In this post, Tim Robinson explains how the meshing of looping gears in the Analytical Engine is achieved. This work is part of ongoing research to demonstrate that Babbage's Analytical Engine was a practical design.
By Tim Robinson
Here we attempt to analyze the accurate meshing of gears forming loops. This is a situation which occurs in many places in the Analytical Engine. At any given time, not all the gears in any particular loop will be participating in motion, because they can also be moved axially in and out of mesh, but for this to be possible, the teeth must accurately align whenever they are to be moved into mesh. The allowed motions of the gears are always through a discrete number of teeth, and at the end of the motion Locks are applied to ensure accurate alignment with no possibility of derangement. If, in this locked condition, the teeth accurately align, then they can be moved smoothly back into mesh. The problem is to determine an accurate set of positions for the axes of the wheels to make this possible. The correct meshing is extremely sensitive to the precise locations of the axes and the original drawings do not allow these positions to be determined with sufficient accuracy. Further the Plan drawings generally show the gears as simple circles representing the pitch circles of the gears with no actual teeth drawn, so they provide no indication of the specific orientation of the individual wheels. Thus if the Engine is to be built, we must determine a precise set of positions for the axes, consistent with the original drawings within the measurement accuracy, while ensuring that all wheels can be oriented so as to correctly mesh.
It is not known how Babbage determined the details of the proposed gear layouts of his many Plan drawings. It is quite possible that he made cardboard mockups of the various wheels to experiment with trial arrangements. While there are many surviving examples of such cardboard trial pieces in the collection of the London Science Museum, these do not include the extensive set of gears which would have been necessary for this purpose, so if he did make them, they have not survived.1 Further, such a trial layout would have been unlikely to deliver the precise positions of the axes which would be needed for actual construction of the Engine, though it undoubtedly offered the possibility of an “existence proof.” Indeed, before embarking on a more detailed theoretical analysis of the problem we undertook something similar, using laser cut wooden mockups for the various wheels, to establish the essential viability of the Plan 27 layout in drawing BAB/A/093. This exercise established that many solutions do exist, but just as would have been the case for Babbage, it does not solve the problem of defining the precise positions required for manufacturing. An example can be seen in Figure 1.
To make progress we will consider one specific instance of a loop, the leftmost arithmetic group in Plan 27 as drawn in BAB/A/093, a sub-section of which is shown in Figure 2.
This consists of the five wheels labeled by Babbage as ″A, ″G, ″S, ″L, and ″J. For simplicity we will refer to these here as A, G, S, L, and J. The axes of these wheels form the vertices of an irregular pentagon AGSLJ. The lengths of the sides of this pentagon are known precisely because each side must be equal to the sum of the pitch radii of the two gears at its ends, and these radii are determined by the number of teeth and the diametrical pitch (DP) of the respective gears. In this particular case the numbers of teeth involved are A: 80, G: 28, S: 40, L: 48, and J: 30, but we can be more general by referring to these numbers as TA, TG, etc. There is one further minor complication in this particular group in that each of the gears S and L is a compound made of two gears permanently attached together. The two gears of each compound have the same number of teeth, but of different pitch, and so therefore of different diameter. The teeth of the two parts of the compound are known to align exactly. This can be seen on Figure 2, where S and L mesh with each other by the large teeth, but with G and L by the small teeth. The only practical effect of this for our purposes is to change the center to center distance of S and L. For reference, the large elements of S and L as drawn are 8 DP, the small elements, and the other three gears are all 10 DP, but as with the number of teeth, this has no effect on the general analysis.
Given a particular placement of the axes, we can determine if the closed loop meshes by counting the teeth around a circuit of the gears. First, we imagine the gears laid out along a straight line, with a second copy of the first gear at the end of the line as in Figure 3.
We imagine a tooth to exist on the line at the left of the first gear. Then, at the right of that gear there will be another tooth if the gear has an even number of teeth, or a space between teeth if the number is odd. The next gear must be placed in mesh with this, and so on down the line. In our particular case, since all gears have an even number of teeth, they will alternate between a tooth and a space at the left down the line. If the line is considered to be made of segments hinged at the axes of the gears, we can bend the segments around to make a loop in which the final gear overlays the first. We can identify the final gear with the first, but only if, in forming the loop, it has rotated through a half integral number of teeth, so the teeth of the two copies exactly coincide. Refer now to Figure 4 in which such a circuit is highlighted.
