Impulse/force in pounds for the time frame

[sophiecentaur]

Leave it, douglis, it ain't werf it.

 

[waynexk8]

@Wayne
I just read your long post. Your insistence on using terms wrongly can only be regarded as arrogant - when you have been told so many times what the right terms are. You equate force / strength / energy at every opportunity which makes the accompanying reams of stuff meaningless.

In this instance, and force is a basically a push or pull, and that’s what strength is in this instance, a push or pull in the form of strength. So I am calling the force that is pushing/pulling the weight force/strength.

And the energy that supplies the muscles for the force/strength is basically calories, broken down into many chemicals.

What's the point in saying that you're sorry if you continue to say the same old rubbish.

If you state what I am saying wrong and why, I will read and use or debate.

You seem to want to pick holes in the explanations given by Douglass and others when they / we have no real idea what question you are actually asking.

If you had/have no idea, you only had to ask.

Question.

Two Clones lift 80% of their 1RM, {Repetition Maximum or the most they can lift up one time} up 20 inch and down 20 inch. So .5/.5 means .5 of a second concentric, {up} and .5 of a second eccentric. {down} One repetition {rep for short} means one concentric and one eccentric, or once up and once down.

Fast,
1 reps at .5/.5 = 6 seconds = distance the weight is moved = 240 inch.
Slow,
1 rep at 3/3 = 6 seconds = distance the weight is moved = 240 inch.

Or the below just for another example.

However this example is a little false, but this needs to be told as it is part of the main of this debate. As if you did do 18 reps fast, you would not only do 3 reps with the slow = 18 seconds, but more like 5 to 6 reps = 3 to 36 seconds, as you always fail faster in the faster reps. This is because in “my” opinion, you have to use more overall or total force to move the weight faster and more distance in the same time frame. Total or overall force means in this instance, as you “only” {and this will apply to a machine or anything as well as human muscles} have a certain amount of available force to use, before it runs out, as you will hit muscular failure quite soon, meaning you cannot pick up the weight anymore as you have ran out of temporary force. So total or overall force means in this instance. As you only have a set amount of force to use in a set time, let’s call that 1000N. So which of the above and below use the most force, or the available force the fastest.

Fast,
18 reps at .5/.5 = 18 seconds = distance the weight is moved = 720 inch.
Slow,
3 reps at 3/3 = 18 seconds = distance the weight is moved = 140 inch.

MY main opinions as to why you must use more total or overall force with the fast.

1,
You fail faster, or hit momentary muscular failure faster on the faster reps, as you “have” to have used up all your temperedly force up and faster, and there is no question that you have not used up all your temperedly force up, because you have, as you hit momentary muscular failure, thus have no force left.

2,
You use more energy, as in calories when doing the faster reps, why do you more energy doing “anything” faster, not just repping weight ? In my opinion it’s because you’re using more force and/or more force faster.

3,
You move the weight with the fast in scenario, 6 times the distance in the same time frame. You accelerate the weight far further then the slow moves it at a constant velocity.

4,
EMG readings state more muscle activity in the faster. This machine is as reliable as a computer in adding equations.

If you say that reading a Physics textbook is beyond you then how could you think that you would understand and be qualified to argue with what you read here?

All I need is, the fast using the same force, less force or more force and why. And if some say they use the same force, how do they account for 1 to 4 ?

I should just turn on your EMG /ESP /SPQR machine and enjoy reading the figures it shows you. That's what you paid your money for. Aim at making them higher (or lower) and see if you feel fitter / stronger as a result. Enjoy things at that level. No one will argue with you or get exasperated.

I have come to the physics forum, as if some think physics states the same force, they why does 1 to 4 differ ? I would have thought most physicists would want to know why.

Have you heard of a force plate or platform ? It’s a machine for taking many force reading, like ground force readings from runners, weightlifters and so on. I will either try to buy one or pay for some tests to be done, if this states the fast uses more force, what then, there will be 1 to 5. However I cannot see how anyone could argue with 1.

My opinion is that the higher forces the fast puts out on the accelerations, when the fast is on its lower force and decelerations, the constant medium forces of the slow, cannot makeup or balance out the fast higher acceleration higher forces. They may add on paper, but that’s not real World practical actions which apply in World.

Again, thank you for your time and help, and I have honestly come here to exasperate anyone, all I wanted was a nice quite polite debate.

Wayne

 

[waynexk8]

Great!For the first time you used some physics!

So what's the change in SPEED when you start and end at rest as it's done when you lift a weight?

You did say “speed” and speed is the rate/time at which an object covers distance, or the rate of change of position.

So your slow rep is more of a speed, or constant speed, so as don’t basically have a change in speed {other than the initial acceleration and closing deceleration, you change in speed is basically zero.

“Please” you reason for this ?

As my faster reps have acceleration, and to accelerate an object is to change its speed, I have basically a constant change in speed on acceleration and deceleration.

I ask again, your reason for this please.

Now my questions,

1,
If you cannot move the/a weight, as in the faster reps when you hit momentary muscular failure, this means you do not have sufficient force to move to move the weight, yes you do not have sufficient force or no you do have sufficient force?

2,
If you can move the/a weight, as in the slower reps when Clone 1 has hit momentary muscular failure, in the faster reps, this means you do have sufficient force to move to move the weight, yes you do have sufficient force to move to move the weight or no you don’t have sufficient force to move to move the weight?

3,
Most people would say, yes you do not have sufficient force to 1, and most people would say, yes you do have sufficient force to move to move the weight to 2.

So this means you use up your temporary force faster on the faster reps, meaning you are using more total or overall force up. And faster, right ?


Wayne

 

[douglis]

So this means you use up your temporary force faster on the faster reps, meaning you are using more total or overall force up. And faster, right ?


Wayne

Everything is perfectly answered at my last post.I don't know what you can't understand but it's becoming ridiculous.
I'll follow sophiecentaur's advice and leave it.

 

[waynexk8]

No sure if this is to me or all.

So it's more like force is the AMOUNT of "pushing", regardless of the time or distance.

Agree. However, for working things out you need to add in other quantities.

Take for example when you press your hands together, there is a certain force, even though there is no distance, and the time has nothing to do with the force, except that it might decrease as you get tired, and will certainly vary somewhat over time.

Agreed.

Consider if you took some dynamic(changing over time) situation and you had a spring in between something apply force and something being moved, you could take a snapshot at any time and determine the force being applied by measuring the spring(assuming you knew it's equilibrium length and spring constant), so time has nothing to do with it.

Time would have something to do with it.

As the more compressed the spring got the more force needed, and vice versa.

Also, if you were taking about applying muscular force, then time again would come into it.

