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Most force/strength used in the same time frame

by waynexk8
Tags: force or strength, frame, time
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waynexk8
#1
Nov21-11, 02:22 PM
P: 399
I have asked similar before, however I don’t think I was going about asking it right, as I never got a fulfilling answer.

Both Men are equal in all ways, their 1RM {repetition maximum, or the maximum they can pick up once is 100 pounds} Reps = repetitions. .5/.5 or similar means the rep took .5 of a second up {concentric} and .5 of a second down. {eccentric} 1m = 1 Meter, mm = Millimetres. Exercise will be the Bench Press, I think all know that.

Man A, 1,
Lifts 80% in this case that’s 80 pounds, he lifts it 1000mm in .5 of a second. As all know, time is needed for the decceleration and stop at 1000mm.
A2,
Lifts 80% in this case that’s 80 pounds, he lifts it up and down 6 times in 6 seconds, lifting it at .5/.5 in total he moved the weight 12000mm. As all know, time is needed for the deccelerations, stop and transistions at 1000mm.

Man B,
Lifts 80% in this case that’s 80 pounds, he lifts it 166 in .5 of a second. No time is needed for the decceleration and stop at 166mm.

B2,
2B,
Lifts 80% in this case that’s 80 pounds, he lifts it up and down 1 time in 6 seconds, lifting it at 3/3 in total he moved the weight 2000mm. As all know, time is needed for the decceleration, stop and transistion at 1000mm.

Which Man uses the most Newtons of force in the same time frame. Or which man uses the most overall or total force/strength, in this same time span??? As in if both Men had only 1000 force/strength reserves in their muscles before force/strength ran out as of fatigue.

Some people seems to think that the same force/strength was used for both Men, I and many other said there was FAR more force/strength used in the faster rep. Now I have more clearly wrote this down, could anyone please comment, thx for your time and help.

I did some tests on a EMG, and every time it states that the faster rep has more muscle activity, as in it uses more force/strength, however some here said the oposite.

Wayne
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Zula110100100
#2
Nov21-11, 02:50 PM
P: 253
It might be easier to consider it in energy(J) also, when getting into the biology part your body has to use energy to keep an object at the same height, due to motion within your body, but in physics this is not represented since not moving the object does not count as work...

To simplify lets consider a weight of 1N, the work done by person A would be 1J(1N*1m) * 6(moving it down took no work from the person, this was done by gravity) so 6J over the course of 3 seconds required a power of 2Watts, The person that went up and down in 6 seconds used 1J over 3 seconds, so .333Watts so the person doing the faster reps took more power and more muscle activity... but these numbers are not really good since it took energy(biologically speaking) to move the weight slowly down over 3 seconds...
waynexk8
#3
Nov22-11, 06:32 PM
P: 399
Hi there, and thanks for your help and time.

Quote Quote by Zula110100100 View Post
It might be easier to consider it in energy(J) also, when getting into the biology part your body has to use energy to keep an object at the same height, due to motion within your body, but in physics this is not represented since not moving the object does not count as work...
We both agree that the Man using the faster reps uses more energy; the debate is just about force/strength. I knew from the start that whenever doing anything faster in the same time frame you uses more energy, the people whom I am debating with agree with this now, but not about the faster rep using more force. One of the things I pointed out, is that if the faster repper is using more energy, then why ??? I said because he must be using more force/strength, if not I asked why else would he use more energy, they have not replied yet.

Quote Quote by Zula110100100 View Post
To simplify lets consider a weight of 1N, the work done by person A would be 1J(1N*1m) * 6(moving it down took no work from the person, this was done by gravity) so 6J over the course of 3 seconds required a power of 2Watts, The person that went up and down in 6 seconds used 1J over 3 seconds, so .333Watts so the person doing the faster reps took more power and more muscle activity... but these numbers are not really good since it took energy(biologically speaking) to move the weight slowly down over 3 seconds...
Yes see what you’re saying these, but as I said, the debate {and this is huge over the internet, going on for 6 years, that’s why I bought the EMG, looks like I will be buying a force plate next, unless you and the other very clever physics can help me out} is specifically about how much force/strength is used. Not sure what you mean by no work done by moving it down ??? As I would say that the person had to do physical work, meaning he had to use energy and force/strength to lower the weight under control, as if he just let it drop, it would drop much faster ???

