Waynes own emg studys fast reps = more force/tension on the muscles

[SumDumGoi]

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.



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



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 ???

Wayne



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

Wayne

Wayne,

Let me clarify my question. Under which condition do you feel would produce the greatest amount of muscular force; 1) Someone who is lifting their 1 RM which is performed at a slow speed, or 2) Someone who is lifting 50% of their 1 RM at a very fast speed?

The problem that you are having regarding the physicist vs. physiologist scenario is that from a physiological standpoint your muscles are simply unable to lift heavy weights quickly. This is because as you increase the speed of the contraction the total number of actin-myosin cross-bridges that are formed at any given point in time are reduced. This needs to happen to allow the muscle to move at a faster velocity. It is only when these actin-myosin cross-bridges are formed that muscular force can be produced (please see my earlier "tug of war" example regarding that force can only be generated on the rope when the feet are in contact with the ground). This is the reason why you are able to hold a weight that is greater than your 1RM statically for a short period of time. This also explains why when you perform a 1 RM why the weight is moved slowly. To handle a weight close to your max you must maximize the total number of cross-bridges that are formed which occurs at a slow velocity. This is also the reason why when you move a weight quickly that less force is produced. The body is not a machine that is capable of producing endless amounts of forces at the velocities you provided in the examples above. These are the principles that your physicist must be aware of. He needs to understand how the body actually works in addition to the math.

In regards to your EMG data, others have provided you a clear rationale as to why your data analysis might be flawed. I am not going to pretend to know exactly how to interpret EMG data analysis, but I do understand basic muscle physiology. The EMG is not a direct measurement of muscular force, but rather the electrical activity associated with the muscle during movement. This could be increased through two different ways; 1) increased force of contraction & 2) increased velocity of the contraction. Your data analysis is focused exclusively on the first aspect while completely neglecting the second. As an example I will provide information from an ECG interpretation of the heart, which utilizes the same principles. An increase in ECG activity can be observed if 1) you were to increase the strength of each heart contraction or 2) if you were to increase your heart rate. I could very easily increase my heart rate without increasing the force of the heart contraction. If I were to solely look at ECG activity it would be increased, but this increase would have nothing to do with the force of each heart contraction.

In the observation I asked you to set up I was trying to test out this theory in regards to your EMG data. I wanted to set up a situation based on physiological principles where the amount of muscular force produced during the fast contraction speed have to be lower than the amount of muscular force produced during a 1RM. I also set up this experiment as a way of maximizing muscular power (which is a completely separate variable). If your EMG data is higher during the fast contraction with a weight that is 50% of your 1 RM compared to the slow contraction performed at your 1 RM your data analysis would be flawed in regards to the faster contraction being capable of producing more force. Instead, what I am suspecting that you are measuring is an increase in muscular power and the higher EMG signal is being caused by an increased rate of cross-bridge cycling due to the faster velocity (see heart rate in the above EKG example).

Let me ask you for the final time. Have you performed this simple observation yet and what are your results? If you don't understand the premise of the experiment and your answer to my initial question in this post is that the person who is lifting 50% of their 1 RM at a fast velocity would be capable of producing more force then your problem is that you simply don't understand physiology as this is not what is observed in reality as shown by the force-velocity curve.

[waynelucky2]

yeh you 2 are basically saying the same thing.

if the 85% load was lifted slowly it would never get the absolute maximum force the muscle is capable of producing. u have to lift it as fast as possible...... therefore faster rep speeds give higher peak forces.

at the start of the 85% rep (say bottom of the bench) the force is maximal as u start accelerating as hard as possible. as the speed of the lift gradually increases, the force produced drops and u essentially slide down the force-velocity curve - u trade force for speed. when the lift returns to the bottom position the cycle starts over. so like morderstwo said - points on the force velocity curve can be specific points in time during the rep.

Yes, BUT how do you think that the median forces make up or balance out the faster peak forces ??? Yes I know the fast has a deceleration, but still how can 100% force for say 60% of the ROM, then this forces goes slowly down from 100 to zero {like the slow rep has to as well} be made up or balanced out by the faster reps, lets recap, there is one slow up at 3/3 and 6 fast reps at .5/.5

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. 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. {bet he’s even got a Christmas present for me this year ROL} 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.
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 ???

