Length of the linear rail for a delta printer?

Hey guys, I'm planning a delta printer and would like to know how i can determin the appropriate length for the linear rails.
In the Kossel design the rails are 400mm long and the vertical beams are 600mm long.
I don't suppose I can directly translate that to any size.

Thanks or the help
Daniel

It seems the optimum angle of the arms for the effector to be centered is 60 degrees and the minimum angle (for accuracy) is 20 degrees. This should be enough to determine the rail length once you decide how tall you want to print.

Ok so I just spent 2 hours in cad quickliy throwing together a makeshift effector, rods and joints.
I think I now know what you mean.

If I were to determin my rod length at a 60 degree angle from the effector to my liniear slide.
The smallest angle between my effector and the slide would be 20 degrees. This would be the case if the center of the effector is closest to a single vertical beam.

Edit: I added an picture of the rough idea in cad.



Edited 1 time(s). Last edit at 01/10/2015 06:49AM by schpongo.

You should probably add a little more distance between the uprights, so the arms are not at 90 degrees. Do a search for Johann Rocholl's openscad drawing of the Kossel. It will help you determine the length of the arms and other parameters.

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etfrench
You should probably add a little more distance between the uprights, so the arms are not at 90 degrees. Do a search for Johann Rocholl's openscad drawing of the Kossel. It will help you determine the length of the arms and other parameters.

I don't quite understand. If I were to increase the distance between the uprights I would also have to increase the rod length and in the end all would be same just larger.

I have been thinking about this fror a while, depending on your size I would say 65-80% of your towerd inner height is what you may want to shoot for as far as linear rail length. Eg: 1000mm inner length how about 750mm linear rails.....

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As far as I have understood the basic geometric concept, the percentage of rail you need only depends on the horizontal frame extrusions.

I was under the impression that the min 20 degrees (for maintained accuracy) was between a tower and it's own arm (at closest position). All usable print radius claims I've seen are 70-65% of the tower to center distance. That only gives the usable print radius (Sin 20 or about 34% of RADIUS (tower to bed center distance)). Accuracy plummets very fast closer to the towers (or out between them).

If 60 degrees from bed to arm at center is optimum then arm should be 2 times the length of bed center to tower distance ( 1 / COS 60).

Pretty sure tower height only effects your max print height (of course there is a minimum that IS effected by the length of the arms).

Edited 2 time(s). Last edit at 01/14/2015 05:52PM by dclarkm.

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dclarkm
Accuracy plummets very fast closer to the towers (or out between them).

I'm skeptical of this assertion. I think accuracy varies as the sine of the angles between pairs of parallel rods - but those angles don't plummet as the effector approaches one of the towers.

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Disclosure: I design Duet electronics and work on RepRapFirmware, [duet3d.com].

I would like to know what the geometric constraints are.
So what are the min and max angles between the components.

Edited 1 time(s). Last edit at 01/16/2015 04:45AM by schpongo.

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dc42

Quote
dclarkm
Accuracy plummets very fast closer to the towers (or out between them).

I'm skeptical of this assertion. I think accuracy varies as the sine of the angles between pairs of parallel rods - but those angles don't plummet as the effector approaches one of the towers.

The effective angle of carriage-effector-carriage equals the sum of the angles of tower base-carriage-effector. So it is related to the sin, but it's really based on the sin of the carriage angles, not the effector.

Accuracy also isn't what plummets. Accuracy (and it's usual companion precision) is how close the positioning is to what the machine is instructed to do, and how repeatable it is. If we consider the printer as a perfect machine that doesn't have slop, backlash, etc the accuracy and precision is the same across the entire print bed. What I think you two are debating is the resolution. The resolution DOES change based on the sin of the carriage angle.

If you consider just a single tower, and constrain the effector so that it can only move perpendicular away from the tower, for each unit of measure the carriage moves up or down, the effector will move towards or away from the tower a distance. The closer the effector gets to the base of the tower, each unit of movement the carriage makes will result in further distance traveled by the effector. And then of course the oppose it true, as the carriage lowers, the carriage moves away at a decreasing distance for each step. This determines the resolution for that particular tower. So in that regard, resolution is worst at the tower.

HOWEVER, we aren't dealing with one tower. We're dealing with 3. If you consider the entire system, resolution is actually highest at the base of each of the towers as you're maximizing the resolution for two rods while minimizing it for the third. You can see this shown in the resolution.jpg attachment image drawing arcs for each unit of movement of the carriage. The relationship between varying resolution you'd think would lead to the lowest resolution being in the center (or at least that's what I first) but the lowest resolution occurs about half way between the center point and each post. That's shown better in the rostock relative step size image attached showing a relative resolution distance as it relates to 1 unit of movement by the carriage.

Here is a simple calculator:
[docs.google.com]