\(r_1 = 6370km\) umhlaba (njengebhola \(r_1 = 6370km\) ) \(r_1 = 6370km\) \(r_2 = 2mm\) ) bese \(r_2 = 2mm\) intambo phezu kwenkabazwe ukuze ilale iqinile ebusweni. Manje uzelula zombili izintambo ngemitha elilodwa lilinye. Zombili izintambo manje kufanele zilele ngokugcwele phezu kwenkabazwe futhi - azisalali ngokuphelele ebusweni, kepha sezindiza ngaphezu kwenkabazwe. Intambo intanta ngaphezu komhlaba iphakeme kangakanani, iphakeme kangakanani ngaphezu kwepea?
Izintambo ezimbili zinobude bokuqala:
$$
l_1 = 2\cdot 6370 km \cdot \pi \Leftrightarrow r_1 = 6370 km = \frac{l_1}{2 \cdot \pi}
$$
njengoba
$$
l_2 = 2 \cdot 2mm \cdot \pi \Leftrightarrow r_2 = 2mm = \frac{l_2}{2 \cdot \pi}.
$$
Kepha manje kungemva kokunwetshwa
$$
r_{1 NEU} = \frac{l_1 + 1m}{2\cdot \pi}
$$
njengoba
$$
r_{2 NEU} = \frac{l_2 + 1m}{2\cdot \pi}.
$$
Kepha manje kuyamangaza
$$
r_{1 NEU} - r_1 = \frac{l_1 + 1m}{2\cdot \pi} - \frac{l_1}{2\cdot \pi} = \frac{l_1 + 1m - l_1}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m
$$
njengoba
$$
r_{2 NEU} - r_2 = \frac{l_2 + 1m}{2\cdot \pi} - \frac{l_2}{2\cdot \pi} = \frac{l_2 + 1m - l_2}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m.
$$
Ngakho-ke ibanga ukusuka ebusweni lizimele nge \(l_1\) noma \(l_2\) , \(l_2\) kuma-radii \(r_1\) noma \(r_2\) emikhakheni. Impendulo \(0.159m\) : Zombili izintambo zintanta ngobude obufanayo \(0.159m\) ) ngaphezulu kobuso.