now that I’ve caught up, has anyone done the force calculations for the pantograph approach? it seems to me that if you are pulling up on both chains, and there is drag (or weight from a dust collector) that would cause the pantograph arms to move in towards each other, making the bit be below the chains.
with the premade stainless rings at <$10 (and especially if we can get away with one bearing) that would probably be far cheaper than the pantograph.
I have doubts about using an inaccurate maslow to cut out the pantograph arms precisely enough, but it may be a case that the exact dimensions matter less than consistency, so we could cut one arm, then move the workpiece and cut the second one in the same space as far as the maslow is concerned, so any errors would be identical
I think this is only true if the pantograph is not ridgid. Slop will tend to cause the bit to sink below the center of the chains.
@bar could you take your latest test and move one brick so they are symmetrical, move the sled up to the top center, rotate it so that one bearing is up against a bracket, and then put a couple of wire ties (or clamp, or whatever) on the other bearing so that it can’t move and test the result?
I believe that the result will work just as well as what you just tested, and if it does, it would significantly simplify things.
If it doesn’t work as well, it would be good to know for sure 
If it doesn’t work well with the dust collector hose, please try it without.
Agreed. However I am curious to see if the sticking that I saw in Bar’s original video has been resolved. The sticking problem of the ring solution is unpredictable, you can’t be sure when the leverage will overcome the friction. In my head, I think the pantograph design has greater leverage to overcome this sticking, but obviously testing is necessary.
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the arms will always point at the bit, but the chain mounts (and chains) can flex. nothing prevents the pantograph from moving without moving the bit. The arms will still point at the bit, but there will be an angle between the chains and the arms
@bar did the new version with the premade ring have the same sticking problems as the earlier one?
one thing to keep in mind about this sticking is how much error it actually produces.
I think we can view the sticking as a right triangle where the hypotenuse is the radius of the circle, the far side is the distance the bearing moves out of place and the near side the effective radius at that point. (well, it’s actually two triangles, one with the angle at the router, one with the angle at the motor, but the longer one to the motor will have significantly less error, so let’s ignore it unless the error looks to be significant)
so if we have triangle with a 12" hypotenuse, a 1/4" opposite, that means the adjacent is 11.9895" for an error of about 1/100", probably not enough to matter (even though it looks scary).
Hmm, I don’t think I understand how this is different than any of the other mounting techniques. Can’t the same flexing be seen in the ring and standard mounts too?
If the chains mount to the pantograph arms with clevis links, they should accommodate the angles.
the ‘sticking’ of the bearings in the ring approach is exactly this sort of error. but the dimensions here are smaller, so there is less room for error.
if anyone actually builds this and it works, that trumps all speculation and discussion 
So far, I’ve seen everyone concetrating on the dimensions of the pantograph, but not how the forces would end up working. for the normal maslow (and the ring approach) the weight of the sled falls out of the calculation, we never need to measure it (just where the CG is to calculate tilt), but in the pantograph design, it seems to me that the weight of the sled matters more. I could be wrong, but nobody has done the math to indicate otherwise.
The chains will flex, it’s not a matter of needing to change the connection to accomodate the angles, it’s that if the pantograph flexes to move the angles we loose accuracy. If pulling down on the router moves the chains closer together then the bit is now below the point where the arms are pointing
Flexing is an issue with all these designs, rings or pantograph or quadrilateral. Designing for the forces is part of the challenge.
I’ve measured 38 pounds of tension on one of the chains with a 24 pound sled at the center of the top edge of the work area, where the angle is most obtuse, the force on the arm most nearly 90 degrees. Someone schooled in the strength of materials could calculate the flex.
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Based on the angles that I saw last night and those reported in the thread, I don’t think there is an orientation (vertical or horizontal) where the pantograms can share a mounting point using straight arms. The mounts would have to be at least 12 inches from the center of the sled to avoid the router.
I am still game to try a 45 degree version this weekend. The crossover will be a bit janky and the whole thing may suck, but whatever.
the motors can produce ~66 pounds of tension on the chains. the flex in the 1/4" steel brackets is small enough to not matter. for a 3/8" stainless steel ring, this calculator http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_display.cfm?case=simple_centerload#target is saying that the flex is ~1x10-7", far too little to matter
If the pantograph approach works, it can be mounted above the router as shown by
ImpetuousWombat about 23 hours ago
The rings that Bar linked to are 3/16", not 3/8". Still a small number for flex, though.
The equation to find the force on two cables under load at a given angle is:
F = (m x 0.5) ÷ cos(a x 0.5)
where:
F is the calculated force on each cable
m is the mass or in our case weight of the router and sled and bricks
a is the internal angle between the two cables
For those of us who are a little more visual I made this:
This simply means that as the sled gets closer to the top the forces on the system increase quite a bit. You could never pull the chains so they are flat but if you could the force on them would spike to infinity. That’s fun.
That is only accurate when the sled is in the center (both angles the same) off
to the side it gets a lot more complex.
More complex, but lower than in the center, yes?
I don’t actually know. If the two chains come to a [theoretical] point then there is only ever one angle between them no matter where the sled is. It seems that if there’s one angle then the math should still pan out. The two chains are always sharing 100% of the strain, I don’t know if it is possible that they are not sharing it 50/50. My heart says “no” but my mind says “probably?” 