Understanding Force-Energy Relation: Linear or Exponential?

[douglis]

Hi all,

let's say someone holds an object for a minute and another one holds a double mass object for the same duration.
If the force-energy relation was linear the second guy would spend exactly the double energy.My intuition says the relation is not linear but is exponential.
Is there a general rule or equation that gives us the magnitude of the exponent or it's different in any case and depends on the source of the force?

 

[tiny-tim]

hi douglis!

If the force-energy relation was linear …

force is related to momentum, not energy

(and if you hold something still, its energy does not change)

 

[douglis]

Hi tiny-tim,

in my example the momentum doesn't change and of course no work is done.However obviously chemical energy is spent.
My question was about the relation of the expenditure of that chemical energy with the force that is used in order to support the load.Is it increasing linearly along with the load?

 

[tiny-tim]

My question was about the relation of the expenditure of that chemical energy with the force that is used in order to support the load.Is it increasing linearly along with the load?

that's really a biology question

(and if the load is too much, it can't be supported anyway)

 

[mishrashubham]

This might be of some use to you. http://ajpendo.physiology.org/content/282/2/E448.full

Here is an extract..

During isometric contractions, no true work is performed, so the force-time integral (FTI) is often used to approximate isometric work. However, the relationship between FTI and metabolic cost is not as linear. We tested the hypothesis that this nonlinearity was due to the cost of attaining a given force being greater than that of maintaining it. The ATP consumed per contraction in the human medial gastrocnemius muscle (n = 6) was determined by use of 31P-NMR spectroscopy during eight different electrical stimulation protocols. Each protocol consisted of 8 trains of a single frequency (20 or 80 Hz) and duration (300, 600, 1,200, or 1,800 ms) performed under ischemic conditions. The cost of force generation was determined from the ATP turnover during the short-duration trains that did not attain a steady force level. Estimates of the cost of force maintenance at each frequency were determined by subtracting the ATP turnover during the shorter-duration trains from the turnover during the long-duration trains. The force generation phase of an isometric contraction was indeed more metabolically costly than the force maintenance phase during both 20- and 80-Hz stimulation. Thus the mean rate of ATP hydrolysis appeared to decline as contraction duration increased. Interestingly, the metabolic costs of maintaining force during 20-Hz and 80-Hz stimulation were comparable, although different levels of force were produced.

And yeah, this should be in the biology forum.

edit: Good question by the way. I had never seriously thought about this before.

 

[douglis]

This might be of some use to you. http://ajpendo.physiology.org/content/282/2/E448.full

Here is an extract..



And yeah, this should be in the biology forum.

edit: Good question by the way. I had never seriously thought about this before.

The link you gave is very interesting and somewhat answers my question.Thanks.

It says that the metabolic cost to maintain the force(like in my above example) is linear with the force.
''During electrically elicited isometric twitches, ATP consumption increases linearly over the range of frequencies''

However the metabolic cost to generate the force is not linear:
''During an isometric tetanus, however, it has been suggested that metabolic cost is not a linear function of contraction duration, because the cost of achieving a level of force is greater than that of maintaining it''.

 

[SW VandeCarr]

The answer depends on what fatigue related metric you're using. The following study showed a generally linear pattern of EMG indicators of large muscle fatigue under isometric stress.

http://www.ncbi.nlm.nih.gov/pubmed/15878288

EDIT: As tiny-tim pointed out, the typical force-energy relation doesn't apply if no work is done. In the purely isometric situation, one could measure the relative time to exhaustion for objects of different masses. I haven't found any such simple studies in a quick search but I suspect it would be close to linear to a critical point followed by a rapid decay of the form F(t)= 1-e^{-kt} and the "survival" fraction (S) is S=1-F(t) . The location of critical point would depend on the strength of subject.

