Re: [Emc-users] Log file
On Sun, Mar 13, 2011 at 12:12:17PM +0330, Farzin Kamangar wrote: We need some way to record login name of machine users, date and time of operations to have the information to refer to at a later time. Having never heard of such functionality, I spent a minute on a crude wrapper: Copy this to /usr/local/bin/run_emc » #!/bin/bash # Wrapper to record who used emc, when, and for how long. record=/var/tmp/emc_usage [ ! -e $record ] touch $record # Nothing fancy == easy for anyone to customise: echo $record whoami $record echo Started EMC2 at: $record date $record /usr/bin/emc echo Session ended at: $record date $record « - That's all. Make it executable: $ chmod a+x /usr/local/bin/run_emc Try it out: $ run_emc OK, now we make swarf, then close the application. To check the record: $ more /var/tmp/emc_usage erik Started EMC2 at: Sun Mar 13 20:41:27 EST 2011 Session ended at: Sun Mar 13 20:41:46 EST 2011 That's an easy way to do it, given that you have individual user accounts for us to pick up user identification. Near enough for starters? Erik -- The game can also serve as a didactic analogy, used to convey the somewhat counter-intuitive notion that design and organization can spontaneously emerge in the absence of a designer. - http://en.wikipedia.org/wiki/Conway%27s_life -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] Chips on Flash
On 3/12/2011 11:09 PM, Kenneth Lerman wrote: A first class parcel costs $1.05 plus .17 per ounce. That's pretty cheap. Ken On 03/12/2011 09:53 PM, Kent A. Reed wrote: On 3/11/2011 11:32 AM, Kirk Wallace wrote: I'm not sure how typical my situation is, but it costs me about $20 to download 1GB of data, so it's not exactly cheap to download the latest EMC2. I wonder if there would be a demand for EMC2 on some sort of flash media? How cheap could it be? I didn't see anyone directly address your question, Kirk. Looking at my latest MicroCenter flyer, 2GB USB flash drives are going for about 5USD, retail. I couldn't find anyone advertising quantity discounts, but I didn't look very hard. I'm not sure about the best method for shipment, but the lowest postal service rate is 2.38USD for media mail up to 1 pound. Let's call it 3USD to account for packaging. Not counting the labor/time to acquire, burn, and test the flash drive, that's about 8USD delivered, or about 40 percent of your current cost. If you provided your own flash drive, the cost would fall to about 5-6USD (e.g., the cost of shipping the drive both ways). Burning CDs instead would result in a cost somewhere around 3-4USD, although more time is required to burn and test the CD. It seems to me this method is most likely to work well by pairing up with a buddy who has better Internet access rather than depending on the EMC2 developers. I'm not sure, though, what you mean by the latest EMC2. If this means the latest LiveCD distribution, then the burden on the buddy is pretty minimal. If you mean EMC2 as of last night along with an fully patched Ubuntu, then the burden on the buddy is greater. Regards, Kent (willing to be a small-time buddy) And I find that I misinterpreted what media mail means. It isn't appropriate for this use in any case. So maybe subtract a dollar each way from my estimate---with the caveat that my USPS office, at least, interprets their regulations scrupulously, which drives my wife crazy when she wants to send those musical greeting cards to our grandkids. I'd want to discuss with them the optimal packaging and rate selection for sending flash drives. Regards, Kent -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
I'd write a small program in c or python to solve this. You only need to solve the problem for 1/4 of the perimeter. After that, rotate and mirror Here are my thoughts on solving it: - find parameteric equations for an arc of the circle in 1st quadrant and the edge of the square - http://en.wikipedia.org/wiki/Parametric_equation - for simplicity, I'm assuming a radius of 1 - circle: x=a*cos(t); y=a*sin(t); in first quadrant, 0tpi/2 - line with slope -1: *x1=-t+1; y1=t; 0t1* - circle with parameter range 0 to 1: *x2=cos(pi*t/2); y2=sin(pi*t/2); 0t1* - assuming material thickness of 1 unit, with the square on the bottom at z=0: *z1=0; z2=1* - compute n values of x1,y1,x2,y2 between 0 and 1 with a for loop - use x2, y2, z2 as X,Y,Z in the gcode - use x1, y1, z1, x2, y2, z2 to find values for A, B, and/or C Mark On Sat, Mar 12, 2011 at 5:30 PM, Viesturs Lācis viesturs.la...