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.