Impulse/force in pounds for the time frame

In summary, the conversation discusses the question of what the maximum impulse force would be on the components/parts of a machine as it lowers and then immediately stops a weight of 100 pounds at 2m/s for 1000mm before lifting it back up at the same speed. The conversation also addresses the force needed to lift the weight from rest and how it would increase every 10th of a second during the lift. The conversation also touches on the issue of force units being measured in both US and SI units. It is suggested that the time interval during the accelerating phase must be about 1 second to stay within the limits of the machine, and it is noted that the conversation involves a man lifting the weight, which can complicate the
  • #1
waynexk8
398
1
A question please in three parts, need the numbers for the first before I can asked the next.

A Machine lowers from rest, 100 pounds under control, at 2m/s, for 1000mm. Then immediately stops the weight, and lifts it back up at 2m/s.

At the transition from negative to positive, what impulse force in pounds, would be the maximum on the Machines components/parts, and for how long, until the normal acceleration forces that would be on the components/parts if it lifted the weight from rest.

Just in case I did not explain right. The force on the components/parts, lifting from rest would keep getting higher, let's go for every 10th of a second. Say from the lift at rest, the first 10th of a second would have at that vilocity ? 105 pounds of force needed to lift the weight at that vilocity, then the next 10th of a second it may need 110 pounds as the vilocity went up, and more and more force would be needed to keep the vilocity going up, until the decceleration. But what would be the impulse/force on/from the components/parts if the weight was lowered at the above vilocity, and for how long in 10ths or whatever in time would that higher force have to be until the normal acceleration forces of the lift at rest.

Thank you for your time and help.

Wayne
 
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  • #2
Sorry, etited the above slightely, as the machine only has a maximum force to use of a 125 pounds. However if the weight has been falling that fast, and is stopped fast, the force on the machine will be well over the 125 pounds, even thou when it does slow down stop and relift, the weight will keep going down for a few mm’s, the force then on it will be far higher than the 125 pounds.

Wayne
 
  • #3
Say Wayne,
1.) Bodies in motion cannot immediately (instantaneously) come to a stop; they must stop (decelerate) over a time interval. The impulse force depends on that time interval.
2.) why is your force measured in USA units and distances/velocities in SI units? You'll have to convert one to the other.
3.) If the time interval to stop is rather small, net impulse force will be very high; if the time is large, impulse force will be much less.
4.) If you are lifting a 100 pound body from rest with your machine rated at 125 pounds, to a constant speed of 2m/s (say 6.5 ft/s), then the max net force is 25 pounds, the acceleration is about 8 ft/s (per Newton 2), and the time interval during the accelerating phase must be about 1 second to stay within the limits of the machine.
 
  • #4
PhanthomJay said:
Say Wayne,
1.) Bodies in motion cannot immediately (instantaneously) come to a stop; they must stop (decelerate) over a time interval. The impulse force depends on that time interval.

Hi PhanthomJay, thank you for your time and help, and to anyone else who joins in the debate.

Yes agreed, there would have to be a deceleration from the eccentric motion to the concentric, and in this/that time frame there would be the highest force used by the machine, but more important, in this very short deceleration phase there would be also the highest force “on” the machines parts/components. I am unable to work out the extra force from the weight coming down as of the acceleration component, or am I able to work out that force that will be on the machine, or the time frame that this deceleration too acceleration would take. I am no physicist, however this is one of a very large and long debate, all I can give are the weights, distances and other times frames.

PhanthomJay said:
2.) why is your force measured in USA units and distances/velocities in SI units? You'll have to convert one to the other.

SORRY about that, I am from the UK, and a lot of us work in both, however it was wrong of me to put both down, imperial or metric is fine.

PhanthomJay said:
3.) If the time interval to stop is rather small, net impulse force will be very high; if the time is large, impulse force will be much less.

Yes very true, as the weight is being moved quite fast at .5 of a second concentric, and point .5 of a second eccentric, the net impulse force will be relatively very high. As we all know, the deceleration usually takes less force, however in this example it will take more, as the machine is try first to stop the fast downward motion of the weight, with its appeared extra weight as of the acceleration, then to acceleration it again, thus throughout the whole deceleration the machine is trying to accelerate the weight.

Some people are saying that the force on the machine will be as much as 300 pounds for a Milly second, I would not say that much. However, if it able to be worked out, it would be “very” interesting to find out it’s as ? 180 pounds for .2 of a second, or what the real numbers are.

PhanthomJay said:
4.) If you are lifting a 100 pound body from rest with your machine rated at 125 pounds, to a constant speed of 2m/s (say 6.5 ft/s), then the max net force is 25 pounds, the acceleration is about 8 ft/s (per Newton 2), and the time interval during the accelerating phase must be about 1 second to stay within the limits of the machine.

Ok, see what you are saying, could we go for the weight being lifted for just 500mm up, and 500mm down instead of 1m up and 1m down ? Not sure if I should be saying this, as it may complicate matters, but actually it’s a Man lifting the weight, and a Human muscle on average can lower 40% more than it can lift its one time maximum. However if you do not need to know this, and can would it out for just the machine that would be better. I said a machine, as I did not want to get involved that a Man could not lift at constant forces, meaning because of the biomechanical advantages and disadvantages thought the ROM {Range of Motion} its impossible to life like a machine.

Wayne
 
  • #5
It really depends on the nature of the movement. If you were to raise the weight halfway up (1/4 m) in 1/4 second at constant acceleration, the acceleration would be 8 m/s^2, and since Fnet=ma, where a is 8 and m is about 4.5 kilos, then Fnet = 36 N up, which means that since the weight acts down at about 450 N, you got to push up at 486 N, or about 110 pounds. For the remainder of the upward thrust, you've got to decelerate to 0, and the weight helps you to decelerate, so the net force is 36 N down, which means you only need to apply a force of 414 N up, or about 90 pounds. Same on the downstroke, 90 pounds to the halfway point and 110 pounds back to the beginning.

But it is more likely in weightlifting that the acceleration phase takes place in a much shorter time period and distance, say 0.1 second for a distance of say 1/8 m; then its constant controlled velocity from this point until you decelerate near the top. In which case your initial acceleration is 25 m/s^2, Fnet = 110 N, or your force is 450 + 110 = 560 N, or about 125 pounds. These are just rough numbers for talking purposes only, but the bottom line is that the initial liftoff and final dropoff requires the greatest force, and perhaps over 150 pounds if you accelerated quickly over an even shorter time.
 
  • #6
PhanthomJay said:
It really depends on the nature of the movement. If you were to raise the weight halfway up (1/4 m) in 1/4 second at constant acceleration, the acceleration would be 8 m/s^2, and since Fnet=ma, where a is 8 and m is about 4.5 kilos, then Fnet = 36 N up, which means that since the weight acts down at about 450 N, you got to push up at 486 N, or about 110 pounds. For the remainder of the upward thrust, you've got to decelerate to 0, and the weight helps you to decelerate, so the net force is 36 N down, which means you only need to apply a force of 414 N up, or about 90 pounds. Same on the downstroke, 90 pounds to the halfway point and 110 pounds back to the beginning.

Very interesting.

So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow;

110
110
110
110
110
110
90
70
50
30

Then zero for the transition from positive to negative.
Total, just for a reference number 900.

PhanthomJay said:
But it is more likely in weightlifting that the acceleration phase takes place in a much shorter time period and distance, say 0.1 second for a distance of say 1/8 m; then its constant controlled velocity from this point until you decelerate near the top. In which case your initial acceleration is 25 m/s^2, Fnet = 110 N, or your force is 450 + 110 = 560 N, or about 125 pounds. These are just rough numbers for talking purposes only, but the bottom line is that the initial liftoff and final dropoff requires the greatest force, and perhaps over 150 pounds if you accelerated quickly over an even shorter time.

Even more interesting.

So am I right in saying again, that on the above example, if we split the whole upward motion into 10 parts again, and I accelerated up for 60% But came off the eccentric to the concentric where the peak forces are, that it “may” look like this if we could plot my force somehow;

150
125
125
125
110
110
90
70
50
30

Then zero for the transition from positive to negative.
Total, just for a reference number 985.

