The pallets are to be equidistant; the wheel teeth of the “club” form; there are to be two rollers; wheel, pallet, and balance centers are to be in straight line. The lock is to be 1½°, the run ¼°, making a total lock of 1¾°; the movement of pallets from drop to drop is to be 10°, while the fork is to move through 10¼° from bank to bank; the lift on the wheel teeth is to be 3°, while the remainder is to be the lift on the pallets as follows: 10¼(1¾+3)=5½° for lift of pallets. The wheel is to have 15 teeth, with pallets spanning 3 teeth or 2½ spaces, making the angle from lock to lock =360÷15×2½=60°, the interval from tooth to tooth is 360÷15=24°; divided by 2 pallets=24÷2=12° for width of tooth, pallet and drop; drop is to be 1½°, the tooth is to be ¾ the width of the pallet, making a tooth of a width of 4½° and a pallet of 6°. The draw is to be 12° on each pallet, while the locking faces of the teeth are to incline 24°. The acting length of fork is to be equal to the distance of centers of scape wheel and pallets; the impulse angle is to be 28°; freedom from dart and safety, roller is to be 1¼°, and for dart and corner of crescent 5°; freedom for ruby pin and acting edge of fork is to be 1¼°; width of slot is to be ½ the total motion, or 10¼÷2=5°; shake of ruby pin in slot=¼°, leaving 5¼=4° for width of ruby pin. Radius of safety roller to be 47 of the theoretical impulse radius. The length of horn is to be such that the end would point at least to the center of the ruby pin when the edge of the crescent passes the dart; space between the end of horn and ruby pin is to be 1½°. It is well to know that the angles for width of teeth, pallets and drop are measured from the wheel center, while the lifting and locking angles are struck from the pallet In the fork and roller action, the angle of motion, the width of slot, the ruby pin and its shake, the freedom between dart and roller, of ruby pin with acting edge of fork and end of horn are all measured from the pallet center, while the impulse angle and the crescent are measured from the balance center. A sensible drawing board measures 17×24 inches, we also require a set of good drawing instruments, the finer the instruments the better; pay special attention to the compasses, pens and protractor; add to this a straight ruler and set square. The best all-round drawing paper, both for India ink and colored work has a rough surface; it must be fastened firmly and evenly to the board by means of thumb tacks; the lines must be light and made with a hard pencil. Use Higgins’ India ink, which dries rapidly. We will begin by drawing the center line A A B; use the point B for the escape center; place the compass on it and strike G H, the primitive or geometrical circle of the escape wheel; set the center of the protractor at B and mark off an angle of 30° on each side of the line of centers; this will give us the angles A B E and A B F together, forming the angle F B E of 60°, which represents from lock to lock of the pallets. Since the chord of the angle of 60° is equal to the radius of the circle, this gives us an easy means of verifying this angle by placing the compass at the points of intersection of F B and E B with the primitive circle G H; this distance must be equal to the radius of the circle. At these points we will construct right angles to E B and F B, thus forming the tangents C A and D A to the primitive circle G H. These tangents meet on the line of centers at A, which will be the pallet center. Place the compass at A and draw the locking circle M N at the points of intersection of E B and F B with the primitive circle G H. The locking edges of the pallets will always Having shown how to delineate the locking face of the engaging pallet when locked, we will now consider how to draft both it and the disengaging pallet in correct positions when unlocked; to do so we direct our attention until further notice to Fig.28. The locking faces Q M of the engaging and S N of the disengaging pallets are shown in dotted lines when locked. We must now consider the relation which the locking faces will bear to E B in the engaging, and to F B in the disengaging pallets when unlocked. This is a question of some importance; it is easy enough to represent the 12° from the 30° angles when locked; we must be certain that they would occupy exactly that position and yet show them unlocked; we shall take pains to do so. In due time we shall show that there is no appreciable loss of lift on the engaging pallet in the We will notice that the face Q M of the engaging pallet when unlocked stands at a greater angle to E B than it did when locked, while the opposite is the case on the disengaging pallet, in which the angle S N F is much less than S N F. This shows that the deeper the engaging pallet locks, the lighter will the draw be, while the opposite holds good with the disengaging pallet; also, that the draw increases during the unlocking of the engaging, and decreases during the unlocking of the disengaging pallet. These points show that the draw should be measured with the fork standing against the bank; not when the locking corner of the pallet stands on the primitive circle, as is so often done. The recoil of the wheel (which determines the draw), is illustrated by the difference between the locking circle M N and the