Average force do not apply here

[waynelucky2]

Average force do not apply here, because the forces are not the same, meaning the concentric force is very loaded because of the sliding filament action of muscle action, but the eccentric is very under loaded because it more like a breaking action of the muscles.

K wrote;
We agree yet you still can't see the obvious, that's why I'm laughing. A fast rep has six times more acceleration, but it aslo has 6 times the deceleration, they cancel each other out giving you an average acceleration of zero, or an average force equal to the load.

Hi K.

I explained how this cannot be below, as the concentric and eccentric are NOT the same or equal.

Thats why weeks ago I tried giving many clues, I said you cannot use the average forces in this instant, as it will not work, its like comparing the average speed of a sprinters forward 100m and his backwards 100m, it tells you nothing of his overall speed is his forwards 100m.

As the eccentric force needed to lower the weight is so low, you can never commper it to the very high force of the concentric.

D. and Ron did not see or understand this, so now they try to ignore it.

K wrote;
The load is 100 pounds,

A fast rep looks like, 160 accel, 40 decel (minus 60 force)= 160+40 divided by 2 = 100 average
A slow rep looks like 110 accel, 90 decel (minus 10 force)= 110+90 divided by 2 = 100 average

The fast reps have 60 acceleration, the slow reps have 10 acceleration, the slow reps have 10 deceleration, the fast reps have 60 deceleration.

Fast reps = 60 acceleration minus 60 deceleration = 0
Slow reps = 10 acceleration minus 10 deceleration = 0

As the eccentric is very under loaded, its like using 40 to 50% on a concentric, there is NO way that you can make up for what you lose out on in the concentric, as the fast concentric = 70% more force used, and that is without adding in the peak forces, shall we call them 35% more ???

If so with the peak forces and the full concentric forces, {we could take off a little for the time the peak forces cut into the concentric forces ???} this = 105% MORE force than your slow rep, but as your very under loaded eccentric does produce more force than mine, lets be very generous and give you 20% back ??? With that still my net or overall force are 85% MORE than yours.

And remember these forces were worked out for 1 of your reps at 2/4 = 6 seconds, and for 1 of my reps at .5/.5 = 1 second. But on the 1 rep to 1 rep you have time on your side, but your concentric is a long low force concentric, mine is a short high force concentric, thats why my force needed is 70% higher. But if you still think you have time on your side, I have another 5 more reps to go.

So do you think these numbers are quite fair ??? And if not ALL, please say why.

Fast rep,
Concentric force =1781 kg m/s^2
Eccentric force = 181 kg m/s^2
Overall force = 1962 kg m/s^2
Average force = 981 kg m/s^2

Slow rep,
Concentric force = 1031 kg m/s^2
Eccentric force = 968.5 kg m/s^2
Overall force 1999.5 kg m/s^2
Average force = 999.75 kg m/s^2



Its like trying to add up the averages of a orange and a lemons juice added together, it does not work, its non scientific.

Wayne

[waynelucky2]

D wrote,
He already asked a physicist and tried to prove him wrong too:
http://physicsforums.com/showthread.php?t=381643

His (main) problem is that he doesn't understand the meaning of average.

D. that is quite a silly thing to say, most know here what average means.


You have to pick 10 runners from 100, to run a 100m sprit forwards, you are given their average running speeds of 100m forwards and 100m backwards for the year, but you do not know that some train just for backward running.

I have to pick 10 runners from 100, to run a 100m sprit forwards, I am given their 100m averages for year, who do you think would win each race, my runners or your runners ??? It would be mine 9 out of 10 would it not.


DO YOU NOT STILL UNDERSTAND THE ECCENTRIC IS UNDER LOADED, THUS YOU CAN NOT USE IT FOR AN AVERAGE WITH THE CONCENTRIC, THIS ALSO GOES FOR YOU RON, YOU TWO ARE VERY BAD SPORTSMEN, YOU ARE QUITE PLAINLY WRONG, AND HAVE NOT ANSWERED OR COUNTER JUST ABOUT EVERY THING I HAVE TOLD YOU, WHY, BECAUSE YOU CAN NOT, AND IF YOU DID YOU WOULD BE PROVING YOURSELF WRONG.

D wrote,
If,for example,you'll prove to him that for both the 1/1 and 3/3 reps the average force is 100 pounds he believes that this means that it takes 1 or 3 seconds of exerting force to reach the 100 pounds.

What are you on about again, you tried this silly thing before, and I do not like people putting words in my mouth. When you use a 100 pounds, it takes milly second to reach this force, then if you choose to keep the force at a 100 pounds it will stay these until you fatigue.

Why would you say a silly think like that ??? Its just because you failed miserably, and also failed to answer just about anything I wrote down.

As John said from the start of the thread over at Ellingtons, your having us on, but why is Ron falling for this I can never understand.

D wrote,
He doesn't understand that if that was true he wouldn't be able to lift the weight at first place. Or else he doesn't understand that same average force also means same average tension per unit of time.

Let me try and make you understand one more time.

Just pretend that your pushing your palm of your hand though water for 1m, its quite hard is it not ??? Then when you move it back in the other direction, you take your hand out of the water, its far easier now is it not ??? NOW TRY AND MORE FASTER THOUGHT THE WATER, IT GETS HARDER AND HARDER THE FASTER YOU TRY AND MOVE IT, does it not ??? This is because when a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.