Thx for all the help and time

Wayne

[SumDumGoi]

Your peak force is still not going to be higher during a faster rep compared to a slower rep. You lack a fundamental understanding of basic physiology and are therefore confused about the issue. But feel free to continue on with make believe numbers while never once attempting to perform the basic experiment I asked you to perform which demonstrates the point I made in my previous sentence.

[N@tural1]

To sum up in an extremely simplistic way that Wayne may understand.

1/ It takes more force to move any given load faster as opposed to slower.
2/ The slower a load moves the more force you generate provided that you are attempting to move the load as fast as you can

[douglis]

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.


Wayne

The whole point of the force curve is that you can use the value of 100(maximal force) ONLY at the first segment where the speed is zero.

[waynelucky2]

Going out, internet seems ok, will get back to all tomorrow.

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, starting at 30.

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

From someone over at the physics forum.
Hmm, well, I believe part of it is that they like to keep this pretty textbook and educational, without a lot of stuff that would confuse students that came here, in my opinion.

If you could please do a few more EMG tests for me to get an idea of what is going on,

Please do a test where you hold your empty hand at that same height(grip your hand as tightly as you would grip the bar, if you can) and also do a slow and fast test with no weight.

This thread has made me realize that if we considered it to be two constant forces(i.e. if you pushed with a constant force, and gravity is the other) Then you should be able to determine the force applied, but it is also a simplification of what is happening, and doesn't directly represent what happens when one lifts weights.

I figure that since as was mentioned many times the impulse must be equal and opposite to have a 0 velocity on both ends, that means that
gT = Ft1
where g is the force of gravity, T is the overall time, F is the force you pushed with and t1 is how long you pushed for

The work done must also be the same so
gX = Fx1
where g is gravity again, X is the overall distance, F is the force applied, and x1 is the distance that it was applied

since f = ma we know that we can get the acceleration from the force, and so it must also be true that

x1 = .5(a)t12

where a = (F-g)/m because we need the net force during the time that we are pushing and gravity is pushing back. You can combine those equations and solve for

F = g2t2/(gt2-2xm)

So this gives you the constant force required to move an object against gravity for a distance of x in a time of t with a mass of mHmm, well, I believe part of it is that they like to keep this pretty textbook and educational, without a lot of stuff that would confuse students that came here, in my opinion.

If you could please do a few more EMG tests for me to get an idea of what is going on,

Please do a test where you hold your empty hand at that same height(grip your hand as tightly as you would grip the bar, if you can) and also do a slow and fast test with no weight.

This thread has made me realize that if we considered it to be two constant forces(i.e. if you pushed with a constant force, and gravity is the other) Then you should be able to determine the force applied, but it is also a simplification of what is happening, and doesn't directly represent what happens when one lifts weights.

I figure that since as was mentioned many times the impulse must be equal and opposite to have a 0 velocity on both ends, that means that
gT = Ft1
where g is the force of gravity, T is the overall time, F is the force you pushed with and t1 is how long you pushed for

The work done must also be the same so
gX = Fx1
where g is gravity again, X is the overall distance, F is the force applied, and x1 is the distance that it was applied

since f = ma we know that we can get the acceleration from the force, and so it must also be true that

x1 = .5(a)t12

where a = (F-g)/m because we need the net force during the time that we are pushing and gravity is pushing back. You can combine those equations and solve for

F = g2t2/(gt2-2xm)

So this gives you the constant force required to move an object against gravity for a distance of x in a time of t with a mass of m.
This would be a good time to reemphasize that this is a simplification, and doesn't directly represent what happens when one lifts weights. When one actually lifts weights I am sure the force applied varies greatly over time, and is NOT constant. This equation gives you what the force would be from a machine that supplied a constant acceleration and then stops instantly at t1

What you are saying is that a greater average speed(in terms of overall distace/overall time) must have had a greater force, and that does make sense
because gravity will slow the object down at 9.8m/s2 no matter the velocity it gets, so to go further in a short time it must build up velocity quickly

Running the data you gave me through that equation gives:

F = g2t2/(gt2-2xm)

A) 80lbs(36kg) is 352.8N for 1m in .5s
F = (352.82 * .52) / [( 352.8*.52) - 2 * 36 * 1]
F = (124467.84 * .25 ) / (88.2 - 72)
F = 31116.96 / 16.2
F = 1920.8N

B) 80lbs(36kg) is 352.8N for .166M in .5s * note all that changes is x
F = (352.82 * .52) / [( 352.8*.52) - 2 * 36 * .166]
F = 31116.96 / 79.848
F = 389.7N

But here is why we say that over the same time the force is the same:

gT = Ft1
F = ma

A) 352.8N(g) * .5s(T) = 1920.8(F) * t1
176.4 = 1920.8t1
t1 = .092s

So the greater force (1920.8) was only applied for .092 of the .5s that is 18.4% of the time.