Ok let’s just concentrate on this part for a moment, say that two machines lifted the weights, and they had 100 pounds of force/strength for maximum, and were using 80%

Machine A,
Lifts 80% and lifts it 1000mm in .5 of a second. As all know, time is needed for the deceleration and stop at 1000mm.


Machine B,
Lifts 80% and lifts it 166mm in .5 of a second. No time is needed for the deceleration to stop at 166mm, as the machine is still lifting the weight at 1000mm every 3 seconds, but I would just like to know the force/strength that it uses to lift the weight 166mm in .5 of a second, as the faster lifts the weight 1000mm in the same time frame.

Hope you get what I mean, and have made myself clear.

Wayne

Zula110100100
#4
Nov22-11, 11:29 PM
P: 253
Most force/strength used in the same time frame

Ok letís just concentrate on this part for a moment, say that two machines lifted the weights, and they had 100 pounds of force/strength for maximum, and were using 80%

Machine A,
Lifts 80% and lifts it 1000mm in .5 of a second. As all know, time is needed for the deceleration and stop at 1000mm.

Machine B,
Lifts 80% and lifts it 166mm in .5 of a second.
So I think here is the problem, if the force is 80lbs, and it is lifting an equal mass, it will lift it in an equal time, as force is by definition acceleration * mass, so in order to lift it slower one would have to use less force, or more mass.
Zula110100100
#5
Nov23-11, 12:07 AM
P: 253
Aaaaaah, I think I get what you are looking for! So lets say the weight is 100kg, gravity pulls that down with 980N of force, so keeping it still, you must still apply 980N, So, if you were to raise it 1m, in .5s, the required acceleration can be found with:

s = 1m
t = .5s
s = 1/2at2
a = 2s/t2 = 2(1)/(.25) = 8m/s2

so the force applied was 800+980 = 1780N for .5s = 890N*S

moving it down at an equal speed would require pushing against gravity to get -8m/s2 980(force to resist gravity) - 800(force going into acceleration) = 180N

So on the way down we apply 180N *.5s = 90N*s
for a sum of 980N*s

For person B, raising it .166m in .5s would be
a = 2s/t2 = 2(.166)/(.25) = 1.328m/s2
Add that to the 980 to keep it still and you get 980+132.8 = 1112.8N * .5s = 556.4N*s
to come down at that same speed would be 980 - 132.8 = 847.2N *.5s = 423.6N*s

for a sum of 980N*s which is equal

Which makes sense because if C is the force required to keep something still against gravity, and x is the force required for a constant acceleration of some set amount, and t is a constant time overwhich the acceleration occurs

we are really looking at
t(C+x)+t(C-x) = 2tC
tC+tx+tC-tx =2tC
tC+tC+tx-tx =2tC
2tC = 2tC
waynexk8
#6
Nov23-11, 08:19 AM
P: 399
Hi there,

This is fantastic, just what I wanted. I am not that good at physics, so may I asked a few laymenís questions please. When you worked out the fast moving the weight 1000mm in .5 of a second, you did add in the deceleration phase for this ??? And if you did, this force/strength used would have been more without a deceleration phase. I imagined you did add these in, but as the physics are over my head, thought I best ask.

What you have done is just what I was hoping, looking for, could we now do one more equation please if you have the time.

As we know you worked out the force/strength to move the weight up from a still start to 1000mm in .5 of a second, but could we please add in what we called the peak forces, or the MMMT {Momentary Maximum Muscle Tensions} this is the transition from negative to positive. And you know FAR more than me, that to stop a weight that is moving down at 1000mm every .5 of a second, and then immediately stop and move it back up to 1000mm at .5 of a second, will take more force than moving it up from a still start, of as the down acceleration components, as the weight will appear the be far more than its still weight of 150kg it would be more like 180kg if it hit a scales. As for every reaction, there is an opposite reaction, and this reaction of moving the weight from the downward eccentric transition back up to concentric, will put/use more force/strength.

Could you please if possible work this Milly second or more peak force and then I could add it to the 890N you worked out the force/strength was for the still start.

SORRY, I always try too over explain things.

Thx for your time and help, itís been great.

