TPI Feeds X-Feeds: Generalized Excel file for your Lathe

[B2]

@BStrickland

Thanks for these charts!

You are most welcome! I am glad they are of use to some folks. An improved version of the Excel work book macros are coming. What model lathe are you using?

Dave L.

 

[verbotenwhisky]

I am hoping we are finished with this project and I will now be home, until the next. I will stick a post-it to my forehead....

 

[verbotenwhisky]

I understand. I hat to keep asking. On the other hand do you know anyone else with a PM1236 (not a PM1236T) that could run this test? It should not take very long to scratch a couple of threads, but is pretty hard to do when you are not near the lathe!

Hey B2, with the following 24T (top) 127T center and 48T (BTM) A-1 produces 50TPI and B-1 is +/- 64TPI. I will run B-2 Tonight for additional information. I ran several of the threads off the chart and from the manual and it was accurate in the setups provided which I ran, all imperial.

 

[B2]

with the following 24T (top) 127T center and 48T (BTM) A-1 produces 50TPI and B-1 is +/- 64TPI. I will run B-2 Tonight for additional information. I ran several of the threads off the chart and from the manual and it was accurate in the setups provided which I ran, all imperial.

Hi

Great to hear from you.

I'm not for sure we are on the same page with the upper and lower designations, but we can work with that! I used the manual's Imperial and the Metric figures, listed on the first inserted table below. The manual had the 24T gear at the gear box (I called it L24 for L=Lower gear) and it had 48TPI at the spindle (I called it U48 for U=Upper). It does not matter what we use in the middle (exchange gear position as long as there is only one gear and not a 120/127 combination exchange gear) as it just transfers the motion between the upper and lower gears. Likewise, I think the middle knobs M1 vs M2 is just a factor of 2x so is no real problem. You seem to have used the opposite top and bottom gear locations for the 24 and 48 gear positions from my table. So I set the spread sheet to L48 and U24 as you used and got some values. Then, my spread sheet predicts A1 to yield 48TIP and B1 to have yielded 72TPI. Close to your findings, but .... Maybe these are as close as you could measure????

If you look at the table inserted again, and the manual figure 3-20, it provides the A4 and B4 thread values to be 10 and 15 respectively. This is a ratio of A4:B4=10:15=2:3. I believe this ratio must be maintained for any of the numbered Knob#s as you go up or down the table. In other words: A1/B1=A2/B2=A3/B3=A4/B4=A5/B5=2/3. Otherwise, the lettered knobs positions changes its gears and would also be needing to change the numbered knobs gears at the same time... a mechanical nightmare.

So the values I got via the spread sheet, using the upper-48 and lower-24 gear settings (your arrangement), gave a TPI ratio of A:B=48:72=2:3. Of course it should as this is the info I used to design it. Your values of 50 and 64 would produce a ratio of 50:64=25:32, which would disagree with the manual's value for the A4:B4 ratio.

What is the explanation for the discrepancy. Possibilities: 1) the wrong number of teeth in either the up or lower gear was used in the cuts, 2) the spread sheet calculations are off, 3) the manual's table is incorrect, or 4) your TPI measurements are off a bit. I am hoping it is the last possibility.

I created a second table for you where I used the 24(Top-Spindle) and 48(Bottom-Gear Box) gear locations as you described. It is below and of course each of the TPI values are larger than my earlier table by the gearing ratios changes. (Ie 48TPI/12TPI=4 between the two tables, since the gears have been changed by (48/24)*(48/24)=4 when the upper and lower gears were switched.)

It is very difficult to determine when the travel has been exactly 1 inch unless you have a DRO and you count the spindle turns.... or unless you scratch a thread for many inches and count a LOT of turns of the spindle (or a lot of scratches). When I was doing this sort of thing I was amazed at how difficult it is to get feed rates accurate ... and I do have a DRO as well as an electronic spindle turn counter on my machine. To get 1% accuracy you need to have MANY more than 100 threads.

So I propose an experiment that you can use to confirm my table..or not. You can use either the 24(Top-Spindle) and 48(Bottom-Gear Box) gear locations, but it maybe easier to observe if you use the L24 and U48 condition. You indicated that A1 produced 50TPI, for your gear arrangement, but maybe it was 48TPI? Scratch the A1 thread (12TPI or 48TPI depending upon the upper and lower gear choice). Then back the tool out a bit and rewind the spindle so that the tool travels back to near the beginning of the scratch without ever disengaging the saddle feed gears. Put the tool back to the scratch position and change the Knob to B2. Scratch the new thread right over the old one. Since the values on the diagonal of the table are all the same (if my sheet is correctly calculating things) the scratches should be the same for any of the diagonal settings you should get the same scratch spacing on top of one another for any the diagonal knob settings. A1, B2, C3,D4 or D5.

