Impulse/force in pounds for the time frame

[waynexk8]

Could you please read page 16, c.

http://homepage.mac.com/wis/Personal/lectures/biomechanics/BiomechanicsLectures.pdf

Wayne

 

[waynexk8]

Wayne. As you do not know how that EMG machine works, you cannot insist about what it tells you.
If you insist "it must" it's up to you to prove it. Our physics is quite self consistent at this level and needs no further justification. It's innocent until YOU prove it guilty.

Wayne. As you do not know how that EMG machine works, you cannot insist about what it tells you.
If you insist "it must" it's up to you to prove it. Our physics is quite self consistent at this level and needs no further justification. It's innocent until YOU prove it guilty.

However I have proved the physics equations are not right, as they have not added all the variables in, if they had, then why/how do you think that {and there are many studies on this} if you fail, being you hit temporary muscular failure roughly 50% faster when doing the faster reps, that the slower reps use the same total or overall force ? Also you have not given me any equations.

As they cannot, as the faster reps have used up this total overall force faster. Let me prove this in more detail.

Clone a lifts slow, he lifts 80 pounds up and down at 3/3 for 60 seconds and let’s say at exactly 60 seconds he hit momentary muscular failure, meaning he cannot lift the weight again.

Clone b lifts fast he lifts 80 pounds up and down for 30 seconds and let’s say at exactly 30 seconds he hit momentary muscular failure, meaning he cannot lift the weight again.

You and D are saying at both 60 seconds and 30 seconds both clones have used the same temporary total or overall force, that is right is it not ?

1,
If both held the weight half way up, they would both use up all their temporary total or overall at the exact same time. So on my faster reps, when the fast hits momentary muscular failure 50% this means he has used up his temporary total or overall force up faster, right ? Actually he’s used up his temporary total or overall force and energy faster, as you cannot have or use one up without the other.

2,
A,
The force at work on a barbell with a weight of 200kg and an Acceleration of 2m/s 200 x 2 = 400N = 400N at every instant in time ?
B,
The force at work on a barbell with a weight of 200kg and no acceleration, the weight is moving at .2m/s 200 x .2 = 40N at every instant in time ?


3,
As you both seem to say that when we both hit momentary muscular failure, whether I am lifting the weight for 30 seconds or 60 seconds, you say I am using the same temporary total or overall force? [/b]So are you also saying, that if I hold the weight half way up for 30 seconds and then 60 seconds, that I am still then using the same temporary total or overall force ?[/b] But if we both use the exact same temporary total or overall force when we both hit temporary muscular failure, what would happen if I {as I have more time then you, as I fail roughly 50% faster} started lifting a lighter weight, then stopped when you hit temporary muscular failure ?

4,
Why do I use more energy the exact same time I use more acceleration, and the exact same time I use more acceleration, I “have” to use more force, but your saying this is not connected. Why is it that you only use the same energy, when you have traveled the same distance as me ? As in 6 seconds I move the weight 6 times further.

5,
Are you saying it takes the same temporary total or overall force to move a weight at a constant speed, and to accelerate it as fast as possible.

6,
How/why does the EMG state more muscle activity in the faster reps ?

Wayne

 

[waynexk8]

Acres of blather, yet again, Wayne.

I have to explain what happens someway. And when/now I do, I have no reply ?

I, and everyone else, understand what RMS means; it is very well defined.

Sorry, I do not know what RMS means, please could you say ?I think it’s a measure of the of a varying quantities that are positive and negative like in the accelerations and decelerations in the EMG ? Is this about right please ? Ignoring if the EMG is right or wrong ?D. you said the RMS did not measure positive and negatives, but its actually meant for measuring this

What your machine does in order to produce the "RMS" figures it displays is a complete mystery to both you and me. We can draw no conclusions.

I just think it’s that the peak highs it measures, do not and cannot be made up by the slows force when the fast is on the decelerations. It’s like in how I explained how a bridge will break under the fast but not the slow.

Btw, douglis, RMS is only your value of 0.7ish in the case of sinusoidal situations. RMS value depends totally on the time profile and, to measure it accurately, you need a good number of samples over the cycle. A weight lifting cycle will be far from sinusoidal.

The machines reads as many samples as I put it to, and in any time frame, and “all” the times I have done the tests, with light of high loads it always reads higher for the faster reps, all samples and times on both fast and slow tests are done by the machine, and are the exact same. Not sure what you mean by this here please ? RMS is only your value of 0.7ish in the case of sinusoidal situations.

Thank you for your time and help again.