As you “said” those numbers are just rough numbers for talking purposes only, but would the above be a rough estimation only ? As some people are saying that if you lift the same weight “very” slow for say 166mm to very fast for say 1000mm in the same time frame, that the same total or overall force will be used by or and on the muscles, we all know that you have to use more energy doing anything faster in the same time frame, and as you fail far far far faster when doing faster repetitions, my opinion is that’s it’s you have to use more total or overall force ? My EMG {a machine that reads the electrical activity in the muscles, the force or strength that the muscles are using} machine also gave me a higher readout force the faster repetitions.

First is the most important, as it shows the EMG average muscle activity, or muscle force/strength used, as we know, the higher the average, the higher the total/overall force/strength used.

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


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

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

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

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

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

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

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

MORE thank a big thank you for your time and help.

Wayne
 
  • #7
There is little value in guessing some possible values, because these guesses are made without knowledge of whether they are even likely or possible. It seems you are understanding the concepts just not the values of the numbers. The only cure for this, I believe, is with actual proven data, not with guesses as to what it could be if it were like this or that. Perhaps it would surprise you, or perhaps not, but guessing is no answer.
 
  • #8
waynexk8 said:
Very interesting.

So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow;

110
110
110
110
110
110
90
70
50
30

Then zero for the transition from positive to negative.
Total, just for a reference number 900.

Wayne

Your numbers are just figments of your imagination.
What part of PhanthomJay's answer didn't you understand?
He told you that for the first half you have to apply force equal with 110 pounds and for the second equal with 90 pounds.

Just to prevent you from posting for pages and pages until the thread will be closed...I'll ask three questions:

1)Do you understand that when you lift a weight the average force is always equal with the weight?
2)Do you understand that when you lift a weight the average acceleration is always zero?
3)Do you understand that when you lift a weight the applied force over the duration of the lifting(or else the "total force" as you call it) is always equal with gravity's impulse?
 
  • #9
waynexk8 said:
Very interesting.

So am I right in saying, that on the above example, if we split the whole upward motion into 10 parts, and I accelerated up for 60% that it “may” look like this if we could plot my force somehow;
Assuming you accelerate up half the distance to the top and reach a speed of 2m/s, then accelerate down from a speed of 2m/s to a momentary stop at the top, then continue accelerating down as you lower the weight to midway to 2m/s, and finally accelerating up as you lower to its start point, then the numbers read more like this:

UPSTROKE
110
110
110
110
110
[Strike]110[/Strike]90
90
[Strike]70[/Strike]90
[Strike]50[/Strike]90
[Strike]30[/Strike]90

DOWNSTROKE
90
90
90
90
90
110
110
110
110
110

Then zero for the transition from positive to negative.
When you are talking force, the force you apply is always positive (upward), whether on the up or down stroke. The force you apply is not zero until you fully release (let go of) the weight (or drop the weight if you're a lunk. :wink:)
Total, just for a reference number 900.
the total of these up forces divided by 10, or down forces divided by 10, or up and down forces divided by 20, gives you the average force during the cycle, or 100 pounds average force. This would be the force required if you lifted up and down very slowly at constant speed without accelerating. In the example acceleration case, the force you apply varies from 110 to 90 pounds.
Even more interesting.

So am I right in saying again, that on the above example, if we split the whole upward motion into 10 parts again, and I accelerated up for 60% But came off the eccentric to the concentric where the peak forces are, that it “may” look like this if we could plot my force somehow;
more like, for the upstroke in the example I gave,

[Strike]150[/Strike]125
[Strike]125[/Strike]100
[Strike]125[/Strike]100
[Strike]125[/Strike]100
[Strike]110[/Strike]100
[Strike]110[/Strike]100
[Strike]90[/Strike]100
[Strike]70[/Strike]100
[Strike]50[/Strike]100
[Strike]30[/Strike]75
Then zero for the transition from positive to negative.
Total, just for a reference number 985.
again average force = 100 pounds.
As you “said” those numbers are just rough numbers for talking purposes only, but would the above be a rough estimation only ?
Yes, rough, actual values depend on how you choose to do the lifting. Move it slowly at constant speed for the duration, and your applying more or less 100 pounds of force throughout
As some people are saying that if you lift the same weight “very” slow for say 166mm to very fast for say 1000mm in the same time frame, that the same total or overall force will be used by or and on the muscles, we all know that you have to use more energy doing anything faster in the same time frame, and as you fail far far far faster when doing faster repetitions, my opinion is that’s it’s you have to use more total or overall force ?
You have to use more force during the accelerating phase up, so if you start off slowly you are using 100 pounds of force, but then as you pick up speed the force you apply rises to 125 or 150 or more, until you decelerate at the top using much less force at the final transition from upward to downward thrust. But you are correct, for most of the duration you are applying a force much larger than 100 pounds, so the average force is also higher than 100 pounds. Better to move slowly in a controlled manner.
My EMG {a machine that reads the electrical activity in the muscles, the force or strength that the muscles are using} machine also gave me a higher readout force the faster repetitions.
yes, good machine!
First is the most important, as it shows the EMG average muscle activity, or muscle force/strength used, as we know, the higher the average, the higher the total/overall force/strength used.

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


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

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

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

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

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

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

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

MORE thank a big thank you for your time and help.

Wayne
I'm not sure of all these numbers, but it shows that the actual motion and muscular activity is complex. In particular, for item 1, it shows that more force is needed when moving 'fast'. I think I'll try this at the gym this weekend...but using 20 pound weights!
 
  • #10
douglis said:
Your numbers are just figments of your imagination.
What part of PhanthomJay's answer didn't you understand?
He told you that for the first half you have to apply force equal with 110 pounds and for the second equal with 90 pounds.

Just to prevent you from posting for pages and pages until the thread will be closed...I'll ask three questions:

1)Do you understand that when you lift a weight the average force is always equal with the weight?
2)Do you understand that when you lift a weight the average acceleration is always zero?
3)Do you understand that when you lift a weight the applied force over the duration of the lifting(or else the "total force" as you call it) is always equal with gravity's impulse?

By even asking those questions you don’t yet see that there are not just the ONE force at work, there are TWO, at least, thus you have to add then up, you seen to be only adding the one force up ? why. If you go over to BB.com, and please look at what I have named a few of my threads, then you might get it.

I have tried to tell you over and over, that {see what PhanthomJay states as well} when you lift a weight with more velocity or and acceleration, you have to use more force, and the more force you apply, the more the {not sure if I am saying this right, however I am try} weight will give an opposite reaction force, as to every action there is always an equal and opposite reaction, this opposite reaction, then puts more force on the pushing force, the muscles.

Or we could say; The g-force, associated with an object in its acceleration. So we have accelerations produced by gravity, and non-gravitational forces, proper accelerations. I explained this in my Clay example, in other simple words, the faster you push up on and object, the more it pushes down on you. From a site; G-forces, when multiplied by a mass upon which they act, are associated with a certain type of mechanical force in the correct sense of the term force, and this force produces compressive stress and tensile stress. Such forces result in the operational sensation of weight.

As there are two or more forces in action here, the force “from” the muscles, and the force
“on” the muscles from the weight being moved.

Wayne
 
  • #11
PhanthomJay said:
Assuming you accelerate up half the distance to the top and reach a speed of 2m/s, then accelerate down from a speed of 2m/s to a momentary stop at the top, then continue accelerating down as you lower the weight to midway to 2m/s, and finally accelerating up as you lower to its start point, then the numbers read more like this:

UPSTROKE
110
110
110
110
110
[Strike]110[/Strike]90
90
[Strike]70[/Strike]90
[Strike]50[/Strike]90
[Strike]30[/Strike]90

However PhanthomJay, what about the opposite reaction forces, the force “ON” the muscles from the weight itself being moved ? As if I lifted a 100 pouinds very slowely, there would be no extra forces on the muscles as I call it, the opposiot reaction forces, but there are all the time reactuion forces on the muscles from the weight itself when moving faster.

