Is Average Force the Same in Fast vs. Slow Weightlifting Reps?

[waynexk8]

D. wrote;
I will try to help Wayne by simplifying his question.
Let's say two weightlifters lift 100 pounds.The first one lifts them in 1sec and the second in 10sec.Obviously the first weightlifter used 10 times greater power than the second.
For that reason Wayne somehow believes that the first weightlifter also used greater force.

This is clearly not true in my opinion.When you lift a weight(regardless if it's in 1 or 10sec) the starting and ending velocity is zero hence the average acceleration is zero too[a=(V2-V1)/t=(0-0)/t=0].So the average force that you use is (F=mg+ma=mg) equal with the weight regardless the lifting speed.

Do you understand what average means ? Its only all the numbers added up. But what it misses, is that the higher high forces in the faster rep, and the higher peak forces are higher.

The higher high forces are 20 or we can say 25% higher, and the peak forces are 60 or we can say 75% higher.

So the higher high forces and higher peak forces are higher, that’s what I have been saying from the start, the total or overall forces are higher, but you will not comment on this or register this ?

D. wrote;
As for the energy that was used...from physics POV the work is identical in both cases.But since the second weightlifter(10sec lifting) used the same average force for 10 times longer I believe it's obvious that he spent more energy.

You have now done something VERY wrong, you have changed the debate.

The debate is, I lift a weight 1 second up, 1 second down 5 times = 10 seconds, you lift the weight 5 seconds up, 5 seconds down.

SO PLEASE, why did you say I only lifted at 1 second up and 1 second down once, but you lifted 5 seconds up and 5 seconds down ? YOU KNOW VERY WELL THIS IS NOT THE DEBATE, AND HAS NEVER BEEN THE DEBATE ?

So now, HOW is the work equal ? If I move the weight 1m up and 1m down 6 times to your 1 time in the same time frame, I have moved the weight 10m more, so more work has been done,

D. wrote;
Now specifically in Wayne's example.
He tries to compare the force and energy in two cases.Either you lift and lower a weight 6 times in 6 seconds or once in 6 seconds again.
As I believe I proved above the average force is the same in both cases.As for the energy that was used...the work done in both cases is zero from physics POV.But since the same average force was used for 6 seconds in both cases my guess is that also the energy expenditure is the same or with biomechanics terms the time-force integral is identical.

I don't know which part of my post you didn't understand to say something like that and to be honest I don't really care.
I won't continue in this discussion.Try to understand what I wrote cause I believe it's really simple.

SORRY, I read it wrong.

Look D. Energy is not all to do with just average force, how are you working out energy ?

And if you look back on this thread, they have worked out quite plain with physics that the faster rep uses more power = more energy, so please tell me why you think their calculations are wrong ?

look on every nutrition site or book. Just borrowed this from your other friend.

Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry. Many Statistics will be measured eg; Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, to determine the effects of exercise at different intensities. On every occasion when an activity has been practiced at a faster rate, the heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, has been higher, therefore energy expenditure, nutrient oxidation will be higher.

The increase in the rate of energy expenditure/cost will be due the increase in the intensity/force of muscle activity. When a muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy.


Wayne

 

[waynexk8]

Look D. all nutritional people, and most other people know you use more energy in the faster reps, and below is what some nice p[people here worked out for me. I cannot understand why when you know that there is more power involved in the faster reps, that from that you cannot understand that there is more work and energy used, all your doing is contradicting yourself.




I quote from what members wrote;
When an object is moving at velocity v with mass m, it has a certain KE dictated by KE=0.5mv2. That is, to move it at the velocity you must give it that amount of kinetic energy.

So in your case, m is constant which means the deciding factor is velocity.

Let's say that in the slow reps the average velocity is 1m/s, that means your muscles must provide 22 joules of energy. Now let's assume the average velocity for the fast reps is 2m/s, that means your muscles must provide 90 joules of energy. This is all because of the v2 term in the KE equation.

The faster you move something, the more energy it takes. As you can see, simply doubling the velocity (which would halve the rep speed) requires over four times more energy. Each time you double speed of the reps, and as such halve the time for the reps, you are increasing the energy requirement in this manner.

That is why you're using more energy.
a = f/m therefore if you double to force to 2f you get 2a = 2f/m.

Let's say we have an object that is 1kg. To move that 1m in 1s (1m/s) requires a KE of 0.5mv2 = 0.5*1*1 = 0.5 Joules of energy.
Now, to move it 100m in 1s (100m/s) requires a KE of 0.5*1*10000 = 5000 Joules. So in the first case I need a tiny amount of energy, in the second I need a huge amount in comparison.

