Impulse/force in pounds for the time frame

[waynexk8]

Wayne, for the 3rd time now, please answer this question:

Do you understand that it is physically impossible to move a weight up and down using a constant force? I need to know if that makes sense to you before we can proceed.

That’s an odd question.

Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.

Now with the human body it’s far harder than with a machine. As take the bench press, there are several biomechanical advantages and disadvantages thought the ROM {Range Of Motion} of the exercise, there is point in the movement, where the leverage of certain muscles acting about the shoulder joint and chest, are at a mechanical disadvantage, and thus creating a disadvantage at about half way to three quarters the way up the lift, this point will decrease the force potential and thus deceleration to the bar.


Note on a 1RM, the acceleration was for roughly 65%, so on using 80% like we are; the acceleration would be at least ? 80%

https://docs.google.com/viewer?a=v&...BmBxS-&sig=AHIEtbT6QUrHzT2mMWFVWF8f8ikRAXeHCA

Also, if you work out in physics how far a barbell will move if you let go when pressing it, and your using 80% for 15 inch, is does NOT go what physics say, as of the biomechanical disadvantages and biomechanical advantages of the body. But it would move the 3 inch if a machine pushed with a constant force of 80% for that force and time.

Jeff Pinter wrote, and he did say I could use this.

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

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

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

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

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

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

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

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

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

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

Jeff

So as I said before, could we not keep this to a machine moving the weight please ?

Big thanks for your time and help DaleSpam, and the rest, thank you.

Now its too late and I have no time to read or answer the others.

Wayne

 

[sophiecentaur]

So as I said before, could we not keep this to a machine moving the weight please ?

Wayne

Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO.
You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

 

[douglis]

Either it's a muscle or a machine...one thing will always be for sure.
The muscle's/machine's impulse will always be equal with gravity's impulse regardless if the lifting is fast or slow for the same duration.The net impulse is always zero.

 

[Dale]

That’s an odd question.

Yes, I know that. So if you are lifting a weight up 500mm in 3 seconds, and going to reverse the direction, as in weightlifting, lifting a weight up and down. First you have to accelerate, then “try” to move it with a constant force, and then decelerate the weight.

OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.

So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you.

The plot is generated in Mathematica using the following code:

y[t_] := -.25 Cos[2 \[Pi] t/3] + .25;
Plot[y[t], {t, 0, 3}, Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], 
   Style["Height (m)", Larger]}, 
 PlotLabel -> Style["Position of Weight During Lift", Larger]]

 

[waynexk8]

Absolutely.

Great. Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

But you still are not accepting that the energy that a machine needs to supply can be near ZERO,

You lost me there ? If the machine is powered by the energy, diesel or electricity, to move 80% of its maximum force up and down, will take a certain amount of this energy, but as we all agree, that the faster/further an object moves in the same time frame the more energy is needed, there is no debate there, or is there ? As something, makes for the more energy used, and I say it’s the higher forces of the accelerations that cannot be made up or balanced out by the more constant forces of the slower repetitions, when the fast is on the deceleration.

Run 1 mile = 3 minutes, walk 1 mile 9 minutes, both use the same energy. Run 3 minutes, walk 3 minutes the run uses more energy, this is what I have said all along.

the average force will be ZERO and the net Impulse will also be ZERO.

Not sure I understand that, as if I lift 80% of my 1RM, {hope we all know what 1RM means by now, if anyone does not, I have explained many times, it’s the 1 repetition maximum a person can lift} I will try to use 100% force thought the ROM/lift, the only time I will use at the beginning and end, but they do not count, as it’s the lifting we are on about, not the times after the lifting.

But thought we all agreed average means nothing in this debate, as if the average force is the same for 1 repetition and a 100 repetitions for the same speed, it means nothing.

You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.

Right, sure we all agree on this, however, as I lift the same weight 6 times further in the same time frame, using 100% force to the slow using 80% force, how can the total overall muscle activity or force be the same, you might say again, but in physics you can’t measure total overall muscle force, well try this.

Set our machine to lift 80% up and down for 60 seconds set at a lifting force of 80% of the machines maximum moving force, and then set the machine to move 80% up and down for 60 seconds set at a lifting force of 100%

Fast, the machine lifted with a set force of 100% efficiency for 60 seconds.

Slow, the machine lifted with a set force of 80% efficiency for 60 seconds.

The fast lift with 20% more force efficiency for 60 seconds

But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

Said this as the muscle will start to complicate things with their biomechanical advantages and disadvantages thought-out the ROM.

There has to be a reason, why the faster makes for the more energy used, I said why I think, no one else does, D, mentions, biological, that’s no answer at all, it’s says nothing.