An impulse has doesn't mean a sudden impact, and doesn't mean its "high"(especially as that is relative to what you are talking about[an elephant can probably handle much higher impulses than a mouse]). You can have an impulse of .0001N*S, taking place over an hour(theoretically), What makes it an impulse is that it is force taking place over time, not just the force we would see in a snapshot, but the measure of that "amount of pushing" for some "amount of time". The same way that acceleration would have a different meaning that acceleration for some amount of time(which is a change in velocity, [itex]/delta v[\itex]).

Agreed.

Wayne

 

[waynexk8]

If you're the Expert now, on what Physics can and can't do, then I suggest you answer the question yourself.

Never said I was an expert.

I have just read your comments on Douglis's post and it is clear that you don't even read his sentences to the end.

I have, and answer everything. Please show me what I have not answered ?

I have to conclude that you find us all to be totally incompetent in the field of Physics so I suggest you go and find a Forum in which the contributors know enough of the right sort of Physics to satisfy you.

I don’t understand why you say that. I asked a polite question, and “gave” 4 reasons why it could be wrong, and NOT once have you had a go at answering them. You seem to want to turn this into a mocking match; I simply do NOT want that. “Please” tell me why you can’t debate/answer my 4 reasons ? You seem to be trying to mock me because I ask a question.

Does it, for one minute, strike you that your whole idea could just be flawed?

Hmm, not sure there, as it does take a certain force/strength to move an object so and so distance in so and so time frame, so there should be an answer.

Your resolute use of the nonsense expression "force / strength"

I explained that, however you do not say why you think this is wrong. A force is basically a push or pull, in this instance the force is my muscle strength. If you say why it’s wrong maybe I will agree.

and others,

What others please. Tell me and I will explain or learn from you.

demonstrates that you just don't really want to get to grips with the real stuff. Just why do you keep posting here?

I do want to learn, honest.

Sophiecentaur, you don’t know me and I don’t know you, but I can honestly say I did or do not want it to turn out like this, all I want is a polite friendly debate, also I do not get why you do not try and debate/answer my 4 reasons, “OR” answer this question to which I asked before and is full physics.

Can you have more power and the same force in the same time frame ?

Please again Sophiecentaur, thank you for your time and help, and sorry if IO upset you, it was/is not intended.

Wayne

 

[sophiecentaur]

There you go again with acres of unreadable (and unread) rambling. State your question like everyone else does: in a couple of lines, using the right terms (which you know by now). No one will make sense of that old ramble above.
Admit it, though, you just want an unending chat and not an answer.

 

[waynexk8]

For God's sake...read my whole sentence and try to make your brain work.
You DON'T use force.You use energy to apply force.

We all know you use energy to apply force, no one is debating this ?

I "bring up" the energy thing because higher energy usage(as in fast lifting) does NOT equate greater force application.

You seem to be tying yourself in knots, as you are constantly contradicting yourself.

1,
We all agree that you have to use more energy the faster you do anything in the same time frame.

2,
When you do something faster in the same time frame you “have” to have high/faster/longer accelerations, these high/faster/longer accelerations “must” have higher forces, as you can’t have high/faster/longer accelerations without initial higher forces for the acceleration.

But you say, I quote; higher energy usage(as in fast lifting) does NOT equate greater force application.

3,
So do you think it’s “just” a coincidence that when you have high/faster/longer accelerations to a constant speed, that the energy needed for the high/faster/longer accelerations goes up ? Or do you think it’s because the force “has” to go up for the high/faster/longer accelerations ?

4,
Tell me, as of several months back you stated that you do not use more energy doing anything fast, you then over the debate got convinced you were wrong by someone from this forum, then agreed that you were wrong, and admitted that you have to use more energy when doing anything fast, and in our case lifting weight. Could you please state why you have to use more energy ? Is it a, when the accelerations and force go up, or b, on the deceleration when the force goes down ?

5,
What/why does equate greater energy application in the faster reps, is its not force ?

First on a physics forum, you “NEED” to “state/say” WHY you “think” higher energy usage does not equate greater force application.

Apples and oranges.Only in Wayne's world those two mean the same thing.

I “never” once said they mean the same thing, odd thing to say ?

Your EMG states greater quadratic mean(RMS) as expected since in fast lifting the force has greater peaks.

I have higher peak, and lower lows, please do not forget about the lower lows, as the EMG WILL take these reading to, and average them all up, or do you want to try and forget about my lows when its convenient ?

The RMS also known as the quadratic mean, is a statistical measure of the magnitude of a “varying” quantity. So YOU state that my high forces and my low forces are all balanced out by your medium forces. So if the EMG takes the highs and the lows on my high and low forcers, as you claim they should all balance out, but they do NOT do they ?

RMS It is especially useful when varieties are positive and negative.

It can be calculated for a series of “discrete” values or for a “continuously” “varying” function. The name comes from the fact that it is the square root of the mean of the squares of the values. It is a special case of the generalized mean/average.

Here it is again!Are you able to understand the difference of these three?

I never said I do not know the difference ? Why would you think or say that ? I said, I quote; I fail faster in the faster reps, thus I “have” used up my temporary energy/force/strength.

I fail in the faster reps because I have used my temporary energy up, as in the muscles energy ATP glucose and creatine. And used my temporary force up, push or pull. And used up my strength push or pull.

The acceleration is always exactly balanced by the deceleration regardless the length of each phase.That's why the average force is always equal with the weight.

Why do you “think” its balanced, you cannot say, if it was balanced, why is you energy and distance not made up or balanced out in your slower reps, if you claim its balanced out ?

NONSENSE.
The force-energy relation is NOT linear.The force "balances out" while the energy NOT.

Again you state something with “nothing” to back it up ? You need to try and explain why the faster reps use more energy, you need to say why and how you think the forces balance out make up but the energy does not ? I say why, but you can not ?

So what if I buy or get tests done on a force plate, and they state what I claim, will you accept your physics equations are missing variables then ?

Wayne

 

[waynexk8]

There you go again with acres of unreadable (and unread) rambling. State your question like everyone else does: in a couple of lines, using the right terms (which you know by now). No one will make sense of that old ramble above.

Please, just say what you do not get or understand and I will try to explain. I am not sure if I can say it short, as its “need” an explanation. As some things I need to explain, and I did my best trying to, what would you think or me if I left things out and told you later ?

Question short,

I lift a weight 1000mm in .5 of a second, you lift a weight 166mm in .5 of a second, which uses the most force, not the peak force used, but the total or overall force.

Question long as I think it needs more of an explanation as I know more about it.

Two Clones lift 80% of their 1RM, {Repetition Maximum or the most they can lift up one time} up 20 inch and down 20 inch. So .5/.5 means .5 of a second concentric, {up} and .5 of a second eccentric. {down} One repetition {rep for short} means one concentric and one eccentric, or once up and once down.

