block. I'm trying to determine the force needed by the actuators so I can size them and I'm not sure whether I'm off by a factor of two and doubling the sizes needed or not.

See the picture at this link:

http://imgur.com/73SXWHW

This particular structure uses TWO pantographs (the x structure is stacked twice). If I simply look at the angle of the actuator(assume theta = 45° in the picture) would the force needed be F/Sin(45°) or TWICE this?

I've used the formula in an older derived analysis available at

http://www.dtic.mil/cgi-bin/GetTRDoc?Location=U2&doc=GetTRDoc.pdf&AD=ADA225220

to come up with my initial design locations. (If anyone wants this prepared in excel I can save you some time)

This derived analysis calculates forces as if the actuator pivots are located at the orange dots (instead of the more favorable locations shown). The end result of these calculations is 109,107lbf. When fully lowered the angle of the cylinders is 10°. F/Sin(10°)= 57,507lbf. Multiply that by 2 and you have 115,175lbf (a difference of only 5%) Is this just a coincidence or is it as simple as checking that the vertical component force from the actuator is double the payload (since there are TWO pantograph)

Please take a look and steer me onto the right direction here. Thanks everyone!





-Kevin

 

[Denial]

Let the vertical force at the top be P;
the height of the structure be H;
the length of the actuator be L; and
the force in the actuator be F.

If you are prepared to ignore the self weight and the flexibility of the structure, then you can apply a simple version of what a structural engineer would call the Principle of Virtual Work.[ ] Envisage the actuator extending by a small amount dL.[ ] This will cause the top platform to rise by dH.[ ] Because the system is in static equilibrium, we can say that
F.dL = P.dH
ie
F = P.(dH/dL)

So you now have a task that is pure geometry:[ ] construct an approximate diagram of H as a function of L.[ ] It will probably require the calculation of only a small number of (L,H) points.[ ] Fit a smooth curve through these points, then you have dH/dL at any actuator extension you want.

This approach is covered in section 3.2 of the document you link to, but in a much more elaborate way.

 

[kgwhipp]

Thanks Denial, that was the logical leap I needed.

I had actually already created a plot of H(L) (knowing there might be some value inherent to it)

http://imgur.com/RdTGmkC

When I take the derivative of this (dH/dL) I get the equation

dH/dL = 7.7901-0.0468*L

(at its minimum (zero extension), the length of the cylinders L is = 78.4*) Thus at zero extension dH/dL = 4.12

This gives me a final value of F = 45,228lbf.

The dH/dL from the earlier calculations is 10.91. I was expecting to see a smaller number from the plotted results because of the favorable angle but I this is such a large difference!

*Should I change this value and use L=0 for the zero extension scenario? This would give me dH/dL at 7.79 (which is closer to the 10.91)









-Kevin

 

[Denial]

I'm getting out of my depth now, so I'll leave the answering of your follow-up question to someone else.[ ] However, at least as far as the mathematics of the problem is concerned, there is absolutely nothing special about the L=0 condition.[ ] The datum for L is completely arbitrary (as is that for H), as exemplified by the fact that the solution involves only dH/dL.

 

[tr1ntx]

I think you would sum the moments about the left pivot point, where the 2 pantographs meet, with the forces translated into Normal and Axial. Your load will act through the middle pivot point of the X. (This ignores friction in the pins and sliding friction at the bottom & top where the "X's" get narrower, and weight of the scissors, but you can add those in later and iterate to narrow it down to the correct answer.)