[psychomill]

Stevo.... the picture is perfect and clears up much... The mechanical configuration is similar to my Mazak eMachines.

Next question is what is the normal axis directions on this machine? On your machine, "up and down" is Y-axis, the table (or the part) is sitting on the B rotary. From I can tell, the machine is a "5-axis HMC" in relation to axis directions correct? So this would make your spindle rotary the A-axis?


From your macro, looks like you program normal to the standard machine axis directions right? (I use 3D Coordinate Rotation a lot so my axes roll around depending on my head angle).



PS: My machine is 'normally' vertical so the axes are like a VMC. My 2 rotary points are the table carries C and the spindle carries B. Up and down with my spindle is Z axis.

[stevo1]

Yes that is correct. The spindle axis rotation is ”A”. Everything else is as you stated above.

When I rotate my spindle to A-90 (vertical) I use the G18 work plane and the Y-axis is up and down from the table as the Z direction would be on a standard VMC. The Z-axis acts as the Y-axis on a standard VMC.

I should have posted my code earlier as it may have made things a bit more clear. It’s really straight forward and that is probably why it drove me nuts at first that it would not work. Now realizing that it needed polar coordinates it makes me fell less like a doofis. I don’t have slightest idea of what is all wrapped into using this G12.1 as this is a first for me.

As you can see by the code I have B0 as the CL directly between 2 scallops. I can rotate to the first scallop right at the .250 radius but I just can’t seem to get around the scallop properly. Once I can then the macro will just repeat that same section of code.

It always looks easier on paper hey!!

Thanks again,
Steve

[psychomill]

I gotta run now because I've got a major fire to put out.. BUT, if you're up to it, you can give this a whirl.....

in Polar programming your rotary axis is simply taking the place of a linear axis. This is true with the code as well. If I remember right, I used to post code then edit the output by swapping my own rotary/linear axes with an editor (this was some years ago before I got posts that supported Polar programming for machines).

In my shriveling mind, this is how I made this work:

Program your profile as "normal" without rotary just as if you had the travel on a big machine to walk the tool around the whole part. Post the code. Now it's just a matter of swapping the stationary linear axis with the rotary axis (keep all signed directions [+ & -] the same).

So, in your case, your "normal" output would be a profile cut with ZX positions. Now on an editor, just simply swap all your "X" letters with the letter "B". Everything else stays the same. Cut this into your program and use G12.1 and it should cut your arcs.

Here's a short snip of what I'm talking about. This is using my machine (I think you'll get the idea for yours):

Normal XY profile output
X16.983Y0.
G12.1
G1X16.983Y0.
G2X16.7277Y-2.9338I-16.983J0.
X16.5771Y-3.6909I-.3162J-.3306
X14.3316Y-9.1119I-16.5771J3.6909
.
.
.
G13.1

For my machine, I swap "Y" with "C"
X16.983C0.
G12.1
G1X16.983C0.
G2X16.7277C-2.9338I-16.983J0.
X16.5771C-3.6909I-.3162J-.3306
X14.3316C-9.1119I-16.5771J3.6909
.
.
.
G13.1


As stated for your machine, I think if you just spit out ZX profile program, then just swap the X output with B, this will work....


Sorry, may or may not be the right way but my brain is at Mach 3 right now...
Gotta run.... My "fire" is going out of control now....

[Kiwi]

...Steve, You can use the code I posted if change your set to run in this direction.....

WARNING: Steve, I would check the code as the lug profile and radius position are not as your drawing.

[angelw]

Stevo,
One other thing you may need to know if you haven’t used polar coordinate interpolation before, is that the linear and rotation axes for polar coordinate interpolation must be set in parameters beforehand. I’m not sure what parameters these are in the series 15 but are 5460 and 5461 in late controls. I suspect they may be the same, as the parameter to specify whether the rotation axis for cylindrical interpolation is the X, Y, or Z, or an axis parallel to one of these axes, remained the same between series 15 and later controls.

