An Account of a Balance of a New Construction, Supposed to Be of Use in the Woollen Manufacture. By W. Ludlam, B. D. Fellow of St. John's College, Cambridge

XXV. An Account of a Balance of a new Construction, supposed to be of Use in the Woollen Manufacture. By W. Ludlam, B. D. Fellow of St. John's College, Cambridge. Read June 6, 1765. It is of consequence in some branches of the woollen manufacture, that the thread of which any piece is woven should be all of the same fineness. After it is spun, it is made into skains of the same length, and these are sorted according to the fineness of the spinning. The manufacturers usually distinguish and denominate the fineness, by the number of skains which go to the pound; the coarsest being about 12 to the pound, and the finest near 60. There is no other method of sorting in use, except by the eye; but it requires great nicety to distinguish the size of threads so small, and long experience to know by the look only, how many skains of any particular sort will make a pound. A method of weighing them readily would save much time: the machine here delineated is for that purpose. It resembles the beam of a common pair of scales; at one end of it is a fixt weight, which I call the counterpoise, at the other a hook: in sorting, the skain to be examined is put upon the hook, and sinks down more more or less, according to its weight, till the counterpoise by rising balances it, and then the index or cock of the beam points out, on a graduated arch, the number of skeins of that sort, which go to the pound. A scale, instead of the hook, might be used for weighing money, if the arch were properly divided for that purpose. Mr. Rouse of Harborough, many years ago, made a machine for sorting woollen thread upon the same principle with this; but as what he did was mostly tentative, he was not aware of some considerable advantages which the theory points out. For the machine will not distinguish with equal nicety the skeins of every size. In Mr. Rouse's machine, the divisions were too small, and the largest chanced to fall at 18 to the pound; but it would have been better if the finer sorts had been more accurately distinguished, as being of greater consequence to be well sorted, and more difficult to be sorted by the eye only, than the coarser ones. This machine distinguishes best the yarn of 36 to the pound, one of the finest sorts, as I am informed, in common use, the largest division lying between 36 and 37; the other divisions are as large, and the whole range of the index as much as can be allowed without other inconveniencies. The theory contains the necessary rules for finding the angle of the beam, for calculating the divisions on the arch, and for placing their largest interval in any part of them. Directions Directions for making the Balance. It consists of a mahogany stand, a steel beam and brass ring for the divisions. Fig. 1. FGH is the triangular base of the stand, having a screw in each angle to set it level: into this is dove-tail'd the upright back KK; the standard board LLL is put into a dove-tail'd groove in the back, and tenoned into the triangular board at the bottom. The two cocks CC, between which the arbor of the beam plays, and the ring RR are screw'd to the standard board LL. The beam A B, with its cock or index E, is broad in the middle to gain strength, and pierced to make it lighter. It is rivetted on a collet soldered to the arbor, as clock wheels are. The pivots run in hard brass, and have plates of hardened steel for their points to bear against. At one end of the beam is the counterpoise A, consisting of two round pieces of brass, screwed together through an hole in the beam. The other end of the beam is made thicker for about half an inch in length, and slit to receive a loop of hardened steel, which hangs on the steel pin B; on the lower part of the loop hangs the hook, which holds the skain. The block of wood N, is screwed to the standard board; the upper part of which is lined with velvet for the counterpoise to rest upon, when the skain is taken off the hook; the brass pin P is for the other arm of the beam to bank against. The angle of the beam A C B is $168^\circ 6'$, the radius C A or C B, 6 inches; the breadth of the beam in the middle $\frac{3}{4}$ of an inch, at the ends $\frac{1}{4}$; the thickness at the end B where it is slit $\frac{1}{6}$; everywhere else $\frac{1}{8}$. The length of the arbor $1\frac{3}{4}$; diameter of the pivets $\frac{1}{5}$. The weight of the counterpoise one ounce avoirdupoise. The hook, with the loop and steel pin included, 0.68 avoirdupoise. The division of 50 is at the top of the ring. In making the beam, the point of the index E must be equally distant from the centers A and B; the whole beam