Now it’s gets even more interesting. As many times I have thought about the deceleration, and would have thought your numbers were more correct, as when I do lift weight in this very explosive style, I would say it’s far more like that. As I “try” to accelerate up all the way, but as its harder to accelerate the more you accelerate, it’s getting harder, thus I have to use less force, and my body is sub concisely telling me I have to decelerate and stop and do the transition to the opposite direction soon in a Milly second.



PhanthomJay said:
DOWNSTROKE
90
90
90
90
90
110
110
110
110
110

When you are talking force, the force you apply is always positive (upward), whether on the up or down stroke. The force you apply is not zero until you fully release (let go of) the weight (or drop the weight if you're a lunk. :wink:) the total of these up forces divided by 10, or down forces divided by 10, or up and down forces divided by 20, gives you the average force during the cycle, or 100 pounds average force. This would be the force required if you lifted up and down very slowly at constant speed without accelerating. In the example acceleration case, the force you apply varies from 110 to 90 pounds.more like, for the upstroke in the example I gave,

[Strike]150[/Strike]125
[Strike]125[/Strike]100
[Strike]125[/Strike]100
[Strike]125[/Strike]100
[Strike]110[/Strike]100
[Strike]110[/Strike]100
[Strike]90[/Strike]100
[Strike]70[/Strike]100
[Strike]50[/Strike]100
[Strike]30[/Strike]75again average force = 100 pounds. Yes, rough, actual values depend on how you choose to do the lifting. Move it slowly at constant speed for the duration, and your applying more or less 100 pounds of force throughout You have to use more force during the accelerating phase up, so if you start off slowly you are using 100 pounds of force, but then as you pick up speed the force you apply rises to 125 or 150 or more, until you decelerate at the top using much less force at the final transition from upward to downward thrust. But you are correct, for most of the duration you are applying a force much larger than 100 pounds, so the average force is also higher than 100 pounds.

Yes this is what I have always said, as these MUST be a higher average force, as you are developing more power, and using more energy, thus WHAT else could you be using more of if you’re traveling far far far more distance in the same time frame and using more energy, it MUST be more “FORCE” if not what ?

A question please, if you use more power, does that mean you use more overall or total force ? Total or overall forces in the meaning as, if you only had a 100 pounds of force, or strength in my case, before your force runs out, and it can and does in the human body quite fast if you are lifting 80%

PhanthomJay said:
Better to move slowly in a controlled manner.yes, good machine!I'm not sure of all these numbers, but it shows that the actual motion and muscular activity is complex. In particular, for item 1, it shows that more force is needed when moving 'fast'.

Yes in my and many other people opinions you have to use more force to lift faster, it’s like trying to lift 100% with just 80% it will just not go up, as you’re not using enough force.

PhanthomJay said:
I think I'll try this at the gym this weekend...but using 20 pound weights!

Sounds cool,

Could I ask another question, it’s sort of on impulse. As my force lifts the weight faster and for more distance in the same time frame, some people think that when you lift slow, that when my force is on the deceleration, its where the slow moving people make up or balance out the forces, however I think no, as if this was true, then they should then also make up the difference in distance moved, but they don’t, meaning the lower force don’t make up or balance out the high force when they are on lower forces as of the decelerations.

That I also think I may have proved that the slow lifting forces cannot make up or balance out the fast lifting forces with my clay theory. Put a piece of clay between your hand and the weight, lift a weight up and down one time for 3 seconds each way, then lift the weight up and six times at .5 of a second each way. The clay will be far more squashed on the faster reps, thus more force, or more important MORE tension is going onto the muscles ? As the clay will feel all the forces of the slow and fast reps.

Wayne
 
  • #12
waynexk8 said:
By even asking those questions you don’t yet see that there are not just the ONE force at work, there are TWO, at least, thus you have to add then up, you seen to be only adding the one force up ? why. If you go over to BB.com, and please look at what I have named a few of my threads, then you might get it.

Wayne

I don't know what forces you fantasize...but there're only two in this example.The force applied by the muscles and the weight.
When you accelerate the muscle force is greater than the weight and when you decelerate is less than the weight.The acceleration is always balanced by the deceleration(average acceleration=zero) since you start and end at rest and on average you apply equal force with the weight.
End of story.

The magnitude that you call "total force" or "effect of force over time" can be described by the impulse and is always equal with gravity's impulse over the duration of the lifting regardless the lifting speed.
 
  • #13
Zula110100100 said:
There is little value in guessing some possible values, because these guesses are made without knowledge of whether they are even likely or possible.

Yes I suppose your right in a way about the guessing, not sure it’s actually guessing mind you, and you and the other Physicist are very clever, and if you know all the variables Physics should be able to locate an answer. Also by chatting about the subject, we call all learn more.

Zula110100100 said:
It seems you are understanding the concepts just not the values of the numbers.

Not sure what you mean there, as the values are the different forces thought-out the ROM {Range Of Motion} of the concentric and eccentric. When I try a fast explosive rep I try to use as much force/strength as I can, however it’s basically impossible to keep this up for 100% of the time, as its too demanding.

Zula110100100 said:
The only cure for this, I believe, is with actual proven data, not with guesses as to what it could be if it were like this or that. Perhaps it would surprise you, or perhaps not, but guessing is no answer.

Yes you are quite right there. As I thought my EMG reading would be enough to show some people they were wrong in there numbers, as they were, or could not be adding in all the variables into the equations, D. does not seem to understand the opposite reaction on the pushing force from the weight being pushed. Then there was my Clay scenario, that also showed in practice that the fast would use more force, thus put more tension on the muscles.

However, I will try and get data from the Myotest, or the Apt from the IPod, or buy a force plate; also we have the device/machine you said you were hoping to make. Yes they might surprise me or perhaps not.

Thank you for your input and help.

Wayne
 
  • #14
Would like to add to this last question.

A question please, if you use more power, in the same time frame with the same weight, does that mean you use more overall or total force ? Total or overall forces in the meaning as, if you only had a 100 pounds of force, or strength in my case, before your force runs out, and it can and does in the human body quite fast if you are lifting 80% Does also in a machine when the energy runs out.

Wayne
 
  • #15
douglis said:
I don't know what forces you fantasize...but there're only two in this example.The force applied by the muscles and the weight.

I have not much time; however there are many more than two forces at work ? The muscles doing the pushing, the weight resisting back more and more the faster you go, gravity on both the force and the weight, air resistance {however we agreed to ignore this one, but was that the right thing to do ?}

Also you do not seem to believe EMG readings that are use all over the World.

douglis said:
When you accelerate the muscle force is greater than the weight and when you decelerate is less than the weight.The acceleration is always balanced by the deceleration(average acceleration=zero) since you start and end at rest and on average you apply equal force with the weight.
End of story.

What I am saying is, that when the fast high force go from the high forces of the accelerations, to the lower forces of the decelerations, that the constant medium forces of the slow, can NOT make up or balance out here those higher forces, and you have no way of knowing this. As of more work done, more energy used, more distanced travelled, and the high forces themselves. This has verified with my clay experiment and EMG.

douglis said:
The magnitude that you call "total force" or "effect of force over time" can be described by the impulse and is always equal with gravity's impulse over the duration of the lifting regardless the lifting speed.

The EMG and many things disagree with this.

Wayne
 
  • #16
waynexk8 said:
What I am saying is, that when the fast high force go from the high forces of the accelerations, to the lower forces of the decelerations, that the constant medium forces of the slow, can NOT make up or balance out here those higher forces, and you have no way of knowing this. As of more work done, more energy used, more distanced travelled, and the high forces themselves. This has verified with my clay experiment and EMG.

Wayne

Same average and equal with the weight force means by definition that the upper fluctuations of force are exactly balanced by the lower fluctuations.
If you don't have the intelligence to understand this...what's the point of the discussion?
 