B) 352.8N * .5 = 389.7N * t1
176.4 = 389.7t1
t1 = .45s
So the lesser force was applied for .45s that is 90% of the time

Maybe this will help though.

Wayne

[douglis]

But here is why we say that over the same time the force is the same:

gT = Ft1
F = ma

A) 352.8N(g) * .5s(T) = 1920.8(F) * t1
176.4 = 1920.8t1
t1 = .092s

So the greater force (1920.8) was only applied for .092 of the .5s that is 18.4% of the time.

B) 352.8N * .5 = 389.7N * t1
176.4 = 389.7t1
t1 = .45s
So the lesser force was applied for .45s that is 90% of the time

Maybe this will help though.

The guy is right!

[waynelucky2]

To sum up in an extremely simplistic way that Wayne may understand.

1/ It takes more force to move any given load faster as opposed to slower.
2/ The slower a load moves the more force you generate provided that you are attempting to move the load as fast as you can

True.

Wayne

[waynelucky2]

The whole point of the force curve is that you can use the value of 100(maximal force) ONLY at the first segment where the speed is zero.

Dont understand what your saying, if the speed is zero, there is no movment, no force ???

Wayne

[waynelucky2]

Your peak force is still not going to be higher during a faster rep compared to a slower rep.

Not sure why you say that ??? We all agree that the peak forces are higher in the faster reps, how could you think other ???

You lack a fundamental understanding of basic physiology and are therefore confused about the issue.

No.

But feel free to continue on with make believe numbers while never once attempting to perform the basic experiment I asked you to perform which demonstrates the point I made in my previous sentence.

I have only just seen that post, will read it and reply.

Wayne

[all pro]

Not sure why you say that ??? We all agree that the peak forces are higher in the faster reps, how could you think other ???
Wayne

No Wayne. We all agree that peak force is LOWER the faster the bar speed, well everyone but you.

pounds of force 400 @ 1000 rpm = 76 horsepower
pounds of force 350 @ 2000 rpm = 133 horsepower
pounds of force 300 @ 3000 rpm = 171 horsepower
pounds of force 250 @ 4000 rpm = 190 horsepower
pounds of force 200 @ 5000 rpm = 210 horsepower
Force is dropping, horsepower is increasing as rpm increases. The human body does the exact same thing.

[waynelucky2]

Let me clarify my question. Under which condition do you feel would produce the greatest amount of muscular force; 1) Someone who is lifting their 1 RM which is performed at a slow speed,

But this slow speed, will be as fast as the person can move the weight, and that’s what this debate is about.

The greatest force and time would be the 1RM.

or 2) Someone who is lifting 50% of their 1 RM at a very fast speed?

But this is not the debate at all ???

The problem that you are having regarding the physicist vs. physiologist scenario is that from a physiological standpoint your muscles are simply unable to lift heavy weights quickly.

Speed here is a relative word. Actually as I said, on heaver weight the muscles are trying to lift them as fast as possible.

This is because as you increase the speed of the contraction the total number of actin-myosin cross-bridges that are formed at any given point in time are reduced. This needs to happen to allow the muscle to move at a faster velocity. It is only when these actin-myosin cross-bridges are formed that muscular force can be produced (please see my earlier "tug of war" example regarding that force can only be generated on the rope when the feet are in contact with the ground). This is the reason why you are able to hold a weight that is greater than your 1RM statically for a short period of time. This also explains why when you perform a 1 RM why the weight is moved slowly. To handle a weight close to your max you must maximize the total number of cross-bridges that are formed which occurs at a slow velocity. This is also the reason why when you move a weight quickly that less force is produced. The body is not a machine that is capable of producing endless amounts of forces at the velocities you provided in the examples above. These are the principles that your physicist must be aware of. He needs to understand how the body actually works in addition to the math.

