PICKING.
As soon as a warp-shed is sufficiently opened by the healds, the shuttle, containing weft, is propelled through it. That operation is termed “picking,” and may be accomplished by either of two types of picking motions known as “over” and “under” picking motions. The “over-pick,” also known as the “cone” and Blackburn pick ([Fig. 54]), is in most general use, especially for narrow and quick-running looms weaving light and medium-weight fabrics; whilst the “under-pick” (Figs. [56] and [57]), of which there are many modifications, is chiefly confined to medium and broad looms, which require a picking motion capable of developing greater force. A shuttle is propelled by a picker made of hide, which is connected by means of a leather strap to the picking stick A ([Fig. 54]). The upright shaft B is the fulcrum of the lever. The cone C is the short arm of the lever which receives the force from the picking tappet D. The tappet is so shaped that as it revolves it gives a sudden quick movement to the cone-shaped stud, and therefore to the shuttle. It is obvious that as the shuttle must move from one side of the loom to the other, and back again, for two revolutions of the crank shaft, the picking tappets must be placed on a shaft whose speed is one-half that of the crank shaft; therefore the bottom shaft in the loom on this account is made to move at the required speed, and the picking tappets are placed on this shaft at opposite sides of the loom.
FIG. 54.
The chief requirement in a good pick is that as little force as possible shall be wasted in the loom. The relative positions of the tappet shaft and cone should be such that the force is exerted as nearly as possible in the direction of the dotted line E at right angles with the upright shaft B. It is impossible to effect this throughout the whole course of the stroke, but it is obvious that if this is approached as nearly as possible, the pick will be smooth, and the wear and tear reduced to a minimum. A very considerable amount of power is wasted if the direction of the force is too much downward.
The direction of the force is at right angles to a line drawn tangent from the cone at the point of connection with the picking tappet. Thus in [Fig. 54] the direction of the force is indicated by the dotted line M, which is at right angles to the dotted line N, drawn tangent to the cone at the point of connection with the tappet.
FIG. 55.
The intensity of the force depends on the length of the stroke of the tappet and on the suddenness of the curve of the working face. If in two looms the length of tappet is the same, but in one the portion of a revolution occupied in making the stroke is less than in the other, there will be a greater intensity of force in the loom with the quicker stroke. In [Fig. 55] the portion of a revolution occupied in making the stroke is indicated by the angle AB. If this angle is increased, the force of the pick will be lessened, and if the angle be decreased, the force of the pick will be augmented. It will be understood from this that if the picking tappets are short the pick is liable to be harsh. If a fair length of tappet is given, a smoother and better-timed pick can be made. The curve on the picking tappet gradually approaches a radial line as it nears the end of the stroke, but the combined influence of the change in the position of the cone and the backward movement of the slay causes the shuttle to move quickest in the early part of its movement in the box.
There is a relation between the length of the shuttle-box and the length of the picking tappet. If the tappet is a short one, the shuttle-box must be short; and if a longer tappet is used—the leverage of the picking arm and other parts being the same—the shuttle-box will be longer.
It is obviously inadvisable to have too short a tappet, as the movement of the shuttle in the box must in that case be extremely sudden, in order to have the necessary force.
FIG. 56.
FIG. 57.
An underpick motion is given at [Fig. 56]. A picking treadle, A, centred at C, is pressed suddenly down by the picking bowl B, which is fastened on to the wheel on the bottom shaft in the loom. A strap, E, connects the treadle and the picking lever. In [Fig. 57] this connection is shown. The strap from the treadle is fastened to the quadrant, and as the treadle is pressed suddenly down, the picking lever H is moved forward. The shape of the curve E, which the picking bowl strikes, regulates the character of the movement given to the lever H, and it is well not to have the curve too small and sudden, or the pick will not be satisfactory. The curve on the treadle in [Fig. 88] ([p. 120]) is perhaps better than the one in [Fig. 56], as it is longer, and is therefore not liable to be so jerky.
There are numerous other picking motions, which chiefly differ in the mechanism for actuating the picking lever.
Beating up the Weft is the third primary movement in weaving. This movement is performed by a crank on the top shaft in the loom and a connecting rod or crank-arm which connects the crank and the slay together. This is shown at [Fig. 38], where the crank C and crank-arm D give a reciprocating movement to the slay S. The slay moves upon a rocking shaft, E, as a fulcrum, and when the crank is at the front centre the slay-swords should be perpendicular, or nearly so. Sometimes the fulcrum is taken a little forward, but it is never advisable to have the slay over the perpendicular when in contact with the cloth.
