Most force/strength used in the same time frame

[sophiecentaur]

=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.

 

[Zula110100100]

Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41

I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.

What are you trying to find out about PS resistance training overall?

 

[Zula110100100]

bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)

 

[douglis]

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for

The answer for the applied force gives exactly what he was looking for.He just don't like the answer!

 

[sophiecentaur]

bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)

Brackets 'of' division, multiplication, addition & subtraction was what they gave me. Then, later, it became BIDMAS, with the I for Indices.

 

[waynexk8]

Hi,

Just did the static test, and some others, they were all with about 80% and for 15 seconds on the leg extension.

Static hold half way up. Average muscle activation = 92.1.

Slow. Average muscle activation = 159.1.

Fast. Average muscle activation = 191.7.

Wayne

 

[waynexk8]

The process that causes it to register more is the "force-applied" impulse being shorter time, more force(higher reading), but this will still be equal to the impulse of gravity if you average in the time that no force was applied. Just before it is moving up again,

Hmm, see you point, but this time when there will be no force, will be very small, about ? Say we semented the whole .5/.5 rep of 1 second in a 100 parts, the rest time maybe just 1 part, so how and why do we need to average in something so small, as this will also be in the slower rep, but for a little longer.

it IS at a still start again. Any extra force applied to "get it going back up" from downward motion can be split between stopping it and starting again. The "stopping it part" is part of the downward rep.

Wayne

 

[waynexk8]

I see the problem is a little deeper than I thought...

Yet again you fail to answer.

=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.

After thinking about this, I make it 1, as what I did was x 80 x 0 the other way around, like this, 0 x 80, so if you x 80 by 0 it says at 80, but if you do it like this, 80 x 0, your timesing 0 by 80, but if you x anything by 0, its always going to be 0.

Not sure how you get 41 ? But no matter, as Douglas did not even reply, it seems like he did not then understand the problem I was trying to get round.

Wayne

 

[Zula110100100]

Static hold half way up. Average muscle activation = 92.1.

Slow. Average muscle activation = 159.1.

Fast. Average muscle activation = 191.7.

If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each. This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.

Also, is not power in mechanics, the combination of forces and movement ? I thought power was the product of a force on an object and the object's velocity ? As we all will agree that the fast uses more power, {even I can work that out with some simple equations} so if power is the of a force on an object and the object's velocity, does that not mean more force more power ?

Lets say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.

 

[waynexk8]

Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41

Don’t get that, but no prob.

I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.

Ok, get you a bit more, will do the tests tomorrow with a few different weights, to late here now,

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.

Great, like you attack, yes I think it’s great to change our angle of attack, and try and look at it from all different angles, and great you’re here, as someone very unbiased, and new to look at the debate.

What are you trying to find out about PS resistance training overall?

PS ?

wayne

 

[Zula110100100]

Not sure how you get 41 ?

Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41

 

[waynexk8]

If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each.

Could not this be not true ? As in my theory, when the fast reps are on their high peak forces, their accelerations, then on the low forces the decelerations, when they are on the low force decelerations, the slow force is still on its medium forces, and the medium forces of the slow reps, cannot make up or balance out the high peak forces, with their medium forces. The only way they could make up these forces, if the slow reps went on longer, but they do not.

Best I say this, as the faster reps will have what 12 v=0 starts and stops, too the slows just 2 v=0.
As the full debate is this; which rep/s use the most overall or total force or strength, using roughly 80% of the persons 1RM, and moving the weight 1m up and 1m down, 6 reps at .5/.5 = 6 seconds, moving the weight 12m in all, or 1 rep at 3/3 = 6 seconds, moving the weight 2m in all

As you know, an is the integral of a force with respect to time. When a force is applied to a rigid body it changes the movement of that body. A small force {slow rep} applied for a LONG TIME can produce the same movement change as a large force {fast reps} applied briefly, because it is the product of the force and the time for which it is applied that is important. The impulse is equal to the change of momentum. However both reps are done in the same time frame.

This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.

Will do.

Let’s say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.

It’s getting late here so not thinking straight, as its 1.30. But to accelerate a weight up, and thus to reach different heights in the same time, you would have to use more Newton’s of force to reach the higher high, or move it at a faster m/s ?

Wrote this earlier.
I can work out the power, learned this from a site, {hope its right} as seen below and to be honest, I thought it would be basically as simple as this, was I wrong.
Calculate how much power/strength I would be used on both rep speeds. Distance weight of 91 kg moved 1.85 M.

To determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/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 91kg barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account.

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

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.