  • #17
waynexk8 said:
However PhanthomJay, what about the opposite reaction forces, the force “ON” the muscles from the weight itself being moved ? As if I lifted a 100 pouinds very slowely, there would be no extra forces on the muscles as I call it, the opposiot reaction forces, but there are all the time reactuion forces on the muscles from the weight itself when moving faster.
If your muscles are pushing on the weight with a force of 100 pounds, the weight is pushing in the opposite direction on your muscles with a force of 100 pounds. If your muscles are accelerating the weight and pushing on it with a force of 150 pounds, the (normal force of) the weight is pushing in the opposite direction on your muscles with a force of 150 pounds. There is no addition of action-reaction forces here. This is Newton's 3rd law where equal and opposite force pairs act on different objects. Note that there are only 2 forces acting on the barbell; it's weight, acting down, which is always equal to 100 pounds, and the muscle force acting up, which is 100 pounds when it is not moving of moving at constant speed, or more than that when you are accelerating it (changing its speed to a higher value), or less than that when you are decelerating it (changing its speed to a lower value).
Now it’s gets even more interesting. As many times I have thought about the deceleration, and would have thought your numbers were more correct, as when I do lift weight in this very explosive style, I would say it’s far more like that. As I “try” to accelerate up all the way, but as its harder to accelerate the more you accelerate, it’s getting harder, thus I have to use less force, and my body is sub concisely telling me I have to decelerate and stop and do the transition to the opposite direction soon in a Milly second.
Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it. Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration, only a force of 100 pounds on your muscles.
Yes this is what I have always said, as these MUST be a higher average force, as you are developing more power, and using more energy, thus WHAT else could you be using more of if you’re traveling far far far more distance in the same time frame and using more energy, it MUST be more “FORCE” if not what ?
Sorry, i misspoke, the average force (for what it's worth) is always just 100 pounds. Even though for most of the time the applied force is 150 pounds while accelerating, there is very little force that you exert when decelerating (the weight does the workfor you), so the average force averages out, if you know what I mean (in eight equal intervals on the upstroke, force you apply might be 125, 150,150,150,150, 25, 25, 25; average = 800/8 = 100 pounds). You can be moving at constant speed slowly (say 1m/s) or at constant speed quickly (say 3m/s),and the force you exert is still the same 100 pounds for both cases. Your muscles are tiring more easily in the high powered fast case, because internal energy is being lost from the complex muscular activity..if your muscles were robotic, it wouldn't matter..or a machine getting powered electrically...but humans are not machines, they need chemical/biological/food energy to operate. Don't confuse the two.
Could I ask another question, it’s sort of on impulse. As my force lifts the weight faster and for more distance in the same time frame, some people think that when you lift slow, that when my force is on the deceleration, its where the slow moving people make up or balance out the forces, however I think no, as if this was true, then they should then also make up the difference in distance moved, but they don’t, meaning the lower force don’t make up or balance out the high force when they are on lower forces as of the decelerations.
You are not correct. However, the power required is greater when you are moving faster with the same applied force.

That I also think I may have proved that the slow lifting forces cannot make up or balance out the fast lifting forces with my clay theory. Put a piece of clay between your hand and the weight, lift a weight up and down one time for 3 seconds each way, then lift the weight up and six times at .5 of a second each way. The clay will be far more squashed on the faster reps, thus more force, or more important MORE tension is going onto the muscles ? As the clay will feel all the forces of the slow and fast reps.

Wayne

More force goes into the muscles only during the accelerating phase. For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound. The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused. :confused:
 
  • #18
PhanthomJay said:
If your muscles are pushing on the weight with a force of 100 pounds, the weight is pushing in the opposite direction on your muscles with a force of 100 pounds. If your muscles are accelerating the weight and pushing on it with a force of 150 pounds, the (normal force of) the weight is pushing in the opposite direction on your muscles with a force of 150 pounds. There is no addition of action-reaction forces here.

Sorry I meant the normal action-reaction forces.

PhanthomJay said:
This is Newton's 3rd law where equal and opposite force pairs act on different objects.

Yes.

PhanthomJay said:
Note that there are only 2 forces acting on the barbell; it's weight, acting down, which is always equal to 100 pounds, and the muscle force acting up, which is 100 pounds when it is not moving of moving at constant speed, or more than that when you are accelerating it (changing its speed to a higher value), or less than that when you are decelerating it (changing its speed to a lower value).

So the weight of the weight acting down is what I call the force of gravity ? And we are not counting the force of air resistance as it’s to small ? Ok.

PhanthomJay said:
Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it.

Yes understand and totally agree.

PhanthomJay said:
Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration, only a force of 100 pounds on your muscles.

Right.

I still have a like problem with this deceleration. There was a “good” study that a person used 80% and the deceleration was for 48% however the full concentric took 1.5 seconds to go about 22 inch. My repetition speed is 3 times more than that.

Here is my problem and questions, or more the way I lift or see both lifts that we are debating.

1,
If I was to lift a weight that was 80% {that’s 80% of my 1 repetition maximum} for a ROM {range of motion} of 20 inch, but just push up as hard as I could and not try to slow down, like if I was shooting the putt, or throwing a stone, thus then would have to be using more force than if I needed to decelerate ? We are doing a little experiment on if I decelerate a lot, a bit or not at all, maybe Zula will fill you in on this.

2,
That’s the way I envisage I do lift these weights on my fast repetitions, {will do some new videos if you like ?} I explode up and keep on pushing and pushing; I don’t or did not seem at all I was or had to decelerate, and until me and D. Found that study, we was trying to work this out with me decelerating for only 10 to 20% of the ROM. So could not I fully accelerate up, and then immediately reverse ? As that’s what it seems I do.

As there is no offloading, as we worked out that if you accelerate 80% up, can’t remember the distance, but if you immediately stopped the weight would move 3 inch, however it does not, it does if a machine pushes it, but not the Human body, as of the biomechanical advantages and disadvantages thought the ROM, meaning the body just can’t produce full constant force thought-out the range of motion.

Will get back to the rest, been very busy and sleepy.

Big thank you for your and the or members help and time.

Wayne
 
  • #19
waynexk8 said:
That’s the way I envisage I do lift these weights on my fast repetitions, {will do some new videos if you like ?} I explode up and keep on pushing and pushing; I don’t or did not seem at all I was or had to decelerate, and until me and D. Found that study, we was trying to work this out with me decelerating for only 10 to 20% of the ROM. So could not I fully accelerate up, and then immediately reverse ? As that’s what it seems I do.
Wayne

The length of the acceleration phase is of little importance.As long as the terminal velocity is zero the acceleration is always balanced by the deceleration.Or...with your own words...the forces "make up".

However experimantally was found the acceleration phase lasts for about the half distance with the 80% of your maximum weight and much less with less weight when you try to lift as fast as possible.
In general,the faster the lifting the less the acceleration phase and the grerater the deceleration phase.
BTW at the study you're referring the lifting lasts ~1sec which is perfectly normal with such weight.Check it again.
 
  • #20
douglis said:
The length of the acceleration phase is of little importance.

It is of all the importance.

I just did the test Zula told of me. I attached a weight to the bar, and curled 80% up and down as fast as I could, the string and weight stayed basically the same, meaning high tension and little deceleration. I then accelerated up fast again, but stopped very suddenly, the string went slack and weight with was half a pound jumped up.

douglis said:
As long as the terminal velocity is zero the acceleration is always balanced by the deceleration.Or...with your own words...the forces "make up".

Do you mean as if with a lighter weight, and I lifted it so fast that is would leave my hand, are you calling the leaving of my hand, terminal velocity is zero ?

On paper it may look balanced/made up, however how do you know it is ? As you will agree that a high force for a short time, can produced the force impulse on a low force for a longer time, so why do you think the forces balance out make up ?

If I was to press up into a very thick piece of clay at a high force velocity with an 80% dumb bell in my hand, my hand could go in 2 inch, then there will be an opposite reaction which it called the tension on the muscles, I do this 6 times in 6 seconds and make a 12 inch hole in the clay, very high tensions on the muscles. I then take the same dumb bell and press is up into a very thick piece of clay at a low force velocity with an 80% dumb bell in my hand, I do this one time, the dumbbell would not even go the first 2 inch, but still lots of tension on the muscles.

I think you coined the makeup.

douglis said:
However experimantally was found the acceleration phase lasts for about the half distance with the 80% of your maximum weight and much less with less weight when you try to lift as fast as possible.
In general,the faster the lifting the less the acceleration phase and the grerater the deceleration phase.