Wayne

 

[waynexk8]

Well...you don't have to be an expert.It's not quantum mechanics.Every manual states that rectifies the raw EMG with the RMS method.
We all learned at school that the RMS is the 70% of the positive peak value.

Show me that anywhere on the net ? WHY would an EMG read out on 70% of the peak ? it’s a measure of the of a varying quantities that are positive and negative like in the accelerations and decelerations in the EMG ? Is this about right please ? Ignoring if the EMG is right or wrong ?D. you said the RMS did not measure positive and negatives, but its actually meant for measuring this


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

It’s the average of the positive and negative muscle activity, forces.

http://www.analytictech.com/mb313/rootmean.htm

No...you didn't meausure any "total or overall muscle activity/ force output".You also did NOT measure the peak.You meusured the RMS(70% of the positive peak).
The Total Muscle Activation is meusured by the integrated EMG which is the area below the EMG curve.

Read over the above and the internet to see what RMS means.

All these "explanations/scenarios"() are the effect of the peak force.

Yes that’s what I have been saying all along, the peak forces of the accelerations do more damage, and your lower forces on the slow, DO NOT and CAN NOT do as much damage in the same time frame CAN THEY ? That is a question, and that means that as they can’t do as much damage, means they cannot balance out or make up the force, right ?

Wayne

 

[waynexk8]

Please this is a pure physics question, could we get an answer please.

Sorry I have been very busy.

One thing, could we possibly answer my original question, before I come back on the rest, as I am not sure if I am getting my point over, As I have and always been on about the extra force, the ground, or in our case the more muscles reaction force, thus more tension on the muscles, that is more from the peak acceleration force from the transition from negative to positive, the MMMT {Momentary Maximum Muscles Tensions}. HOPE this question. Newton’s Third Law, is more of a physics equation you can easy work out and get your teeth into.

I think/hope we will all agree, in that if you lift a weight up, then immediately lower, then in a completely different lift, you lower, then immediately lift the weight, there will be more force needed on the second lift, as the moment I lifted the first 80 pounds, there would be just 80 pounds to lift, however, if 80 pounds is traveling down 20 inch in .5 of a second, the acceleration components will appear to make the 80 pounds, far far far heaver that it actually is. Could we please try and work out how much more heavy the weight would be, or register on a scales if you let it drop, and then the extra force that is needed on the second lift ? Say in % ways.

A,
You lift 80 pounds up, from a still start, to 20 inch in .5 of a second; the weight is moving at 40 inch per second, and you then immediately lower the weight back down in .5 of a second.

B.
You lower 80 pounds down 20 inches; you immediately lift the weight back up 20 inch.

Thank you all for your time and help.

Wayne

Wayne

 

[sophiecentaur]

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

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. My physics does not have a "temporary" force, it has a strict definition for RMS and only uses the term "total force" to describe the sum of forces applied at one instant of time.
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?

 

[douglis]

Show me that anywhere on the net ? WHY would an EMG read out on 70% of the peak ? it’s a measure of the of a varying quantities that are positive and negative like in the accelerations and decelerations in the EMG ? Is this about right please ? Ignoring if the EMG is right or wrong ?D. you said the RMS did not measure positive and negatives, but its actually meant for measuring this


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

It’s the average of the positive and negative muscle activity, forces.

http://www.analytictech.com/mb313/rootmean.htm

Read over the above and the internet to see what RMS means.


Wayne

In your second link you can see the most simple explanation of what the RMS means.

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.

 

[waynexk8]

Waiting for an E-mail back from roger bartlett.

Introduction to Sports Biomechanics: Analysing Human Movement Patterns.

Wayne

 

[sophiecentaur]

And you seriously think that his reply will bring in a whole "new" lot of Physics into the topic?
The amount of Physics needed to answer your original question is minimal because it can only answer a question that is actually asked 'with Physics'. Whilst you ask for average forces that are not weight and total force over time no one can give you an answer because the questions lie outside the grammar and vocabulary of Physics.

 

[waynexk8]

And you seriously think that his reply will bring in a whole "new" lot of Physics into the topic?
The amount of Physics needed to answer your original question is minimal because it can only answer a question that is actually asked 'with Physics'. Whilst you ask for average forces that are not weight and total force over time no one can give you an answer because the questions lie outside the grammar and vocabulary of Physics.

And you seriously think that his reply will bring in a whole "new" lot of Physics into the topic?
The amount of Physics needed to answer your original question is minimal because it can only answer a question that is actually asked 'with Physics'. Whilst you ask for average forces that are not weight and total force over time no one can give you an answer because the questions lie outside the grammar and vocabulary of Physics.