Wayne
sophiecentaur
#7
Nov23-11, 09:41 AM
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This question has no answer from Physics. It's been posed many times and in many different forms but anyone who goes away thinking they have an answer is deluding themselves. This is well demonstrated by the (mis-)use of non-Physics terms throughout threads like this one.
All Physics can tell you is the Work Done on the objects that are being lifted. A simple subject like Physics does not attempt to discuss how much actual Energy the muscles 'Use'. It is up to Biologists and their ilk to discuss the details of the metabolism in muscles.
You'd still be using a lot of energy just keeping a heavy weight at a constant height.
'Common sense' doesn't apply here but you can say that the person who is more knackerd after the exercise may possibly have been expending more energy - all other things being equal.
Zula110100100
#8
Nov23-11, 02:12 PM
P: 253
you did add in the deceleration phase for this ??
No, I did not, the answer I gave is for a constant smooth acceleration up until the force would be instantly cut down to the lowering force.

we often have to simplify things to get a workable problem, which is why sophiecentaur is pretty much right that physics cannot easily model this situation with a high level of accuracy, I disagree with that it is beyond the scope of physics, it is just some really really complicated physics for which we cannot get enough data to make an accurate model. Surely beyond my level of physics.

If we knew more about the bio-physics going on then it would be answered better.

and for the mis-use of terms is due to the fact that words like energy have meanings in physics and another meaning in general. From what I can tell the terms I used is an accurate physics description of the force that would be measured between the weights and your hand given a constant acceleration each time. It is however, unlikely that this is what really happens when someone lifts weights.

for instance, the time that the "deceleration phase" begins make a difference, and for the peak force applied to get it to turn around depends on the velocity it gets up to and how long it takes to stop it and re-accelerate it

What was the time resolution of the EMG data? How accurate was the timing of the reps done in those tests? It could possibly give an idea of the time in the upward phase that the muscles begin pushing less....I am not sure how that would work, is there any way you could send me that data?

To re-emphasize: It will be very difficult to get an accurate picture of what goes on in terms of getting tired out and I probably can't give you a better answer, but It would be cool to think about.
douglis
#9
Nov23-11, 04:11 PM
P: 148
Quote Quote by Zula110100100 View Post
No, I did not, the answer I gave is for a constant smooth acceleration up until the force would be instantly cut down to the lowering force.
Hi Zula....Wayne has asked this many times before and probaby he'll keep asking until he'll get the answer he likes.

When you lift a weight...the motion starts and ends at rest so the average acceleration is always zero(a=(v2-v1)/t=(0-0)/t=0).So the average net force is zero too and the applied force equal with the weight regardless the lifting speed.
But....Wayne can't accept that.
sophiecentaur
#10
Nov23-11, 04:55 PM
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Until Wayne actually learns some of the 'Physics Rules' he won't accept the consequence of them. What I can't understand is why he wants an answer from Physics when the answer is that there isn't one.
The daft thing is that he'd probably accept some old bollocks answer that some non-Physicist made up. No where's the sense in that?

I suggest that he should exercise his reason a bit instead of just his muscles.
Zula110100100
#11
Nov23-11, 06:45 PM
P: 253
This is very much a physics problem, it is just that he is not asking what you want him to ask.


sophiecentaur: as your signature says, "it is all models"

It is only a system that we set up that says Work is net force over distance, yet you feel his question and answer must reside in that one model, but like you said, he is not looking for that answer, so he will not be satisfied with it. This is in the general physics forum, not classical physics, so we should not be constrained to work in only ONE system.

"physics is the study of matter and motion" - wikipedia

This is a problem involving matter and motion, so it is indeed a physics problem.

So instead of telling him he's got his words all wrong, I will instead try to answer the question I think he is asking, and if that is what he is asking, he will maybe learn about the physics going on here.

Do you not agree that in physics we have to simplify the situation so that we are looking at the aspect that we want.

Instead of looking at the net forces, we are looking at force-applied, while this does not all go into the kinetic energy, one can assume that biologically speaking energy is used when the weight is not being moved, this is as I said due to activity within your muscles and so on...

We do not know the conversation he is having, but it seems his question is about force-applied, NOT energy.

But I actually got on here to tell him that if 980N is 80% of the max, the max is 1225N, so you could not raise it 1m in .5s as that would take 1780N, so we must reword the problem so that 1780 is the max, or is 80% of max.