@gmail.comwrote: Hello, gentlemen! I would like to ask, if anyone has an idea, how to create a code to produce this kind of part (both files contain the same model, I just saved it in 2 different formats): http://www.cutting.lv/fileadmin/user_upload/Test.IGS http://www.cutting.lv/fileadmin/user_upload/Test.STL My goal would be cutting this part from thick slab of material. Those flat planes represent top and bottom surfaces of the slab. Basically I would like to get code, in which the waterjet (or any 5 axis plasma or cuts a circle on the top, but rotary joints that tilt the head would move so that something like a square is on the bottom. So the problem is finding out the necessary tilt angle, which corresponds to the slope of the edge. I thought that it could be something like dividing the top contour (in this case - the circle) in 0.1 mm segments and then getting the slope angle, but I have no idea, how to do that. Can anyone recommend some kind of solution? Is there some _affordable_ CAM application that can do that (I have found one that costs 12K EUR, but I do not even consider that to be an option)? Or can I calculate that myself with some trigonometry? Since I know the distance from one plane to another (that is the thickness of material), I would need only horizontal distance from one line to another to get the angle with atan function. The distance between both lines could be calculated in 0,1 - 0,2 mm increments. That would not affect the quality of the result and probably would not create insanely long code for such a small part. I would appreciate any ideas on this matter. I think that any solution that works will do! Thanks, Viesturs -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
Suggest a search on the keyword lofting, this is similar (maybe) to your problem. -Original Message- From: Viesturs Lācis [mailto:viesturs.la...@gmail.com] Sent: den 12 mars 2011 23:30 To: Enhanced Machine Controller (EMC) Subject: [Emc-users] CAM-related question Hello, gentlemen! I would like to ask, if anyone has an idea, how to create a code to produce this kind of part (both files contain the same model, I just saved it in 2 different formats): http://www.cutting.lv/fileadmin/user_upload/Test.IGS http://www.cutting.lv/fileadmin/user_upload/Test.STL My goal would be cutting this part from thick slab of material. Those flat planes represent top and bottom surfaces of the slab. Basically I would like to get code, in which the waterjet (or any 5 axis plasma or cuts a circle on the top, but rotary joints that tilt the head would move so that something like a square is on the bottom. So the problem is finding out the necessary tilt angle, which corresponds to the slope of the edge. I thought that it could be something like dividing the top contour (in this case - the circle) in 0.1 mm segments and then getting the slope angle, but I have no idea, how to do that. Can anyone recommend some kind of solution? Is there some _affordable_ CAM application that can do that (I have found one that costs 12K EUR, but I do not even consider that to be an option)? Or can I calculate that myself with some trigonometry? Since I know the distance from one plane to another (that is the thickness of material), I would need only horizontal distance from one line to another to get the angle with atan function. The distance between both lines could be calculated in 0,1 - 0,2 mm increments. That would not affect the quality of the result and probably would not create insanely long code for such a small part. I would appreciate any ideas on this matter. I think that any solution that works will do! Thanks, Viesturs --- --- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
[Emc-users] Icon for pyVCP
I have a stand alone pyVCP application that configures a Modbus device. I would like to have an icon and script that invokes the application. The problem is that halrun requires the -I (interactive) option. When I close the application, a terminal is left open with the halcmd prompt. From there 'quit' or 'exit' is needed to completely close the session. Does anyone have a script that can bring up pyVCP and close it completely, or some other solution? I suppose the 'emc' script does this for EMC2 but it would take me a while to figure out how emc works and then to modify it to my application. -- Kirk Wallace http://www.wallacecompany.com/machine_shop/ http://www.wallacecompany.com/E45/index.html California, USA -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
