Frame Geometry Question

About to build my frame soon and have read most of these posts, and want to get the best geometry for a 4’ by 8’ sheet total accuracy. I realize their are many issues, but getting the best geometry in my mind gets the better shot at it.

For my shop I do not have any constraints on L, W, or H. The final design will be on wheels to move it around. I don’t anticipate cutting anything larger than 4x8’. I do plan a center horizontal slot to rout mortises on up to 6x6’s, but usually 2x2’s to 4x4’s. That slot will be recessed into the “spoil board”.

But this discussion is about geometry. I’ve made a dynamic sketch in F360 that models your spreadsheet geometry. For instance, this is the “Standard” as referenced by the spread sheet. It dimensions all 9 key cornerpoints and centers with distances and angles.

It correlates with the top center angle of 17.2º, the “lift angle” for top dead center, and 8.6 º as the “pull angle” on the lower left corner. After reading many of the posts I believe these are the two that are causing the most accuracy issues, as far as geometry is concerned.

I have focused on improving both of these angles to get more “lift” and “pull”.

By changing the height from 18 to 30, and the width from 116 to 156, both of the lift and pull angles increase to 21º. This should increase the lift in the top center position, and the pull in the lower corners.

Note: The 12’ (140" top centers) standard pull angle in the spreadsheet is 18.4º. So the increase over that is just 2.6º.

So my question to you is based on your experiences so far, do you think 21º is enough? Lift / or Pull?

I am avoiding chain weight / sag for this discussion. The longest distance from sprocket is 148 3/16" and the shortest is 42 27/64", so the chain motion length minimum is still about 105" plus some extra, so the first 40" or so from the Maslow ring can be an aluminum rod or something light.

Thanks for your thoughts. Do you agree increasing the lift and pull angles will help?



Boookmarked this one for later read. Thank you!

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I’m about to build my own frame and am thinking along the same lines as you. Will hold off commenting for a while so as to give those with a more intimate knowledge of the Maslow an opportunity to give there input first.

In short, YES, 100%. you will increase the ‘parabola of accuracy’ to encompass most if not all of the 4x8 sheet.

we’re still waiting on more imperical data in the forums on those with large format builds, but just about all reports are good so far and its more a matter of lack of people reporting than what we expect the results to be.

as maslow is designed to modularly scale up or down, you’re invited to experiment with your build’s outer boundaries and to report back here anything you wish to share!

my own 10’ build is about to be upgraded to a more robust 12’ frame in a more suitable layout for my shop, so I cant report on it just yet, but your assumptions above are spot on.

hope this helps,

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Yes, but, unless you have a local aluminum or steel distributor getting a 156" continuous bar/tube/channel is pretty much impossible.
you really have to ask your self how often you are going to be cutting an item that is over 6 feet long and 3 feet tall? Because anything smaller than that will fit fine in the “normal” cutting area.
One can just easily move the material to the left so it is in the center, cut half the sheet, then
scoot it back to the right so that it is again centered and cut the rest out.

yes it is convienient to just set up an entire 4x8’ sheet of cut parts and let it go. but if you don’t do that often then cutting the sheet’s center area might be an easier solution.

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Thank you. I most certainly will share my results once I get it built and tested.

Thank you for your reply. I would like to cut full sheets without repositioning the target board.

I’m thinking that if the lift angle is increased to be greater than the lift angle of the original standard frame, then the tension on the chains, and therefore compression on the top bar, will be reduced as compared to it. I’m not sure as I am not capable of calculating that, but it makes commonsense to me that if the angle is taller than the original Maslow design, then the tension on the chains won’t be as much as the original design, not like the low angle on the 12 foot bar is being built now.

The original standard frame had a top center-point lift angle 17.2°, the 12 foot standard frame with 140 inch center distance has a lift angle of 14.4° and the proposal I have here is 21°. I guess I’ll have to build it to find out.

also if you go with a bigger design you will have to make the 6 wire motor control cables longer by a few feet, so grab some heat shrink and 20 gauge extention wires I think you have a good idea, let us know how it goes.

Thanks for the tip. I knew there would be some other complications/hurdles somewhere along the line.

I tend to look far more at the tension than the angles. the angles affect the
tension, but if you reduce the max tension enough, you can add weight to the
sled to get even better min tension.

angles aren’t the entire picture.

for reference, the spreadsheet is at

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Maybe it’s different in Canada; the box stores don’t carry Unistrut, but all of the electrical suppliers do.
Everyone I called stocked 120" and 240" in similar quantities. The place I went (Eecol Electric) had nearly 4000 of each in stock in my city of 200k people.

FYI, my 20 foot unistrut was CAD$31 (approx USD$25).
(I had to cut it down to 13’ in the parking lot so it would fit in my little VW microbus.)


I immensely appreciate the work you did on this spreadsheet - Thanks!

In related maths - do you know if anyone made a spreadsheet that calculates the min/max forces on the cutting tool? Primarily, the lateral forces in the bottom corners.
If it exists, It would be nice to add that to a new column on the chain-forces spreadsheet.

This, as well as in the center at the top of the sheet. The force exerted by the bit has a definite effect, and workarounds were thoroughly discussed by a forum member earlier this year. Wish I could provide a link to that. In short, on the sides of the workarea, choosing the cut direction so that the force from the bit pulls away from the weaker, more distant motor improves accuracy. In the top middle area lowering the feed rate reduced the effect of the bit force adding to the sled weight to drag the bit lower than desired.

I would say that the force needed to move the cutting tool through the material
(including the friction of the sled) should be pretty constant, no matter where
you are on the material. It may vary based on how fast you are trying to move.

note that while running the motor, the vibration of the motor will jar the sled
and bounce it loose if you are right on the edge

If we could get an idea of what this force is, we could subtract it from the
tension figures and find out how much force is left.

We know that we are right on the edge of usability, as a common problem is
people getting squiggly lines when cutting down (the sled sticks and then comes
loose and drops, wandering a bit). We also know that people have trouble cutting
into the bottom corner.

how much of this is the sled friction, vs how much is cutting force is hard to

could someone who has a way to measure tension measure how hard you have to pull
to move the sled up near edge, and then do the same movement again with the
motor running, a third time with the bit into the wood 0.2" and a forth time
with it going much deeper into the wood.

to find the worst case, I would say move at full speed.

Will do.
I have scales to put in-line with my chains - I will try to connect them tomorrow,
I think I can do this without the motors running (I’m still waiting for a H-bridge board to arrive).

As a starting point to calculating the budget for the weakest lateral forces in the bottom corners, I think that the following sketch is correct. It calculates the largest lateral force (Gravity) before accounting for friction, router torque inertia, bit pull/bite, momentum, etc…
Does anyone see any errors here?

I’m not thinking clearly about this right now - but I believe the next biggest force is the lateral component of the chain tension, which would add 17.3lb*sin(21°)=6.2lbs in the same direction… That adds up to 12.9lbs of force, which seems like a lot -

I would sure appreciate any validation or correction if someone has thought this through.

Maybe I can figure out a setup to check this with the fish scales tomorrow.

you are double counting the force.

it’s actually what you list as tangent force that matters (lateral force would move up the chain), but both of those are based on the tension on the far chain.That tension is all that prevents the sled from swinging on the near chain.

7 pounds sound plausible for a 13’ top beam.

Yes - sleep helps;I was conflating applied force and reactive tension. This makes it clear that the 17.3 lbs on the chain is actually the same applied force as the 20lbs due to gravity.
(Also, what I was calling lateral force, is actually apparent lateral force.)

Thanks @dlang !

(I will post again once I get my scales set up)