Impulse/force in pounds for the time frame

[waynexk8]

But in full tomorrow night and I must apologise for not answering some posts yet, I will get back to them, and thank you all again.

The below is again for D. it’s from “another” EMG expert. And would be interested to see how/or why he was wrong, and thought my EMG was wrong, mind you it’s not the first time you have been wrong, you had metabolism wrong, energy wrong, your rocket theory, now EMG, however we all get things wrong and you’re a very clever person. I got my training wrong for 15 years, doing the slow, so I cannot pass judgment.

Hi Wayne,

I got the email you sent to Motion Labs - we have a cooperation with them. The company I work for - prophysics AG in Zurich, Switzerland - market and sell the myon EMG system as well as the proEMG data capture and analysis software.

You can check out details of our prophysics on www.myon-prophysics.ch

I read through your email. EMG measures the electrical activity of muscles using electrodes attached to the skin surface above the muscle's belly. To answer your direct question, whether to use Integrated or RMS calculation, the answer is probably RMS. The reason for this is that the RMS calculation is the "standard" way of representing the average voltage amplitude of the signal, and this is correlated with the muscle power.

Wayne

 

[sophiecentaur]

Wayne
For someone who claims to know nothing of Physics, you have become very assertive and a bit too stroppy for my liking.
Nothing I have ever written about the Physics of this topic is wrong. It's just too basic and glaringly obvious. The same goes for Douglis's statements about Physics.

The stuff about EMG results is not the Physics of moving masses up and down. I's about what your Muscles are doing. The fact is that, even when lowering something, your muscles are using energy (unless you just drop it on the floor). Hence, there is no direct relationship between Energy Expended in your muscles and Work Done on the Weights you lift. That is the beginning and end of the whole topic and neither you nor your fellow EMG users can change it.

The readings on your EMG machine can give you a lot of useful information about your Body and that's all.

I can't actually understand what you are trying to achieve here and slagging people off for making statements that you clearly don't actually understand is not getting us anywhere.

 

[douglis]

it’s not the first time you have been wrong, you had metabolism wrong, energy wrong, your rocket theory, now EMG, however we all get things wrong and you’re a very clever person.

Show me where I was wrong.
From physics point of view...you can not possibly know if you spend more energy by lifting and lowering a weight than just holding it for the same duration.Only biology has the answer...so I was never wrong.It was supposed to be a physics discussion.

Hi Wayne,

I got the email you sent to Motion Labs - we have a cooperation with them. The company I work for - prophysics AG in Zurich, Switzerland - market and sell the myon EMG system as well as the proEMG data capture and analysis software.

You can check out details of our prophysics on www.myon-prophysics.ch

I read through your email. EMG measures the electrical activity of muscles using electrodes attached to the skin surface above the muscle's belly. To answer your direct question, whether to use Integrated or RMS calculation, the answer is probably RMS. The reason for this is that the RMS calculation is the "standard" way of representing the average voltage amplitude of the signal, and this is correlated with the muscle power.

Wayne

Check the graph.Anyone who has the intelligence to breathe can understand that the RMS amplitude is NOT the average.

 

[waynexk8]

Wayne
For someone who claims to know nothing of Physics, you have become very assertive and a bit too stroppy for my liking.

Sorry about that, what part did you mean please ? I am not that type honest, it’s just me a D. have been debating so long, we have often said the odd thing to each other, nothings meant by it honest.

I did not say I knew nothing of physics, it’s more the physics theory I do not know.

Ok, “please” sophiecentaur, one of the things that could have made me stroppy, is that I do not get why you or D. cannot answer this one, please “could” you ?

1,
I lift a weight up from a motionless start, up 500mm in .5 of a second.

2,
A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second.

3,
As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ?} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ?

4,
How much more force please.

5,
If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ? 50% more force for .2 of a second, then the same force as the first lift ?

6,
When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ? I would say it’s the motionless lift, am I right or wrong please ?

Nothing I have ever written about the Physics of this topic is wrong. It's just too basic and glaringly obvious. The same goes for Douglis's statements about Physics.

How is it obvious ? As when using 80% of your 1RM {Repetition Maximum} you fail to lift the weight again, or you hit momentary muscular failure 50% faster on the faster reps, this must surely mean you have used up your temporary force faster, it can’t mean you have used up your temporary force slower, or you would still have force left, but you don’t, you don’t have any force left, so that can mean one think only, that on the fast you are using more force per unit of time, if not please explain your way of thinking. I use more energy in the same time frame; I move the weight far far far further in the same time frame.

I never said you did wrote anything wrong, I might have said I think you could have left something out, like not adding in all the variables, or kinology, or like in my 1 to 6 above. But what I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ? Its impulse, force with respect to time. A 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. But in our case the reps, lifts are done for the same time frame.

The stuff about EMG results is not the Physics of moving masses up and down. I's about what your Muscles are doing. The fact is that, even when lowering something, your muscles are using energy (unless you just drop it on the floor). Hence, there is no direct relationship between Energy Expended in your muscles and Work Done on the Weights you lift. That is the beginning and end of the whole topic and neither you nor your fellow EMG users can change it.

The EMG does not compute energy ? The EMG computes the electrical signals the muscles give out, the more electrical signal = more muscle activity = more muscle force, of can we say muscle strength used. So I think it is the physics of the forces, and its adding in, or saying that the faster you move, the more force is used, its adding in the peak forces of the faster accelerations of the faster reps, and in my opinion it shows that the slow forces cannot make up or balance out the higher force or tensions that the faster has put on the muscles, thus it reads out higher for the fast. That is why more energy is used; it has to use more energy as of the higher forces from the higher velocities and accelerations. Also that’s why the fast mores the weight 6 times further, as it takes more force to move a weight further in the same time frame, yes ?

The readings on your EMG machine can give you a lot of useful information about your Body and that's all.

I can't actually understand what you are trying to achieve here and slagging people off for making statements that you clearly don't actually understand is not getting us anywhere.

Sorry honestly did not try to slag you off, not sure where you think I did that, but if you think that, I will, and will do it now, I apologise for anything I have said, and it was not meant at all.

Ok, I imagine you know or have heard what a force plate/platform is ? If I buy or have tests done, and that state the average and total or overall forces are higher in the fast, like the EMG states, what would you say then ?

http://en.wikipedia.org/wiki/Force_plate

Will get back to the others tomorrow. Again, thank you for your time and help; I only want a friendly debate.

Wayne

 

[sophiecentaur]

I did not say I knew nothing of physics, it’s more the physics theory I do not know.


Wayne

This absolutely sums up what's wrong with your approach. Physics IS THEORY. If you don't use the theory then anything you have to say is just idle chit chat.

