# Muscle force/energy relationship

I found the Hill muscle model, which helps relate the force a muscle can generate to its change in length over time (its "contraction velocity"). Is there a similar model to get an idea of the energy consumed during a contraction as a function of the force generated?

You can calculate the energy consumption using basic physics formulas.

Work = Force x Distance

and the unit of work is the Joule. Therefore if force is a function of change in length then substituting this function in for force in the above calculation we have that work is a function of length.

If you want to talk about energy consumed in terms of calories rather than joules,

1 calorie = 4.184 joules

• That's part of it, but what about isometric exercise? – Jay Lemmon Jun 16 '14 at 21:41
• The same physics applies, you are buttressing your muscles but they will still be exerting a force. The resulting distance of movement is small so that you don't notice it, however the force is greater. Work is still force x distance. – Spinorial Jun 16 '14 at 21:54
• But that doesn't help in building a mathematical model for muscle activation. I want a predictive model for the energy consumed during a muscle contraction, whether it's isometric or isotonic. Even for only isotonic contractions, if you only account for mechanical work you're entirely ignoring the heat generated. You'll only get a lower bound on the amount of energy consumed. – Jay Lemmon Jun 16 '14 at 22:00
• Your question asked for energy as a function of force generated. Not as a function of heat and force generated. If you want the total energy used then yes you would have to take into account the inefficiencies of muscle contraction which generates heat. The question then is how efficient are myocytes in converting energy from ATP to force. This a variable quantity see the following paper ncbi.nlm.nih.gov/pubmed/11350777 – Spinorial Jun 16 '14 at 22:08
• Yes, I asked for energy as a function of force generated. Because muscles can produce force with no net movement, mechanical work is simply not a complete answer to the question :( Did you read the body of my question, or just the title? – Jay Lemmon Jun 16 '14 at 22:24