Wayne

 

[waynexk8]

Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41

Ok thx, will read that a few more times tomoorow, to late here now, half aslepp.

Wayne

 

[waynexk8]

Why can't anyone asnswerr this please ?

1,
We both have a 100 pounds of maximum strength {or a machine is lifting} that we can bench press. We use 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.

Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moré distance than the slow.

2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Question, basically the same as the above.

As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1Newton, if you move it futher in the same time from you have to use more force, N’s.

Wayne

 

[douglis]

Why can't anyone asnswerr this please ?

I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Wayne

Wayne...you got your answer.What's clear is that you don't understand the answer or you don't like it.

In both cases the applied force is equal with gravity's impulse over 6 seconds...so it's the same.

40 + 40 x 0 + 1 = ?
Yet again you fail to answer.

Yes...I failed to answer so I asked my 6 years old son and told me it's 41.

 

[sophiecentaur]

Why can't anyone asnswerr this please ?

1,
We both have a 100 pounds of maximum strength {or a machine is lifting} that we can bench press. We use 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.

Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moré distance than the slow.

2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Question, basically the same as the above.

As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1Newton, if you move it futher in the same time from you have to use more force, N’s.

Wayne

You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 2.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.

 

[waynexk8]

You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 3.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.

Hi sophiecentaur,

I don’t understand what you’re saying, to me the questions I asked on distance are just basic questions, and I put them down as plain as I can. I will give an inch or more. I am not from the pub or locker room.

I have made videos too which you have seen that you fail faster in the faster reps, thus why else would you fail if it was not because of you had used more force.

I bought an EMG that shows more muscle activity in the faster reps, = more force used.

I asked quite plain and polite questions on distance, and also proved with this little scenario that more force was used.

You place a piece of clay between your hands and the weight, THE CLAY WILL REPRESENT THE FORCE AND THUS TENSION FROM AND TO THE MUSCLES. The clay would be squashed MORE when doing the faster reps in the same time frame, YES ? What you miss is that the high peak forces of the faster reps, can and do not make up or be balanced out by the medium forces of the slower reps when the faster reps are on the deceleration.

HERE IS WHAT A PHYSICIST ON ANOTHER FORUM SAID,

1. The first man uses the most force to do that, since he must push stronger for the weight to move a larger distance within the same time interval (acceleration is greater). Power is larger, energy is larger.

2. The fast man again. Since the time is smaller, there is more acceleration required and thus, higher magnitude of force. Power would be larger, energy would be larger.

3. Same as 2 since this is merely 4 times the work they are doing.

I honestly have not tried in any way to offend you, if I have I am all to sorry, that was not my intentions, I did say I would try to learn. What gets me is why you or some others can’t answer my distance questions ? I don't get why people are getting uptight, all I ever wanted is to ask polite questions, and trieed my best to lay them out right.

If you can in my subjects, I would try to help you.

Wayne

 

[waynexk8]

Think this may help the physicists here understand more, and no, this is not in any way to sound sarcastic, as I need to explain in more my layman’s terms.

Wayne,

As asked above, could you please clarify what your argument is? Are you saying that the maximum possible forced produced by a muscle will occur with faster vs. slower contractions?

How can I put this, say a muscle has a 100 pounds of force/strength to use up in a rep or in a set, the faster reps, I am saying that they use say a 100 of that total overall force/strength to the slow using say 70%, thus more total overall tension will go on the muscles, when doing the faster reps.

Does not my more distance moved in the same time frame, more EMG activity, muscles fail faster in the faster reps, and some more things show that.

Also, in regards to your physicist, what is his background in physiology? From a mathematical standpoint some of the things you have said are true.

Thank you, which ones please ? Now we are getting somewhere.

However, when you are presented with what happens from a physiological standpoint your muscles would not be able to produce the increase in force required for the math to work. In other words there are physiological reasons why you cannot move a heavy weight very quickly and why when you move a light weight quickly muscular force is reduced. You need to apply the math within the constraints of muscle physiology. The person you are talking with needs to know both physics and physiology to understand the problem.

Well said. As because of the biomechanical advantages and disadvantages thought the ROM, the muscles cannot push up with full force like a machine could, but I don’t mind keeping it as a machine, as it still holds true. I mean how can a machine push with a force a weight 1000mm to 166mm in the same time frame, and not use more total or overall force. Even if you just could the accelerations of the machine pushing fast, that’s say 600mm the machine has accelerated the weight to 166mm of the slow, and SOME here are trying to tell me that they use the same overall force accelerating the weight 600mm to 166mm ?