Slow rep,
Lets say forward though the water puts 120T {tension on the muscle} on the muscles for 3 seconds, and backwards though the air puts -20T on the muscles for 3 seconds

Fast rep,
Lets say forward though the water puts 205T {tension on the muscle} on the muscles for .5 of a second, and backwards though the air puts -90T on the muscles for .5 of a second. This stroke is done 6 forward and 6 backwards.


The going backwards thought the air puts so very little T on the muscles compeered to going forward though the water, its not much point in adding the T to the equation, but lets be fair and say it adds 20% But to be fair we have to add in the peck forces from the fast transition, from air to the water, and this goes up to about 270%, so lets add on 35% for this for .2 of a second, so its 205 – off 20% = 164 + on 35% = 220T.


So we have T on the muscles, thats tension on the muscles.

Fast reps = 220T.

Slow rep = 140T.


Thus the faster reps = 55% more tension on the muscles for 6 seconds of each rep/reps, 12 seconds = 110% more tension on the muscles, and 18 seconds = 165% more tension on the muscles for the faster reps.


This is as we know not an exact science, but its not a bad estimate, unless you think otherwise ???


Wayne

[waynelucky2]

Muckle_Ewe wrote,
Sorry I meant You have proven yourself wrong but you just don't know it.

The 100 > 199 thing is the 2 reps at 100 vs the one at 199.

You say you dont add up the forces but many times you have repeatedly said 'I need another 5 more forces' or 'I use that force 5 more times'. Which is the basic idea of your argument.

If the area under the graphs are the same we are right.

Not sure what graph you are on about ???


You think you can not add up force ??? But what happens when you lift a 100 pounds at a constant speed ??? You are using and using a 100 pounds of force every milly second are you not ??? As you cannot keep lifting this weight if you do not keep using force can you, so you are continually using force, this force is happening over and over again repeated right ??? And when I change direction I use less force, then change direction again and use more force. And if I stop after 4 reps and put the weight down, then pick it up several seconds later I have to use this force again do I not ???

And when you accelerate a weight when lifting it, your using many different amounts of forces, are you not ???

So I do need many more forces do I not ???


But that was not so much the point, that was more to try and get people to see that,

1,
If you use a force of 120 for 2 seconds, I COULD USE A FORCE OF 205 FOR 1 SECOND, AND IT OVER THE SAME TIME FRAME, PUT THE SAME AMOUNT OF TENSION ON THE MUSCLES AS YOUR 2 SECOND REP, WOULD IT NOT ??? AS ITS MOVED THE WEIGHT DOUBLE THE DISTANCE IN LESS TIME, THUS IT MUST HAVE PUT THE EXACT SAME TENSION ON THE MUSCLES, MUST IT NOT ??? IF NOT THEN SAY WHY MY HIGH FORCE DOES NOT PUT THE SAME AMOUNT OF TENSION ON MY MUSCLES IN LESS TIME.

Do not some still not get it, higher force for a shorter time can and does put the same tension on the muscles as a low force for a short time, BUT I HAVE 5 MORE REPS = 5 MORE TENSIONS ON THE MUSCLES.

In classical mechanics, an 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. The impulse is equal to the change of momentum.

And before someone says it, NO, you cannot gain all this back from your eccentric, as its under loaded, you can only gain 20% back, so in my.

Wayne

[waynelucky2]

K wrote,
He needs to understand that the average force per second is the same, distance matters not. I doubt he's ever going to get it, probably best if we all just ignore him otherwise he'll never stop.

K, why do people keep saying that ??? Of course I know what average means, just look at what I have put below many times. But do some of you know where and where not you can use averages ???

K or any of you, I have a little question for you, you can see below that I have worked and put down the average forces, thus tension on the muscles.

Question,

My average force was worked out for 1 second, the slow rep was worked out for 6 seconds, so does this mean our average forces thus tensions on the muscles are the same ???

Fast rep,
Concentric force =1781 kg m/s^2
Eccentric force = 181 kg m/s^2
Overall force = 1962 kg m/s^2
Average force = 981 kg m/s^2

Slow rep,
Concentric force = 1031 kg m/s^2
Eccentric force = 968.5 kg m/s^2
Overall force 1999.5 kg m/s^2
Average force = 999.75 kg m/s^2








Let see if I can first tell you how the graph would look like.

Say we have 300 going up for the tensions hitting the muscles. 100 and above being force needed to move the weight up, and 6 across for the seconds.

Your rep would say go from 100 up to 120 up, right across in a straight line to the 2 across, then for the transition to concentric, it would drop down to about 80 up right across in a straight line to the 1, well sorry 2 pass 1, the other side of the line going up, expect you get what I mean. So it 2 across, and 4 back across.

My rep would go from a 100, in a very steep angel straight up to 205 across, until it reached .5 across, then it would take a dive down to about ??? 20, until .5 across the other way, then for the transition to concentric it would shoot up to 270, until say .2 across, then go back down to 205 until it again reached .5 this would have each time for the 6 reps.

HOW DO YOU ALL SEE IT ???