It explains why you seem incapable of stating a question in less than ten paragraphs and incapable of understanding an answer that takes up a single line.
Would you ever ever ever challenge a mathematician about something on the grounds that you know Maths but not the 'theory'?
You do not believe that Physics cannot answer your question because you just don't know the Physics. Until you are prepared to learn some of the 'theory' you will never get this. Stick to arm waving.

 

[waynexk8]

This absolutely sums up what's wrong with your approach. Physics IS THEORY. If you don't use the theory then anything you have to say is just idle chit chat.

It explains why you seem incapable of stating a question in less than ten paragraphs and incapable of understanding an answer that takes up a single line.
Would you ever ever ever challenge a mathematician about something on the grounds that you know Maths but not the 'theory'?

Sorry, yes my writings are long.

If I had four marbles on the table, and the mathematician took 2 away, and said I still have 4 there, then yes I would challenge his theory with my practical World experiments.

You do not believe that Physics cannot answer your question because you just don't know the Physics. Until you are prepared to learn some of the 'theory' you will never get this. Stick to arm waving.

I am trying to learn, however I think you could also try to see how theory does not always provide the truth, as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ? Also, how does the fast move the weight 6 times futher in the same time span, how does the fast use more energy in the same time span, if its not as you two say using more force up per unit of time.

Physics, or right/correct physics, is only right/correct physics, when the theory is proven with practical experiments, unless that, it stays an unproved theory.

I am not arm waving, I have come here asking and hoping for answer to questions, so I ask quite simple questions like the below, but I get no answer, please I do not understand why you can’t answer the below. Not being sarcastic here, but it’s the only thing I can think of, you do understand what I am saying/asking below ? If not please say, as its looks very straight forward to me, please tell me if the below is the wrong way to ask, or just say why you will not answer, as I do not get it. I can here to ask some friendly polite questions, and it seems more of a match to put me down, I just don’t understand this.

1,
I lift a weight up from a motionless start, up 500mm in .5 of a second.

2,
A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second.

3,
As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ?} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ?

4,
How much more force please.

5,
If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ? 50% more force for .2 of a second, then the same force as the first lift ?

6,
When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ? I would say it’s the motionless lift, am I right or wrong please ?

Nothing I have ever written about the Physics of this topic is wrong. It's just too basic and glaringly obvious. The same goes for Douglis's statements about Physics.

How is it obvious ? As when using 80% of your 1RM {Repetition Maximum} you fail to lift the weight again, or you hit momentary muscular failure 50% faster on the faster reps, this must surely mean you have used up your temporary force faster, it can’t mean you have used up your temporary force slower, or you would still have force left, but you don’t, you don’t have any force left, so that can mean one think only, that on the fast you are using more force per unit of time, if not please explain your way of thinking. I use more energy in the same time frame; I move the weight far far far further in the same time frame.

I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ? Its impulse, force with respect to time. A 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. But in our case the reps, lifts are done for the same time frame.

Originally Posted by sophiecentaur
The stuff about EMG results is not the Physics of moving masses up and down. I's about what your Muscles are doing. The fact is that, even when lowering something, your muscles are using energy (unless you just drop it on the floor). Hence, there is no direct relationship between Energy Expended in your muscles and Work Done on the Weights you lift. That is the beginning and end of the whole topic and neither you nor your fellow EMG users can change it.

The EMG does not compute energy ? The EMG computes the electrical signals the muscles give out, the more electrical signal = more muscle activity = more muscle force, of can we say muscle strength used. So I think it is the physics of the forces, and its adding in, or saying that the faster you move, the more force is used, its adding in the peak forces of the faster accelerations of the faster reps, and in my opinion it shows that the slow forces cannot make up or balance out the higher force or tensions that the faster has put on the muscles, thus it reads out higher for the fast. That is why more energy is used; it has to use more energy as of the higher forces from the higher velocities and accelerations. Also that’s why the fast mores the weight 6 times further, as it takes more force to move a weight further in the same time frame, yes ?

Wayne

 

[waynexk8]

Show me where I was wrong.
From physics point of view...you can not possibly know if you spend more energy by lifting and lowering a weight than just holding it for the same duration.Only biology has the answer...so I was never wrong.It was supposed to be a physics discussion.

D. you can surely work out the power used in both ? Surely you must know that if you move something further in the same time frame you need more energy, that’s goes for everything, then it’s in every calorie counting book and site and nutritionist have know this for 100s of years, I also showed you how they work it out in a sealed room. Then after years it was a physicist who worked it out for you with physics and told you.

LOOK, please let’s forget about that, as you did immediately say you was wrong when you was, I salute you for that, I will and do not want to rub it in every again, so sorry.

No its not just a physics debate, it’s a World wide debate, and whatever needs to be added in, to answer the question, will be.

Problem is, you do not answer questions.

Check the graph.Anyone who has the intelligence to breathe can understand that the RMS amplitude is NOT the average.

So now you are saying that two people that make work with and use EMG machines are wrong ? Please tell me what your graph is supposed to mean and why you made it that way, in other words please explain it in full, you never explain anything, then please send an E-mail to these people with my return headings, and tell them how/why you think they are wrong.

E-mail, sales.i@motion-labs.com

Subject, Re: [From Website] to Edmund Cramp.

E-mail, roren@prophysics.ch

Subject, Re: Your email regarding EMG systems

Then please try to answer the below, and if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ? Did I use up more because I was using more force per unit of time, the same, or less. But the main issue is the force, the energy is “only” the supply, and we are on about the force, the energy supply. as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ? Also, how does the fast move the weight 6 times further in the same time span, how does the fast use more energy in the same time span, if it’s not as you two say using more force up per unit of time.

Wayne

 

[sophiecentaur]

Only an instrument that measures what is happening to the weights themselves can measure the work involved and the forces involved. The EMG doesn't do this so it is not contradicting the Physics.
You are clearly not "trying to learn". Not even one small step at a time. Because you disregard every basic part of theory, preferring the 'but surely' argument and 'it stands to reason'.
You dismissed my analogy concerning a mathematician by quoting an example where the Mathematician was not right and had no proper reasoning behind his result. But you argue against the RMS thing and that is totally Maths.

 

[douglis]

LOOK, please let’s forget about that, as you did immediately say you was wrong when you was, I salute you for that, I will and do not want to rub it in every again, so sorry.

As always you misunderstood what I admited I was wrong.
I was wrong from biology point of view.From physics point of view I was perfectly right and you're saying nonsense.You can't possibly know if you spend more energy if you lift a weight 10 times up and down or you just hold it for 10 seconds.

No its not just a physics debate, it’s a World wide debate, and whatever needs to be added in, to answer the question, will be.

OMG...you're totally delusional!What world wide debate you idiot?
It's just a couple of us with superhuman patience trying to explain basic physics to you.

So now you are saying that two people that make work with and use EMG machines are wrong ? Please tell me what your graph is supposed to mean and why you made it that way,

The graph is not mine it's from the first site I found when I googled "rms amplitude".