The movement of the slay should be eccentric. It is obvious that when the slay is at the back of its stroke its movement should be sufficiently slow to allow time for the shuttle to pass through the shed; and that when beating up, the speed of the slay should be sufficient to knock the weft firmly into the cloth. A crank and crank-arm give the kind of movement required.
The eccentricity of the slay’s movement depends upon the length of the crank and crank-arm, and upon the position of the crank-shaft in relation to the point of connection of the crank-arm with the slay. The position of the crank-shaft in relation to the connecting pin varies in different makes and widths of looms. We shall see that the position of the shaft and the direction in which the loom runs have an important bearing on the force exerted by the slay in beating up the weft. For ordinary looms the usual position of the shaft is a little below the level of the connecting pin when at the front centre, and when the shaft is in this position the movement of the slay is the most even and least eccentric. To obtain this position of the crank-shaft in a diagram, first draw the line SA ([Fig. 58]) to represent the slay-sword when the reed is in contact with the cloth; this we will assume is perpendicular. We will suppose SA to be 24 inches, S being the rocking shaft and A the connecting pin which connects the crank-arm with the slay. Suppose the loom we are dealing with to have a 3-inch crank and a 12-inch crank-arm. Describe the arc AN from the centre S, and mark off on the arc a distance from A equal to twice the length of the crank. As the crank is 3 inches long, mark off the point B, 6 inches from A. This point B represents the position of the connecting pin when the slay is at the back of its stroke.
FIG. 58.
From A, rule the line AX in such a position that the arc AB makes the least possible departure from it. It will be found that this necessitates AX cutting the arc AB at a point a little past the middle of the arc. With the length of the crank and crank-arm, viz. 15 inches, in the compasses, from A as the centre cut the line AX at E, and this gives the position for the crank-shaft which will give the least possible eccentricity to the slay. This will be obvious, as the nearer the connecting pin moves on the straight line AX, the less will be the eccentricity of the slay.
That the movement of the slay in the back half of its stroke is slower than in the front half can easily be proved by taking the length of the crank-arm in the compasses, and, after bisecting the arc AB at C, from C marking off the points D and H on the crank circle. It will be seen that both these points are somewhat inside the top and bottom centres of the crank indicated by the dotted line, and therefore the slay moves from C to A and back, the front half of its stroke, in less time than it moves from C to B and back to C. The reason for this eccentricity or unevenness in the movement of the slay is that when the crank is moving from the back centre to the top centre the crank-arm is oscillating and opening an angle with AX while the slay is moving forward, and therefore while the crank is making this quarter of a revolution, the connecting pin of the slay will move something less than from B to C; and while the crank is moving from the top centre to the front centre, the crank-arm is straightening or closing the angle while the slay is moving forward, and thus the connecting pin will move a greater distance than from C to A while the crank is making this quarter of a revolution. When the crank moves from front to bottom centre the angle is opening while the slay is moving backwards, and therefore the pin will move a little more than from A to C; and when the crank moves from bottom to back centre the angle is closing while the slay moves backwards, thus retarding the velocity of the slay.
FIG. 59.
This will be better understood from [Fig. 59], where CD is the crank-arm, and ED the crank at the top centre. A is the position of the connecting pin when at the front of its stroke, and B its position when at the back of its stroke. The dimensions are as in [Fig. 58]—viz. 12-inch arm, 3-inch crank—and for simplicity we will assume the connecting pin moves on the straight line AE. From A to B is 6 inches, and therefore it is obvious that AC is something over 3 inches, and that the connecting pin moves this distance whilst the crank is making the quarter of a revolution from top to front centre. The distance AC can be obtained as follows. It is obvious that CDE is a right-handed triangle, and therefore CD2 is equal to CE2 + ED2. Therefore CD2-ED2 = CE2, and having obtained the length of CE, we can subtract this from AE, which leaves the length of AC. The formula will stand thus—
CD2
−
ED2
=
CE2
122
−
32
144
−
9
∴ 135
=
CE2
length of AE
=
15·0000
inches
11·6189
inches
AC
=
3·3811
inches
The answer may be obtained in one calculation as follows:—
AE
−
√CD2 − ED2
or
15
−
√122 − 32
15
−
√144 − 9
15
−
√135
15
−
11·6189 = 3·3811
We thus see that the connecting pin moves 3 inches + 0·3811 inch while the crank is moving from the top to front centre. It will also move the same distance while the crank moves from front to bottom centre.