On Sun, 2011-03-13 at 11:45 -0400, Mark wrote: I'd write a small program in c or python to solve this. You only need to solve the problem for 1/4 of the perimeter. After that, rotate and mirror Here are my thoughts on solving it: - find parameteric equations for an arc of the circle in 1st quadrant and the edge of the square - http://en.wikipedia.org/wiki/Parametric_equation - for simplicity, I'm assuming a radius of 1 - circle: x=a*cos(t); y=a*sin(t); in first quadrant, 0tpi/2 - line with slope -1: *x1=-t+1; y1=t; 0t1* - circle with parameter range 0 to 1: *x2=cos(pi*t/2); y2=sin(pi*t/2); 0t1* - assuming material thickness of 1 unit, with the square on the bottom at z=0: *z1=0; z2=1* - compute n values of x1,y1,x2,y2 between 0 and 1 with a for loop - use x2, y2, z2 as X,Y,Z in the gcode - use x1, y1, z1, x2, y2, z2 to find values for A, B, and/or C Mark On Sat, Mar 12, 2011 at 5:30 PM, Viesturs Lācis viesturs.la...@gmail.comwrote: Hello, gentlemen! I would like to ask, if anyone has an idea, how to create a code to produce this kind of part (both files contain the same model, I just saved it in 2 different formats): http://www.cutting.lv/fileadmin/user_upload/Test.IGS http://www.cutting.lv/fileadmin/user_upload/Test.STL My goal would be cutting this part from thick slab of material. Those flat planes represent top and bottom surfaces of the slab. Basically I would like to get code, in which the waterjet (or any 5 axis plasma or cuts a circle on the top, but rotary joints that tilt the head would move so that something like a square is on the bottom. So the problem is finding out the necessary tilt angle, which corresponds to the slope of the edge. I thought that it could be something like dividing the top contour (in this case - the circle) in 0.1 mm segments and then getting the slope angle, but I have no idea, how to do that. Can anyone recommend some kind of solution? Is there some _affordable_ CAM application that can do that (I have found one that costs 12K EUR, but I do not even consider that to be an option)? Or can I calculate that myself with some trigonometry? Since I know the distance from one plane to another (that is the thickness of material), I would need only horizontal distance from one line to another to get the angle with atan function. The distance between both lines could be calculated in 0,1 - 0,2 mm increments. That would not affect the quality of the result and probably would not create insanely long code for such a small part. I would appreciate any ideas on this matter. I think that any solution that works will do! Thanks, Viesturs Hi Viesturs, That must have been a pretty clean iges file as it imported into synergy just fine. Much better than some of the Catia stuff I've gotten. Unfortunately, I don't have the slightest how to CAM it in 4 axis or 5 axis. It is rather small so 4 axis should do it. Just plain calculations do make sense and even if the code is 10K lines or so you really don't care. Have fun. Dave -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
2011/3/13 Mark mpic...@gmail.com: I'd write a small program in c or python to solve this. You only need to solve the problem for 1/4 of the perimeter. After that, rotate and mirror Here are my thoughts on solving it: - find parameteric equations for an arc of the circle in 1st quadrant and the edge of the square - http://en.wikipedia.org/wiki/Parametric_equation - for simplicity, I'm assuming a radius of 1 - circle: x=a*cos(t); y=a*sin(t); in first quadrant, 0tpi/2 - line with slope -1: *x1=-t+1; y1=t; 0t1* - circle with parameter range 0 to 1: *x2=cos(pi*t/2); y2=sin(pi*t/2); 0t1* - assuming material thickness of 1 unit, with the square on the bottom at z=0: *z1=0; z2=1* - compute n values of x1,y1,x2,y2 between 0 and 1 with a for loop - use x2, y2, z2 as X,Y,Z in the gcode - use x1, y1, z1, x2, y2, z2 to find values for A, B, and/or C Thanks! I am not sure that I understand correctly all the calculations You offer, but it gave me an idea, that since I am using circle as the basic shape, I could paste both contours in polar coordinate system and the difference of radius between both contours on each theta value would give me the horizontal distance between them. Adding vertical distance and I get value of the slope with atan function, theta value gives direction - they both are position commands for A/B and C axis respectively. Looks pretty simple on paper, but I do not know, how do I get proper g-code from it. I will spend some more time with Your formulas, maybe I will manage to understand them completely. Addressing these calcs with parametric equation approach seems promising to me. Viesturs -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
On 12 March 2011 22:30, Viesturs Lācis viesturs.la...@gmail.com wrote: So the problem is finding out the necessary tilt angle, which corresponds to the slope of the edge. I think that it is non-trivial. The problem is deciding which point on the bottom curve matches which point on the top curve (and vice-versa). Once you have pairs of points then the mathematics is relatively simple. The solution probably involves finding the set of shortest lines from top to bottom, and then bottom-to-top. You will often find a situation where a sharp corner at the top matches up with an entire arc below, and vice-versa, so need to solve in both directions. I think I would do it in Matlab. -- atp Torque wrenches are for the obedience of fools and the guidance of wise men -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