Wayne

 

[waynexk8]

Absolutely. But you still are not accepting that the energy that a machine needs to supply can be near ZERO, the average force will be ZERO and the net Impulse will also be ZERO.
You would be right in saying that the Maximum force for a fast repetition rate would be greater than the maximum force for a slow rep rate.
But how does this machine help you to analyse what you think your muscles are doing? Where's the essential connection in your mind between the two entirely separate things?

OK, I am glad you understand that. I apologize that you think it was an odd question, but from several of your previous comments it was not clear that you understood that point. I am glad that you do and it was just a miscommunication.

K great, thank you.

So, let's take the scenario of a machine or a person lifting a 50 kg weight up and down 500 mm in 3 s as smoothly as possible. I generated the attached plot to represent a typical position over time graph for such a lift. Look at it and see if it seems like a reasonable approximation to you.

The plot is generated in Mathematica using the following code:

y[t_] := -.25 Cos[2 \[Pi] t/3] + .25;
Plot[y[t], {t, 0, 3}, Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], 
   Style["Height (m)", Larger]}, 
 PlotLabel -> Style["Position of Weight During Lift", Larger]]

Yes that look quite ok, I think. We are lifting 80% are we not, or close too ?

Wayne

 

[waynexk8]

This demonstrates well how you have got this wrong from beginning to end. Your muscles are not a machine. I can design a machine that uses NO energy whilst doing 'reps' at any rate and for any length of time (except to overcome some friction - and we could reduce this to an arbitrarily small amount).

How do you work that out ? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ?

The way the forces vary as the weights accelerate will depend on the rate

Right, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension, move/accelerate very fast very high forces, = very high tensions on the muscles, move very slow, very low force, = very low tensions on the muscles, move very slow, very low force.

but the mean velocity is zero for each cycle, and so is the energy transfer.

Not sure what this has to do with the debate, as we need to know about the velocities themselves, that are produced by the person/machine creating these forces, as I just said, the faster you try a move/accelerate the weight, the more action reaction forces go back on your muscles as tension.

btw, when you describe your machine as having a certain "lifting force" you do not specify for how long or over what distance this force is applied - nor what happens on the way down; the model is incomplete.

Ok sorry there.


The machine has a lifting force of 100 pounds, it can lift 100 pounds up 1m in .5 of a second, and any speed slower, and can lower it 1m in .5 of a second, and any speed slower, for an unlimited time, for debates sake, the weight the machine will use will be 80% that’s 80 pounds.


The machine first lift the 80% up 1m in 3 seconds and lowers it 1m in 3 seconds, 1 time, = 2m in 6 seconds.


Then the machine lifts the 80% up 1m in .5 of a second, and lowers it 1m in .5 of a second, 6 times = 12m in 6 seconds.

On the way down it lowers it in control 1m in .5 of a second, I say in control, as if let to drop, it would get to the ground far faster. Also, as of the acceleration components, when the weight is being lowered, the faster it’s lowered the more force it puts on the machine/muscles, like in the faster it hits the ground the more impact force.

All this machine needs to consists of is a wheel, with the weights hung on the periphery. You could modify it, if you like, so that the weights are on a crank and push rod so that they are constrained to go just up and down. Over one cycle of operation, ZERO work has been done on the weights and Zero energy is expended by the energy source. All that is necessary is to start the machine up and bring it to the desired speed. This (kinetic) energy could all be reclaimed at the end of the exercise.
Perhaps this will help you to see that there is absolutely no parallel between what your muscles are doing and a simple mechanical model. Anything you may 'think' that your muscles are doing is entirely subjective. However many times you repeat a story about reps and rates and what you reckon you can tell us about what your muscles feel like, there is no direct correspondence between your body and simple mechanics. I believe that is what you have been trying to show all along.
__________________

Wow that sounds interesting, but its far too late to comment on that now, will get back to it tomorrow.


Wayne

 

[waynexk8]

To all the other that have posted here, I will get back to them tommorrow, and thanks for your time and help.

Wayne

 

[sophiecentaur]

How do you work that out ? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ?


Wayne

Plus all the rest
That comment shows that yo don't get the mosgt basic part of all this thread and others.

The machine doesn't need to use use "very little energy" on the way down. IT GETS ALL THE ENERGY BACK! Unlike your muscles, which don't have Energy Recovery. So the two cannot be compared.
You insist that this problem can be solved your way and you have the nerve to hang onto the idea in the face of people who know much more basic Physics than you. The only hope you have is to do a Physics course at some level which may help you understand what you need to know in order to grasp how crazy your idea is.
If someone told you that swimming the Pacific is a no no, how long would you not believe them?