When the crank is moving from the bottom to the back centre, the connecting pin will move 3 inches − 0·3811.
3·0000
0·3811
2·6189
inches,
and the same distance when the crank moves from back to top centre.
It is often necessary in comparing looms to obtain the distance travelled by the connecting pin for a smaller movement of the crank than a quarter of a revolution.
Suppose it is desired to find the distance travelled by the connecting pin while the crank moves through 30 degrees to the front centre.
Take a 4-inch crank and 12-inch crank-arm. In [Fig. 60], ED is the crank, 4 inches, and DC the crank-arm, 12 inches, the angle O = 30 degrees. P is the position of the connecting pin when at front of its stroke.
FIG. 60.
To find the distance CP. From a table of natural sines we can obtain the sine of an angle of 30 degrees, viz, sin 30° = 0·5, and therefore, knowing the length of ED, viz. 4 inches, we can obtain the length of DN, it being 0·5, or half of ED, in an angle of 30 degrees.
Having two sides of a triangle, we can obtain the third side thus:
ED2 − DN2
=
EN2 and
CD2 − DN2
=
CN2
Having obtained the length of CN and EN, we can easily obtain CP by subtracting CF from the length of crank and crank-arm together. Working out the problem in figures we get—
ED2 − DN2
=
EN2
42 − 22
=
EN2
16 − 4
=
EN2
12
=
EN2
CD2 − DN2
=
CN2
122 − 22
=
CN2
144 − 4
=
CN2
140
=
CN2
Therefore CE = 15·2962 inches, and subtracting this from PE, which is 16 inches (12 + 4 = 16), we get 16 − 15·2962 = 0·7038 as the distance CP, which is the distance moved by the connecting pin for the 30 degrees movement of the crank.
The complete formula is as follows:—
PE − [√(ED2 − DN2) + √(DC2 − DN2)] = CP, or distance moved by the connecting pin for the given number of degrees through which the crank moves, ND being obtained from a table of sines.
To find the distance moved by the connecting pin while the crank moves through 5 degrees—say, from 30 degrees to 25 degrees in beating up.
To solve this it will only be necessary to subtract the length of CE when the crank is forming an angle of 30 degrees from the length of CE when the crank forms an angle of 25 degrees. In the previous example we found that for 30 degrees, CE = 15·2962 inches, and therefore proceeding in the same manner for 25 degrees, we get from table of sines, sin 25° = 0·4226, and 0·4226 of 4 inches 1·69; therefore ND =1·69 inches, and
42 − 1·692
=
EN2
16− 2·856
=
EN2
13·144
=
EN2
and 122 − 1·692
=
CN2
∴ 144 − 2·856
=
CN2
141·144
=
CN2
therefore CN = 11·88 inches, and CE will equal 11·88 + 2·626, or 15·506 inches, when the crank is forming an angle of 25 degrees.
15·506
length of CE for 25 degrees
15·296
length of CE for 30 degrees
inches
0·210
distance moved by pin whilst crank moves through 5 degrees, from 30 degrees to 25 degrees, in beating up.
In this manner it is easy to calculate the distance travelled by the pin for any number of degrees moved by the crank, and by comparing the velocity of the slay in different looms, the force of the beat up can be compared.
The force exerted by the slay varies as the square of its velocity. Thus, if in two looms where the weight of the two slays and the tension on the two warps are the same, the velocity of the slay in one loom is twice that of the other at a certain point in the beat up, the force of the former slay at that particular point will be four times the force of the latter, 12 : 22
We can thus compare the force exerted by the slay in different looms at any point of the beat up.
The force of the beat up is chiefly exerted upon the pick when the crank is nearly at the front centre, and the force exerted will also depend considerably upon the tension on the warp; but the slay is doing some work in beating up from the moment the reed begins to move the pick forward.
Possibly the most reliable method of comparing the force of the beat up in different looms is to calculate the time occupied by the slay in moving through a specified distance at the front of its stroke in beating up. This necessitates a rather different calculation to the preceding examples, but is equally as simple.
FIG. 61.