This has basically nothing to do with the debate, think you may have the debate back to front, sorry there, could be part my fault, bet D. has had it wrong as well.

In regards to your EMG data, others have provided you a clear rationale as to why your data analysis might be flawed.

Sorry have not read all the posts yet, but most should know that an EMG takes the muscle activity, I mean what more do you want, as more activity more force/strength used right ??? I mean all have seen EMG test on bench press, incline bench press and decline bench press, which shows which has the most pec activity.

I am not going to pretend to know exactly how to interpret EMG data analysis, but I do understand basic muscle physiology. The EMG is not a direct measurement of muscular force, but rather the electrical activity associated with the muscle during movement.

Right.

This could be increased through two different ways; 1) increased force of contraction & 2) increased velocity of the contraction.

But to increase velocity, YOU MUST HAVE MORE FORCE/STRENGTH used.

Your data analysis is focused exclusively on the first aspect while completely neglecting the second.

No its not, force and velocity are sort of related, you can’t have one without the other. More force more velocity, more velocity more force, how can you have it another way ??? To change a body's velocity there must be a net force acting on that body. A car will decelerate when you take your foot off the pedal, because there will be forces acting to decelerate it, and when you hit the pedal again, the car accelerates with its opposing forces. Force has no direct relationship with velocity, only acceleration.

This is taken from a site;
The way I visualize it is to imagine a force acting on a particle for a particular amount of time--say 1 second.

The work done on the particle is Force × the distance it moved.

Power is work/time, so that is:
P = (force × distance) / time

But that's the same as:
P = force × (distance / time)

And since (distance / time) = speed,
P = force × speed

D. seems to this more power = more force is not true, but he can NOT explain why.
__________________

As an example I will provide information from an ECG interpretation of the heart, which utilizes the same principles. An increase in ECG activity can be observed if 1) you were to increase the strength of each heart contraction or 2) if you were to increase your heart rate. I could very easily increase my heart rate without increasing the force of the heart contraction. If I were to solely look at ECG activity it would be increased, but this increase would have nothing to do with the force of each heart contraction.

How could you increase your heart rate without increasing the force of the heart contraction ???

In the observation I asked you to set up I was trying to test out this theory in regards to your EMG data. I wanted to set up a situation based on physiological principles where the amount of muscular force produced during the fast contraction speed have to be lower than the amount of muscular force produced during a 1RM. I also set up this experiment as a way of maximizing muscular power (which is a completely separate variable). If your EMG data is higher during the fast contraction with a weight that is 50% of your 1 RM compared to the slow contraction performed at your 1 RM your data analysis would be flawed in regards to the faster contraction being capable of producing more force. Instead, what I am suspecting that you are measuring is an increase in muscular power and the higher EMG signal is being caused by an increased rate of cross-bridge cycling due to the faster velocity (see heart rate in the above EKG example).

It’s not measuring power, its measuring muscle activity, force/strength.

Here are a few tests I did, two pads on biceps and two pads on foramens.
Held out 15 pounds in the middle of a concentration curl, then added 10 pounds and then another 10 pounds.
67.6
148.7
228
Held out arm only in the middle of a concentration curl. Held a 3.5 pound DB bar out only in the middle of a concentration curl. Did reps as fast as possible with the 3.5 pound DB bar.
8.5
17.2
123

Let me ask you for the final time. Have you performed this simple observation yet and what are your results? If you don't understand the premise of the experiment and your answer to my initial question in this post is that the person who is lifting 50% of their 1 RM at a fast velocity would be capable of producing more force then your problem is that you simply don't understand physiology as this is not what is observed in reality as shown by the force-velocity curve.

Will do your test, but it will show the 1RM higher. But this is not the debate, as the weight is different and the time is different.

Wayne

[waynelucky2]

Zula wrote;
If you could please do a few more EMG tests for me to get an idea of what is going on,

Please do a test where you hold your empty hand at that same height(grip your hand as tightly as you would grip the bar, if you can) and also do a slow and fast test with no weight.

Sorry, on the first test the weight was 15 pound to start not 30.

