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The Athlete's Equations · Pacing

Rapoport's wall

lesson 2 of 325 minnot yet done

A marathoner carries a finite store of carbohydrate: muscle glycogen in the working legs, liver glycogen, and what is eaten on the way. The fraction of energy that comes from carbohydrate rises with intensity, so the faster the pace the more carbohydrate each kilometre costs. Rapoport (2010) set the arithmetic out:

need(I) = cost · f_CHO(I) · distance ≤ muscle + liver + intake · hours

The wall is where need exceeds supply. The Factor Lab draws both curves against intensity; the crossing is the fastest pace the store can pay for. Eating 60 g of carbohydrate an hour moves it right; a bigger leg muscle mass moves it right; a hotter day moves the whole time up through Ely's heat curve.

Drafting

Air resistance is about 6% of the cost at two-hour pace, growing with the square of speed. A runner ahead removes half of it, a rotating formation most of it. Through the cost curve that is worth about three and a half minutes to a 2:20 marathoner in a formation.

Rapoport's wall: the carbohydrate a marathon needs against the carbohydrate on board, byintensity (VO₂max 70, 62 kg)020040060050%60%70%80%90%intensity, fraction of VO₂maxgrams of carbohydrateeating nothing: wall at 76% · 2:46:34eating 60 g an hour: the aerobic ceiling, not the store · 2:20:39needed at this paceon board, eating 60 g/hon board, eating nothing
Two curves and their crossing. Eating nothing, the store runs out at 76% of VO₂max, which is a 2:46 marathon. Eating 60 g an hour, the carbohydrate stops being the binding constraint at all.
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