 

[douglis]

sophiecentaur...I think that he first needs to grasp more simple things.For example...he doesn't seem to understand the meaning of average force.
Otherwise he wouldn't say nonsense like "...the forces don't make up..." or "...average means nothing in this debate...".
More important he would understand that same average force equates same impulse(what he calls "total/overall force") when is applied for the same duration.

 

[Dale]

Yes that look quite ok, I think. We are lifting 80% are we not, or close too ?

I don't know about percentages, but from this motion diagram we can easily use Newton's laws to calculate the force exerted on the 50 kg mass by the lifter. I have plotted the force over time in the attached image.

At each point in time, the height of this plot represents the force applied by the lifter at that time. That has units of N. On this plot, we can also see a shaded area. This shaded area represents the impulse. The shaded area has units of Ns = kg m/s which is the same units as momentum. The average force is the impulse divided by 3 s. It has units of N also.

Note, as we had discussed before, the force must change over the course of the lift, it is higher when the weight is accelerating upwards, and lower when accelerating downwards. Also, although the force gets lower when accelerating downwards, it is never 0 in this case.

Does all of this make sense, in particular the impulse and average force?

Plot[Evaluate[50 D[y[t], {t, 2}] + 50 9.8], {t, 0, 3},  
 Frame -> True, 
 FrameLabel -> {Style["Time (s)", Larger], Style["Force (N)", Larger]}, 
 PlotLabel -> Style["Force from Lifter", Larger], 
 PlotRange -> {0, 560}, 
 Filling -> 0]

 

[sophiecentaur]

I don't think that Wayne actually knows what question he is actually asking any more. The only way he can ask about anything is in the form of a two page story instead of a concise question. That shows he has not actually formulated a question but just wants to chat about Reps and all that stuff. He just feels that there must be 'some simple Physics' involved.

(I'm right, aren't I, Wayne?)

 

[waynexk8]

DaleSpam...it doesn't have to be for over the whole rep.
Even if you examine separately the lifting and the lowering phase the average force is always the weight.In both phases the weight starts and ends at rest so the average acceleretion is always zero.

Look D. You say you will not let me apologise for repeating, but I have too, as I said over two years ago on a thread I actually called; average force means nothing here. So I say it again, if the average force is the same for 1 rep at 1/1 and 100 reps at 1/1 and 1 rep at 5/5 and 100 reps at 5/5, it means nothing in this debate.

As we are debating which rep with the same weight, in the same time frame puts the most tension on the muscles, as you fail at roughly 50% faster with this % and rep speeds on the fast, that’s MORE then obvious it’s the fast rep, or do you think you fail faster on the fast because it puts less tension on the muscle ? If so please say why.

Also, you said there was only one way to sort this out, with EMG, I bought one, and it showed after three experts on EMG said RMS was about the best way to find this out, that there is more overall/total activity on the fast reps. But you still insist when a real World practical experiment has proved you wrong, but you can’t state why.

You keep going on about average, when as I said it means nothings, and you also can't not tell me how you or why you work out this average, what’s average got to do with this.

Wayne

 

[waynexk8]

No Wayne...we will not forgive you for repeating yourself.You ask again the same nonsense ignoring all the answers.

For some strange reason you're unable to understand that it's impossible to use the 100% of your force for the whole set.Regardless

No D. I TOTALLY understand this, I know the force/velocity curve, and I know I can’t use 100% force on 80% that’s why I always say, I try to use 100% of my force. But the point is, for the set amount of time, let’s call it 20 seconds, I am try to use as much force as I can, that’s quite close to 100%, you on the other had for the set time of 20 seconds, are not trying to use 100% force, you “are” only using 80% of your force, 20% less than me for the set time of 20 seconds.

Fast = as much force as he can exert for 20 seconds, let’s just call that 100.

Slow, = 80% of his maximum force for 20 seconds.

How can 80% for the same time frame be as high as 100% ?

Same in a car going uphill and down, I hit the gas 100% say this speed = 100mph, in one hour I have traveled 100 miles, you hit the gas at 80%, = 80mph, in one hour you have traveled 80 miles. [b/]You “only” traveled 80 miles as you DID NOT hit the gas with ENOUGH force, as I hit the gas with MORE force for the same time frame,[/b]

if you lift fast or slow you use the same average force for the same duration.

Explained why/why this average means, and explain why/how you have worked it out ? As I don’t know why you bring it up, or what it means. Sorry there, but I just don’t understand.

The impulse(what you stupidly call "total/overall force") is always the same.

I/we have been told now that physics can’t seem to measure overall/total force, or there can’t be one ? But the EMG can measure this, and it states categorically you are wrong, as it take into account the higher high force on the fast as the higher peak forces of the accelerations, and what I have said all along, is that your medium force cannot make up balance this out, if so, why would the EMG state higher for my fast ?

