Impulse/force in pounds for the time frame

[sophiecentaur]

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.




Wayne

The point you are trying to make is not to do with Physics. It's to do with what you feel in your muscles and how you are trying to fit it to some kind of 'Mechanical Science', that you've invented. You have absolutely no idea of what your muscles are doing at the time they're doing it. There is a delay, corresponding to a large fraction of the time of one lift, in what you do and your consciousness of what you did. That's no basis for an experiment because the evidence is fundamentally flawed.

You must have spent hours and hours on this and it really isn't a good way to learn anything about a subject.
This is not uncommon. Many people write posts about their pet approach to some topic and insist on having it explained in their terms only. It so often fails because they are not prepared to start from scratch and get the subject properly sorted out. Douglis has very patiently, point by point, taken you through your problem and you just reply with remarks like:
"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."
His conclusion would be absolutely blindingly obvious to you if you had taken the trouble of approaching this from the beginning instead of doggedly insisting that your view is the right one. Science doesn't necessarily 'make sense' when you try to reconcile it with ideas which, themselves, don't make sense.
You could have completed a whole GCSE Physics Course in the time you have wasted trying to justify your misplaced views on these threads. If you had, then you would have no trouble understanding this really elementary bit of Classical Science. You would also see how simple Physics can't help you with this 'undefined' question that you keep trying to ask.

As I hinted before, I really question your motives here. I suspect you are just attention seeking.
I'll bet you haven't read all the words in this post, too.

 

[waynexk8]

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.

Ok I will be able to except that the total or overall force will not be able to be equated by physics, I just find it odd, as this the below F = Ma seems so easy, but then when I ask for the force for 80 pounds moving 20 inch in .5 or a second, and 80 pounds moving 3.3 inch in .5 of a second, it can’t be done ?

In total or overall force, I mean the muscles only have a temporary limited amount of force before its used up, so let’s say I can lift 200 pounds 10 x at 1/1 = 20 seconds, then I could not lift it a full rep again, but maybe half a rep, but let’s forget about the half a rep, could not we work out how much force I have, or much I have used ?

F = Ma.
200kg and an Acceleration of 20m/s or 10m/s then we can work out the Force pushing the car by multiplying the weight by the Acceleration like this,

Fast rep,
200 x 20 = 4000N.

Slow rep,
200 x 10 = 2000N

Yes I know for my reps there will be a deceleration, but don’t see how the above 2000N can make up or balance out the 4000N ?

PLEASE, could explain this too me. So this means that I would have to use a force of 4000N to be able to move the 200kg 20m in 1 second, so that means I would have to apply the 4000N for 1 second only for it to move 20m in 1 second ? And what ?

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.

Ok Maximum force would be ok for a start, how would you do that please ? And total work done on the weights.

Believe it or not I can work the power out quite easy, that’s one of the first things that got me back interested in physics. I know this is like 1 and 1 too you, but hope I am doing it right here ?

Calculate how much power I would be used on both rep speeds. Distance weight moved 1.85 M.

Determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s = 892 kg.m/s or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

We can calculate that lifting a 200 lb barbell overhead a distance of 1.85 m required 1650 J of work.

Let us only add up the positive part of the lift.

The concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.

If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s.

Slow set,
825 x 6 reps = 4950Joules.

Fast set,
3300 x 25 reps = 82500Joules

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.

Big thank you for that as at first I thought you was trying to make out I was a crank, as imagine you going to a place where no one knows you, and they say you know knowing of physics and don’t know what you’re talking about. But I have been training for nearly 40 years, and the Wrought Iron is my business, thanks again for saying that, as was getting a bit up tight.

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!

Ok yes I will have to go along with the rules, but I know I go on about it, but the EMG ? Or do you know what a force plate is ? I imagine so, what if I bought one, or got someone to do some tests for me, and just say, and please just say, and I am not saying you are, but let’s just say that like the EMG the force plate readings come back stating the fast do have a higher average force reading, what would you do, would you be very interested, and very interested to find out not how the physics equations were wrong, as they are not, but would you be very interested to find out what variables you left out ?