Then just to convince me that my table is correct you could do the same for A1 and B1 and confirm that there are 3 (for B1) scratches per 2 (for A1) scratches. This should be very visible and clear that the phase of the scratches is not changing when you observe this over a distance. It maybe difficult to tell when the TPI is 48 and 72 as the threads are so closely spaced. Hence, the L24 and U48 condition (12TPI) would be easier to observe this ratio.

Likewise you could compare the ratios of A1 to A2. 12TPI to 8TPI. Again 3:2 scratches.

I am not concerned about the metric threads unless we find that the above is not correct. After all the 127T:120T gear is just causing the table's TPI values to be divided by 25.4 mm/inch and inverted to yield mm/T. The A4 table position is 10TPI which is then (25.4/10TPI)=2.54mm/T

Looking forward to hearing about your results.

Dave L.

 

[verbotenwhisky]

Hi

Great to hear from you.

I'm not for sure we are on the same page with the upper and lower designations, but we can work with that! I used the manual's Imperial and the Metric figures, listed on the first inserted table below. The manual had the 24T gear at the gear box (I called it L24 for L=Lower gear) and it had 48TPI at the spindle (I called it U48 for U=Upper). It does not matter what we use in the middle (exchange gear position as long as there is only one gear and not a 120/127 combination exchange gear) as it just transfers the motion between the upper and lower gears. Likewise, I think the middle knobs M1 vs M2 is just a factor of 2x so is no real problem. You seem to have used the opposite top and bottom gear locations for the 24 and 48 gear positions from my table. So I set the spread sheet to L48 and U24 as you used and got some values. Then, my spread sheet predicts A1 to yield 48TIP and B1 to have yielded 72TPI. Close to your findings, but .... Maybe these are as close as you could measure????

If you look at the table inserted again, and the manual figure 3-20, it provides the A4 and B4 thread values to be 10 and 15 respectively. This is a ratio of A4:B4=10:15=2:3. I believe this ratio must be maintained for any of the numbered Knob#s as you go up or down the table. In other words: A1/B1=A2/B2=A3/B3=A4/B4=A5/B5=2/3. Otherwise, the lettered knobs positions changes its gears and would also be needing to change the numbered knobs gears at the same time... a mechanical nightmare.

So the values I got via the spread sheet, using the upper-48 and lower-24 gear settings (your arrangement), gave a TPI ratio of A:B=48:72=2:3. Of course it should as this is the info I used to design it. Your values of 50 and 64 would produce a ratio of 50:64=25:32, which would disagree with the manual's value for the A4:B4 ratio.

What is the explanation for the discrepancy. Possibilities: 1) the wrong number of teeth in either the up or lower gear was used in the cuts, 2) the spread sheet calculations are off, 3) the manual's table is incorrect, or 4) your TPI measurements are off a bit. I am hoping it is the last possibility.

View attachment 441916

I created a second table for you where I used the 24(Top-Spindle) and 48(Bottom-Gear Box) gear locations as you described. It is below and of course each of the TPI values are larger than my earlier table by the gearing ratios changes. (Ie 48TPI/12TPI=4 between the two tables, since the gears have been changed by (48/24)*(48/24)=4 when the upper and lower gears were switched.)

View attachment 441922

It is very difficult to determine when the travel has been exactly 1 inch unless you have a DRO and you count the spindle turns.... or unless you scratch a thread for many inches and count a LOT of turns of the spindle (or a lot of scratches). When I was doing this sort of thing I was amazed at how difficult it is to get feed rates accurate ... and I do have a DRO as well as an electronic spindle turn counter on my machine. To get 1% accuracy you need to have MANY more than 100 threads.