I don't care if you don't want to accept the truth or you just don't have the intelligence to understand it.
It's a fact...accept it.

That’s quite odd ? What truth have you ?

1,
EMG states fast,

2,
You use more energy in the fast,

3,
You do more work in the fast.

4,
So that’s more power in the fast,

5,
You move the weight 6 times further in the fast,

6,
You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

7,
More speed, velocity and acceleration on the fast.

Is that enough truth, not sure what you have ?

Wayne

 

[waynexk8]

The same as always!You ignore everything that has been answered.

No I do not; you are the one doing that.

Let's redefine the question.You asked which lifting speed has greater effect of force over time(impulse) which in Wayne's world is defined as "total/overall force".

Everyone explained to you that the impulse is identical regardless the lifting speed but you deny that fact based on some "practical proofs".Let's see them once again.[/quyote]

Before I answer this, you NEED to state why lift you are referring to, as the two different lifts, MUST have a different impulse.

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Also, “IS” the impulse the same on 1 rep and a 100 ? As I don’t think you are measuring the impulse with respect to time, as if you think you are, you are then saying f=ma is wrong, as your saying you don’t need a higher force to move something further in the same time frame.

Slow,
Car has a weight of 400kg and an Acceleration of 30m/s force pushing the car 400 x 30 = 12000N of force for 1 second.

Fast,
Car has a weight of 400kg and an Acceleration of 100m/s force pushing the car 400 x 100 = 40000N of force for 1 second.

Fast = 2800N more for 1 second.

“NOT” sure why I am wrong there, is it the deceleration ?

The EMG reads the Root Mean Square of the values.
Does the higher RMS somehow change the fact that the impulse is the same?NO!

The EMG show the total/overall muscle activity = force = tension on the muscles, its show what we are debating about, which has the most total/overall muscle activity, = the fast says the EMG.

Does the higher rate of energy expenditure somehow change the fact that the impulse is the same?NO!

That is NOT an answer, you need to say and explain why there is more energy used in the fast, I have told you, and you need to counter.

Does the greater distance somehow change the fact that the impulse is the same?NO!

Again, that’s no answer. Yes it does, I move the weight further in the same time frame, of course that needs more force, more acceleration needs more force, I mean its simple physics, its common sense. Newton's Law that force is equal to mass times acceleration, but we know in non-relativistic limit mass is invariant so if we apply more force it causes greater acceleration. The net force acting upon the object will be equal to the rate at which its momentum/movement change. When the object's velocity increases, so does its energy and hence it’s mass equivalent. It thus requires more force to accelerate it the same amount than it did at a lower velocity. Newton's Second Law.

Does the higher rate to fatigue somehow change the fact that the impulse is the same?NO!

All these prove what you say wrong; actually it would be nice if you gave me readable equations that state the impulse is the same per time, using the same weight over different distances and with diffract velocities and accelerations ?

So Wayne...quit these nonsense and see what everybody's telling you.Your above pseudoarguments are just a sad attempt to ignore the facts and keep leaving in your own world.

Odd think to say, why 1 to 7 prove you wrong. D. for once shows me proof, I showed you on 1 to 7 of the last post.

Do you admit you only use 80% force for the set time ? I try to use 100% force, 80 – 100 = 20 more force used, quite simple, I use more force, thus more tension on the muscle = you fail faster.

Wayne

 

[waynexk8]

Please, for all to answer this one.

I was just wondering and thinking, ARE you adding “all” the force, I mean with the fast there are NOT just force being exerted by the muscles, there are HUGE forces on them, for say .1 of a second x the 6 reps = high forces on the muscle for maybe .6 of a second. Have you added these in ?

I mean the peak force from the transition from negative to positive, the force on the muscle, NOT given out by the muscles ? We call them the MMMTs {Momentary Maximum Muscle Tensions} these forces “ON” the muscles can be as high as 140%

Have you added these on ?

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Wayne

 

[Dale]

Wayne, please focus. I am trying to help you learn some physics, that is what you came here for, right? Do you understand the force-time diagram. Do you see how the force is not constant, but that it varies over the lift? Do you graphically see what the impulse is?

 

[Dale]

1,
EMG states fast,

EMG measures electrical activity in the muscles, not force nor energy.

2,
You use more energy in the fast,

Yes.

3,
You do more work in the fast.

No, you do 0 work over a rep regardless of if you do it fast or slow.

4,
So that’s more power in the fast,

Peak power, yes, average power is 0.

5,
You move the weight 6 times further in the fast,

OK, this can be interpreted more than one way, but at least one of them is correct.

