Part 11 (1/2)

As the carriage _C_ and the tool are eared to the spindle, the number of threads to the inch that are cut depends, in every case, upon the nu the carriage one inch If the lead-screw has six threads per inch, it will e and the thread tool travel one inch along the piece to be threaded Now if the change gears _a_ and _c_ (see also sketch _A_, Fig

29) are so proportioned that the spindle iven time, it is evident that the tool will cut six threads per inch If the spindle revolved twice as fast as the lead-screw, it would make twelve turns while the tool moved one inch, and, consequently, twelve threads per inch would be cut; but to get this difference in speeds it is necessary to use a co that will cause the lead-screw to revolve once while the lathe spindle and work29 (A) Lathe with Si (B) Compound Geared Lathe]

Suppose that nine threads to the inch are to be cut and the lead-screw has six threads per inch In this case the work must make nine revolutions while the lead-screw e and thread tool to move one inch, or in other words, one revolution of the lead-screw corresponds to one and one-half revolution of the spindle; therefore, if the lead-screw gear _c_ has 36 teeth, the gear _a_ on the spindle stud should have 24 teeth The spindle will then revolve one and one-half times faster than the lead-screw, provided the stud rotates at the same rate of speed as the ears that is required for a certain pitch can be found bythe number of threads per inch of the lead-screw, and the number of threads per inch to be cut, by the same trial multiplier The formula which expresses the relation between threads per inch of lead-screw, threads per inch to be cut, and the nuears, is as follows:

threads per inch of lead-screw teeth in gear on spindle stud ------------------------------ = ----------------------------- threads per inch to be cut teeth in gear on lead-screw

Applying this to the exaiven, we have 6/9 = 24/36 The values of 36 and 24 are obtained by6 and 9, respectively, by 4, which, of course, does not change the proportion Any other nu 24 and 36 teeth were not available, this ears of this size, soht be used

Suppose the nuears supplied with the lathe are 24, 28, 32, 36, etc, increasing by four teeth up to 100, and assume that the lead-screw has 6 threads per inch and that 10 threads per inch are to be cut Then,

6 6 4 24 -- = ------ = -- 10 10 4 40

Byboth nuears having 24 and 40 teeth, respectively The 24-tooth gear goes on the spindle stud and, the 40-tooth gear on the lead-screw The nuear _b_, which connects the stud and lead-screw gears, is not considered as it does not affect the ratios between gears _a_ and _c_, but is used siear to the other

We have assuear _a_ is eared in the ratio of one to one and make the same number of revolutions In some lathes, however, these two ears were placed on the lead-screw and spindle stud, the spindle would not make the same number of revolutions as the lead-screw

In that case if the actual number of threads per inch in the lead-screere used when calculating the change gears, the result would be incorrect; hence, to avoid ives the correct result, regardless of the ratios of the gears which connect the spindle and spindle stud:

_Rule--First find the nuears of the same size are placed on the lead-screw and spindle, either by actual trial or by referring to the index plate Then place this number as the numerator of a fraction and the number of threads per inch to be cut, as the denominator; multiply both numerator and denominator by some trial number, until nuears that are available_ The product of the trial number and the nuear _a_ for the spindle stud, and the product of the trial nuear for the lead-screw

=Lathes with Coed as shown at _A_, Fig 29, it is referred to as siears between the stud and screw as at _B_, which is ter is practically the saears used in co, place the ”screw constant”

obtained by the foregoing rule, as the numerator, and the number of threads per inch to be cut as the denominator of a fraction; resolve both numerator and denominator into two factors each, and multiply each ”pair” of factors by the sa nuears (One factor in the numerator and one in the denominator make a ”pair” of factors)

Suppose the lathe cuts 6 threads per inch when gears of equal size are used, and that the nuears available are 30, 35, 40 and so on, increasing by 5 up to 100 If 24 threads per inch are to be cut, the screw constant 6 is placed in the numerator and 24 in the denominator The numerator and denominator are then divided into factors and each pair of factors is ears, thus:

6 2 3 (2 20) (3 10) 40 30 -- = ----- = ------------------- = ------- 24 4 6 (4 20) (6 10) 80 60

The last four nuears which should be used The upper two having 40 and 30 teeth are the _driving_ gears and the loo having 80 and 60 teeth are the _driven_ gears The driving gears are gear _a_ on the spindle stud and gear _c_ on the interear, and the driven gears are gears _b_ and _d_ It ears is placed on the spindle stud, or which of the driven is placed on the lead-screw

=Fractional Threads=--Soiven as a fraction of an inch instead of stating the number of threads per inch

For exa 3/8-inch lead

The expression ”3/8-inch lead” should first be transformed to ”number of threads per inch” The nule) equals:

1 3 8 --- = 1 - = - = 2-2/3 3/8 8 3

To find the change gears to cut 2-2/3 threads per inch in a lathe having a screw constant of 8 and change gears varying fro by 4, proceed as follows:

8 2 4 (2 36) (4 24) 72 96 ----- = --------- = ----------------------- = ------- 2-2/3 1 2-2/3 (1 36) (2-2/3 24) 36 64

As another illustration, suppose we are to cut 1-3/4 thread per inch on a lathe having a screw constant of 8, and that the gears have 24, 28, 32, 36, 40 teeth, etc, increasing by four up to one hundred Following the rule:

8 2 4 (2 36) (4 16) 72 64 ----- = --------- = ----------------------- = ------- 1-3/4 1 1-3/4 (1 36) (1-3/4 16) 36 28

The gears having 72 and 64 teeth are the _driving_ gears, and those with 36 and 28 teeth are the _driven_ gears