Your above examples violate what the muscle is capable of doing in the physiological world so they are not worth commenting on.

Yes I know that, but are just basic examples, I think all here know that ? Big thanks for your time and help.

Wayne

 

[waynexk8]

Wayne...you got your answer.What's clear is that you don't understand the answer or you don't like it.

In both cases the applied force is equal with gravity's impulse over 6 seconds...so it's the same.

Add in the acceleration conponants please, How can the applied force be equal with gravity's impulse over 6 seconds all the time ? Add in time and distance.

Or try and tell me how you move a weight 166mm and I move a weight 1000mm in the same time frame, and I accelerate for about 600mm of that distance. Your trying to tell me that you use the same force moving a weight far less disstynce in the same time frame



Yes...I failed to answer so I asked my 6 years old son and told me it's 41.[/QUOTE]

Let’s see you try and prove this, with basic numbers

40 + 40 x 0 + 1 = ?

Is not 40 + 40 = 80 ? Now we x 80 by 0, and whatever we x 0 by it will always be 0 right ? So we are now at 0 right ? So add 1 to 0 = 1 right ?

Wayne

 

[douglis]

Add in the acceleration conponants please, How can the applied force be equal with gravity's impulse over 6 seconds all the time ? Add in time and distance.

Or try and tell me how you move a weight 166mm and I move a weight 1000mm in the same time frame, and I accelerate for about 600mm of that distance. Your trying to tell me that you use the same force moving a weight far less disstynce in the same time frame

Irrespective of any of the details along the way (acceleration, deceleration, fraction of time spent accelerating, distance covered, etc.), the average force is always equal to the weight of the object, so long as you start and finish at rest.

Let’s see you try and prove this, with basic numbers

40 + 40 x 0 + 1 = ?

Is not 40 + 40 = 80 ? Now we x 80 by 0, and whatever we x 0 by it will always be 0 right ? So we are now at 0 right ? So add 1 to 0 = 1 right ?

Wayne

Instead of trying to prove nonsense learn to do the mathematic acts at the right order or find a kid to show you.

 

[Zula110100100]

@wayne, That is basically what we have been saying, the answer we gave you in not taking into account physiological factors. That is the whole reason you simplified it to a machine, in the machine, the OVERALL impulse is equal.

In the case of the clay, yes, the fast one would make more of a dent in the clay due to higher peak forces, but as you said, this is a greater acceleration and a greater veloicty in a small time, as a result, in order for gravity to cause it to stop again it must act longer, so when you average them together, they are equal.

Also, as mentioned before, you get different mechanical advantage at different anges with your arms.

 

[douglis]

I really can't figure out what Wayne can't understand.
If we compare 6 reps with .5/.5 and 1 rep with 3/3 the only difference is that the 6 reps have higher fluctuations of force.But in both cases the same average force(the weight) is applied for 6 seconds.

What's so hard in the above to understand?Why does he keep posting the same irrelevant examples?Do we have to use superior maths for something obvious even to a 10 years old child?

 

[waynexk8]

I really can't figure out what Wayne can't understand.
If we compare 6 reps with .5/.5 and 1 rep with 3/3 the only difference is that the 6 reps have higher fluctuations of force.But in both cases the same average force(the weight) is applied for 6 seconds.

What's so hard in the above to understand?Why does he keep posting the same irrelevant examples?Do we have to use superior maths for something obvious even to a 10 years old child?

How do you work out my EMG states you are wrong ?

And how do you work out my clay example proves you wrong ?

Irrespective of any of the details along the way (acceleration, deceleration, fraction of time spent accelerating, distance covered, etc.), the average force is always equal to the weight of the object, so long as you start and finish at rest.

Could you do me a favour please, could you tell me again how you work out this average force. Let’s say I am moving a weight at 1m/s and move it for 1 second, I then move a weight 100m/s and move it for 1 second, how do you work out the averages ? AND WHAT YOUR SAYING IS THAT YOU USE THE SAME TOTAL OR OVERALL FORCE TO MOVE THE SAME WEIGHT 1M IN 1 SECOND, AND TO MOVE IT 100M IN ONE SECOND ?

You always do the same thing; you say something without an explanation.

Don’t you understand you are moving against gravity, which is a force, and the faster you move against gravity the more force you need, yes you need less force for deceleration, but if you’re covering more distance in the same time frame, you HAVE to have more total or overall force.

And for the Hundredth time, I am not on about average force; I am on about total or overall force.

My more distance in the same time frame.