This debate is HUGE on the internet. Over several different forums, by 1000s of different people, and lots or separate debates, and it’s been going on for well over ten years.

Not that well, back to answer the rest tomorrow.

Wayne

 

[sophiecentaur]

The basic rules of all this are so simply stated that they don't justify all this 'chat'. All the blather is just clouding the issue.
I shall not bother to repeat, yet again, the very few relationships and definitions.
All I can say is that, if you cannot state your question in just a few words with no numbers and just the right terms, used correctly, there isn't an answer. No one wants to read another list of numbers and jargon terms. We have established that it doesn't get us anywhere.

 

[waynexk8]

This is what I want to do, but not on a jump, but 6 reps at .5/.5 = 6 seconds and 1 rep at 3/3 = 6 seconds. And find the average force used, and the peak, but the total or overall force used.

http://people.brunel.ac.uk/~spstnpl/Publications/VerticalJump(Linthorne).pdf

Wayne

 

[douglis]

This is what I want to do, but not on a jump, but 6 reps at .5/.5 = 6 seconds and 1 rep at 3/3 = 6 seconds. And find the average force used, and the peak, but the total or overall force used.
Wayne

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

 

[sophiecentaur]

He has been told this many times but just does not choose to take it on board. Until he does, we are all wasting our time.

 

[waynexk8]

Just composing my next post.

In the meantime ou might enjoy the below from some recent communications I had from about the best known muscle physiologists, Roger M. Enoka, Ph.D. Professor and Chair Department of Integrative Physiology University of Colorado And one of his colleagues Per Aagaard Professor, PhD Institute of Sports Science and Clinical Biomechanics University of Southern Denmark.


Rogers Neuromechanics of Human Movement is out on [Audiobook] (CD-ROM)

https://www.amazon.com/dp/0736002510/?tag=pfamazon01-20

The number of muscle fibers activated to lift a weight depends on two factors:

(1) the amount of weight; and (2) the speed of the lift. Although more muscle fibers are activated during fast lifts, they are each generating MORE force. We know this because the rate at which the muscle fibers are activated by the nervous system increases with contraction speed.

Although your question seems relatively straight forward, it is not. Despite the popularization of the terms slow and fast muscle fibers, the characteristics of muscle fibers are not so black and white. Human muscle fibers are often classified as types I, IIa, and IIx.

This distinction is NOT based on contraction speed (slow or fast) but is based on the activity of an enzyme that is related to contraction speed. When the enzyme activity is assessed with an histochemical stain, the fiber types appear quite distinct: black, grey, and white.

When the enzyme activity is quantified, however, there is a continuous distribution of enzyme activity across the population. Furthermore, muscle fiber size (a measure of force capacity) varies continuously across the population and in some cases type I ("slow") fibers are actually the biggest.

I do not know how much work is performed by the different fiber types in the two scenarios you describe. I don't think this has been measured. The closest muscle physiologists have come to answering your question is to measure the size of muscle fibers in individuals who perform different types of training.

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

Cheers.

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

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

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

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

The all-or-nothing principle only refers to the discharge of action potential by a motor neuron; either it discharges an action potential or it does not. It is not correct to apply this principle to the force generated by a muscle fiber, which depends on the action potential-mediated level of calcium within the fiber.

Each muscle fiber action potential releases a certain amount of calcium from the storage site (sarcoplasmic reticulum) that enables the contractile proteins to interact and produce force. The amount of calcium released by a single action potential is less than that required to produce maximal muscle fiber force. Consequently, the force produced by a muscle fiber depends on action potential rate.

Cheers.

