Damn... If they OP can't respond to this, you are wasting your spoonfeeding....
Oh, wait, he hasn't been back in the thread since the post! People are f'ing lazy...
Sorry for the radio silence, real life has been calling.
I picked the X,Y zero to be the front and center of the pocket. You can locate that off the mag well and/or the bolt catch slot with dimensions from a drawing.
I've labeled each endpoint of every segment along the toolpath. I propose to start at point A as it will not touch the final surface at that position so any funny marks from plunges or dwells will not affect the final finish.
The next step is to find the coordinates of all of these labeled points. Anyone care to give it a shot? It is just basic addition and subtraction using the given dimensions...
Edit: BTW, I am using CAD to make all this but pen and paper is equally effective. I am using CAD as it is a lot easier to make a legible copy ready for upload.
Damn... If they OP can't respond to this, you are wasting your spoonfeeding....
Oh, wait, he hasn't been back in the thread since the post! People are f'ing lazy...
CAD, CAM, Scanning, Modelling, Machining...
mcphill
The OP got his machine setup & made the part within 2 days of posting, I don't think he has had the need to come back
Mactec54
My effort was directed at thisdjj. Hopefully he responds.
You see now why no one wants to waste their time with these people...
I'm sorry Keebler I will try this tomorrow been real busy at work and won't have a day off until tomorrow
I thank you for your time and it will not be in vane
So I've been looking at the drawing and would like to know what points do you need the math for I'll be more than happy to do it for the learning experience as well as you helping me
The easiest way to program a toolpath is with centerline programming, that meaning you program the path that the center of the tool will take. Each line of G code moves the tool from one point to another point. These points are the ones I have labeled A,B,C,etc. You need all of them to write a program. I'd suggest an excel-like table with point label, X coordinate and Y coordinate listed out, you can do it on a piece of paper or something too but if you keep it digital it is easier for others to check your work. I have already given you point A in the drawing.
Matt
If you choose not to complete the exercise I have presented then my efforts were indeed in vain. Regardless, I am glad you are making an effort to learn something.
You are right I took a look at it printed it out and couldn't figure start to finish points for the math and had just figured to try and get some help that's why I signed up for the class.
The location I marked on the drawing as the origin is the place where X = 0 and Y = 0. You will move your machine until the center of the spindle is at that point and then zero the X and Y axes so the machine reads (0,0) when it is at that point.
To get to point A (where we want to start cutting) we have to move 1.8925" in the negative X direction and .0005" in the negative Y direction. These numbers are ones I already provided for point A.
To find these coordinates, you just need some addition and subtraction.
For the x coordinate of point A, we will take the given length of the slot (2.080") and subtract the radius of the tool (we already decided on a 3/8" diameter endmill).
2.080 - 0.1875 = 1.8925". Since it is left of zero, it should be negative so -1.8925 is the coordinate we need. It is better practice to always add when you move from left to right and subtract when you move right to left. The proper math is:
0 - 2.080 + 0.1875 = -1.8925"
For the Y coordinate it is a bit more math but still straightforward. Starting at zero, we will go up .345 (half of .690) which is the distance from zero to the top of the pocket. Now we go down 0.095 to get to the top of the smaller left pocket, go down another 0.438 for the width of that pocket, then finally we add the tool radius back on.
0 + 0.345 - 0.095 - 0.438 + 0.1875 = -0.0005"
Now we have the coordinates for point A (-1.8925,-0.0005)
Moving on to point B, Same deal, just add and subtract to get there, I'm not going to hold your hand...
For X: 0 - 3.512 + 1.080 - 0.1875 = -2.6195"
Y is obviously the same value as for point A, it is a straight line.
Point B is (-2.6195, -0.0005)
So now we are well on our way...
A (-1.8925,-0.0005)
B (-2.6195, -0.0005)
This will be my last post in this thread until some brain power is expended...
I'm going to print this out so that I can study and understand it thank you
Hi keebler303, allow me to pick up the slack. I want to learn this. So D is 0.1875 from the lower edge, the width of the part is 0.500 assuming center line splits it symmetrically (judging by the fact that you didn't leave any measurements to show otherwise.. we have an Y for the D at - (0.25 - 0.1875) = - 0.0625, now X is a bit funky for me, correct me if i'm wrong it says Radius of the curve is at 0.218, so it should be 0.218 from D to the right edge of the cut, the cut is at 0.1875 from B on X, so for D, X= -2.6195 - (0.218 - 0.1875) = -2.65. So final result for D (-2.65, -0.0625). Now for C, X is the same as for B since it is a straight down path, but Y = -0.0005 - 0.062/2 = - 0.0315, so the final answer for C(-2.6195,-0.0315)
Am I on the right track, just want to confirm before continuing.
And thank you for doing it. I'm sure it will help a lot of noobs out here, me inclusive.
Coredump
Thanks for coming.
I wouldn't assume anything on a drawing. I previously mentioned that my origin (0,0) is centered on the .690 dimension in the Y axis, so I'll go off that.
Y for point D goes like this: 0+(.690/2)-.095-.5+.1875 = -.0625 Looks like your ASSumption was ok.
X for point C goes like this: 0-3.512+1.0800-.1875 = -2.6195
The difference between Xc Xd and Yc Yd should be the same since they are on an arc 90 degrees apart. To figure it out, I subtracted the tool radius from the big radius: .218-.1875 = .0305
Xd = Xc-.0305 = -2.6195-.0305 = -2.6500
Yc = Yd+.0305 = -.0625+.0305 = -.0320
So your Y coordinate for C was a bit off, this is why: If you start at your assumed centerline, which is at Y=0, then your Yc should be 0-0.062/2 NOT -0.0005-0.062/2. That gives you Yc = -0.0310. That is not what I just calculated above so what is going on? I checked my drawing and the two radii on either side of the .062 are actually .219 radius and not .218. To make things easier, lets just make all the radii .218. If we do that, the .062 dimension grows to .064. Just cross out the.062 dimension on your copy and make it 0.0640. Now Yc = 0-0.064/2 = -0.0320. We got it!
Keep up the good work
Matt
One note about working with numbers:
If you are working to 4 decimal place accuracy, you should state all your numbers to that accuracy. If you don't, it is hard to tell if 1 is actually .9900 rounded, or if you just typed the 1 and forgot to enter the rest of the number. If you see 1.0000 then you KNOW it is 1 for sure.
It is a good habit to get into to help avoid errors and mistakes.