Skeletal Muscle: Length, Load and Speed

An activated muscle shortens fastest when it lifts nothing and not at all when the load equals the force it can hold, so at both extremes it does no external work. The first look at muscle followed a contraction from T tubule to troponin. This lesson puts numbers on length, load, speed, fuel and fatigue.

Overlap sets the force a fibre can hold

Hold a muscle at one length, stimulate it, and the force it develops is its isometric force. That force depends on sarcomere length, the distance from one Z line to the next. In 1966 Gordon, Andrew Huxley and Julian held the length of a short segment of single isolated frog fibres under servo control and measured tension at each sarcomere length, an experiment told in molecular motors.

In those frog fibres, active force was greatest on a plateau at about 2.05 to 2.2 µm, where thin filaments reach every myosin head. On the descending limb, at longer lengths, force fell roughly in line with lost overlap, extrapolating to zero near 3.65 µm. At short lengths force also fell, more steeply below about 1.67 µm; that loss is usually read as interference, with thin filaments crossing into the opposite half-sarcomere and thick filaments meeting the Z lines, not as too little overlap. Picture rowers whose oars must reach the water: pull the boat ends apart and fewer oars reach; push them together and the oars clash.

Beyond the plateau, even resting muscle resists with passive tension from titin and connective tissue. Active tension is total tension minus passive tension at the same length, the definition used by Hessel and colleagues in 2024. Force also depends on history: after an active stretch, a fibre settles above the isometric force at the same final length, residual force enhancement. In permeabilized fibre bundles from engineered mouse psoas muscle, partly activated at 25 °C, cutting half of the titin lowered that enhancement from 30.7 to 17.4 percent of the reference force (Hessel and colleagues, PNAS 2024). A static overlap diagram misses this. Heart muscle normally works below its optimal length, where stretch raises force, part of the behaviour read from pressure-volume loops.

The heavier the load, the slower the shortening

Isometric: length held while force develops. Isotonic: constant load while length changes. Concentric: the active muscle shortens. Eccentric: the active muscle lengthens because the load exceeds the force it can hold, as the thigh muscles do when you walk downhill.

The classic experiment. In the 1920s W. O. Fenn had reported that a frog muscle allowed to shorten and do work releases more energy than one held isometric, which a stretched spring in a viscous fluid, its energy fixed at stimulation, could not explain; a viscous brake would also make speed fall linearly with load. A. V. Hill (profiled in protein binding and allostery) asked what sets how fast an activated muscle shortens. For his 1938 paper in the Proceedings of the Royal Society B he used isolated frog sartorius muscles at about 0 °C, tetanized each, and let it shorten against a series of loads on a lever while a thermopile recorded the heat. He read speed from the record of shortening, not from the total distance shortened.

Shortening released extra heat in proportion to the distance shortened (the heat of shortening, with a coefficient he called a; Hill himself found in 1964 that this heat coefficient rises with load, so it is not the same quantity as the constant a in the speed equation below). Speed fell with load along a rectangular hyperbola, not a line. External power peaked at an intermediate load. He fitted the constants a and b to frog sartorius at 0 °C; the a/F0 of 0.25 used below is close to the usual value he gave for frog sartorius, though the other numbers in the problem are model values. This ruled out a passive viscous-elastic muscle and ruled in a quantitative, load-dependent rule for speed and power. It gave no molecular mechanism (A. F. Huxley’s 1957 cross-bridge model later reproduced the hyperbola), and its constants are frog values at 0 °C, not human ones.

Hill’s equation: (F + a)(v + b) = (F0 + a) b, so v = b (F0 − F) / (F + a)

F is the load, v the speed, F0 the isometric force at that length and activation; a and b are constants. Extrapolated to zero load, the equation gives the maximal speed, vmax = b F0 / a. Real fibres depart somewhat from a single hyperbola at very light and very heavy loads, so a fitted vmax and a directly measured unloaded speed can differ.

P = F x v, in watts (1 W = 1 N x 1 m per s)

Power is zero at zero load and at the isometric load. When a/F0 is about 0.25, it peaks near 0.31 F0, a fraction that depends on the curvature.

Worked problem. In this made-up example the values are illustrative, not measured human constants; only the frog length landmarks, Hill’s equation and a typical frog a/F0 are borrowed. A model fibre bundle produces 100 N on the plateau (2.05 to 2.20 µm). Assume the descending limb is a straight line from full force at 2.20 µm to zero at 3.65 µm, a/F0 = 0.25 and vmax = 0.40 m per s at every length, and quasi-steady shortening at one length with constant activation and temperature. At a sarcomere length of 3.00 µm, find speed and power against 20 N and the best load; then find peak power on the plateau.