Around a circuit such as this, depending on whether the gears have an odd or even number of teeth, and whether there is an odd or even number of gears in the loop, the total number of teeth passed over must be either an integer or a half integer number. In the case at hand, all the gears have an even number of teeth, and there is an odd number of gears, so we require a half integral number of teeth in the loop, so that a space on gear G will mesh with the tooth assumed on the left of gear S.
Now, we should point out here that an exact mesh is always possible for an odd number of gears in the loop, whatever the positions of the axes (assuming only correct center to center distances). By turning, say, gear S clockwise, gear L will turn counterclockwise, J clockwise, A counterclockwise, and G clockwise. (We are ignoring here that ultimately G is going to be meshed with S which would preclude any motion.) Since both S and G are turning clockwise, there must be a positions in which the teeth exactly align to produce the mesh we desire. Note that this would not be the case with an even number of gears in the loop, for then, when the first turns clockwise the last would turn counterclockwise, and any mis-alignment of the teeth would persist for any amount of rotation.
Since our particular example has an odd number of gears, why then does this not solve our problem? i.e., after placing the gears in arbitrary positions satisfying the center to center distance requirements, then just turn the gears to find a point at which they mesh. Unfortunately, we have another requirement which is that some particular gear must be in a fixed orientation, because in practice, in the full system, there are additional gears involved with which this group must also mesh. So, in order to fix the position of a reference gear and still obtain the correct mesh, without being able to rotate the gears arbitrarily, we must slightly adjust the positions of the axes. It should be noted here that if we were dealing with a loop including an even number of gears and if we choose to fix the first gear in a position other than the one in which a tooth aligns with the initial lineup, we would need to adjust the fraction of teeth around the loop accordingly.
Suppose we know the interior angles of the pentagon. Label them a, g, s, l, and j, following the naming of the axis/gear at the corresponding vertex. Then we can compute the number of teeth around the circuit. We’ll start at the left of gear S where it meshes with G. The number of teeth passed over around the loop as highlighted in Figure 4 will be:
Since gears by definition have a whole number of teeth around the whole circumference, and we are only interested in the fraction of a tooth left over around the loop, we can drop the 360 in two of these terms, simplifying it to (where here we are representing the angles as fractions of a full circle):
Positive terms in this case correspond to clockwise transit around the gear, and negative terms to counter-clockwise transit. We need to relate this formula algebraically to the assumed positions of the gears and then find solutions to the resulting equation which satisfy the half tooth offset required round the loop.
Without loss of generality, place gear L at the origin (0, 0) and gear S vertically above on the y-axis at (0, rL + rS) where rL and rS are the radii of the large tooth elements of gears L and S respectively. If necessary, we can later rotate and translate the entire arrangement to accommodate it in the overall machine. Next, fix the location of gear J, with reference to the original drawing, from which we measure ∠JLS = l. This is acceptable, since the final intent is to come up with positions which match the drawing within measurement accuracy. Simple trigonometry then provides the coordinates (xj, yj) for the axis J.
We are now left with four unknowns, being the x and y co-ordinates of the axes A and G, and we need to choose these to precisely meet the meshing criterion. They cannot be determined with sufficient accuracy from measurements of the drawing alone, though we will require the final positions to be consistent with the drawing within the measurement accuracy. We can find three equations with four unknowns. The first two place the axes A and G on circles centered on the axes of their mating gears J and S respectively, while the third sets the distance between the axes A and G:
Here, all the radii rn are known.
Given we have three equations in four unknowns it should be possible to introduce a single parameter t and then express the four unknown coordinates in terms of t. Further, we should then be able to determine all the interior angles of the pentagon in terms of t, and from that compute our expression for the number of teeth around the loop in terms of t and the coordinates of the remaining axes. Then, by plotting this function against t, we could read off possible values of t which satisfy the half tooth criterion. This seems excessively onerous.