The plane used before G12.1 is specified, selected by G17 ect, is canceled and restored when G13.1 is interpreted.

You will get some strange results if all the rues aren’t followed. Your tool path may look perfect on a PC based Back Plot but be rubbish at the machine if plane selection is neglected.

Regards,

Bill

[stevo1]

Psycho, Kiwi, Bill,

Thanks for all the input. I will give this a go tomorrow at the control. I don't use any Cadcam, backplot, PC stuff at all. I use long hand, Acad, and math to figure everything out first then code it at the control. I know I need to get out of the stone age but this has worked thus far all these years.

I really never thought that using Polar coordinates would seem this far advanced then what I am use to. I have written some crazy indepth macros but I speak that like another language.

I have to be honest Psycho the code you posted makes no sense to me but I am going to see what I get when I plug it in it's not you I just don't have a clue on polar. My wife says I am bipolar.....that I understand but I still think shes crazy

No worries Kiwi.....I know to prove everything out first. Been at this for quite some time now, but thank you for the warning.

Steve

[psychomill]

Good points by Bill.... I take this for granted on other controls...

Kiwi makes a good point about the program as well and I agree with him. Be careful with codes pulled from online or outide sources. Of course IMO, that's a given. And as Kiwi noted, the lug profile isn't correct. Being in a rush, I used a diameter of .415 instead of a radius.

So, thanks to Kiwi's sharp eye, I'll formally correct this (again, this is using my machine's configuration)


disclaimer: Below code is an example

G1X16.983C0.
G2X16.7666C-2.7022I-16.983J0.
X16.5245C-3.9198I-.3552J-.5622
X14.4563C-8.9129I-16.5245J3.9198
X13.7666C-9.945I-.5433J-.3835
X9.945C-13.7666I-13.7666J9.945
X8.9129C-14.4563I-.6487J-.1464
X3.9198C-16.5245I-8.9129J14.4563
X2.7022C-16.7666I-.6553J.113
X-2.7022I-2.7022J16.7666
X-3.9198C-16.5245I-.5622J.3552
X-8.9129C-14.4563I3.9198J16.5245
X-9.945C-13.7666I-.3835J.5433
X-13.7666C-9.945I9.945J13.7666
X-14.4563C-8.9129I-.1464J.6487
X-16.5245C-3.9198I14.4563J8.9129
X-16.7666C-2.7022I.113J.6553
C2.7022I16.7666J2.7022
X-16.5245C3.9198I.3552J.5622
X-14.4563C8.9129I16.5245J-3.9198
X-13.7666C9.945I.5433J.3835
X-9.945C13.7666I13.7666J-9.945
X-8.9129C14.4563I.6487J.1464
X-3.9198C16.5245I8.9129J-14.4563
X-2.7022C16.7666I.6553J-.113
X2.7022I2.7022J-16.7666
X3.9198C16.5245I.5622J-.3552
X8.9129C14.4563I-3.9198J-16.5245
X9.945C13.7666I.3835J-.5433
X13.7666C9.945I-9.945J-13.7666
X14.4563C8.9129I.1464J-.6487
X16.5245C3.9198I-14.4563J-8.9129
X16.7666C2.7022I-.113J-.6553
X16.983C0.I-16.7666J-2.7022

[psychomill]

AND......

Steve, the way I came to understand Polar programming is this..... The only thing really happening is the "axis substitution" between a rotary axis and a linear axis. The control does all the calculating.

So in polar, you no longer use an angle input for A, B or C (or other rotary axis that is valid). You actually use inputs as if you're walking around the part when not using a rotary. Then, it's just a matter of swapping out the axis name of the rotary with the stationary axis.

Take a look at my recent post where I described the axis swapping. On my machine, all that is happening is I swapped XY positioning with XC (My Y became C). Look at the values, there is no difference. In your case, your ZX positioning would now be ZB positions (You X becomes B).