Late here, will answer the rest later

Wayne

[morderstwo]

Average force do not apply here, because the forces are not the same, meaning the concentric force is very loaded because of the sliding filament action of muscle action, but the eccentric is very under loaded because it more like a breaking action of the muscles.

The averages we have spoken about apply within the positive stroke,
then within the negative stroke.

! We are not averaging pos & neg together !

During the pos u get peak in tension from acceleration at the bottom (start of the rep) then a dip in tension due to deceleration as the weight approaches top of rep.

Similar fluctuation on the neg. A dip due to 'dropping' of the weight at the top, then a peak to bringing it to a stop at the bottom.

I do not like people putting words in my mouth.

Lulz

[douglis]

[b]Let me try and make you understand one more time.

Just pretend that your pushing your palm of your hand though water for 1m, its quite hard is it not ??? Then when you move it back in the other direction, you take your hand out of the water, its far easier now is it not ??? NOW TRY AND MORE FASTER THOUGHT THE WATER, IT GETS HARDER AND HARDER THE FASTER YOU TRY AND MOVE IT, does it not ??? This is because when a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.


Wayne

I wanted to ignore you but this is a really stupid example.Water resistance(just like air resistance) is not fixed like the resistance provided by the weight in weight lifting.So faster means harder.
I don't believe I have to expand on this any more.

Morderstwo saved my time and answered the average nonsense thing.

[N@tural1]

douglis, nwlifter morderstwo etc..

Its best to just ignore Wayne on this subject now. It's clear he doesn't get it and what more is there to say? Wayne will try and provoke a response by bringing up stupid invalid examples and claiming everyone but he is wrong but guys.. ignore it.

[Kelei]

Yeah good idea, Wayne is now on ignore.

[waynelucky2]

Back later with the rest. And thx for the input.

Nwlifter wrote:
But how do accelerations cancel ??? If in my fast rep I had constant acceleration, starting and ending at zero, how can accelerations cancel out ???

Because, If you start at zero and end at zero, as much as you accelerate, you MUST decelerate the same to equal 0.

Start at 0
Add 12
to get back to 0 you have to subtract 12

Start at 0
Add 99
to get back to 0 you have to subtract 99

Cancels, see?

Hi Ron,

Not sure where all this subtracting happens in the real World, no I am honestly not mocking you or being sarcastic, as I cannot see what you are on about yet, so stay with me.

D. said the below were not right, but all the physics people said they were right, so shall we use these for now ??? Thus I can show you how I see it and no cancelling out. Lets just ignore the jokers.

Fast rep,
Concentric force =1781 kg m/s^2
Eccentric force = 181 kg m/s^2
Overall force = 1962 kg m/s^2
Average force = 981 kg m/s^2

Slow rep,
Concentric force = 1031 kg m/s^2
Eccentric force = 968.5 kg m/s^2
Overall force 1999.5 kg m/s^2
Average force = 999.75 kg m/s^2






Fast rep,
I first am trying to accelerate 100kg from a standing start to 1m at .5 of a second, or as we need a decelerating phase, we established this to be 20% so I actually am accelerating, or trying to accelerate the weight .8m at .4 of a second and this takes 1424.8. {taken 20% off the above} Of force/strength of my muscles.


I then still need to use a force/strength to lower the weight back down, for 80% of the lowering phase I use 144.8 of force/strength from my muscles. Now at this transition, which is from eccentric to concentric, I was going to use the last 20% of the eccentric, and first 20% of the concentric for the peak tensions, sound ok ???


And for this shall we say I add 40% on the full Concentric force = 1424.8 + 40% = 1994.72. Therefore, for the whole rep starting with the eccentric I have used 2139.52F/T {force tension on the muscles}


Now I know all the above is not spot on, but if we work out the slow rep like that we can have a good comparison.




Now can you see where I do not see any cancelling out ??? As you MUST use a force/strength to lift the weight which = 1994.7F/T and then you MUST also use a force/strength to lower the weight, which = 144.8F/T.


Thus how can to forces/strengths that HAVE to be used, be cancelling each other out ??? This may work in physics, but not in the weight room, as you have to add in biomechanics. I like as I said, D. said that physics says when you lift a weight up then 9down, that both cancel each other out and now work have been done, but we all know that in the real World and in the weight room we HAVE done work, up and down.

Is this the sort of point you are trying to make, or am I missing something, if so could you please say.

Ho and what about as the eccentric is so under loaded, the slow negatives higher force/strength used, can NEVER make up for the fast concentric force/strength used now can it ???

Wayne

[waynelucky2]

douglis wrote:

Waynes wrote:
Nwlifter wrote:
Because we established a while ago, that you do not get how force works, you think it takes time for force to develope, instead of how it really is, it's always there.

Ron, this it yet another issue you have got wrong, I cannot understand why you possibly think I think like that ??? Just PLEASE tell me why you think I think like that.


Hi D.

Why is that D. ??? I did not add any averages up there ??? So could you please explane your way of thinking ???