Fast,
1 reps at .5/.5 = 6 seconds = distance the weight is moved = 240 inch.
Slow,
1 rep at 3/3 = 6 seconds = distance the weight is moved = 240 inch.

Or the below just for another example.

However this example is a little false, but this needs to be told as it is part of the main of this debate. As if you did do 18 reps fast, you would not only do 3 reps with the slow = 18 seconds, but more like 5 to 6 reps = 3 to 36 seconds, as you always fail faster in the faster reps. This is because in “my” opinion, you have to use more overall or total force to move the weight faster and more distance in the same time frame. Total or overall force means in this instance, as you “only” {and this will apply to a machine or anything as well as human muscles} have a certain amount of available force to use, before it runs out, as you will hit muscular failure quite soon, meaning you cannot pick up the weight anymore as you have ran out of temporary force. So total or overall force means in this instance. As you only have a set amount of force to use in a set time, let’s call that 1000N. So which of the above and below use the most force, or the available force the fastest.

Fast,
18 reps at .5/.5 = 18 seconds = distance the weight is moved = 720 inch.
Slow,
3 reps at 3/3 = 18 seconds = distance the weight is moved = 140 inch.

MY main opinions as to why you must use more total or overall force with the fast.

1,
You fail faster, or hit momentary muscular failure faster on the faster reps, as you “have” to have used up all your temperedly force up and faster, and there is no question that you have not used up all your temperedly force up, because you have, as you hit momentary muscular failure, thus have no force left.

2,
You use more energy, as in calories when doing the faster reps, why do you more energy doing “anything” faster, not just repping weight ? In my opinion it’s because you’re using more force and/or more force faster.

3,
You move the weight with the fast in scenario, 6 times the distance in the same time frame. You accelerate the weight far further then the slow moves it at a constant velocity.

4,
EMG readings state more muscle activity in the faster. This machine is as reliable as a computer in adding equations.

Admit it, though, you just want an unending chat and not an answer.

What I will admit, is I am sick of people not trying to give me an answer, because I seem to have proved them wrong, if you think I have not, then please post an answer and try and refute 1 to 4, is that too much to ask.




Let’s change this around for a minute, and YES I know this is different, but let’s say you ran up to the shop and back, which is uphill, 100m away, as fast as you could 6 times in 300 seconds = 1200m, most people would either not make it, or be totally exhausted. You then walk very slowly to the shop and back one time in 300 seconds = 200m, do you honestly think you have used the same overall or total force, or the same overall or total muscle force ?

I say again, I have not started this mocking type match; I have been polite and only asked questions, and said if anyone does not understand them to ask. All I want is some kind of answer.

Wayne

 

[douglis]

I have higher peak, and lower lows, please do not forget about the lower lows, as the EMG WILL take these reading to, and average them all up, or do you want to try and forget about my lows when its convenient ?

Wayne

I didn't even read all the rest you wrote because you're repeating the same NONSENSE that has been answered dozens of times.

The above is the only one that worth answering.
The RMS takes the square root of the mean of the squares of the values.By taking the squares of the values you turn the negative values into positives.
I know you have problems with basic maths but you must have been taught that in high school.For example...-2^2=4 not -4.

So...your EMG turns the "lower lows" into "higher peaks".The RMS is NOT the technical mean that we've been discussing so far...it's the quadratic mean...huge difference.
The RMS(quadratic mean) is exactly the 70% of the peaks...the positive peaks and also the negatives that turned into positive.
Since fast lifting has greater peaks will definitely have higher RMS.

Your EMG maybe fine for comparing the activation of a muscle group between different exercises but is NOT offered for a fast vs slow comparison.
That comparison could only be possible if you'll find the integrated EMG(iEMG) like they did in the push ups studies(and found greater iEMG for the slow push ups!) or as I did at the graphs I mailed you.

 

[waynexk8]

Please see the video stating that more overall, total force was used with the fast rep, but this time the more force was used in "less" time, go from 5 min.



Fast
P = 695
F = 579
V = 192

Slow
P = 649
F = 546
V = 161

Wayne

 

[douglis]

Please see the video stating that more overall, total force was used with the fast rep, but this time the more force was used in "less" time, go from 5 min.

That's an accelerometer.From the change in speed estimates the peak value of the acceleration and then from the equation F=mg+ma finds the peak value of force(Fmax).
Check their site..."Concentric strength(N) = Fmax in the push"
https://docs.google.com/viewer?a=v&...Vseg--&sig=AHIEtbSVGzCSQRrN1crmIrlXybsDwA_Pwg

The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies.

 

[waynexk8]

sophiecentaur, I was woundering why you did not replly to this post ? As without sounding sarcastic, you got the force produced by the agonist and antagonist wrong, thus i think we need to clear thins up, and the other things.

https://www.physicsforums.com/showpost.php?p=3725969&postcount=63

Could you go from here please;

sophiecentaur wrote;
You claim that your machine tells you the force involved (in N) but then say that it reads electrical activity in μV. When you 'tense up' your arm, there is no net force at all (it stays still, in its original position) but there is loads of muscle activity.

Not fully with you on that one sorry, or maybe reading you wrong. As when I tense up my arm with the pads on the moving muscles, let's say the biceps and foararm in the curl or arm flextion, if I just tence those muscles the reading on the machine tells me, if I just hold the weight half way up the machine tells me, and when I miove the weight up and down the machine tells me, and tells me the high signals, with my muscles are producing high force, and the lower end of the signals where my muscles are producing the lower force. So this is a net force, when you tense and when you lift.

sophiecentaur wrote;

as your machine would show, but the antagonistic muscles are producing equal and opposite force.

No, the antagonistic do “not” produce equal and opposite force in any barbell exercises, they produce very little force, maybe as little as 5 or 10% the agonist muscles do all the lifting and all the lowering. {actually the lowering or eccentric portion of the lift which the biceps and forearm does, {as well as the concentric} able a person to lower 40% more under control, say lower in 6 second, than the person can lift for their 1RM.

So no the antagonistic muscles are not producing equal and opposite force. As the biceps are the curling or flexing muscles, and the triceps are the extending muscles.

sophiecentaur wrote;

So there's no direct connection between muscle activity and force produced.

Yes there is, it’s a direct comparison. As if I lift 30% then 80% the readings like the peak force and average force will be far higher.

sophiecentaur wrote;

That is unless you have electrodes on every muscle group and the machine can do some complicated 'addition' of the effects of all the muscles.

You only need the pads on the lifting muscles, as these are the ones producing the force for both up and down.

Yes the machine does do very complicated equations instantly all the time.

sophiecentaur wrote;

You are still after some relationship between that muscle activity and the measurable work done on a weight when lifting it. But if it's possible to have loads of muscle activity and Zero work done, then there clearly is not one. Can you not accept that?