Suppose it is required to compare the force exerted by the slay in beating up (say the front 1 inch of its stroke) in two looms, one with a 12-inch crank-arm and 3-inch crank and the other with an 11-inch arm and 4-inch crank. The weight of the slays, the speed of the looms, the tension on the warps, and the timing of the primary movements, the same in each case.
In [Fig. 61] the smaller circle represents the 3-inch crank and the larger one the 4-inch crank. CP = 1-inch, CB = 11-inch arm, and CD = the 12-inch arm. It is obvious that if we can obtain the two angles made by the cranks, viz. ∠ CAB and ∠ CAD, we shall be able to get the time, or fraction of a revolution, occupied in moving the slay from C to P. As we know the three sides of the triangle we can obtain the angle enclosed by any two sides, and what is required in this case is to obtain the angles BAC and DAC. In triangles of this kind where there is no right angle, we can obtain the cosine of the angle as follows:—
| CA2 + AD2 − DC2 | = AQ, the cosine of angle DAC, | |
| 2CA.AD | ||
| and | CA2 + AB2 − BC2 | = AN, the cosine of angle BAC. |
| 2CA.AB |
The proof of this formula is given in Euclid, Book 2.
Having obtained the cosines of the two angles, we can find the angles themselves by referring to a table of sines and cosines.
Then as AP = 15 inches,
CA = 14, AD 3 inches, DC 12 inches, BA 4 inches, BC 11 inches; and reducing the formulæ to figures, we get:
| 142 + 32 − 122 | = | 196 + 16 − 144 | = | 0·7262 cosine, |
| 2 × (14 × 3) | 84 |
and by referring to a table of sines, we find that cosine 0·7262 = angle 43° 26´, therefore angle DAC = 43½°, about. Also
| 142 + 42 − 112 | = | 196 + 16 − 121 | = | 91 | = | 0·8125, |
| 2 × (14 × 3) | 112 | 112 |
and by referring to a table of sines and cosines, we find cosine 0·8125 = angle 35⅔°.
We thus find that to move the connecting pin 1 inch to the front of the stroke, in the loom with 11-inch arm, the 4-inch crank will move through 35⅔°, and in the loom with the 12-inch arm the 3-inch crank will move through 43½° for the same movement of the slay. Assuming the force exerted by the latter to be 1, the force of the former will be as 35⅔ squared : 43½ squared
It may be as well here to give a short explanation of the system of obtaining angles by sines and cosines.
As the crank moves forward it is obvious that the line DQ will become shorter, and as the angle becomes larger the line DQ will increase in length. In trigonometry, the ratio between the length of the line DQ and the radius AD is called the sine of the angle, and if the radius is 1, the length of DQ will be the value of the sine. In an angle of 30° the sine is exactly ½ the radius, and the relation between the radius and the sine for every angle is known, and arranged in “tables of sines.” The length of AQ will also vary with the angle, and the length of this line is called the “cosine” of the angle QAD. The cosine of an angle of 30° is therefore the same as the sine of an angle of 60°. When the sine is known it is easy to obtain the cosine as follows:—
Cosine = √1 − sin2. Thus for an angle of 30°, cosine = √1 − 0·52, or cos2 = 1 − 0·52, therefore cos2 = 1 − 0·25, or cos2 = 0·75, ∴ cos = √0·75 = 0·866. By reversing, the sine may be obtained from the cosine.
The value given to the sines and cosines must not be taken for the actual length of the lines; they are simply the ratio to the radius. Thus in an angle of 30°, if the radius is 1 inch the length of the sine will be ½ inch and the cosine 0·866 inch. If the radius is 2 inches, the actual length of the sine will be 1 inch and of the cosine 1·732 inches.
TABLE OF SINES AND COSINES.
Angle.
Sine.
Cosine.
Angle.