6,
You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

While I am sure that you do fail faster with fast reps I don't think that your conclusions follow. Something is exhausting the muscle's ability to function, but why MUST it be the tension/time. Why couldn't it be the energy expenditure/time, or the oxygen debt, or ATP depletion, or lactate buildup, or temperature rise? I can think of lots of things that it could be, so the MUST just isn't true. Just because they fail faster does not imply that there is more tension/time.

7,
More speed, velocity and acceleration on the fast.

Definitely.

Please get back on track, if you want to get anything out of this you need to actually challenge yourself mentally and learn a bit. Do you understand the previous graph, in particular, do you see what is meant by impulse?

 

[douglis]

That’s quite odd ? What truth have you ?

1,
EMG states fast,

2,
You use more energy in the fast,

3,
You do more work in the fast.

4,
So that’s more power in the fast,

5,
You move the weight 6 times further in the fast,

6,
You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

7,
More speed, velocity and acceleration on the fast.

Is that enough truth, not sure what you have ?

Wayne

O.K. Wayne...so you claim that all the above somehow prove that more force per unit of time is applied for the fast reps.
The least thing you should do is to try to prove with physics that those variables have the effect you claim.You obviously can't even try because you lack of even basic physics knowledge so you only have two options:

1)DaleSpam has the superhuman patience to help you learn step by step some basic physics.Shut up and read carefully what he writes.

2)A better and more time saving option...read DaleSpam's last post and just accept it as a fact with no question asked.Especially this part:

so the MUST just isn't true. Just because they fail faster does not imply that there is more tension/time.

 

[waynexk8]

Wayne, please focus. I am trying to help you learn some physics, that is what you came here for, right?

Ok.

Yes, and thank you

Do you understand the force-time diagram.

Yesish.

Do you see how the force is not constant, but that it varies over the lift?

Yes I understand that all too well.

Do you graphically see what the impulse is?

Not really sure what you are getting at here ? Are you saying/showing that my force will go up and up, as I am trying to accelerate the weight as much as I can, then as I have to decelerate it to stop for the transition in the repetitions, I have to use less force ? If so yes I see that. However I think the area for the acceleration deceleration is a bit out. We found in a study that when using 81% the bar actually decelerated for 52% of the ROM. {range of motion} However, there was a flaw in that lift, as the concentric part of that lift, took 1.5 seconds. So I would say one of my lifts would have more an acceleration of 80% so I am using close to 100% force for 80% of .5 of a second.

Wayne

 

[sophiecentaur]

Wayne - you really can't bring yourself to believe that you are on a hiding to nothing and that what goes on in your arms is to do with your Muscles and their non-ideal behaviour. Muscles are not simple machines or springs and cannot be modeled as such. If they were, you would use no energy / force / strength / bananas if you just stood and held something stationary. You know that doesn't happen. Why keep ignoring this?

 

[Dale]

Yesish.

OK, that doesn't sound very confident. Can you explain what makes you hesitate or a little unsure?

Not really sure what you are getting at here ?

I am not getting at anything yet, I am just teaching you the meaning of the various important quantities. Are you unsure about what it means to have an area under a curved line, or is there something else bothering you about the area under the force v time diagram?

Are you saying/showing that my force will go up and up, as I am trying to accelerate the weight as much as I can, then as I have to decelerate it to stop for the transition in the repetitions, I have to use less force ? If so yes I see that.

That is correct. The part where you are accelerating it up is the "peak" on the force v. time diagram, and the part where you are decelerating it is the "valley".

However I think the area for the acceleration deceleration is a bit out. We found in a study that when using 81% the bar actually decelerated for 52% of the ROM. {range of motion}

OK. So in the approximation I am using the bar is decelerating for 50% rather than 52%. We could certainly use more accurate representations of the motion, but if we did so the math would quickly get more complicated. I just used the simplest function that I thought was close. I am actually very glad to know that it is only ~2% off by that measure.

 

[waynexk8]

Will get back to the older questions.

EMG measures electrical activity in the muscles, not force nor energy.

