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Equation 7 · The Logistics Tail: Sustainment as the Binding Constraint on Organised Force

What does this equation mean?

P(r)=L0−2rfv,rmax⁡=v L02f.P(r) = L_0 - \frac{2rf}{v}, \qquad r_{\max} = \frac{v\,L_0}{2f}.

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

Start with2rf
Divide byv
This relates toP(r)
How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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PP

Symbol P

P is the quantity selected or evaluated by the optimization written on the right.

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rr

Symbol r

the radius.

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L0L_0

Symbol L_0

L0L_0 occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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ff

Symbol f

f appears in the objective or constraint used by the optimization on the right.

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vv

Symbol v

v appears in the objective or constraint used by the optimization on the right.

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rmax⁡r_{\max}

Symbol r_max

rmr_max appears in the objective or constraint used by the optimization on the right.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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fraction

fraction

Divide the expression above the line by the one below it.

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subtraction

subtraction

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

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subscript

subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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2rf2rf

Numerator: 2rf

The complete quantity above the fraction bar.

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v L0v\,L_0

Numerator: vL_0

The complete quantity above the fraction bar.

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2f2f

Denominator: 2f

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

The structural consequence is a hard radius rather than a gradual decline. Consider a schematic model, offered to expose an assumption rather than to reproduce any historical campaign. A transport team departs a base carrying load L0L_0 , consumes f units of that same load per day for its own animals, and covers v units of distance per day. To reach a force at radius r and return, it is on the road for 2r/v days, so the load it can actually hand over is P(r)=L0−2rfv,rmax⁡=v L02fP(r) = L_0 - \frac{2rf}{v}, \qquad r_{\max} = \frac{v\,L_0}{2f}. Beyond rmax⁡r_{\max} the convoy arrives having eaten everything it set out with. Adding a second convoy to supply the first does not defeat the limit; it recurses, and the tonnage required grows far faster than the tonnage…
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The structural consequence is a hard radius rather than a gradual decline. Consider a schematic model, offered to expose an assumption rather than to reproduce any historical campaign. A transport team departs a base carrying load L0L_0 , consumes f units of that same load per day for its own animals, and covers v units of distance per day. To reach a force at radius r and return, it is on the road for 2r/v days, so the load it can actually hand over is P(r)=L0−2rfv,rmax⁡=v L02fP(r) = L_0 - \frac{2rf}{v}, \qquad r_{\max} = \frac{v\,L_0}{2f}. Beyond rmax⁡r_{\max} the convoy arrives having eaten everything it set out with. Adding a second convoy to supply the first does not defeat the limit; it recurses, and the tonnage required grows far faster than the tonnage delivered. This is why the practical answer for two millennia was not better convoys but a different mechanism entirely: draw supplies locally, move along navigable water, or stop.

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