Making the Maslow Bigger - MaslowMega

Projects
1

We have had our Maslow up and running a few weeks and its great! Now we want to make it bigger. The idea is we want to use ALL of the 4x8 space for table surface. We are going to extend the bed size by 2ft on all sides and extend the chains as well. The 2ft buffer area will be fixed with mdf so basically the 4x8 area will be a cutout that you can pop your board into. The idea is this would give us support for the sled all the way to the edges. The top bar will probably end up being a steel box beam so we can get the length but we are thinking of experimenting with staggered 2x4’s screwed together.

I will let you know how we progress.

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I am also considering just how BIG to make my Maslow. I do not have a space issue in length or height. I have been using the spreadsheet for chain tensions (static tension due to sled weight, not reduced for the 15 Degree slope which must take some of the weight, and does not include any “dynamic” calculations due to friction of the sled on the material, acceleration and deceleration, or force of cutting any of the various materials with various bits.)

Since most of the reported accuracy issues are in the lower corners and top center, increasing those tensions, horizontal component on the lower ones, and reducing chain tensions on the top one, can be achieved by wider and taller locations of the motors. I have one design plan for a 176" motor separation, with a motor 40" above the top of the sheet. (96" x 48"). Those locations provide more pull and lift where needed, but introduce additional issues for chain length, chain sag, chain stretch, etc. as being discussed in this forum.

There is a great deal of work still being done and I have been waiting and watching this evolve.

I had another sketch the other day on how to eliminate “chain sag”. The chain, from the closest corner to the furtherest corner, moves about 105", plus or minus. If that movement were vertical, no chain sag would occur. Replacing the chain in the active work area with a thin lightweight cable would suffice. (Of course there will be “cable sag”, but choosing a thin lightweight cable, and having sufficient counterweight and sled weight, you could play a piano string, with very little, if measurable, sag.)

IMG_6718%20(1)
IMG_6718 (1).jpg4032Ă—3024 1.76 MB

In this sketch the “red” is the chain, and the black would be cables. As the sled moves from one extreme to the other, the chain moves vertically from one side of the motor to the other. (The red circle on the bottom would be the placement of the Maslow motor. Down where you don’t need a ladder to work on them.). The sled weight and counterweight oppose each other, and I would load them up to eliminate “cable sag”.

This still leaves “chain stretch” as an issue, and perhaps adds “Cable stretch” as another issue, but I think would kill the “chain sag” issue.

The “static forces” at the key points are as follows:

For the “standard Maslow” (116" Centers, 18" above board, 20# sled):

At Top Center, Tension 17.05#, Angle 17.24 Deg., Vertical Lift 10#
At Lower Corner, Tension 3.26#, Angle 81.38 Deg., Horizontal Pull 0.49#

For the 12’ Maslow"(140" Centers, 18" above board, 20# sled):

At Top Center, Tension 40.15#, Angle 14.42 Deg., Vertical Lift 10#
At Lower Corner, Tension 6.44#, Angle 71.57 Deg., Horizontal Pull 2.04#

For this Monster Maslow(216" Centers, 60" above board, 20# sled):

At Top Center, Tension 20.59#, Angle 29.05 Deg., Vertical Lift 10#
At Lower Corner, Tension 9.76#, Angle 60.95 Deg., Horizontal Pull 4.74#

Obviously, the higher and farther away from the workpiece, the better. But, WHAT IS PRACTICAL?

This configuration would also need some other calibration process. (Hopefully one that can eliminate the sag calculations and eliminate the left to right motor chain stretch.

I’m still watching and reading the posts, hoping for a better solution.

3

there is slop in the chain (similar to stretch in the cable, but probably more consistant)

figuring out the sag and the stretch is what the calibration is supposed to do, you don’t want to eliminate these even if you reduce the amount of sag.

one other problem you are introducing with steel cable is the memory in the cable as it spends most of it’s life spooled tightly, and then as it’s spooled out to reach the far corner it’s under the lowest tension

one thing to remember is that increasing the tension drastically lowers the sag

with 10.5 ft out at 3.2 lb tension (roughly stock, and not accounting for the angles), you have:
https://www.spaceagecontrol.com/calccabl.htm?F=3.2&a=10.5&q=.09&g=32.18503937&Submit+Button=Calculate
total chain length 10.53822377 ft (error of 11.6mm)
with 11.5 ft out and 6.5 pounds force (roughtly 12’ top beam) you have:
https://www.spaceagecontrol.com/calccabl.htm?F=6.4&a=11.5&q=.09&g=32.18503937&Submit+Button=Calculate
total chain length 11.51254491ft (error of 3.7mm)

so merely going out a foot on each size cuts your error 2/3

your monster maslow, without the calbe trickery, just the distances would be in a similar ballpark:
~16 ft of chain at a tension of 9.7 pounds
https://www.spaceagecontrol.com/calccabl.htm?F=9.7&a=16&q=.09&g=32.18503937&Submit+Button=Calculate
total chain length 16.01470713ft (error of 4.48mm)

so I think this is a case where adding more distance isn’t helping you (I think you are trying to go too high)

14’ top beam at 28" is better than your monster maslow (in the corners as far as sag goes, just a little better along the top)

min force 8.22
max force 30.95

https://www.spaceagecontrol.com/calccabl.htm?F=8.2&a=12.5&q=.09&g=32.18503937&Submit+Button=Calculate
12.5 ft of chain, 8.22 pounds
final chain length 12.50981288ft (error of 2.9mm)

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4

Not by 2ft but I already made a design that does this.

http://www.indyplanet.us/80/60-frame-manual-1/

Thank you

*One other note ,you might call it colossus or something else the fact the Maslow uses a part named “Mega” makes the subject confusing.

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5

Wouldn’t changing the angle of incline increase the tension on the chain and there by reduce chain sag?

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some, but it also reduces the force available to push the bit into the material.

How significant this is depends on the bit in use

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Are not the motors worm gear driven? If so, the counter balance would not do anything except equalize the load on the motor and/or minimize backlash. Please tell me where I am going wrong.

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if the counter balance is less than the tension on the sled, it reduces the
load. (backlash doesn’t matter here as the chain is always tensioned in one
direction)

But if it’s more than the tension on the chain, it invokes the backlash, so
you loose accuracy.

David Lang