Electrical muscle activity in the muscles is the force/strength they are using for the set time. Why/how could you think other, and what did you think it was ?

http://www.actabio.pwr.wroc.pl/Vol4No2/2.pdf

For the evaluation of muscle activities associated with force exertion the surface
electromyography method is well established. The amplitude of the EMG signal
quantitatively expresses muscle activity [16], [18], [32], [40] and has been used in
studies of various vocations to estimate muscle loads in tasks involving upper limbs [9],
[17], [38].
As maximum force exerted by the hand depends on upper limb location, for
musculoskeletal load assessment it is important to determine how the value of
maximum force changes in relation to upper limb location. Although studies which
considered this problem (as cited above) have been performed, taking into account
variety of upper limb locations, further research is still needed for normalisation
purposes.
The force which the muscle exerts as well as muscle tension expressed by the
amplitude of the EMG signal depend on muscle length (upper limb location) [3]. Also
the study of DUQUE, MASSET and MALCHAIRE [7] confirmed that differences in EMG
signal amplitude in the flexor carpi radialis muscle should occur according to wrist
flexion and extension, and the study of Wright (as cited in [6]) showed that the activity
of the long head of the biceps brachii depends on the arm abduction and arm rotation.
Muscle activity during force exertion can be spread up between muscle activity for
upper limb stabilisation in a defined upper limb location and activity connected with
the external force exertion. It should be expected that not only the component of
muscle activity, which is responsible for upper limb stabilisation, depends on upper
limb location but muscle activity associated with force exertion is influenced by upper
limb location as well. Therefore, it is also an interesting problem to see whether the
component of muscle activity, which is associated with handgrip force exertion, varies
according to upper limb location.

This is why I bought the EMG, to show what I call the total/overall muscle force or/and strength used in a set time will be different.

Let me try and prove my total/overall muscle force theory. Lift a very light weight up and down for 10 seconds, lift a very heavy weight up and down for 10 seconds, and the very heavy weight will need more total/overall force.

You use more energy in the fast,

=DaleSpam;3819491]Yes.

You do more work in the fast.

No, you do 0 work over a rep regardless of if you do it fast or slow.

As I move the weight 12m to the slow 2m, that’s more work done ? Or are you saying, that if I move up, and then back down to the starting position I have done no work ? Still don’t get that, as I thought work was force times the distance through which goes, thus 12m = more distance the force was used for than the 2m ?

So that’s more power in the fast.

Peak power, yes, average power is 0.

Don’t get that sorry ? Let's calculate how much power I would be used on both rep speeds. Distance weight 91 kg 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. 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.

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 = 4950Joules.

Fast set,
3300 x 25 = 82500Joules

You move the weight 6 times further in the fast,

OK, this can be interpreted more than one way, but at least one of them is correct.

K.

You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

While I am sure that you do fail faster with fast reps I don't think that your conclusions follow. Something is exhausting the muscle's ability to function, but why MUST it be the tension/time. Why couldn't it be the energy expenditure/time, or the oxygen debt, or ATP depletion, or lactate buildup, or temperature rise? I can think of lots of things that it could be, so the MUST just isn't true. Just because they fail faster does not imply that there is more tension/time.

Basically you fail faster because your muscles are working harder, and why would your muscles need to work harder ? Because they were using more force/strength per unit of time. As the muscles don’t work harder because things are getting easier, but they work harder like machines or anything else, when the situation gets harder, and its far harder to accelerate a weight 6 times in 6 seconds, and to move a weight 12m to 2m in the same time frame.

It will be also energy expenditure/time, oxygen debt, ATP depletion, lactate build-up, and temperature rise, but the question is, “why” do you use more of these ? I would think there can only be one answer, as the muscles are using more total/overall force, and having more total/overall force exerted on them, thus more tension on the muscles.

Question,
1,
I lift and fail at 24 seconds, I lift the weight 24 times and 48m, you lift the weight for 24 seconds, you lift the weight 4 times and 8m, if wanted you could lift the weight for 48 seconds, but you stop at 24 seconds, which muscle has worked the hardest ? Which do I use more force.

2,
I lift and fail 50% faster, at 30 seconds, I then lift up a lighter weight and lift it for another 30 seconds. which muscle has worked the hardest ?

Which do I use more force, 1 or 2 ? Which lifters muscles work the hardest ?

More speed, velocity and acceleration on the fast.

Definitely.

Please get back on track, if you want to get anything out of this you need to actually challenge yourself mentally and learn a bit. Do you understand the previous graph, in particular, do you see what is meant by impulse?[/QUOTE]

Right answered that one.

Wayne

 

[Dale]

Electrical muscle activity in the muscles is the force/strength they are using for the set time. Why/how could you think other, and what did you think it was ?

I have a PhD in biomedical engineering and did some coursework and research in functional electrical stimulation for neuro-prosthetic applications, so I know a thing or two about EMG, EEG, EKG, muscle recruitment, electrical stimulation, pacemakers, etc. Electrical muscle activity measures voltage changes due to the depolarization of the muscle cell membranes during the muscle's action potential, not the force exerted by the muscle.

The amplitude of the EMG signal
quantitatively expresses muscle activity [16], [18], [32], [40] and has been used in
studies of various vocations to estimate muscle loads in tasks involving upper limbs [9],
[17], [38].