face Q M for the engaging, and S N for the disengaging pallet, and along the acting surface it is alike on each pallet, showing that the draft angle should be the same on each pallet. A number of years ago we constructed the escapement model which we herewith illustrate. All the parts are adjustable; the pallets can be moved in any direction, the draft angles can be changed at will. Through this model we can practically demonstrate the points of which we have spoken. Such a model can be made by workmen after studying these papers. The adjustable model escapement. We will again proceed with the delineation of the escapement here illustrated. After having drawn the locking face Q M, we draw the angle of width of teeth of 4½°, by planting the protractor on the escape center B. We measure the angle E B K, from the locking face of the pallet; the line E B does not touch the locking face of the pallet at the present time of contact with the tooth, therefore a line must be drawn from the point of contact to the center B. We did so in our drawing but do not illustrate it, as in a reduced engraving of this kind it would be too close to E B and would only cause confusion. We will now draw in the lifting angle of 3° for the tooth. From the tangent C A we draw T A at the required angle; at the point of intersection of T A with the 30° line E B we have If the real or outer circle of the wheel intersects both the locking circle M N and the path O of the discharging edge at the points where T A intersects them, then there will be no loss of lift on the engaging pallet. This is precisely how it is in the diagram; but if there is any deviation, then the angle of loss must be measured on the real diameter of the wheel and not on the primitive, as is usually done, as the real diameter of the wheel, or in other words the heel of the tooth, forms the last point of contact. With a wider tooth and a greater lifting angle there will even be a gain of lift on the engaging pallet; the pallet in such a case would actually require a smaller lifting angle, according to the amount of gain. We gave full directions for measuring the loss when describing its effects in Fig.8. Whatever the loss amounts to, it is added to the lifting plane of the pallet. In the diagram under discussion there is no loss, consequently the lifting angle on the pallet is to be 5½°. From V A we draw V A at the required angle; the point of intersection of V A with the path O will be the discharging edge O. It will now only be necessary to connect the locking corner M with it, and we have the lifting plane of the pallet; the discharging side of the pallet is then drawn parallel to the locking face and made a suitable length. We will now draw the locking edges of the tooth by placing the center of the protractor on the locking edge M and construct the angle B M M of 24° and draw a circle from the scape center B, to which the line M M will be a tangent. We will utilize this circle in drawing in the faces of the other teeth after having We now construct W A as a tangent to the outer circle of the wheel, thus forming the lifting angle D A W of 3° for the teeth; this corresponds to the angle T A C on the engaging side. W A touches the outer circle of the wheel at the intersection of F B with it. We will notice that there is considerable deviation of W A from the circle at the intersection of J B with it. At the intersecting of this point we draw U A; the angle U A W is the loss of lift. This angle must be added to the lifting angle of the pallets; we see that in this action there is no loss on the engaging pallet, but on the disengaging the loss amounts to approximately ° in the action illustrated. As we have allowed ¼° of run for the pallets, the discharging edge P is removed at this angle from U A; we do not illustrate it, as the lines would cause confusion being so close together. The lifting angle on the pallet is measured from the point P and amounts to 5½°+ the angle of the loss; the angle W A U embraces the above angles besides ¼° for run. If the locks are equal on each pallet, it proves that the lifts are also equal. This gives us a practical method of proving the correctness of the drawing; to do so, place the dividers on the locking circle M N at the intersection of T A and V A with it, as this is the extent of motion; transfer this measurement to N, if the actual lift is the same on each pallet, the dividers will locate the point which the locking corner N will occupy when locked; this, in the present case, will be at an angle of 1¾° below the tangent D A. By this simple method, the correctness of our proposition that the loss of lift should be measured from the outside circle of the wheel, can be proven. We often see the loss measured for the engaging pallet on the primitive circumference G H, and on the real circumference for the disengaging; if one is right then the other must be wrong, as The space between the discharging edge P and the heel of the tooth forms the angle of drop J B I of 1½°; the definition for drop is that it is the freedom for wheel and pallet. This is not, strictly speaking, perfectly correct, as, during the unlocking action there will be a recoil of the wheel to the extent of the draft angle; the heel of the tooth will therefore approach the edge P, and the discharging side of the