The RMS is not the average.This is not offered for discussion...it's basic maths.You don't have to mail professors!

Check the below numbers:
-2, 5, -8, 9, -4

Their average is 0 but their RMS is 6.16.
The RMS is NOT the average.It's the quadratic mean.You probably don't have a clue what that means so you have to trust the link and me.
http://www.analytictech.com/mb313/rootmean.htm

Then please try to answer the below, and if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ?

I've tried so many times to explain that the force-energy relation is not linear and greater energy expenditure doesn't equate greater force.
I'm sure by now that that's way beyond your intelligence.

 

[douglis]

I stop here any physics related discussion with Wayne because obviously is a waste of time.
What's the point anyway?The below study is exactly what he's looking for.

as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time

Wayne

I have shown you the absolute proof that this is nonsense with a real study but you choose to ignore it.

Take a look again:

The fast push ups failed at 81.2 sec while the slow push ups failed at 101.2 sec.It's obvious that the fast push ups had greater rate of energy expenditure(see table 1).

BUT the Total Muscle Activation(for the pectoralis major for example) for the fast push ups is 2114.23 while for the slow push ups is 3121.81(see table 3).

So the muscle activation per second for the fast push ups is 2114.23/81.2=26.04
and the muscle activation per second for the slow push ups is 3121.81/101.2=30.85

http://jmbe.bme.ncku.edu.tw/index.php/bme/article/viewFile/635/839
Conclusion:
greater rate of energy expenditure does NOT equate greater muscle tension per unit of time.In fact,in the case of push ups,the exact opposite is true.

End of discussion.

 

[waynexk8]

Only an instrument that measures what is happening to the weights themselves can measure the work involved and the forces involved. The EMG doesn't do this so it is not contradicting the Physics.

It’s the muscles themselves that are exerting the force. For every reaction, there is an opposite reaction, so if you measure the work or the force from the muscles or the force moving/on the weight, you will find the reading.

But it’s the muscles forces we are after, or the tensions on the muscles, that’s the forces from the muscles forces onto the weights, with the opposite reaction forces creating tensions on the muscles, and as the EMG takes the muscle activity = muscle force, more muscle activity = more muscle force, yes ?

When you say work, do you mean the mechanical work, the product of a force times the distance through which it acts, work of the force, we know the fast does more work, or more work of the force, so is not that more force if it’s done more work, work of the force ?

You are clearly not "trying to learn". Not even one small step at a time. Because you disregard every basic part of theory, preferring the 'but surely' argument and 'it stands to reason'.

Sorry, when I say surely, I am more thinking out loud; please don’t think anything of it.

You dismissed my analogy concerning a mathematician by quoting an example where the Mathematician was not right and had no proper reasoning behind his result. But you argue against the RMS thing and that is totally Maths.

See your point on the analogy concerning a mathematician; yes suppose I was wrong there, but just making a point.

Another thing, I have not been explained what RMS is ? But then again, I could say that you are now disagreeing with people who make and work with EMG machines, and two of these stated to measure the total or overall force output of the muscles, RMS is about the best.

"Please" could you ansner this ? If not please state why.

1,
I lift a weight up from a motionless start, up 500mm in .5 of a second.

2,
A weight is being lowered under control, moving 500mm in .5 of a second, the mm before it hits the ground, I then have to try and lift it up 500mm in .5 of a second.

3,
As of the acceleration components, when moving down with that acceleration, the weight will appear {not sure of the right physics saying for this sorry ?} to be heaver, or have more force than if it was motionless. So it is going to be heaver/harder to lift, thus I “will” need more force than the first lift, think we all agree there ?

4,
How much more force please.

5,
If you can’t work out how much more force, and please you can add in your own accelerations for this if you need to. But if you can’t work it out, do you agree that the force needed would be about, and this is a very rough guess ? 50% more force for .2 of a second, then the same force as the first lift ?

6,
When you two, or anyone else are compeering my fast lifts force, and your slow lifts force, which modal are you compeering your lift two, the lift started from a motionless start, or the lift that is being lowered first then lifted ? I would say it’s the motionless lift, am I right or wrong please ?


Wayne

 

[sophiecentaur]

I am not disagreeing with what the EMG people are saying. The problem is that you do not seem to understand what they are saying and how it applies to your question.
Why should you need to have RMS explained any more. It isn't an explanation that you need. What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

I cannot be bothered to answer that question because it is just a smoke screen to protect you from going to the trouble of working stuff out for yourself.

 

[waynexk8]

As always you misunderstood what I admited I was wrong.
I was wrong from biology point of view.From physics point of view I was perfectly right and you're saying nonsense.You can't possibly know if you spend more energy if you lift a weight 10 times up and down or you just hold it for 10 seconds.

YES you can, I don’t understand why you don’t know this or could think other, you use more force on the acceleration and on the higher vilocity, will explain this in full tomorrow. Do a search on physics and energy used uin distance and or disstnce per unit of time.

Then do another sherach, on room calorimetry. the 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.

jarednjames wrote;Let's say it takes 10N to lift the weight in 1 second (a = f/m), if you apply 20N it would only take 0.5 seconds and so on. But the force applied per unit time is the same.

The difference comes when you look at energy. The kinetic energy of moving the weight (100lbs) at an average velocity of 1m/s is 22 joules, but the KE of moving the weight at 2m/s is 90 joules. So the higher the average velocity of the weight, the more the energy use increases - which is why you get tired quicker.

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.
* All figures are for guidance only.

jarednjames wrote;
To reduce the time to move an object a distance of 1m, you have to increase the force. By increasing the force, the acceleration becomes larger:

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.

OMG...you're totally delusional!What world wide debate you idiot?

This has been going on for many years on different forums with different people.

It's just a couple of us with superhuman patience trying to explain basic physics to you.

If you as you do, claim you are right, then how do you account for the following.
1,
As when using 80% of your 1RM {Repetition Maximum} when you fail to lift the weight again, or you hit momentary muscular failure 50% faster on the faster reps, this must surely mean you have used up your temporary force faster, it can’t mean you have used up your temporary force slower, or you would still have force left, but you don’t, you don’t have any force left, so that can mean one think only, that on the fast you are using more force per unit of time. I use more energy in the same time frame; I move the weight far far far further in the same time frame.

And if you say I run out of force because of I used up the energy fast, I know that, the question is sort of why did I use up more energy doing something faster ? Did I use up more because I was using more force per unit of time, the same, or less. But the main issue is the force, the energy is “only” the supply, and we are on about the force, the energy supply. as in the fast rep fails at lifting first, that “must” mean, that the person doing the fast, used more muscle force per unit on time, if not why does the fast have no force to pick up the 80 pounds, but the slow does ? Also, how does the fast move the weight 6 times further in the same time span, how does the fast use more energy in the same time span, if it’s not as you two say using more force up per unit of time.