The only thing I'm not sure is how your machine will interpret this in G18. Just be sure you plane select before going into G12.1


Trust me Steve, it's not all that advanced once you realize what the scenery looks like.... and I know you will...

[Kiwi]

G1X16.983C0.
G2X16.7666C-2.7022I-16.983J0.
X16.5245C-3.9198I-.3552J-.5622
X14.4563C-8.9129I-16.5245J3.9198

This code starts at X16.983 when Steve wants X0
Also path arcs 9.1554deg but Steve's code shows 9.2025deg.

[psychomill]

Kiwi....

I think you missed the parts of my posts where I said the code is for my machine's configuration and not Steve's. That's the subtle difference. My rotary axis is not on the same linear axis as Steve. My machine coordinate system grid isn't even the same as Steve's. It's mainly to get the idea of what Polar programming is and Steve is a sharp guy.

As for the angular difference.... Would depend on where Steve is "blending" the arcs when the math was done. The program I posted is off of my system. If I draw this based on Steve's drawing info, I get 9.1554 degrees. None the less, the "error factor" here is pretty small even at that part's diameter (unless he needs to holds tenths). However, I will say that if this were to be looped for each lug segment, then the stack up by the time you get to lug #16 would be significant. The code I posted goes all the way around the part though....


I would like to add... the actual position values is not what's important. It's the format and the math (equation) for Polar Coordinate Rotation that needs to be there. The rest is gravy...
I'm not programming the part for him, it's meant as an example. Sorry if that wasn't clear before...

[Kiwi]

...I think you missed the parts of my posts where I said the code is for my machine's configuration and not Steve's. ..

I understand your machine is a different set up and you have used C where he wants B.
I would have thought you would have posted some code that resembled what he needs for his setup after reading his post. Your code starts at 3 o'clock when he wants 12 o'clock.

..I have to be honest Psycho the code you posted makes no sense to me but I am going to see what I get when I plug it in....

[angelw]

Kiwi....

I think you missed the parts of my posts where I said the code is for my machine's configuration and not Steve's. That's the subtle difference. My rotary axis is not on the same linear axis as Steve. My machine coordinate system grid isn't even the same as Steve's. It's mainly to get the idea of what Polar programming is and Steve is a sharp guy.

As for the angular difference.... Would depend on where Steve is "blending" the arcs when the math was done. The program I posted is off of my system. If I draw this based on Steve's drawing info, I get 9.1554 degrees. None the less, the "error factor" here is pretty small even at that part's diameter (unless he needs to holds tenths). However, I will say that if this were to be looped for each lug segment, then the stack up by the time you get to lug #16 would be significant. The code I posted goes all the way around the part though....


I would like to add... the actual position values is not what's important. It's the format and the math (equation) for Polar Coordinate Rotation that needs to be there. The rest is gravy...
I'm not programming the part for him, it's meant as an example. Sorry if that wasn't clear before...

Hi Psycho,
Based on Stevo's drawing I get the same numbers you get, and at the end of the day its the part drawing that needs to be adhered to.

My approach would be the same as yours, that is, generate the code to go all the way round; there would be no accumulation of error. Irrespective of the diameter, any error caused by a large diameter is not going to be with the maths, or where the part origin is. Unless there is something about the part we don't know that would influence where origin is, I would program it as you have with the center as origin.

Regards,

Bill

[psychomill]

Your code starts at 3 o'clock when he wants 12 o'clock.

That depends on who's machine you're looking at. My code on my machine is "12 o'clock".

Remember, the rotary axes aren't the only difference... I also said that our coordinate system is different (as in the orientation is different).


Tell you what though, I used have a HMC post that used Polar commands. I'll see if I can dig that up. That would put the coordinate system orientation and the rotary axis the same as Steve's. The only difference then should be the plane select for G18 with Steve because he's running his A-axis @ -90.°

[Kiwi]

....Hi Psycho, Based on Stevo's drawing I get the same numbers you get, and at the end of the day its the part drawing that needs to be adhered to.

