Gearbox Failure - Are We Exceeding Motor Design Limits?

I have a hypothesis I wanted to share with the community. I’m hoping this doesn’t cause too much of a stir. I intended to wait until I had a more detailed write up with diagrams put together to better illustrate my point before I started a discussion. However, I sense a failure cluster is beginning to form and I decided that getting this out in the community is more important to try to prevent further accidents/lost work.

I believe the recent problems we are experiencing with drive motor gear failures are related to the load the drive motors are experiencing at certain locations on the cutting surface. I also believe that the default frame design is the root cause of the problem. It is also a problem with custom frame designs where they exhibit similar characteristics to the default Maslow frame in one regard; angle of the drive chains as measured from the horizontal. I further believe that operating time at certaing locations of the cutting surface is a variable that needs to be considered.

Here is a crude example of what I am trying to say. It’s intended to describe a load suspened from a single point to two points on the load but it works for loads suspended from two points two a sigle point on the load if we flip it upside down.

As the angle decreases below 30 degrees, the load on the chains increases at an alarming rate. In our example, at 30 degrees the load on each chain is 14kg for a 14kg sled. At 27.8 degrees, the load on the each chain is aproximately 30kg. This is also the rated load of the drive motors. So, the load on the chains almost doubled for a decrease of only 2.2 degrees.

Again, here is a crude example. When fliped, it works to illustrate my point. The important takeaway from the sketch is the loads expressed for various angles as some fraction of or multiple of the total load:

As an exercise, I did some rough calculations using the design parameters of a users maslow. Keep in mind that my calculations do not include the amount of reduced load that is carried by the frame due to the incline of the frame. They also do not factor in the amount of increased load due to the friction of the sled on the cutting surface. They also do not factor in any counter weight that may be used. So take all I am about to say with a grain of salt.

Based on the reported numbers of the user, when his sled is at the top center of the cutting surface the chains had an angle from the horizontal of aproximately 16 degrees. I calculated that the load placed on EACH CHAIN at this location is aprox. 50.1kg.

It’s much worse when the sled is at the top left or right corners of the cutting surface. In these areas, the far side chain has an even shallower angle than when the sled is in the top center. Here, the chain angle is only 9.3 degrees. I had to calculate the 9.3 degrees based on the information the user provided. The resulting load placed on this chain, in this position, is aproximately 88kg PER CHAIN. I suspect (and it’s still a hypothesis) that this user is exceeding the load rating of the motors by a factor of 2.93 when the sled is in the top left and right corners of the cutting surface. This is probably why this users gear box failed when and where it did. The motors are experiencing the highest loads when the sled is in the top corners. As an asside, I am curious to know where the sleds were at the time of failure for the users who are experiencing them. My guess is that the sleds were at the upper half of the cutting surface and more likely in the upper corners. It’s not scientific. Once worn the gears could fail anywhere, but most likely will when they are under the highest loads.

I suspect the default frame design is the root cause of the problem. The user I spoke with and the numbers I used in my examples above came from his machine which he has reproted to be the default maslow frame. If so, the design of the frame is allowing the chains to regularly exceed the 27.8 degree limitation. Of course this is true for ANY user designed frame that allows for the exceeding of these limits and therefore may not be limited to the default Maslow frame. Time of operation in the areas where this limitation is exceeded is a secondary consideration. More time spent at these locations, the more wear on the gears. This also explains why ALL of the teeth are severly worn. Some may eventualy fail first but every picture I have seen shows teeth that are worn to knife edges and lubricant that is full of fine particle debris. This is indicative of gears under prolonged, high loads, not suffering singular, high impact loads.

Unfortunately, It is not practical to design a frame that prevents operation with a chain angle that does not exceed the 30 degree limit. I tried a couple of month ago. The physical space and chain lengths requied in order to cover the 4x8 cutting surface is not practical for most of us. Therefore, here are some potential fixes and workarounds:

1. Avoid areas where the 30/27.8 degree limits are exceeded.
2. Limit the time of operation in the areas where the limits are exceeded as much as possible.
3. Shrink the size of the cutting surface to allow for a reasonable physical footprint to acomplish #1.
4. Employ a counterweight system to better balance the loads.
5. Source replacement gears made of a harder material that can withstand the forces experienced.
6. Source motors/gearboxes with more powerful and robust units.