It may be that I am just the non-physicist bollock to give him an answer he is content with, but if that is the case, tell me where my physics is wrong as I am indeed not a physicist, more of a hobbyist(though I am considering switching majors), but I will try to keep it within the constraints of what I know about physics.

Where is the error in knowing the impulse of force-applied? we could subtract the impulse from gravity to get the net impulse and still figure out the change in momentum...it is as if the fact that we can use net force for a net impulse makes the force-applied impulse forbidden.


If you refer to the OP
Which Man uses the most Newtons of force in the same time frame.
it should be apparent that he is asking about the impulse of the force-applied

More specifically, Which force-applied impulse is greater in each situation, is that not an answerable question?
douglis
#12
Nov24-11, 01:44 AM
P: 148
Quote Quote by Zula110100100 View Post

It may be that I am just the non-physicist bollock to give him an answer he is content with, but if that is the case, tell me where my physics is wrong as I am indeed not a physicist, more of a hobbyist(though I am considering switching majors), but I will try to keep it within the constraints of what I know about physics.
Zula....your physics is not wrong.Just your assumption of a constant smooth acceleration up until the end of lifting is wrong.

What really happens is this:
You start from rest, accelerate, then decelerate to rest. The net change in momentum is therefore zero, which is equal to the net impulse delivered. Therefore, the average net force on the object is also zero. But since the net force is equal to the vector sum of your upward force and the downward weight of the object, the average upward force should equal the average downward force, i.e., the weight.
sophiecentaur
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Nov24-11, 03:48 AM
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@Zula110100100
You are clearly in as more patient mood than I. But I have read virtually the same question from Wayne before. The reason that I say his question has no answer is because he is mostly concerned with 'Energy Used' and not 'Work Expended On'. Work is easily defined and, as long as all the variables are specified (force and distance at all times) the total work done can be calculated. This is a pretty trivial exercise because any A level student could do it, with a little prompting. By dividing the Work Done by the time taken, you can also get the Power.
But what you cannot do is tell the Energy Used by the body. You can possibly measure it over a longish time interval by analysing the gases in the air breathed in and out but it is different from person to person and from activity to activity. I mean, here, that the efficiency is an imponderable and that Physics cannot tell you how two different people's performances will differ in terms of Energy Used.

Science tries to be reductionist for a reason. It aims to be able to predict future behaviour of systems, based on evidence drawn from the past. Merely going over a mass of anecdotal evidence gets us nowhere. It may be fun (post-footie match discussion) but it is not going to help with accurate predictions. I can't see why it isn't important to use the 'right' terms in analysing what goes on in a Gym any less than it is important to use the correct Arithmetic when adding up the bill at the end of one's Gym session. You don't want the wrong answer in either case.

So I point out when the 'wrong question' is asked because only a 'right question' can actually yield a meaningful answer. One essential part of a 'right question' is that the right and appropriate terms should be used. That is not idle pedantry; it's a way of ensuring that the answer you get applies to the actual question that was asked. Swap the +-X and ų keys on your calculator around and you will get absolute nonsense answers out of it - the same thing applies in Science.

If you really think that you can calculate the Energy Used then you are welcome to try but I really don't think it is a calculation that Physics, alone has any hope of achieving. Would Wayne be prepared to learn and use the large number of new terms and ideas involved? (I guess Wayne can answer that for himself)
douglis
#14
Nov24-11, 04:25 AM
P: 148
Quote Quote by sophiecentaur View Post
I mean, here, that the efficiency is an imponderable and that Physics cannot tell you how two different people's performances will differ in terms of Energy Used.
sophiecentaur....you're right that physics can not give a definite answer for the energy expenditure but I think Wayne is now concentrating on the calculation of the applied force....which of course has also been answered dozen of times.
waynexk8
#15
Nov24-11, 07:30 AM
P: 399
Not much time now, but back in full later, but big thx for all the help.

First I am not interested in Energy, as I know that the fast will use more, itís been know and worked out for over 50 years using a Calorie chamber, were they hook you use to many things and take many readings. Anything that is done faster uses more energy, but why is that ??? In my opinion it is and can only be you have used more force/strength in the same time frame, how else would you cover more distance in the same time frame ??? D. admittedly admits he was wrong about this, as before he thought that the same energy was used, however the fast uses far more energy.