Here's another great quote from you that clearly shows that Ron is right when he says that you think it takes time for force to develop and clearly you don't have a clue what average means:

"Fast rep = 1 second,
Total force = 1962 kg m/s^2
Average force = 981

Slow rep = 6 seconds,
Total force = 1999.5 m/s^2
Average force = 999.75

WHAT WE COULD SAY IS YOUR AVERAGE FORCE PER SECOND =999.75 DIVIDED BY 6 SECONDS = 166.625.

Wayne"



I did not work out the forces, now did I ??? NO, actually I do not know how to work them out, but would like to know, hint hint.

So as you seem to think your the Greek God of physics ROL. {only friendly leg pulling} Why do you not work out the forces needed for each rep then ??? And with the peak forces added into the fast reps. Best if we work the fast rep out from the eccentric, we lower the weight 100kg in .5 and then 20% from the end to 20% of the concentric, we have the peak forces, then the forces to keep it moving at .5 of a second up to 1m, are there any other things you need to know, we will leave up air resistance but as you know we will have to add gravity in. Your rep as its just about constant should be quite easy.

Could you do this D. ??? If not my Mothers cousin only lives a short walk away, hes a retired physicist, I could ask him ???

Wayne...you're driving me insane!We worked out the forces more than a hundred of times.
The formula is the most simple in the world.F=mg+ma.If the average acceleration is zero(a=0,starting and ending speeds zero) the formula becomes F=mg regardless the speed or if it's the negative or positive part.
So in your example for 100kgr the force is 981N.That's it...it's really simple...and the peak and lows are included.

How can the acceleration be zero ??? I am accelerating, or trying to accelerate as much as possible. Why are you just quoting the acceleration due to gravity is -9.81m/s2 ??? On this Planet ???

On the physics forms they said the above was right, D. what are you on about not ??? As HIT83 worked them out, here is it in full; please say what you think is wrong with it, as I am not understanding you ???

Posted by hitman83
I read everything but feel your misunderstanding of the above is why you're so confused. Your physics isn't as strong as you think it is and you've made several mistakes in the analysis' you've used to validate your argument.

Here's a very simplified analysis of what's going on, but it will work to explain the concept without the use of calculus. Let's assume we're comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.

Known: Mass(m) =100kg (220lbs) Acceleration(a)=?? Distance(d)=1m

First let's solve for the acceleration required to move the weight 1m in the time frames of the sets.

Calculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):
d=1/2at^2
1m=1/2*a*(0.5s)^2
a=2(1m)/(.25s^2)
a=8 m/s^2

Calculate a, to travel 1m in 2s (Raising set 2):
d=1/2at^2
1m=1/2*a*(2s)^2
a=2(1m)/(4s^2)
a=0.5 m/s^2

Calculate a, to travel 1m in 4s (Lowering set 2):
d=1/2at^2
1m=1/2*a*(4s)^2
a=2(1m)/(16s^2)
a=.125 m/s^2


Now lets solve for the forces required to accelerate the weight

Calculate the force required to raise 100kg, 1m, in 0.5s:
Sum of forces=ma
F1-mg=ma
F1-(100kg)(9.81m/s^2)=(100kg)(8m/s^2)
F1=(981+800) kg m/s^2
F1=1781 kg m/s^2 required to raise the weight 1m in 0.5s

Now compare it to the force required to raise 100kg, 1m, in 2s
Sum of forces=ma
F2-mg=ma
F2-(100kg)(9.81m/s^2)=(100kg)(0.5m/s^2)
F2=(981+50) kg m/s^2
F2=1031 kg m/s^2 required to raise the weight 1m in 2s

***As predicted, it takes more force to raise the weight 1m in 0.5s than it does to raise it in 2s. It takes F1/F2=1.73 times as much force to do so. You're right about this but nobody is disagreeing with you here

However, now let's look at what happens on the way down.

Calculate the force required to lower 100kg, 1m, in 0.5s:
Sum of forces=ma
mg-F3=ma
(100kg)(9.81m/s^2)-F3=(100kg)(8m/s^2)
F3=(981-800) kg m/s^2
F3=181 kg m/s^2 is required to lower the weight 1m in 0.5s

Now compare it to the force required to lower 100kg, 1m, in 4s
Sum of forces=ma
mg-F4=ma
(100kg)(9.81m/s^2)-F4=(100kg)(.125m/s^2)
F4=(981-12.5) kg m/s^2
F3=968.5 kg m/s^2 is required to lower the weight 1m in 4s

***Contrary to your belief, it takes MORE force to lower the weight 1m in 4s than it does to lower it in 0.5s. In fact, it take A LOT more. It takes F5/F3=5.35 times as much force to do so! That's 5.35 times more force to go slow than fast!!!!

douglis wrote:
Needless to say there's no such thing as "overall force".

Sorry but there is.

A net force, (also known as a resultant force) is a vector produced when two or more forces act upon a single object. on it calculated by vector addition of the force vectors acting upon the object. A net force can also be defined as the overall force acting on an object.

If I use a force that is more than the force you are using, and I use it again, I have used this force/strength twice, have I not, I mean if I do not use the force/strength again, how can I move the weight ???


The definition of net force has nothing to do with adding forces rep after rep or "gross force" or any other twisted thing that you fantasize.
It simply means that if the average acceleration is zero the net force is zero too or else the muscle force is equal with the weight.