Ah, got it. rotate and tilt is almost all it needs. Is this for mental exercise or does this part have a purpose? Dave On Sun, 2011-03-13 at 18:13 +, Dave Caroline wrote: Dave its water jet/plasma so the head has all the rotation and 5 axis needed, Viesturs, I cannot see the item as heekscad bombed on the stl file but gcode can handle the maths so you can can just do the maths in the gcode program. Also then is another way which you can develop from sheet metal practice http://search.sheetmetalworld.com/news/articles/197.cfm and develop the paths in gcode break into as many triangles as needed for accuracy Dave Caroline On Sun, Mar 13, 2011 at 5:30 PM, dave dengv...@charter.net wrote: On Sun, 2011-03-13 at 19:02 +0200, Viesturs Lācis wrote: 2011. gada 13. marts 17:45 Lars Andersson l...@larsandersson.com rakstīja: Suggest a search on the keyword lofting, this is similar (maybe) to your problem. Unfortunately majority of results were talking about boat hulls, rest of them - how to use lofting function in 3ds max. But I already know, how to create such model, problem is to calculate, how that shape on the sides is created. It consists of straight lines, connecting top and bottom contours, but I do not know, how to determine them, how to find a way to calculate each of them. 2011/3/13 dave dengv...@charter.net: Hi Viesturs, That must have been a pretty clean iges file as it imported into synergy just fine. Much better than some of the Catia stuff I've gotten. All the compliments go to SW2010, in which I created the model. Unfortunately, I don't have the slightest how to CAM it in 4 axis or 5 axis. It is rather small so 4 axis should do it. No, it is 5 axis stuff, because the basic contour is not a straight line, it has turns and thus the plane, in which the tool would be tilted, also turns. It it takes 5 axis then I'm missing something because it looks like a truncated cone which intersects with a cube. If so, then along the axis of rotation it is a pair of straight lines. So you calculate and machine one pass then rotate and do another. Dave Viesturs -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] Draftsight for Linux
Can this be true?? Hopefully SW is next http://www.fcsuper.com/swblog/?p=2662 -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
On Sun, Mar 13, 2011 at 12:51 PM, Viesturs Lācis viesturs.la...@gmail.comwrote: Thanks! I am not sure that I understand correctly all the calculations You offer, but it gave me an idea, that since I am using circle as the basic shape, I could paste both contours in polar coordinate system and the difference of radius between both contours on each theta value would give me the horizontal distance between them. Adding vertical distance and I get value of the slope with atan function, theta value gives direction - they both are position commands for A/B and C axis respectively. Looks pretty simple on paper, but I do not know, how do I get proper g-code from it. I will spend some more time with Your formulas, maybe I will manage to understand them completely. Addressing these calcs with parametric equation approach seems promising to me. Viesturs (5 axis cut - arc at top, straight line at bottom) (commented code is for max5kins - note, no A axis) (tested on 5axis sim) (assuming table at Z=0) #10 = 1 (material thickness) #11 = 2 (radius, and the square's diagonal) G10 L2 P2 Z[-1*#10] G55 G00 Z[#10] (X[#11] Y0 A0 B0) X[#11] Y0 B0 C0 M3 #1 = 0 (#1 is used as the parameter for the equations) O101 while [#1 lt 1] G1 Z[#10] F10 #21=[cos[#1*90]*#11](X) #22=[sin[#1*90]*#11](Y) #24=[atan[[[1-#1]*#11]-#21]/[#10]] (A, for AB) #25=[atan[[#1*#11]-#22]/[#10]] (B, for AB) #26=[#24+#25] (B, for BC **pretty sure this is not right - how do we really calculate this?**) (X[#21] Y[#22] A[#24] B[#25]) X[#21] Y[#22] B[#25] C[#1*90] #1 = [#1+.01] O101 endwhile M2 -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] CAM-related question
On 13 March 2011 22:26, Viesturs Lācis viesturs.la...@gmail.com wrote: Honestly - I just would like to produce some range of samples, that would demonstrate, what a 5 axis waterjet machine is capable of. There was a time when I worked somewhere with a 4-axis wire EDM machine where lots of people had cubes with one of their initials on the top, and the other on the bottom. -- atp Torque wrenches are for the obedience of fools and the guidance of wise men -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] Draftsight for Linux
On Sun, 13 Mar 2011 20:43:15 +0200, you wrote: Can this be true?? Hopefully SW is next Uggg - why? Steve Blackmore -- -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
[Emc-users] DraftSight for linux, free beta