So are you saying that what you and D. said, in that the forces needed to moved a weight up 1m and down 1m 6 times = 12m in 6 seconds, and that the forces needed to moved a weight up 1m and down 1m 1 time = 2m in 6 seconds, are not the same, because it can’t be worked out ? But what about the below.

What if we changed it to Tension, that’s the tension on the muscles that the force creates, as that is the main.

Sophiecentaur is the below right please ? I did not do it.

Here's a very simplified analysis of what's going on, but it will work to explain the concept without the use of calculus. Let's assume we're comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.

Known: Mass(m) =100kg (220lbs) Acceleration(a)=?? Distance(d)=1m

First let's solve for the acceleration required to move the weight 1m in the time frames of the sets.

Calculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):
d=1/2at^2
1m=1/2*a*(0.5s)^2
a=2(1m)/(.25s^2)
a=8 m/s^2

Calculate a, to travel 1m in 2s (Raising set 2):
d=1/2at^2
1m=1/2*a*(2s)^2
a=2(1m)/(4s^2)
a=0.5 m/s^2

Calculate a, to travel 1m in 4s (Lowering set 2):
d=1/2at^2
1m=1/2*a*(4s)^2
a=2(1m)/(16s^2)
a=.125 m/s^2


Now let’s solve for the forces required to accelerate the weight

Calculate the force required to raise 100kg, 1m, in 0.5s:
Sum of forces=ma
F1-mg=ma
F1-(100kg)(9.81m/s^2)=(100kg)(8m/s^2)
F1=(981+800) kg m/s^2
F1=1781 kg m/s^2 required to raise the weight 1m in 0.5s

Now compare it to the force required to raise 100kg, 1m, in 2s
Sum of forces=ma
F2-mg=ma
F2-(100kg)(9.81m/s^2)=(100kg)(0.5m/s^2)
F2=(981+50) kg m/s^2
F2=1031 kg m/s^2 required to raise the weight 1m in 2s.




So does it mean that the forces needed to raise a weight 1m in 2 seconds = 1031 and to raise the weight in .5 of a second = 1781 ?

Wayne

 

[douglis]

So are you saying that what you and D. said, in that the forces needed to moved a weight up 1m and down 1m 6 times = 12m in 6 seconds, and that the forces needed to moved a weight up 1m and down 1m 1 time = 2m in 6 seconds, are not the same, because it can’t be worked out ? But what about the below.

The forces "can be worked out" only if you know what kind of forces you're looking for.
In physics there're two values of forces.The average and the momentary value at each specific instant of time.The first is the weight in both cases of your example and for the second we need more data.

Here's a very simplified analysis of what's going on, but it will work to explain the concept without the use of calculus. Let's assume we're comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.

Wayne

The assumption is wrong hence the calculations are wrong too.You start and end at rest so you don't accelerate for 100%.Every acceleration is accompanied by a deceleration of equal magnitude.The average acceleration is zero.

 

[waynexk8]

The forces "can be worked out" only if you know what kind of forces you're looking for.
In physics there're two values of forces.The average and the momentary value at each specific instant of time.The first is the weight in both cases of your example and for the second we need more data.

Sophiecentaur seemed to say what I ask can’t be worked out; this is why I asked the above.


There must be an overall or total force over time ? Not just average and momentary value at each specific instant of time ? This is where the EMG or/and force plate comes in, as I think it seems impossible to work out other wise. How in more data, thought I gave you all that, why don’t you put this in for both lifts ?

Why do you honestly think, or are you just maybe say, this as we have debated for so long, {by the way, someone put me onto someone who works with EMG, am showing him mine} but why do you think that your lower forces can make up or balance out when mine are on the decelerations ?

Let me put it very simple, when I or you press up with a high or low force, there is an opposite reaction on the muscles, right ? This we call tension. So my press puts a 100 tension on the muscles for say 60% your forces never put a 100 forces on the muscles. My force on the transition, could put as high as 140 force thus tension on the muscles, your never puts a 140 tension on the muscles, so has not my forces made more of a dent in the muscles, or a tear ? Yes they have, have they not, so what gets me, is how you think that your lower forces done over the same time frame can put/do this same dent or tear in the muscles ? With your lower forces ? They CANT made the same DEPTH of dent tear can they ?

Just go back and see what I said on this thread on my bridge theory, and the others.

Then PLEASE say is and WHY you do or don’t agree, but please JUST say that you do see the point I am trying to make, if you agree or not, can you two see the point I am trying to make, then please say why you think it’s wrong, then think of why I hit failure faster, as it must be these higher peak forces, and the very high peak force from the transition from negative to positive, the MMMT. {Momentary Muscular Maximum Tensions}

The assumption is wrong hence the calculations are wrong too.You start and end at rest so you don't accelerate for 100%.Every acceleration is accompanied by a deceleration of equal magnitude.The average acceleration is zero.