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


Roger M. Enoka, Ph.D. Wrote;
One important point to emphasize in these types of discussions is the concept of slow and fast twitch muscle fibers. Unfortunately, this terminology is misleading because there are not two (or three) types of muscle fibers; rather, there is a continuous distribution in every muscle from the fibers with slow contractile kinetics through to those with fast kinetics.

Because there are not distinct types of muscle fibers, it is not possible to design an exercise program that stresses either "fiber type".

A more appropriate functional distinction between muscle fibers is the force at which the motor units are activated during a muscle contraction, which is known as recruitment threshold.

Motor units with low recruitment threshold can be either slow or fast twitch, whereas motor units with high recruitment thresholds are all fast twitch. But, recruitment thresholds decrease with contraction speed so that all motor units in a muscle are activated when rapid contractions are performed with loads 40% of maximum.

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

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

Hi Roger,
The part on recruitment threshold, is a tricky one to get your head around. I think it means the faster you lift, the muscle fibers lowers their activation recruitment force, so that more can be recruited faster, and are thus recruited faster, as more are needed faster.

Am I right or half right or wrong ?

Hi Wayne,

You were right.

Cheers.


Wayne

 

[sophiecentaur]

Fascinating stuff but what has it to do with the quasi Physics that you want to apply to the problem, Wayne?
Where does he mention average or total force? This guy knows what he's talking about and you should pay attention to what he says. He talks of muscle fibre activity and energy - which is just what I should have expected - but none of your old rubbish.

 

[waynexk8]

1)The average force is the weight.

Get back to that one.

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

We all agree that the peak will/must be higher in the faster reps.

3)Total or overall force doesn't exist in physics.

I am sure it must ? As they work the average ground reaction force of exercises out on a force plate ?

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).

Why do you seem to think you are such an expert on EMG now ? As you did not know anything before, and why do you keep say my EMG is RMS ?

D. you said my EMG showed only the peak forces, I said it measured the total or overall muscle activity/ force output. I showed you this before but you did not comment, could you please, 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 something ? BUT the average was a 140 !

http://www.youtube.com/user/wayneroc...36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

 

[sophiecentaur]

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.

 

[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.

Fascinating stuff but what has it to do with the quasi Physics that you want to apply to the problem, Wayne?
Where does he mention average or total force? This guy knows what he's talking about and you should pay attention to what he says. He talks of muscle fibre activity and energy - which is just what I should have expected - but none of your old rubbish.

sophiecentaur, ok it’s up to me to provide physics is not wrong, but I think it’s not adding in all the variables. As I think the medium forces of the slow, cannot make up or balance out the higher forces of the faster when they are on the decelerations. And in my opinion this is what the machies adds in, the ground reaction higher peck force.

But for the moment could you please clear up and explain what you think RMS, {Root mean square}
means please after looking at these two very short videos.

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

Fast test.
http://www.youtube.com/user/waynerock999?feature=guide#p/u/37/pd0ZAm_2ioY

As if you look at the low rest reading you can see {not saying this is at all the exact lowest reading} on the slow, its 70, and highest 196, average = 133, machines states 140. As if you look at the low rest reading on the fast, its 155, and highest 226, average = 190, machines states 196. So it can not be the pack muscle activity, it’s the average muscle activity, force output.

By the way this weight was far too light.

1, Fast 409, Slow 349,
2, Fast 437, Slow 346,
3, Fast 0.1, Slow 0.3,
4, Fast 0.6, Slow 0.7,
5, Fast 1104, Slow 1114,
6, Fast 146.0, Slow 193.4,
7, Fast 175.0, Slow 173.0,


1/ WRK This is the work average for the session measured in [?V]
AVG microvolts. The average readings will vary from one patient to
another.

2/ RST This is the rest average for the session measured in ?V - Below
AVG 4 ?V a muscle is beginning to rest.

3/ AVG This is the average onset of muscle contraction measured in
ONST seconds, readings below 1 second can be considered normal.

