Machsupport Forum
Mach Discussion => General Mach Discussion => Topic started by: GRAYHIL on March 12, 2014, 05:02:42 AM
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Hi All
On my lathe/mill set up I can rotate the top slide through through 90deg. manually so that x and z both cut in the same direction.
If I program
G1 X6 Z3 F.01 it will traverse 9 inchs in the same direction.
What will the actual feed rate be.
Graham
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You don't say if you are using it as a lathe or mill and also what feed rate mode is in effect.
I may also add that Mach would not know that the axes are moving in the same direction and would assume that X is perpendicular to the Z ( if a lathe ) and as such the distance would be 6.7".
RICH
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Rich
I can assure you the x axis moves 6 inchs and the z moves 3 inchs in the mill mode and using G95 S460 F.01
I do it to increase the x axis total travel for facing long pieces.
I do not understand where you get 6.7 inchs from.
When I do this I have to move the saddle towards the chuck manually using the travi dial to put on a cut depth.
Graham
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Graham,
G1 X6 Z3 is a a co-ordinated linear move. Yes, X will move 6" & Z will move 3" as you noted, BUT, that gcode is telling Mach to provide motion along the hypotenuse of those two moves ( along the programmed linear XZ path) and that distance is 6.7" .
Mach or any contoller would not know that you doing something goofy / tricking the software to accomplish what only you know.If you coded for individual axis moves then the feedrate would be as coded.
Attached is info from the manuals for your reading,
RICH
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Rich
I see where the 6.7" comes from, its the hypotenuse as you say.
So it will do 6" at .01/rev and 3" at .005/rev to arrive at the same spot.
So in my case of the axis both moving in the same direction it will be a combined feed of .015/rev???
Thanks
Graham
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So it will do 6" at .01/rev and 3" at .005/rev to arrive at the same spot.
So in my case of the axis both moving in the same direction it will be a combined feed of .015/rev???
Actual feedrate calculation is done internaly by Mach and do not know how it's being programmed internaly.
It's not that simple as the "guess" you posted.
Depends on what feedrate mode you are using, if starting from current velocity or no axis motion, your accel / velocity
/ steps per unit settings for the motor, motor rpm if G95 mode, etc.
The x & Z motors will need to accelerate and de-celerate and move at different feedrates to reach the same end
point which is based on 6.7" where F.01 is inches per spindle rev for G95. Each axis motor is pulsed differently to move the controlled point to the final location over a period of time and distance and acceleration may be modified ( if possible) depending on time to achieve. You would need to calculate each axis based on coded move and Mach's instruction for that move then apply to the individual axes for what you are doing.
Per above you have not provided all the info required to answer you question:
What will the actual feed rate be.[/quote)
RICH
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Rich
I understand what you mean re accelaratin/de acceleration etc
But from a practical point of view .015 is right?
G95 S460 F.01 M4
G1 X6 Z3
When x and z go in the same direction on a modified mill.
Graham
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But from a practical point of view .015 is right?
If that's what your "calculations" show it to be.
RICH
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So it will do 6" at .01/rev and 3" at .005/rev to arrive at the same spot.
So in my case of the axis both moving in the same direction it will be a combined feed of .015/rev???
No - it will do 6.7" at 0.01/rev. So will do the 6" at ~0.0089 and the 3" at ~0.0045
So an additive feedrate of ~0.0134
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Since the individual axes move different distances the feedrates are different. The only time the feedrates for each axis would be the same is if the distances were the same. Both axes must change to maintain the controlled path ie; the linear move angle and for the above the change in x with repsect to z would be a value of 2 or otherwise the controlled point would not follow the coded path.
I agree with ya Stirling....got to go,
RICH
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The more I learn the harder it gets.
So a circle must have hundreds/thousands of calcultions?.
But I get the principle.
Graham
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So a circle must have hundreds/thousands of calcultions?.
To get the feedrate of each axis at any given time? Well it depends on how you command your circle. If it's "hundreds/thousands" of G1 moves then yes but if it's G2/G3 then no.
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I hesitate to ask but how does it do a circle under g2/g3?
Graham
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If the circle is made up of straight lines then Mach has no idea it's a circle and plans each individual move just as if it was a "random" sequence of moves. If it's a G1/G2 then by definition Mach knows it has to produce a circle (or part of i.e. arc). In that case the relationship between the X and Y axes is a simple formula that holds for the whole arc/circle. Each axis simply describes simple harmonic motion (http://en.wikipedia.org/wiki/Simple_harmonic_motion) where one is a sine and the other a cosine.
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Thanks Rich and Stirling for your explanations.
It helps me to understand the complexity of what goes on when I enter a command into Mach 3.
Graham