and index made so as to poise itself, and remain at rest in any position, before the counterpoise steel pin and loop are put on. The counterpoise being at first made too heavy, to adjust it, take off the beam, but let the two cocks and ring remain screwed on; hang then a fine wire and plummet on the top division, and with a watch glass look through one pivot hole till you see the wire against the other; turn the screw G till the wire bisects the hole, and the top division is then perpendicularly over it. Take away the plumb line and put on the beam and counterpoise; on the loop hang a weight, which together with the steel pin and loop makes one ounce avoirdupoise; turn now, by degrees, some metal off the counterpoise, till the index points at the top division, and the counterpoise will be truly adjusted, whether the arms of the beam, C A, C B, are precisely equal or not: for though the arms should be nearly equal, it is not necessary they should be exactly so, as in scales. The divisions are set on the ring by an instrument made on purpose, which will very readily cut them, tho' unequal, with great exactness, on any circle large or small. To prove the beam, put a weight into the scale (that of 28 to the pound is the best in this case) and see if it brings the index down to the proper division on the ring: if it carries it too far, the angle of the beam ACB is too great; if the contrary, too little; and the arms CA, CB, must be set a little in, or out, till the angle is right: or the angle of the beam may be first found experimentally, by the rule hereafter given, and the divisions calculated to it, which is not much trouble; for having a table ready made for the intended angle, the alterations in that table occasion'd by a small variation from that angle will be easily found by the rules at the end of the following Theory. When the balance is to be used, a weight of $\frac{1}{5}$ of a pound avoirdupoise is to be put on the hook: The screw G must then be turned, till the index E points at the division of 50: The machine is then properly adjusted, and the weight may be taken off, and a skain put on. Let ACB be a bent lever moveable about the angular point C, B a scale at the end of the arm B, A a counterpoise at the end of the other arm A; Given the angle of the lever, the length of the arms, the respective weights of the scale and counterpoise, to find the position of the lever when at rest. Produce the arm BC to I, so that CI may be to CB as the weight of the scale to the counterpoise; join AI, turn the lever about till AI is perpendicular to the horizon, and in that position it will be at rest. For if HB be drawn parallel to IA, and FCH perpendicular to HB and IA, then CF : CH : CI : CB, or as the scale to the counterpoise; that is, perpendiculars from the center of motion to the lines of direction are reciprocally as the forces applied, therefore in that case the forces balance each other. Cor. Join AB and draw CG parallel to AI, and AG : GB :: CI : CB, or as the scale to the counterpoise, therefore G is their center of gravity, which lying in a perpendicular to the horizon passing thro' the center of motion, again shews the lever to be at rest. Scholium. There are indeed two positions in which the lever will be at rest, one when the center of gravity is above, and another when it is below the center of motion; the latter only is consider'd in what follows. Let the arms CA and CB be equal, let the weight of the counterpoise be fixt, but that of the scale variable; then if the fixt line CA (=CB) represent the former, the variable line CI will represent the latter. In the line CI, take CD = CA and join AD; thro' the center C, draw ECK parallel to AD, which therefore bisects the angle of the lever ACB, and the part EC may stand for the index or cock of the lever. Let CP be perpendicular to the horizon, and ECP the angle of the index with that perpendicular is equal to DAI; but ACK is the semi-sum, and DAI the semi-difference of the angles angles C A I, C'IA: Therefore CI + CA : CI - CA :: or the sum of the weight of the scale and counterpoise is to their difference, as the tangent of ACK, half the angle of the lever, to the tangent of DAI = ECP, the inclination of the index to the perpendicular when the lever is at rest: and the angles at A and I, or the inclinations of the arms, will be the sum and difference of ACK and DAI. Hence the scale and counterpoise being given, the inclination of the index may be found from the angle of the lever; or the angle of the lever from the inclination of the index. The weight of the counterpoise being fixt, let that of the scale vary uniformly; it is required to find when the angular motion of the index is greatest. All things remaining as before, draw AS