Users/Telanoc/Gears

A Beginners Guide to Gearbox Design. You can find the computed Gear Ratios here: Gearbox Ratios

Notation Conventions:

A shaft location will be denoted by its row and column, for example 8C is the top row, third from left
When discussing gear ratios a “/” denotes one gear that meshes with and turns another. For example 5/3
When discussing gear ratios a “*” denotes two gears that share a single shaft. For example 3*5
An example would be 5/3*6/5/3. This means a 5 tooth gear turning a 3 tooth gear. On the same shaft as that 3 tooth on a different layer is a 6 tooth gear, which turns a 5 tooth gear, which then turns a 3 tooth gear

Every gearbox must start at “1A” (the lower left hand corner as you face the gearbox)
Every gearbox must have at least one, but no more than three outputs in the top row (Row 8)
You may use up to, and no more than, thirty (30) gears in your gearbox
You may use up to, and no more than, fifteen (15) shafts in your gearbox. A shaft is a row and column intersection that has one or more gears (e.g. B2)
There are three layers in a gearbox. Gears can only mesh when they are on the same layer. The input and output layers don’t matter.
There are eight rows numbered 1 to 8
There are eight columns numbered A to H
Gears come in 1,3,4,5,6, and 7 tooth.
- 1 tooth is a “Spacer”, and costs no materials. However, it does count for your gear total
- 3 and 4 are considered “Small Gears”, and will cost 2 debens of metal to cast.
- 5 and 6 are considered “Medium Gears”, and will cost 15 debens of metal.
- 7 is considered a “Large Gear”, and will cost 100 debens of metal

The sum of adjacent gears in the horizontal or vertical axis must be 6. Therefore only 3/3 is valid when moving one space in either the horizontal or vertical axis
The sum of adjacent gears in the diagonal axis must be 8. So 4 and 4, 5 and 3, and 7 and 1 are valid configurations.
The sum of the gears in the horizontal or vertical axis that are separated by one space (so not adjacent) must be 11. Therefore valid configurations are 6 and 5, 7 and 4
The sum of the gears in a “Knight’s move”, which is to say two space away and one off the axis, must be 12. Therefore 6 and 6 or 7 and 5 are valid pairs.

Whenever a gear turns another gear, it is a ratio based on the fraction created from the number of teeth.
- If a larger fear turns a smaller gear, the output value increases. For example, if a 5 tooth turns a 3 tooth, the output will be increased by 67% because five thirds (5/3) is equal to 1.67.
- If a smaller gear turns a large gear, the output value decreases. For example, if a 3 tooth turns a 5 tooth, then the output will be 60% of the input because three fifths (3/5) is equal to 0.6.

If you have a multiple output gearbox, see if one or more the ranges overlap. For example, if the requirement is A90-A120 C95-C125 G80-G110, a single ratio of 100 will satisfy all three requirements. This will greatly simply your design process
If you have a multiple output gearbox, and not all of the ranges overlap, or none of them do, look for ratio chains that use the same sets of gears. For example, you have a gearbox that is A150-A180 and F310-F340. Clearly, neither of those ranges overlap. But if we lookup at the ratios in those ranges we see that 167 uses a 5/3 gear set and 333 uses a 5/3*6/5/3 gear set. Therefore, we can use a 5/3 set for both ratios, thus saving us gears and shafts.
Since the gears are simple math, you can increase ratios and then decrease them. To give an example, say we had the reserve of the above gearbox requirements, A310-A340 and F150-F180. You could use the chain of 5/3*6/5/3 to get to A333, and then from the last 3 tooth gear put a set of 3/5/6, giving a total of 5/3*6/5/3/5/6. This actually a horrible example, and is used only to illustrate the point.
Try to avoid using 7 tooth gears. Not only are they much more expensive than other types, they are harder to work around. If you must use one, try to put it on either layer 2 or 3 where it is more likely to be out of the way.