Are you saying the force, the Newton is wrong ? It is equal to the amount of net force required to accelerate a mass of 1kg at a rate of 1m/s.

Are you saying that to accelerate a mass of 1kg at the rate of 1m/s, or 5m/s or 10m/s, all you need is the exact same force as 1 Newton ? As that’s what it sounds like. So the question why are you saying this ?
F = ma let's us work out the forces at work on objects by multiplying the mass of the object by the acceleration of the object.
Example:
The force at work on a Formula 1 car as it starts a race! If the F1 car has a Mass of 600kg and an Acceleration of 20m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 20 = 12000N

Wayne says, If the F1 car has a Mass of 600kg and an Acceleration of 40m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 40 = 24000N. But your saying that’s wrong ?


If the car has a mass of 700kg And a driver pushes the car with an acceleration of 0.05m/s/s Then:
F = MA
Force = 700kg x 0.05m/s/s
Force = 35kN (kiloNewtons)

Wayne says, If the car has a mass of 700kg And a driver pushes the car with an acceleration of 0.05m/s/s Then:
F = MA
Force = 700kg x 1m/s/s
Force = 700kN (kiloNewtons)
But you’re saying that’s wrong ?

http://www.racemath.info/forcesandpressure/what_is_f=ma.htm

Instead of trying to prove nonsense learn to do the mathematic acts at the right order or find a kid to show you.

D. basically it was a sort of trick question for you, as the way you think now is VERY wrong.

LOOK, you CANT change the order of the question to suit you, WHY, and when I say WHY, I mean please give me an answer as why you THINK you can change the other of the question ? IF you change the order of the QUESTION, it NO longer because the same question, thus you CAN NOT changed the dammed order of the question, I can’t believe anyone can THINK they can answer a question like this, with changing the order of the question.

He is what happens when you change questions, ONE, YOU HAVE NOT ANSWERED THE ORIGINAL QUESTION, HAVE YOU ? YES OR NO ?

TWO, I ask you this question, 4 + 4 = 8 x that number by 2 = 32, YOU might think I am wrong, but ho not, well I am not, well I am but I am now doing what you did, {and what you’re doing in our big physics debate} I changed the question to suit myself, I changed it to this, 4 x 4 = 16 x that number by 2 = 32.

I cannot believe anyone could try and change the question, have you a good explanation for this ?

Back later for you post Zula110100100.

Wayne

 

[douglis]

How do you work out my EMG states you are wrong ?

Your EMG is not suitable for lifting that involves SSC...I explained why.

And how do you work out my clay example proves you wrong ?

Your irrelevant clay example just proves that higher peak force is produced with fast lifting.Nothing more.

Could you do me a favour please, could you tell me again how you work out this average force. Let’s say I am moving a weight at 1m/s and move it for 1 second, I then move a weight 100m/s and move it for 1 second, how do you work out the averages ? AND WHAT YOUR SAYING IS THAT YOU USE THE SAME TOTAL OR OVERALL FORCE TO MOVE THE SAME WEIGHT 1M IN 1 SECOND, AND TO MOVE IT 100M IN ONE SECOND ?

Wayne

I must have said this more than a hundred of times.
I work out the average from the net impulse delivered.Either you move the weight 1 or 100m the change in momentum is zero so the net impulse delivered is zero too.Concequently the average net force is zero and the applied force equal with the weight.
End of story.
It's obvious that you're not able to understand the answer or you don't want to.Either way continue alone.I'm fed up with that nonsense.

 

[waynexk8]

D, are you saying the average, total, overall forces here are the same ? I am.

https://www.physicsforums.com/attachment.php?attachmentid=783&d=1079995820

But on the above I rep once and you rep once, so let’s show you where you ARE wrong on averages, or should I say thinking as averages are the same, the total overall force is the same. Too which it is NOT.

Slow reps,
1 + 1 + 1 + 1 + 1 = 5 / BY 5 = 1

Fast reps,
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10 / BY 10 = 1

NOW don’t you get it ? YES the averages ARE the same in both speeds, BUT please NOTE the faster reps had a higher total or average force, that’s what I am trying to tell you, DON’T TAKE TO MUCH NOTICE ON THE AVERAGE.

Wayne

 

[waynexk8]

Your EMG is not suitable for lifting that involves SSC...I explained why.

I did not read why ?