Not the case. As I could accelerate the weight nearly right up if I wanted, as if I was throwing the weight, and if I did this with a light weight, there would be no deceleration until I left go, this is what I do with a weight that is more heavy, about 80%

douglis said:
BTW at the study you're referring the lifting lasts ~1sec which is perfectly normal with such weight.Check it again.

Sorry it “is” 1.5 seconds lifting 81%
There is something wrong with this study, or they were not testing for maximum force bench press, as I or most could lift 80% 20 inch in .5 seconds, so the question is why did they take 1.5 seconds, but as they did take 1.5 seconds, I can see why the large deceleration.

Page 455.

http://www.scribd.com/doc/2528195/Elliot-et-al-A-biomechanical-analysis-of-the-sticking-during-the-bench-press [Broken]

Wayne
 
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  • #21
waynexk8 said:
Do you mean as if with a lighter weight, and I lifted it so fast that is would leave my hand, are you calling the leaving of my hand, terminal velocity is zero ?

Terminal velocity is the velocity at the end of ROM which is zero in the case of lifting(not throwing)

On paper it may look balanced/made up, however how do you know it is ? As you will agree that a high force for a short time, can produced the force impulse on a low force for a longer time, so why do you think the forces balance out make up ?

The change in momentum is zero so the average net force is zero...hence the forces "make up".Nothing more to talk about...I don't know what you can't understand.

If you're going to doubt it without any argument you'll just be ignored.

If I was to press up into a very thick piece of clay at a high force velocity with an 80% dumb bell in my hand, my hand could go in 2 inch, then there will be an opposite reaction which it called the tension on the muscles, I do this 6 times in 6 seconds and make a 12 inch hole in the clay, very high tensions on the muscles. I then take the same dumb bell and press is up into a very thick piece of clay at a low force velocity with an 80% dumb bell in my hand, I do this one time, the dumbbell would not even go the first 2 inch, but still lots of tension on the muscles.

I also don't understand why you're repeating that irrelevant clay example.It just shows the higher peak force that causes the deformation of clay...nothing more.

Not the case. As I could accelerate the weight nearly right up if I wanted, as if I was throwing the weight, and if I did this with a light weight, there would be no deceleration until I left go, this is what I do with a weight that is more heavy, about 80%

The length of the deceleration phase is defined by the laws of physics...not by you.


Sorry it “is” 1.5 seconds lifting 81%
There is something wrong with this study, or they were not testing for maximum force bench press, as I or most could lift 80% 20 inch in .5 seconds, so the question is why did they take 1.5 seconds, but as they did take 1.5 seconds, I can see why the large deceleration.

Page 455.

http://www.scribd.com/doc/2528195/Elliot-et-al-A-biomechanical-analysis-of-the-sticking-during-the-bench-press [Broken]

Wayne

Sorry it "is" 1 sec lifting.Check page 459 figure 6a...although it's of little impotance.
 
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  • #22
douglis said:
Terminal velocity is the velocity at the end of ROM which is zero in the case of lifting(not throwing)

Please explain how “my” lifting it different to throwing ?

Yes, but please let us know “your” actual meaning of terminal velocity here, also why you include it, if what you think it is, is zero. As in weightlifting and physics it’s different.

Lifting and throwing a weight up in say the bench press, in my form of explosive lifting with 80% is basically the same. As I try to get the weight from a to b as fast as possible, that the same as if I was trying to throw the weight on the bench press.

douglis said:
The change in momentum is zero so the average net force is zero...hence the forces "make up".Nothing more to talk about...I don't know what you can't understand.

The change in concentric momentum/movement is from zero velocity, to acceleration, {80% moved 20 inch in .5 of a second} to a deceleration, then zero for the transition, then reverse down for the eccentric.

We are not looking for the net force the way you are saying it, and never have been, why do you mention this ? You are saying if both forces are equal they cancel each other out the net force is zero. The force we are talking about are the forces from the muscle for holding the weight, for moving the weight, for accelerating the weight and for decelerating the weight, then add in the opposite reaction force, as in the more force you produce the more the clay would get squashed, or the more tension on the muscles. Like my EMG reading proved.

douglis said:
If you're going to doubt it without any argument you'll just be ignored.

Doubt what, I don’t understand what you think I will not doubt ? What would be nice if you understood my far higher reading in a practical World experiment, my EMG reading ?

douglis said:
I also don't understand why you're repeating that irrelevant clay example.It just shows the higher peak force that causes the deformation of clay...nothing more.

You are quite wrong there, the deformation of clay in both lifts, will show all forces, or if you would like to state how/why you think that the force of you moving very slowly will not be shown in the clay ? As the law states that for every reaction there is an opposite reaction. Hold the weight with the clay, then move the weight very slow up and down with the clay, you will see the clay is deformed more when you move the weight. It has to show all, please explain why you think it will not ?

douglis said:
The length of the deceleration phase is defined by the laws of physics...not by you.

I am, and we are all physics. I can, as I did earlier, choose the length of the acceleration and deceleration. Simple test, punch with as much force and velocity upwards as you can, and slow and stop as close to arms length as you can, now with as much force and velocity upwards as you can, but try and stop half way up, you have just changed both acceleration and deceleration.

The strength of the person will work out the acceleration and deceleration. As I said, I lift explosively and it’s basically the same as if I was throwing explosively, I would try to do both as fast as possible.

douglis said:
Sorry it "is" 1 sec lifting.Check page 459 figure 6a...although it's of little impotance.

Sorry, on the graph it clearly sates 1.5, also on your page, where are you looking ?

It’s of the greatest impotence, why do you think not ? If I move the weight 1m in .5/.5 of a second 6 times to your 1m in 3/3 seconds, I have created 6 large force accelerations and peak forces. I will tell you why more, but no time now.

Why do you think your lower forces can make up or balance out my higher force ? You still have not explained this, how can say 80 force make up or balance out a 100 force, if both forces are used for close to the same time frame ?

Did you try the weight on a barbell with string, to show very little deceleration ?

Wayne
 
  • #23
waynexk8 said:
Please explain how “my” lifting it different to throwing ?

Yes, but please let us know “your” actual meaning of terminal velocity here, also why you include it, if what you think it is, is zero. As in weightlifting and physics it’s different.

This can be understood even by common sense which is something you're not familiar with.
You throw an object when it's leaving your hands with velocity.You lift an object when it doesn't leave your hands so by definition at the end of lifting the velocity is zero and that's called terminal velocity.

The change in concentric momentum/movement is from zero velocity, to acceleration, {80% moved 20 inch in .5 of a second} to a deceleration, then zero for the transition, then reverse down for the eccentric.

Exactly!So you go from zero velocity to zero velocity...the change in momentum is zero.So the net impulse is zero and the net force zero too.
The forces "make up"...the average applied force is equal with the weight regardless the length of the acceleration/deceleration phases and the distance covered.

The ONLY forces available are the muscle force and the weight.There're no additional action-reaction forces to add as you fantasize.

Read carefully Phantomjay's last post.He's explaining perfectly that and your irrelevant clay example.

I'm out of this thread cause it's obvious you have some kind of obsession and I'm wasting my time.I hope Phantomjay will have the patience to explain it further to you.

Sorry, on the graph it clearly sates 1.5, also on your page, where are you looking ?
Wayne

page 459,figure 6a.
PM me for further explanation cause that's irrelevant with the thread.
 
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  • #24
douglis said:
This can be understood even by common sense which is something you're not familiar with.
You throw an object when it's leaving your hands with velocity.You lift an object when it doesn't leave your hands so by definition at the end of lifting the velocity is zero and that's called terminal velocity.

Yes right.
As even with 80% moving it for 30 inch at 2m/s the bar will not leave your hands, so yes terminal velocity is zero.

douglis said:
Exactly!So you go from zero velocity to zero velocity

Here is where I cannot understand why you think this. As in our case, when you move a weight from a to b, you “must” start at zero, accelerate/decelerate go back to zero, for the transition, the accelerate/decelerate and so on. You can’t go from zero to zero. You “are” missing out the “actual” momentum/movement that we are debating/concerned about, the forces we are talking about are in-between both zero movements, and you talk about the zeros ?

douglis said:
...the change in momentum is zero.So the net impulse is zero and the net force zero too.

