Gearing Gearwheels are used in clocks when one shaft is required to rotate at a different speed from another.
	In this example a gearwheel with 32 teeth drives one with 16. When the driving wheel rotates once the driven wheel rotates twice, so that the rotational speed of the output shaft is twice that of the input shaft. Thus the formula for shaft speed is:
No. of teeth on driving wheel No. of teeth on driven wheel ……….Output shaft speed =…………………………………………….. x Input shaft speed Gearwheels of a small diameter are known as pinions and are usually thicker than normal gearwheels.
Rack and Pinion A rack is a piece of flat material with teeth cut into one face. The teeth are meshed with a gearwheel or pinion such that when the wheel turns the rack slides along or vice versa.
Changing direction Where the input and output shafts are at right angles special gearwheels are required. Two types are used in this clock. 1. Bevel gears. The teeth are cut at 45 degrees. 2. Contrate gearwheels. These are wheels whose teeth are perpendicular to the wheel. The teeth are meshed with a normal gearwheel or pinion to produce the right angled drive.
Worms These are special gears where the teeth are cut spirally on the outside of a cylindrical core, rather like the threads of a screw. The worm is meshed with a normal gearwheel or pinion. Typically if the worm makes one complete revolution the meshed gear will only rotate by one tooth. So if the gearwheel has 50 teeth then the worm must be turned 50 times to cause one revolution of the gearwheel.
The worm may be regarded as equivalent to a normal gearwheel with only one (or sometimes two or three) teeth. An important property of a worm is that whereas a worm can drive a gearwheel, the gearwheel cannot drive the worm - it is a mechanical impossibility. This is a property which is put to good use in this clock.
Transported Gear Train We know that gears are used for speed change, e.g. a gearwheel with 10 teeth driving one with 30 teeth will result in a rotational speed reduction of 10/30, or 1/3. I.e. the output shaft will rotate at 1/3 of the speed of the input shaft. Suppose that the output shaft were required to turn 10 times when the input turns 11^1/4 times. The formula is now 10/11^1/4. But this would require a gearwheel with 11^1/4 teeth - which is impossible. The solution is to multiply everything by 4, giving a formula of 40/45. Thus the desired result could be obtained by using a gearwheel with 40 teeth driving one with 45. Easy. Now suppose that the output shaft is required to turn 366^1/4 times when the input turns 365^1/4 times. This is an actual requirement for one stage of the clock. Using the above method we would require a gearwheel with (366^1/4x 4) = 1,465 teeth meshed with one of 1,461 teeth. Even if such wheels could be made, they would be impractically large. Richard had an ingenious solution to this problem - the transported gear train. This is a system where shaft mountings are actually fixed to gearwheels rather than to the frame of the clock. Here is an example similar to the above, but with slightly changed values, to make it easier to understand. Suppose that we have a gearwheel which is required to make 361 turns when another makes 360. In other words, when yellow wheel makes 360 turns, the blue wheel is required to make that number plus 1. We could start by devising a gear train which gives an output of 1 turn for an input of 360 turns. One way to do this (admittedly not the simplest way) is with the following gear train.
If we could now devise some way of mounting the whole geartrain on the yellow wheel, then when the input shaft and the geartrain rotate together 360 times, the output shaft will rotate 360 +1 = 361 times. Let’s do it…. In the diagram below the gearwheels are identified by the number of teeth they each have, rather than being labelled A,B,C etc. Wheel 190 is driven by a pinion such that it rotates 360 times each year. The objective is to make wheel 144 rotate 361 times in the same period. A bracket is fixed to the face of wheel 190. Pinion 9 meshes with wheel 144. When wheel 190 rotates there are two possibilities:
1). Pinion 9 rotates and wheel 144 doesn’t 2). Pinion 9 does not rotate, so wheel 144 is dragged around at the same rate as wheel 190. Pinion 9 is on the same shaft as wheel 20 which engages with a worm. We already know that a gearwheel cannot drive a worm which means that pinion 9 is unable to turn. So option 1) is ruled out and option 2) applies. Wheel 16 is a contrate gearwheel which cannot turn as it is bolted to the clock frame. It is meshed with wheel 18.
Each time that wheel 190 and the bracket make one revolution, wheel 18 is carried once around wheel 16. The result is that wheel 18 makes 16/18 of a revolution. This means that when wheel 190 turns 360 times wheel 18 makes (360 x 16/18) = 320 revolutions. Wheel 18’s shaft connects to a worm having the equivalent of one tooth. The worm meshes with wheel 20 which therefore turns (320 x 1/20) = 16 times. Wheel 20 drives pinion 9 which in turn meshes with wheel 144 with the result that wheel 144 makes (16 x 9/144) = 1 revolution. Only one? Yes - remember that this is with respect to the bracket, which has made 360 revolutions. So wheel 144 makes (360 + or - 1) turns. Whether it is + or - depends on which way the spiral teeth are cut on the worm - left hand thread or right hand thread. By making the correct choice wheel 144 can be made to rotate 361 times. Another way to look at this is that each time wheel 190 makes one complete revolution of 360 degrees, wheel 144 is dragged around with it but the transported gear train causes wheel 144 to creep ahead by a small amount (actually 1 degree). After wheel 190 has made 360 revolutions wheel 144 will have made 360 turns plus (360 x 1 degree) I.e. 361 revolutions. If you had difficulty following the above please read it through again, and persevere, as there are five separate applications of transported gear trains in the Wallingford clock.