If not please explain your way of thinking, please for once answer.


2,
I don’t understand, and no one have tried to explain, is that when my peak acceleration forces are say a 100% or 100 pounds of force, how do you think that your 80% or 80 pounds of force can make up the 100 pounds of force, when its only 80 pounds, how can 80 ever be as high as a 100 ? The only way the 80 pounds of force could make this extra force up, or the shall we say the 100 pounds of tension on the muscles to the 80 pounds of tension on the muscles, is if the 80 pounds of tension was on the muscle far far far longer, please do you agree there, if not, how does the 80 pounds of tension make up the higher 100 pounds of tension on the muscles, as a 80 pounds tension, can never be as high as the 100 tension, as how could 80 pounds feel like 80 pounds on the muscles ? Its impulse, force with respect to time. A 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. But in our case the reps, lifts are done for the same time frame.

If not please explain your way of thinking, please for once answer.

The graph is not mine it's from the first site I found when I googled "rms amplitude".

Yes but what do you think it means, and why did you show it ?

The RMS is not the average.This is not offered for discussion...it's basic maths.You don't have to mail professors!

I never said it was.

Check the below numbers:
-2, 5, -8, 9, -4

Their average is 0 but their RMS is 6.16.
The RMS is NOT the average.It's the quadratic mean.You probably don't have a clue what that means so you have to trust the link and me.
http://www.analytictech.com/mb313/rootmean.htm

GOD, I “think” I know what you two are doing wrong. No wonder its what or why the EMG experts use it, no time to explain now, as not 100% sure, but you should be able to see why, if you try to understand why am EMG expert would ”have” to use the RMS or quadratic mean. Problem is, why are you not using this ? As we are on about the total or overall muscle activity, force.

RMS or quadratic mean will used in the situations, where it is the square of the values that matters; an electrical current squared will be proportional to the power. A quadratic mean, RMS should be used for periodic data (i.e., over time, a graph describes a sine curve), or when both positive and negative data are included, and what you seek is an "average distance from zero" for the various data point. The quadratic mean, a statistical measure of the magnitude of a varying quantity. It is especially useful when varieties are positive and negative.


You times it by two; you're multiplying a number by itself. Five squared is 25, five times five is 25.


The problem is, why are you two not adding in the negative force ?

I've tried so many times to explain that the force-energy relation is not linear and greater energy expenditure doesn't equate greater force.
I'm sure by now that that's way beyond your intelligence.

I know it’s not liner, try and explain on this physics forum then.

Wayne

 

[waynexk8]

I am not disagreeing with what the EMG people are saying. The problem is that you do not seem to understand what they are saying and how it applies to your question.
Why should you need to have RMS explained any more. It isn't an explanation that you need. What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

I cannot be bothered to answer that question because it is just a smoke screen to protect you from going to the trouble of working stuff out for yourself.

I will tomorrow, thanks for your time and help, just going out.

This should interest you, see table 2.

Force-velocity, impulse-momentum relationships:

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf


Wayne

 

[sophiecentaur]

I have read the paper an I think I can see your problem. I find little to disagree with what is written.
They are not considering the whole cycle (lift/lower) when they refer to mean force. They are not only discussing free lifts but machines that present RESISISTIVE loads. Everything changes in that case because you are not just changing gravitational potential energy in that case but work is being done in overcoming friction.
If you had read what they say then you would not think they are disagreeing with established physics at all. You did not understand what you were arguing about because you insisted on giving details instead of condensing your questions into something meaningful.

 

[Dale]

Hi waynekx8, I didn't notice that you were back.

It looks like you are still confusing "work done" and "energy expended" as we discussed last year: https://www.physicsforums.com/showthread.php?p=3190515#post3190515

 

[douglis]

They are not considering the whole cycle (lift/lower) when they refer to mean force. They are not only discussing free lifts but machines that present RESISISTIVE loads.

sophiecentaur...if you read the ''methods'' paragraph the authors admit that they record ONLY the ''propulsive'' force and not the whole average force for the lift.By ''propulsive'' force they mean the additional than the weight force in order to accelerate.With the method they use they can measure only the positive value of the propulsive force.The negative values in order to decelerate are not recorded.

 

[sophiecentaur]

So the whole thread is GIGO, as usual. And I wonder why they use a term like "propulsive force" which has no strict definition faik.

You need to read more carefully, Wayne.

 

[waynexk8]

Will be back later to get back to some of the old questions I have missed.

Here D, I double checked, what do you say now on this and the other mails regarding RMS and EMG ? I understand why they say to usde RMS, and the other methords, do you now understand why you use ROM ?

Hi Wayne,
There are a few research papers floating around that suggest that the RMS envelope of EMG activity approximates to the force exerted by a muscle. There are several different methods of integrating and averaging muscle activity - all produce estimates of muscle force, the RMS method appears to be slightly more favored but I don't believe that there's any solid proof that it's any more accurate than any other method - the different methods are discussed in the EMG section of the Craik and Otis book which goes into a reasonable amount of detail while remaining understandable.
Regards,
Edmund Cramp
--
<< Personal contact information removed by Moderators >>
Wayne

 

[sophiecentaur]

Wayne still doesn't understand that answer.
I notice he doesn't ever quote from a long correspondence with any of his other sources.

 

[waynexk8]

That's interesting. The TMA is the sum of all activity over a period of time and that is going to depend on a lot of things, I reckon. It seems to me that we evolved with a certain minimum level of activity when doing heavy work so 'slow' work may just not be as efficient. In fact the whole system is not efficient; if it were, we'd need to have ratchets to lock our limbs in elevated positions once they were up there - rather than burning up energy and hurting ourselves as we do. (I believe horses lock their legs when they sleep standing up)
As you say, very little of this is Physics and Wayne seems to think that one can be 'bent' to fit the other.
It's been entertaining, though, at times.

Dont understand why you don't think lifting and lowering weight and the human movment is not physics ? Its physics, biomacanics and Kinology.

Check out page 49 and 50, and have a look at the physics equastions there.

http://books.google.co.uk/books?id=2JI04kdV9isC&pg=PA41&source=gbs_toc_r&cad=4#v=onepage&q&f=false

The muscles have several different muscle fibers, but basically there are the fast and slow. The fast muscle fibers are used for more explosive movements, like heavy lifting for 10 to 20 seconds, 100m sprint. Slow muscle fibers are for the more endurance, lifting very light weights for several minutes, marathon running.

However, the slow muscle fibers have very little capacity for growth, if any. But the fast muscle fibers have far great capacity for growth, these are the ones you need to train and focus on. And these muscles fibers are best trained in the manner they evolved for. There are many ways to do this, but on average the time spent training these will be with 80% training them for multiple tension with reps from 6 to 20 fast explosive reps. At times you can go as low as 3 or maybe up to 30. One other thing about using high force acceleration explosive reps, is you can use far higher weights, and I think that’s self explanatory.