The change in momentum/movement can “not” be zero, as there “is” momentum/movement in-between both zeros, or from a to b.

What you say net force, as I said last time, you are on about the force moving the weight and the weight pushing back, the opposite reaction forces. We are “not” concerned with both forces cancelling each other out, we are concerned with the forces from the pushing force the muscles, which causes tension on the muscles, and the opposite reaction forces from the weight. So in our case they are both creating force from and on the muscles, thus we need to add these forces up, as of the tension on the muscles. You seem to be saying they cancel each other out, thus not force was applied from the muscles or from the weight and not tension on the muscles.

You hold a weight half way up, you then move the weight up 20 inch in the time you held it, in say 20 seconds, basically the same forces from your muscles, thus tension on your muscles. However when I am accelerating the weight up and down 20 times in the same time frame, I have {if we say for at least the acceleration is for 60% of the concentric} accelerated the weight 240 inch for just the concentric, how on Earth you can think you have used the same total or overall force in doing this is beyond me. As I asked before, if you claim to use the same total or overall forces, and they make up or balance out in the end, why does not the distance make up or balance out ? Why do I use far far far more energy, why do I do more work, and work is the product of a force times the distance through which it acts, and it is called the work of the force. Also why do my EMG reading produce higher reading for the faster repetitions done in the same time frame.

douglis said:
The forces "make up"...the average applied force is equal with the weight regardless the length of the acceleration/deceleration phases and the distance covered.

You say this without any proof, evidence or an explanation, as we are on the physics forum, I think you need to at least try and prove this with proof, evidence with an explanation. I also repeat, why is the EMG higher ?

douglis said:
The ONLY forces available are the muscle force and the weight.There're no additional action-reaction forces to add as you fantasize.

Maybe you misunderstood, as the weight with the muscle force pushing up it, is the action-reaction forces !

douglis said:
Read carefully Phantomjay's last post.He's explaining perfectly that and your irrelevant clay example.

Will do that right now……….you mean this, I quote;
Phantomjay wrote;More force goes into the muscles only during the accelerating phase.
True.
Phantomjay wrote;For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound.

Yes right, more squash, or should I say more compression on the clay for the fast. No, the clay does not rebound; it does not whatever you do to it, not sure why you say this.

Phantomjay wrote;The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused.

Yes this what it seems to say on paper, “however” I have contended from the beginning of this debate on other forums, “how” can you know, that as we all agree the fast repetition is using more and a LOT more force on the first say 60% {remember the fast does 6 repetitions in the same time frame as the slow moving the weight 6 times more the distance, using far more energy} than the slow, then when the fast is on its deceleration, it’s using less force than the slow, {and here is my main point} “how” do you know/think or work out, that the slow force makes up or balances out the forces here ? as the slow force is “never” as high as the higher force of the fast acceleration forces, in my opinion the whole forces are not linear and cannot make up or be balanced out, as of my EMG reading, the more work/distance moved and more energy used.

It’s like a impulse; a small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. So I say the small force would need far far far more longer time to make up or balance out the large force made by the faster accelerations of the faster repetitions.

That’s why you “always” fail far far far faster doing faster repetitions to slow, as the high acceleration forces, put far far far more tension on the muscles than the slow in the long run.

douglis said:
I'm out of this thread cause it's obvious you have some kind of obsession and I'm wasting my time.I hope Phantomjay will have the patience to explain it further to you.

I think/know we both have an obsession with this debate. I like to learn, and have learned very much. I would have thought that as my other things as well as my EMG reading prove you very wrong, and “you” would like to know “WHY and HOW” they show you are wrong, do not you like the truth ? Zula is constructing a machine to find out the forces that both velocities use.


douglis said:
page 459,figure 6a.
PM me for further explanation cause that's irrelevant with the thread.

It’s says nothing about the repetition taking 1 second on 6a, but does state very clearly on the page I gave in a graph that the whole concentric took 1.5 seconds, it’s as clear as day.

If I move a weight faster in the same time frame, and then I move the weight far far far faster in the same time frame, the acceleration will be higher “and” longer making for lager force impulse that in my opinion cannot be made up by the slower.


Yet again you miss out very many of the main issues of this debate, what about the weight and string test on the weight for deceleration ?


Wayne
 
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  • #25
Wayne...let's start it all over again.
Do you understand that the average force that's applied by the muscles is always equal with the weight regardless the repetition speed?Yes or no?
 
  • #26
Phantomjay wrote;More force goes into the muscles only during the accelerating phase.[/i]
True.
Phantomjay wrote;For the slow case, you might have 110 pounds at the beginning during the accelerating phase...for the fast case, it might be 150 pounds during the accelerating phase, so more squooshh...and clay doesn't rebound.

Yes right, more squash, or should I say more compression on the clay for the fast. No, the clay does not rebound; it does not whatever you do to it, not sure why you say this.

Phantomjay wrote;The averrage force is still 100 pounds for both cases. But yes, at some point during the fast case, your muscles are receiving (and giving) more force during the accelerating phase, because it requires more force to accelearte to the higher speed in the same distance. Now I bet your really confused.

Yes this what it seems to say on paper, “however” I have contended from the beginning of this debate on other forums, “how” can you know, that as we all agree the fast repetition is using more and a LOT more force on the first say 60% {remember the fast does 6 repetitions in the same time frame as the slow moving the weight 6 times more the distance, using far more energy} than the slow, then when the fast is on its deceleration, it’s using less force than the slow, {and here is my main point} “how” do you know/think or work out, that the slow force makes up or balances out the forces here ? as the slow force is “never” as high as the higher force of the fast acceleration forces, in my opinion the whole forces are not linear and cannot make up or be balanced out, as of my EMG reading, the more work/distance moved and more energy used.

It’s like a impulse; a small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. So I say the small force would need far far far more longer time to make up or balance out the large force made by the faster accelerations of the faster repetitions.

That’s why you “always” fail far far far faster doing faster repetitions to slow, as the high acceleration forces, put far far far more tension on the muscles than the slow in the long run.

PhanthomJay said:
Your body is not telling you to decelerate it; the laws of physics demand it if you are to slow it.

Yes very true.

PhanthomJay said:
Your body tells you that your muscles are getting strained with too much force, and it will tell you that even if you were to hold the weight still and motionless for a long period of time, without speed , acceleration, or deceleration,

Again true.

PhanthomJay said:
only a force of 100 pounds on your muscles.Sorry, i misspoke, the average force (for what it's worth) is always just 100 pounds.

I do not think average means much.


The average net force is equal to the weight, regardless to duration and time ? So if I move a weight for an hour, for a 1000m and then for 1 sec, 1mm. And if the net force is the same, the moving of the weight for an 1 hour will put a very lot of force out from and onto the muscles, even thou the average is the same.


PhanthomJay said:
Even though for most of the time the applied force is 150 pounds while accelerating, there is very little force that you exert when decelerating (the weight does the workfor you),

I contend this, the weight does “no” work for me, I am moving the weight, nothing else is, what else could be moving the weight, I do not understand what you mean or think ? If I stopped at anytime in the lift, the weight “would” stop moving, it would “not” travel or move on its own, how could it ? Someone worked out below, that if you lift 200 pounds 15 inch at 1m/s that the weight itself would {but it does not, please read on as too why} move 20% or 3 inch if you immediately stopped the pushing force at the top. However this does “not” happen, because of the biomechanical advantages and disadvantages thought the ROM, {Range Of Motion} so this means that there in “not” very little force, but a “very” lot of force being used thought the ROM. Even if a machine was pushing it, the weight cannot move on its own ?

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.


PhanthomJay said:
so the average force averages out, if you know what I mean (in eight equal intervals on the upstroke, force you apply might be 125, 150,150,150,150, 25, 25, 25; average = 800/8 = 100 pounds). You can be moving at constant speed slowly (say 1m/s) or at constant speed quickly (say 3m/s),and the force you exert is still the same 100 pounds for both cases.