Alternatively, we can just accept that these are cheap motors and cheap gears and we get what we pay for. Consider the gears to be a consumable like toilet paper. Have a ready stock of replacement gears available and be prepared to replace them often and prophylactically to avoid a failure induced work piece spoilage.

Again, all of the above is based on some rough calculations that did not take into account several other factors. I am working on a white paper to illustrate my point better and to include the missing variables. Given the rash of failures being experienced by the community, I felt that it was more important to have this discussion sooner rather than later.

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one thing that you are missing is that the sled is not applying full force to
each chain, part of it’s weight is held by the other chain

so your ‘worst case’ of the top corners, is not actually the worst case. Yes,
the angle of one chain is very shallow, but the vast majority of the sled weight
is supported by the other chain, so the tension on the shallow chain is
substantially less than the tension when the sled is in the top center.

Second, it’s not a case of the chains get to an angle and the sled appears
between them, the sled only moves based on the force the motors apply, so when
you exceed that, the motors should stall.

David Lang

I menioned in my post that I hadn’t taken that into account and that it was a missing variable. Sense you mentioned it, using our 14kg sled as an example, the amount of vertical load decreased due to the frame supporting the load is somewhere around 3-4kg. Per my post I also didn’t factor in the increased load to to the friction of the sled on the material being cut. That value can be calculated but is going to be dependent heavily on the construction of the individual sled. Which is why I didnt include it in my post. Conservatively, that will ADD 1kg back into the total load. I can provide a more detailed number when I complete my write-up. Also, per my post, I did not include counterweight effects as not all uers have converted to these and still use elastic cords for tension control.

In order to maintain a position anywhere on the cutting surface, both motors are working to move the sled +x/-x or +y/-y. Which means, both motors are carying a share of the load at all times. To maintain a given position, each motor has to provide a counter force in the X/Y direction to the other motor. Therefore the far side motor has to contribute an X/Y counter force for the system to remain in balance. if it didn’t the near side motors horizontal force would drop to zero and the sled would swing into a vertical orientation. At this point ALL sled vertical load would then be carried by the near side motor.

Per your example, with the sled in a top corner, the near chain is indeed carrying the majority of the vertical load and a little bit of the horizontal load. But it’s not carrying ALL of the vertical load or the horizontal load. The far side motor is still carrying a portion of each. In order to maintain position the far chain has to provide a counter force in each vector in order for all froces to balance out. This results in the far side chain having a very small share of the vertical load BUT a massive share of the horizontal load. The end result is a chain tension in the far chain that is orders of magnitude greater than the total load of the sled. This was what I was trying to illistrate with the diagrams. Unfortunately they only describe loads centered between the suspension points. I will have a free body diagram in my write up to explain this in greater detail too.

The sled is in constant movement during an operation cycle. It may be that at times one motor is not moving for many moments during the cycle. However, both motors are in constant movement almost all of the time in order to maintain the desired position. Even with a static condition with the sled not moving at all, the gears in both motors will have to support the sled load continuously. Which means the teeth of the gears are having to support the sled load at all times, wether the motors are turning or not. It’s not the actual motors capability of moving the load that’s the problem. It’s the capability of the gears to withstand the loads being placed on them.

When turning, the high loads are causing excessive wear of the gears. The gears start to break down, contaminating the lubricant, decreasing its effectiveness and exacerbating the problem of gear wear. Gears wear to the point of failure. I believe this is all happening before the motors ever have a chance to stall. Maybe the motor specs are not correct. Maybe the actual stall point is much higher than advertised. Maybe the manufacturer posted the lower torque limit to keep people from operating in a range where they knew the gears can’t sustain the forces invovled. I can’t say.

