Example

If length of a train car $(L)=80ft$

Case $1$: Suppose the train has only $1$ car, its speed is $30m\frac{i}{h},$ and at that speed the minimum distance between two successive trains must be $d=220ft.$ In this case the line capacity will be equal to:

Capacity $=\frac{30**5280}{80+220}=528$ cars $\frac{?}{h}$

Example

Case $2$: Suppose the train has only $1$ car, but its speed is $60m\frac{i}{h},$ and at that speed, the minimum distance between two successive trains must be $d=850ft.$ In that case the line capacity will be equal to:

Capacity $=\frac{60**5280}{80+850}=341$car$\frac{s}{h}$

Example

Case $3$: Now, suppose the train has $4$ cars, but its speed is $30m\frac{i}{h},$ at which the minimum distance between two successive trains must be $d=260ft.$ In that case the line capacity will be equal tc

Capacity $=4^{**}[\frac{30**5280}{4**80+260}]$

$=1092$ cars $\frac{?}{h}$

Example

Case $4$: Now, suppose the train has $4$ cars, but its speed is $60m\frac{i}{h},$ at which the minimum distance between two successive trains must be $d=1030ft.$ In that case the line capacity will be equal to

Capacity $=4^{***}[\frac{{60}^{**}5280}{4^{**}80+1030}]$

$=939$car$\frac{s}{h}$

Example

Case $5$: Now, suppose the train has $80$ cars, but its speed is $60m\frac{i}{h},$ at which the minimum distance between two successive trains must be $d=3280ft.$ In that case the line capacity will be equal to:

Capacity $=80**[\frac{60**5280}{80**80+3280}]$

$=2618$car$\frac{s}{h}$

Example

Case $5$: Now, suppose the train has $80$ cars, but its speed is $60m\frac{i}{h},$ at which the minimum distance between two successive trains must be $d=3280ft.$ In that case the line capacity will be equal to:

Capacity $={80}^{**}[\frac{60**5280}{80**80+3280}]$

$=2618$car$\frac{s}{h}$

Example Results Summary

$\backslash$table$[[\backslash$table$[[$# of$],[$cars$]],\backslash$table$[[$Speed$],[($mi$/$h$)]],\backslash$table$[[$Min distance$],[($ft$)]],\backslash$table$[[$Capacity$],[($cars$/$h$)]]],[1,30,220,528],[1,60,850,341],[4,30,260,1092],[4,60,1030,939],[80,60,3280,2618]]$

Such a versatility of rail transport enables one to play with different variables to achieve the desired goal, subject to the available tractive force $($locomotive power$)$ and station length.

Assume that a train must have one locomotive plus a number of cars. The capacity of the train will be proportional to the number of cars only $($ie$,$ do not include the locomotive in capacity calculations$).$ If the length of the train locomotive $(L)=80ft$ and the length of a train car$(L^{\text{'}})=50ft,$ and the spacing between any two successive units $=3ft,$ redo examples on pages $15-20$ in Lecture $9.$ In doing so$,$ use the same given values for speeds, numbers of cars, and the minimum distances required between two successive trains $($front of a train to the end of the train ahead of it$),$ find the capacities in each case, and present them in a table.

Use the above cases to solve the problem