As I said above, I can’t see how the weight would move on its own ? it would be more like 125, 150, 150, 150, 150, 125, 125, 80, 20, zero movement for the transition.

“If” the forces averaged out, “why/how” is the EMG readings higher, why/how is power higher why/how do you use more energy, why/how does the fast move the weight 6 times further, why/how do you fail on the fast far faster ?

Play from 5.00

Note that the differences.
Slow,
Power 649
Force 546
Velocity 161.

Fast,
Power 829
Force 579
Velocity 192.
Imagine if the person had done 6 repetitions fast and one slow, the force on the fast would/could be far far far higher, as on the video, the speeds were basically quite close.

PhanthomJay said:
Your muscles are tiring more easily in the high powered fast case, because internal energy is being lost from the complex muscular activity..

Yes, that’s because I have to be using more force ? If not how/why else would I need to use more energy ?

PhanthomJay said:
if your muscles were robotic, it wouldn't matter..or a machine getting powered electrically...but humans are not machines, they need chemical/biological/food energy to operate.

The machine too would need more energy to use more force, its impossible otherwise to use more force without more energy, is it not ?

PhanthomJay said:
Don't confuse the two.You are not correct.

So why is more energy used when force goes up then, as its “not” a coincident that the “exact” moment you use more force you use more energy, is it ?

PhanthomJay said:
However, the power required is greater when you are moving faster with the same applied force.

You would have to use more force to move faster. Where Fnet is the total external force.

Wayne
 
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  • #27
douglis said:
Wayne...let's start it all over again.
Do you understand that the average force that's applied by the muscles is always equal with the weight regardless the repetition speed?Yes or no?

I understand to well what you are thinking/saying. Your saying that, {these numbers are just for me showing you this} if I moved a weight 1000mm to you moving it 166mm in the same time frame, that all the higher force I use acceleration the weight for say the first 60% you make up or balance out this force when I am on the deceleration. Or if I use 100N 100N 100N 100N 100N 100N for the first 60% then use 50N 50N 50N 50N = 800N. You use 80N 80N 80N 80N 80N 80N 80N 80N 80N 80N = 800N.

1,
This seems to work on paper, but not in practice.

2,

However, the following states the force do not balance out or make up.

3,
However not on EMG reading, which test the electrical activity of muscles.

4,
Not on distance as the weight is moved six times further in the same time frame.

5,
I accelerate 600mm to you moving at a constant velocity for only 166mm.

6,
You always fail far faster in the faster reps.

7,
Why do you use more energy the moment you use more force ? If as you claim your forces make up or balance out, should not the energies ?


8,
I know we went over this once, but there is still things bugging me, could we go over here on the physics forum, as other members can comment.

As you know, if I immediately stopped at any part of the acceleration, the weight would stop immediately, and would not carry on at all, so “HOW” do I as you claim I use less force than the weight ? As if you less force, does that “NOT” mean that I would be then reversing the direction, not still going forward but backwards ? As deceleration is negative acceleration or decreasing velocity over time.

Why can not the fast repetition, coming off the eccentric portion, hitting the peak forces on the transition from negative to positive. Could it not be like this, maximum force 100 pounds, weight used 80%

130, 100, 100, 100, 100, 100, 95, 90, 85, 80, zero, for the transition from positive to negative.

These next numbers mean nothing, and are just my example.
.5m/s, 2m/s, 2m/s, 2m/s, 2m/s, 2 m/s, 1.6m/s, 1.2m/s, .8m/s, .6m/s, zero, for the transition from positive to negative.


I can’t seem to find any studies of moving 80% at 2m/s for say 20inch, or there about, will look more under weightlifting.

9,
What I have said all along, is my higher force impulses, cannot be made up or balanced out by your lower force impulses, and you have no way to prove it, however I do as in all the examples above.

10,
Just thought of something, eccentric muscular action for the deceleration, this will take force from the muscles, muscles are able to withstand greater eccentric than concentric loading due to increased fast-twitch muscle fibre and motor unit recruitment. Will have to think about this one.

Wayne
 
  • #28
douglis wrote;
I'm not saying that your EMG is wrong.What I say is what I know from physics.The RMS {Root Mean Square}is NOT the technical average.It's the ~70% of the peak values.When we say the average force is always equal with the weight we're talking about the technical average which is shown only from the EMG normalized force-tension graph.


Quite wrong I think, please read this. Seems like it’s the average.


http://www.raeng.org.uk/education/diploma/maths/pdf/exemplars_engineering/8_RMS.pdf



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


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

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

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

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

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

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

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

Wayne
 
  • #29
Hi D.


Hope we both are both are seeing understanding force/strength in the same way ? As I do, as do explain things to you, but at times you never acknowledge that you understand or agree. We all agree that when using 80% for the fast at .5/.5 for 15 repetitions = 15 seconds, and the slow at 3/3, that you will be able to go for longer using the 3/3, as you fail faster doing the .5/.5 yes ?


So this means I “HAVE” used up my force and faster, yes ? Or as I call it strength faster, so if we are Clones, and have a 1000N of force to use up, at using 80% as explosively/dynamically, I “have” used up “all” my force, “but” as you are “STILL” exercising, repping the weight, you “MUST” still have force/strength left, right ?


doulas wrote;
Hi Wayne...I believe you got all the answers you wanted on the physics forum.It's clearly stated by the physicists that you are not correct.I believe that you just don't want to accept the truth and that's why nobody(including me) is bothered to answer.


I have not had many answer yet, waiting for the next reply, but as I know these people are busy.


Zula is constructing a force machine.


If I fail faster, as told you above, if I also use more energy, more the weight more distance, I fail understand how or why you think you are right ?



Please read chapter 4, just the first 2 pages, it says like I have been trying to say all along, these forces that I talk about cannot be easily equated with physics.

One more,
If I try and lift as much as I can fast, to you lifting slow, I will lift more, thus I create more force and for longer.

http://www.findphysio.com/E-books/Biomechanical%20Evaluation%20of%20Movement%20in%20Sport%20and%20Exercise.pdf [Broken]


doulas wrote;
Just one thing is left to clarify.

Check carefully the force-time graph(figure 6a) at page 459.It's clearly shown that at the ascent phase lasts from 1.2sec to 2.2sec.The duration of the ascent is ~1sec which is perfectly normal for the 6-7RM(81% of 1RM)http://www.exrx.net/Calculators/OneRepMax.html.

Also you can clearly see that the acceleration phase is only for the first half(maybe even less) of the ascent.It's also shown perfectly how the forces "make up".For the first half the force is above the bar weight and for the second is below the bar weight.


Not sure what you want me to look at here ??


doulas wrote;
This EMG graph is the perfect answer for everything you ask.Nothing more to add.


http://www.scribd.com/doc/2528195/Elliot-et-al-A-biomechanical-analysis-of-the-sticking-during-the-bench-press [Broken]

I “have” told you and all that look at this can quite easy see, on page 455, the this bench press took 1.5 seconds this kind on very slow bench press world have a acceleration and deceleration phase like that, as of the sticking point in the bench press. As if the press was done that slow. My bench press, as I keep telling you, and you don’t want to listen, lasts as you know, for three times less than this, as mine goes three times faster, its move the whole concentric repetition in .5 of a second.

Wayne
 
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  • #30
This looks like the same question, yet again Wayne.
Will you keep asking it until you get the answer you want? :wink:
 
  • #31
sophiecentaur said:
This looks like the same question, yet again Wayne.
Will you keep asking it until you get the answer you want? :wink:


Hi sophiecentaur,

I am honesty not like that, as I recently bought an EMG machine, {the results are on this thread} it reads the electrical muscle activity, the more average or total force/strength you use it reads out, and I did many tests on it, and everyone came out stating that you use more force when moving faster.

Also we all agreed that you use more energy when moving a weight faster, so as soon as the force and average force goes up, the energy does also, this can’t just be a coincidence ? Then you move the weight 1000mm moving fast, and only a 166mm moving slow, that means “if” I accelerated for only 60% I would have accelerated the weight 600mm to the slow only moving the weight at a constant velocity for 166mm.