Do we all agree that there will be more force/strength used in the second and succeeding reps than the first ??? As the first is from a still start, and the next are when the weight is coming down from the negative, and has gathered more force/strength on the way down. So could we work out the force/strength and add in the peak forces/strengths for the others reps.

By the way all, I have bought an EMG this shows that in the fast there was more muscle activity in the fast, more muscle activity = more force/strength, thatís in my opinion because the peak force was not added in. Also we all know you fail far faster in the faster reps, in my opinion itís because you have used more force/strength, hereís me showing this.

http://www.youtube.com/user/wayneroc.../0/sbRVQ_nmhpw

Wayne be prepared to learn and use the large number of new terms and ideas involved. Yes I am all to ready to learn, and donít want any old B.

So could we work out the force/strength and add in the peak forces/strengths for the others reps.

Wayne
douglis
#16
Nov24-11, 07:45 AM
P: 148
Quote Quote by waynexk8 View Post

Do we all agree that there will be more force/strength used in the second and succeeding reps than the first ??? As the first is from a still start, and the next are when the weight is coming down from the negative, and has gathered more force/strength on the way down. So could we work out the force/strength and add in the peak forces/strengths for the others reps.

Wayne
The force is not gathered and you can not add forces the way you.
Either in 10 seconds you do 1 or 5 repetitions you apply the same average and equal with the weight force for 10 seconds.
End of story.

You're not here to learn anything.You've been explained the above hundreds of times and you continue the same nonsense.
Zula110100100
#17
Nov24-11, 01:14 PM
P: 253
The force is not gathered and you can not add forces the way you.
Either in 10 seconds you do 1 or 5 repetitions you apply the same average and equal with the weight force for 10 seconds.
End of story.
The average is the same, but that is not the end of the story.

I guess every post I should include: This is not likely to be how lifting weights actually works, at all.

But...lets take for example, the lowering of 100kg, over 1m, starting at the top, and at rest.

If I apply no force for .5m then it will gain a KE of mgh 100(-9.8).5 = -490J

The instantaneous velocity will be Ke = .5mv2
sqrt(490*2/m) = -3.13m/s
avg velocity -1.56
s/v = t
.5/1.5 = .319 seconds = t

In order to stop it in .5m requires an equal change in KE over the same distance, so yes, it is also 490J, and the acceleration is 9.8, showing a NET force of 980N, however, to achieve this net force requires force applied to be (980+980) = 1960N, times another .319seconds equals an applied force of
Gravity impulse = -626N*s
Force-applied impulse = 626N*s
Net Impulse = 0;
Total time: .638s
Total dist: 1m
Δv = 0

Now the same setup but it will drop freely for .75, in that situation it gets 745J, this time we must counter than in .25m, so the required net force over that .25m is 2940N
at the .75mark we would have an instantaneous velocity of 3.86m/s, that is a momentum of 386N*m/s, Δp = Ft so the time to come to a stop is Δp/F = 386/2940 = .13seconds
The time to traverse the .75 and get to that speed is Δp/g = 386/980 = .39
so:

Gravity impulse = 509.6N*s
Force-applied impulse: = 509.6N*s
Net impulse: 0
Total time: .52s
Total dist: 1m
Δv = 0

So this is the same weight, time difference is .118s (A: .638s, B:.52s), now the force-applied impulse does equal the gravity impulse, but are not equal with each other, the slower rep(A) does take more Newton*seconds, now, I agree, the wording is bad, and Wayne, you must realize that by applying that many newtons, for a seconds, you don't divide it like meters/seconds, newtons are a force that can be applied over some seconds to give a multiplied result of Newton*seconds, this is called an impulse and is responsible for a change in momentum.

I believe the point being made by Douglis is that in order to have 0 momentum on the top and bottom the sum of the impulses must be 0, otherwise it would still have momentum in one direction or another, however, the slower you go down, the greater the time that gravity is acting on the weight, this creates a larger impulse from gravity, in order to get that sum of 0, this in turn, requires a greater impulse of force-applied, so going slower does require more FORCE be APPLIED each SECOND

So that is showing that on the way down(including deceleration time) a slower lowering requires a greater force-applied impulse.