perpendicular to CI; and the variation of CI will be the same as that of SI (the line CS being constant): For a like reason the variation of the angle DAI is the same with that of SAI. Now the variation of SI, is to that of the arch which measures SAI, (AS being radius) as AI² to AS²; if therefore CI or SI flows uniformly, the fluxion of SAI or DAI, or the angular motion of the index, will be greatest when AI is least, that is, when AI coincides with AS, or when the inclination of the index is equal to DAS, or when the arm carrying the scale is parallel to the horizon. The angle DAS is the complement of ADS (=ACK) or of half the angle of the lever. Again CA : CS: or radius, to the co-sine of the angle of the lever as the counterpoise to the weight of the scale, when the angular motion of the index is greatest. Let now the weight of the scale as well as the counterpoise be fixt, but into it let there be successively put variable weights or skains of thread, to be denominated by the number of them which (at that size) go to the pound: Let the weight of these skains vary at each successive change in such a manner, that the number which goes to the pound may be increased each time by unity: It is required to find when the angular variation of the index, in one exchange, will be greatest. All things as before, let CL represent the fixt weight of the scale, LI the variable one of the skain, CI that of both together. Call AS, s: SL, d: put \( p = \text{one pound}, x \) the number by which any one skain is denominated, consequently \( \frac{p}{x} \) its weight = LI; whence SI = LI - LS = \( \frac{p}{x} - d \), whose fluxion (putting the uniform fluxion of \( x \) to be unity) is \( \frac{-p}{xx} \): but the fluxion of SI is to that of the arch which measures SAI, as AI\(^2\) to AS\(^2\), or as AS\(^2\) + SI\(^2\) to AS\(^2\), or as ss + \( \frac{p}{x} - d^2 \) : ss, that is ss + \( \frac{p}{x} - d^2 \) : ss :: \( \frac{-p}{xx} \) to the variation of the index sought; whence that variation is greatest when \( x = \frac{pd}{ss + dd} \), that is when \( \frac{p}{x} \) or the weight of the skain = \( \frac{ss + dd}{d} \): wherefore if CA represent the counterpoise, and CL the scale, join AL and draw AF perpendicular pendicular to it, cutting CL produced in F, then will LF represent the weight of the skain, and DA F be the inclination of the index, when its variation is greatest. Fig. 4. On FL as a diameter describe a semicircle, and let CA intersect it again in a: Join La, aF, then will LaF be a right angle as well as LAF; therefore with the given angle of the lever ACB and given scale CL, we have the same maximum FL, either with the counterpoise CA, or Ca; either of which gives the other; for CA : CL :: CF : Ca. Fig. 4. In like manner, if the angle of the lever, the counterpoise CA, and FL the weight of the skain at the maximum are given, with Fig. 5. the center A and radius AE = \(\frac{1}{2}\) FL describe a circle cutting BC produced in E and e, towards C set off EL and el, equal to the radius aforesaid, and either CL or Cl will be the weight of the scale that will give the proposed maximum. It is manifest that if AE or \(\frac{1}{2}\) FL is less than AS, the problem is impossible; therefore the maximum is limited, and must be proposed such, that the weight of the skain in that case be less than 2 AS. **Example:** Let the semi-angle of the lever be 84°: 03', the counterpoise one ounce, let the division for 50 to the pound, stand at the top of the graduated arch perpendicularly over the center of the lever, that is, let LD = \(\frac{1}{5}\) oz. consequently LC or the scale = CD - LD, or CA - LD = 1 - \(\frac{1}{5}\) = 0.68 0.68 oz. then is \( CS = 0.9785090 \), \( LS \) or \( d = 0.2985090 \), \( AS \) or \( s = 0.2062042 \), and \( \frac{ss + dd}{d} \) or \( \frac{ss}{d} + d \left( = \frac{AS^2}{LS} + LS \right) = LF = 0.440951 \) oz. and \( \frac{16}{0.440951} = 36.2852 \); therefore the greatest motion of the index will be when a skain of 36 to the pound is changed for one of 37, and the largest interval on the graduated arch will be that which lies between 36 and 37. CA : CL :: CF : Ca or 1 : 0.68 :: 1,120951 : 0.762247 ounces, the other counterpoise which gives the same maximum. Let the maximum be assumed at 36 skains to the pound exactly, or when \( FL = 0.4444444 \), then is \( \frac{1}{2} FL \) or \( AE = 0.2222222 \), and \( SE \left( = \sqrt{AE + AS \times AE - AS} \right) = 0.0828404 \), whence \( CL \left( = CS - AE + SE \right) = 0.8391272 \), and \( CL \left( = CS - AE - SE \right) = 0.6734464 \): if this last quantity be taken for the weight of the scale, the division for 49 in the pound would fall at the top of the arch very nearly, therefore assume the division of 50 in the pound for the