When I say overall force, I mean the force from the concentric, the eccentric and the peak forces added all together. And when I say you add the forces up for the reps I do, I mean I am trying to simplify things, as say you lifted a weight up and down, and you added the forces up to be 1500, then the next day you lifted the same weight a there forces came to 1500 again, thus add them together and you get 3000, and if we do that for the slow rep all works outish. But the 1 low force slow rep is for a longer time, but its must be for a longer time because the force is lower, thats why its the same force thus tension on the muscles in the end, as its a an 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 {slow rep} force applied for a long time can produce the same momentum change as a large {fast rep} force applied briefly, because it is the product of the force and the time for which it is applied that is important. I not sure if you and Ron get the product of the force and the time for which it is applied that is important ???

Wayne

[HelloBiceps]

I ain't reading all that ****. I just stick to basics.

[waynelucky2]

douglis wrote:
Waynes wrote:

How can the acceleration be zero ??? I am accelerating, or trying to accelerate as much as possible. Why are you just quoting the acceleration due to gravity is -9.81m/s2 ??? On this Planet ???

On the physics forms they said the above was right, D. what are you on about not ??? As HIT83 worked them out, please say what you think is wrong with it, as I am not understanding you ???


No...at the physics forum they said the average acceleration is zero and the culculations are nonsense.The below quotes are not mine but belong to physicists:
"This is all non-sense though since your initial assumption was wrong. You need to recalculate your accelerations, before you can re-visit the resulting force requirements."

"Are we talking about weight lifting here?If so I don't think we can consider that we accelerate for all the way up or down.
In fact the weight always starts and ends at rest.That means the average acceleration is always zero and the force is always equal with the weight(F=mg) both in the way up and down regardless the speed."
http://www.physicsforums.com/showthr...61#post2676961

You are right they did say that.

I did not say I could accelerate all the way up, I said 80%

And why if it starts at rest and ends at rest can be the average acceleration zero ??? FIRST, are you talking about the milly second that the weight stops at each transition ??? I do need to know that first.

So you are saying if I jump in my car, or rocket, that is at rest, and I hit the gas, and watch my car/rocket accelerate to 60, then slam the brakes on until its at rest, that no acceleration has been ???

douglis wrote:
Or read what the other physicist said(Got it?The physicist...not me):
"What you need to do is to calculate the acceleration. Acceleration is the change in velocity divided by the change in time, and the velocity is the change in distance divided by the change in time. So let's go back up to the top. Your mass is moving at a constant 2 s for 1 m, so v = Δx/Δt = 1m/2s = 0.5 m/s. Now, you say that the speed is constant so the change in v is Δv = 0, so the acceleration is also 0.

D. you are the one into physics, you are going to have to do this for me, I do not know how to do that, I said that before.

I did not say my speed was constant, I said I was trying to accelerate 100kg up .8m in .4 of a second, using the other 20% for the deceleration phase, then the eccentric, will be down .8m in .4 of a second, using the other 20% of the eccentric, and 20% of the concentric for the deceleration and peak force phase, and what gets me is no one seems to be able to work this out ???

douglis wrote:
Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0."

http://www.physicsforums.com/showthr...=381643&page=3

Now why are you saying my .5/.5 is constant ??? As its such a short distance, and its only 80% so I have an extra 20% more force to use, so I would think I could easy accelerate these 6 reps, and find it quite hard to believe that I could only move them at a constant speed, that would be just about imposable, would not you think ???

Waynes wrote:
douglis wrote:
When I say overall force, I mean the force from the concentric, the eccentric and the peak forces added all together. And when I say you add the forces up for the reps I do, I mean I am trying to simplify things, as say you lifted a weight up and down, and you added the forces up to be 1500, then the next day you lifted the same weight a there forces came to 1500 again, thus add them together and you get 3000, and if we do that for the slow rep all works outish.

Wayne

I know very well that this is what you mean but it's ridiculous and has nothing to do with the meaning of net force.

Well I am glad you know what I meant and was getting at. But if you want to find the average force, you need the force of the concentric, eccentric, and the peak forces, do you not ??? Thats why I said overall force. Maybe you should look outside physics at times, as at times you just have to do that, as physics does not seem to have an easy answer to this like adding up 2 and 2 = 4 dose it ??? So we need to think outside physics or use something else.


Like when physicists use imaginary time, it is another direction of time moving at right angles to ordinary time. It is used to avoid singularities, Singularities are a problem for physicists because the known physical laws do not apply. However, they do not know if it exists, they just need things like this to make things fit.

douglis wrote:
Here's what the physicist(not me) said about net force:
"Even if you change the amount of time the average net force remains the same,

If I move a weight 100m in 1 second, to you moving it 100m in 10 seconds, I will be using more force.

douglis wrote:
namely 0. Remember, we are starting and stopping at v=0. So vi=0, vf=0, and Δv=vf-vi=0. Average acceleration is Δv/Δt, which is 0/Δt=0 regardless of Δt."
http://www.physicsforums.com/showthr...=381643&page=2

And here's the opinion of another physicist about "overall force":
"And terms like "overall force" and "strength" do not exist in Physics. If you wish to calculate something, it needs to have a definition, or you need to define it in terms of already well-defined quantities."