Wayne

 

[waynexk8]

Wayne.
You produce too much writing for anyone to be able to read.
You claim to have "proved" that Physics is wrong. Really??

No not physics is wrong; it’s just that all the variables have not been added in, like the EMG says the average/mean forces are higher in the fast.

YOUR Physics may be wrong but you do not explain what yor Physics actually is.

My (the Physics that everyone else recognises) says that the "average" (=mean?) force is equal to the weight.

Why is your average/mean force equal to the weight being moved ? The average/mean force would be equal to the force moving the weight ? So if the weight is 80 pounds and the force 100 pounds, why do you say the force is only 80 pounds and not a 100 pounds ? Are you taking about a certain time frame, as if I exerted 100 pounds of force on 80 pounds for .2 of a second and immediately stopped your saying that my average/mean force was only 80 pounds, when I am saying I exerted 100 pounds, so the average/mean force should be 100 pounds ?

My physics does not have a "temporary" force,

What I meant is that if I pushed with a force of 100 pounds for .1 of a second, that would be a temporary force. And if I tried to push the same weight for 10 seconds with a 100 pounds of force, then the force I tried to use, would have many Milly seconds of temporary force at different forces, as I could not keep up of push with 100 pounds for the ten seconds, so all forces are temporary.

it has a strict definition for RMS

Right, if I push with a positive force of over 80 pounds on the 80 pounds that’s a positive force, but if I push with a positive force of under 80 pounds, that’s still a positive force, and this force that’s under 80 pounds HAS to be taken account and added in for the total force, what it seems you and D. are doing, is leaving out this force for some unknown reason, I don’t get this could you explain, as all forces big or small, must be added in, as they are forces used by the muscles, thus tension are from big and small forces, you can’t leave them out.

and only uses the term "total force" to describe the sum of forces applied at one instant of time.

Right, so one instant could be .1 of a second, thus you would have to add all the forces up over the say 6 second time frame.

The above applies to PF as a whole. If you want a PF discussion then I suggest you use PF terms and not use your own terms, which have no meaning.
You claim you are not being arrogant but what other word is there for it?

It’s hard for me, it’s like you coming to my Wrought Iron Workshop, you or anyone would need a 5 year course, same as me learning this, I am honestly trying.

Wayne

 

[sophiecentaur]

I get the impression that the above is an attempt to justify yourself more than anything else.
You make the point that Physics is involved in all of this. Well, that is obvious; I am aware of the Physics involved in pretty much the whole of my life. The point is that your contributions have involved virtually no real physics and the questions have been posed in 'not-real-physics[ terms. You prove my point in what you write about muscle fibres and their behaviour - interesting stuff but you quote no Physics in there because it is not relevant to the information.

You seem to have failed to pick up on the point that the forces dealt with in the information you have quoted to us with do NOT relate to weight-only situations. Any reciprocating movement which involves friction will involve extra work on top of weight times height lifted. This is why you seem to think that what I, for one, have been saying, is wrong. That extra consideration of friction changes everything. Do you understand that? If you were constantly to measure and record the force between your hands and the handles / bars you are moving and you also measured the changing position then you could calculate accurately, the actual Work Done on the load. But it would still not tell you how hard your muscles have been 'working'. (I have made this point several times already and the EMG information says as much, too).

What the EMG is trying to do is to tell you about is muscular behaviour. It also mentioned that RMS values give a good indication of that. Fair enough but, as you don't seem to cotton on to what RMS actually involves, then you may not understand what they have written.

There really is no hope for you, I fear. You are still using this expression "total force" but you haven't been consistent in what you mean by it. The fact is that the force you exert (that is a total force) is sometimes more and sometimes less than the weight on a free lift. If you examine what a whole lift actually feels like, you will see that's true - particularly with an 'easy' weight, which will feel lighter as you near the top and it slows down. I(f you moved up fast enough, you could actually throw the weight up and there would be no force on your hands.) The mean force, including all parts of the lift, will be just the weight. If you put a brake on the process (in some sort of machine) then this will no longer be true as you'd need a force even without the weight being there.
btw do you write Milly seconds just to wind people up? Or don't you know what ms are?

If I came into your wrought iron workshop then I would not have the nerve to tell you how to do your job and I would make every attempt to talk 'your language' about your work. You could, at least, be polite enough to do that where Science is concerned. Your use of 'mangled terms is very disrespectful of the Subject as you can see the language used correctly on every response you get. Despite your protestations, you have demonstrated that you have not in fact, been prepared to learn as you still maintain the use of your nonsense terms.

I think we have been very patient with you aamof. I cannot imagine you would be similarly indulged by the other correspondents sustaining such a long thread of conversation. They would soon get too exasperated to respond.

 

[waynexk8]

I do not subscribe to the term "total force". There is no such quantity in Physics. So the rest of your post means nothing, I'm afraid.

Hmm, ok what could or can we call this ? Is there a word for this in the other branch of physics, kinology or biomechanics ? And if there is no meaning of this in physics, neither of us are right then ? As we can’t be if it has no meaning. Funny, but you say that the average force is equal to weight, so if there is an average, there must be a total, or please, what is your meaning by average ? And what’s it the total of ? D. has been debating this against me, thought he would know these, then why is he saying be total or overall forces are the same ?

So could not we say, that if I could hold 80 pounds half way up for 500 seconds, that my overall or total as in overall force was 80f/500s ? And if you could only hold the weight for 400 seconds, yours was 80f/400s ? Yes I know this is new, but that’s how physics evolved, equations of force, power, and work and so on were added thought out time.

So if I moved 80 pounds up 1m and down 1m 20 times in 20 seconds at .5/.5, I could say 80f/20s/.5.5 ? And the person moving the weight 1m and down 1m 2 times in 20 seconds at 5/5, could say 80f/20s/55.

Ok, maybe I am going in the land of fantasy here, but it’s nice to try and think of new things.

BUT there has to be an equation for this, as of the EMG, and a force plate can calculate this ?

Wayne

 

[sophiecentaur]

Funny, but you say that the average force is equal to weight, so if there is an average, there must be a total, or please, what is your meaning by average ?

Wayne

OK. You go from London to Bristol at an average speed of 62miles per hour. You could work that out easily enough. But how would you, Wayne, work out what your 'total speed' was? That's how daft your idea of total force is.

 

[waynexk8]

Here is what Roger Enoka said to me, when I asked him the question.

http://www.amazon.com/dp/0736066799/?tag=pfamazon01-20

Roger Enoka wrote;
I do not think it is appropriate to perform such comparisons in terms of average or peak force as the force varies continuously during each action. The relation f = ma applies only at one instant in time. When force (f) varies, the force applied to the object must be expressed as either the work done (force x distance) or the impulse applied (force x time). Because of these requirements, I cannot answer your question without knowing how all the forces vary over time.