Looks like you are assuming the centre of the lug is on the same radius as the OD (33.466)

[Kiwi]

.... I want to keep my cutter/table at X0 and only rotate B-axis

Psycho..Looks like you missed this point.

[stevo1]

I just wanted to jump in and say thanks at the moment for the entire dialog as it is starting to make more sense. I only have a few moments as the wife is out of town for work, the teenagers are screaming for pizza(almost 9pm at night)….the baby is crying and I have to be up at 4am. I know….I know…preaching to the choir :violin:

I just wanted to state a few things. The reason I was only looking to move the B-axis instead of interpolating all the way around the part is because I do not have Z-axis stroke to reach past the center of the B-axis but by about an inch or 2. I really just thought that I could plug a few lines of code where my G3’s where at and everything would be good.

Kiwi…..you bring up a very good point about the lug being on the same radius of the OD. That is my fault. The reason that the numbers are different is because there are small holes in the center of the lug that the radius of the lug is based off. I had no idea that this was going to be as involved as it is so I left some of the details out (like the macro) just so things did not get cluttered up. So my apologize go to everyone. I of all people know better than to leave out details. All I need is an inch and I will turn it into a mile. It however does not seem so in this case and my biggest problem is I just don’t have the time to stand in front of the control to figure this out like I use to yrs ago.

Tomorrow morning I will post the drawing out with the details for you guys.

Anyhooo. I really do appreciate the help immensely as most of you know its pretty rare that I start threads asking for help. Most is only here helping where I can. So once again thank you all and I will catch you in the am.

Steve

[angelw]

The reason I was only looking to move the B-axis instead of interpolating all the way around the part is because I do not have Z-axis stroke to reach past the center of the B-axis but by about an inch or 2. I really just thought that I could plug a few lines of code where my G3’s where at and everything would be good.

Steve

Steve,
When you use polar coordinate interpolation, you create your program the same as if the the tool is interpolating around the part, but the result is what you have stated you want to achieve. It is really is quite easy so long as the rules regarding plane selection are attended to. Its much the same as Cylindrical interpolation where circular interpolation can be carried out on the circumference of a part as if it were a flat surface. The program is created as if the cylindrical surface has been unwrapped and placed flat on the table.

Regards,

Bill

[Kiwi]

....Kiwi…..you bring up a very good point about the lug being on the same radius of the OD.

No. Others are assuming the cente of the lug is on the OD radius. I believe from your info that the lug centre radius is shorter.

[stevo1]

No. Others are assuming the cente of the lug is on the OD radius. I believe from your info that the lug centre radius is shorter.

You would be correct!!! I do not have the drawing here at home but the BC diameter of the holes that the lug radius are based off are slightly smaller then the OD of the part. Again my apologize for leaving out these details.

You do however have to understand that when helping here in the forums that some assumptions are made. Psycho knows me and probably realizes that in most cases all I need is to be pointed in the right direction and most details are not needed. Everyone who helps here is working on some assumptions, if we did not then most help would be scarce. Myself like Psycho are members at many other forums (5 to be exact), along with the fact that I moderate 1 of them.....I make many assumptions. This is however small in comparison to the size of the business that Psycho is running so believe me when I say he is allowed to make all the assumptions that he wants with me. These sites are a really good source of "pay it forward" mentality which I truly believe in.

This is probably about my 25 thread at this site to where I actually ask for help here in this forum and the rest of posts I have made are all about helping others where I can. So I really don't want to see this turn into the "who's got a bigger MT". No offense Kiwi but Psycho has some pretty big MT's....literately.

I really gotta crash now....starting to talk stupid.

Night all,
Steve

[Kiwi]

... I don't use any Cadcam, backplot, PC stuff at all. I use long hand, Acad, and math to figure everything out first then code it at the control...

As it appears that nobody is able to supply accurate usable code, and take Psycho advise and code the complete circumference, it looks like you will be "plotting point code all day long....".