Yes I know that, please I just want to know what I asked above so I can go on with something else.

It can’t be a deceleration of equal of the acceleration, I or you need to find out more on this, we need to look at things going up very fast and stopping very fast ? What about a grasshopper, let’s see if they say anything anywhere, or a ? Piston in an engine, we should be able to find some information on this, surely.

One thing, could we possibly answer my original question, why can’t anyone answer this ? Its not that hard, if I lower 80 pounds at 2m/s how much force is needed to bring it to a stop, and how much forces will go onto my muscles, it’s a lot more than 80 pounds I know that. As I have and always been on about the extra force, the ground, or in our case the more muscles reaction force, thus more tension on the muscles, that is more from the peak acceleration force from the transition from negative to positive, the MMMT {Momentary Maximum Muscles Tensions}. HOPE this question. Newton’s Third Law, is more of a physics equation you can easy work out and get your teeth into.

I think/hope we will all agree, in that if you lift a weight up, then immediately lower, then in a completely different lift, you lower, then immediately lift the weight, there will be more force needed on the second lift, as the moment I lifted the first 80 pounds, there would be just 80 pounds to lift, however, if 80 pounds is traveling down 20 inch in .5 of a second, the acceleration components will appear to make the 80 pounds, far far far heaver that it actually is. Could we please try and work out how much more heavy the weight would be, or register on a scales if you let it drop, and then the extra force that is needed on the second lift ? Say in % ways.

A,
You lift 80 pounds up, from a still start, to 20 inch in .5 of a second; the weight is moving at 40 inch per second, and you then immediately lower the weight back down in .5 of a second.

B.
You lower 80 pounds down 20 inches; you immediately lift the weight back up 20 inch.

Thank you all for your time and help.

Wayne

 

[sophiecentaur]

Wayne (yet again)
I have looked at some of your last comments and I think I am getting to understand your problem. I should have got a clue from the extreme length of your posts. Your problem is that you don't actually know what question to ask so you just relate case after case of what you have experienced, in great detail, and want someone to make up a suitable question and then answer it for you.

I think that you may, in fact, have begun to accept that the mean force on an object, that starts and ends stationary, is just its weight. Good.

You still seem to think that there is a definite answer about Maximum force, related to rate of lifting. But you haven't accepted that it depends solely on the dynamics of what your muscles happen to be doing. It will depend on Energy / Food supply to the cells and all sorts of other chemical matters that Mechanics (which is the bit of Physics you are invoking) can discuss. There are many different combinations of force and time that can take your weight up to the top. A good lifter will make the best of what he's got and have found some sort of optimum.

There is, of course, an unspecific connection between lifting rate and the force needed and also the Power needed but you're far more dealing with Energy flow than just forces. Once the weight is back down again, all the energy you put into the exercise is lost (ignore a bit of bounce in your tendons) and the subsequent lifts will need equal amounts of energy. You will 'fade' when the muscle cells run out of temporary food and the input rate can't keep up and also the waste products build up (I imagine it's all Anaerobic?). You know all this, I'm sure, so why not apply it to your question?

What you CAN do, is to calculate the least force needed for a certain mass - height - time cycle. That assumes a constant lifting force for a while, followed by zero lifting force until the weight is brought to a halt by gravity. There's a bit of algebra which gives you a quadratic equation to solve. I tinkered with it enough to see that it can be done and the details seem to check out. BUT this is totally irrelevant to a real lift because the force is changing all the time (your machine tells you that). You can Measure this force in any case, so isn't that good enough for you? Perhaps a good lifter manages to keep his maximum force as low as possible and spread the demand over the lift. That would be most efficient, I think.

I think we have finally put to bed the concept of "Total Force".

 

[douglis]

Sophiecentaur...I'll try to help you get into Wayne's incredible mind!

In biomechanics there's a term called Total Muscle Activation(TMA) that is a product of the processing of the EMG signal and is used by some scientists.The TMA is calculated by the integration of the EMG curve.
If you have the time...take a look at the "Materials and methods" paragraph at the below study.The procedure is described pretty well:
http://jmbe.bme.ncku.edu.tw/index.php/bme/article/viewFile/635/839

I'm sure that the TMA is exactly what Wayne describes as "total/overall force".

Obviously the TMA is not a physics term and should not be discussed here.
But since the average force is the same(the weight) in fast and slow lifting we can assume that also the average muscle activation is the same.Hence the integration of muscle activation in respect of time(TMA) must also be the same.