Sorry.

So can we work this out on work done (force x distance) or the impulse applied (force x time) ?

Roger Enoka wrote;
I'm sorry, but I do not understand the question. My guess is that you are asking about the relative muscle force when lifting and lowering a weight either quickly or slowly. Even if the amplitude of the average acceleration of the weight was the same for both phases of the lift, net muscle force would differ. To lift a weight, muscle force must be greater than the weight. In contrast, the weight can only be lowered when muscle force is less than the weight. As a result of these constraints, the muscle uses more energy to lift the weight than it does to lower the weight.

I hope this is helpful.
Wayne,

Fortunately, physics does work on the human body, we just need to formulate the question precisely. The problem with the rationale provided by Jeff is that there are too many assumptions, such as the use of presumed average values. As indicated by John Casler, a 250 lb maximum does not mean that this force is applied throughout the lift. Indeed, this represents the force at the weakest point in the lift. Physics does apply, but solutions require real data that varies over time.

Cheers.

Roger M. Enoka, Ph.D.
Professor and Chair
Department of Integrative Physiology
University of Colorado

Wayne

 

[sophiecentaur]

@ Wayne - don't forget to look at and respond to post 176.

Those replies are what I might have expected and I have said the same on many occasions. So who will you apply to next to get your non real answer to your non real question?

"So can we work this out on work done (force x distance) or the impulse applied (force x time) ?" Those two formulae relate to two different quantities. Which one did you want to work out and why? (and how many times a second?)

 

[waynexk8]

Wayne wrote:

Jeff could I have a look at the calculations for the offloading in the last 10% of the lift in a 1/1 you worked out please, could you as well keep it as layman’s terms as poss, if you can’t do that could you then in brackets put down what some of the equations/math/abbreviations are, what so and so actually mean. Hi Wayne,

OK, let's see if I can reproduce this. Let's assume we're doing a bench press with 200 pounds, and our 1RM is 250 pounds (80% 1RM). Furthermore, the ROM is 15 inches.

The first thing we must do is convert pounds into the English unit of mass - the "slug". That is - 200 pounds/32 ft/sec^2=6.25 slugs.

Now, from F(net)=ma, we have 250-200=6.25 x a, or a=8ft/sec^2. This is the acceleration of the bar during the concentric.

Now, let's assume that we will push with our maximal force (250 pounds) up to the 12" point. We next need to find the velocity of the bar at this point from the equation v^2=2ad. Plugging in "a" from above and "d"=1ft, v=4 ft/sec. Note that this is not the top of the lift, but 3" from the top.

Next we want to find the time it takes to get to the 12" point, from the equation d=1/2at^2. Plugging in our values of "a" and "d", t=.7 seconds (again not the top).

Now, we assume that we stop pushing the bar at the 12" point, and let gravity slow it down, so that it comes to rest at the top. From v^2=2ad, we plug in our value of "v", but here we use a=32 ft/sec^2 (acceleration of gravity). From this we find that d=3", so that the bar comes to a perfect halt right at the top of the ROM - the 15" point.

Finally, we want to find out how long it takes gravity to stop the bar, from d=1/2at^2. Plugging in d=3" and a=32 ft/sec^2, t is found to be about .1 seconds.

Therefore, the total time for the concentric in this case would be .7 seconds (for the acceleration phase, or "onloading") plus .1 seconds (for the decceleration or "offloading" phase), for a total of .8 seconds. This is a bit faster than 1/1, but the best I could do this late at night. If your ROM were a bit longer then 15", then the speed would be closer to 1/1.

So, we see that in this case the offloading relative to the ROM is 3" out of a total of 15", or 20%. The offloading relative to the time is .1 sec out of .8 sec, or about 12%. Note that this is worst case...that is the first rep. As the set progresses, the offloading will reduce with each rep, due to fatigue, and towards the end of the set will be negligible. Therefore, you could state that the "average" offloading of the entire set relative to the ROM would be about 10%.

Note that this also assumes we can push with maximal force all the way to the 12" point. If our strength curve is such that our force output diminishes towards the top, then the offloading will be less than given above.

Jeff

Wayne again, but we had a problem here, as in real life, there was no 3 inch offloading, the weight did, and does NOT move out of your hands, however there was an answer, as it would move this far if a machine that could use the same force over the lift lifted and stopped. The reason that the physics did not work on the muscles and the force they produce, is that the physics LEFT OUT variables, as of the muscles have biomechanical advantages and disadvantages thought the range of motion, thus could not push up with full force all the time.

John wrote;
One thing for certain however, and that is, if the SLOWING is performed purposely, it WILL reduce muscle tension levels, and increase duration. That is not to say these are not valuable to some goals, but it is VERY difficult for many (if not most) people to maintain an awareness of how this affects the stimulus.

Also without discussing Power Expressions, Intensity Levels, and Work Output, we again cannot make any meaningful determinations as to what relationships the stimuli have.

Here is a video of a fellow Benching (FAST) over 400# 20 reps.

http://www.youtube.com/...rom=PL&index=20

It represents a significant level of Intensity, Power, and Work. It also due to the speed represents some EXTREMELY HIGH muscle tensions, and many exposures to them.

If this same fellow used the same load and increased the TUT x 2 or x 4 (twice or four times as long) by reducing the speed, the stimulus would have been FAR different, and the total stimuli except for fatigue, would have decreased.

Now it would be a difficult case to prove that the faster 20 RM effort would not offer superior results to a Slower 5 - 10 rep slower set with the same load.

If one finds a purposely slower rep speed "increases" hypertrophy it could be for several reasons, but seldom would they be related to "increased muscular tension force", but more an adaptation due to fatigue based stimuli.
Wayne

 

[waynexk8]

I was trying to get thought the old posts, have not read some of the new ones yet, but for now let’s jump to this, hopefully physics question. The concepts of work and energy are closely tied to the concept of force because an applied force can do work on an object and cause a change in energy. Energy is defined as the ability to do work, work is done on an object when an applied force moves it through a distance, mathematically, work is W = F • x, where F is the applied force and x is the distance moved, that is, displacement.

When I run 50m as fast as I can, then run 100m as fast as I can, I have used more overall, total force from my muscles, but let’s just say legs, running the 100m, do we agree there ? Now am I saying this right, or should I be asking which did the most work ?

As when energy is transferred either, to or away from an object by a force acting over a distance/displacement, work will be done on that object. The net work can also be expressed as the work done by the net force acting on an object,. W = Fd cos 0. The net work done on an object is the sum of the work done by each individual force acting on that object. The net work can also be expressed as the work done by the net force acting on an object. Can we work anything out on the work done by the constant and non constant force for the .5/.5 for 6reps moving the 80 pounds 12m and the 3/3 for 1 rep moving the 80 pounds 2m.