Here are a few tests I did, two pads on biceps and two pads on foramens.
Held out 15 pounds in the middle of a concentration curl, then added 10 pounds and then another 10 pounds.
67.6
148.7
228
Held out arm only in the middle of a concentration curl. Held a 3.5 pound DB bar out only in the middle of a concentration curl. Did reps as fast as possible with the 3.5 pound DB bar. Just held my hand out, and lightly griped. Did a slow curls with no weight. Did fast curls with no weight.
8.5
17.2
123
15.2
16.5
80.7

Zula wrote;This thread has made me realize that if we considered it to be two constant forces(i.e. if you pushed with a constant force, and gravity is the other) Then you should be able to determine the force applied, but it is also a simplification of what is happening, and doesn't directly represent what happens when one lifts weights.

Yes my force/strength pushing a weight and gravity pushing back, and the harder you push the more resistance there will be.

Zula wrote;I figure that since as was mentioned many times the impulse must be equal and opposite to have a 0 velocity on both ends, that means that
gT = Ft1
where g is the force of gravity, T is the overall time, F is the force you pushed with and t1 is how long you pushed for

The work done must also be the same so
gX = Fx1
where g is gravity again, X is the overall distance, F is the force applied, and x1 is the distance that it was applied

since f = ma we know that we can get the acceleration from the force, and so it must also be true that

x1 = .5(a)t12

where a = (F-g)/m because we need the net force during the time that we are pushing and gravity is pushing back. You can combine those equations and solve for

F = g2t2/(gt2-2xm)

So this gives you the constant force required to move an object against gravity for a distance of x in a time of t with a mass of m

This would be a good time to reemphasize that this is a simplification, and doesn't directly represent what happens when one lifts weights. When one actually lifts weights I am sure the force applied varies greatly over time, and is NOT constant.

I can’t pretend that I understand the physics equations, but get the jest of what your say.

Yes there will be many different forces thought the rep, as of the biomechanical advantages and disadvantages. However, I don’t mind working it out on a machine lifting the weights, as it’s the same, the faster you lift the more the opposing force, thus more and more force will be needed the faster you go, also the faster you go the more distance you will move the weight in the same time frame, and to move a weight further in the same time frame needs more force.

Zula wrote;This equation gives you what the force would be from a machine that supplied a constant acceleration and then stops instantly at t1

What you are saying is that a greater average speed(in terms of overall distace/overall time) must have had a greater force, and that does make sense
because gravity will slow the object down at 9.8m/s2 no matter the velocity it gets, so to go further in a short time it must build up velocity quickly

Running the data you gave me through that equation gives:

F = g2t2/(gt2-2xm)

A) 80lbs(36kg) is 352.8N for 1m in .5s
F = (352.82 * .52) / [( 352.8*.52) - 2 * 36 * 1]
F = (124467.84 * .25 ) / (88.2 - 72)
F = 31116.96 / 16.2
F = 1920.8N

B) 80lbs(36kg) is 352.8N for .166M in .5s * note all that changes is x
F = (352.82 * .52) / [( 352.8*.52) - 2 * 36 * .166]
F = 31116.96 / 79.848
F = 389.7N

But here is why we say that over the same time the force is the same:

gT = Ft1
F = ma

A) 352.8N(g) * .5s(T) = 1920.8(F) * t1
176.4 = 1920.8t1
t1 = .092s

So the greater force (1920.8) was only applied for .092 of the .5s that is 18.4% of the time.

B) 352.8N * .5 = 389.7N * t1
176.4 = 389.7t1
t1 = .45s

So the lesser force was applied for .45s that is 90% of the time

Maybe this will help though

Sorry I don’t get why the two are different ??? ANYONE HERE PLEAE.

If it were over 60% of the ROM then A) it would probably not be a constant force at all, and B) It would have to be less than the high force, and more than the low force.

Hmm, sorry I still don’t get it, still don’t get why you say that over the same time the force is the same, As I don’t understand the two physics equation, don’t understand why or how you worked out two different equations for the same lifts, could you please try and tell me why they worked out different in more layman’s terms ??? Also still not sure why you say that the fast greater force (1920.8) was only applied for .092 of the .5s that is 18.4% of the time, Am I not Appling, {that’s if it’s a machine moving the weight} 1920.8 for just about the whole 60% of the rep ??? And if it’s the second rep, the forces must be bigger as of the higher peak forces.

Put this over at BB.com, would you like to join and post over there, or carry on like this if you like and I will post these over there, as there a lot very clever people over there.


