Velocity Ratio and Number of Teeth: the Continued-Fraction Method#
The discretization problem#
As we have seen, the velocity ratio in gears is given by the equation:
\(i = \frac{\omega_2}{\omega_1} = \frac{Z_1}{Z_2}\)
Since the numbers of teeth are integers, we cannot obtain just any velocity ratio, but only a discrete number of them. Moreover, this ratio between integers cannot be arbitrary, because there is usually a limit on the maximum and minimum number of teeth we can use (for example, \(Z_{\min} = 20\) to avoid undercutting without correction, and \(Z_{\max} = 100\) due to space or cost limitations).
Approximation using continued fractions#
To approximate a given velocity ratio (which may be irrational or have many decimal digits) we can use the continued-fraction method, which provides the best successive rational approximations.
Practical example: approximating \(\pi/10\)#
Suppose we want to approximate a velocity ratio of \(i = \pi/10 = 0.31415926...\)
Step 1: We express the number as a continued fraction through successive divisions:
\(0.31415926 = \frac{314.15926}{1000} = \frac{1}{\frac{1000}{314.15926}}\)
We keep expanding:
\(= \frac{1}{3 + \frac{57.5224}{314.15926}} = \frac{1}{3 + \frac{1}{\frac{314.15926}{57.5224}}}\)
\(= \frac{1}{3 + \frac{1}{5 + \frac{26.5472}{57.5224}}} = \frac{1}{3 + \frac{1}{5 + \frac{1}{\frac{57.5224}{26.5472}}}}\)
\(= \frac{1}{3 + \frac{1}{5 + \frac{1}{2 + \frac{4.428}{26.5472}}}} = \frac{1}{3 + \frac{1}{5 + \frac{1}{2 + \frac{1}{5.99...}}}}\)
Step 2: The successive approximations (convergents) are:
Continued fraction |
Simple fraction |
Decimal value |
Relative error |
|---|---|---|---|
\(\frac{1}{3}\) |
\(\frac{1}{3}\) |
\(0.3333\) |
\(6.10\%\) |
\(\frac{1}{3+\frac{1}{5}}\) |
\(\frac{5}{16}\) |
\(0.3125\) |
\(0.528\%\) |
\(\frac{1}{3+\frac{1}{5+\frac{1}{2}}}\) |
\(\frac{11}{35}\) |
\(0.3143\) |
\(0.040\%\) |
\(\frac{1}{3+\frac{1}{5+\frac{1}{2+\frac{1}{5}}}}\) |
\(\frac{60}{191}\) |
\(0.314136\) |
\(0.007\%\) |
Practical solutions with tooth-count constraints#
The possible solutions to our problem, if we cannot go below 20 teeth (to avoid undercutting), will be multiples of the fractions obtained:
Fraction \(\frac{1}{3}\):
\(\frac{20}{60}\), \(\frac{21}{63}\), \(\frac{22}{66}\), \(\frac{23}{69}\), …
Fraction \(\frac{5}{16}\):
\(\frac{20}{64}\), \(\frac{25}{80}\), \(\frac{30}{96}\), …
Fraction \(\frac{11}{35}\):
\(\frac{22}{70}\), \(\frac{33}{105}\), \(\frac{44}{140}\), …
Fraction \(\frac{60}{191}\):
\(\frac{60}{191}\), \(\frac{120}{382}\), …
Final decision: If, for example, we could not exceed 100 teeth (due to space limitations), the optimal solution would be:
\(Z_1 = 22, \quad Z_2 = 70 \quad \Rightarrow \quad i = \frac{70}{22} = \frac{35}{11} = 0.31429\)
with a relative error of only 0.040%, which is excellent for industrial applications.
Advantages of the method#
It provides the best rational approximations at each step (convergents).
It allows choosing the optimal solution according to design constraints (\(Z_{\min}\), \(Z_{\max}\)).
It is especially useful when irrational transmission ratios are required (such as \(\pi\), \(\sqrt{2}\), the golden ratio, etc.) in special applications (astronomical instruments, precision instrumentation).