It’s a problem ripe for study.
One thought is that sag is affected by direction of travel, friction and location - in the lower corners the long chain can sag a lot if the sled is moving horizontally and the long chain is feeding out. Not sure how to capture that, though.
It’s a problem ripe for study.
What we need first is to figure out sag when static or moving very slowly.
The problem feeding out the long chain in the lower corners is that the
available force of gravity to accelerate the sled is very low, not a problem of
friction being higher there.
The first problem is that every example we have found for a formula for chain
sag assumes that the low point of the sag is below both endpoints.
In our case, the low point of the sag curve would be past the lower endpoint.
This needs a question to a math guru (stack overflow possibly??) to figure out.
After that we can tak about implementing the resulting math and see where we
One variable that I do not see addressed is that the current recommendation is to cut your final sled with a temporary sled on a questionable frame and suspicions calibration.
The variations of software to create the gcode is a different thing.
My first sled i cut with a simple router jig and was within 0.5mm.
(make an angled cross from the centre of your sled and measure from there)
The second sled was cut on my desktop cnc.
Given all the variables i would would not be happy with that recommended sled if it includes mounting positions.
The question with this is how to bootstrap an inaccurate machine to an accurate
I posted an example of how you could make a 45 linkage kit that is horribly
inaccurate to any design, but as long as the holes are drilled per instructions,
results in a usable/accurate linkage kit.
the only thing on the sled that need to be accurate is the mounting of the
linkage kit relative to the center of the bit.
With the 45 linkage and top mount designs, you can do this by using one of
the arms to mark the hole locations after the router is mounted to the sled. you
then need to make the standoffs from the sled with holes that are truely
perpendicular to the surface of the standoffs.
With the ring kit it’s a bit harder.
Do you know how much of that is chain vs. frame flex? I could see a 2x4 or 2x6 flexing a few mm under a 30 kg load.
The last chain-sag calculation is robust to scenarios in which the min-point is both above and below the endpoints. The only time that becomes an issue is when using the simplified calculation which assumes a parabolic shape.
Here is the thread where this was discussed.
The equation on post 42 is the one to use.
However, it has to be manipulated to solve for s.
Rotate the center point perpendicular to the bit by the radius of the sprocket.
it is? Could you point me at the source of that? in our last discussion on the
topic, I don’t remember anyone finding the math for when the midpoint is between
the endpoints in height.
don’t you need angles to do that? and you also need to know how much chain is
used up wrapping around the sprocket to that point.
But give it a try, I may be wrong. See if your math is more accurate than the
The equation used is about half-way down the web page. The only other tricky thing is assigning the weight of the chain in lb/f to (lambda x g), due to nuances of english units.
There are a lot of equastions about half-way down the page, which one are you
the weight of the chain is easy, 0.09 lb/ft (0.13Kg/m)
About rotational radius. How about mounting the router upside down in the sled with a 1/4" or 1/2" rod in the collet that is long enough to stick up past the top of the ring. Then the distance could be measured accurately and easily. This assumes little or no runout in the router collet and shaft.
It’s also test to see that ring and router are concentric.
Sometimes I get in a hurry, and it is not as clear as I thought it was. I edited my original post. Hopefully it is more clear.
your updates don’t go out in e-mail
This of course assumes that the router base is symmetrical, but that’s I think a safe assumption. Good idea. Not sure what I’d do if I’d do if I found it to not be centered. Maybe patch the router holes and redrill them a tad in the right direction? You might also get some slight variation from mounting and remounting the router base.
What I did to check is turn on the router, plunge the router bit 1/8” and rotate the ring and see how big the hole ended up. But I learned from optical calibration that I’m not able to perfectly rotate the ring without slight (mm scale) nudges in the translation, so I’m not super confident in this method either. My hole ended up not much bigger than 1/4” (maybe 1mm bigger) so I think I’m pretty close to centered.
I have a 1/2" rod I use to test runout and I set it up. It’s still not easy to get it to stay in place while screwing the ring down. But I got it to about a half a mm all around.
To make sure the router is centered I went ahead and added the router holes to the sled nc file, concentric. Then I countersunk both sides of the holes and used screws with conical heads. So the router should be pretty well centered whether it’s mounted right side up or upside down.
I’m getting close to 143 from the center of the bit to the outer extremity of the metal carriage. If that rotational radius is supposed to be measured from the center of the bit to the center of chain link hole that the cotter pin goes through then I think I just verified the 140 everybody has been using. It’s a good number.
Another thought is to hang the sled from a rope, rotated sideways such that the two bearings are on opposite ends of the ring. One bearing is on top and the other is on bottom. You could measure the distance between the surfaces and use 1/2 that as your measurement.
That’s the best thing I’ve heard. Give it a try. You could tie a big knot in the rope and put it through the chain hole and hang it from the bearing. Then do the same on the bottom bearing with a weight on the end of that rope. That would stretch it out nicely.
And I bet it ends up measuring 280!
I was thinking about some additional calibration mechanisms. I like @madgrizzle’s visual calibration, and I really like the functionality that it provides: calibrate the machine with no user-input. I thought of one or more option for calibrating the machine with no user-input. Here is one.
Cut holes in the face of the Maslow, and mount straight bars on the back-side of the plywood. Straight bars could be simple cables that are stretched tight. These bars/cables could be mounted in a known location. If they are not in a known location, their location could be estimated alongside the other machine parameters. Use a touch-sensor on a router-bit which extends through the plywood face. Identify where the bars are using the touch-sensor. The delta between where the machine thinks the bars should be, and where they actually are provides information necessary for the calibration. Here is a drawing of the system from the back-side. The black lines are the cables or bars, and the white parts are holes in the plywood. This is an idea. I don’t have the time to actually implement, so I am just trying to stimulate some creativity.