1. Length. Fraction of plateau force = (3.65 − 3.00) / (3.65 − 2.20) = 0.65 / 1.45 = 0.448, so F0 = 44.83 N.
2. Constants. a = 0.25 x 44.83 N = 11.21 N. Because vmax = b F0 / a, b = 0.40 x 0.25 = 0.100 m per s.
3. Speed. v = 0.100 x (44.83 − 20) / (20 + 11.21) = 0.100 x 24.83 / 31.21 = 0.0796 m per s.
4. Power. P = 20 N x 0.0796 m per s = 1.59 W.
5. Check. (F + a)(v + b) = 31.21 x 0.1796 = 5.61; (F0 + a) b = 56.04 x 0.100 = 5.60, equal within rounding (both are 5.603 with unrounded values).
6. Best load. With f = F / F0 and q = a / F0, power is proportional to f (1 − f) / (f + q). Setting the derivative to zero gives f² + 2qf − q = 0, so f = √(q² + q) − q = 0.559 − 0.25 = 0.309. Best load 0.309 x 44.83 = 13.85 N; speed 0.100 x 30.98 / 25.06 = 0.1236 m per s; peak power 13.85 x 0.1236 = 1.71 W.
7. Plateau. F0 = 100 N, a = 25 N, best load 30.9 N, speed 0.100 x 69.1 / 55.9 = 0.1236 m per s, peak power 30.9 x 0.1236 = 3.82 W.

Stretching from 2.20 to 3.00 µm cut F0, and peak power with it, by the same factor of 0.448, while the speed at peak power stayed at 0.309 of vmax. This is external power, not ATP turnover: an isometric muscle does no external work yet keeps splitting ATP.

Model muscle with frog length landmarks; not human data.

current load best load (most power) load above F₀

Sarcomere length 3.00 µm Load F 20 N Curvature a/F₀ 0.25 Maximal speed v_max 0.40 m per s

Active isometric force F₀ at this length

44.83 N

Constants a and b

—

Shortening speed at this load

—

External power at this load

—

Best load

—

Peak power at the best load

—

—

Assumptions, made up for this sketch (not measured): a model fibre bundle that holds 100 N on the plateau, 2.05 to 2.20 µm; a straight descending limb from full force at 2.20 µm to zero at 3.65 µm (the frog length landmarks); the fall in force at lengths shorter than 2.05 µm (interference, not modelled); a/F₀ and v_max are the same at every length; quasi-steady shortening at one length with constant activation and temperature. The values are illustrative, not human constants; a/F₀ (0.25) is the usual frog sartorius value Hill gave in 1938; v_max (0.40 m per s) and the slider ranges are model values. Rule used: F₀ = 100 N × min(1, (3.65 − L) / 1.45), a = (a/F₀) F₀, b = v_max (a/F₀), v = b (F₀ − F) / (F + a), P = F × v, best load = F₀ (√(q² + q) − q) with q = a/F₀.

Rule to keep: the sarcomere length sets the isometric force F₀ the muscle can hold; against any load below F₀ Hill’s hyperbola sets the speed; power F × v is zero at zero load and at F₀ and peaks at a load in between, whose fraction of F₀ depends on the curvature.

One impulse twitches; many fuse

A twitch is the mechanical response to a single action potential. If a second action potential arrives before relaxation is complete, cytosolic calcium is still raised when more is released, more troponin stays occupied, and force climbs higher: summation. At moderate rates the force record shows ripples, an unfused tetanus; at higher rates it becomes smooth, a fused tetanus. How the nervous system recruits motor units and sets their firing rates belongs to motor units and reflexes.

Human fibres fall into three main types: type I (slow oxidative), type IIA (fast oxidative-glycolytic) and type IIX (fast glycolytic); why human “IIb” fibres are really IIx is told in muscle and movement. Contraction speed follows the rate of the myosin’s ATPase and cross-bridge cycle. Type I fibres twitch more slowly, so their twitches fuse at lower firing rates. They are red from myoglobin (oxygen transport) and dense capillaries, rich in mitochondria, and resistant to fatigue; IIX fibres rely mostly on glycolysis and fatigue quickly. Force per cross-sectional area differs much less between types than speed does, and since peak power in Hill’s model scales with F0 x vmax at fixed curvature, a fast fibre with several times the maximal speed gives several times the peak power at similar force.

Three ATP supplies, running at once

Resting ATP would last only seconds at a high rate of use, yet it barely falls in most exercise. The first buffer is phosphocreatine (PCr), present at several times the ATP concentration in resting human muscle (metabolism: basic concepts):

PCr + ADP + H+ → creatine + ATP (creatine kinase)

The reaction consumes a proton, which matters below. Glycolysis, from muscle glycogen or blood glucose, makes ATP without using oxygen (glycolysis). Oxidative phosphorylation needs a continuing oxygen supply and burns carbohydrate and fat (oxidative phosphorylation). All three run together, with shares that shift with intensity and duration, so fixed “10-second” or “2-minute” windows misdescribe them.

The 1920s oxygen debt idea of Hill and colleagues held that the extra oxygen taken up after exercise paid for clearing the lactate made during it: about a fifth of the lactate was to be oxidized to fuel rebuilding the rest into glycogen. In rats exercised to exhaustion, Brooks and Gaesser (1980) found lactate removed within 15 minutes while oxygen consumption stayed above rest for 120 minutes. The time courses came apart. The descriptive term is excess post-exercise oxygen consumption (EPOC), which has several sources: resynthesis of phosphocreatine and refilling of oxygen stores early in recovery, and raised temperature and hormones over a longer time (Gaesser and Brooks, 1984). Lactate is a fuel: in six active men cycling at 55 percent of peak oxygen uptake, infused lactate raised lactate oxidation and lowered glucose oxidation (Miller and colleagues, 2002); the whole-body story belongs to exercise physiology.