0°
0·00
1·00
90°
1°
·0175
·9998
89°
2°
·0349
·9994
88°
3°
·0523
·9986
87°
4°
·0698
·9976
86°
5°
·0872
·9962
85°
6°
·1045
·9945
84°
7°
·1219
·9925
83°
8°
·1392
·9903
82°
9°
·1564
·9877
81°
10°
·1736
·9848
80°
11°
·1908
·9816
79°
12°
·2079
·9781
78°
13°
·2250
·9744
77°
14°
·2419
·9703
76°
15°
·2588
·9659
75°
16°
·2756
·9613
74°
17°
·2924
·9563
73°
18°
·3090
·9511
72°
19°
·3256
·9455
71°
20°
·3420
·9397
70°
21°
·3584
·9336
69°
22°
·3746
·9272
68°
23°
·3907
·9205
67°
24°
·4067
·9135
66°
25°
·4226
·9063
65°
26°
·4384
·8988
64°
27°
·4540
·8910
63°
28°
·4695
·8829
62°
29°
·4848
·8746
61°
30°
·5000
·8660
60°
31°
·5150
·8572
59°
32°
·5299
·8480
58°
33°
·5446
·8387
57°
34°
·5592
·8290
56°
35°
·5736
·8192
55°
36°
·5878
·8090
54°
37°
·6018
·7986
53°
38°
·6157
·7880
52°
39°
·6293
·7771
51°
40°
·6428
·7660
50°
41°
·6561
·7547
49°
42°
·6691
·7431
48°
43°
·6820
·7314
47°
44°
·6947
·7193
46°
45°
·7071
·7071
45°
Angle
Cosine
Sine
Angle
We see from [Fig. 61] that in a loom with a 4-inch crank and 11-inch arm, the velocity of the slay is much greater when beating up than with the 3-inch crank and 12-inch arm.
The effect of the length of the crank-arm on the velocity of the slay can easily be shown by a diagram or by calculation. If the length of the crank-arm be altered without altering the length of the crank, there will be found a somewhat quicker movement of the slay at the beat up in the loom with the shorter arm. The difference is not so great when the crank-arm is a long one in proportion to the crank. The chief cause of the difference in the velocity of C in [Fig. 61] is the difference in the length of the crank. It is obvious that the longer the crank the greater the angle which it will cause the arm to make, and therefore the greater will be the acceleration of the velocity of C when the angle is closing and the slay moving forward. Likewise, it is obvious that the shorter the arm the larger will be the angle to close, but the principal thing to notice is that an increase in the length of the crank causes an increase in the velocity of the slay owing to the extra distance which it has to travel in each revolution; so that even if the crank-arm were lengthened in exact proportion to the increase in the length of the crank, so as to keep the angle to be closed in beating up the same, there would still be a considerable increase in the velocity of the slay, caused by the extra distance it has to travel. This lengthening of the crank has obviously much more to do with the increase in velocity of the slay than the shortening of the arm has.
The longer the crank the further back from the cloth will the slay be taken, and assuming that the shed is open for the shuttle when the crank is at the bottom centre, a long crank is obviously more suitable for a wide loom, as, having to move further back, it will allow a longer time for the shuttle to pass through the shed than a short crank would; therefore the wider the loom, the longer the crank is required to be to allow time for the shuttle to pass.
The time allowed for the passage of the shuttle may also be increased by using a short arm so as to increase the eccentricity of the slay.
The longer the crank, the greater the velocity of the slay, therefore a long crank is suitable for heavy work, as it stores up more force in the slay than a short one. The force may also be increased by shortening the crank-arm, thus increasing the eccentricity of the slay.
FIG. 62.
The position of the crank-shaft in relation to the connecting pin has some effect upon the eccentricity of the slay’s movement. [Fig. 62] shows this, but to see clearly the effect it would be advisable to make an accurate drawing to a large scale. Four positions of the crank-shaft are shown. The one on the line A is just a little below the level of the connecting pin, so that the pin moves as nearly as possible on the line A when making the front quarter of its stroke. The circle on the line B is the position where the pin moves as nearly as possible on line B when at the back quarter of its stroke; D is any higher plane, and C any lower one. Divide the stroke of the connecting pin LR into four equal parts, and from S, with the crank-arm in the compasses, cut the circles with the arc E, and from T cut the circles with the arc F. It will be found that in the circle A, OP is slightly longer than in any of the other circles; therefore this is the position where the beat up is slowest. It will also be found that in the circles B and C there is scarcely any difference in OP, therefore sinking the crank-shaft from within reasonable limits makes very little difference; if anything, there is a slight decrease in the size of OP as the plane is lowered, but it is very slight, and the increase in the velocity of slay would also be very slight. On the other hand, by raising the crank-shaft to D a considerable increase in the velocity of the slay in beating up takes place, as it will be found that in this circle OP is much less than in the others.
At the back of the stroke it will be found that in the plane B the distance XY is least; therefore there is here the least dwell of the slay at the back of its stroke with the shaft in this position. This is because the pin moves as nearly as possible on the line B whilst the crank is at the back part of its stroke. As the crank is raised or lowered the dwell at the back increases slightly.