Well I gave them all the data, I do not see why they then could not calculate the net force or overall force, and overall force is in physics.

I mean I move a weight 100m up in 1 second, then down in 1 second, to you moving it 100m up in 10 seconds, and then down in 10 seconds, I will be using more force than you. Say mine was 500N to your 10N.

So I used 500N ok ??? So if I did the same thing the next day I would use another 500N, and you did the same, so why for arguments sake can we not say if I moved the weight 3 times up and down, I have used an accumulated 1500N ??? To your 30N ???

As every time I do lift a weight up and down, that force is constantly being used, its not like I just use it one time and it keeps lifting the weight. I have to use a large force to lift the weight, then ease of to lower the weight and use less force, then a large force again, and do this over and over. But they sort of call this mechanical work, mechanical work is the amount of energy transferred by a force acting through a distance.

But you say mechanical work does not count HOW/WHY I mean if I move something using my energy, which is a force though a distance, and over a far bigger distance and far faster, why do you think this does not count ??? Its like you trying to tell me that if I climbed a rope with a 80 pound sake on my back, 5 more times than you in the same that it means nothing, I mean I have and must have used more force {strength} and more power {work energy} but you seem to think that I have not ???

Now my go.

You still have not addressed why I said you can not add up the average forces up in this issue, as the negative is so under loaded, thus your little bit of extra force you use in the eccentric, can never make up for the lose in the concentric. And I know none of you thought of that, and nether you no Ron or anyone else has answered this.

Wayne

[waynelucky2]

douglis, nwlifter morderstwo etc..

Its best to just ignore Wayne on this subject now. It's clear he doesn't get it

You said that you did not say that there was more tension in a 2/4 to 6 at .5/.5 so it seems you on my side, so what exactly do you not think I get ???

As if you think, and I do mean think I do not get anything, please state this, otherwise, why say things like the hitters say ???

and what more is there to say? Wayne will try and provoke a response by bringing up stupid invalid examples and claiming everyone but he is wrong but guys.. ignore it.

I mean I suppose this is an invalid example ??? For example, subjects performed 96% more pull-ups in16% less time, and 145% more push-ups in 51% less time, when performing the fast repetitions than when performing repetitions with a 2/4 cadence.

And my video made me fail 55% faster.

And why are all these people failing faster ??? I expect its because they are using less overall force, strength, and there is less tension on the muscles ???


I mean come on all of you that have been debating with me, do not you not think like me, the people in the studies, and everyone else fails FAR FAR FAR faster because they use more overall force/strength, and put more overall tension on the muscles, IF NOT PLEASE SAY WHY YOU THINK THEY FAIL FASTER ???

Wayne

[waynelucky2]

I wanted to ignore you but this is a really stupid example.Water resistance(just like air resistance) is not fixed like the resistance provided by the weight in weight lifting.So faster means harder.
I don't believe I have to expand on this any more.

So are you saying that its not harder, or you not use more force/strength to move a weight 1m up in .5 of a second, to moving it 1m up in 2 seconds ??? If you are saying that, and have been all along, everyone knows your wrong, and it takes more force to get a weight up to the same point in 4 times less the time. Remember to move a 100kg at is 80% up just 1m in .5 of a second, takes huge accelerations, does it not ???

Morderstwo saved my time and answered the average nonsense thing.

Wayne

[jb4476]

Well I gave them all the data, I do not see why they then could not calculate the net force or overall force, and overall force is in physics.

As far as I'm aware, the term "overall force" does not exist in mechanics. The term net force exists. The net force on an object is simply the vector sum of the forces acting on that object. But it is only meaningful to speak of the net force acting upon an object at a given instant in time (unless the net force is constant). If we want to talk about the net force acting on an object over a given interval of time, then we write out a function F(t). We can, of course, take the integral of the force acting upon an object with respect to time, which gives the net change in momentum over the interval of time under consideration (that is effectively the integral form of Newton's Second law).

[waynelucky2]

The averages we have spoken about apply within the positive stroke,
then within the negative stroke.

! We are not averaging pos & neg together !

Well I thought you were, why did you not say before ??? But why are you not saying this ??? As both are part of the rep.

During the pos u get peak in tension from acceleration at the bottom (start of the rep) then a dip in tension due to deceleration as the weight approaches top of rep.

If as we are doing, moving out from the eccentric into the positive, then there will be the peck tensions, then the force needed to accelerate it up the 1m in .5 of as second, take off 20% of this for the deceleration phase.

So without any numbers, you are now trying to tell me that it takes exactly the same force to accelerate the same weight up .8m in .4 of a second, that it takes to move at a constant speed up 1m in 2 seconds ???

So your saying HIT83 calculations are not right, even as he says, that for simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.

Fast rep,
Concentric force =1781 kg m/s^2
Eccentric force = 181 kg m/s^2
Overall force = 1962 kg m/s^2
Average force = 981 kg m/s^2

Slow rep,
Concentric force = 1031 kg m/s^2
Eccentric force = 968.5 kg m/s^2
Overall force 1999.5 kg m/s^2
Average force = 999.75 kg m/s^2


So we could take 20% of the fast rep concentric like I said, so its 1781 – 20% = 1424.8, but then we need to add something on for the peak forces, this would be about 35% more. But if we just say its 1424.8, its still more than the slow concentric, and with this rep you would have to have a deceleration phase at say at least 5% so its 962.35.