Let us take a different approach, looking directly at the angles involved. The pentagon is uniquely defined by the lengths of the five sides and two of the five interior angles, from which the remaining three angles can be calculated. So, given we know the lengths of the sides, if we start by fixing one of the angles to a value which closely matches the drawing (within measurement error), and parameterize a second centered around the value again read from the drawing, we can calculate the other three. From these five angles we then compute the tooth offset. Assume we fix angles a and l. Divide the pentagon into three triangles, so that:
Applying the cosine rule to triangle AGJ we compute:
Similarly for triangle JLS:
Hence, by further application of the cosine rule, we obtain the three remaining angles:
Using these expressions for the angles, we can calculate the tooth count around the loop for two values of the angle l which straddle the approximate value determined from the drawing (73.5°). Then, even though the tooth count is a non-linear function of the angle, if our initial range for the angle l is very small, a simple linear interpolation allows us to determine a value for the angle which provides a tooth count extremely close to the required half-tooth offset. If necessary the process could be iterated using a smaller range around this newly determined value. This is trivial to program, and the resulting angles are shown in Table 1.
| Angle | Value (°) |
|---|---|
| a | 50 |
| l | 73.5759 |
| j | 165.9831 |
| g | 157.118 |
| s | 93.3231 |
The anomalously round value of the angle a here is because this was arbitrarily chosen by measurement of the drawing, while the other four are all calculated to meet the tooth count constraint. Once the angles are all fixed, simple trigonometry provides the xy coordinates for the locations of the axes.
In the actual Plan 27 layout there is a second symmetrical loop to the right of the group we have been considering, reflected in the line joining the ″L and ″S pinions. (For clarity here we return to using Babbage’s naming.) Our choice of starting point for computing the number of teeth in the loop is a poor one. With the choice to align teeth at the ″G ″S interface, this second group will not mesh because ∠GSL differs from a right angle by other than a half tooth, so there will not be a tooth at the corresponding ′S1 ′G1 interface in the reflected group. This can be addressed by choosing instead to align the teeth at the ″L ″S interface. To do this we need to rotate ″S a small amount. Now, we know that this will cause ″L to rotate to a new position which will not mesh. Thus, we must choose slightly different angles for the pentagon. At first sight this seems to indicate an inconsistency, since the tooth count round the loop only depends on the number of teeth on the gears and the angles of the pentagon, none of which have changed. How can this be? The answer is that when changing the starting point of the loop, we need to change the direction the loop passes around gear ″S as is shown in Figure 5. This in turn changes one term in the formula for the tooth count from sTs to (360 − s)Ts. Thus, there is no inconsistency. For this arrangement the calculation results in the revised values for the angles as shown in Table 2. Note that in order to get an acceptable solution in this case, it was necessary also to adjust the assumed value of the angle a slightly. The tooth count round the loop is particularly sensitive to this angle because of the large number of teeth on this particular gear and, without this adjustment, even though we can find an viable solution, the calculated value of the angle l falls well outside the range consistent with the drawing.
| Angle | Value (°) |
|---|---|
| a | 50.45 |
| l | 74.14 |
| j | 165.1086 |
| g | 156.9588 |
| s | 93.3426 |
The revised arrangement now enables the reflection across SL necessary to provide for both the ″A and ′A groups. However we also have to accommodate the third arithmetic group around axis A. We cannot simply perform a further reflection across ′A, because this wheel is itself not aligned with teeth along a vertical line. However, we have two angles which can be chosen to define the pentagon, and so far we have set a to an arbitrary value and then used only l to obtain the half tooth offset round the loop. By using both these degrees of freedom we can arrange to have A symmetrical across a vertical line, while simultaneously meeting the half-tooth offset requirement around the loop. Figure 6 highlights in blue a second path from the tooth at the bottom of L to the bottom of A.
In this case since we only require A to be symmetrical about the vertical line the terminal point of this path may be either a tooth or a space between teeth, whichever can be most easily accommodated. A very small adjustment to the angle a results in the angles in Table 3.
| Angle | Value (°) |
|---|---|
| a | 50.4479 |
| l | 74.2879 |
| j | 164.9346 |
| g | 157.1564 |
| s | 93.1731 |
This solution delivers 162.4999 teeth around the complete loop, and 65.0001 teeth from L to A.