Wayne

 

[waynexk8]

I am not disagreeing with what the EMG people are saying. The problem is that you do not seem to understand what they are saying and how it applies to your question.

Ok, here is what I think. When you or the EMG takes the force readings, there are values of force above the weight of the weight being moved, for the accelerations, and values below for the decelerations, what the EMG and RMS does is adds all these up in the fast lifting and averages them out, and does the same on the slow, what D. and maybe you have done, is canceled the forces out of the values under the weight, and took them of the values above the weight ?

Why should you need to have RMS explained any more. It isn't an explanation that you need. What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

First I need to know if you agree or why you think they used the RMS ? And the same for D. as he seemed, or did think the RMS was wrong, I need to know why he thought it was wrong, and now what he thinks as they say it’s the best way to average the forces out. I can’t just go away from this; I need to understand what he thought before and now, and why he was wrong, and why he tried to say RMS can not work.

What you need to do is to use the definition to work out some numbers yourself. Douglis has given you a perfect worked example.

Ok will have tried. What numbers has he given ? Or maybe I have not come to it yet.

=sophiecentaur;3783708]I cannot be bothered to answer that question because it is just a smoke screen to protect you from going to the trouble of working stuff out for yourself.

I know this is easy for you, but when I started and can now work out the power used in different reps with different weights speeds and times, I really enjoyed it, but I don’t know where to start on this. Please I need an example. But first I need to know if you agree or why you think they used the RMS ?

Wayne

 

[Dale]

Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.

 

[douglis]

Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.

Those discussion are from many years ago and have been posted in more than 10 different forums around the internet(without any permission of course).The same is also true for the discussions here!

Twelve pages and so many threads for a question that can be answered in just one line!
The average force is always the weight regardless the lifting speed.

 

[waynexk8]

What forces have you the slow got that can make up the propulsive forces in the studys, there are no extra force in the slow reps, are there ?

Back later to catch up on all posts.

Wayne

 

[jmmccain]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

 

[sophiecentaur]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

I'd advise not getting too involved with this. All your suggestions have been made many times before, in this thread and earlier threads. Wayne does not believe in the accepted ideas of Physics. He has his own models and vocabulary of Physics.

Also, your thought experiment on the bathroom scales would not show the sampled forces accurately or frequently enough to convince Wayne. There would be errors which he would jump on and claim that the experiment showed him to be right. The scales and camera would lie in practice unless a much more sophisticated system were used.

In any case, what you say would only apply to free lifts and not to exercises on machines that introduce friction. That also confuses Wayne and strengthens him in his misconceptions.

 

[waynexk8]

I wish you and D. would do the same, and answer some of my questions ?

OK. You go from London to Bristol at an average speed of 62miles per hour. You could work that out easily enough. But how would you, Wayne, work out what your 'total speed' was? That's how daft your idea of total force is.

You got a point on that one, see what you mean, WILL think more on this one, BUT why then are you, and D. saying, the average force is the same ? I think that you have taken my maximum force from the acceleration, and then took away my minimum force from the deceleration, but “why” did/do you do this ? However ever, speed is a little different to force, take a look on this video, they measure the maximum force and minimum force, and then they calculate the average force, which will be the same, but we do NOT want this, this is what I keep ON telling D. over and over, we do not want the average force, we want all the three average force, the forces from the negative, the force when lowing the weight, the peak force, the huge peak force from the transition from negative to positive, and the positive force, the forces when lifting the weight.

NOW, if the weight was 80 pounds and my 1RM was a 100 pounds, if I start the lift from the top, lower in .5 of a second and lift in .5 of a second for the fast, and lower in 3 seconds and lift in 3 seconds.

1,
The fast,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ? 140 ? Then the tensions on the muscles to accelerate the weight up would be close to 100. 79 + 140 + 100 = 319/3 = 106 over 1 second.

2,
The slow,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ? 85 ? Then the tensions on the muscles to accelerate the weight up world be close to 80. 79 + 85 + 80 = 244/3 = 81 over 6 seconds.

3,
Divide the slow of 6 seconds by the fast of 1 second, 81/6 = 13 over 1 second.

Ok that’s wrong, but how would you do this please ? What if we added them up ? As in this debate/test/study, average means nothing, so it’s fast, 79 + 140 + 100 = 319 over 1 second. Slow 79 + 85 + 80 = 81 over 6 seconds, 13 over 1 second.

Take a look at this video, they say how they work out the average force on it, and they show the maximum force and minimum force, click on click to preview. What you and D. are doing, is taking to points, and taking the maximum and minimum forces and averaging them up, well of course the average force is going to be the same, but it means nothing here, you are just taking the maximum and minimum, and adding them together and averaging them out, that’s what we don’t want here, we NEED to add in the maximum acceleration force, the maximum peak force, {of the transition from negative to positive, this maximum peak force, can put up to 140 pounds of force on the muscles} and the maximum deceleration force, what you are doing is leaving out the most important, the maximum peak force from the transition, why are you leaving this force out, which put the huge tensions on the muscles, and the debate is, which puts the most over the same time frame, the most overall or total tension on the muscles.

http://webcache.googleusercontent.com/search?q=cache:Nze_pWKG-40J:www.fittech.com.au/products/ForcePlate.htm+force+plate+average+force&cd=2&hl=en&ct=clnk&gl=uk

Wayne

 

[waynexk8]

I have read the paper an I think I can see your problem. I find little to disagree with what is written.

They are not considering the whole cycle (lift/lower) when they refer to mean force.

Right could we get to this point, as D. would not explain, could you explain on what you think they are leaving out ? as I do not get it.

They are not only discussing free lifts but machines that present RESISISTIVE loads. Everything changes in that case because you are not just changing gravitational potential energy in that case but work is being done in overcoming friction.

Please forget the machines, this at this moment is just about free weights.

If you had read what they say then you would not think they are disagreeing with established physics at all. You did not understand what you were arguing about because you insisted on giving details instead of condensing your questions into something meaningful.

I did not think they are disagreeing, or never said they were disagreeing with established physics.

What I said, was what forces do you think you have that can make up of balance out the higher propulsive forces of the fast in the studies ?

Let’s take the mean propulsive forces, slow 6.2mean in 10.9 seconds. Fast 45.3mean 2.8 seconds, now let’s divided the mean slow of 10.9 seconds by the fast 2.8 = 3.8, so now let’s divide the slow mean by 3.8 = 1.6.

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.

Please what forces have I left out that the slow has to make up or balance out these ?

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf

Wayne

 

[waynexk8]

Will get back to jmmccain and Dalespam tomorrow, funny I have two very good high definition cameras.