For your case to move that 1m in 1s (1m/s) requires a KE as above (0.5 Joules).
Now, to move it 1m in 0.5s (2m/s) requires a KE of 0.5*1*4 = 2 Joules. So again you can see how simply halving the time of the repetition requires you to use more energy to complete it. The time applied is considered in the velocity figure.

For you to move the weight 1 rep in 1s requires 0.5 Joules - that is the energy you must provide to do it.
For you to move the weight 1 rep in 0.5s requires 2 Joules - again, that is the energy you must provide to do it.

If you do not provide that energy, you can't complete the rep in the required time.

This has been the root of your problem and mistake all along.

Wayne

 

[douglis]

Do you understand what average means ? Its only all the numbers added up. But what it misses, is that the higher high forces in the faster rep, and the higher peak forces are higher.

The higher high forces are 20 or we can say 25% higher, and the peak forces are 60 or we can say 75% higher.

So the higher high forces and higher peak forces are higher, that’s what I have been saying from the start, the total or overall forces are higher, but you will not comment on this or register this ?

Wayne

Wayne...you came to a physics forum and you keep using terms and forces from your imagination.How can you expect to have a serious answer?

There's no such thing as "higher high forces" or "total or overall forces".
There're only two forces.The force that the muscles exert and the weight(force due to gravity).If you try to lift the weight in an accelerative manner the muscle force will have higher fluctuations BUT since the average acceleration is zero(as I showed in my last post) the average net force will be zero too.Therefore the average muscle force will be equal with the weight.

 

[sophiecentaur]

@wayne
You are still trying to force some Physics terms into your ideas about exercise and to get a meaningful conclusion. This is a non-starter. Why do you think we're telling you this?
You insist that "average force" has a worthwhile meaning and that it is the answer to your question. Why isn't it used throughout Mechanics? For a start, there are many measures of 'average'. Just one of them (the Mean) is defined something like you say. But Force could be 'averaged' over time or over distance, for a start and you'd normally get different answers in each case.

This simple example should show you that there is no clear relationship between average force and work done.
Take two identical masses, suspended on springs. They are bobbing up and down with the same maximum displacement. One happens to be going at twice the rate of the other. The mean force on the faster moving mass is twice the mean force on the slower one. There is more energy in the faster system. But, in neither case is any external work being done on either, since the instant when springs were initially stretched!

Pretty well everyone who has replied to you has been using the correct terms and the correct formulae, relating the correct quantities, and has told you that you are onto a loser with this. Does it not strike you that it could be you who has got it wrong and that you need to try to look at this in a different way if you want some proper understanding?

 

[waynexk8]

Wayne

@wayne
You are still trying to force some Physics terms into your ideas about exercise and to get a meaningful conclusion. This is a non-starter. Why do you think we're telling you this?

Maybe I am a little, but it’s difficult for me, as sort of you going into a bar full of labours, and telling them if they just hold a very heavy cement bag, and after 10 minutes when they can hardly hold it, that no work has been done.

You insist that "average force" has a worthwhile meaning and that it is the answer to your question.

Sophiecentaur, this is where you get “ME” wrong, the fact is I think the opposite of that, and always have done, I think the average force means not much, I do NOT go with the average strength, I feel it’s very wrong in this debate.

This is why I keep saying, that in the below, with is the force % used for 5 segments of the slow and fast reps, that 140 is 60 or 75% more than 80. 100 is 20 or 25% more than 80. And again 100 is 20 or 25% more than 80. So that means that the faster reps have a higher high force, and a higher peak force. So the faster reps have a higher total or overall force. And average means absolutely nothing.

Slow rep,
80, 80, 80, 80, 80.

Fast reps,
140, 100, 100, 60, zero.

Why isn't it used throughout Mechanics? For a start, there are many measures of 'average'. Just one of them (the Mean) is defined something like you say. But Force could be 'averaged' over time or over distance, for a start and you'd normally get different answers in each case.

This simple example should show you that there is no clear relationship between average force and work done.
Take two identical masses, suspended on springs. They are bobbing up and down with the same maximum displacement. One happens to be going at twice the rate of the other. The mean force on the faster moving mass is twice the mean force on the slower one. There is more energy in the faster system. But, in neither case is any external work being done on either, since the instant when springs were initially stretched!

Yes very true and I understand that example. Howe ever, if the one mass was going at twice the rate of the other, would have more initial force and move further ? But hats what I am getting add, in the faster reps the higher force is higher. When you say mean, you mean average ?

Pretty well everyone who has replied to you has been using the correct terms and the correct formulae, relating the correct quantities, and has told you that you are onto a loser with this. Does it not strike you that it could be you who has got it wrong and that you need to try to look at this in a different way if you want some proper understanding?

I do appreciate what you say above, and for you all helping me and your time, but if you could stay with this a little longer, but then you could be right, I could need to look into it in a different way. Or maybe I am saying or explaining wrong.