It's interesting that in the above study,if you check the tables,the TMA per second is greater for slow push ups!Who knows why?

 

[sophiecentaur]

Sophiecentaur...I'll try to help you get into Wayne's incredible mind!

In biomechanics there's a term called Total Muscle Activation(TMA) that is a product of the processing of the EMG signal and is used by some scientists.The TMA is calculated by the integration of the EMG curve.
If you have the time...take a look at the "Materials and methods" paragraph at the below study.The procedure is described pretty well:
http://jmbe.bme.ncku.edu.tw/index.php/bme/article/viewFile/635/839

I'm sure that the TMA is exactly what Wayne describes as "total/overall force".

Obviously the TMA is not a physics term and should not be discussed here.
But since the average force is the same(the weight) in fast and slow lifting we can assume that also the average muscle activation is the same.Hence the integration of muscle activation in respect of time(TMA) must also be the same.

It's interesting that in the above study,if you check the tables,the TMA per second is greater for slow push ups!Who knows why?

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.

 

[waynexk8]

Wayne (yet again)
I have looked at some of your last comments and I think I am getting to understand your problem. I should have got a clue from the extreme length of your posts. Your problem is that you don't actually know what question to ask so you just relate case after case of what you have experienced, in great detail, and want someone to make up a suitable question and then answer it for you.

Maybe I am not asking it right in the scientific way, however I do know what I am asking, and what I mean, and how in my opinion and a EMGs machine that I am right. A EMG is a well recognized machine in the sports and medical industries, tell me why you don’t go along in real World practical tests/studies, that have show there to be more muscle activity {force} when the muscle is used faster ? Do you both know what a force plate is ? It’s another well known machine in the sports and medical industries, if I buy one of those, or get professional tests done, and it goes the same way as my EMG, will you then look very deep into your equations ? Another thing that gets me, is I gave you may simple scenarios like the “bridge” that shows there must be more overall or total force when moving fast, but you did not comment on any of them ?

You muscles have a temporary limited force output, right ? I am asking which repetition speed uses that faster, or puts more tension on the muscles in the same time frame, so here I show you how the faster must put out more force per the same time, thus more tension on the muscles.

1,
We all agree that the faster repetition on using 80% of your 1RM, {repetition Maximum} that you fail rough 50% faster doing the fast.

Question,
As I fail 50% faster than the person moving the weight slow, are you two saying that after I lift the weights for 15 seconds and you for 30 seconds, that we have both used the same total or overall force ? Yes you have both been saying this all along.

Answer,
If I fail 50% faster, that must mean that I am using up more force per unit of time, right ? As I “have” used up my temporary muscular force, so as I “have” used it up, “how” can you say I have not used up my temporary force not faster than you, when I have ? So if we both lifted for 15 seconds, I would hit momentary muscular failure, meaning I “have” used up all my temporary muscular force, right ? But you have not used it up, thus I “must” be using more overall or total force per unit of time than you, right ? As the EMG states, so does the distance the weight is moves says, so does power say.

Also, if I hit momentary muscular failure at 15 seconds with that amount of weight, what would happen then if I picked up a lighter weight and keep on lifting it, at the 30 seconds when you hit temporary muscular failure, would I still have used the same total or overall force as you the slow, or used more ? Which let's me use more total or overall force, or the same as you doing the slow for 30 seconds ? Doing the lifts for 15 seconds, or doing the lifts for 15 seconds to momentary muscular failure and then lifting a lighter weight and keep going until 30 seconds ?

Or am I asking the wrong question, should I be asking the same of strength not force, but force is a push or pull, using force, that’s strength.

It could just be, that you thinking mistaken certainties, as you have not given any kind of equation that you are right. Mistaken certainties are things you're sure are true but which, in fact, are not, like WHY do the fast use more energy ? Just for once try and think and prove how that is, if as you think I am not using more force per unit of time ?

Also, you shot putt a weight, my weight goes further than yours, are you telling me, that to putt a shot further, you don’t use more overall or total force, but the exact same force as to move it less distance, then what do you use more of the move it further ? More acceleration, but that uses more force.

So are you saying that when something accelerates, and accelerates only for 600mm to something moving at a constant speed for 166mm, that both use the exact same total or overall force ?

I think that you may, in fact, have begun to accept that the mean force on an object, that starts and ends stationary, is just its weight. Good.

You are taking of the average force right ? Average in this instance/question/debate means nothing, will show you why tomorrow.

Sorry its 2.15, will have to get back to the below.