Wayne

 

[waynexk8]

No time just now, to read the posts and answer.

Just wanted the state why average force means nothing in this debate

I lift 80 pounds at “any” speed, let’s go for 1/1, {1 second positive and 1 second negative} you all are saying that the average force for 1 repetition, and the average force for a 100 repetitions are the exact same ? However, we all know that you will exert far more total or overall force doing the 100 repetitions. You will exert a force the same as you do for 1 repetition {forgetting fatigue here} = 2 seconds, for 200 seconds, so its overall or total force output x 200 seconds.

So the question is, why does everyone keep commenting on the average force is the same, when it has no reverence in this debate ? What we are looking for, is in which repetition speed, in the same time frame, puts the most overall or/and total tensions on the muscles ?

As I have said, as you fail 50% faster using the .5/.5 to the 3/3 using 80% this does mean thet you have put more tension on the muscles faster, right ? And to put more tension on the muscles, and faster, you will have to use more force, if not please say why you think not ? PLEASE will someone try to answer this.

Are there anyone out there that thinks if you fail to lift a weight up after say 10 seconds, as you are lifting it faster, that it is not because you have put more tension on the muscles than the slow lifting, if so say why please.

Wayne

 

[sophiecentaur]

You were the one who asked what average force was. You also asked about Total Force.
Do keep up!

 

[sophiecentaur]

Wayne
Can you quote me a single statement, from anyone else, on this forum or in the communications with others that you have shown us, that indicates what you are talking about makes any sense in terms of the Physics? Everything I have read looks like non-committal and polite put-downs. When will you get the message?
Have you not realized that no one wants to discuss your endless, blow-by-blow descriptions of a lifting session? If you cannot be bothered to condense your questions into a digestible form then I, for one, can't be bothered to read them. I (/we all) have given you all the facts about the very straightforward Physics that relates to the process.
You haven't been prepared to accept it because it "doesn't feel right to you", somehow and you haven't been prepared even to use 'our' vocabulary. Can you be surprised that you aren't getting satisfaction?

 

[waynexk8]

Hi waynekx8, I didn't notice that you were back.

It looks like you are still confusing "work done" and "energy expended" as we discussed last year: https://www.physicsforums.com/showthread.php?p=3190515#post3190515

Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ? And we know for a fact that the fast does more work and uses more energy.

Average means nothing in this debate. What I am looking for is the higher overall or total forces used, thus more overall or total force on the muscles, just take this example.

Lift a weight up and down for 10 seconds, lift a weight up and down for 6o seconds, the average force is the same, but lifting the weight for 60 seconds WILL and DOES put more total or overall tension on the muscles, this means the total or overall force was more.

What we WANT to know is the same for lifting the weight up and down 6 times in 6 seconds, and lifting the weight up and down 1 time in 6 seconds.

Now I say without a shadow of a doubt, it’s the fast, and here is why. If you took both repetitions speeds to momentary muscular failure, meaning if you lifted the weights at the two given speeds until you could not move the weight, you would fail about 50% faster lifting the weight the fast way. This can ONLY mean one thing, you put more tension on the muscles faster, thus the muscles failed faster, thus more force must have been used in not only the same time frame, but in less time frame, you also use more energy in the fast, why ? It’s because your putting out more force and tension on the muscles, if not, why please ? As no one here is giving me a direct answer, they seem to not want to, or can't ?

Wayne

 

[sophiecentaur]

Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ? And we know for a fact that the fast does more work and uses more energy.

Wayne

That statement is meaningless. There is no "debate" possible on that basis. Why not do us all a favour and use PF language?

Total force is as daft as total speed. Come to terms with that.

 

[Dale]

Actually I am not confusing work done and energy expended ... And we know for a fact that the fast does more work and uses more energy.

Yes, you are still confusing them. In fact, this quote proves that you are confusing them since you finish with the confused statement that the fast does more work and uses more energy.

The correct statement is that the fast does the same amount of work and uses more energy (i.e. the human machine is less efficient at doing fast work). If you were not still confusing the concepts then you would not still be making the same incorrect statements.

What I am looking for is the higher overall or total forces used,

Let f(t) be the force exerted by the human on the weight at time t. Please define overall or total force in terms of f(t).

For example, average force from time t_i to time t_f is:
\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}

Please provide a similar rigorous definition for total or overall force.

 

[sophiecentaur]

@Dalespam
I see you are being drawn into this. Beware - madness lies there.

 

[Dale]

@Dalespam
I see you are being drawn into this. Beware - madness lies there.

Thanks for the warning sophiecentaur, but I have worked with wayne before, same topic different thread. I will only stick around while it is fun.

 

[sophiecentaur]

It's quite incredible how he bounces back with the same question, which, in itself, has no meaning.

 

[Dale]

I was trying to think of weight lifting analogy to describe his behavior in terms that he could understand, but my weight lifting vocabulary is fairly limited.

Basically, I want to get across the point about how stupid I would be if I were to go to him and a bunch of other weight lifting experts and ask for their advice on a weightlifting problem and when all of them tell me the same thing I then ignore their expertise and continue doing the opposite. Particularly if they are united in their opinion and persist in their opinion after detailed questioning.

They are experts in weight lifting, I know that I am ignorant on the subject, and since they all agree with each other, and my opinion disagrees with them, the only reasonable thing to do is to recognize that my uninformed opinion is almost certainly wrong and try to understand their advice.

Also, it wouldn't be helpful for me to call a weight machine a dumb bell, nor would it be helpful for me to talk about things they have never heard of like the supercalifragilistic rep without describing in detail what I mean by that. Furthermore, if they corrected me on my incorrect use of the term weight machine and dumb bell and I continued to use it incorrectly (for years) they would reasonably become frustrated.

 

[sophiecentaur]

Precisely. For "force" substitute "bannana" and his physics would make the same amount of sense.

 

[waynexk8]

Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Hi jmmccain,

Yes,

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

How do I do this please ?

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

However, as I said, there is a problem using average force, as if you do 1 repetition at any speed, say 1/1, that’s 1 second up and 1 second down, you will get the same average force if you do 1 repetition or a 100 repetitions. So average will not tell us anything, or will it ?

1 repetition = 100 force up and 60 force down, 100 + 60 = 160/1 = 160, average force = 160.
100 repetitions, 100 x 100 = 10000, 100 x 60 = 6000, 10000 + 6000 = 16000/100 = 160.

Some the average force is the same, BUT we all know that the doing the 100 repetitions is going to use more muscle force and put more tension on the muscles

I made a thread on this and told D. this average means nothing in this debate, but he and some others still insist it does, but can’t not say ?

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.

This seems very interesting, and have heard of this before, but not sure this will give use the right results. PLEASE what results will this tell us, I don't understand what this will show ?

Wayne