Unfortunately, the full Plan 27 imposes even more constraints. In particular, the O pinions which interface between the S pinions and the racks. For this meshing to work out perfectly, the horizontal distance between any two of the O pinions must be a whole number of teeth on the rack. From the known angles and sides of the pentagon, we can compute this distance, in teeth as
where the denominator is the tooth to tooth spacing of the rack of 10 DP. For our chosen angles, this turns out to be 53.983 teeth, very close to a whole number of teeth. This is quite by chance. To highlight just how extraordinary this coincidence is, simply changing the linear pitch of the rack from π/10 to 0.3141, which is almost certainly within the manufacturing tolerance, we obtain 53.99. We note however, that had this not been a fortuitous alignment, there is still the possibility that an alternate set of angle values might be found. This is possible because it is not the absolute number of teeth in the constraining paths which matters, only the fact they are an appropriate multiple of a half tooth. In hindsight, it is probable that Babbage fixed the number 54 before embarking on the process of finding a satisfactory layout of the loops, and we find the happy coincidence, simply because our solution has been specifically chosen to match the drawing within measurement accuracy. However, since the O pinions are widely spaced, and the rack never moves by more ±9 teeth from its neutral position, an exact solution could always be obtained by segmenting the rack, so that each of the O pinions acts on its own segment, which segment may then be perfectly aligned to its mating pinion. As a final option for fine adjustments, profile shifting one or more of the gears would permit small changes in the lengths of the sides of the pentagon.
Sadly, we are still not done, because there are additional gears, the three B pinions, which have to be accommodated between the A gears and the rack, which form additional loops. First we note a small anomaly. The center of the S pinions lies 5 inches below the rack. From this, and the known angles and sides of the pentagon we compute the centers of the A gears to be
below the rack, and this number appears consistent with measurements from the drawing. Yet pinion ′B, of 26 teeth, is slightly offset from the vertical line from ′A to the rack. If it were in line and meshing perfectly, the centre of ′A should be exactly 6.6in below the rack. With the offset of the pinion, this number would then be reduced slightly, so it clearly does not match the computed and measured 6.629in. This is certainly a place where a profile shift of the pinion will solve the problem, and indeed the measured diameter of that pinion from the drawing does seem to be slightly greater than what would be expected for one cut without offset. However, why is it displaced to the right? This is easily explained. The pinion has an even number of teeth, but ′A has a tooth at the top, directly below a space on the rack, the opposite of what would be needed. However, a very small displacement of the gear to the right of 0.079in provides a perfect mesh. This requires the line between the centers of ′A and ′B to be 0.85° from the vertical. This appears consistent with the drawing. A profile shift of 0.015in to increase the diameter of the gear by 0.030in results in a perfect mesh.
Finally, we have the loop from ′A, through ′B1 and the rack to the point directly above ′A. Before any profile shift, a 36 tooth gear delivers a loop of 39.06 teeth, whereas we require a half integral number of teeth. A very slight profile shift of 0.0215in increases the diameter by 0.043in, which, with the centre appropriately offset, results in the loop being exactly 39.5 teeth. The line between the centers of ′A and ′B1 then makes an angle of 34.33° to the vertical, in agreement with the drawing.
In the full Plan 27 layout there is one more independent case to consider: the loop of three wheels L, I, and I1. (The additional loop ′L, ′I, and ′I1 is just a trivial reflection of this.) Here, since there are only three wheels, the axes form a triangle with zero degrees of freedom. We know, since the number of gears is odd, a solution is possible by an appropriate rotation of the gears themselves. The only remaining issue is to orient the triangle as a whole around the vertex L so as to achieve this, given that we have already fixed the orientation of L to have a tooth at the bottom. In this case, since the number of gears is odd, we require a half integral number of teeth around the loop from the bottom of L, counter-clockwise around ′I, clockwise around I, then back to the bottom tooth of L. A measurement from the drawing suggests the triangle should be offset by approximately 19.5° from the vertical, but the calculation shows we actually need 18.78° to get the correct mesh.
A fully constrained CAD layout which confirms these choices can be seen in Figure 7. This drawing defines the precise locations of all the principal axes of the Mill. As a final check we constructed a CAD assembly of placeholder gears. It turned out this revealed a careless error in the original calculation of the placements of the ′B and ′B1 axes, which prevented the correct meshing and requiring a small correction. We now have high confidence in this final layout, which can be seen in Figure 8. Here all gears, including the rack, are simultaneously in mesh.
1 Babbage’s “CAD” system — Cardboard Aided Design.
Figures 2, 4, 5 and 6 are based on the BAB/A/093 image in the Science Museum Group collection The Babbage Papers which are licensed under CC BY-NC-SA 4.0.
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