You still seem to think that there is a definite answer about Maximum force, related to rate of lifting. But you haven't accepted that it depends solely on the dynamics of what your muscles happen to be doing. It will depend on Energy / Food supply to the cells and all sorts of other chemical matters that Mechanics (which is the bit of Physics you are invoking) can discuss. There are many different combinations of force and time that can take your weight up to the top. A good lifter will make the best of what he's got and have found some sort of optimum.

There is, of course, an unspecific connection between lifting rate and the force needed and also the Power needed but you're far more dealing with Energy flow than just forces. Once the weight is back down again, all the energy you put into the exercise is lost (ignore a bit of bounce in your tendons) and the subsequent lifts will need equal amounts of energy. You will 'fade' when the muscle cells run out of temporary food and the input rate can't keep up and also the waste products build up (I imagine it's all Anaerobic?). You know all this, I'm sure, so why not apply it to your question?

What you CAN do, is to calculate the least force needed for a certain mass - height - time cycle. That assumes a constant lifting force for a while, followed by zero lifting force until the weight is brought to a halt by gravity. There's a bit of algebra which gives you a quadratic equation to solve. I tinkered with it enough to see that it can be done and the details seem to check out. BUT this is totally irrelevant to a real lift because the force is changing all the time (your machine tells you that). You can Measure this force in any case, so isn't that good enough for you? Perhaps a good lifter manages to keep his maximum force as low as possible and spread the demand over the lift. That would be most efficient, I think.

I think we have finally put to bed the concept of "Total Force".

when a muscle fails faster, why does it fail faster ? As its working harder or easier ? I and most would say harder, so that would mean using more force, yes ? If no, then why ?


Thank you for the rest, that sounds very interesting, dammed wish I had time to answer, as love what your saying/thinking. So please how do we calculate the least force needed for a certain mass.

Wayne

 

[waynexk8]

When I say which uses the most overall/total force, I mean, if you lift 80 pounds for 10 seconds, and then lift 40 pounds for 10 seconds, you will have used more overall/total force lifting the 80 pounds for 10 seconds.

And you will in the above use more energy lifting the 80 pounds, right ? Thus EVERY time you use MORE energy, you have to use MORE force, right ? As you are using more acceleration and velocity = more force/energy.

You can not show me a faster lift, where you don't use more energy, when you are not using more acceleration/velocity and these = force,

Wayne

 

[sophiecentaur]

You are asking about Work Done and forces. We have told you all we can about those strictly defined quantities. You have your own special set of meanings for those and other words you use.
Your machine tells you about your muscle activity. You ask why your musclus fade / tear under some extreme use. Clearly some forces and energy is involved but your muscles don't study Physics. You can't rely on them to behave as you want or even to do the best thing possible. They give up because you try to make them do more than they're prepared to do. They break because you abuse them into doing more than they're designed to do. What has that to do with Physics? They aren't like a dumb electric motor that will work or fail according to simple rules (that can be modeled fairly accurately). You will probably never know all the details of the biochemistry and biomechanics involved so stop trying to explain it all in terms of simple Physics.
Until you state your questions in the right words, used the right way, you can't get an answer.
Check up on and learn the accepted meanings of the terms you use. This 'Question and Answer' method that you insist on is just not working. You have to give just a little if you want a result.

 

[douglis]

When I say which uses the most overall/total force, I mean, if you lift 80 pounds for 10 seconds, and then lift 40 pounds for 10 seconds, you will have used more overall/total force lifting the 80 pounds for 10 seconds.

And you will in the above use more energy lifting the 80 pounds, right ? Thus EVERY time you use MORE energy, you have to use MORE force, right ?

Wayne

No...it's possible to use the same and even more energy by lifting explosively the 40 pounds even though the average force is the half.The huge fluctuations of force are very energy demanding.
In fact,there was a study that compared the energy expenditue of light weighted jump squats vs heavy squats.The jump squats used more energy but obviously their "total/overall force" was less.

For the last time.Greater energy expenditure doesn't equate greater force.

 

[waynexk8]

Back in full tomorrow, been very busy.

First, we all agree that there is far far far more Power in the faster, right ? But is not Muscular power, the combination between force and velocity ? So more Power = more force and velocity !

Wayne

 

[sophiecentaur]

Power = Force TIMES Velocity
Forget whether it's "muscle power" or power from anything else. Also, forget words like"far, far" - these things are just proportional. And Power is an instantaneous value, varying as force or velocity varies.
If you can accept this, then we have started talking the same language.