So you’re now saying that all the sport science people all over the World are wrong when they use EMG ? D. you know very well the EMG shows the muscle activity, and it makes no difference if I have more SSC than you, as that’s part of the lift, YOU CANT TAKE IT OUT ? God, you don’t like something because its right, so you say it’s wrong, pathetic. I paid about 200 Euros to prove you wrong and I did, please don’t try and get out of an EMG test. Ok what do you want me to do then ? Take a reading of me doing full stops at both ends of the reps ? Would that be fair ?

Look D. I told you why your equations are wrong, ITS BECAUSE YOU ARE NOT ADDING IN THE PEAK FORCES FROM THE TRANSITION FROM NEGATIVE TO POSITIVE, But these are part of the rep/sets, you can’t ignore them like your physics does, you have to add them in, or don’t you think they count ? If not why ?

When I buy my force plate, you will see results like my EMG.

Your irrelevant clay example just proves that higher peak force is produced with fast lifting.Nothing more.

HOW do you work that out ? Please dammed you explain the way you think, as it seems you’re making things up, as you have no explanation of your theory.

LOOK the clay will feel all the forces of the reps right ? If you think no, say why. This means that the clay is the tension on the muscles, how can you say/think it’s just the peak forces ? If I lift a weight slow in 2 seconds and then at 6 seconds, the clay will not be the same at the end.

I must have said this more than a hundred of times.
I work out the average from the net impulse delivered.Either you move the weight 1 or 100m the change in momentum is zero so the net impulse delivered is zero too.Concequently the average net force is zero and the applied force equal with the weight.
End of story.

YES, but the total force can not be the same, you cannot move 100 pound a 100m in 1 second and move a 100 pounds 1m in 1 second and use the same force, you would need a huge amount of force to move the weight a 100m.

You mean like this, If not explain why, don’t just say things without an explanation, it’s not fair.

But on the above I rep once and you rep once, so let’s show you where you ARE wrong on averages, or should I say thinking as averages are the same, the total overall force is the same. Too which it is NOT.

Slow reps,
1 + 1 + 1 + 1 + 1 = 5 / BY 5 = 1

Fast reps,
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10 / BY 10 = 1

NOW don’t you get it ? YES the averages ARE the same in both speeds, BUT please NOTE the faster reps had a higher total or average force, that’s what I am trying to tell you, DON’T TAKE TO MUCH NOTICE ON THE AVERAGE.

It's obvious that you're not able to understand the answer or you don't want to.Either way continue alone.I'm fed up with that nonsense.

An EMG puts pads on you and takes the average muscle force, activity, and your saying that’s nonsense.

I am the one answering your questions and counting you back, it’s not the other way about, it’s you who not only do not understand, PLEASE for once answer with a good layman’s teams, not just words that mean nothing.

Wayne

 

[Zula110100100]

@Wayne - If you do the experiment I asked about using a static hold with different weights you would either, A) See why the EMG results don't work in this situation, or B) Prove to Douglis that they do work in this situation.

My hypothesis is that when you double the force required to keep it still(by doubling the mass) you will see less than a double result from the EMG. This may or may not be true, I don't have a machine to test it but I feel it would bring us much closer to agreement on this issue.

F = ma , Force = mass * acceleration, The mass is constant to it really comes down to the acceleration to decide the force.

Do you agree that in order to have a velocity that begins and ends on 0, it must both accelerate and decelerate?

So let's say we accelerates it with a₁ acceleration for t₁seconds, it's velocity, v₁ is then given by
v₁ = a₁t₁

Now, it must decelerate back to 0, so gravity accelerates is a₂ for t₂, so this time we are starting at v1 velocity and going to 0, well, that is really the same amount as starting at 0 and going to v1, so the equation

v1 = a₂t₂ must also be true.

If it gave a different velocity it would not cancel the other velocity right to 0, it would still be in motion.

With that being the case, how can a₁t₁ not equal a₂t₂


The clay situation doesn't work for total force because like you said, it is a measure of peak force, it will only measure what the most force applied for an instant was, not in the same time frame. that is the difference, when asking for who applies most in the same time frame you are requiring that we average in the time that less or no force is applied, had the question been, "Which person generates a larger force" the answer is the fast one, hands down, it must have been a larger acceleration to still have time to decelerate and make it the same distance faster.

But that was not the question. Which is why I say you should re-evaluate what you are trying to learn or show. There may be another factor to look at.

The force at work on a Formula 1 car as it starts a race! If the F1 car has a Mass of 600kg and an Acceleration of 20m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 20 = 12000N

Wayne says, If the F1 car has a Mass of 600kg and an Acceleration of 40m/s/s then we can work out the Force pushing the car by multiplying the Mass by the Acceleration like this 600 x 40 = 24000N. But your saying that’s wrong ?