Wayne

[waynelucky2]

No Wayne. We all agree that peak force is LOWER the faster the bar speed, well everyone but you.

NO NO NO, WHAT ARE YOU TWO ON ABOUT NOW, WE ALL AGRRED FROM THE FIRST SECOND OF THE DEBATE THE FAST REP CREATS MORE PECK FORCE.

pounds of force 400 @ 1000 rpm = 76 horsepower
pounds of force 350 @ 2000 rpm = 133 horsepower
pounds of force 300 @ 3000 rpm = 171 horsepower
pounds of force 250 @ 4000 rpm = 190 horsepower
pounds of force 200 @ 5000 rpm = 210 horsepower
Force is dropping, horsepower is increasing as rpm increases. The human body does the exact same thing.

What you some of you people on about ??? Think you got the debate the wrong way round, sorry that could be part on me, but don’t see how. I am not talking about the force/velocity curve for the last time, I am not on about that you can generate more force on a very heavy weight than a light one, I know and agree with all that, that has nothing to do with this debate.


We lift the same weight, let’s say 80% for 1m up, I lift the weight fast as I can, this will create the highest force possible, I mean just jump on a scales lift the weight slow, then lift it as fast as possible, you will then see the reading go up by a long way, as when you push up the is an opposite reaction. The Newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared. In dimensional analysis, F = ma, multiplying m (kg) by a (m/s2), SO THE MORE NEWTON FORCE YOU USE ON THE SAME OBJECT THAT’S THE SAME WEIGHT, YOU HAVE MORE ACCELERATION AND VELOCITY, AND MORE PEAK FORCE, AND MUCH MORE PEAK FORCE ON THE TRANSITION FROM NEGATIVE TO POSITIVE.


F = ma lets 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. SO IF IT MOVES FASTER LIKE 40M/S/S ITS 600 X 40 = 24000N.


A force however will produce acceleration, according to Newton's Second Law: F=ma, or force = mass x acceleration. Newton's Second law of motion states that the net force on an object is directly proportional to its rate of change of momentum. Thus F= m(v-u)/t where F = Force, M = mass, v = final speed, u = initial speed and t is the change in time. This is because a change in momentum is m(v-u). (v-u)/t is the rate of change of velocity which is acceleration. Thus, F=ma. Acceleration = Force / Mass.


Wayne

[waynelucky2]

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

D. wrote;
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.

D. YOU CAN NOT CHANGE THE QUESTION NOR THE ORDER OF THE QUESTION, WHY DO YOU DO THIS ??? HAVE YOU BEEN DOING THIS IN THIS DEBATE TOO SUIT THE ANSWER YOU WANT. PLEASE TELL ME WHY YOU CHANGED MY QUESTION AROUND, THAT JUST CANT BE DONE, AS I ASK THE QUESTION, YOU CAN NOT CHANGE IT


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 order 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 ???

Wayne

[waynelucky2]

Wayne, here is a simple experiment for you to try with your EMG machine you purchased. Using a weight that is equal to your 1RM for biceps curls, measure the amount of muscular activity that is produced during a contraction using your 1RM. Then using a weight that is equal 50% of your 1RM force, perform the biceps curl as quickly as possible.

Based on the force velocity curve, the highest force you will produce is when you are using slower contraction speeds. (actually eccentric contractions are capable of producing higher forces if you are using a weight > 1RM, but let's leave that out of it for now). If muscular activity is higher during the fast contraction speeds compared to the slow contraction speeds, then you can not use the EMG data as a proxy to measure muscular force, as this is impossible. That is unless you would like to re-write every Exercise/Muscle Physiology textbook as you prove every researcher in this area to be incorrect.

I am by no means an expert on EMG analysis, however I believe what you are actually measuring is muscular power and not force. If your measurements are higher during the experiment above you will have confirmed that you increase muscular "power" at higher velocities, however the actual force of the contractions would be reduced. Let us know how your data turns out.

As I said, this has nothing to do with the debate, and is not the debate.

Wayne

[waynelucky2]

Wayne,

I am not trying to prove anything in relation to the debate. The purpose of the above-mentioned study would be to test whether or not you can use your EMG data as a proxy to measure muscular force for comparison at two different rep speeds. If you can't, then the results of your studies are pointless.