4/ AVG This is the average muscle release measured in seconds,
RLSE readings below 1 second can be considered normal.

5/ W/R This is the average peak value measured in ?V - The value will
PEAK vary from one patient to another.

6/ WRK This is the average muscle deviation when contracting the
AVDV muscle. Readings of below 20% of WRKAVG can be
considered adequate, below 12% can be considered good.

7/ RST This is the average muscle deviation when the muscle is at rest.
AVDV Below 4 ?V a muscle is beginning to rest.

Wayne

 

[waynexk8]

The point you are trying to make is not to do with Physics. It's to do with what you feel in your muscles and how you are trying to fit it to some kind of 'Mechanical Science', that you've invented.

Here is another way of seeing why in my opinion the faster peak force made in the accelerations of the faster reps can not be made up by the mediam forces of slower reps when the faster reps are on the lower forces in their deccelations. As in my opinion the higher peak forces of the faster reps, will do more damage

Could I just for thins post, post some explanations/scenarios of what I mean in laymans terms when I mean that the fast reps will do more damage, thus use more overall or total force.

You drive {ONE SLOW REP = 6 MINUTES} a 80 pound car over a bridge that has a length of 6 miles and 101 pound breaking force {the car is the force created by the muscles, and the bridge in the muscles, and the breaking force in the breaking force, or the force that would tear the muscles, a tear means the muscles have taken more than they can stand, and are totally damaged} You drive your car over in 6 minutes and all is fine with the bridge {muscles} structurally.

I on the other hand, I drive {SIX SLOW REPS = 6 MINUTES} my 6 cars at a 100 pound each over the 6 mile bridge one at a time at 1 minute each, the first car generates a force very close to the breaking force, so do the second and third cars, but the time the fourth, fifth and sixth cars have gone over the bridge, they will most probably caused enough damage that the bridge breaks, why ? Because each was generating MORE force.

Say a Bird flapped his wings once very slowly for 30 minutes, he would not fly, flapped his wing very fast for 30 seconds, he would be flying very high and far. Thus is that not more total or overall force used ? As the Bird has moved his bodyweight far far far further then the slow flapping Bird, thus more force, more acceleration, more velocity, more energy, more disstyance.

My point, the peak force of the faster reps as of the accelerations, are far far far higher than those of the slow reps medium forces, thus in the peak forces small time frame but done 6 times as 6 reps, will put more tension on the muscles, causing them to tear.

Just imagine an imaginary rubber band going across your room in front of you. You roll a light weight over it in 6 seconds, the rubber band bend/dip slightly, as its having a constant light force on it, now roll along it a heaver weight, but each weight is rolled over 6 times in the same time frame as the lighter weight was rolled over once. On each of the fast rolling weight, you see a far far far larger bend/dip. That’s because there is more overall/total force on the rubber band, thus more tension, damage is being done.

If I arm wrestled you, and you put 80 pounds of force and I put 100 pounds of force, I would beat you the 6 times I went again you, and the time frame would be the same. Even if I decelerated for the last 30% then put the 100 pounds of force on again, I would still beat you on my second pull/push of force.

Is not there a way to work out my higher impulse impact forces ?

 

[douglis]

Why do you seem to think you are such an expert on EMG now ? As you did not know anything before, and why do you keep say my EMG is RMS ?

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.

D. you said my EMG showed only the peak forces, I said it measured the total or overall muscle activity/ force output. I showed you this before but you did not comment, could you please, 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 something ? BUT the average was a 140 !

http://www.youtube.com/user/wayneroc...36/B8gtpp8ozvU

What do you say to that my friend D. ?

Wayne

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.

You drive {ONE SLOW REP = 6 MINUTES} a 80 pound car over a bridge blah...blah...blah

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

 

[sophiecentaur]

Acres of blather, yet again, Wayne. I, and everyone else, understand what RMS means; it is very well defined. 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.
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.

 

[waynexk8]

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

 

[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