However, the below questions seems straightforward to me.

As we all agree that more power is used, and I will work that out after, {please all I know any of you could work this out easy, but please let me do it, then you can comment on how and if I worked it out right} energy is used in the faster rep, done in the same time frame, and if work is the amount of energy transferred by a force acting through a distance.

This is my way of thinking, if we have used more energy, we must have used more force ? And I think that more force is the higher high forces, and higher peak forces in the first half or three quarters of the faster rep, as I explained above. The high forces are 25% higher, and the peak forces are 75% higher.

So basically I am saying, is if you use more energy, there HAS to have been MORE muscle activity, or more force/strength used in the muscle, thus more tension on the muscle, as otherwise why would the muscle use more energy ? Also in my example, the fast rep has moved the weight 10m more in the same time frame.

Example 1,
Take a bridge. {this is the muscle} The bride has a strain or braking point of 141. At most of the day the bridge has a certain number of cars going over, and the strain is 80, {the slow reps} then you get more and more cars going over the bridge at once {these are the faster reps higher high forces} the strain goes up to 100. Then at one point you have even more cars going over the bridge, the strain gets up to 140. {the faster reps higher peak forces} Both the strains last the same time.

Now let’s say the bridge had a steel structural strain test after the strain got up to 80, they x-ray the steel for small cracks or tears.

They then had a steel structural strain test after the strain got up to a 100 and 140; they x-ray the steel for small cracks or tears.

I and most if not all on here, would say the bridge, or the muscle in this case, got far more cracks and strains when the force/strength was 100 and 140. THUS if the scenario had been a muscle, it would have had far far far MORE TENSION on it. So as I sais; if you use more energy, there HAS to have been MORE muscle activity, or more force/strength used in the muscle, thus more tension on the muscle, as otherwise why would the muscle use more energy ?

Wayne

 

[sophiecentaur]

A bridge puts no energy into the system so how is this, in any way, analogous to a muscle? The cars are putting in the energy by going across and deforming the bridge. Are you just trying to wind people up with this?

I don't care about your pub full of navvies. They can take Physics or leave it and you don't have to talk to them if you don't want to get duffed up for being a boffin.

 

[JaredJames]

Woah there! Where did bridges come into this?

I haven't read through in a while but there is a lot of junk in your last post that makes absolutely no sense under the context of the OP.

Wayne, your posts are getting longer and making less sense. I recommend you get a very clearly defined question - as simple as possible - and try asking again.

 

[waynexk8]

A bridge puts no energy into the system so how is this, in any way, analogous to a muscle? The cars are putting in the energy by going across and deforming the bridge. Are you just trying to wind people up with this?

Sorry, maybe I explained wrong, the bridge is the muscles, and the cars are the forces putting the tensions on the muscles.

So yes the cars {forces that put the tension on the muscles} are also using more energy, and this is my main point, as if the muscles are using more energy, and we all agree on this, except for D. then why are they using more energy ? It must be because there is more muscle activity, which is higher forces used thus higher tensions on the muscles then the slower rep

NO, please I can show you from training related forums that this debate has been going on for over two years, so I thought let us come over here and have a new independent and non biased view.

So to all, do you agree. that as there is more energy used in the faster reps, that’s because there is more muscle activity ? Which is becouse higher forces are used thus higher tensions on the muscles than the slower rep, if not please why and how.

I don't care about your pub full of navvies. They can take Physics or leave it and you don't have to talk to them if you don't want to get duffed up for being a boffin.

Sorry, maybe I should not have said that, I know physics is right, and can work out all things if all the variables are known. I was only trying to say it’s hard at times for people that are 49 like me, and only started very slowly to learn it again.

OK, here is my Power equations.

Fast rep,
150kg moved up at .5 of a second for 1.2m

Slow rep,
150kg moved up in 3 seconds for 1.2

To determine the force we will need to figure out what the weight of the barbell is. (W = mg = 91 kg x 9.81 m/s² = 1471.5 kg.m/s² or 1471.5 N) Work is equal to Force x distance, U = 1471.5 N x 1.2m = 1765.8 Nm.

Fast rep,
.5 divided by 1765.8 = 3531.6

Slow rep,
3 divided by 1765.8 = 588.6

Wayne

 

[waynexk8]

Woah there! Where did bridges come into this?

I haven't read through in a while but there is a lot of junk in your last post that makes absolutely no sense under the context of the OP.

Wayne, your posts are getting longer and making less sense. I recommend you get a very clearly defined question - as simple as possible - and try asking again.

Ok sorry, does my last post clear this up a little ? Hope so.

and again all thanks for you help and time, I have a little more to write on impulse a little later.

Wayne

 

[JaredJames]

So yes the cars {forces that put the tension on the muscles} are also using more energy, and this is my main point, as if the muscles are using more energy, and we all agree on this, except for D.

