Can a component know the diameter of an arc its cutting
01 Mar 2019 10:46 #127397
by rodw
Andy thanks. I had a few minutes this morning and threw halscope into the mix so will look a bit further before I give up. I think the rate of change might remain at 0 which kinda makes sense as the delta changes should be identical each servo cycle given a constant velocity and a constant circular shape. The other algorithm looks interesting.. I just thought of it but maybe skipping a random number of servo cycles would mean the time period would vary so its not all identical..
I had some issues today which shot my plasma plans for the weekend apart. Business comes first.
Replied by rodw on topic Can a component know the diameter of an arc its cutting
It might be better to treat this as a geometry problem rather than circular motion.
Andy thanks. I had a few minutes this morning and threw halscope into the mix so will look a bit further before I give up. I think the rate of change might remain at 0 which kinda makes sense as the delta changes should be identical each servo cycle given a constant velocity and a constant circular shape. The other algorithm looks interesting.. I just thought of it but maybe skipping a random number of servo cycles would mean the time period would vary so its not all identical..
I had some issues today which shot my plasma plans for the weekend apart. Business comes first.
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01 Mar 2019 21:57 - 01 Mar 2019 22:04 #127465
by rodw
Well I've had a look at this using the latter interesting version that looked simple until I got to calculating the angle a.
Imagine we are traversing a circle, we end up with the latest side A, the previous side B and a chord from the earliest point to the current point being Side C.
The approach I took was to calculate the slopes for B and C convert them the inclinations (angle for the slope in radians). By subtracting the inclinations (atan of the slopes), I think we end up with angle a (aa in my code). Heres what I came up with:
So am I on the right track with angle aa?
Replied by rodw on topic Can a component know the diameter of an arc its cutting
Yeh, I think there is an error in the way I calculate the difference between the heading rate of change but I've not found it yet as I ran out of time util tonight
It might be better to treat this as a geometry problem rather than circular motion.
The needs three points, but we are already considering three points anyway to get heading rate of change.
stackoverflow.com/questions/22791951/alg...ngles-given-3-points
And a rather interesting version:
math.stackexchange.com/questions/133638/...points-actually-work
Well I've had a look at this using the latter interesting version that looked simple until I got to calculating the angle a.
Imagine we are traversing a circle, we end up with the latest side A, the previous side B and a chord from the earliest point to the current point being Side C.
The approach I took was to calculate the slopes for B and C convert them the inclinations (angle for the slope in radians). By subtracting the inclinations (atan of the slopes), I think we end up with angle a (aa in my code). Heres what I came up with:
FUNCTION(_)
{
static float last_x1 = 0.0;
static float last_y1 = 0.0;
static float last_x2 = 0.0;
static float last_y2 = 0.0;
static int cutting_arc = 0; /// True if cutting an arc
float a,b,c,k,aa; // length a,b,c; area k; angle aa
mtype = (Motion_Type_T) type_in;
switch(mtype){
case ARC:
if(cutting_arc > 1){
a = sqrt(pow((xpos_in - last_x1),2) + pow((ypos_in - last_y1),2));
b = sqrt(pow((last_x1 - last_x2),2) + pow((last_y1 - last_y2),2));
c = sqrt(pow((xpos_in - last_x2),2) + pow((ypos_in - last_y2),2));
aa = atan((last_y1 - last_y2)/(last_x1 - last_x2)) -
atan((ypos_in - last_y2)/(xpos_in - last_x2));
k = 0.5 * b * c * sin(aa);
arc_radius = (a * b * c)/ (4 * k);
last_x2 = last_x1;
last_y2 = last_x1;
last_x1 = xpos_in;
last_y1 = ypos_in;
}
else{
last_x2 = last_x1;
last_y2 = last_x1;
last_x1 = xpos_in;
last_y1 = ypos_in;
last_heading = heading;
}
cutting_arc++;
is_active = (cutting_arc > 1: 1,0);
break;
default:
cutting_arc = 0;
is_active = 0;
break;
}
}So am I on the right track with angle aa?