Although this spreadsheet can calculate X/Y loads at various points on the cutting surface, I think its missing the point when it comes to the total load on the motors.

A simple test to confirm my theory is to attach scales in between the chains and the sled and just measure the chain loads at various points on the cuting surface.

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I think this is a simple vector addition. For the simple case where the sled is static, the sum of all forces must equal 0.

This means that, assuming a vertical frame for the moment, the sum of the vertical forces in the chains are equal to the force of gravity (mass in kg * 9.81). The exact distribution between each depends on their respective angles, with a sine or cosine factor depending on how you measure your angles.

The horizontal forces, however, must be equal as there’s no other force acting sideways on the chains.

The weight taken up by the slanted frame can also be easily calculated, you just have to move your frame of reference to look at the machine sideways, where you have the force of gravity pointing straight down, the force from the frame pointing normal to the workpiece, and the combined force of the chains pointing parallel to the workpiece.

When in movement, you have to add the force of friction, which is always opposite the direction of movement, and all these forces must now equal mass * acceleration instead of zero. I suspect these are an order of magnitude less than the static forces, unless you have a lot of back-and-forth movement in your gcode.

Thanks for the explanation. I’m learning each day from this project!
14kg… Quite a big sled.
During assembly I read somewhere to aim for 8 to 8,5kg. And that weight won’t be a problem if I do some quick calculations.

Correct, I just haven’t done this yet.

Agree with this also. However it’s dependent of the parameters on the individual sled and therefore difficult to calcualte for ALL sleds. It would have to be done on an individual basis.

Thinking about a bit more, the forces created by the router bit could get pretty high; I think we should take that into account.

The motors that came with the kit are also rated for a torque of 30kg*cm, which gives a max load of 25.64kg on the chain (given the sprocket radius). A 14kg sled will easily bust this near the top of the workpiece.

I menioned in my post that I hadn’t taken that into account and that it was a
missing variable. Sense you mentioned it, using our 14kg sled as an example,
the amount of vertical load decreased due to the frame supporting the load
is somewhere around 3-4kg. Per my post I also didn’t factor in the
increased load to to the friction of the sled on the material being cut.
That value can be calculated but is going to be dependent heavily on the
construction of the individual sled. Which is why I didnt include it in my
post. Conservatively, that will ADD 1kg back into the total load. I can
provide a more detailed number when I complete my write-up. Also, per my post,
I did not include counterweight effects as not all uers have converted to
these and still use elastic cords for tension control.

no, I’m not talking about weight supported by the frame tilt, I’m talking about
when the chain is on one edge and one chain is steep and the other chain is
shallow

in the extreme case of the bottom corners (where the chain is the steepest to
the near motor), there is only about 3 pounds on the chain to the other motor

Per your example, with the sled in a top corner, the near chain is indeed
carrying the majority of the vertical load and a little bit of the horizontal

I never said they did, but I did say that the top corner is not where there is
the most tension on the chains, even though the angle is the shallowest.

The far side motor is still carrying a portion of each. In order to maintain
position the far chain has to provide a counter force in each vector in order
for all froces to balance out. This results in the far side chain having a
very small share of the vertical load BUT a massive share of the horizontal
load. The end result is a chain tension in the far chain that is orders of
magnitude greater than the total load of the sled.

no, do the math, when one chain has the vast majority of the vertical load, the
horizontal load is actually small, so even though the other chain has a massive
share of it, it’s still a trivial amount of load

This was what I was trying
to illistrate with the diagrams. Unfortunately they only describe loads
centered between the suspension points. I will have a free body diagram in my
write up to explain this in greater detail too.

see the spreadsheet that shows the load in the top, middle, bottom and center
and edge matrix

The sled is in constant movement during an operation cycle. It may be that at
times one motor is not moving for many moments during the cycle. However, both
motors are in constant movement almost all of the time in order to maintain
the desired position. Even with a static condition with the sled not moving at
all, the gears in both motors will have to support the sled load continuously.
Which means the teeth of the gears are having to support the sled load at all
times, wether the motors are turning or not. It’s not the actual motors
capability of moving the load that’s the problem. It’s the capability of the
gears to withstand the loads being placed on them.

yes, but the ‘sled load’ is never going to be higher than the pull the motors
can apply. So if the motors only provide 30Kg of force, the total load on the
chains is never going to exceed 60Kg (the force of both motors pulling against
each other)

Although this spreadsheet can calculate X/Y loads at various points on the cutting surface, I think its missing the point when it comes to the total load on the motors.