 

[waynexk8]

This was from an EMG expert. It shows my EMG results are quite right, and D. does not know the difference between the two EMGs, as basically there is no diffrence. What is basically proves, is that the on the faster reps, the higher peak forces from the accelerations, cannot be made up balanced out with the constant medium forces of the slower reps, when the faster reps are producing less force on the decelerations. This seems quite obvious to me, as if I struck something very hard like a heavy barbell, it’s going to put a huge jolt/tension on my muscles, do this 6 times in the same time frame as someone moving the barbell slow, and the muscles will, as they do feel very high tension on them and fatigue far far far faster than when moving the barbell slow.
Hi Wayne,

Thanks for considering our products - while our EMG systems are research orientated and will give you the best quality data, realistically in this situation the quality of your results will be far more dependent on the way that the system is used and the way that the data is interpreted. My advice would be to record the data as raw data - obtain the best raw EMG data that you can and then process the data to determine the results. By recording raw data you have the opportunity to reprocess the data in different ways to discover the best processing methods.

The difference between integrated mean EMG data and RMS EMG data is simply mathematical - the end results are simply derived from slightly different processing methods. I'd suggest a couple of books that you might want to obtain and read that discuss the different methods:

Cram JR, Kasman GS, Holtz J: Introduction to Surface Electromyography. Aspen Publishers ISBN 0-8342-0751-6
Craik RL, Otis CA : Gait Analysis:Theory and application. Mosby ISBN 0-8016-6964-2

Wayne

 

[waynexk8]

Power = Force TIMES Velocity
Forget whether it's "muscle power" or power from anything else. Also, forget words like"far, far" - these things are just proportional.

Right.

And Power is an instantaneous value,

So "if" we were using the exact same power over 10 seconds, and could mesure/read this power to be the exact same every .1 of a seconds, could we say I 100J in 10 seconds, and for 20 seconds would be 100J

This equation for power states that a powerful machine/muscle is both strong and fast ?

varying as force or velocity varies.
If you can accept this, then we have started talking the same language.

Do you mean if force and velociry goes up so does power ? Or when power goes up so must force and velocity.

Wayne

 

[waynexk8]

When I say which uses the most overall/total force, I mean, if you lift 80 pounds for 10 seconds, and then lift 40 pounds for 10 seconds, you will have used more overall/total force lifting the 80 pounds for 10 seconds.

And you will in the above use more energy lifting the 80 pounds, right ? Thus EVERY time you use MORE energy, you have to use MORE force, right ? As you are using more acceleration and velocity = more force/energy.

You can not show me a faster lift, where you don't use more energy, when you are not using more acceleration/velocity and these = force,

Wayne

douglis;3769291]No...it's possible to use the same and even more energy by lifting explosively the 40 pounds even though the average force is the half.The huge fluctuations of force are very energy demanding.

Sorry you got that wrong, I said if I lift 80 pounds for 10 seconds and 40 pounds for ten seconds, what I meant as I did not state speed or distance, was both amounts lifted the same speed and thus distance. This then would mean I have used more total or overall force and energy lifting the heaver weight.

In fact,there was a study that compared the energy expenditue of light weighted jump squats vs heavy squats.The jump squats used more energy but obviously their "total/overall force" was less.

You are one about different weights used, I am not.

For the last time.Greater energy expenditure doesn't equate greater force.

I am on about the same weight, the above was just to explain to you how you will use more total or overall force, it had nothing to do with the actual debate.

I move 80 pounds up and down 20 times in 10 seconds, you lift 80 pounds up and down 1 time in 10 seconds, I use more energy.

Greater energy expenditure does equate greater force, if you move the same weight faster in the same time frame.

Wayne

 

[douglis]

I move 80 pounds up and down 20 times in 10 seconds, you lift 80 pounds up and down 1 time in 10 seconds, I use more energy.

More energy yes...more peak force yes...but the average force is 80 pounds in both cases hence the forces "make up".

Greater energy expenditure does equate greater force, if you move the same weight faster in the same time frame.

Wayne

Well...since you discovered a new law...all you have to do is to prove it somehow or else it will remain a figment of your imagination.

 

[sophiecentaur]

Wayne
You still seem to be insisting that your muscles and the way your body controls them are much simpler than they are. No one has argued that doing things faster can knacker you more and can damage muscles. It's just that you seem to insist on having a simple Physics formula - but not real Physics - your brand of Physics, in which all the names are jumbled up and re defined. No wonder we can't help you.

 

[waynexk8]

1)The average force is the weight.

What your trying to say, or think, is that if you hold 80 pounds half way up for 10 seconds, and I move the weight up and down 20 times in 10 seconds, that we have both put out the exact same total/overall force thus tension on the muscles, right ?