Yes you can measure the above.

You are trying to test whether or not faster speeds produce more force. You can't move a large weight with a high velocity. The point was that the greatest force produced during a muscular contraction is going to occur at a low velocity and a high amount of weight. By performing a repetition at your 1RM you would ensure that you are producing the highest amount of muscular force you can, which you could then measure using your EMG machine.

Right.

The maximum amount of muscular power you can produce is going to occur using a weight that is ~50% of your 1RM speed at a high velocity. If you are moving the weight at a high velocity the amount of force you can produce is going to be less than the force achieved at your 1RM.

All depends on the amount of time you do both lifts.

What I am contending is that the high velocity/low weight (increased power) will result in an increase in muscular activation beyond what will be observed for your 1RM when you are moving a high amount of weight slowly. This is because despite the fact that your overall force produced will be less, due to a reduction in the total number of actin-myosin cross-bridges that are simultaneously binding, the rate of cross-bridge cycling would be increased. In other words the high power rep would result in a higher level of muscular activation despite the fact that a lower amount of force is being produced.

The high velocity/low weight, it would only be higher force if it was dome for a longer time span. But in my debate we are talking of the same time span, so if it took 4 seconds to do your 1RM, then you would have to do the other reps for 4 seconds, but then the weight is different, and that’s not the debate.

If this is what you observe it would indicate that the increase in muscular activation you are observing with your high rep speeds is not due to an increase of force being produced. Instead what you would be measuring is an increase in muscular power and the increase in activation is caused by an increase in velocity, not force, of the contraction.

Again, I am not trying to prove/disprove anything you have said. What I am trying to do is verify whether or not the EMG machine is measuring what you think it is measuring.

Yes, what you say was good, and I thank you.

Wayne

[waynelucky2]

OOO OOO I can answer this question. The answer is NO! The EMG is only measuring the CNS output. To measure force out put you have to 1. be lifting your 1 rep max
2. be standing on a force plate
3. have your bench on a force plate.
4. measure the distance the weight moved and time it. Then do the math.
5. ALL of this has been explained to Wayne millions of times add nauseum.

EMG does NOT measure the CNS.

Electromyography (EMG) is a technique for evaluating and recording the electrical activity produced by skeletal muscles. thus more activity more force, just like I told you years ago.

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

Wayne

[waynelucky2]

Let him run the experiment I suggested. If muscular activity is higher during the fast contractions when muscular force is lower then he will see why his analysis is flawed with his own eyes.

I can run the tests if you like, but as they are not in tyhe same time span, what is the point ???

Wayne

[waynelucky2]

His analysis is certainly flawed.Here's a study where you can find a raw and also a normalised EMG.
It is clearly shown in the normalised EMG(figure 6) that the average muscle activity with fast lifting with 81% of 1RM is exactly equal with the load.Just like if the load was lifted slow or if it was staticaly held.
http://www.scribd.com/doc/2528195/El...he-bench-press

D. I thought you was going to comment on this, as still dont understand what you think you see ???

Wayne

[waynelucky2]

Do you think one can reach maximum possible force production using 85% of 1RM?

At times on the transistion from negative to posative I would say you can reach higher force then your 1RM. As the 85 coming down will have more weight then your 1RM.

wAYNE

[waynelucky2]

I'm still looking for Louie's article on force plates. It's old so it might take a while to come up with it. In the mean time this one should piss Wayne off.