This is a completely different situation, if you include a requirement that they both came to rest again from the force of friction and air-resistance, the again, the overall force would be 0, this example does not look at that, it looks only at the force to accelerate it, not decelerate it, if the question was about two cars accelerating an unknown amount, but both coming to rest after an equal time of an equal force acting on them the answer would be that the same force was applied.

The important this is doing that experiment to see what the correlation between force and EMG activity is.

 

[waynexk8]

Hi, still need to answer your other post, but connection went again, no time to answer this one in full.

@Wayne - If you do the experiment I asked about using a static hold with different weights you would either, A) See why the EMG results don't work in this situation, or B) Prove to Douglis that they do work in this situation.

Yes will definitely do, it’s been a busy.

@My hypothesis is that when you double the force required to keep it still(by doubling the mass) you will see less than a double result from the EMG. This may or may not be true, I don't have a machine to test it but I feel it would bring us much closer to agreement on this issue.

F = ma , Force = mass * acceleration, The mass is constant to it really comes down to the acceleration to decide the force.

Yes.

@Do you agree that in order to have a velocity that begins and ends on 0, it must both accelerate and decelerate?

Yes.

However why is this zero movement at both ends of the transitions so important in this debate ? As they will be so small compeered to the rest of the reps, they will be like ? 1 part in 10,000.

@So let's say we accelerates it with a₁ acceleration for t₁seconds, it's velocity, v₁ is then given by
v₁ = a₁t₁

Now, it must decelerate back to 0, so gravity accelerates is a₂ for t₂, so this time we are starting at v1 velocity and going to 0, well, that is really the same amount as starting at 0 and going to v1, so the equation

v1 = a₂t₂ must also be true.

If it gave a different velocity it would not cancel the other velocity right to 0, it would still be in motion.

With that being the case, how can a₁t₁ not equal a₂t₂


The clay situation doesn't work for total force because like you said, it is a measure of peak force, it will only measure what the most force applied for an instant was, not in the same time frame.

Hmm, not sure on this, as I did not say the clay would just measure peak force; I think it would measure all the forces, why don’t you or anyone think it will measure all the forces ? As a scales would if you stood on them, and if you were lifting overhead and put the clay under your hands and feet, would not this measure all the forces ? As soon as you lift the weight, the clay would squash a little, then if you accelerated it slow or fast, or very slow, the clay would then squash more. Dammed it getting late.

Here is a quick way on how I think the clay and in the lifting the muscles will feel more tension on the muscles, thus the muscles are putting out more force.

Stand up, hold a light weight in your open palm, with hand and arm out stretched and turn around fast, and the weight will stay in your hand with any other force. The weight is now pressing into your hand, and the faster you turn, the more it presses into your hand, this is exactly the same when you exerted more force from your muscles, in turn they have to get more tension on them. Yes I know there is a deceleration, but still in the deceleration I am still trying to push as hard as possible.

Let’s look at this situation without deceleration, as I don’t think there will be much, as the studies said using 80% there would be 40% deceleration, but the lifter took 1.5 seconds, I am talking about .5 of a second, and at the last moment I will be accelerating, basically is more of a jerk at the top on the lift to reverse the rep, if you get what I mean, then this happens over and over. What would you say about the both forces if there was no deceleration ? And there are lifts like this, which I will try out on my EMG. I lift up with full acceleration, and then stop, then I rep down and do the same thing, actually that’s what I basically do now, without the stop.

Sorry I am rambling and its late, will get back when I am fresh.

that is the difference, when asking for who applies most in the same time frame you are requiring that we average in the time that less or no force is applied, had the question been, "Which person generates a larger force" the answer is the fast one, hands down, it must have been a larger acceleration to still have time to decelerate and make it the same distance faster.

But that was not the question. Which is why I say you should re-evaluate what you are trying to learn or show. There may be another factor to look at.

Hmm, will have to read that one and get back to it, not sure what you mean by saying I am asking; that we average in the time that less or no force is applied ? All I am asking is that we add in all the force/forces that the muscles generate lifting the same weight for different distances and velocities.

This is a completely different situation, if you include a requirement that they both came to rest again from the force of friction and air-resistance, the again, the overall force would be 0, this example does not look at that, it looks only at the force to accelerate it, not decelerate it, if the question was about two cars accelerating an unknown amount, but both coming to rest after an equal time of an equal force acting on them the answer would be that the same force was applied.

The important this is doing that experiment to see what the correlation between force and EMG activity is.