The cars are not expending any energy to create tension on the cables.

Gravity on acting on the car creates a downwards force (the weight) which puts tension on the cable. No energy is expended in this process and the cables don't expend energy supporting the cars.

 

[waynexk8]

The cars are not expending any energy to create tension on the cables.

Gravity on acting on the car creates a downwards force (the weight) which puts tension on the cable. No energy is expended in this process and the cables don't expend energy supporting the cars.

The cars are not expending any energy to create tension on the cables.

Gravity on acting on the car creates a downwards force (the weight) which puts tension on the cable. No energy is expended in this process and the cables don't expend energy supporting the cars.

Would not the cables {the muscles} not have to use energy to just hold the cars {weight} and use a force to keep that at a position ? However that was just a simple scenario to try and make more/different sense of this problem.

However all, could we go back to the muscles are using more energy when lifting the weight up and down faster in the same time frame, thus there must be a scientific reason for this, I say it’s because the muscles have more activity, and are using more overall or total force, otherwise, why are the muscles use more energy ? We could put this debate to a machine if its easier !

Wayne

 

[JaredJames]

Would not the cables {the muscles} not have to use energy to just hold the cars {weight} and use a force to keep that at a position ?

Force yes, energy no.

However all, could we go back to the muscles are using more energy when lifting the weight up and down faster in the same time frame, thus there must be a scientific reason for this, I say it’s because the muscles have more activity, and are using more overall or total force, otherwise, why are the muscles use more energy ? We could put this debate to a machine if its easier !

The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

 

[waynexk8]

Here is what someone wrote from this forum.

{Make some assumptions and see what the numbers say. You are moving with peak to peak amplitude of 1m, in either 6 sec (3 up 3 down) or 1 sec (0.5 up 0.5 down).

Assume the weight is doing simple harmonic motion up and down. That is probably not a very good approximation but it's easy to calculate

Amplitude = 0.5m, frequency = 0.167 Hz or 1 Hz.

Maximum acceleration = a times (2 pi f)^2
= 0.55 m/s or 19.7 m/s
= about 0.06G, or 2G
So the maximum force would be 1.06 times the weight lifted for the slow case, and 3 times for the fast case.

Don't take those numbers as "accurate" but they do suggest there would be an effect.

The lifter might apply a large force for a short time and then ease off, rather than a smaller force for the whole 3 seconds. That would reduce the difference in the peak force.

So what some of you are saying is, that force must be only expressed as either the work done (force x distance) or the impulse applied (force x time).}

So as we all know, the higher high forces and higher high forces are higher, thus these would be why the energy used was more, and thus put a greater ammout of tension on the muscles




So let's see if I get this right, force can = work and impulse.

So I think we all can agree, that the faster rep will have a greater quantity of work, and if work = force, then that must mean more force ? As the faster reps have moved the weight far far far further then the slow rep in the same time frame, work = force x distance.

And

As impulse = force x time. The small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. However as I have applied a higher force for the same time, would not the impulse be higher ?

Wayne

 

[JaredJames]

So as we all know, the higher high forces and higher high forces are higher, thus these would be why the energy used was more, and thus put a greater ammout of tension on the muscles

I have no idea what you are talking about. Complete gibberish.

I explained in my previous post why more energy was used. Period. End of story.

To reiterate for you, you are imparting KE on the weight from CE in your body. The more KE you impart the more CE your body must use. Therefore, the faster you move the weight the you get significant energy increases (as I showed before with the difference in KE from 1m/s to 2m/s goes from 0.5 Joules to 2 Joules, so you need more chemical energy to gain a faster rep speed).

 

[waynexk8]

Force yes, energy no.


The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

Great, that’s basically what I think.

What some other people think, and they are in the very very very lower of the training World, they are the one grain of sand in the World, the rest of us are the rest of the grains of sand, not sure how low that is of a percentage, but as you can see its low, ROL.

Here are how the top people train, trying to move a heavy weight as fast as possible.
Derek Poundstone,

Mariusz Pudzianowski,

Ronnie Coleman


They think that when we are deceleration in our faster reps, that then is where their more constant lower force make up the force, but as I sort of proved with my percentage theory, it’s not so. fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and the lower forces of the slower reps can not make up the force in the same time frame.

Wayne

 

[sophiecentaur]

All this about bridges is irrelevant nonsense.
Also, as has been said many times, if you don't use the right definitions and the right terms there is NO Physics that can apply.

No one has disagreed about the fact that you expend more Power doing faster reps. But there is no simple relationship between this and the Work Done (defined, as it is, within Physics). You can keep on about your numbers and stats but that won't alter this fundamental problem. Why do you keep trying? It just gets up people's noses. Discuss it as much as you like in a non-technical group but don't expect to get anything out of this forum if you're no prepared to go along with the basic rules of Physics.

