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Guides > Gearbox Design > Tutorial > Solutions

Solutions for the Gearbox Designs from the Tutorial

First Design: B143-B172 E233-E280 G437-G542

Start with the gearbox finder:
B167 = 53
E278 = 53 * 53
G463 = 53 * 53 * 53

You can see that two of the 53 combinations can be shared: All outputs can use the first 53 combination, and E8 and G8 can share the second 53 combination. You can do this with only 5 medium gears total:

Result: 14 small gears, 3 medium gears

Second Design: A206-A249 E131-E164 G275-G333

A240 = 65 * 653
E144 = 65 * 65
G288 = 65 * 65 * 653

I have chosen these ratios (they are not the cheapest possible for each output) because they can share the most things: 65 * 65 will be common for all outputs, then A branches off starting with a 3-tooth, E continues straight away from there, and G gets another 653 combination after that:

Result: 13 small gears, 6 medium gears


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Last edited February 9, 2005 11:55 am by LittleCleo (diff)
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