“If” that were so, then why does anyone that uses 80% of their 1RM {repetition maximum} fail in lifting the weight, or/and hit momentary muscular failure say 50% faster ? I will tell you why. Its because the higher force needed for the accelerations, and for the decelerations, as on the decelerations you are still pressing with as much force as you can, its only on the very last portion of the lift that you immediately lower force very fast for the transition from positive to negative, then immediately on the negative there is force from the muscles thus tension on the muscles.

So there we have it, you fail faster, and this is basically because the faster reps “are” putting more tension on the muscles, so they fatigue/tire 50% faster. But your saying they don’t they fatigue/tire faster or put more tension on the muscles, your saying that they fatigue/tire 50% faster, because they use up more energy, and they use up more energy not because they are moving with greater acceleration = more force, or traveling with more velocity = more force, your saying they just they fatigue/tire faster because of no reason than they use more energy for some unknown reason, is that right ?

You also think that your medium forces can make up or balance out my higher forces, but you cannot say why ? However I can say why I say it’s not, as I move the weight 6 times more distance, use more power, that’s more energy and more work done.

Question to all please
If I was to lift and try to throw the same weight with open hands, are you saying then I would use more force ?

2)The peak force requires more data.If you use an accelerometer you can find the peak acceleration and then the peak force.

Yes.

3)Total or overall force doesn't exist in physics.What you're looking for is the integrated EMG or maybe you can describe it with gravity's impulse which is the same in both cases of your example(weight X 6 seconds).

So what about tension, can physics find out how much tension was put on so and so ?

No, that’s where the EMG and force plate come into action, these work out the total or overall force, or and average force.

If you think total or overall force does not exist in physics, why are you saying both lifts have the same total or overall force ?

We could call this total of overall force, total force with respect to time, so if I put 100 force on 80 pounds for 4 seconds, and you put 80 force on 100 pounds for 4 seconds, who put out the total or overall force ? As I would have moved the weight further, that would categorically say I did, right ? If not why ? If I put 100 force on 80 pounds for 4 seconds, and you put 80 force on 100 pounds for 6 seconds, who put out the total or overall force ? As I would have moved the weight further, that would categorically say I did, right ? if not why ? If I put 100 force on 80 pounds for 4 seconds, 79 pounds of force for 2 seconds, and you put 80 force on 100 pounds for 6 seconds, who put out the total or overall force ? As I would have moved the weight further, that would categorically say I did, right ? If not why ? And ? Means I am please asking you a question.

You seem to think, that as I move the weight further, I don’t use more total or overall force to do this ?

Wayne

 

[sophiecentaur]

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.

 

[waynexk8]

Wayne
You still seem to be insisting that your muscles and the way your body controls them are much simpler than they are.

That are far more complicated, and am trying to simplify things, as instead of muscles, we can just say this is a machine pushing the muscles with force, and if you put too much strain on that machine, it will fail faster than the machine you don’t put as much strain, and strain in this case is higher force output, and thus more tension on the machines moving parts. Put to more strain on the machine, as in making it move as fast as it can with very high accelerations, and there will be too much tension on the machine, and it will fail. Same are the car that have been driven further, its parts fail faster, and that’s because more acceleration force thus tensions on the moving parts.

No one has argued that doing things faster can knacker you more and can damage muscles.

Ok thanks for that.

But I see it as the higher acceleration forces, anytime you move faster, it could be 3/3 against 10/10 in the same time frame, but every time you try and move faster, there has to be more acceleration, thus higher forces on the muscles.

It's just that you seem to insist on having a simple Physics formula - but not real Physics - your brand of Physics, in which all the names are jumbled up and re defined. No wonder we can't help you.

Sorry I am trying, but if D. is right, then physics does not have an equation for total or overall force, in the same time frame “could not we be the first to try and find one ?” As physics is physics, we should be able to work this out somehow, or is it the power equation, power = force acceleration ? Thus we all know there is more power in the faster, if more power and more power = more force acceleration, then am I not right ?

Will read the other posts later and get back to them, thank you for your time and help again.

What if a bought a force plate ?

Wayne

 

[douglis]

your saying they just they fatigue/tire faster because of no reason than they use more energy for some unknown reason, is that right ?

No...the reason is very well known.Biology studies tell us that the fluctuations of force are more energy demanding.

You also think that your medium forces can make up or balance out my higher forces, but you cannot say why ?

Everyone has told you why.Because the average force is the weight in any case.This by definition means that "your" higher peaks are exactly balanced by "your" lower peaks and the result is "my" medium force(the weight).

“could not we be the first to try and find one ?”
Wayne

First learn to use the basic physics terms and leave your big plans for the much later future.