And how can you say (above) "if work = force" in the same sentence as "work = force times distance"? Can you not see the contradiction? Also, Impulse has nothing to do with Work.

Just read a basic Mechanics / Dynamics book and get some discipline in your Science and you may have a chance of getting this.

 

[JaredJames]

They think that when we are deceleration in our faster reps, that then is where their more constant lower force make up the force, but as I sort of proved with my percentage theory, it’s not so. fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and the lower forces of the slower reps can not make up the force in the same time frame.

Wayne

Please stop using these numbers, I don't think they mean anything and I don't think anyone else knows what they mean either.

Frankly, from the explanations you have given it is clear that not even you are that sure of what you're talking about.

Sorry to sound so blunt, but I don't think dragging this out any further is going to be productive.

 

[waynexk8]

I have no idea what you are talking about. Complete gibberish.

Sorry, I meant the higher forces for the first part of the faster reps, please see.

fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, thuis I can not see how the forces are the same.

I explained in my previous post why more energy was used. Period. End of story.

Yes ok thank you.

To reiterate for you, you are imparting KE on the weight from CE in your body. The more KE you impart the more CE your body must use. Therefore, the faster you move the weight the you get significant energy increases (as I showed before with the difference in KE from 1m/s to 2m/s goes from 0.5 Joules to 2 Joules, so you need more chemical energy to gain a faster rep speed).

Yes get you, to use more KE, you have to have more muscle activity and to do this you have to use a higher force/strength to move an object from a to b faster, as moving the object 1m in 1 second, but to move the object 6m in 1 second takes more higher high force/strength and higher peak force/strength. As if you do not use a higher force, the object will not move so fat in the same time frame. Thus because the more KE, from the higher high and peak forces the more energy is used.

Big thank you for your help and time, but maybe more for your patience.

Ho and for the person that thought I maybe having you on, I was definitely not, as I have made many videos proving you fail far far far faster in the faster reps, here one of me showing this.

http://www.youtube.com/user/waynerock999?feature=mhum#p/u/0/sbRVQ_nmhpw

Wayne

 

[JaredJames]

I have made many videos proving you fail far far far faster in the faster reps, here one of me showing this.

Why would you need a video showing this?

It is clear from even basic energy calcs that you use significantly more energy to go faster. So anyone who questions this clearly doesn't understand the physics and frankly, doesn't have a say. Especially not to dismiss it.

 

[waynexk8]

Please stop using these numbers, I don't think they mean anything and I don't think anyone else knows what they mean either.

These numbers were just meant as a basic guideline, nothing more, just somthing to get people to see more into the debate. As we were using 80 pounds as 80% of or RM {repetition maxumun} the most we could lift for once. So we just said for the first part of the rep we were useing a 100 as in pounds, then still a 100, then still a 100, then down to 80 for the start of the deceleration, then down to 20.

Frankly, from the explanations you have given it is clear that not even you are that sure of what you're talking about.

Well I do know what I am talking about, as all the below agree with me, lots are biomacanics, kinsologists and have master in physics, but I find it hard to put into words. Momentum, or as I like to say, off loading may be an issue, if the weight is to light, and you are not accelerating the weight enough.

Acceleration is not momentum, its the opposite of momentum, acceleration requires more force/strength, where momentum requires less force/strength.

Roger Enoka,
The number of muscle fibers activated to lift a weight depends on two factors: (1) the amount of weight; and (2) the speed of the lift. Although more muscle fibers are activated during fast lifts, they are each generating MORE force. We know this because the rate at which the muscle fibers are activated by the nervous system increases with contraction speed.

The force that a muscle must exert to move a load depends on two factors: the mass of the load and the amount of acceleration imparted to the load. The number of muscle fibers recruited during the lift increases with the speed the lift.


The rate at which any motor unit, low or high threshold, can discharge action potentials is not maximal during slow contractions. As contraction speed increases, so does discharge rate for all motor units.

The most common finding is that it is the intermediate fiber type, the fast muscle fiber (type IIa) that experiences the biggest increase in size (strength) in individuals who perform conventional weight lifting (heavy loads,) and body building (lighter loads, fast/explosive reps) training. Neither type of training appears to have a significant effect on the size of types I and IIx fibers.

William Kraemer,
http://www.education.uconn.edu/direc...ails.cfm?id=44

Steven Plisk,
http://www.excelsiorsports.com/files...e_Training.pdf

It’s generally understood that a certain threshold of training intensity is needed to effect positive adaptation, but many athletes and coaches still believe that resistance must be sufficient that the weight can’t — or shouldn’t — be moved very fast. I intend to challenge this proposition, and to make a case for the fact that acceleration is the name of the game even when executing basic structural movements (e.g. the squat and deadlift). It’s really just a simple matter of understanding the fundamental nature of force, and of putting this concept into practice regardless of task or workload.