Contracting calf muscles squeeze the deep veins between them, and one-way valves let blood move only toward the heart; on relaxation the veins refill from below. This muscle pump aids venous return from the legs (cardiac output and venous return).

Muscle is also the main adjustable heat source. Only a minority of the energy turned over appears as external work (in cycling, at most about a quarter); the rest is heat. Shivering is involuntary, rhythmic contraction that does almost no external work, so nearly all its energy becomes heat (skin and human heat).

What fails in fatigue

Fatigue is a fall in force or power during repeated activity that recovers with rest. Intense glycolysis is accompanied by lactate ions and protons (where the protons come from is debated), but the evidence points away from the lactate ion. In mammalian muscle, the traditional account of lactate and protons impairing the contractile proteins is probably of limited importance. Leading alternatives are inorganic phosphate (Pi), which acts on cross-bridges and on calcium release, failure of calcium release from the sarcoplasmic reticulum, and ionic changes that affect the action potential (Allen, Lamb and Westerblad, 2008). Westerblad, Bruton and Lännergren (1997) acidified intact single mouse fibres by about 0.5 pH unit with carbon dioxide: tetanic force fell 28 percent at 12 °C but only 10 percent at 32 °C, with no significant change in maximal shortening velocity at 32 °C. Cold preparations overstate the effect of acidosis: the authors concluded that direct inhibition of force by acidification is not a major factor in fatigue at physiological temperatures, although the loss at 32 °C was not zero.

The modern experiment. Do protons and phosphate act directly on the motor, and by the same step? Woodward and Debold (Frontiers in Physiology, 2018) purified fast skeletal myosin and actin from chicken pectoralis and let a small ensemble of myosin heads pull on one actin filament in a laser trap (stiffness about 0.02 pN per nm), at 100 µM MgATP, in three conditions (pH 7.4; pH 6.5; pH 6.5 with 15 mM added phosphate), and no membrane, sarcoplasmic reticulum or lactate. If both ions acted at the same step, they would be expected to change the pattern of force events in similar ways. Lowering pH to 6.5 cut the ensemble’s average peak force by about 20 percent. At pH 6.5, added phosphate largely abolished the high-force events above 2 pN, and events became brief, suggesting faster detachment of myosin from actin. Protons lowered force directly at actomyosin, and phosphate added at low pH removed the high-force events, so the two ions appear to act by distinct mechanisms. It does not show how much each contributes in working human muscle (chicken proteins, low ATP, a fixed buffer), and lactate was simply absent.

In seven anesthetized rats, Lewis and colleagues (2025) found twitch force fell beyond a fitted phosphate breakpoint of 17.1 ± 2.9 mM, measured by phosphorus magnetic resonance spectroscopy, and tracked phosphate more consistently than protons, an association the authors state they did not test causally.

A disease as a test. In McArdle disease (glycogen storage disease type V), inherited loss of muscle glycogen phosphorylase (gene PYGM) stops muscle glycogen from feeding glycolysis. During a brief intense contraction, lactate in muscle and venous blood barely rises. Intracellular pH does not fall, and because creatine kinase consumes protons it may rise slightly. Phosphorus spectroscopy in patients reports an abnormally low phosphocreatine to phosphate ratio at the end of exercise, with no acidosis. Yet force fails early, and painful contractures can occur. After several minutes of moderate exercise, more work is often tolerated (the “second wind”), consistent with rising delivery of blood glucose and fatty acids to oxidative metabolism. Fatigue here develops without acidosis or a rise in lactate, so lactate accumulation cannot be its necessary cause.

Design problem

You have isolated mouse soleus (a slower muscle, with roughly 40 percent type I and about half type IIA fibres) and extensor digitorum longus (EDL, mostly fast type II fibres) muscles in oxygenated solution at a fixed temperature. Design an experiment to compare their force-velocity and power curves, and predict the result.

One way to do it

At the length giving the largest twitch (Lo), record a reference tetanus for F0. Tetanize and release onto loads from about 5 to 90 percent of F0 (afterloaded isotonic contractions), measuring speed from the early steady part of shortening, after the brief transients at release. Isovelocity ramps work the other way round: impose each speed and measure the force.

Normalize force to physiological cross-sectional area (mass divided by density and by optimal fibre length, not whole-muscle Lo) and speed to fibre lengths per second. Fit Hill’s equation for F0, a/F0 and vmax, and compute power as F x v at each load.

Controls: same temperature and stimulation; random load order; a repeated reference tetanus, discarding runs where it falls; a check for an anoxic core.

Prediction: similar force per area but higher vmax in EDL, so its power curve peaks higher. Do not predict a direction for a/F0 without data. Limits: whole muscles mix fibre types, one temperature, mouse not human; confirm fibre types by myosin heavy-chain staining or gels.