Wayne

 

[Zula110100100]

However why is this zero movement at both ends of the transitions so important in this debate ? As they will be so small compeered to the rest of the reps, they will be like ? 1 part in 10,000.

The velocity being 0 on both ends are what make us so confident that the impusle(force * time) of the machine is equal to gravity, if it were not the same number on both ends, the impulses would not be equal. This was in the study you linked to "Linear impulse equals a change in linear momentum" Momentum is mass * velocity, so in order for the velocity to start at one number( 0 ) then increase( change in velocity = change in momentum = impulse) and then decrease exactly back to ( 0 ) requires an equal and opposite change in velocity, thus, an equal and opposite change in momentum, and thus, and equal and opposite impulse.

Hmm, not sure on this, as I did not say the clay would just measure peak force; I think it would measure all the forces, why don’t you or anyone think it will measure all the forces ? As a scales would if you stood on them, and if you were lifting overhead and put the clay under your hands and feet, would not this measure all the forces ? As soon as you lift the weight, the clay would squash a little, then if you accelerated it slow or fast, or very slow, the clay would then squash more. Dammed it getting late.

Lets say you place the clay on the table, and place a weight on it, it causes it to depress a certain amount. You wait 5 hours, the amount does not change. Now you add more weight, again it depresses further, you wait 5 more hours, it doesn't change. Now you take all the weight off, it does not return to it's original shape, it stays depressed the amount it was when it was under it's peak weight.

 

[douglis]

I did not read why ?

So you’re now saying that all the sport science people all over the World are wrong when they use EMG ? D. you know very well the EMG shows the muscle activity, and it makes no difference if I have more SSC than you, as that’s part of the lift, YOU CANT TAKE IT OUT ? God, you don’t like something because its right, so you say it’s wrong, pathetic.

Wayne

All the serious studies that use electromyography in dynamic contractions normalize their raw EMG with the MVIC method...and of course the effect of force over time is given by the integration of the EMG(iEMG).

I'll quote the post you missed:

Ha...just saw that...no wonder why in the study you post they use isometric contractions.The RMS is accurate here since with isos the force doesn't fluctuate.
In dynamic contractions where the stretch-shortening cycle is used the raw EMG must be normalized by using maximal voluntary isometric contractions (MVIC) just like they did with the two push ups studies and the Elliot et al study(the one that found 52% deceleration).

Here's explained better:

"EMG signal amplitude normalization technique in stretch-shortening cycle movements

Normalization using maximal voluntary isometric contractions (MVIC), irrespective of rms processing (total, mean or peak), demonstrated greater CV above the raw data for both muscle actions. "
http://www.sciencedirect.com/science/article/pii/105064119390013M

For all the rest...I don't think there's any point to continue.

 

[waynexk8]

Hi all, just done the experiment, these were static holds with my EMG machine pads on my biceps and forearm. For those new into this thread, an EMG reads the muscle or in this case the average muscle activity for a certain time frame.

I was as I would have thought, not sure what other here was thinking, and looking forward to what this proves, me right in part of this EMG debate or me wrong. The machine times itself, and every time I had the weight in the midway point before I pressed the button.

First best say just in case people here do not know. You can lower more under control than you can lift. So if my repetition maximum was 100 pounds, I could lower say 130 to 140 pounds under control. This is because the muscle fibers are more efficient at lowering, {not 100% this is right} as think of a fish and its scales, you can slide you hand over the smooth fish, but try going backwards and it’s not the same story. Also the fast muscle fibers, the ones that are used for the repetition maximums, or the heaver lifting compeered to the slow muscles that are called upon for endurance work, well the lower calls upon these fast muscle fibers more efficiently.

Static holds, with weight getting 10 pound heaver every time.

1,
67.6
2,
105.6
3,
148.7

Then I did a slow and then fast with the last weight, on the fast I was still able to rep, but not full reps, meaning it was getting very hard.
4,
228.1
5,
257.

Wayne

 

[Zula110100100]

What was the weight, you just say it gets heavier by 10.

 

[waynexk8]

Equation 3 from that source points out that a linear impulse is equal to change in linear acceleration

If we take an force of 6.2N over 10.9 seconds, that is an impulse of 67.58N*s
This means we must see a change in momentum of 67.58kg*m/s The weight presumably started at a velocity of 0, so this means it is left with a velocity of .4m/s, thus, it has not yet stopped moving, if you stop applying force, it will take gravity another .04s to stop it, and during that time there is a negative propulsive force, an applied force less than the force required to hold the object up.

