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10th June 2015, 07:24 AM #16Pink 10EE owner
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At the end of the day, extra change gears will be required.. There is no way that can be got around...
Light red, the colour of choice for the discerning man.
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10th June 2015 07:24 AM # ADSGoogle Adsense Advertisement
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10th June 2015, 09:12 AM #17Member
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RC,
My initial checks into possible compound gear trains suggests that your assessment is correct, depending on the degree of accuracy required to cut this worm.
Andrew.
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10th June 2015, 11:56 AM #18Senior Member
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I am more used to working with DP gears and not thinking very clearly this morning, but I think RC is correct.
In that case a very good approximation of pi/2 is given with either a 55T/35T combination gear or a 44T/28T combination gear.
Pi/2= 1.570795..... and 55/35 or 44/28 = 1.5714.... so no need to go into very large gears like 157 and 100 teeth, and a slightly better approximation,
probably plenty close enough for a slow speed worm.
If you are using one of these compound gears you would probably need a suitable sized idler gear in the train to allow them to mesh with the spindle gear and the gear on the gearbox.
Frank.
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10th June 2015, 10:36 PM #19Intermediate Member
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Thanks for all the feedback from everyone, I was indeed expecting to need other gears to what I have, which is why I want to learn all about calculating pitches etc. ( think a 66 tooth gear may be what I need from other calculators).
I want to learn all this so I can cut my own gear to get the 3.1416mm pitch, so I can cut a worm and a matching wormgear to make my own rotary table, but I first need to make (even a crude one) a dividing head, I have a straight shank ER25 collet on order to start that, I can cut the dividing plates easily as I built a quite large and strong gantry CNC so the dividing plate will be easy to make from aluminium plate.
This is about learning how to do things for myself rather than simply buying everything and it may save money along the way which will of course be handy. (I'll probably end up buying more tools anyway)
mahgnia, thanks especially for the info, it has helped clear some of my cerebral fog, I never thought of dividing A/B or I/II etc. so that was enlightening. I would like to take you up on your offer of including B&C gears in the spreadsheet if that's OK with you.
I'm going over everything again and again to get my head around it.
Thanks again everyone, it has been great to get such a response.
Eric
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11th June 2015, 01:38 AM #20Member
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Eric,
Spreadsheet updated to allow selected gear to be installed in position C.
Position B is an idler gear by the original chart and would therefore not affect the ratio. It would be changed to allow various gear combinations to reach the spindle/reversing tumblers.
Weiss Pitch Chart.xls
From this spreadsheet, possible combinations are:
C, D, E
54,96,56 with QCGB in CIII
63,100,70 with QCGB in BI
Both of these have an error that may be acceptable for your purpose.
Andrew.
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11th June 2015, 01:54 AM #21Member
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Eric,
Does the gear shaft at position B allow for two gears on the shaft, or is it a single gear idler only?
Andrew
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11th June 2015, 09:14 AM #22Intermediate Member
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Thanks very much Andrew, that was quick, you are obviously good with spreadsheets.
Still trying to get my head around the ratio's you provided and how each ratio is chosen as I can't 'see' the gears and the manual is not clear. That is, I can't work out which is the driver or the driven gears, but getting there.
The B gear (30T) is on it's own, it cannot be ganged.
Eric
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11th June 2015, 07:30 PM #23Member
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Eric,
If you know the number of teeth on the drive gear on the lathe spindle, we can work back through first principles to determine the change gears you need to cut the worm most accurately.
We would start with pi being very closely approximated by the fraction 22/7,and with the leadscrew being 2mm pitch, the leadscrew would need to rotate 11 times for every 7 spindle rotations.
Actual gearing numbers can be worked out from there, if the size of the spindle drive gear is known.
Real QCGB ratios can also be worked out from this.
Andrew.
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11th June 2015, 07:41 PM #24Pink 10EE owner
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Th eprogram in this thread is handy for this sort of thing
https://www.woodworkforums.com/f65/lathe-thread-cutting-change-gear-calculator-118398Light red, the colour of choice for the discerning man.
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12th June 2015, 10:24 AM #25
I just realised that the link in that other thread is no longer valid, however I can't edit or post to that thread because of it's age ... so here is a valid link if you wish to use it - https://www.dropbox.com/s/837nflzn6y...GV1.2.exe?dl=0
Cheers.
Vernon.
__________________________________________________
Bite off more than you can chew and then chew like crazy.
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12th June 2015, 11:03 AM #26Intermediate Member
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Hello again everyone, I have been reading my manual on the lathe and physically checking the gears and now I can at least understand the manual.
the drive train from the spindle to the LS is.. spindle 40T, forward engagement 30T-40T-40T- (reverse removes 30T, puts 40T-40T in), banjo 30T-100T-30T-96T then I II III and A B C
I'll checkout the dropbox stuff soon thanks.
