Is this drill chuck out of spec?

D

drcrea

Registered
Registered
Joined
Jan 27, 2024
Messages
11
Greetings all,
New member and a bit of a novice. I'm trying to understand TIR (total indicated runout) and whether or not my new drill chuck is within the manufactures specs. I purchased a Llambrich JK-160 MT2 drill chuck that gives a "certified" max TIR of 0.0016". I've read a number of online posts talking about runout and a few talked about TIR as being either the stated value or +/- (double) the stated value. This is where I'm a bit confused.

Is the TIR, in this case 0.0016" or double 0.0032"?

After installing the new chuck, I measured 0.0005" at the quill and on the arbor itself. On the drill body I measured 0.001". When measuring at the chuck, I am getting 0.0065" consistently. I chucked several different items including a Starrett edge finder and several 1/2" router bits having long smooth shanks. All gave the same results when measured at the chuck. I also rotated the chuck 180 degrees (it's tabbed) and repeated the measurements with similar results. Rotating my have corrected ~0.00025" or so. Hard to measure that.

I wasn't expecting this much runout given the very low runout at the quill/arbor and the spec's of the drill chuck. My novice expatiation is that my quill is introducing 0.0005" and the chuck is certified to have 0.0016" thus the total runout I should measure at the chuck would be around ~0.0021" - 0.0026". However I'm getting much more at 0.0065".

Am I misunderstanding how to measure for TIR or is this chuck just out of spec?
 
Welcome, the gurus will be along shortly. Edit, more gurus. :)
 
Runout is comprised of two parts; radial and angular runout. Radial simply means tha at the measurement point, the runout in the radial direction indicates how far the center of the part being measured is from the axis of rotation (x2). A part can have zero radial runout at the measurement point but the center of the part measured can be tilted relative to the spindle axis. If that is the case, moving to a different measurement point will give a non zero TIR value. A typical method of measurement of runout is to measure close to the chuck and also at some specified distance from the chuck. This gives some indication of the angular contribution. In my experience, the value distant from the chuck is always greater than the value close to the chuck.

Another factor to consider is that TIR is a composite of the individual TIR's of the various components. The spindle runout measured at the MT socket,the Morse taper arbor , the drill chuck, and the pin being measured all have their individual runout. Furthermore, the individual runouts don't simply add but depend on their individual rotation angle with respect to the other components.

Finally, the pin used to measure runout may not be perfectly straight. This can be assessed by marking the high spot and rotating the pin 180º and remeasuring though.
 
TIR, Total indicated reading, (often called total indicated runout) means that if you put a gauge up to something that rotates, and measure the difference between the extream high and extreme low reading, that will be the "TOTAL" indicated runout.

Because that part you're measuring is rotating, if it's out 0.005 inches, when you read the HIGH, it'll be the diameter PLUS five thousandths, and when you read the LOW, it'll be the diameter MINUS five thousandths. So five thousandths off of center will show a TIR of 0.010 inches. Two times the actual deviation from "true".

If the product calls for 0.0016 TIR as the spec, then that is what you should directly measure. If they list 0.0016 RUNOUT as the spec, then you should expect up to 0.032 to show up on your indicator.

You said you measured at the quill AFTER installing the new chuck. The first thing you should have done is measured inside of your drive feature. The female MT2 taper. Anything you measured with the chuck installed will NOT be orienting that male morse taper on the chuck. You'd need to measure that at two points as a minimum, as far to opposite ends of the "socket" as possible, although absolutely avoiding any pre existing damage or marks... You've got to clock the high spots and the low spots foor BOTH measurements there, as you'll need to determine if the taper in the spindle is off center, and you'll have to determine if it's bored in there straight. Both of those conditions will exist to some degree in ANY spindle, no matter how precise it is. The only question is, is your metrology department good enough to find it.... The next question is of course, how much the drill chuck is running out. Nothing on the chuck it's self will tell you anything useful, the critical surfaces there are the taper (buried up in the spindle), and the inside clamping face of the chuck jaws. So all of the measurements of the chuck will require measuring multiple points on a round shaft that is plenty long, and is more accurate than the rest of your metrology department. Again, you are measuring two points, clocking positions of the high and low from those measurements, and working out how much is runout, and how much is being crooked.

Other notable things- To check the runout on a morse taper assembly it needs to be FIRMLY seated. That's what trues one up to the other. The other thing is that these errors can "stack". If your MT2 spindle socket is "angled" by 0.0002 inches, then at the end of the drill chuck, that's gonna show up more like half to a whole thousandth at the end of a theoretically perfect drill chuck, and at the far end of a half inch drill, you're looking at a runout you can actually see. Not like you have to duck when the drill bit comes around kind of runout, but if you're looking for it, you'll see it.

One thing to try, minimal measuring, no math... Document your current measurements AND positions. Mark the taper where it is. Knock the chuck out of the spindle, clock it 120 degrees, re-install the chuck in that position, and repeat your measurements. Then do that a second time. Clock it another 120 degrees, and document a third set of measurements.

If those numbers repeat PERFECTLY in all three positions, your drill spindle is perfect. If there's any deviations there, you have just proved that you have errors in both the chuck AND the spindle. In which case, you'll see that in some positions the "tolerances" stack in your favor, minimizing the runout of the whole assembly, to where it's less then the error of the two parts individually....
 
Jake M said:
If they list 0.0016 RUNOUT as the spec, then you should expect up to 0.032 to show up on your indicator.
Click to expand...
Has to be a typo Jake . .0032 sounds a bit better . ;)
 
Just because nobody said it yet; it’s a drill chuck. If it looks to the naked eye like it’s spinning straight, then it’s as precise as it needs to be. Drill chucks are for punching semi precise holes in things. If you want precision, then you follow up that operation (hole making) with a boring, and/or reaming, and or grinding operation to get your precision.
For THIS hobby machinist anyways, .0016 to .0032 is one decimal place beyond my measuring capability anyways. It’s very likely beyond most people’s measuring capabilities.
Sorry I didn’t answer your question.
 
Back up one step for us....is the MT2 integral to the chuck, or is it an arbor pressed into the chuck?
 
You must log in or register to reply here.
Back
Top