Brocot’s tables

Before the advent of computers computing gear trains could be a messy business. however there was one really elegant way of doing it. This method is due to Brocot.

He starts with the assumption that the user has a (hopefully)  full set of gears from 1(sic) up to 120.

He produces two tables

The first is the ratios produced by any permutation of two gears, N and D, from this set. They are in order of the ratios. The list actually starts with many entries where one gear cannot exist, ie it has too few teeth.

A typical part of the table would be

N          D            ratio

85       109          0.77 981 651

39        50           0.78 000 000

71        91           0.78 021 978

32        41           0.78 048 780

Of course all the inverses of these ratio also exist.

This table is used by finding the pair that gives the nearest to what we are after. At any time we can always modify this by a factor of, say, 10 by popping in another gear.

Suppose we want a ratio of 0.78 010 000

It can be seen in the table shown we do not have the ratio we want. We have one will be just a bit higher that that we wanted. The other will be just a bit lower. And, of course we have done this with just two gears.

Which ever one of these we choose there will be a slight difference between this ratio and the ratio we actually wanted. This leads to two sets of calculations. For the high and low values we have to find the error –  the twiddle factor.

 

For the high value

The high value is  0.78 021 978

We can find the difference by dividing the ratio we actually want by the one (of the two) we have just got.

TFhigh =   0.78 021 978 / 0.78010000 = 1.0001534

If I look up this in the table “Ratios of consecutive and factorable numbers I find the following entries

A/B                     A                     A factors      B                   B factors        B/A

0.99984615    6499               67.97            6500            2.2.5.5.5.13  1.00015387

If I take the high value, 0.78021978, and multiply it by A/B this will give me the final ratio.

FinalR = 0.78021978 * 0.99984615 = 0.7800997

This is correct to 0.3 parts in one million. It might appear that this is good enough but there might be problems over whether we have the gears or whether they can be made to fit together etc.

The end result will usually require a gear train of not more than six gears.

These tables can be found in:

Revised Manual of Gear Design, Section 1, Gear Ratio and Mathematical Tables

revised by Eliot Buckingham

published by Buckingham Associates Inc