Eric
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12th June 2015, 07:12 PM #27Member
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Eric,
The 40T spindle gear and the 100T gear at position C would be the first calculation, as the 30T and tumbler gears are all idler gears in the train, so they do not affect the gear train ratio.
So, from the original chart, we have the following:
spindle 40
C 100
D 63
E 63
The gears so far give a speed reduction ratio of 5:2
With the QCGB in BIII the leadscrew is turning the same speed as the spindle (giving same pitch as the leadscrew)
Therefore, the QCGB in BIII must be giving a speed increase ratio of 2:5.
From this the QCGB actual ratios are:
BIII = 2.5
AIII = 1
BI = 1.75
BII = 1
CIII = 1.25
AI = 0.7
AII = 0.4
CI = 0.875
CII = 0.5
To get the leadscrew to rotate 11 times for each 7 spindle rotations (pi/2 ~ 1.5708) could potentially be done as follows (given the gears needed):
C, D, E, QCGB
100, 66, 42, BIII (40/100*66/42*2.5)=1.5714
40, 66, 42, AIII (40/40*66/42*1)=1.5714
80, 88, 28, BII (40/80*88/28*1)=1.5714
40, 88, 28, CII (40/40*88/28*0.5)=1.5714
The less close (but available gears) possibilities from the spreadsheet are:
54, 96, 56, CIII (40/54*96/56*1.25)=1.5873
63,100,70, BI (40/63*100/70*1.75)=1.5873
72, 96, 30, CI (40/72*96/30*0.875)=1.5556
54, 63, 30, AIII (40/54*63/30*1)=1.5556
Unfortunately, none of the gears you have is divisible by 11, and so you will need to accept a less accurate pitch, or obtain/make a gear with the required tooth count.
Hope this all makes sense.
Andrew.
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13th June 2015, 12:03 PM #28Intermediate Member
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Thanks again Andrew, everything you have written makes sense to me except for the part where you derived the original ratios, I cannot get that straight in my head, sorry.
the ratio's I/II to A/C I get, but why C/B instead of B/C ?
and the following ratio's, how do you know which ratio's to multiply together such as ...
AIII = 0.4 (BIII ratio x the A/B ratio)
BI = 0.7 (BIII ratio x the I/III ratio)
etc.
I'm not fathoming how these were chosen.
Eric (the dumb)
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13th June 2015, 01:04 PM #29Intermediate Member
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oops, sorry double up on this post - eric
Thanks again Andrew, everything you have written makes sense to me except for the part where you derived the original ratios, I cannot get that straight in my head, sorry.
the ratio's I/II to A/C I get, but why C/B instead of B/C ?
and the following ratio's, how do you know which ratio's to multiply together such as ...
AIII = 0.4 (BIII ratio x the A/B ratio)
BI = 0.7 (BIII ratio x the I/III ratio)
etc.
I'm not fathoming how these were chosen.
Eric (the dumb)Last edited by ericg; 13th June 2015 at 01:05 PM. Reason: doubled up
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13th June 2015, 01:57 PM #30Member
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C/B was used as it was convenient for the calculation description. It is simply the inverse of the B/C ratio. (B divided by C = 2, therefore C divided by B = 0.5).
In the earlier post, the QCGB has two sets of selectable gears, the A, B, C set and the I, II, III set.
The final ratio through the whole gear box is the two selected part ratios multiplied together.
For example:
BIII = 1 can also be written as B * III = 1
If you change only the ABC part of the QCGB from B to A , and keep the I,II,III part the same, then the change in the output ratio will be the same as the A/B ratio.
If the A/B ratio is 0.4,
then for AIII
A*III = A / B * B * III {multiply the equation by B / B which equals 1}
= (B*III ratio * A/B ratio) = 1 * 0.4 = 0.4 ......{the B cancels out in the equation to give A*III = 0.4}
Similarly, if you change only the I,II,III part of the QCGB from III to I, and keep the A, B, C part the same, the the change in the output ratio will be the same as the I/III ratio.
If the I/III ratio is 0.7 and AIII = 0.4 {from above}
then for AI
A*I = A * III * I / III {multiply the equation by III / III which equals 1}
= (A*III ratio * I/III ratio) = 0.4 * 0.7 = 0.28 ......{the III cancels out in the equation to give A*I = 0.28}
The above numbers are relative to one another, the actual QCGB ratios (output speed /input speed) are listed below:
BIII = 2.5
AIII = 1
BI = 1.75
BII = 1
CIII = 1.25
AI = 0.7
AII = 0.4
CI = 0.875
CII = 0.5
I hope this has not muddied the workings further....
Andrew.
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