pallet approaches the tooth, as only the discharging edge moves on the path P. A good length for the teeth is 110 the diameter of the wheel, measured from the primitive diameter and from the locking edge of the tooth. The backs of the teeth are hollowed out so as not to interfere with the pallets, and are given a nice form; likewise the rim and arms are drawn in as light and as neat as possible, consistent with strength. Having explained the delineation of the wheel and pallet action we will now turn our attention to that of the fork and roller. We tried to explain these actions in such a manner that by the time we came to delineate them no difficulty would be found, as in our analysis we discussed the subject sufficiently to enable any one of ordinary intelligence to obtain a correct knowledge of them. The fork and roller action in straight line, right, or any other angle is delineated after the methods we are about to give. We specified that the acting length of fork was to be equal to the center distance of wheel and pallets; this gives a fork of a fair length. Having drawn the line of centers A A we will construct an angle equal to half the angular motion of the pallets; the latter in the case under consideration being 10¼°, therefore 5° is spaced off on each side of the line of centers, forming the angles mAk of 10¼°. Placing our dividers on A B the center distance of ’scape wheel and We will put it in the following proportion; let Ac equal acting length of fork, and x the unknown quantity; 2810.25Acx; the answer will be the theoretical impulse radius. Having found the required radius we plant one jaw of our measuring instrument on the point of intersection of c c with kA or mA and locate the other jaw on the line of centers; we thus obtain A the balance center. Through the points of intersection before designated we will draft X A and Y A forming the impulse angle X A Y of 28°. At the intersection of this angle with the fork angle kAm, we draw i i from the center A; this gives us the theoretical impulse circle. The total lock being 1¾° it follows that the angle described by the balance in unlocking =1¾×2.7316=4.788°. According to the specifications the width of slot is to be 5°; placing the center of the protractor on A we construct half of this angle on each side of kA, which passes through the center of the fork when it rests against the bank; this gives us the angle sAn of 5°. If the disengaging pallet were shown locked then mA would represent the center of the fork. The slot is to be made of sufficient depth so there will be no possibility of the ruby pin touching the bottom of it. The ruby pin is to have 1¼° freedom in passing the acting edge of the fork; from the The radius of the safety roller was given as 47 of the theoretical impulse radius. They may be made of various proportions; thus is often used. Remember that the smaller we make it, the less the friction during accidental contact with the guard pin, the greater must the passing hollow be and the horn of fork and guard point must be longer, which increases the weight of the fork. Having drawn in the safety roller, and having specified that the freedom between the dart and safety roller was to be 1¼°, the dart being in the center of the fork, consequently kA is the center of it; therefore we construct the angle kA X of 1¼°. At the point of intersection of X A with the safety roller we draw the arc g g; this locates the point of the dart which we will now draw in. We will next draw dA from the balance center and touching the point of the dart; we now construct bA at an angle of 5° to it. This is to allow the necessary freedom for the dart when entering the crescent; from A we draw a line through the center of the ruby pin. We do not show it in the drawing, as it would be indiscernible, coming very close to A X. This line will also pass through the center of the crescent. At the point of intersection of Ab with the safety roller we have one of the edges of the crescent. By placing our compass at the center of the crescent on the periphery of the roller and on the edge which we have just found, it follows that our compass will span the radius of the crescent. We now sweep the arc for the latter, thus also drawing in the remaining We have the acting edges of the fork on cc and have also found the position of the ends of the horns; their curvature is drawn in the following manner: We place our compasses on A and r i, spanning therefore the real impulse We think the instructions given are ample to enable any one to master the subject. We may add that when one becomes well acquainted with the escapement, many of the angles radiating from a common center, may be drawn in at once. We had intended describing the mechanical construction of the escapement, which does unmistakably present some difficulties on account of the small dimensions of the parts, but nevertheless it can be mechanically executed true to the principles enumerated. We have evolved a method of so producing them that young men in a comparatively short period have made them from their drafts (without automatic machinery) that their watches start off when run down the moment the crown is touched. Perhaps later on we will write up the subject. It is our intention of doing so, as we make use of such explanations in our regular work. |