Last edit: 01 Mar 2019 22:04 by rodw.
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01 Mar 2019 22:05 #127466
by tommylight
Replied by tommylight on topic Can a component know the diameter of an arc its cutting
G2, G3, I, J, K ????
Could any of those be used to drop the feed rate ?
Even if that works ( it should, just did not think about it, just got the idea now ), there remains the problem of small segments that are short straight lines ( i do have the gcodetools for inkscape set like that on one of the machines ). Not very important for now.
The benefit of this would be that all the radius’s would be cut slower, the downfall is even large radius’s would be slower, unnecessarily.
Could any of those be used to drop the feed rate ?
Even if that works ( it should, just did not think about it, just got the idea now ), there remains the problem of small segments that are short straight lines ( i do have the gcodetools for inkscape set like that on one of the machines ). Not very important for now.
The benefit of this would be that all the radius’s would be cut slower, the downfall is even large radius’s would be slower, unnecessarily.
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01 Mar 2019 22:33 #127467
by Mike_Eitel
Replied by Mike_Eitel on topic Can a component know the diameter of an arc its cutting
I have one ( silly?) question:
Isn't it typical for small holes that the acceleration of both axes is quite high and with sinosoid relation. Big holes need more time.
m5c
Mike
Isn't it typical for small holes that the acceleration of both axes is quite high and with sinosoid relation. Big holes need more time.
m5c
Mike
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01 Mar 2019 22:57 #127468
by rodw
Tommy, not so. Andy's idea was to use lincurve to set a table of feed rates which would be used to control adaptive feed so we can modify the result based on an algorithm I think lincurve allows 16 different levels. eg. for this radius range , use this speed override.
Possibly, I'm not sure but for small holes, with plasma you might just spot the holes and drill to size anyway.
Replied by rodw on topic Can a component know the diameter of an arc its cutting
Trying to do this at the HAL level not in Gcode so it just happens in the background according to some sort of AlgorithmG2, G3, I, J, K ????
Could any of those be used to drop the feed rate ?
The benefit of this would be that all the radius’s would be cut slower, the downfall is even large radius’s would be slower, unnecessarily.
Tommy, not so. Andy's idea was to use lincurve to set a table of feed rates which would be used to control adaptive feed so we can modify the result based on an algorithm I think lincurve allows 16 different levels. eg. for this radius range , use this speed override.
I have one ( silly?) question:
Isn't it typical for small holes that the acceleration of both axes is quite high and with sinosoid relation. Big holes need more time.
m5c
Mike
Possibly, I'm not sure but for small holes, with plasma you might just spot the holes and drill to size anyway.
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02 Mar 2019 00:13 #127470
by tommylight
I never said it was a good idea, just an idea.
Then again i might give it a try to modify gcodetools to do just that. To many projects, not enough time.
Replied by tommylight on topic Can a component know the diameter of an arc its cutting
Far be it for me to keep you from doing that !Trying to do this at the HAL level not in Gcode so it just happens in the background according to some sort of Algorithm
I never said it was a good idea, just an idea.
Then again i might give it a try to modify gcodetools to do just that. To many projects, not enough time.
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01 May 2019 12:55 #132431
by rodw
Replied by rodw on topic Can a component know the diameter of an arc its cutting
Hmm, I finally revisited this problem and came up with the solution. I was a bit stuck on the Area of the circle
Now I just have to code it!
(And Hope it Works!)
Radius of an arc (R) with 3 points on the circumference of an arc (x1,y1; x2,y2; x3,y3)
R = (A* B * C) / (4 * K)
Where:
A = distance between point 1 and 2
B = distance between point 2 and 3
C = distance between point 3 and 1
K = area of the triangle
K = (x1 * ( y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) /2
A = sqrt(pow(x2-x1),2) + pow(y2-y1),2)
B = sqrt(pow(x3-x2),2) + pow((y3-y2),2)
C = sqrt(pow(x3-x1),2) + pow((y3 - y1),2)
Now I just have to code it!
(And Hope it Works!)
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