But lets say for a second your right, and it takes ??? 1400T {tension on the muscles} for both reps to move 1m, your using a low force for a long time. Right ??? And I am using a high force for a short time, so in they average out on each other ??? Is that what you are saying ??? If so we HAVE counted your 2 seconds in the equations as its a low force for a long time right ??? Remember now we have counted your 2 seconds in this. And in my .5 of a second in the equations as its a high force for a short time right ??? Remember now we have counted your 2 seconds in this.

Now we come back to what I said I the first place, I can do 2 concentric and 2 eccentrics, to your 1 positive thus I have, yes I have to use this force or 1400T all over again, AND PLEASE DO NOT SAY YOU ARE USING THIS 1400T ALSO FOR THE WHOLE 2 SECONDS, AS THIS IS WHERE YOU ARE GOING WRONG, YOUR NOT USING 1400T FOR THE WHOLE 2 SECONDS, YOUR USING LESS THAN THIS FOR THE WHOLE 2 SECONDS, IT ONLY ADDS UP TO THIS IN THE END BECAUSE YOU ARE USING LESS FORCE A LONGER TIME THAN ME, THATS WHAT I HAVE BEEN TRYING TO SAY, I MEAN READ THIS.

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.

Let me rephrase it,

When a force is applied to a rigid body it changes the momentum of that body. A small force like in the low/small force slow rep, applied for a long time can produce the same momentum/movement change as a large force like in the high/large force in the fast rep applied briefly, because it is the product of the force and the time for which it is applied that is important.

My force was short and high, in .4 or a second I raised the force up to 1400. I did not use 1400 all the time as you seem to think, I used many different forces as I was trying to accelerate the weight as fast as possible, and as you know, acceleration is the change in velocity over time. Acceleration is the rate at which something speeds up, the acceleration is proportional to the overall force acting on it {Newton's second law}

So if we could graph this force it might go something like this, in .5 of a second, this is without the peak forces, zero, 400, 900, 1200, 1400, 400, zero transition. And if we could add that force up as tension on the muscles, it would equal 4300.

Your force was long and low for a long time, but in the end it added up to the same as my fast rep, as it was a low force for far longer. YOU FORCE WILL NEVER REACH TO 1400. It might be something like this, zero 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, 215, its 4300 again.

[b]Its like me hitting a piece of steel with one hard blow that take .5 of a second, and I bend the steel at right angles, but you hit the piece of steel with 4 easier blows that takes 2 seconds, and in the end you bend the steel also to right angles, THUS IN THE END YOU HAVE USED THE SAME OVERALL FORCE AS ME, AND PUT THE EXACT SAME TENSION ON THE STEEL.

BUT I CAN HIT THE PIECE OF STEEL SAY AT LEAST 1 MORE TIME IN THE SAME TIME FRAME AS YOU, bending the steel the exact same distance again, making it touch, THUS PUTTING TWICE THE AMOUNT OF TENSION ON THE STEEL IN THE SAME AMOUNT OF TIME AS YOU, BUT IN OUR CASE ITS THE MUSCLES WE PUT THE TENSION ON, AND I PUT THE EXACT SAME AMOUNT OF TENSION ON THE MUSCLES THAT YOU DO ONCE IN 2 SECONDS CONCENTRIC, 2 TIMES, OR ONE MORE TIME THAT YOU.

A small force like in the low/small force slow rep, applied for a long time can produce the same momentum/movement change as a large force like in the high/large force in the fast rep applied briefly, because it is the product of the force and the time for which it is applied that is important.

AS YOU DO NOT THINK YOU ARE ACTUALLY USING THE SAME AMOUNT OF FORCE AS ME PER UNIT OF TIME IN YOUR 2 SECONDS DO YOU ??? AS I THINK YOU MIGHT THINK THIS, IF SO THATS WHERE I SAID YOUR MISTAKE WAS WEEKS AGO. No your are using a lower force for a longer time, you all think you are using the same force as me over your 2 seconds, do you not ???.

Similar fluctuation on the neg. A dip due to 'dropping' of the weight at the top, then a peak to bringing it to a stop at the bottom.
Lulz

And thats without adding the peak forces in.

Wayne

[waynelucky2]

As far as I'm aware, the term "overall force" does not exist in mechanics. The term net force exists.

Hi jb4476,

A net force, (also known as a resultant force) is a vector produced when two or more forces act upon a single object. on it calculated by vector addition of the force vectors acting upon the object. A net force can also be defined as the overall force acting on an object. A net force can also be defined as the overall force acting on an object, when all the individual forces acting on the object are added together.

The net force on an object is simply the vector sum of the forces acting on that object. But it is only meaningful to speak of the net force acting upon an object at a given instant in time (unless the net force is constant).

Right.

If we want to talk about the net force acting on an object over a given interval of time, then we write out a function F(t). We can, of course, take the integral of the force acting upon an object with respect to time, which gives the net change in momentum over the interval of time under consideration (that is effectively the integral form of Newton's Second law).

Yes.