Do we all agree that there will be more force/strength used in the second and succeeding reps than the first ??? As the first is from a still start, and the next are when the weight is coming down from the negative, and has gathered more force/strength on the way down.
I do not agree with this, on the second upward rep, you would have once again brought it to a stop at the instant before the rep begins, so it should be equal. The extra force to stop it goes into the previous downward rep. this really depends on when you view a rep beginning but I feel that is the most accurate is that the next upward rep begins once you are at the bottom of the range, which we agreed in this system you have decelerated by the time you get there.

The force is not gathered and you can not add forces the way you.- douglis
This is wording too really, I would almost define a force as a factor of gathering, a net force defines: 1: The rate at which energy is gathered(or negatively gathered, lost) over distance, 2: The rate at which momentum is gathered(or negatively gathered, lost) over time

For example, in gravity a 100kg weight would gather kinetic energy at a rate of 980N/m

It seems the same would be true for the way up, so slower reps = MORE force-applied impulse, but also MORE gravity impulse balancing it out, so less work being done since less NET force over the same distance
waynexk8
#18
Nov24-11, 07:54 PM
P: 399
Quote Quote by Zula110100100 View Post
The average is the same, but that is not the end of the story.

I guess every post I should include: This is not likely to be how lifting weights actually works, at all.

But...lets take for example, the lowering of 100kg, over 1m, starting at the top, and at rest.

If I apply no force for .5m then it will gain a KE of mgh 100(-9.8).5 = -490J

The instantaneous velocity will be Ke = .5mv2
sqrt(490*2/m) = -3.13m/s
avg velocity -1.56
s/v = t
.5/1.5 = .319 seconds = t

In order to stop it in .5m requires an equal change in KE over the same distance, so yes, it is also 490J, and the acceleration is 9.8, showing a NET force of 980N, however, to achieve this net force requires force applied to be (980+980) = 1960N, times another .319seconds equals an applied force of
Gravity impulse = -626N*s
Force-applied impulse = 626N*s
Net Impulse = 0;
Total time: .638s
Total dist: 1m
Δv = 0

Now the same setup but it will drop freely for .75,
However in the faster rep, itís not dropping freely, I am lowering it under control, as if it dropped freely, it would drop far faster.

Quote Quote by Zula110100100 View Post
in that situation it gets 745J, this time we must counter than in .25m, so the required net force over that .25m is 2940N
at the .75mark we would have an instantaneous velocity of 3.86m/s, that is a momentum of 386N*m/s, Δp = Ft so the time to come to a stop is Δp/F = 386/2940 = .13seconds
The time to traverse the .75 and get to that speed is Δp/g = 386/980 = .39
so:

Gravity impulse = 509.6N*s
Force-applied impulse: = 509.6N*s
Net impulse: 0
Total time: .52s
Total dist: 1m
Δv = 0

So this is the same weight, time difference is .118s (A: .638s, B:.52s), now the force-applied impulse does equal the gravity impulse, but are not equal with each other, the slower rep(A) does take more Newton*seconds, now, I agree, the wording is bad, and Wayne, you must realize that by applying that many newtons, for a seconds, you don't divide it like meters/seconds, newtons are a force that can be applied over some seconds to give a multiplied result of Newton*seconds, this is called an impulse and is responsible for a change in momentum.
Hmm sorry about this, but I think itís best to ask questions I think if I donít get something.

So the slow eccentric seemed to take .638s to be lowered 1m and the force/strength used was 626N ??? But if we are talking of slow rep, it was lowered in 3 seconds.

And the fast eccentric .52s that basically .5 of a second, to be lowered 1m and the force/strength used was 509.6 ??? But if we are talking of fast reps, it was lowered in .5 of a second, yes, but lowered 6 times in all, thus we have to add the force/strengths up, as it takes more force/strength to lower something 6 times to 1 time.


Quote Quote by Zula110100100 View Post
I believe the point being made by Douglis is that in order to have 0 momentum on the top and bottom the sum of the impulses must be 0,
Not sure the point here to be honest.
However at both transitions the zero movement is so small of a time, do we need to bother with this ??? As at that time, which would be a Milly of a second, would have no force/strengths ???

Quote Quote by Zula110100100 View Post
otherwise it would still have momentum in one direction or another,
But we do have movements in both directions, its only the grip of the hands and the muscles forces/strengths that stops this on both transitions ???