Yes I understand the concept of work, if I do not move the weight no work has been done, but there is still muscle activity, energy and force being used.

sophiecentaur wrote;

There is really no more to be said on the topic (except for another acre of figures about rep rates and pounds lifted).

I find it odd you say that, but after reading the about you might differ.

What about this question then ? I am sure you can talk or understand scenarios outside of physics ? Like other branches of physics, kinology, biomechanics. Or, just thought of this, let’s say your back at university, and you have to do a practical test and scenario for your PhD, and the Lecturer asks you the below, and you have to answer. Or could you just have a go for me, or suggest another way I can ask it, but as a physics adviser, I thought you would like to try and work out how the equations do not add up in the real World tests/experiments as in below ? Please ? As I don’t see how you can step outside the box, as I thought all physics should be tested in the real World after they have been calculated on paper.

The Question.
You always fail faster about 50% faster in the faster repetitions, so if you fail faster, you “HAVE” used up your temporary force/strength, as you hit muscular failure faster thus you cannot lift the weight anymore. To me, if both Clones started the fast and slow lifting at the same time, as the slow Clone are still lifting the weights when the fast Clone has hit muscular failure and unable to lift the weight anymore, this “is/seems irrefutable, or categorically right to me, and I am not trying to sound smug or anything here, but if the slow Clone is still moving the weight, then the fast Clone “has” used up more overall or the total force/strength they had faster, if you see my point, please do you see what I am saying here ?

sophiecentaur wrote;

I can only suggest that you approach the manufacturers of your machine and ask them for their opinion.

These machines are/have been used in Hospitals and sports facilities for years, they are as efficient as your calculator, actually they are calculator/computers in another form. The machines are used as much as the everyday car; they are very well known and used.

http://medical-dictionary.thefreedic...ectromyography


Originally Posted by sophiecentaur
They may well be more prepared to speak you language as it is in their interest to sell as many of those machines as they can. I think they will tell you that the machine gives a good indication of how much energy the muscles are transferring and / or the forces. I have no problem with that (the neurological application of the machine seem very worth while). They may even launch into some link between that and the mechanical work done. That will make you happy. Great, but it won't make the Science any more valid.

IF, you want to prove your physics equations right, you would need to “try” and answer the question above, and say why the below happens in the real World, as I am in contact; New Scientist Magazine, Physicstoday Magazine and Physics World Magazine. If you could be the first to solve the puzzle of how and why the equations and real World tests on this Phenomena is, maybe you could be in these Mags.

Thank you for your time and help.

Wayne

 

[sophiecentaur]

How can the machine distinguish between situations where there are and there aren't antagonistic muscles at work? Are you saying that your machine 'knows' when your arm is moving and when it's stationary? Does it know then you are holding, lifting and lowering the weights?

How can you expect to get a proper answer whilst you still insist on using the term force/strength? Do me a favour and look the individual words up. They describe entirely different aspects of Physics. Which one do you mean when you use that term? Why do you hang on to these deliberately nonsense terms instead of using the right one? You are saying black is white on every occasion.

The very least you could do, if you really do want some sense, is to use the correct terms. Imagine you had a calculator and the + key sometimes gave you a - and the X key sometimes gave you a ÷. You would say it was rubbish, wouldn't you? Constantly using confusing terms is the equivalent to a dodgy calculator.

At least, you could do us all the courtesy of talking the right language. (The language that 11 year old kids are quite happy to learn to use in School.) Don't ask me to help you with this - just look up any word you want to use and see if it actually fits what you want to say. I get the impression that you are not looking anywhere else for your information but just want to be spoon fed by PF. This is unreasonable.

btw, I don't have to justify the Physics Equations in this context. They work fine for everyone but you so I am not out of step here. Give me a proper question (not ten pages of weight lifting jargon) and I can guarantee a good answer. Give me another 'Wayne' question and there will be no available answer from Physics.

 

[waynexk8]

That's an accelerometer.

It a accelerometer, with built in velocity, power, force and other, the unit uses three-dimensional accelerometer. By measuring body movement in three dimensions (backwards and forwards, up and down, side to side), the body can be tracked in space. The speed (and change in speed) of movement can be used to calculate such things as flight time, number and speed of repetitions and power. Many physical parameters can then be calculated from these parameters.

From the change in speed estimates the peak value of the acceleration and then from the equation F=mg+ma finds the peak value of force(Fmax).
Check their site..."Concentric strength(N) = Fmax in the push"
https://docs.google.com/viewer?a=v&...Vseg--&sig=AHIEtbSVGzCSQRrN1crmIrlXybsDwA_Pwg

Not sure if I get you there, as the fast produced more Newtons, yes ? So more Newtons is more total or overall force, right ? As how can more N be the same ?

The "total/overall force" can be found ONLY by using integrated electromyography(iEMG) like they did in the push ups studies.

integrated electromyography uses RMS, like my machine, I can set it for as many samples as I want. You have to use RMS to perform the integration. Do you know what Integrated means ? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.

Calculate the integrated EMG envelope on- and off-line. The integration function incorporates an RMS (Root Mean Square) feature set to operate over a user-specified number of samples. Adjust the RMS time constant by increasing or decreasing the number of samples used to perform the integration. The number of samples used in the RMS integration divided by the sample rate is proportional to the time constant of the integration.

http://www.biopac.com/researchApplications.asp?Aid=41&AF=61&Level=3

Do a search for integrated below, you will find RMS is part or integrated, as RMS has to be used, as RMS = Averaged or root-mean-square (RMS)

Integrated EMG (iEMG) important for quantitative EMG relationships (EMG vs. force, EMG vs. work) best measure of the total muscular effort, useful for quantifying activity for ergonomic research, various methods: mathematical integration (area under absolute values of EMG time series) root-mean-square (RMS) times duration is similar but does not require taking absolute values.

http://uk.search.yahoo.com/r/_ylt=A7x9QV9meiRPvTQAmu9LBQx.;_ylu=X3oDMTE0cTc5aTMzBHNlYwNzcgRwb3MDMgRjb2xvA2lyZAR2dGlkA1VLQzAwMV83MQ--/SIG=12jemp264/EXP=1327819494/**http%3a//www.health.uottawa.ca/biomech/courses/apa4311/emg-p2.pps

Why do you refer to RMS on my machine ? Show me the machine they used in the press up ?

I showed you the those press up studies were flawed, in that it was total muscle activity they took, and as the slow went on for longer.

Wayne

 

[sophiecentaur]

Do you know what Integrated means ? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.
Wayne

So that is your definition of 'Integration'? It is not the Mathematical definition that is used (and works) in Physics so there is no point in your using the term.

 

[douglis]

Not sure if I get you there, as the fast produced more Newtons, yes ? So more Newtons is more total or overall force, right ? As how can more N be the same ?

