But if your gantry is has a .001 twist and the shims don't work, its not really going to make a difference in any work your doing.
I decided to make myself feel better about the twist in the gantry beam. What I'm seeing could well be what came from 80/20. Their stuff isn't perfect, and the folks at 80/20 don't claim it is. If the twist is "baked in," I hope that shims and bolt tightening are sufficient to remove the twist. I guess we'll see.
Gary
The Old Man and the C -----NC
But if your gantry is has a .001 twist and the shims don't work, its not really going to make a difference in any work your doing.
Retired Master Electrician, HVAC/R Commercial. FLA Saturn 2 4x4 CNC Router Mach4 Kimber 1911 45ACP
If the ridges in the spoilboard are .001", wouldn't that mean that the beam is off by much less?
Imo, .001" steps with a 60mm bit, is perfectly acceptable. That's only .0002" with a 1/2" bit.
Gerry
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Not necessarily. The beam is 6" high and the bit sets out in front of the beam by about the same amount. So, there is a 1:1. The fact that the bit sets lower when cutting should make no difference. it's just an extension of the gantry beam face.
What you ask about would occur, if the gantry beam height was less than the distance from the beam to the cutter. In that case, small changes at the gantry would maker larger changes at the cutter. Conversely, if the gantry beam height was greater than the distance to the cutter, you would need more shims to correct. With a 1:1, the ridge error=shims at the gantry. If the gantry height was 2X the distance to the cutter, it would take shims equaling 2X the error to correct.
As noted above, if the bit position is just an extension of the gantry beam's face (and it is), it doesn't matter how large the cutter's diameter. For a given depth of cut, the ridges will be the same, except with a smaller cutter, the ridges will be closer together. FWIW, I confirmed this by running a 1/2" cutter. Exactly the same ridges, only closer together.
Remember, you have to zero the cutter to the top of the spoilboard. When zeroing, the edge that contacts spoilboard first is what sets zero. That means the error will be the same with any cutter - only closer together.
Gary
The Old Man and the C -----NC
Actually, I had that backwards.
Say you have a 6" beam, 6" above the table.
Draw a line straight down and you have a 12" leg of a 90° triangle. The bottom of the triangle is your 60mm bit.
Now, rotate the triangle so that one edge of the bit is .001" higher, and the top moves over about .005".
But since the beam is only half that, it would mean the beam is off by about .005"
Thinking again, you don't need to account for the distance to the spoilboard. Just the beam height, and the bit diameter.
It's not 1:1.
Gerry
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(Note: The opinions expressed in this post are my own and are not necessarily those of CNCzone and its management)
Your spoil board can move X much when humidity changes?
Retired Master Electrician, HVAC/R Commercial. FLA Saturn 2 4x4 CNC Router Mach4 Kimber 1911 45ACP
I guess I'm just dense, but I'm not following. I don't understand what the bit diameter has to do with the issue.
Full Disclosure: I inadvertently misreported that my spindle set out in front of the gantry beam at about the same distance as the height of the gantry beam. When I wrote that, I was out of town and reporting from a faulty memory. The gantry beam is about 6" high and and the center of the spindle is about 8.5" from the face of the beam.
I believe I did a bad job of explaining the 1:1 ratio thing. It had nothing to do with the bit. I was referring to the ratio of the height of the gantry beam to the distance from the face of the gantry beam to the center of the spindle. If the two distances are equal, there is a 1:1 ratio. With a 1:1 ratio, if I added a shim to rotate the gantry beam .001" forward, the spindle would rotate downward by .001"
In my mind, it's really all about angles. To explain where I'm coming from, let's take the .001" deviation out of it. Rather, let's look at it from an angles perspective. With a 1:1 ratio, rotate the gantry 5 (or any) degrees, and the spindle angle rotates 5 (or any) degrees. Now that I have the true measurements, I dealing with a roughly 1:1.42 ratio (8.5/6=1.42) The angular relationship does not change, but the amount of movement at one end affects the amount of movement at the other end differently. In this case, a .001" deviation at the spoilboard translates to a .0007 at the gantry to get things right. (1" at the spoilboard divided by 1.42= about .0007" at the gantry). Thought of another way, for every .001" rotation at the gantry will result in .00142" rotation at the spindle.
Here's why I'm having trouble relating the bit diameter to the issue at hand. First let's make the assumption there is no error to correct. Second, let's assume that I rotate the gantry by 45 degrees. Finally, let's assume that my DOC is .005". The gantry is at 45 degrees relative to the spoilboard, and so is the cutting face of the bit (in a perfect world, the face of the cutter is parallel to the top and bottom of the gantry beam). Everything rotates the same. Now, I use my touch plate to zero the 60mm bit. Only the lowest edge of the bit can/will make contact with the touch plate. Start the surfacing routine. The bit lowers to -.005" and cuts. Only the leading edge will cut. Most of the bit won't make contact at .005" DOC. We end up with a series of ridges with a .005" deviation. The real deviation will be much more. Now let's change to a 1/2" bit. Same routine. The cut will be .005" deep and most of the bit won't make contact. We end up with a series of ridges with .005" deviation, same as with the 60mm bit. .005" won't tell us the true deviation, because there isn't full contact. Assuming a 60% stepover for both bits, the ridges are exactly the same, except the ridges are closer together with the 1/2" bit.