So I propose an experiment that you can use to confirm my table..or not. You can use either the 24(Top-Spindle) and 48(Bottom-Gear Box) gear locations, but it maybe easier to observe if you use the L24 and U48 condition. You indicated that A1 produced 50TPI, for your gear arrangement, but maybe it was 48TPI? Scratch the A1 thread (12TPI or 48TPI depending upon the upper and lower gear choice). Then back the tool out a bit and rewind the spindle so that the tool travels back to near the beginning of the scratch without ever disengaging the saddle feed gears. Put the tool back to the scratch position and change the Knob to B2. Scratch the new thread right over the old one. Since the values on the diagonal of the table are all the same (if my sheet is correctly calculating things) the scratches should be the same for any of the diagonal settings you should get the same scratch spacing on top of one another for any the diagonal knob settings. A1, B2, C3,D4 or D5.

Then just to convince me that my table is correct you could do the same for A1 and B1 and confirm that there are 3 (for B1) scratches per 2 (for A1) scratches. This should be very visible and clear that the phase of the scratches is not changing when you observe this over a distance. It maybe difficult to tell when the TPI is 48 and 72 as the threads are so closely spaced. Hence, the L24 and U48 condition (12TPI) would be easier to observe this ratio.

Likewise you could compare the ratios of A1 to A2. 12TPI to 8TPI. Again 3:2 scratches.

I am not concerned about the metric threads unless we find that the above is not correct. After all the 127T:120T gear is just causing the table's TPI values to be divided by 25.4 mm/inch and inverted to yield mm/T. The A4 table position is 10TPI which is then (25.4/10TPI)=2.54mm/T

Looking forward to hearing about your results.

Dave L.

I will look at this tonight and run the passes. I had issues counting the scratches and can easily see there being the error you are finding, 48 vs. 50 that would be my old eyes and a magnifying glass and the 64 vs. 72.... same same. I did not realize my pitch gauges were so limited until I looked at them last night so I had to count them manually, sent the wife to the machinist tool store and solved that issue today so I can get better accuracy tonight.

Best regards VW

 

[B2]

Good Morning @verbotenwhisky ,

Wow! I am not for sure my wife would know where there would be a Machinist tool store!

Anyway, the scratch test of comparing two supposedly equal TPI values will be more accurate in determining if they are exactly equal. I think that if you just run a scratch for one set of knob settings (A1) over the scratch made with the different set of knob settings on the diagonal (B2, C3, D4, or E5) any errors will be quite observable ... as the two scratches will not exactly match if the table is not correct. To make this even more dramatic just increase the scratch length as any error will just appear to as a larger deviation the longer the scratch. In fact you should be able run all 5 of these knob settings generated scratches over each other an it would only appear as a single TPI. This will work for any upper and lower set of gears so choose some where there is observable spacing between the scratches marks.

Mathematically it is kind of interesting, but you may not want to read below this line:
As I noted before the ratios between any two of the table entries will be the same for any set of upper and lower gears. Since the ratios are constant it means that there is a common multiplicative factor which will make all of the table entries integers. For the table where the A1 =12TPI this match number is 210=14*5*3 So if you multiply each of these table entries by 210 you get integers. i.e A1, 12*210=2520.

Divide any of these entries by 210 and you get the table where the diagonal is 12TPI. Divide these entries by (210/4) and you get the table where A1 is 48TPI. Because in these two cases the ratios of the upper and lower gears were just 2 or 1/2, but not simply 1, it makes it just a bit more confusing. However, if the upper gear were set to 105 = 210/2 and the lower gear were 1 (a strange looking gear) the table would appear just as shown! You can test this on the spread sheet. Just make a the gear entries(105 at column "O", "Screw2C" and 1 at column "V", "spindle") in the spread sheet columns and then select them. For A1 you should get the 2520 etc.

I think this table, the 105 number, and the 8TPI of the lead screw tells us all of the over all gear ratios, (or their reciprocals?), of the gear box.

Good Luck
Dave L.

 

[verbotenwhisky]

Good Morning @verbotenwhisky ,

Wow! I am not for sure my wife would know where there would be a Machinist tool store!

Anyway, the scratch test of comparing two supposedly equal TPI values will be more accurate in determining if they are exactly equal. I think that if you just run a scratch for one set of knob settings (A1) over the scratch made with the different set of knob settings on the diagonal (B2, C3, D4, or E5) any errors will be quite observable ... as the two scratches will not exactly match if the table is not correct. To make this even more dramatic just increase the scratch length as any error will just appear to as a larger deviation the longer the scratch. In fact you should be able run all 5 of these knob settings generated scratches over each other an it would only appear as a single TPI. This will work for any upper and lower set of gears so choose some where there is observable spacing between the scratches marks.