No...the total/overall force(effect of force over time...impulse) is given by N*s...NOT by N.
The more N means greater peak force as they also state by themselves("Concentric strength(N) = Fmax in the push")

integrated electromyography uses RMS, like my machine, I can set it for as many samples as I want. You have to use RMS to perform the integration. Do you know what Integrated means ? Combining or coordinating separate elements so as to provide a harmonious, interrelated whole = RMS.

Yes...but you must also normalize the raw EMG data in order to integrate.Check the paragraph " materials and methods".It's described pretty well.The equation (1) shows that the integration is done for the normalized data.
http://jmbe.bme.ncku.edu.tw/index.php/bme/article/viewFile/635/839

Anyway...this discussion is meaningless.The fact is that the RMS is the 70% of the peak and naturally higher in fast lifting.End of story.

I showed you the those press up studies were flawed, in that it was total muscle activity they took, and as the slow went on for longer.

Wayne

Those press up studies are perfectly designed...your mind is flawed.
For example...compare the durations and the Total Muscle Activations.

Slow push ups:Duration=101.2 sec...TMA(triceps)=3145.29
Fast push ups:Duration= 84.2sec...TMA(triceps)=2138.91

The duration of the slow push ups is only ~20% greater but the Total Muscle Activation is ~47% greater.This is a direct PROOF that slow push ups have greater muscle activation per unit of time.
We have such a well designed study examining exactly what you're looking for and you're still around forums saying NONSENSE.

 

[waynexk8]

Back a little later.

This will please S. and D and the rest of the forum, some pure physics for a change.

Could anyone state the Newton’s needed for the below,

1,
From a still start, 80 pounds is moved up 20 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

2,
From a still start, 80 pounds is moved up 3.3 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

3,
80 pounds is being lowered under control at 20 inches in .5 of a second, then when still in full downward motion, it’s moved upward 20 inches, in .5 of a second, stops and reverses. Only the Newton’s for the immediate deceleration, {wherever this may be I would say the last 5%}
stop and forward/upward to stop.

Wayne

 

[sophiecentaur]

Can't ba answered if you don't know the acceleration, I'm afraid.
Force=Mass X Acceleration
(Then add the weight, in this case)

Are you assuming constant force all the time?

 

[douglis]

Back a little later.

This will please S. and D and the rest of the forum, some pure physics for a change.

Could anyone state the Newton’s needed for the below,

1,
From a still start, 80 pounds is moved up 20 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

2,
From a still start, 80 pounds is moved up 3.3 inches, in .5 of a second, stops and reverses. Only the Newton’s for the forward.

Wayne

Again that question.Define the value of force you're interested in.The peak...the average or the "total" force(integration of force in respect of time)?

For the peak force...we need more data.

For the average force...in both cases you start and end at rest.The net change in momentum is therefore zero, which is equal to the net impulse delivered. Therefore, the average force is equal with the weight in both cases...80 pounds or 356N.

For the "total" force...again in both cases is equal with gravity's impulse for .5sec.So...356 X .5=153N*s.

 

[sophiecentaur]

Calculate the peak force needed to throw a computer out of a window in sheer exasperation!

 

[waynexk8]

Can't ba answered if you don't know the acceleration, I'm afraid.

Will change to Metric.

Hmm, thought I had put enough information in, as if I am moving the weight 500mm at every .5 of a second, I decelerate for say the last 20% thought you would/could work this out, or am I doing it wrong ? Seems I was wrong interesting.

I am moving the weight on 1, at 1m/s. {m/s 1 meter per second}

I am moving the weight on 2, {which is now 83mm} in .5 of a second. Moving it at 83m/s. {83mm per second} This is basically going at a constant speed, as its accelerated to moving at 83m/s.

So are you saying that I could have different accelerations and decelerations ? This is as I thought, but D. {douglis} seems to think other. {that is right is it not D. ?}

Force=Mass X Acceleration
(Then add the weight, in this case)

Misunderstand that a little, so best say and ask.

Are you assuming constant force all the time?

Hmm, with a muscle it would not be a constant force, but let’s says it’s a machine moving the weights and the force is constant.

Wayne

 

[waynexk8]

Again that question.Define the value of force you're interested in.The peak...the average or the "total" force(integration of force in respect of time)?

Impulse force, total, overall, integration of force in respect of time.

For the peak force...we need more data.

Enough now for 1 and 3, as 2 will not be much higher than the weight.

For the average force...in both cases you start and end at rest.The net change in momentum is therefore zero,

How do you work out the net change in momentum/movement is zero ? As the net force in 1 = 500mm and in 2 = 166mm. So the net, total momentum/movement will be 500 and 83 ?

I start at rest and end in rest, but the distance move on 1 = 500mm and 2 = 83mm

which is equal to the net impulse delivered. Therefore, the average force is equal with the weight in both cases...80 pounds or 356N.

For the "total" force...again in both cases is equal with gravity's impulse for .5sec.So...356 X .5=153N*s.

See, I don’t get how you come to this, as the below, shows the total or overall force to be higher for the faster moving in less time ? And its not the peak. Please see the video stating that more overall, total force was used with the fast rep, but this time the more force was used in "less" time, go from 5 min.



Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

Wayne

 

[waynexk8]

Calculate the peak force needed to throw a computer out of a window in sheer exasperation!

WaynesWorldphysics ?

Please sophiecentaur, could you work out what the Newont force is below is, peak, average or total ?

Go from 5 min.



Fast
P = 695
F = 579Newtons
V = 192

Slow
P = 649
F = 546Newtons
V = 161

This one may help.

http://www.youtube.com/watch?v=Ycu6y4EHfog&feature=related

Wayne

 

[waynexk8]

So that is your definition of 'Integration'? It is not the Mathematical definition that is used (and works) in Physics so there is no point in your using the term.

I will write to an EMG expert, and ask then on intergration and RMS.

I will also write to the makers of my machine, and ask them what actualy my machine reads out on the average.

Wayne

 

[waynexk8]

D. I just thought of somthing, I can video the machine and prove its NOT the peak force, but the average. Or you should be able to work it out here.

Take a look at the slow rep video, you will notice the peak force = ? was it a 190 somthing ? BUT the average was a 140 !

http://www.youtube.com/user/waynerock999?feature=guide#p/u/36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

 

[waynexk8]

How can the machine distinguish between situations where there are and there aren't antagonistic muscles at work?

If I am curling, arm flexion, then I will but the four pads on my biceps, as these are the prime moves.

Are you saying that your machine 'knows' when your arm is moving and when it's stationary? Does it know then you are holding, lifting and lowering the weights?

Yes the machine know when my arm is moving and when it's stationary, holding, lifting and lowering the weights. As the pads detect “all/every” of the electrical signals in my muscles the pads are on.

Here take a look at the makers video, go to 1.40min



Or I could make a video if you want.