You also fail far far far faster on the faster reps, you hit muscular failure faster.

http://www.youtube.com/user/waynerock999?feature=guide#p/a/u/0/sbRVQ_nmhpw

We as we all agree that when using 80% {of your 1RM repetition maximum} for the fast at .5/.5 {.5 of a second concentric and .5 of a second eccentric} for 15 repetitions = 15 seconds, and the slow at 3/3, that you will be able to go for longer using the 3/3, as you fail faster doing the .5/.5 yes ? And you fail roughly 50% faster on the fast repetitions.


So this means I “HAVE” used up my force and faster, yes ? Or as I call it strength faster, so if we are Clones, and have a 1000N of force to use up, at using 80% as explosively/dynamically, I “have” used up “all” my force, “but” as you are “STILL” exercising, repping the weight, you “MUST” still have force/strength left, right ?

So, as well as 99% of the Worlds bodybuilders, strongmen, weightlifters, and all sportspeople of every sport train in the faster repping way, and the evidence above, I just don’t see how some here think other ? It’s basically physics equations have not got all the variables, thus they are adding it up wrong. Bottom line is, the slow constant medium forces cannot balance out or make up the high peak forces of the fast on their accelerations, in the same time frame, it’s like impulse, impulse is defined as the integral of a force with respect to time. When a force is applied to a rigid body it changes the momentum of that body. A small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important.

We are now constructing a force plate.

Thank you for your help on the last thread, and hopefully on this one, as maybe you can spot where the equations are going wrong.

Wayne
 
  • #32
The equations are 'gong wrong' because you can't expect them to apply in a situation like this (as in all of your questions). You confuse the terms you use and seem to think that 'work done on' is necessarily related to 'energy transferred from muscles'.
I really don't know why you want to include Physics in your training if you won't learn Physics to an appropriate level.
If you continue to talk in the wrong language, how can you expect to have a proper conversation . Quote "I have used up all my force" - this is a meaningless statement in terms of Physics. Your posts are full of such statements but you refuse to modify them or learn the appropriate way of putting things.

If someone can only play the Piano with one finger then it's no surprise that a Piano Concerto gives them a problem.

Like I have often said, you can't bend Physics to your will. You just have to follow where it takes you. You don't seem to want to, though.
 
  • #33
Would anyone else agree that if you apply only an upward force(against gravity) during the upward rep, then yes, it is exactly equal to gravity once you come to a stop. Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity. I saw wayne mentioned the experiment we are throwing together, and I was hoping this made sense to you sophie and douglis. It seems that is where this conversation is hung, in the assumption that you are only working against gravity. With a fast enough rep it is possible that is not the case. I am getting together a cheap accelerometer with which we will be testing this theory out.
 
  • #34
Zula110100100 said:
Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity.

Zula...we have discussed this case before in another of dozens Wayne's same threads.
You're referring to the case where you also use the antagonists muscles in order to stop the weight and you don't rely only on gravity for the deceleration.
Of course you're right...in that case the overall force is greater than gravity and probably that's what he's doing in his experiments.

However...Wayne denies that case.He insists(without any reasonable argument) that without any "help" from the antagonist muscles(muscles apply only upward force) the overall force is greater with fast lifting ignoring the fact that the net impulse is always zero.
Please,if you have the patience(cause I don't) explain to him that he's totally wrong.
 
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  • #35
Hi all, and thank you for the replies, “no” time to read any of the posts, but this one came up on my mail.

---Quote (Originally by Zula110100100)---
Would you also agree that if however, you apply a sharp upward force and then ALSO a force WITH gravity, the net impulse applied to the weight is still 0, consisting of +your force against gravity, and -gravity and -your force with gravity, thus, the overall force you are applying is greater than gravity.
---End Quote---
Zula...we have discussed this case before in another of dozens Wayne's same threads.
You're referring to the case where you also use the antagonists muscles in order to stop th weight and you don't rely only on gravity for the deceleration.
Of course you're right...in that case the overall force is greater than gravity and probably that's what he's doing in his experiments.

However...Wayne denies that case.

NO I do not, why would I ? I have been training for about 40 years, and it’s well known that you have to use antagonist’s muscles in order to slow, stop and very slightly reverse the weight

He insists(without any reasonable argument) that without any "help" from the antagonist muscles(muscles apply only upward force)

I never said this ?

the overall force is greater with fast lifting ignoring the fact that the net impulse is zero.
Please,if you have the patience(cause I don't) explain to him that he's totally wrong.
He insistswithout any reasonable argument.

I have more than a reasonable debate, see above posts and below. Please remember, I am doing 6 more repetitions in the same time frame, and moving the weight 6 times as far.

I “proved” in my opinion I was right in my last post. Or could anyone here please try to explain that when you fail {and you do this is fact} in the faster repetitions, too which that means you have no force/strength left, but the person doing the slower repetitions does have force/strength left, how can I have not used all my force/strength up faster when I have used my force/strength up 50% faster ? Proof, evidence and facts, I have used my force/strength up first, thus if we both had 1000N of force/strength, I have used it up faster, to say other would be a full English and physics contradiction, it would be like say the Man who ran the 100m the slowest was the fastest.


Meaning the physics calculations need to add in the other branch of physics, Kinology kinology/physics which treats of the laws of motion, of moving bodies to the calculations. And Biomechanics the study of the structure and function of biological systems. These with the fact that you use more energy the instant you use more force/strength, also that you move the weight further in the same time frame, and that you fail faster.


I repeat, the physics calculation cannot be right, as the situation cannot exist, as the faster repetitions do indeed finish way before the slow, this is why my EMG readings in the same time frame, put the fast repetitions higher, because on the tests I did not go to failure, as we needed the results of the same time frame. Please take a look at the videos, the weight here was too light.


http://www.youtube.com/watch?v=B8gtpp8ozvU&feature=player_profilepage#t=3s


http://www.youtube.com/watch?v=pd0ZAm_2ioY&feature=player_profilepage#t=1s


Wayne
 
<h2>1. What is impulse?</h2><p>Impulse is a measure of the change in momentum of an object. It is equal to the force applied to an object multiplied by the time for which the force is applied.</p><h2>2. How is impulse calculated?</h2><p>Impulse is calculated by multiplying the force applied to an object by the time for which the force is applied. The unit of impulse is Newton-seconds (N·s) in the metric system, or pound-seconds (lb·s) in the imperial system.</p><h2>3. What is the relationship between impulse and force?</h2><p>Impulse and force are directly proportional to each other. This means that the greater the force applied to an object, the greater the impulse will be. Similarly, the longer the force is applied, the greater the impulse will be.</p><h2>4. How is impulse related to change in momentum?</h2><p>Impulse is equal to the change in momentum of an object. This means that the impulse applied to an object will result in a change in its momentum. The direction of the change in momentum will depend on the direction of the force applied.</p><h2>5. How is impulse measured in pounds for a specific time frame?</h2><p>In the imperial system, impulse is measured in pound-seconds (lb·s). This means that the force applied must be measured in pounds (lb) and the time for which the force is applied must be measured in seconds (s).</p>

1. What is impulse?

Impulse is a measure of the change in momentum of an object. It is equal to the force applied to an object multiplied by the time for which the force is applied.

2. How is impulse calculated?

Impulse is calculated by multiplying the force applied to an object by the time for which the force is applied. The unit of impulse is Newton-seconds (N·s) in the metric system, or pound-seconds (lb·s) in the imperial system.

3. What is the relationship between impulse and force?

Impulse and force are directly proportional to each other. This means that the greater the force applied to an object, the greater the impulse will be. Similarly, the longer the force is applied, the greater the impulse will be.

4. How is impulse related to change in momentum?

Impulse is equal to the change in momentum of an object. This means that the impulse applied to an object will result in a change in its momentum. The direction of the change in momentum will depend on the direction of the force applied.

5. How is impulse measured in pounds for a specific time frame?

In the imperial system, impulse is measured in pound-seconds (lb·s). This means that the force applied must be measured in pounds (lb) and the time for which the force is applied must be measured in seconds (s).

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