top, which will make but little alteration in the maximum, and the weight of the scale will be 0.68, the maximum between 36 and 37, as before. \( 2 AS = 0.4124084 \), and \( \frac{16}{0.4124084} = 48, 8419 \); therefore with the angle of the lever and counterpoise as before, the maximum cannot be higher than at 48 skains to the pound. In Mr. Roule's machine the angle of the lever was $165^\circ$, the counterpoise 1.46057 ounces, the scale 1,19543 or AS = 0.378024, therefore his machine did not admit of a maximum higher than between 26 and 27 skains to the pound. Fig. 3. To give one example of finding the inclination of the index, let it be required to find the angle for a skain of 32 to the pound, whose weight is 0.5 ounce; whence CI = 1.18, CA = 1, and CI + CA : CI - CA; or $2.18 : 0.18 ::$ Tang. ACK = $84^\circ : 03'$. Tang. ECP = $38^\circ : 23'$. This rule for the inclination of the index, suggests an easy way of finding the alterations in it, occasion'd by any small variation in the angle of the lever, other things remaining the same. For the change in the Logarithmic Tangent of ECP, the fourth term in the proportion, is equal to that made in the Logarithmic Tangent of ACK, the third term; the other two terms being the same. Take therefore the difference of the Logarithmic Tangents of the two semi-angles, and divide it by the natural numbers 1, 2, 3, 4, &c. successively, and their quotients among the differences of the Logarithmic Tangents, in Tables for each minute, will point out the quantity of the angle ECP, when the alteration in it is 1, 2, 3, 4, &c. minutes successively; by which means a table may be easily formed for the correction of that first computed. Suppose, for instance, the semi-angle of the lever to be only $84^\circ$: instead of $84^\circ : 03'$, for which a table has been made, it is required to correct this table, so as to suit it to this semi-angle $84^\circ$. The difference of the Log. Tangents of these semi-angles is 36.608; divide this by by any of the natural numbers, for instance by 10, the quotient 3660, standing among the Tabular differences (in Tables for each minute) at $21^\circ : 50'$ and $68^\circ : 10'$, shews that when the value of E.C.P in the table already made, is either $21^\circ : 50'$ or $68^\circ : 10'$, 10 minutes must be subtracted to adapt it to the lever whose semi-angle is only $84^\circ$: and so of the rest. The greatest difference of the inclination of the index will be when E.C.P is, $45^\circ$: the Tabular differences being then the least. All this is not indeed quite accurate, but sufficiently so for practice. To find accurately when this difference is greatest, and what that greatest difference is; add (or subtract as the case requires) half the difference of the Logarithmic Tangents of the semi-angles aforesaid, to the Logarithmic Radius, and it will give the Logarithmic Tangent of E.C.P, in that case; and the difference between E.C.P thus found, and its complement, is the greatest difference sought. Thus half 36,608 added to 10.000000 is 10.0018304, the Log. Tangent of $45^\circ : 7' : 14''$, the inclination of the index, when its change by reducing the semi-angle of the lever to $84^\circ$, is greatest, and the quantity of that change is $14' : 28''$, 8. A Table A Table of the inclination of the index, when the semi-angle of the lever is $84^\circ : 03'$, the counterpoise 1 ounce and the scale 0.68 ounces avoirdupoise, for any number of skains to the pound from 10 to 70. | Skains to the pound | 0 | 0 | 0 | 0 | |---------------------|---|---|---|---| | 10 | 75 : 03 | 25 | 52 : 56 | 40 | 20 : 15 | 55 | 8 : 04 | | 11 | 73 : 56 | 26 | 51 : 00 | 41 | 18 : 02 | 56 | 9 : 30 | | 12 | 72 : 47 | 27 | 49 : 01 | 42 | 15 : 50 | 57 | 10 : 53 | | 13 | 71 : 35 | 28 | 46 : 59 | 43 | 13 : 41 | 58 | 12 : 13 | | 14 | 70 : 20 | 29 | 44 : 54 | 44 | 11 : 35 | 59 | 13 : 30 | | 15 | 69 : 02 | 30 | 42 : 46 | 45 | 9 : 31 | 60 | 14 : 44 | | 16 | 67 : 40 | 31 | 40 : 36 | 46 | 7 : 30 | 61 | 15 : 55 | | 17 | 66 : 16 | 32 | 38 : 23 | 47 | 5 : 32 | 62 | 17 : 03 | | 18 | 64 : 48 | 33 | 36 : 09 | 48 | 3 : 38 | 63 | 18 : 08 | | 19 | 63 : 17 | 34 | 33 : 54 | 49 | 1 : 47 | 64 | 19 : 11 | | 20 | 61 : 42 | 35 | 31 : 37 | 50 | 0 : 00 | 65 | 20 : 12 | | 21 | 60 : 04 | 36 | 29 : 20 | 51 | 1 : 44 | 66 | 21 : 10 | | 22 | 58 : 22 | 37 | 27 : 03 | 52 | 3 : 24 | 67 | 22 : 06 | | 23 | 56 : 37 | 38 | 24 : 46 | 53 | 5 : 01 | 68 | 23 : 00 | | 24 | 54 : 48 | 39 | 22 : 30 | 54 | 6 : 34 | 69 | 23 : 52 | | | | | | | | 70 | 24 : 41 | N. B. According to the report from the committee appointed by the House of Commons to enquire into the original standard weights, one pound Avoirdupoise is 1 lb. 2 oz. 12 dwt. Troy; consequently one ounce is 18 dwt. 6 gr. and 0.68 oz. is 12 dwt. 9.84 grains.