Hi all; I just downloaded and installed the mdv rpm on this pclos box, and the rpm went it except for the k-menu additions, so I have to run it from a user terminal session until I figure out how to edit the kmenu's on pclos. Has anyone else played with it yet? I ran it, selected 'new' as it said in the help, then went to the next section to try and setup a properly scaled gridded view port for the next project, in this case the 8 sticks it will take to make 2 overlapping at the center doors for this cabinet I'm building. Unforch, nothing in the Properties column on the left appears to be editable, so its not possible to proceed further using known dimensions for the snap to grid etc etc. I'm a dummy I guess. This thing isn't in english as its first language, so some of the terms I see in the help file are probably going to need some explaining here and there. I did finally get it to take a filename in the first box, but when I continue to adjust other things, the filename disappears and when I get frustrated and quit, the project is still called NO_NAME.dwg. I think I need to start from square one. But, before I put a lot of effort into this, is there such a thing as a .dwg to .ngc converter? Thanks everybody. -- Cheers, Gene There are four boxes to be used in defense of liberty: soap, ballot, jury, and ammo. Please use in that order. -Ed Howdershelt (Author) http://tinyurl.com/ddg5bz http://www.cantrip.org/gatto.html It is impossible to travel faster than light, and certainly not desirable, as one's hat keeps blowing off. -- Woody Allen -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] Draftsight for Linux
Viesturs Lācis wrote: Wow, that is great news! I think that first few precedents are the hardest part here - we should see even more apps on Linux once the ice has started to move. Unfortunately I am afraid that SolidWorks licence will not be cheaper for Linux users and still will be in +10K range. Please correct me, if I am wrong with the numbers here. BTW does anybody know, if SolidWorks has a licence for students? And what is approximate amount payable for that? Yes they do. It looks like it's $150 for a 12-month license. They do mention that the student version is unsuitable for commercial use, but I don't know what that means. They say it's fully functional, so maybe it watermarks files or something ... Information is from this page: http://www.solidworks.com/sw/education/student-software-3d-mcad.htm (the pricing is available if you click the buy student edition button) - Steve -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] DraftSight for linux, free beta
I think I need to start from square one. But, before I put a lot of effort into this, is there such a thing as a .dwg to .ngc converter? If you can output dxf files, Inkscape may be able to read it in and output gcode for you. There is even a page in the linuxcnc wiki that shows how to get gcode out of Inkscape. Kyle -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
Re: [Emc-users] Single to Three Phase Rotary Converters
On Wed, Mar 9, 2011 at 12:35 AM, Kirk Wallace kwall...@wallacecompany.comwrote: Let me try to provide more details on my understanding of the phase timing of DIY converters. Attached is a schematic of a common rotary converter. The source is here: http://www.metalwebnews.com/howto/ph-conv/ph-conv.html http://www.metalwebnews.com/howto/ph-conv/fig1.html I used this to make my converter and is the only design I have some understanding of. As the schematic shows L1 and L2 go straight to the output and are unaffected. Since L1 and L2 are single phase, L1 is a mirror of L2, therefore 180 degrees out of 360 apart. But, but, you are committing a tautology. If you reference L1 to L2 and vice versa, of course L1 is a mirror of L2. In your scheme there is no 'ground' or reference point, so you only have one phase coming in, and the motor generates the second (and third) phase L3, shifted with respect to L1/L2. One way to see it comes from the trigonometric identity holding that sum of three sines shifted by 0, 120 and 240 degrees is zero: sin(x)+sin(x+2*%pi/3)+sin(x+4*%pi/3)=0 Let's say we take some arbitrary reference point, so that L1(t) is the voltage on the first leg. Assume L2(t)-L1(t) = V sin(2 pi f t). If the converter generates L3(t)=L1(t)-V sin(2 pi f t + 2/3 pi), then L1-L3 is V sin(2 pi f t + 2/3 pi), i.e. shifted by 120 degrees, and L3-L2 is -V sin(2 pi f t + 2/3pi) - V sin(2 pi f t) which reduces to V sin(x+4*%pi/3) i.e. the same AC shifted 240 degrees. Note that it doesn't matter what the reference point is, because its potential L1(t) drops out --- all that matters are voltage differences between the legs of the circuit, not the potentials of the legs themselves. -- Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d ___ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