So in my opinion the data is not good, because they did not wait for it to stop to take their readings.

Ahh, now I get this part a little more. So you’re saying if the ROM was 1m, what part do you think they took the readings ? But again, the slow was done for 10.9 seconds, what difference could this make ? How much more force would that add to the slow ? As remember the lift did take 10.9 seconds. So I now don’t get you again.

If you take 1672.2N(the applied force required to get a propulsive force of 6.2N) * 10.9+ 0N*.04s = 1672.2 and then divide that by the total time 10.94 = 1666.0859N avg which is really close to the 1666N that were required to hold it up.

YES, but that’s just about RIGHT. As basically if a lift took 10 seconds, all you are doing is moving it very slow, so all you’re doing is basically holing it up. This is my basic debate, as to move the weight much faster, you have to use more force, and more, and then a very fast deceleration.

For the faster rep, we have an avg propulsive force of 45.3N * 2.8s = 126.8N*s, so we know it is left with a velocity of .75m/s which would take gravity .076s to decelerate, so again, add the 45.3 to 1666 = 1711.3*2.8s = 4791.64N*s from the force applied
+0n*.076s =4791.64/(2.8+.076) = 1666.078 once again really close to the force required to hold it still in the first place.

Have to read that over, getting late, and don’t get what each numbers are for yet.

So overall, sure, it is more force to get the same weight going faster, anyone will agree there,

Yes.

the problem is that unless you apply a force to pull the weight down toward you before it stop moving up, then comparing the force for the whole cycle should be the same.

But I am using the force of the opposite muscles to bring it to a stop and reverse the weight, but up until the last Miily second, I use as much force as I can from the muscles.

If I used a light weight and accelerated it fast, and was not holding it, it would move out of my hands, I would have to wait Milly second for it to return, but that’s BASICALLY what happens in my reps, but as I am holding onto the weight it does not go out of my hands. But I am using a weight that is too heavy to move out of my hands anyway, thus I have to push will full force right and UNTIL THE VERY END of the lift. Will video some very heaver fast lifts on a Smith machines with my hands open, that a machine that like this.



What even if I accelerate for say 3 times the distance as the other, and accelerate longer ? Sorry don’t get that, let me read and think it over, my fault for starting so late again.

in other words, if you apply more than required to hold it up, and then never less than required to hold it up, it will keep moving up, since that is not the case there is more going on than listed in that expiriment

That sounds interesting, need to read this all over.

ANOTHER THING TWO THINGS WANT TO CLEAR UP, we did go over this before, but could we clear this up for once and for all, deceleration. These numbers are just for this, so don’t take them as 100% fact. We break the fast concentric up into 15 parts of force, and if D. is right on deceleration it goes like this, as in a person using 80% {80 pounds} and the rep taking 1.5 seconds, the study says they decelerate at 60% of the ROM.

60% is on part 9.
100 100 100 100 95 90 85 80 75 70 65 60 30 10 zero on transition and reverse for the eccentric.

What about this. As your accelerating, and then accelerating slower, that your still going faster and faster.
100 100 100 100 100 100 100 100 90 80 70 50 40 20 zero on transition and reverse for the eccentric.

As does deceleration in EVERY case, mean you are then using less force than the weight, in this case 79 pounds, or can it mean you are not accelerating so fast, thus decelerating, so using less and less force as in 2.

Slow would be more like this.
80 80 80 80 80 80 80 80 80 80 80 70 60 40 zero

No need for an explanation, just say 2 can happen or it can’t. However to be honest I always would have thought 2 would have been deceleration, but I never thought about it or read about it before.

So next, the above on the deceleration at 60% was taken from a study that used 80% but the concentric rep took 1.5 seconds, what would happen to the deceleration if the rep took 3 seconds, and what would happen if it took .5 of a second, opinions and facts please. AND WITH THE SLOW REP, HOW DO WE NOT KNOW THIS REP IS DECELERATING FOR 40% OF THE REP AS WELL ?

http://books.google.co.uk/books?id=...the final 52% of the range of motion.&f=false

Thx for all the help and time

Wayne

 

[waynexk8]

Tuesday is my arms night, so I had just done a full arm workout, so what I used on the test, was not that much as when fresh say 40% less, and I did dumbbell concentration hold at midpoint, let me go check the first weight, It was just 30 pounds, then 40 then 51. I could hold these weight quite easy for 15 seconds, do you want heavier weights ?

Here is a concentration curl, but as I said, I just curled half way up, hit the button and held until the machine stopped.



Wayne

 

[berkeman]

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