So if they say the fast as a higher average, as a Physicist, thought you would be very interested, that is, if you do agree with D ? But to me he has not added in all the variables, like ground reaction muscle force and so forth, and left out Kinology and Biomechanics.

How can you expect to get a proper answer whilst you still insist on using the term force/strength? Do me a favour and look the individual words up. They describe entirely different aspects of Physics. Which one do you mean when you use that term? Why do you hang on to these deliberately nonsense terms instead of using the right one? You are saying black is white on every occasion.

Ok sorry, I will try the stick with force.

The very least you could do, if you really do want some sense, is to use the correct terms. Imagine you had a calculator and the + key sometimes gave you a - and the X key sometimes gave you a ÷. You would say it was rubbish, wouldn't you? Constantly using confusing terms is the equivalent to a dodgy calculator.

Ok get your point.

At least, you could do us all the courtesy of talking the right language. (The language that 11 year old kids are quite happy to learn to use in School.) Don't ask me to help you with this - just look up any word you want to use and see if it actually fits what you want to say. I get the impression that you are not looking anywhere else for your information but just want to be spoon fed by PF. This is unreasonable.

btw, I don't have to justify the Physics Equations in this context. They work fine for everyone but you so I am not out of step here. Give me a proper question (not ten pages of weight lifting jargon) and I can guarantee a good answer. Give me another 'Wayne' question and there will be no available answer from Physics.

Ok, I will try and use physics, if I can't will ask.

Wayne

 

[sophiecentaur]

WaynesWorldphysics ?

Please sophiecentaur, could you work out what the Newont force is below is, peak, average or total ?

1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

 

[waynexk8]

Big thank you for staying with me.

1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the ENG states more average force used in the faster reps.

3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

So let’s “just” {please this is just an example to get my point over} say for an example you had 1000 forces to lift the weight, in the fast, you used up this force far far far faster, meaning if you both lift the weight for a set time, and do “not” lift until momentary muscular failure, the faster reps “must” be using up more force faster, as you fail faster lifting faster. Example of how the fast are using more force and faster, more energy used, more distance the weight has been moved, faster to muscular failure, the EMG states more muscle activity or muscle force. I know all the above sounds a bit complicated, but there is total since in there.

Thank you again for you time and help, not sure about the acceleration, hope to learn more on that.

Wayne

 

[douglis]

Big thank you for staying with me.

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

O.K. I'll try to help you understand once again.
Let's say in your above example the load is 800N(81.5kg) while your maximum force ability is 1000N.

For the first 0.4sec you're applying your Fmax (although in reality that's biomechanically impossible) so the net force is 1000-800=200N.
So your acceleration for those 0.4sec is a=F/m=200/81.5=2.45m/s^2

For the last 0.1sec you let the gravity decelerate the load so the net force is -800N
So your acceleration for those 0.1sec is a=F/m=-800/81.5=-9.81m/s^2 and obviously it's equal with g.

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

As I proved you above you used positive acceleration equal with 2.45m/s^2 for the 80% and 4 times greater negative acceleration equal with -9.81m/s^2 for the last 20%.
That's why we're telling you the acceleration is offset by the deceleration.The average acceleration is always zero.
In the equation of force F=mg+ma the "ma" part is always zero so F=mg or else the force is equal with the weight.No Mean additonal force is required.The forces "make up","balance out" or call it whatever you like.

I don't believe it's possible to get a more simple explanation.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.

No...your EMG states that greater RMS is used in faster reps...not average force.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

Wayne

This is the last time I bother to even read that particular question.

You "fail" in a weight lifting set when you have no longer enough energy to apply force equal with the weight.
"Failing" faster with a certain kind of lifting means that with this kind of lifting you spend energy at a higher rate...NOT that you use more force.In all cases the average force is equal with the weight.
It's the fluctuations of force that are more energy demanding.That's the only scientific answer you can get.

 

[waynexk8]

sophiecentaur I see you’re an Engineer, so hopefully you will be able to understand what I am saying, even thou a lot is not in pure physics talk, please, and I am not being sarcastic, as I know people find it hard the way I explain things, but please do you understand what I am trying to say in the below please ?

And please, if you agree with me or not, {AND YOU D.} do you understand what I am trying to say convey too you ? As this is very important, as it makes no difference if you agree or not now, only that you both and other members actually understand what I am “trying” to say ?

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.


Big thank you for staying with me.

Originally Posted by sophiecentaur
1. If you can't tell me the acceleration (i.e. what is the variation of velocity with time during the lift? There are infinite possible combinations that will get the weight from bottom to top of the lift in a given time.) I cannot tell you the force. (Did you not read my F=mA formula?)

Hmm, I thought I told you this on my other post ? Ok, this is where your help will have to come in, as I am not sure how to work this out. Can we call it for now a constant acceleration ? If so, on 1, the weight accelerates from rest to 400mm in 0.4 of a second, then decelerates the last 100mm in .1 of a second

Originally Posted by sophiecentaur
2. Because the weight starts and ends stationary, the Mean additional force must be zero and the mean force will be the weight.. (I think this had been pointed out several times already.)

So as I push up with the force of the weight to move the weight, and then the weight cancels that force out, then you call the mean additional force {that my muscles creating force} must be zero ? If I am right, sort of get that, however, I have used forces, from a 100 to 1 pounds, and that’s all we are concerned about ? Or am I missing something.

My other point is, that I don’t think that when I am using far higher forces, like 100 pounds for the accelerations, that when I am then using slightly less force than 80 pounds for the decelerations, that when the slow rep is roughly using a constant 80 pounds, that the slow reps medium forces can NOT make up or balance out the very high force {the very high tensions that the very high forces have put on the muscles, as of the action reaction force thus tensions on the muscles} that the fast reps are putting out, that’s why the fast reps have used more energy and moved the weight 6 times further in my opinion.

Not sure you get what I say there, please say if you don’t, it’s like a person hits you with great force, and it hits you down, but it would take far more lower force hits and more “time” {but in this debate the time is the same} to hit you down with lower force hits. THIS is why you always fail at lifting the weight, or you hit momentary muscular failure faster with the faster reps, as they ARE doing more damage with the higher forces.

I am not saying the physics equations are wrong, it’s just they cannot tell the full story, as all the variables as liker the high force to medium force has not been worked out, that’s why the EMG states more average force used in the faster reps.

Originally Posted by sophiecentaur
3. What does "total force" mean? Do we add up the forces, measured every second, every tenth of a second, every 100th of a second? It's a nonsense concept as you can only validly add forces that operate at the same time.. Ask the makers of your machine for an answer. There is not a PF answer for you.

It matters not whether you are working in Imperial or Metric - all three questions are either nonsense of indeterminable.