F=m•a Revisited

At first glance, “force is the product of mass and acceleration” appears to imply that there is no force without motion (or vice-versa), but that’s not necessarily the case. For example, since gravity is expressed as an acceleration constant [~9.8 m/sec2], a vertical force of ~980 kg•m/sec2 (or Newtons) would be required to hold a 100 kg barbell in place statically.

Despite the apparent simplicity of F=m•a, the inability or unwillingness to grasp its functional
significance is an underlying cause of the nonsense taking place in many weightrooms. This concept is neither contrived nor trivial, and shouldn’t be tucked away in a physics textbook until needed to support some abstract opinion. In fact, it’s a foundational principle upon which all motion is based (with strength training being no exception). When you consider that any movement is essentially an act of defying gravity — which itself is an accelerative force — the central issue becomes: What is being moved, and how fast?

Vladimir M. Zatsiorsky,
http://www.hhdev.psu.edu/kines/facul...orsky%20CV.pdf

Westside Barbell,
http://www.westside-barbell.com/articles/

Dr. Yuri Verkhoshansky,
http://www.verkhoshansky.com/

Dr. Hatfield, (aka Dr. Squat)
http://drsquat.com/content/knowledge...-look-strength

Per Aagaard Professor, PhD
Institute of Sports Science and Clinical Biomechanics
University of Southern Denmark

When a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

For instance, a 120 kg squat can easily produce peak vertical ground reaction forces (beyond the body mass of the lifter) of 160-220 kg's when executed in a very fast manner! Same goes for all other resisted movements with unrestricted acceleration (i.e. isokinetic dynamometers (and in part also hydraulic loading devices) do not have this effect).

This means that higher forces will be exerted by MORE muscles fiber when a given load is moved at maximal high acceleration and speed - i.e. contractile stress (F/CSA) will be greater for the activated muscle fibers than when the load is lifted slowly...
best wishes
Per

Sorry to sound so blunt, but I don't think dragging this out any further is going to be productive.

That’s ok; you are a gentleman and a scholar, as I said before, and to all that have helped, thank you for your time and help.

Wayne

 

[JaredJames]

Well I do know what I am talking about,

Really? Then you follow with:

Acceleration is not momentum,

Good

its the opposite of momentum,

Bad

acceleration requires more force/strength, where momentum requires less force/strength.

Ugly

Momentum does not require force in any form. Any moving object has momentum but it does not need to be subject to any force.

The more you accelerate a mass, the more momentum it gains. That's the only link there is between them.

That was enough for me to not read on.

 

[douglis]

The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

jarednjames,
If we want to calculate the total energy we must take into account the whole lifting of the weight(let's say it's been lifted for a height h) which ends with zero velocity.So the whole KE,that was gained from the conversion of CE,is finaly converted to PE equal with mgh.
So the theoretical minimum of the energy expenditure when you lift a weight is mgh assuming 100% efficiency.Here's the tricky part and I'll try to make my point with a more simple example.

In the lifting weight example the starting and ending velocity is zero hence the average acceleration is zero so,for simplicity's sake,I'll use constant speeds in my example(average acceleration is zero too).
Two rockets are moving upwards with constant speed.The first with 50mph and the second with 100 mph(the air resistance is excluded).Since the speed is constant the engines of both rockets are using force equal with the rocket weight.
After an hour the engine of the the fast rocket has produced exactly double work than the slow rocket.But the total fuel expenditure will be exactly the same for both rockets.It has to be this way since the rockets use the same force for the same duration.Both the engines were just balancing the weight of the rocket for an hour.The only difference is that the engine of the fast rocket worked with double efficiency.

So I believe that in all cases where the average acceleration is zero(like in weight lifting and in the rockets example) and the same average force is being used,the rate of energy expenditure does not depend on speed and distance.

 

[JaredJames]

Two rockets are moving upwards with constant speed.The first with 50mph and the second with 100 mph(the air resistance is excluded).Since the speed is constant the engines of both rockets are using force equal with the rocket weight.
After an hour the engine of the the fast rocket has produced exactly double work than the slow rocket.But the total fuel expenditure will be exactly the same for both rockets.It has to be this way since the rockets use the same force for the same duration.Both the engines were just balancing the weight of the rocket for an hour.The only difference is that the engine of the fast rocket worked with double efficiency.

So I believe that in all cases where the average acceleration is zero(like in weight lifting and in the rockets example) and the same average force is being used,the rate of energy expenditure does not depend on speed and distance.

This is non-sense.

What you have just said is that two vehicles traveling at constant velocities (different velocities) use the same fuel. Don't be ridiculous.