Mathematically it is kind of interesting, but you may not want to read below this line:
As I noted before the ratios between any two of the table entries will be the same for any set of upper and lower gears. Since the ratios are constant it means that there is a common multiplicative factor which will make all of the table entries integers. For the table where the A1 =12TPI this match number is 210=14*5*3 So if you multiply each of these table entries by 210 you get integers. i.e A1, 12*210=2520.

View attachment 442011

Divide any of these entries by 210 and you get the table where the diagonal is 12TPI. Divide these entries by (210/4) and you get the table where A1 is 48TPI. Because in these two cases the ratios of the upper and lower gears were just 2 or 1/2, but not simply 1, it makes it just a bit more confusing. However, if the upper gear were set to 105 = 210/2 and the lower gear were 1 (a strange looking gear) the table would appear just as shown! You can test this on the spread sheet. Just make a the gear entries(105 at column "O", "Screw2C" and 1 at column "V", "spindle") in the spread sheet columns and then select them. For A1 you should get the 2520 etc.

I think this table, the 105 number, and the 8TPI of the lead screw tells us all of the over all gear ratios, (or their reciprocals?), of the gear box.

Good Luck
Dave L.

I owe you an apology, I had family stuff last night that I had not expected; so, take 2 I will do the scratch passes tonight. Your math makes sense, and yes a very strange gear.

 

[B2]

Hi @verbotenwhisky

No real reason to rush! Its tax time and I have been procrastinating. I guess I should get started.

 

[verbotenwhisky]

Good Morning @verbotenwhisky ,

Wow! I am not for sure my wife would know where there would be a Machinist tool store!

Anyway, the scratch test of comparing two supposedly equal TPI values will be more accurate in determining if they are exactly equal. I think that if you just run a scratch for one set of knob settings (A1) over the scratch made with the different set of knob settings on the diagonal (B2, C3, D4, or E5) any errors will be quite observable ... as the two scratches will not exactly match if the table is not correct. To make this even more dramatic just increase the scratch length as any error will just appear to as a larger deviation the longer the scratch. In fact you should be able run all 5 of these knob settings generated scratches over each other an it would only appear as a single TPI. This will work for any upper and lower set of gears so choose some where there is observable spacing between the scratches marks.

Mathematically it is kind of interesting, but you may not want to read below this line:
As I noted before the ratios between any two of the table entries will be the same for any set of upper and lower gears. Since the ratios are constant it means that there is a common multiplicative factor which will make all of the table entries integers. For the table where the A1 =12TPI this match number is 210=14*5*3 So if you multiply each of these table entries by 210 you get integers. i.e A1, 12*210=2520.

View attachment 442011

Divide any of these entries by 210 and you get the table where the diagonal is 12TPI. Divide these entries by (210/4) and you get the table where A1 is 48TPI. Because in these two cases the ratios of the upper and lower gears were just 2 or 1/2, but not simply 1, it makes it just a bit more confusing. However, if the upper gear were set to 105 = 210/2 and the lower gear were 1 (a strange looking gear) the table would appear just as shown! You can test this on the spread sheet. Just make a the gear entries(105 at column "O", "Screw2C" and 1 at column "V", "spindle") in the spread sheet columns and then select them. For A1 you should get the 2520 etc.

I think this table, the 105 number, and the 8TPI of the lead screw tells us all of the over all gear ratios, (or their reciprocals?), of the gear box.

Good Luck
Dave L.

You were correct as is the table, the passes were on top of each other.

 

[B2]

@verbotenwhisky

Great! I will take the "Draft" off of your uwPM1236 TPI sheet. I will let you know when I posted the new version of the Workbook with the new and revised Macros.

I decided that the Macro I had developed to remove unwanted duplicate lines/rows in the AllTPI sheet had some errors. It seemed to be deleting too many rows. So it needs to be rewritten. Mean while, the simplest case to remove is to delete one of the two cases where the exchange gear is either 120:120 or 127:127. Clearly the results are the same and this represents 25% of all of the lines. If you want to do this by hand just use the sort multiple columns macro and sort on the two exchange gear columns. This will group the 120:120 together and the 127:127 together. Then just select, say all of the 127:127 rows and delete them. You can then resort by the TPI column etc to get the TPI values in sequence.

Thanks again,
Dave L.