Revision as of 16:14, 12 March 2018

5 thousandths, or .12 mm - is enough to cause trouble. Depth of cut is .12mm for circuits.
See Shapeoko discussion - [1]
Latest Shapeoko uses GT2 9 mm
Most RepRaps use GT2 6 mm

Calculations

Positioning accuracy of GT2 belt - Say 1/2" pulley. Circumference about 40 mm. 200 steps per revolution = .2 mm positioning - but with 16 microstepping RAMPS standard - we get down to 0.01 mm positioning accuracy, or 0.5 thousandths. Positioning Accuracy does not stand in the way of a successful build - therefore only backlash and belt stretching need to be resolved. For belt stretching, the solution is proper tensioning. So this looks optimistic, gien a solid frame."
Chart at [2] shows positioning accuracy of GT2 belt to be 0.0003 - about half that of the RAMPS 16 microstepping calculation above - meaning that GT2s can accept up to 1/32 microstepping.
The question is - how much does the belt stretch upon acceleration (force)?
Why is there no resource that states belt stretch as a function of force?

Based on 9 lb tooth jump force for a 20 tooth pulley - 1/2" pulley - if that is for 5 mm of belt width - we could do up to 4 15 mm belts per 2" universal axis. That is 60 mm of belt - or 12*9 lb force = 108 lb of force per axis. Doubled axes mean that we have about 220 lb of tool force for the super heavy mill. That is not a lot, but sufficient for many jobs including milling with up to 1/2" mills.
Cutting force - take a lathe - and let's see how much force is required for a cutting operation. If steel is 50,000 psi strong, then cutting a 0.01 by 1/8" sheath of metal takes that fraction defined by the surface area of the cut - 1/8"*.01" - where it's the width of the cut * depth of cut. So 1/1000 of 50,000 psi is 50 lb force on the tool. If we are doing 50 lb of force on the tool tip - then we should be ok. But catching peak loads would be 500 lb - insufficient for our system to hold. Possible solutions are workpiece cooling and higher tool speeds. In the limit - we can do grinding operations, which require very little force. So precition grinding would be at least one application of a lower force system.

Contact Application Engineering Department - of SDP-SI - this belt bible - [3]- Tel: 516-328-3300

Ebay - 15 mm belt - $1/foot at 100M quantity - [4]. 500 lb per inch breaking strength. Operate at 10% of breaking strength?
SDP-SI has up to 9 mm belts - [5]
$40 for 10m of 9 mm - [6]
$32 Amazon prime for 9mm, 10m - [7]

@@ Line 9: / Line 9: @@
 *The question is - how much does the belt stretch upon acceleration (force)?
 *Why is there no resource that states belt stretch as a function of force?
+==Force==
+*Based on 9 lb tooth jump force for a 20 tooth pulley - 1/2" pulley - if that is for 5 mm of belt width - we could do up to 4 15 mm belts per 2" universal axis. That is 60 mm of belt - or 12*9 lb force = 108 lb of force per axis. Doubled axes mean that we have about 220 lb of tool force for the super heavy mill. That is not a lot, but sufficient for many jobs including milling with up to 1/2" mills.
+*Cutting force - take a lathe - and let's see how much force is required for a cutting operation. If steel is 50,000 psi strong, then cutting a 0.01 by 1/8" sheath of metal takes that fraction defined by the surface area of the cut - 1/8"*.01" - where it's the width of the cut * depth of cut. So 1/1000 of 50,000 psi is 50 lb force on the tool. If we are doing 50 lb of force on the tool tip - then we should be ok. But catching peak loads would be 500 lb - insufficient for our system to hold. Possible solutions are workpiece cooling and higher tool speeds. In the limit - we can do grinding operations, which require very little force. So precition grinding would be at least one application of a lower force system.
 =SMEs (Subject Matter Experts=