looking for a Formula

[rusticr6]

we have 1 TM-2 that we just got and i have been running a Sl-40 for a good while and i am familiar with G42/G41 on the lathe. But on the mill we have not used cutter comp yet and i was looking for a formula to compensate my endmill radius when i go from a linear to an angle.

something like: Radius x tan(angle)

is there anything like that?

in my haas SL-40 lathe book there is a couple of pages that give you the compensation for 1/64 and 1/32 radius. they give 1-89 angles and the x and z amount to compensate for it. i have been trying to figure a formula that would match these but i cant get it spot on.

thanks

[CNCRim]

You don't have cad/cad?. Hint: add half(radius) of the endmill to the geometry you are doing then calculate from there.

[SeymourDumore]

Rustic

I'm either stupid or just really confused. If you're using G41/42 on the lathe, then why do you need to worry about the tool radius? As long as the control has the proper tip rad and direction, then you don't need to worry about anything other than making sure the ramp-on/off distance equals at least 1xRad? If you have the proper tip radius defined, then all your blocks should contain real part dimensions.
The formulas in the HAAS book refers to programming angles and radiuses without tool comp.
The same applies to milling as well. The ramp-on/off move must be at least 1/2xcutter diameter. In case of a 1/2" endmill it must be .2501 or more.

[rusticr6]

Sorry seymour, i wasnt clear on that. i was only saying that we use tool comp on the lathe, not on the mill.
i want to be able to figure out how much to move past the real part dimension at a specific angle in relation to my endmill radius.

Newtexas- we have auto cad and i can draw the part at scale, then i can see where my tool will be on the drawing. but i want to be able to figure all this myself. when you say add half the radius that is only partially right because you will need to move a distance further than that. it will be specific to to that angle.

am i making any sense?? i dont want to sound crazy.


kind of like if you want to put a radius at the beginning and the end of a taper on a lathe- i have a set of formulas that will let me get all the dimensions needed. i have tried to apply these formulas but i cant seem to get the right answer.

so lets say you have a linear that is 2.000 long and then it ramps up at a 15 deg. angle. we are using the side of the endmill, not the face.
you would have to move to the center of the endmill plus a certain amount that would be specific to your endmill radius and the angle.

[SORCHEROR]

rusticr6,if you have cad,than its simple,draw your part first,than offset all the part lines by half the cutter your using,now you have the exact path your cutter should follow,just analzye all the points on the new geometry and use those to program,at least until you get a cam program,this should work well on almost all your parts,asuming your only using lines,arcs etc.for me i cut surfaces and slpines too so i ahd to get cam package

[Jarwalcot]

Rusti,

See if this information helps you any...

[SeymourDumore]

"""
kind of like if you want to put a radius at the beginning and the end of a taper on a lathe- i have a set of formulas that will let me get all the dimensions needed. i have tried to apply these formulas but i cant seem to get the right answer.
"""

That's the part I don't get. If you're using tool nose comp ( G41/G42 ) on the lathe, then there is no formula, nor do you need one. You just program the radius tangent to the taper and done. All you need to worry about is to move off the part enough at the end of the path as to not cause overcut during ramp-off. No formula.

Ditto on the mill. If you have the radius drawn, then you program the linear-radius-linear moves, and then move the tool off the path by at least .0001 + 1/2 diameter.
Here is a quick example with a .01 radius on a 45deg angle:

G00 X-1. Y0. Z1.
G01 Z-0.5 F10.
G01 G41 X0. Y0.
G01 X0. Y0.9959
G02 X0.0029 Y1.0029 R0.01
G01 X0.9971 Y1.9971
G02 X1.0041 Y2. R0.01
G01 X2. Y2.
G01 G40 X2. Y3.


This is a cut for 3 lines having coords:
(0,0 : 0,1 ) (0,1 : 1,2) (1,2 - 2,2)
that are filleted by a .01 radius.
Ramp on is done @1" long from X-1., ramp off is also 1" long to Y3., in each case perpendicular to the previous path.

[Webderland]

E=mc²

[rusticr6]

seymour: i am not using G41/G42 on the mill.
you are most likely more advanced in programming than i am.

how would you program the 2" line then transition to 15 deg. angle with a 1/2" endmill. Without tool compensation. and still receive a 2.000" line.

thanks for the laugh, web

[wganders]

Rusticr6, if you are really interested in trying to figure out how to calculate your cutter pathes, versus plotting it in CAD, you may need to take a Trig class. Before I discovered CAD I would have to trig all my moves.

[SeymourDumore]

Ah Rustic. What you're trying to do is handcode mill programs with nothing but a calculator and a piece of paper.
I could ask why, but won't. I won't because there is no acceptable explanation I could imagine.
But if you must handcode, then:
Suggestion A: Use G41/G42 with full radius comp. This way the only thing need to be figured out is the ramp-on/off move, which is easy.

Suggestion B: If you're not using comp, then draw the part in CAD, offset the path by the cutter radius and then handcode the offsetted path.

I'd strongly recommend the former approach.

[gar]

070928-2000 EST USA

rustricr6:

Suppose you have a line segment from -4, 0 to 0, 0. This lies on the x axis from X=-4 to the origin. At the origin the line direction changes to -15 deg. The slope of the -15 deg line is TAN (-15 deg) = - 0.2679.

If your cutter is above the X-axis, then it starts at -4, 0.25 and moves to 0, 0.25. Now the cutter has to move to a point on a line perpendicular to the -15 deg line at a radial distance of 0.25 from the -15 deg line. This can be done as an arc or a straight line but such as to not cut into the apex of the two intersecting lines.

Using the straight line method you continue on the same X-path until you intersec the bisector of the 15 deg angle. This point is at X1 and Y1 where Y1 = 0.25, and X1 = 0.25 TAN (15/2) = 0.25 TAN (7.5) = 0.25 * 0.1317 = 0.0329. From here your path is at a negative slope of 0.2679. If you go a path length of 1 along the -15 deg line, then the end point of the cut is X = 1 COS (15) = 0.9659 and Y = SIN (15) = 0.2588. Next you have to calculate the center of the cutter which is at a point 0.25 on a perpendicular line to the -15 deg line from the end point of the cut. See if you can calculate this point, and see if I made any mistakes up to this point.

Do you really want to hand calculate the cutter path? For a learning and understanding experience it is good exercise. But, you should have a background in geometry, trig, college algebra, and maybe a little calculus.

.