 

[douglis]

The answer depends on what fatigue related metric you're using. The following study showed a generally linear pattern of EMG indicators of large muscle fatigue under isometric stress.

http://www.ncbi.nlm.nih.gov/pubmed/15878288

EDIT: As tiny-tim pointed out, the typical force-energy relation doesn't apply if no work is done. In the purely isometric situation, one could measure the relative time to exhaustion for objects of different masses. I haven't found any such simple studies in a quick search but I suspect it would be close to linear to a critical point followed by a rapid decay of the form F(t)=1-e^{-kt}. The location of critical point would depend on the strength of subject.

So...we have evidence that at least the rate of fatigue is linear.I don't have any reason to believe that the rate of fatigue differs from the rate of energy expenditure.

 

[SW VandeCarr]

So...we have evidence that at least the rate of fatigue is linear.I don't have any reason to believe that the rate of fatigue differs from the rate of energy expenditure.

Why not? First of all, the above "survival" function can be nearly linear to out to some point and then become frankly exponential, decaying to 0 . On other hand, the metabolic cost function ramps up early in a non-linear fashion to reach some level of demand and then stays relatively flat per unit time. I suspect the cost function will also plummet near the point of exhaustion.

BTW: I edited the above my previous post to show the "survival" fraction S. Also, the article I cited is based on EMG data, not observed physical exhaustion. You need to be very careful how you define the experiment in order to demonstrate specific claims.

 

[mishrashubham]

The link you gave is very interesting and somewhat answers my question.Thanks.

It says that the metabolic cost to maintain the force(like in my above example) is linear with the force.
''During electrically elicited isometric twitches, ATP consumption increases linearly over the range of frequencies''

However the metabolic cost to generate the force is not linear:
''During an isometric tetanus, however, it has been suggested that metabolic cost is not a linear function of contraction duration, because the cost of achieving a level of force is greater than that of maintaining it''.

Actually its the other way round. A twitch contraction occurs when the a short signal is given to the muscle and it contracts quickly only for a short period of time. Isometric implies that the muscle length remains constant all the while generating tension. This is what you might call creating a force.

Tetanus however is sustained contraction, occurs when too many signals come one after the other so that that the muscle is continuously contracted. This is what you might call maintaining the force.

 

[douglis]

Actually its the other way round. A twitch contraction occurs when the a short signal is given to the muscle and it contracts quickly only for a short period of time. Isometric implies that the muscle length remains constant all the while generating tension. This is what you might call creating a force.

Tetanus however is sustained contraction, occurs when too many signals come one after the other so that that the muscle is continuously contracted. This is what you might call maintaining the force.

So in the following conclusion:
''During an isometric tetanus, however, it has been suggested that metabolic cost is not a linear function of contraction duration, because the cost of achieving a level of force is greater than that of maintaining it''.

...by ''not a linear function'' do they mean that the exponent is less than 1?

 

[mishrashubham]

So in the following conclusion:
''During an isometric tetanus, however, it has been suggested that metabolic cost is not a linear function of contraction duration, because the cost of achieving a level of force is greater than that of maintaining it''.

...by ''not a linear function'' do they mean that the exponent is less than 1?

Read the next line...

If the cost of attaining a given force level is greater than that of maintaining it, then the net rate of ATP hydrolysis should decrease as the duration of a tetanus increases.

rate of hydrolysis decreases as duration increases. Though they have not given a concrete equation(nor do I expect them to), this should mean that the exponent is less than 0 (not 1).

 

[Andy Resnick]

Hi all,

let's say someone holds an object for a minute and another one holds a double mass object for the same duration.
If the force-energy relation was linear the second guy would spend exactly the double energy.My intuition says the relation is not linear but is exponential.
Is there a general rule or equation that gives us the magnitude of the exponent or it's different in any case and depends on the source of the force?

Are you asking a question about muscle physiology?

 

[douglis]

Are you asking a question about muscle physiology?

Yes...the question is about muscle physiology since in physics the force is not directly related with energy.

 

[Andy Resnick]

Hrm- my post ended up here well after many other people started to answer your question. There's some good info already presented; do you understand the material now?

 

[douglis]

do you understand the material now?

Well...I got too many informations in too little time.I have to think of it a little.Thanks.