Reversing the direction of the loom makes a difference in the beat-up.
It will be found that in the circle A, OP and ON are about equal, therefore there will be scarcely any change in the velocity of beat-up by reversing the loom; but as the shaft is lowered ON will be found to become less than OP, and therefore a quicker blow is given by reversing the loom if the shaft is in this position. If the shaft is raised, as in the case of circle D, it will be found that ON becomes greater than OP; therefore with the crank above A, reversing the direction of the loom will cause a slower and weaker beat-up.
In the diagram, [Fig. 62], the crank and crank-arm are the same length for each position, the centre of the shaft being indicated by the dotted arc.
Timing of the Primary Movements.
FIG. 63.
The primary movements, shedding, picking, and beating up, are timed differently in relation to each other in weaving different classes of fabrics. For plain cloths, or other cloths where a good cover is required—that is, where the warp has to be spread—the crank should be set about the top centre when the healds are crossing each other. At [Fig. 38] the loom is timed in this manner. When so timed it is obvious that the shed will be considerably or altogether open when the reed is in contact with the cloth. By sinking the centres of the healds below a line drawn from the temple to the back rest, the upper portion of the shed is always slack, and if the pick is beaten up in a crossed shed, the loose ends of the warp are spread between the taut ones. In [Fig. 63] the straight line AB is drawn from the front carrier A to the back carrier B. The centres of the healds when level are on the line ACB, the point C being a little way below the line AB. When one stave is lifted a certain distance and the other goes down the same distance, it is obvious that the upper portion of the warp will be slacker than the lower portion, because the line ADEFB is shorter than ADGFB, and when the reed beats up with the warp in this position the slack ends are spread between the taut ones, thus giving a good cover to the cloth and preventing the reed marks from showing. Each set of ends alternately becomes slack.
Another advantage of beating up when the shed is crossed or partly open for the succeeding pick is that the pick is held more firmly in position than when the shed is not crossed, and therefore the picks can be got in better.
In twilled cloths the boldness of the twill is somewhat affected by the warp being spread, and these cloths are often preferred when made without the healds having been sunk.
FIG. 64.
If the dwell on the tappet is equal to one-third of a pick, as in [Fig. 64], the line D will mark the point of the tappet when the crank is at the top centre. When the crank has made one quarter of a revolution and is at the front centre with the reed in contact with the cloth, the point E will be acting on the treadle bowl. It will be seen that here the shed is almost fully open. When the crank is at the bottom centre the point G will be acting on the bowl, and the shuttle should just be entering the shed. When the point H of the tappet is acting on the bowl the shed will be commencing to close, and the shuttle must be just leaving the shed. When the point I is acting on the bowl the crank will be at the back centre, and when the crank reaches the top centre the healds will be again level.
If the dwell on the tappet is more than one-third pick, and at the commencement the crank is set on the top centre with the healds level, the shed will keep open longer for the shuttle to pass through, and would be more open when the crank reached the front centre. It will be obvious that for a wide loom a longer dwell is required than for a narrow loom.
By having the shed fully open before the shuttle enters the shed, the warp is spread and a good cover put on the cloth, but all this dwell is taken off the time which would otherwise be allowed for opening and closing the shed, and therefore means extra strain on the warp.
If it is not necessary to spread the warp, the shed need not be fully open until the shuttle is entering the shed. In this case the greatest possible amount of time is allowed for opening and closing the shed, thus putting as little strain as possible on the warp.
Speed of Tappets.
As previously stated, the bottom shaft in the loom, being the one used for picking, revolves at one-half the speed of the crank-shaft, and therefore plain cloth tappets may be fastened on the bottom shaft. Tappets of more than two picks to the round are usually fixed on a counter-shaft, S ([Fig. 65]), in looms with inside tappets. Sometimes the wheel E is geared directly into the wheel C on the bottom shaft, but usually a carrier-wheel, D, is used to convey the motion from the bottom shaft. The number of teeth in the carrier wheel has no effect on the speed of the tappets, as it is used simply to fill up the space between the bottom and counter-shafts.
FIG. 65.
