Time of Least Distance

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This page explains how you can estimate the time and value of the least distance from a series of measurements or calculations of how the distance varies with time, of something that moves in a straight line.

Fig. 1: Elongation of Saturn (15 days)

If you look at the position of Saturn in the sky relative to that of the Sun, then Saturn passes the Sun along a roughly straight line, and the elongation is the distance between the Sun and Saturn.

If you know the distance \( r \) at three times \( t_0 − τ \), \( t_0 \), and \( t_0 + τ \), then you can estimate the smallest distance \( d \) and the time \( t_\text{min} \) at which it is attained, from

\begin{eqnarray} v & = & \frac{\sqrt{r(t_0 − τ)^2 − 2 r(t_0)^2 + r(t_0 + τ)^2}}{τ\sqrt{2}} \\ l_0 & = & \frac{r(t_0 + τ)^2 − r(t_0 − τ)^2}{4vτ} \\ d & = & \sqrt{r(t_0)^2 − l_0^2} \\ t_\text{min} & = & t_0 − \frac{l_0}{v} \end{eqnarray}

For example, the measurements from the figure for \( t \) from 328 to 330 are

\({t}\) | \({r(t)}\) |
---|---|

328 | 0.8990879 |

329 | 0.6809476 |

330 | 1.1668921 |

With these values we find

\begin{eqnarray*} t_0 & = & 329 \\ v & = & \frac{\sqrt{0.8990879^2 − 2 × 0.6809476^2 + 1.1668921^2}}{1\sqrt{2}} = \frac{\sqrt{1.12426171}}{\sqrt{2}} = 0.7882313 \\ l_0 & = & \frac{1.1668921^2 − 0.8990879^2}{4 × 0.7882313 × 1} = \frac{0.5532783}{3.1529251} = 0.1754889 \\ d & = & \sqrt{0.6809476^2 − 0.1754889^2} = \sqrt{0.4328960} = 0.6579483 \\ t_\text{min} & = & 329 − \frac{0.1754889}{0.7882313} = 328.7773738 \end{eqnarray*}

so the estimate is that the smallest distance is 0.6579483° and is reached when \( t \) is 328.7773738.

Fig. 2: Elongation of Saturn (2 days)

The estimate of the time and value of the smallest distance is accurate if the motion can be described as having a constant speed along a straight line. If the speed varies slowly or if the trajectory is nearly but not exactly a straight line, then the estimate will be reasonable but may not be exactly right.

For comparison: The estimate of the smallest distance based on the three nearest data points with an interval of 0.1 days (\( τ = 0.1 \)) instead of 1 day is that the smallest distance is 0.6579986° and is reached when \( t \) is 328.7775848, so that differs only 0.00005° in the distance and only 0.0002 dagen or 18 seconds in time. This gives an impression of the accuracy of the estimate based on the interval of 1 day.

Suppose that a straight line runs past but not through point A. The point where the line comes closest to point A is called point B. The distance between A and B is \( d \): That is the shortest distance that a point on the line can have to A. We turn the line into a number line, with the 0 at point B and with the same unit of measurement as for \( d \). A point at number \( l \) on that line then has a distance \( r \) to point A that is equal to

\begin{equation} r = \sqrt{d^2 + l^2} \end{equation}

An object moves along the line at fixed speed \( v \). We have three successive measurements of the distance \( r = r(t) \) for times \( t \) equal to \( t_0 − τ \), \( t_0 \) en \( t_0 + τ \). The corresponding positions on the line are

\begin{equation} l = l(t) = l_0 + v(t − t_0) \end{equation}

Then

\begin{equation} \begin{split} r(t) & = & \sqrt{d^2 + l(t)^2} \\ & = & \sqrt{d^2 + (l_0 + vt)^2} \end{split} \end{equation}

so

\begin{eqnarray} r(t_0 − τ) & = & \sqrt{d^2 + (l_0 − vτ)^2} \\ r(t_0) & = & \sqrt{d^2 + l_0^2} \\ r(t_0 + τ) & = & \sqrt{d^2 + (l_0 + vτ)^2} \end{eqnarray}

From this we derive \( d \), \( l_0 \), and \( v \). We find

\begin{eqnarray} 2 v^2τ^2 & = & r(t_0 + τ)^2 − 2 r(t_0)^2 + r(t_0 − τ)^2 \\ 4 l_0vτ & = & r(t_0 + τ)^2 − r(t_0 − τ)^2 \\ d^2 & = & r(t_0)^2 − l_0^2 \end{eqnarray}

so

\begin{eqnarray} v & = & \frac{\sqrt{r(t_0 − τ)^2 − 2 r(t_0)^2 + r(t_0 + τ)^2}}{τ\sqrt{2}} \\ l_0 & = & \frac{r(t_0 + τ)^2 − r(t_0 − τ)^2}{4vτ} \\ d & = & \sqrt{r(t_0)^2 − l_0^2} \end{eqnarray}

The moment \( t_\text{min} \) at which the smallest distance is attained has by definition \( l = 0 \), so

\begin{eqnarray} 0 & = & l_0 + vt_\text{min} \\ t_\text{min} & = & −\frac{l_0}{v} \end{eqnarray}

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Last updated: 2019-02-05
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