Note, the key word "estimate". If you know the EMG and you know the tension v recruitment curve and you know the position of the limb and you know the force v tension curve for that position then you can use the EMG to make a good estimate as to what the force is. An estimate and a measurement are not the same thing. Force is measured with a force transducer, an EMG is a voltage transducer. The units of the EMG are μV, not N.

This is why I bought the EMG, to show what I call the total/overall muscle force or/and strength used in a set time will be different.

Let me try and prove my total/overall muscle force theory. Lift a very light weight up and down for 10 seconds, lift a very heavy weight up and down for 10 seconds, and the very heavy weight will need more total/overall force.

Before you can prove your total/overall muscle force theory you need to define it, otherwise there is no theory to prove or disprove. That is the purpose of teaching you about the standard physics concepts. I am hoping that as you learn what is meant by them you can express your concepts in the standard language, clearly define your theory, and then we can see the implications.

So, let's talk a little more about impulse and see if it has the properties that you expect for "total/overall muscle force".

Impulse has the property that if you exert twice the force for the same amount of time you have doubled your impulse. So, for example, if you exert 100 lbs for 10 s and I exert 50 lbs for 10 s you have exerted twice the impulse that I have. Does this agree with your concept of "total/overall muscle force"?

Impulse also has the property that if you exert the same force for twice as long you have doubled your impulse. So, for example if you exert 100 lbs for 10 s and I exert 100 lbs for 5 s you have exerted twice the impulse that I have. Does this also agree with your concept of "total/overall muscle force"?

 

[douglis]

Note, the key word "estimate". If you know the EMG and you know the tension v recruitment curve and you know the position of the limb and you know the force v tension curve for that position then you can use the EMG to make a good estimate as to what the force is. An estimate and a measurement are not the same thing. Force is measured with a force transducer, an EMG is a voltage transducer. The units of the EMG are μV, not N.

Hi DaleSpam...if you have the time check the paragraph "2.1 Participants and experimental protocol" in http://jmbe.bme.ncku.edu.tw/index.php/bme/article/viewFile/635/839 .Do you think that the magnitude total muscle activation (TMA) can give a good estimation of the impulse?It's basically the integration of the EMG curve in respect of time.

It's interesting that the TMA per second is greater for the slow puh ups(in contast of what Wayne claims).For example if you check the tables 1 and 3 for the pectoralis major:
for the slow push ups the TMA is 3121.81 for 101.2 sec.(TMA/t=30.85)
while for the fast push ups the TMA is 2114.22 for 84.2 sec.(TMA/t=25.11)

 

[waynexk8]

O.K. Wayne...so you claim that all the above somehow prove that more force per unit of time is applied for the fast reps.
The least thing you should do is to try to prove with physics that those variables have the effect you claim.You obviously can't even try because you lack of even basic physics knowledge so you only have two options:

Hi D. and all.

Are you trying to be sarcastic or something ? I mean we are on a physics forum, I have all ready proved this, ITS UP TO YOU TO “TRY” AND DISPROVE IT, the cards are in my hands, the EMG states you wrong.

Don’t you understand, it’s up to you to try and prove me wrong and you right, WHAT PROOF AND EVIDENCE HAVE YOU ?

1,
EMG states fast,

I have the videos to prove it. All EMGs state this; it was MORE than obvious to me and most.

I mean walk up a 1 mile very steep hill with a pack on your back, then “try” and run up it as fast as you can. Which is the hardest on the muscles or/and which physiologically causes the greatest stimulus, as they are fast and slow actions causing far far far different stimulus even though they produce the same mechanical work, as computed by moving the same load through the same distance. 2,
You use more energy in the fast,

This has been known for a 100 years, I knew this 40 years ago. And this physics site showed and proved this to you.

3,
You do more work in the fast,

Work is the product of a force times the distance through which it acts, I move the weight 12m you move it 2m in the same time frame.

4,
So that’s more power in the fast,


Lets calculate how much power I would be used on both rep speeds. Distance weight 91 kg 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. 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.

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 = 4950Joules.

Fast set,
3300 x 25 = 82500Joules

5,
You move the weight 6 times further in the fast,

I move the weight 12m you move the weight 2m

6,
You fail with these variables, 50% faster in the fast = there MUST be more tension on the muscles per unit of time to make them fail faster, = more tension = there must be more total/overall force if there is more tension as on failing faster.

There is a huge study proving this, have no time to find it right now, but you have seen it.

And there is my video.

http://www.youtube.com/watch?v=sbRVQ_nmhpw&list=UUTeoEssmCPZycmfODrHvM2w&index=71&feature=plcp7,
More speed, velocity and acceleration on the fast.