If the wheel on the crank-shaft A contains 45 teeth, and the wheel B 90 teeth, C 40 teeth, and E 60 teeth, the tappet-shaft S will be making one revolution for three revolutions of the crank-shaft; therefore these wheels will do for three-end twill tappets. This may be proved by multiplying the drivers together and the drivens together, and dividing one by the other, thus—
| 90 × 60 | = 3 |
| 45 × 40 |
It is usual to place two or three wheels on the bottom shaft of the loom, so that any one of them may be geared into the carrier wheel D, each giving the required speed for different tappets. If a 40 wheel, a 30 wheel, and a 24 wheel are placed on the bottom shaft in such a manner that they can be moved along the shaft and any one of them be geared into the carrier wheel, any 3, 4, or 5 pick tappets can be driven with these wheels. We have seen that a 40 wheel at C gives three picks to the round.
Suppose the 30 wheel at C is geared into the carrier wheel, we get—
| drivens | 90 × 60 | = 4 |
| drivers | 45 × 30 |
or the relative speed of the tappets and crank-shaft are as 1:4; therefore these wheels may be used for any tappets with four picks to the round.
If the 24 wheel is at C, we get:
| drivens | 90 × 60 | = 5 |
| drivers | 45 × 24 |
and thus we get the proper rate of speed for tappets five pick to the round.
Some loom makers use the wheel E as a change wheel. With a 24 wheel C and a 36 wheel E we get three picks to the round, thus—
| drivens | 90 × 36 | = 3 |
| drivers | 45 × 24 |
With a 24 wheel C, a 48 wheel E gives 4 picks,
With a 24 wheel C, a 60 wheel E gives 5 picks,
With a 24 wheel C, a 72 wheel E gives 6 picks.
Example.—Find the number of teeth for the wheel C on the bottom shaft to drive tappets seven picks to the round, wheel on tappets 63 teeth.
| 90 × 63 | = 18 wheel. Ans. |
| 45 × 7 |
Woodcroft’s tappets, as a rule, are driven directly from the crank-shaft. As these tappets are usually of a large circumference, a large wheel on them is of no disadvantage, although sometimes intermediate wheels are used.
If the tappets are twelve to the round, and the wheel on the tappets contains 192 teeth, a driving wheel of 16 teeth will be required on the crank-shaft.
| 192 | = 12 picks to the round |
| 16 |
For driving outside tappets, as in [Fig. 39], a driving wheel on the crank-shaft and two intermediate wheels are generally used. The tappets are placed on the bottom shaft outside the loom, but they are loose upon the shaft, and can, of course, be made to revolve at a different speed to the shaft, either in the same or in the opposite direction. This system of driving the tappets is shown at [Fig. 66]. The wheel A, on the crank-shaft, drives the wheel B, on an intermediate stud; the wheel C, on the same centre, drives the tappet wheel D.
FIG. 66.
To find the wheel on the crank-shaft, or the first driver, the other wheels being as follows: first driven wheel, B, 36 teeth; second driver, C, 12 teeth; tappet wheel, D, 120 teeth.
Multiply the two driven wheels together, and divide by the given driver multiplied by the picks to the round, thus—
| 36 × 120 | = 40 first driver, A. |
| 12 × 9 |
To find the second driver for eight picks, the other wheels being: first driver, A, 20; first driven, B, 40; second driven, D, 60.
The given driver multiplied by the picks to the round, 20 × 8 = 160; the drivens multiplied together, 40 × 60 = 2400; 2400 ÷ 160 = 15 wheel required.
To find either of the driven wheels, multiply the two drivers and the picks together, and divide by the driven given wheel, thus—
Example.—Find the wheel for the tappets, D, for 10 picks to the round, the other wheels being: first driver, 16 teeth; first driven, 32 teeth; second driver, 20 teeth.
| 16 × 20 × 10 | = 100 wheel required |
| 32 |
To find both intermediate wheels, multiply the given driver by the picks to the round, and as the product is to the teeth in the tappet wheel, so is the required driven to the required driver.
Example.—Find the two intermediates for 10-pick tappets, if the wheel on the crank-shaft has 18 teeth, and the wheel on the tappets 120 teeth. The 18 × 10 = 180, and therefore the two required wheels must be in the proportion of 180 to 120, the former being the driven wheel. Thus a 36 driven and a 24 driver will give the required speed to the tappets. That this is correct may be seen from the following:—
| 18 × 24 | = 10 picks |
| 36 × 120 |
That the required wheels must be in this proportion will be apparent from the fact that if the wheel B has ten times the number of teeth in A, then B is revolving at the speed at which the tappets are to move; therefore if the wheel C has the same number of teeth that D has, the speed of the tappets will remain the same.