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Higher inverse time feed rates = faster feed rates.
Hence the "inverse" part, which I think they did just so that it would be more intuitive (bigger number == faster). The basic concept is simple enough, but the inverse business tends to confuse the hell outta folks.
That's what I thought!
I only question that because it seems that my Genos goes slower on higher G93 values. I've been pulling my hair out trying to figure out why it's crawling around my part. But on a straight A move it's super fast.
In G93 the control doesn't need to know where the part is relative to the dynamic center of movement.
So what about this makes a better/faster part than G94?
This might be a more complex explanation than needed. The normal feed is distance over time. The inverse is time over distance. Viola, simple answerThe older methods of getting around that are Inverse Time Feed or Degrees Per Minute programming. ...
Inverse Time Feed is really just "time feed," where the CAM figures out exactly how many seconds the cut should take for all axes, and the control just moves them at the proper velocity so they all reach the block end point at the same time. The inverse part is where this gets confusing; the CAM software figures out the speed in minutes, then divides that by 1 to inverse the numbers. The inverse part happens because otherwise, smaller numbers would be faster, and really short blocks would have F values of like 0.00002302.
If you program in inverse time, 9,999 minutes would not normally be an issue It seems to me that helical interpolation in a K&T C control had to be inverse time. A 4 mhz PDP-8 was not exactly ground-shaking speed.Both DPM and Inverse TIme Feed already run into issues where most controls can only take F values at a maximum of 9999.9999.
You mean, like APT did in 1967 ?The modern solution on any serious multi-axis control is Tool Center Point Control. With this, the machine controller knows exactly where it's centers of rotation are, and has the horsepower to do all these back calculations ...
It might not be G93 that is the problem, it could also be HiCut Pro clamping down feed rates, or getting some funky motion, around the very short path segments that drive the higher feed rate numbers.
The funky thing about G93 is that the relationship between velocity and the feed rate number isn't constant. In standard G94, the velocity of the tool on a .001" block versus a 10.000" block is identical. In inverse time, the feed rate is a function of the path length. Then you add path smoothing and accel/decel control (HiCut Pro) to the mix, and you add a lot more complexity.
N100 (CUT SLOT)
(COMPENSATION TYPE - COMPUTER)
/M08
G15 H1
G56 H1 X.4744 Y0. Z4.1875 S6112 M03
(Z STOCK TO LEAVE = 0.)
(XY STOCK TO LEAVE = 0.)
A.611 X.4744 Y0. Z4.1875 S6112 M03
Z3.1875
G94 G01 Z.8375 F150.
G93 X.4814 A.27 F5666.7
X.4836 A.113 F15402.75
X.4844 A-.072 F17425.04
A-47.54 F70.47
X.4828 A-48.124 F5620.8
X.478 A-48.669 F5270.01
X.4706 A-49.135 F4844.56
X.4608 A-49.493 F4413.08
X.4495 A-49.717 F4164.37
X.4374 A-49.793 F4024.04
X.4253 A-49.715 F4016.82
X.414 A-49.488 F4149.53
X.4043 A-49.129 F4432.8
X.3969 A-48.66 F4836.43
X.3922 A-48.115 F5284.95
X.3906 A-47.531 F5634.4
A47.539 F35.19
X.3922 A48.125 F5607.69
X.397 A48.669 F5269.73
X.4045 A49.136 F4834.28
X.4142 A49.494 F4438.71
X.4255 A49.718 F4151.39
X.4377 A49.793 F4008.05
X.4498 A49.714 F4030.88
X.4611 A49.487 F4162.98
X.4708 A49.126 F4445.06
X.4782 A48.657 F4854.71
X.4828 A48.111 F5297.85
X.4844 A47.523 F5599.85
A47.47 F40000.
A-.554 F69.65
X.4836 A-.741 F17202.97
X.4814 A-.895 F15649.51
X.4744 A-1.236 F5669.5
G94 Z3.1875 F350.
G00 Z4.1875 M09
X-10. Y20. Z20. S50
G90
M278(CHIP AUGER OFF)
VC1=VC1+1
M30
I turned Hi cut off, and on that page set the max feedrate to 3937.
It's still not acting right.
I guess it's time to get Gosiger out here.
Here is a pic of my part and my program.
1/4" end mill programmed to 48.8 ipm
One thing to note, the higher feeds are EXTREMELY slow, and the lines with just an A move, they are super fast, almost like it's as fast as the rotary can go.
View attachment 276904
Code:N100 (CUT SLOT) (COMPENSATION TYPE - COMPUTER) /M08 G15 H1 G56 H1 X.4744 Y0. Z4.1875 S6112 M03 (Z STOCK TO LEAVE = 0.) (XY STOCK TO LEAVE = 0.) A.611 X.4744 Y0. Z4.1875 S6112 M03 Z3.1875 G94 G01 Z.8375 F150. G93 X.4814 A.27 F5666.7 X.4836 A.113 F15402.75 X.4844 A-.072 F17425.04 [B]A-47.54 F70.47[/B] X.4828 A-48.124 F5620.8 X.478 A-48.669 F5270.01 X.4706 A-49.135 F4844.56 X.4608 A-49.493 F4413.08 X.4495 A-49.717 F4164.37 X.4374 A-49.793 F4024.04 X.4253 A-49.715 F4016.82 X.414 A-49.488 F4149.53 X.4043 A-49.129 F4432.8 X.3969 A-48.66 F4836.43 X.3922 A-48.115 F5284.95 X.3906 A-47.531 F5634.4 [B]A47.539 F35.19[/B] X.3922 A48.125 F5607.69[/QUOTE] Speed seems right. That's programmed from centerline? The first A-only move is a distance of 47.468° @ Z.8375, arc length would be .6938 * F70.47 = 48.9 ipm Pretty close with my rounding errors along the way. Second A-only move is approx double the angular distance, but half the feed. Looks correct too. All the moves between theoretically should take approx .155 seconds to complete. Assuming an 'arc' of 180°, dia .0908, distance would be roughly .14" at .155 seconds would be 50-something ipm. Seems right.
Seems right.
K&T C, PDP-8 and APT.... you are officially an old fart.
Check the calendar and century.
And got the grey hair to prove itK&T C, PDP-8 and APT.... you are officially an old fart.
This might be a more complex explanation than needed. The normal feed is distance over time. The inverse is time over distance. Viola, simple answer
If you program in inverse time, 9,999 minutes would not normally be an issue It seems to me that helical interpolation in a K&T C control had to be inverse time. A 4 mhz PDP-8 was not exactly ground-shaking speed.
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