It is showing the total force on each chain under static conditions.

When moving, the difference between the static chain tension and the force the
motor can apply is available to accelerate the sled. We have examples where a
heavy sled (or too low top beam) means that when doing a horizontal cut along
the top, the motor retacting the chain cannot keep up and the cut ‘hooks’ (sags
during the cut and then curves up at the end to the final location)

David Lang

the static forces can be as low as 3.2 pounds tension on the stock frame. The
motors can easily apply FAR more than this.

We also know that the ‘force’ of friction can be higher than this, because
that’s when the sled won’t swing away from the far chain into the corner.

We’ve also seen people have machines where the force of gravity (weight of the
sled) is not enough to have it slide vertically down the machine and the chains
both go slack (until the motor vibration jogs it loose), this results in a
squiggly vertical cut.

This is a combination of friction and the cutting resistance, and the general
fix is to reduce friction by sanding/waxing the sled.

these forces are significant, and this is why going to a 12’ beam intead of a
10’ beam is so valuable, it increases the force available almost double, without
increasing the friction, so the net force increases far more than double.

David Lang

This is why a 14Kg sled isn’t recommended with a stock frame (see the
spreadsheet for the example of the 30 pound sled, which we know doesn’t work)

David Lang

I want to make sure I am understanding what the spreadsheet is giving me for an answer, given the default conditions that are currently entered into the spreadsheet as I have downloaded it today.

For a sled weight of 20 lbs, at the top center position, with equal chain angles of 19.21 degrees, I should see a chain tension of 30.39 pounds in each chain? I’m looking at row 12, cells H&I.

17.24 degree chain angle, 33.74 pounds top center (the settings were not the default)

see the ‘maslow presets’ menu to pre-fill the machine dimensions with some common variables

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Ok, Thank You.

In the new kits shipping November we are including two spare metal gears. They are less than \$1 each. I will also have some spares for exsisting owners that need them. It would be best to figure out what is causing the issue but until then at least everyone will have access to cheap replacement gears

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In the event more are needed, will you be the source going forward or are others taking this on as well?

I actually wanted to find other people that wanted to sell kits but unfortunately I don’t think there is anyone else. The price to make 100 kits is very expensive. For most people a kickstarter would be the best way to fund the campaign. We are self funded. If kit sales go well we will continue to supply needed parts.
If kit sales don’t go well hopefully someone else can step up to the plate because selling parts for a few dollars it’s not very profitable

100 kits?? Is that all that will be available in the first batch?

Also, has your full kit with all the improvements been tested and built in real life? Would love to upgrade but feel skiddish about replacing a good machine with one that appears better but has some major kinks.

100 kits takes over \$30,000 to make plus labor since some parts we are making in house. and since we have not even gotten 100 people saying they are interested we are being financially conservative. Hopefully I’m wrong and demand is greater. We plan to make videos in the future showing improvements.

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Sorry, I just re-read my post and realized I should have clarified that I was referring to the supply of replacement gears. Although the info on the supply of whole kits is pretty informative too.

Ok so I have a couple of questions:

Cell B3, Motor Seperaation.
The spreadsheet note says it’s measured between the chain attachment points but I’m not clear where these are. Is this dimension measured between the center of the motor drive shafts? Distance between sprocket teeth at the twelve oclock position? I just want to make sure that I use the dimension properly.

Location References
Am I correct in interpreting these to mean the location of the center of the router bit is above the referenced location? For example, top right means the center of the router bit is over the top right corner of the work area.