Quote Quote by Zula110100100 View Post
however, the slower you go down, the greater the time that gravity is acting on the weight, this creates a larger impulse from gravity, in order to get that sum of 0, this in turn, requires a greater impulse of force-applied, so going slower does require more FORCE be APPLIED each SECOND. So that is showing that on the way down(including deceleration time) a slower lowering requires a greater force-applied impulse.
Hmm, I canít really see that, as of below, but if youíre saying the 3 second lowering to the .5 of a second lowering uses more force/strength, than yes this must be so, as the slow is using force/strength for 2.5 seconds longer.

But what if we added all the 6 lower rings of the faster reps each at .5 of a second ???



Quote Quote by Zula110100100 View Post
I do not agree with this, on the second upward rep, you would have once again brought it to a stop at the instant before the rep begins, so it should be equal. The extra force to stop it goes into the previous downward rep. this really depends on when you view a rep beginning but I feel that is the most accurate is that the next upward rep begins once you are at the bottom of the range, which we agreed in this system you have decelerated by the time you get there.
Hmm, I see your point, however when lifting at say .5/.5 you do the transition so fast from negative to positive, that the faster you do it the harder it is, and itís done very fast. Lets exaggerate, or put up a scenario, letís say I lift 80% from a still start, on the bench press, think we all know what the bench press in, but the next time, the person lifts it for me, and then I take my arms away, and lover them, then the person drops the weight, and I catch it three quarters of the way down and try the press it back up, do we all agree that the drop and press would be far harder, than the still start ??? Actually that drop might even put about 40% more weight on the bar. As thatís more what happens, when repping the weight up and down, you immediately go from lowering the weight fast under control, to lifting it back up before it gone down fully.

If you hold a heavy weight, best if itís at least 20 pounds, then lift it and lower quite fast, you will see that the moment you stop it the reading goes up.

Also, if I am doing squats, say I am doing reps with 300 pounds, its far harder to get the weight moving back up on the concentric, if you lower it very fast, as it gathers speed, thus it appears to be heaver. So if I was doing 12 reps at a slow eccentric, say 1 second, and a fast as possible concentric, I would be able to do the 12 reps, BUT if I lowered the weight much faster, it would make the exercise far harder, and I might only get 6 reps

Quote Quote by Zula110100100 View Post
This is wording too really, I would almost define a force as a factor of gathering, a net force defines: 1: The rate at which energy is gathered(or negatively gathered, lost) over distance, 2: The rate at which momentum is gathered(or negatively gathered, lost) over time

For example, in gravity a 100kg weight would gather kinetic energy at a rate of 980N/m

It seems the same would be true for the way up, so slower reps = MORE force-applied impulse, but also MORE gravity impulse balancing it out, so less work being done since less NET force over the same distance
Going to have to read that part over a few times. However if youíre saying what I just said above, that the 3/3 = 1m, has to use more force/strength than the .5/.5 = 1m, then yes, however the faster reps at .5/.5 are done 6 times, to be done in the say time frame, thatís 6 seconds like the 1 slow rep = 6 seconds.

One thing no one seems to answer, how can you move the same weight more distance in the same time frame without using more force/strength ??? Itís impossible. As in my fast rep moving the weight 1m in .5 of a second, the slow have only moved its weight 166mm.

And if we take the 6 reps at .5/.5 moving the weight 1m up and 1m down 6 times in 6 seconds, its moved the weight 12m to the slow rep of moving the weight 1m up and 1m down 1 time in 6 seconds, its moving the weight only 2m.

Also my EMG reading state that there is more force/strength/more/muscle/activity in the faster reps ???
. First I did right leg, leg extension, quite fast, something like .5/.5
Then I did left leg, leg extension, quite slow, something like 3/3.
Both obviously were to the set time, which was 25 seconds.

Fast = average work = 295 iV
Slow = average work = 177 iV

The fast was 118 IV or 61% higher than the slow, as of MORE average/total force/strength is used.
So what have the few that said other have to say about this ???

Also you fail faster in the faster reps, in my opinion that would say you are using more force/strength faster ???

http://www.youtube.com/user/wayneroc.../0/sbRVQ_nmhpw

Thx for your time and help all.

Wayne


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