Well if you don’t understand that ?

1)DaleSpam has the superhuman patience to help you learn step by step some basic physics.Shut up and read carefully what he writes.

I am listening and answering, and all here are very intelligent and polite, and I thank them.

But no one is disproving what I say, like the 1 to 7 above, just try and say prove the EMG is wrong, but that’s not possible, as it’s a test that’s been done over and over by me, and 100,00 around the World, it’s one of the first tests people do when learning EMG, it’s a standard test.

Then there is you fail 50% faster, I mean that’s more obvious than 1 + 1 = 2, I mean you fail faster because the faster reps, like running to walking are far hider, they are harder because your putting more total/overall tension per unit of time on the muscles, that more tension = more total/overall force per unit of time on the muscles, or what do you think it means.

Why do you think your using more energy ? Look, what happens when you run faster and faster, you use more and more force, vertical and horizontal and a little different, but basically the same, you have to use more force to run faster, and more force to rep the weight faster.

2)A better and more time saving option...read DaleSpam's last post and just accept it as a fact with no question asked.Especially this part:

Will read his posts now.

Wayne

 

[waynexk8]

Could someone please answer this.


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

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

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.

The fast has nearny 3000% more mean propulsive force as in N's in the same time frame.

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

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

Wayne

 

[waynexk8]

Wayne - you really can't bring yourself to believe that you are on a hiding to nothing and that what goes on in your arms is to do with your Muscles and their non-ideal behaviour. Muscles are not simple machines or springs and cannot be modeled as such. If they were, you would use no energy / force / strength / bananas if you just stood and held something stationary. You know that doesn't happen. Why keep ignoring this?

Not sure what you mean here ? I know its to do with my muscles, but I thought we were not debating this out on a machine repping the weight ?

I need to get back to your other post from a few days ago, sorry there.

Wayne

 

[waynexk8]

How do you work that out ? What machine can lift say 80% up, and then lower it down using very little energy, and what energy is this ?

Wayne

Plus all the rest
That comment shows that yo don't get the mosgt basic part of all this thread and others.

The machine doesn't need to use use "very little energy" on the way down. IT GETS ALL THE ENERGY BACK! Unlike your muscles, which don't have Energy Recovery. So the two cannot be compared.

Where and how does the machine get its energy back ? It will be powered by say diesel or electricity, let’s say diesel, so it uses, let’s just say a half a pint of diesel to lift the weight, you tell me where and how the machine gets that diesel back ? You know very well that when the half a pint of diesel is gone, has been used to lift the weight, you can “never” get it back.

And when it lowers the weight, it will have to use energy again, as in the diesel, a little less this time, but it has to use energy/diesel to move in any direction, as its using force, and this force is putting tension on the machine.

You insist that this problem can be solved your way and you have the nerve to hang onto the idea in the face of people who know much more basic Physics than you.

You and all here know physics far better than me, but I give you 7 real World practical points proving your theory does not fit, and as you know, a theory, is just a theory until you can prove it with a practical experiment, and I have proved it wrong and few times.

Just for now take the EMG, and the fact that you fail 50% faster.

Also, what have you proved ? I see no equations, what you did say that there is no such thing as total/overall force in physics, yet the EMG reads out a higher average reading on the faster, and the EMG work with the equations of physics, they are put in the EMG.

I thought is was the physics job to do the theory, and when the practical proves this wrong, the physics needs to be looked at.

The only hope you have is to do a Physics course at some level which may help you understand what you need to know in order to grasp how crazy your idea is.
If someone told you that swimming the Pacific is a no no, how long would you not believe them?

I do see what you mean, but what about my points, like the EMG, and the fact you fail faster, they can only mean one thing, the fast is putting more tension on the muscle, and the only way to put more tension on the muscle is by putting out more force, and taking more force on the muscles.

Wayne

 

[waynexk8]

Could anyone please answer this ?

I was just wondering and thinking, ARE you adding “all” the force, I mean with the fast there are NOT just force being exerted by the muscles, there are HUGE forces on them, for say .1 of a second x the 6 reps = high forces on the muscle for maybe .6 of a second. Have you added these in ?

I mean the peak force from the transition from negative to positive, the force on the muscle, NOT given out by the muscles ? We call them the MMMTs {Momentary Maximum Muscle Tensions} these forces “ON” the muscles can be as high as 140%

Have you added these on ?

Lift 1,
You lift 80% of the ground, up 1m and then down 1m all in 1 second, .5/.5

Lift 2,
You start at the top, lower the weight down 1m, and then lift it back up 1m all in 1 second, .5/.5

On lift 2, on the transition from negative to positive, there will be huge force on the muscles and coming from the muscles.

Wayne