# Impulse/force in pounds for the time frame

1. Jan 2, 2012

### waynexk8

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, untill 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, lets 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, untill 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 untill the normal acceleration forces of the lift at rest.

Thank you for your time and help.

Wayne

Last edited: Jan 2, 2012
2. Jan 2, 2012

### waynexk8

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. Jan 3, 2012

### PhanthomJay

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. Jan 3, 2012

### waynexk8

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.

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.

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.

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. Jan 3, 2012

### PhanthomJay

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. Jan 5, 2012

### waynexk8

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.

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. Jan 5, 2012

### Zula110100100

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. Jan 6, 2012

### douglis

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. Jan 6, 2012

### PhanthomJay

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

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. )
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]75
again 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. 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'. I think I'll try this at the gym this weekend....but using 20 pound weights!

10. Jan 6, 2012

### waynexk8

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. Jan 6, 2012

### waynexk8

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.

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 travelling 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%

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.

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. Jan 7, 2012

### douglis

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. Jan 7, 2012

### waynexk8

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.

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.

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. Jan 7, 2012

### waynexk8

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. Jan 7, 2012

### waynexk8

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.

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.

The EMG and many things disagree with this.

Wayne

16. Jan 7, 2012

### douglis

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. Jan 7, 2012

### PhanthomJay

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).
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.
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.
You are not correct. However, the power required is greater when you are moving faster with the same applied force.

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.

18. Jan 12, 2012

### waynexk8

Sorry I meant the normal action-reaction forces.

Yes.

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.

Yes understand and totally agree.

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. Jan 13, 2012

### douglis

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. Jan 13, 2012

### waynexk8

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.

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.

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%

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

Last edited by a moderator: May 5, 2017
21. Jan 13, 2012

### douglis

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

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.

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.

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

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

Last edited by a moderator: May 5, 2017
22. Jan 13, 2012

### waynexk8

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.

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.

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 ???

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 ???

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.

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. Jan 14, 2012

### douglis

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.

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.

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

Last edited: Jan 14, 2012
24. Jan 14, 2012

### waynexk8

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.

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 ???

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.

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 ???

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

Will do that right now……….you mean this, I quote;
True.
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.

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.

I think/know we both have an obsession with this debate. I like to learn, and have learnt 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.

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

Last edited: Jan 14, 2012
25. Jan 14, 2012

### douglis

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?