If we change to a 5 degree rotation, everything works out the same way, except more of the bit (maybe all) will make contact with the spoilboard. Ridges from both bits will be the same, except for the ridge separation distance and a longer taper angle with the 60mm bit.
Note: I ran both the 60mm and 1/2" bits last week. What I observed comported with what I've described. Note also that I had full contact with both bits, which came as no surprise at a .005" cut depth. FWIW, I took a pencil and scribbled lines all over the spoilboard. Surfacing removed all the lines.
So, I'm left not understanding how bit diameter plays into this. I may be overlooking something, but I don't know what that something might be.
Any help you can offer to help clear this up is much appreciated.
Thank you,
Gary
The Old Man and the C -----NC
You can't look at it as an absolute distance.
You have to use the angle deviation, and use trig (or CAD) to calculate the distances.
And you also can't use an example of 45° angle, which makes everything the same.
Say the gantry is tilted 1°.
At the tip of a 1/2" bit, tilted 1°, the difference between the front and back is .0087".
At the tip of a 60mm bit, tilted 1°, the difference between the front and back is .0412".
If you were to make a cut .002" deep, about 1/4 of the 1/2" bit would be cutting (roughly 1/8")
With the 60mm bit, the same amount of the bit would be cutting, only about 1/8".
When surfacing the spoilboard, you'll be cutting deep enough so that the entire bit is cutting.
Assuming 100% stepover, the 60mm bit will leave ridges nearly 5x taller than a 1/2" bit, when both are tilted at a 1° angle.
So it's not distance you are correcting, but angle.
As you reduce the stepover, the ridges get smaller, but only because subsequent passes are removing the high spots left from the previous pass.
The only thing I'm saying about bit diameter, is that if you're spoilboard has .001" ridges with a 60mm bit, you're not going to notice any error during normal cutting.
Gerry
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(Note: The opinions expressed in this post are my own and are not necessarily those of CNCzone and its management)
Thank you, Gerry. Cloud lifted. Your statement, "[a]s you reduce the stepover, the ridges get smaller, but only because subsequent passes are removing the high spots left from the previous pass" really brought it home for me. As a threshold matter, my measurement of the amount of deviation was flawed. I failed to account for stepover and the fact that subsequent passes would be removing high spots. So, at the bottom line, I really don't know the amount of deviation, and without having that number, I can't calculate the deviation angle. That also means I have no idea how much I need to shim.
The biggest problem is trying to calculate the deviation angle. I suppose I could make one pass at a depth sufficient to ensure full bit contact, measure the difference in the DOC at each end with a dial indicator, and use trig to get to the deviation angle. From that, I can use trig to calculate how much I need to shim to correct the deviation at the gantry level. Another option would be to level my glass plate on the spoilboard, and use my dual indicator tramming gauge to measure the height deviation. I'm thinking the tramming gauge should yield more accurate measurement. Trig does the rest. I'm not comfortable just laying the glass on the spoilboard, because there is twist involved and it will probably result in an inaccurate measurement without leveling.
You may be correct about not seeing the error in normal cutting. I'll have to crunch the numbers and see how it works out with the right ones. I am now confident that the deviation is more than .001", since I was reducing the true deviation (cutting off the high spots) with each successive surfacing pass. I'm pretty sure the deviation is at or less than .005", since I had full contact at .005" DOC. but if it's as much as .005, that's 5X more than I originally thought it was.
Does the methodology outlined above make sense, or am I still missing something?
Thank you so much for getting me on the right track.
Gary
The Old Man and the C -----NC
It would be simple just to do a test surface say area with your spoil board bit and another area the same size with common 1/2 end mill. A test plot in both directions.
Retired Master Electrician, HVAC/R Commercial. FLA Saturn 2 4x4 CNC Router Mach4 Kimber 1911 45ACP
Thanks, Bill. Certainly a possibility. To make it work, I'd have to use 100% (or maybe 90%) stepover. Otherwise, I would risk cutting down some of the angle with each succeeding pass. As I noted earlier, a single pass would probably do the trick as well. The 3rd option was to use my tramming gauge. My issue is with the ridges I get when surfacing side-to-side parallel with the X axis. I was able to handle running parallel with the Y axis with side-to-side tramming.
I decided to use a leveled piece of 3/8" thick piece of plate glass and my tram gauge to get a more accurate handle on the deviation and angle. Turns out the degree of deviation changed a bit, partly because this time I took my measurements at the extreme ends. I didn't do that when I measured before, but I was within a few inches. Turns out the deviation roughly doubled on the left side, and decreased by a little more than half on the right. Percentage wise, big differences. However, in raw numbers, we are talking about very small changes.
In my next post, I'll write more about methodology and get into the weeds on the numbers. I'll also discuss the pros and cons of what to do next. My intent is to document solutions for myself, and to provide others with some troubleshooting tools. I'm just happy that Gerry helped me see where I was going wrong. He changed my direction, and I found a decent methodology to for getting to the relevant numbers. A short, but very productive collaboration. I might have figured it all out on my own, were it not for it being some 50 years since I studied trigonometry in school, and my not having much use for it during my working years. I used integral calculus now and then, but not trig.
I should add that there are some very nice trig calculators online. They make make number crunching a breeze. Here's the one I used: Right-Angled Triangle Calculator
Gary
The Old Man and the C -----NC