By: Louie Simmons

The Advanced Squat Cycle

WWhen designing a yearly model for the squat, many things must be considered, most importantly, the level of preparedness. This article pertains to the very highly trained squatter, i.e., those that squat 900 or more.
One must develop speed strength, which is the ability to accelerate with light to medium loads, creating explosive force.
Strength speed is a learned process to push maximum weights as fast as possible. This increases the powerful stretch reflex system and can be accomplished only by accelerating eccentrics and progressive concentrics. Strong bands must be used here. The bands will drive you down at a much faster rate than gravity alone, thus creating a great amount of kinetic energy, which is transferred into the muscle and connective tissues, causing a strong stretch reflex and providing an equally fast concentric phase.
The bands solve the problem of accommodating resistance. A load may be heavy at the bottom but light at the top. Thus, half the exercise maybe wasted. Fred Hatfield talked about compensatory acceleration. He was on the right track. By pushing as fast as possible against a light or heavy load, more force would be developed. However, if the weights are too light, the bar moves too fast and force is not developed.
Dr. Squat used a very fast eccentric phase that contributed to his very fast recovery in the squat. We have added two important elements.
(1) Bands greatly increase the stretch reflex through accelerated eccentrics. The bands also create a greater load at the top of the lift, thereby accommodating resistance, but more importantly (2) the time under tension is lengthened. This time is necessary for the development of maximum force. Max force is reached in 0.3 to 0.4 seconds. However it usually takes longer to complete a lift. Can the time to fully reach max force be increased with just the barbell? No. But with the addition of bands of adequate strength, the deceleration phase of the bar is greatly reduced on the ascent. You must push as hard as possible for a greater length of time.
Max force is, of course, highest at the start of the ascent, with starting strength being employed. But, by using a large load consisting mostly of band tension and a small amount of barbell weight, I believe the duration of maximum force and muscle tension can be lengthened, thus producing strength speed, the ability to push heavy resistance at a fast rate.

http://www.deepsquatter.com/strength/archives/ls31.htm

This does not piss me off, I agree with it, its not part of my debate ???

Wayne

[waynelucky2]

Here's another one Wayne wont like.
http://www.deepsquatter.com/strength/archives/ls31.htm

Not sure what you are on about, I toaly agree with this, and love things like this.

Wayne

[waynelucky2]

HEY WAYNE! GAME OVER! I WIN!

Siff, M.C. and Verkhoshansky Y.V. (1994) Supertraining. Special Strength
Training for Sporting Excellence. School of Mechanical Engineering.
University of the Witwatersrand. South Africa.

---------------40% IRM-------60% IRM-----------80% IRM
Peak Force (N) 1568----------- 1693.4------------ 1939.9
Peak Power (Watt) 6341.9------ 4828.7---------- 4493.1

http://w4.ub.uni-konstanz.de/cpa/art...File/2960/2806

I agree with this, you dont win anything, if you want more force lift the heavist weight as fast as possable, just like the above.

Wayne

[Orlando1234977]

Well an EMG signal will not necessarily reflect the total amount of force (or torque) a muscle can generate. The number of motor units recorded by electrodes will be less than the total number of motor units that are firing.
Electrodes can't pick up all motor units.

So, if a newly recruited motor unit is close to the electrode the relative increase in the EMG signal amp will be greater than the corresponding increase in force. Likewise I'm sure if a motor unit is too far from the elctrode the ampl will not change but the force will increase.

[waynelucky2]

I wanted to post something similar to this, but I wanted to see where morderstwo was going with this. If you are lifting a weight @ 85% of your 1RM you won't be able to move the weight any faster.

He was trying to tell you that you and All-pro have got the debate the wrong way around, sorry this could be part my fault whoknows.

Wayne

[waynelucky2]

There is no debate. School is in session and I'm instructing. Now if I could just get that Wayne kid to sit down and shut up............

I am agreeing with you. You do NOT know what the debate is, you have what I am trying to debate all wrong.

Wayne

[waynelucky2]

An 85% of 1RM can be a 6-8 rep max depending on the person, you can do at least one deliberately slow rep with this.

all pro - does 85% lifted as fast as possible reach force levels of a 1RM? As per Simmon's force plate tests?

Show the studys please, as yes I would say 85 could go higher then 1RM.

Wayne

[waynelucky2]

Concentrically as you raise the weight you won't be able to move the weight any faster/slower at this intensity.

You have the ability to move the weight as fast as you want, in this case you try to move it as fast as possible, and on the transition from negative to positive, if you do it as violent as possible, you create very high peck forces, how high will they be, we shall see on the test.

Wayne

[waynelucky2]

Yes. At 80% all fibers are recruited and respond on the very first rep. At 85% with the fastest contraction speed you can muster you're putting out 1 rep max force. But it wont last long. Maybe 5 seconds. After that ATP is depleted and the 2X fibers start dropping out because they're out of fuel. But you can do a hell of lot of damage in those 5 seconds!

All-pro, I totally agree. But I can move 80% for say 15 seconds.

And it’s the transition from negative to positive were the highest forces are, as cos a huge weight is being lowered in about .5 of a second, thus it will appear to weigh more than you 1RM.

Wayne