Most force/strength used in the same time frame

[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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