You can't just exclude air resistance - obviously if you throw out the variables you get rubbish like this because your resisting force only comes from gravity which is constant. Even then you need to realize that it takes more fuel to produce the thrust in the high speed scenario.

Fuel use is not linear and so a vehicles traveling at twice the speed of the first will use significantly more fuel - even if it is at constant speed.

 

[douglis]

This is non-sense.

What you have just said is that two vehicles traveling at constant velocities (different velocities) use the same fuel. Don't be ridiculous.

You can't just exclude air resistance - obviously if you throw out the variables you get rubbish like this because your resisting force only comes from gravity which is constant. Even then you need to realize that it takes more fuel to produce the thrust in the high speed scenario.

I excluded the air resistance to create an equivalent example with weight lifting where obviously the air resistance is negligible.I thought there was no need to explain that.
So the thrust is always equal with the weight in any constant speed scenario and so is the fuels expenditure.

Fuel use is not linear and so a vehicles traveling at twice the speed of the first will use significantly more fuel - even if it is at constant speed.

Only if you take into account the air resistance.

 

[JaredJames]

I excluded the air resistance to create an equivalent example with weight lifting where obviously the air resistance is negligible.I thought there was no need to explain that.
So the thrust is always equal with the weight in any constant speed scenario and so is the fuels expenditure.

You're ignoring initial energy expenditure. The initial acceleration (as per my previous example) uses significantly more energy to achieve the required rocket velocity (or rep time) and so already you have used more energy than the slower rocket (or rep). Even if they use the same energy at constant velocity it takes significantly more to achieve (and stop) the faster one.

You can't just wave this off.

It doesn't matter what happens to the KE after the weight stops. You must provide significantly more KE in the faster rep and so that means more CE. As none of that KE gets translated back into CE it's irrelevant and you have the energy expenditure being higher for the faster rep.

Fuel use is not the same.

Only if you take into account the air resistance.

As above re energy expenditure.

Douglis, I'm slightly concerned your post came almost perfectly from here:

http://lofi.forum.physorg.com/Energy-In-Bench-Press_28867.html

Same debate by the look of it. Certain posts / responses are identical to here. Only we seem to be on a one month delay.

 

[douglis]

You're ignoring initial energy expenditure. The initial acceleration (as per my previous example) uses significantly more energy to achieve the required rocket velocity (or rep time) and so already you have used more energy than the slower rocket (or rep). Even if they use the same energy at constant velocity it takes significantly more to achieve (and stop) the faster one.

You can't just wave this off.

Again to create an equivalent example with weight lifting we must assume that the rocket stops at the end only due to gravity and not with opposite thrust.
So...the initial acceleration(which requires more fuels) is exactly balanced by the final deceleration(requires less fuels).On average the fuels are identcal as if the motion would be constant.After all that's what practically means zero average acceleration.

It doesn't matter what happens to the KE after the weight stops. You must provide significantly more KE in the faster rep and so that means more CE. As none of that KE gets translated back into CE it's irrelevant and you have the energy expenditure being higher for the faster rep.

Fuel use is not the same.

Again the same thing with other words.
The KE is gained during the acceleration phase.The more CE that is spent in a fast rep(and is converted to KE) during the initial acceleration,compared with the slow rep,is exactly balanced by the less CE that is spent during the final deceleration(where the gained KE is converted to PE).
The less mechanical energy(PE+KE) that is produced per unit of time in a slower rep is balanced by more heat production.It MUST be this way since the same average force is used per unit of time.
That's also the case with the constant speed rockets.The more mechanical energy that the faster rocket uses is exactly balanced by the more heat that the slow rocket produces.They spend exactly the same fuels with the difference that the engine of the slow rocket works less efficiently.

As above re energy expenditure.

Douglis, I'm slightly concerned your post came almost perfectly from here:

http://lofi.forum.physorg.com/Energy-In-Bench-Press_28867.html

Same debate by the look of it. Certain posts / responses are identical to here. Only we seem to be on a one month delay.

Yes...an old debate.I wasn't the one who brought it back.

 

[Dale]

Hi Wayne,

I just noticed this thread despite that it has been ongoing for a while. Some time back you and I discussed the concept of "average force" in quite some detail. If you will recall, as long as you start and stop in the same location each rep the average force is always equal and opposite to the gravitational force (remember that force is a vector quantity). The speed at which you do a rep does not influence the average force at all. I think that the conclusion from that previous thread was essentially that average force is not a useful measure for what you are really interested in.

You may be interested in a more realistic model of muscle contraction:
http://en.wikipedia.org/wiki/Hill's_muscle_model

 

[JaredJames]

Again to create an equivalent example with weight lifting we must assume that the rocket stops at the end only due to gravity and not with opposite thrust.
So...the initial acceleration(which requires more fuels) is exactly balanced by the final deceleration(requires less fuels).On average the fuels are identcal as if the motion would be constant.After all that's what practically means zero average acceleration.