What I mean with total or overall force, is that if you lift say 80% of your 1RM, you can only lift it for a certain amount of times in a time frame at a certain speed. Let’s say you could lift it up and down 10 times at 1 second up and 1 second down, then at 20 seconds you could not lift it again, so you had in your muscles 20 seconds or 10 lifts in you at that rep speed, of force, after that your force was temporary no longer. Yes I know that sounds a bit daft, but that is actually what happens, and if you had lifted the same weight up and down in .5 of a second up and .5 of a second up, you would have most probably failed to lift the weight in 10 to 12 seconds. Meaning you have used up your temporary force up far faster.

So let’s “just” {please this is just an example to get my point over} say for an example you had 1000 forces to lift the weight, in the fast, you used up this force far far far faster, meaning if you both lift the weight for a set time, and do “not” lift until momentary muscular failure, the faster reps “must” be using up more force faster, as you fail faster lifting faster. Example of how the fast are using more force and faster, more energy used, more distance the weight has been moved, faster to muscular failure, the EMG states more muscle activity or muscle force. I know all the above sounds a bit complicated, but there is total since in there.

Thank you again for you time and help, not sure about the acceleration, hope to learn more on that.


Wayne

 

[sophiecentaur]

Wayne
You still don't get it, do you?
Whatever you 'feel' in your arms and whatever you think that machine is telling you, my points 1,2 and 3 still apply.

1. You can't say it's accelerating constantly all the time. If it were, then it would still be moving at the top. It isn't because, of course, it stops at the top. And, in practical terms, you have no way of knowing without measuring the velocity at very short intervals over the whole of a lift.

2. For the weight to start at zero velocity and to end at zero velocity then the average force Has to be weight. This may annoy or confuse you and it may not be what you think the machine is telling you but is true beyond any shadow of a doubt. You could ask Sir Isaac Newton himself and he would tell you the same.

3. You actually don't know what you mean when you talk of "total force" because you are after an arm waving description of what it feels like to do a lift. I really don't know why you won't accept (from not just me but others, too) that your idea can't be stated in Physics.

What you could find out (but only by measurement) would be the maximum force, the total work done on the weights (thats N lifts times weight times height) and the Power developed (dividing the total work done by the time for the whole exercise).

I don't understand why that isn't enough.
All the other acres and acres of figures you have written on these pages have been pretty much wasted. I'd bet that no one has actually read more than a few sample lines of it and then given up.

How can you be so sure that your question is reasonable when, in the same breath, you admit to not knowing much real Physics?

I have seen some of the movies on your website and I am really impressed by your dedication to your sport. You are clearly a bit of an expert on the practical aspects of it and your advice on how to do the exercises and how to body build is, no doubt, correct. But, when it comes to the Physics of the situation, you just have to be much more rigorous and 'go along with the rules'. Those rules say that you are on a fruitless quest and have been talking mostly rubbish. Why are you bothering and why can't you just accept what you are being told? Do you just like a friendly chat - is that what it's all about?

That is a DEAD PARROT!

 

[waynexk8]

O.K. I'll try to help you understand once again.

Ok thx.

Let's say in your above example the load is 800N(81.5kg) while your maximum force ability is 1000N.

Right.

For the first 0.4sec you're applying your Fmax

Right.

(although in reality that's biomechanically impossible)

Right.

so the net force is 1000-800=200N.

This is where you sort of lose me. As you saying ? The net force on me and the weight = 200N ? If so I am still with you, but why say the net force on my muscles and the weight ? AND WHY take 800 from 1000 ? As we are only concerned with the weight/force on me, my muscles.

So am I not right in saying what physics states, is the net force on me, my muscles is only the weight = 800N ? But that’s “only” when you are holding the weight of moving it very slow.

Now here’s where it get a little more complicated. As if you agree that there is 800N pushing down and me pushing up with 800N, the moment I try too accelerate it as fast as possible I will be pushing up with a 1000N, but not only that, but the weight of 800N will not only be pushing back down with just the 800N, but as of the acceleration components, the weight will pushing down more than 800N as of the action reaction law ?

So you you weight me and the weight on the scales, it could register ? 400 pounds, but the faster I try to accelerate, the more weight/force is pushing back onto my muscles as the reading on the scales goes up and up the faster I try to accelerate.

So your acceleration for those 0.4sec is a=F/m=200/81.5=2.45m/s^2

Right.

For the last 0.1sec you let the gravity decelerate the load so the net force is -800N

Not sure on you here, are you saying that only gravity does the work and force on the weight for the last 0.1 of a second ? If so I do not agree there, as I have to keep pushing for the weight to keep moving. I decelerate the weight, if the weight was heavier or lighter, or if gravity had more force or less force, I my muscles decelerate the weight, I am in full control, especially on 80% if we were talking of 20%, then the accelerating components will be helping moving the weight after a certain speed/force is applied to the weight, as if I immediately stopped, the weight would keep moving.

So your acceleration for those 0.1sec is a=F/m=-800/81.5=-9.81m/s^2 and obviously it's equal with g.

Sorry you lose me a little there ? Equal with g ? I know what g is, but not sure why you are saying equal with it.

As I proved you above you used positive acceleration equal with 2.45m/s^2 for the 80% and 4 times greater negative acceleration equal with -9.81m/s^2 for the last 20%.
That's why we're telling you the acceleration is offset by the deceleration.The average acceleration is always zero.

Right get what you’re saying here, and have done all along, your sort of saying, and I know this is different, but want you to understand that I understand. You’re saying that when my accelerations are higher than your forces, that your forces make up or balance out when my forces are on the decelerations ? BUT what I am saying, is that it may look that way on paper, but how do you think your 80 medium force, can make up for/to my 100 force ? I am saying to do this; you would need FAR more time using your 80 force for the tension to build up on the muscles.

As “IF” you did make up or balance this force up, thus make up or balance this opposite action reaction tension on the muscles, HOW and WHY do I use “more” overall/total energy ? How/why do I move the weight “more” overall/total distance ? How and why do I fail roughly 50% faster “more” ? How and why does the EMG read out far more muscle activity, force on the faster rep.

Four definite things prove you “CAN’T” be making up the force, see my point ?

Don’t you get, that your measure or force does and can never equally my higher force ! Thus you have to need more time to let your force put more tension on the muscles. A small force applied for a LONGER TIME 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.

In the equation of force F=mg+ma the "ma"

You lost me on ma, I need to know what that is for me to understand you point.

part is always zero so F=mg or else the force is equal with the weight.No Mean additonal force is required.The forces "make up","balance out" or call it whatever you like.

I don't believe it's possible to get a more simple explanation.

This is the last time I bother to even read that particular question.

You "fail" in a weight lifting set when you have no longer enough energy to apply force equal with the weight.

You can NOT just say it’s the energy that fails or runs out, that is non scientific, very biased and unfair.