What are you talking about? In both cases (fast and slow rocket) if they use gravity only to stop they don't use any fuel. Meaning the only important parts are initial acceleration and energy requirement for constant velocity. Hence the faster rocket requiring more fuel.

If you have a fast rep and a slow rep, both covering the same distance, you will need to use energy to stop the faster one if the slower one is only using gravity to stop. Meaning the fast rep requires additional energy to stop.

Why are you trying to say the fast rep uses less energy to stop than the slow rep? That's the only way the energy values would balance out.

Slow rep: CE to KE to get the weight moving. KE used to overcome resistance. Gravity decelerates the weight to a stop, but CE is still used to support it at the peak.

Fast rep: More KE required so more CE to get the weight moving. KE used to overcome resistance. Gravity plus additional CE required to decelerate the weight over the same distance to peak, again CE is still used to support it at the peak.

The fast rep requires energy to slow it. More KE is dissipated as heat in the fast reps.

Either way, the fast rep requires more CE (significantly more) and the two don't balance out at the end. Don't confuse the final GPE values - which are equal with both weights - with the fact the fast rep has more energy during travel and as such has to dissipate it as heat to bring it to a halt.

If allowed to stop solely using gravity, the faster rocket / rep will travel further. You must introduce resistance (from thrusters / muscles) in order to stop them in the same distance.

 

[douglis]

What are you talking about? In both cases (fast and slow rocket) if they use gravity only to stop they don't use any fuel. Meaning the only important parts are initial acceleration and energy requirement for constant velocity. Hence the faster rocket requiring more fuel.

The deceleration phase(where the gravity stops the rocket) is much longer at the fast rocket and proportional with the acceleration phase.

If you have a fast rep and a slow rep, both covering the same distance, you will need to use energy to stop the faster one if the slower one is only using gravity to stop. Meaning the fast rep requires additional energy to stop.

Why are you trying to say the fast rep uses less energy to stop than the slow rep? That's the only way the energy values would balance out.

Slow rep: CE to KE to get the weight moving. KE used to overcome resistance. Gravity decelerates the weight to a stop, but CE is still used to support it at the peak.

Fast rep: More KE required so more CE to get the weight moving. KE used to overcome resistance. Gravity plus additional CE required to decelerate the weight over the same distance to peak, again CE is still used to support it at the peak.

The fast rep requires energy to slow it. More KE is dissipated as heat in the fast reps.

Either way, the fast rep requires more CE (significantly more) and the two don't balance out at the end. Don't confuse the final GPE values - which are equal with both weights - with the fact the fast rep has more energy during travel and as such has to dissipate it as heat to bring it to a halt.

jarednjames...have you ever tried to lift weights?You always rely on gravity in order to stop the weight.You never use force from the antagonist muscle(opposite thrust) to do the job.
That means the greater initial acceleration in faster lifting is ALWAYS accompanied by greater final deceleration.The average acceleration is ALWAYS zero or else the weight will fly from your hands.

If allowed to stop solely using gravity, the faster rocket / rep will travel further. You must introduce resistance (from thrusters / muscles) in order to stop them in the same distance.

Here's another example to help you understand my point.

Let's say a rocket starts accelerating from the Earth upwards and after a while the engine shuts down.The rocket decelerates and reaches a maximum height before it will start falling.
Let's say it reached the max H in exactly one minute.The average acceleration for that trip is zero(starting and ending velocity zero) hence the average force is exactly equal with the weight of the rocket.
Now if the rocket was just standing still in the air for that minute again its engine would have used force equal with the rocket's weight for 1min.

In both cases the rocket will use exactly identical fuels since it used the same average thrust for 1 min.

 

[JaredJames]

douglis, I fully understand that two rockets moving at different speeds will stop in different distances when subjected to the same deceleration (gravity).

However, if you have two rockets moving at different speeds and constrain the stopping area to the distance the slowest takes to stop, you must apply additional deceleration to the faster one in order to achieve the required stopping distance.

In this case with the reps, the constraint placed earlier was that they are both the same distance traveled for the weight (my example was 1m).

In which case applying equal deceleration won't stop them in the same time.

A basic example is a car at 30mph will coast to a stop in Xm. To get a car going 60mph to stop in Xm you need to apply the brakes.

That means the greater initial acceleration in faster lifting is ALWAYS accompanied by greater final deceleration.

Gravity only applies one acceleration. So whether you are traveling at 1m/s or 100m/s it is always the same acceleration due to gravity.

The only way to increase the final deceleration is to apply additional force. If you don't, the faster rep will continue until it stops on its own - not within the described limit.