You have no temporary energy left and NO temporary force left. As we are not in this case taking about energy are we ? So if I cannot lift 400 pounds above my head, you will say I have not enough force, right ? And if I can lift up 200 pounds 30 times you will say in the end you did not have enough force left right ? Well actually I did not have {and I am being scientific, non biased and fair here} force “and” energy left, you can’t have one without the other, you “HAVE” to include force and energy when you say I hit momentary muscular failure faster than you.

I ran out of temporary force because I used up all my temporary energy, or/and I used up all my temporary energy because I used up all my temporary force. BOTH are used, and used up. I STILL have energy and force left, but I need to recuperate to get them both back.

"Failing" faster with a certain kind of lifting means that with this kind of lifting you spend energy at a higher rate...NOT that you use more force.In all cases the average force is equal with the weight.

Ok if you want to think backwards, tell me “why” I use up my energy faster ? Or with this kind of lifting why do you think you spend energy at a higher rate ?

More acceleration/force ? if not what and why ?

It's the fluctuations of force that are more energy demanding.That's the only scientific answer you can get.

But this is what I “have” be saying all along, the high fluctuations of force that are more energy demanding, meaning if they are more demanding as you just said, means your forces can not make up or balance out, right.

Wayne

 

[waynexk8]

Wayne
You still don't get it, do you?
Whatever you 'feel' in your arms and whatever you think that machine is telling you, my points 1,2 and 3 still apply.

1. You can't say it's accelerating constantly all the time. If it were, then it would still be moving at the top. It isn't because, of course, it stops at the top. And, in practical terms, you have no way of knowing without measuring the velocity at very short intervals over the whole of a lift.

Right.

2. For the weight to start at zero velocity and to end at zero velocity then the average force Has to be weight. This may annoy or confuse you and it may not be what you think the machine is telling you but is true beyond any shadow of a doubt. You could ask Sir Isaac Newton himself and he would tell you the same.

Ah, but that’s what I am not saying is not true, meaning you’re not understanding what I am trying to say, I am saying the above is right, I always have.

I am saying that my higher acceleration forces, and they are higher and always higher than the slow, as my higher force on the transition from negative to positive could be as high as 140% but my normal higher forces are say a 100 to your 80, thus higher.

My point is, when you push out a 100 force there is a 100 force/tension on the muscles, right ? Now you only push out a 80 force there is a 80 force/tension on the muscles, right ?

You have never put a 100 force/tension on the muscle, right ? So why and how do you think that your 80 force/tension can equal a 100 ? As 80 = 80 not a 100. It would take far far far longer in time for your 80 to make up the force/tension on the muscles/object

I am saying I am doing more damage with my higher forces, and your lower forces can NOT do the same damage in the same time frame, that’s why the below happens;

1,
I use more energy,

2,
I move the weight more distance,

3,
My muscles fail faster in the faster reps, proving they must be doing more damage,

4,
EMG agrees.

5,
I am NOT saying your questions are wrong, I am just stating my higher impulse, my impulse force with respect to time, YOUR SMALL FORCE applied for a long time can produce then same momentum change as MY LARGE FORCE applied briefly, because it is the product of the force and the time for which it is applied that is important.

Please if you do not agree, all I would like is that you do/can understand what I am just trying to say ? I am not at this moment in time concerned who is right or wrong, just that you see my point, as it’s quite frustrating when something argues over you, but they are not actually seeing my point, as they keep repeating what I agree with.

No time for the below its late here.

3. You actually don't know what you mean when you talk of "total force" because you are after an arm waving description of what it feels like to do a lift. I really don't know why you won't accept (from not just me but others, too) that your idea can't be stated in Physics.

What you could find out (but only by measurement) would be the maximum force, the total work done on the weights (thats N lifts times weight times height) and the Power developed (dividing the total work done by the time for the whole exercise).

I don't understand why that isn't enough.
All the other acres and acres of figures you have written on these pages have been pretty much wasted. I'd bet that no one has actually read more than a few sample lines of it and then given up.

How can you be so sure that your question is reasonable when, in the same breath, you admit to not knowing much real Physics?

I have seen some of the movies on your website and I am really impressed by your dedication to your sport. You are clearly a bit of an expert on the practical aspects of it and your advice on how to do the exercises and how to body build is, no doubt, correct. But, when it comes to the Physics of the situation, you just have to be much more rigorous and 'go along with the rules'. Those rules say that you are on a fruitless quest and have been talking mostly rubbish. Why are you bothering and why can't you just accept what you are being told? Do you just like a friendly chat - is that what it's all about?

That is a DEAD PARROT!

Wayne

 

[douglis]

This is where you sort of lose me. As you saying ? The net force on me and the weight = 200N ? If so I am still with you, but why say the net force on my muscles and the weight ? AND WHY take 800 from 1000 ? As we are only concerned with the weight/force on me, my muscles.

I stopped here.It was my mistake to try to make my point with physics equations.It's obvious that you can't follow them.That's not a bad thing but I don't have the time and the patience to explain every single line that I write.

I suggest to leave out physics.We have two push ups studies that examine exactly you're looking for.In physics there's no such thing as "total force" but in biomechanics there's a magnitude called Total Muscle Activation.
Let's discussed them via mail because obviously they're unrelated with physics.

You can NOT just say it’s the energy that fails or runs out, that is non scientific, very biased and unfair.

What makes you think that you're able to judge what's scientific and what's not?
But even common sense tell us that energy runs out and force is applied.Force doesn't run out.It's either applied or not.

You have no temporary energy left and NO temporary force left. As we are not in this case taking about energy are we ? So if I cannot lift 400 pounds above my head, you will say I have not enough force, right ? And if I can lift up 200 pounds 30 times you will say in the end you did not have enough force left right ? Well actually I did not have {and I am being scientific, non biased and fair here} force “and” energy left, you can’t have one without the other, you “HAVE” to include force and energy when you say I hit momentary muscular failure faster than you.

I ran out of temporary force because I used up all my temporary energy, or/and I used up all my temporary energy because I used up all my temporary force. BOTH are used, and used up. I STILL have energy and force left, but I need to recuperate to get them both back.

You can't lift 400 pounds above your head because you don't have the ability to apply that force.
When you "fail" after you have lifted 200 pounds 30 times,even though you have the ability to apply that force,you have no longer the required energy to apply that force.You run out of energy.You didn't run out of force.

Ok if you want to think backwards, tell me “why” I use up my energy faster ? Or with this kind of lifting why do you think you spend energy at a higher rate ?

More acceleration/force ? if not what and why ?

Not more acceleration...neither more force.The mean force is the weight which means "the forces balance out...make out" or whatever.Accept that fact.
The generation of force(fluctuations are generation/degeneration cycles) is more energy demanding than the maintenance of force(as if holding the weight).There're biology studies that prove that.
That's a perfectly scientific answer.