The time integral. We can make two different statements for the effect of a force acting on a particle for a certain length of time, each of which leads each to the relationship having meaning. First, we can multiply the force by the time on which it acts; secondly, we can take the scalar product of the force by the vector displacement which is courses.

http://books.google.co.uk/books?id=v...ime%2C&f=false

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. The impulse is equal to the change of momentum.

Wayne

[jb4476]

A net force, (also known as a resultant force) is a vector produced when two or more forces act upon a single object. on it calculated by vector addition of the force vectors acting upon the object. A net force can also be defined as the overall force acting on an object. A net force can also be defined as the overall force acting on an object, when all the individual forces acting on the object are added together.

Yes, yes, I know all this. I was merely pointing out that terminology "overall force" is not usually encountered, or at least, I've never encountered.

The time integral. We can make two different statements for the effect of a force acting on a particle for a certain length of time, each of which leads each to the relationship having meaning. First, we can multiply the force by the time on which it acts;

That only works when F(t) is a constant k in time, in which case the integral of F(t) resp. dt over a given interval [t0,t1] is simply k(t1-t0). Otherwise, this notion isn't even sensible, let alone meaningful.

secondly, we can take the scalar product of the force by the vector displacement which is courses.

That's something different altogether. If we let x(t) be the path of the particle, where x(t)=(x1(t),x2(t),x3(t)) in a generalized 3D coordinate system, and F(t) is the force acting on the particle at a given time then we can indeed take a path integral of F(t) resp. dx over a spatial interval [x0,x1] by taking the dot product of F with dx and integrating. This quantity, however, is the net work done on the particle, which is different from the net change in momentum, which is the integral of F(t) resp. dt over a time interval [t0,t1].

[douglis]

And why if it starts at rest and ends at rest can be the average acceleration zero ???

Because that's what the basic physics tell us.

I did not say my speed was constant, I said I was trying to accelerate 100kg up .8m in .4 of a second, using the other 20% for the deceleration phase, then the eccentric, will be down .8m in .4 of a second, using the other 20% of the eccentric, and 20% of the concentric for the deceleration and peak force phase, and what gets me is no one seems to be able to work this out ???

Sometimes I believe that you can't even read.
The physicist(not me) said:
"Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0."

Got it?Even though the speed is not constant the average acceleration is zero(even if you accelerate for 80%) and the force equal with the weight regardless the speed and the distance.

So are you saying that its not harder, or you not use more force/strength to move a weight 1m up in .5 of a second, to moving it 1m up in 2 seconds ???

Wayne

I'm saying exactly that and it's been proven to you many times by the above.
But what am I doing here?Clearly there's no hope for you.Ignored from now.

[Atavis]

LOL. Your Threads...

[waynelucky2]

Yes, yes, I know all this. I was merely pointing out that terminology "overall force" is not usually encountered, or at least, I've never encountered.

K.

That only works when F(t) is a constant k in time, in which case the integral of F(t) resp. dt over a given interval [t0,t1] is simply k(t1-t0). Otherwise, this notion isn't even sensible, let alone meaningful.

So you are saying that impulse cannot be calculated on a object thats accelerating ??? As I know you can calculate the 2 forces like on my video below, first I am trying to accelerate the weight as fast as possible, and second I am trying to move the concentric at a constant slow speed. You take the frames from the video somehow, but I do not know how to do this.

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

That's something different altogether. If we let x(t) be the path of the particle, where x(t)=(x1(t),x2(t),x3(t)) in a generalized 3D coordinate system, and F(t) is the force acting on the particle at a given time then we can indeed take a path integral of F(t) resp. dx over a spatial interval [x0,x1] by taking the dot product of F with dx and integrating. This quantity, however, is the net work done on the particle, which is different from the net change in momentum, which is the integral of F(t) resp. dt over a time interval [t0,t1].

That a bit over my head sorry.

The main of the debate is, 2 men the same strength use 80% first moves the weight for 1 rep at 2/4 = 6 seconds. Second moves the weight at .5/.5 for 6 reps = 6 seconds. Which puts the most overall tension on the muscles, or most tension on the muscles, and which in turn uses the most overall force/strength.

We need to work out the slow reps forces, starting with the eccentric to the concentric, and the same for the 6 fast reps. I am starting at the eccentric, as we need to add in the peak forces at the transition from eccentric to concentric on the fast reps. We roughly said that these peak tensions would be happening from the last 20% of the eccentric to the first 20% of the concentric.

HIT83 worked these out, but without the peak forces, and he assumed the weight accelerates 100% of the way up and down for both sets.

Fast rep,
Concentric force =1781 kg m/s^2
Eccentric force = 181 kg m/s^2
Overall force = 1962 kg m/s^2
Average force = 981 kg m/s^2

Slow rep,
Concentric force = 1031 kg m/s^2
Eccentric force = 968.5 kg m/s^2
Overall force 1999.5 kg m/s^2
Average force = 999.75 kg m/s^2

If you note both of the concentric and eccentric reps had an average force the same, but I pointed out that as the eccentric is so under loaded, there is no way that the slow eccentrics more force used can make up the 70% extra force the fast rep used.

BUT, thats for just 1 rep at 2/4 and 1 rep at .5/.5, there are another 5 reps to add on for the fast reps, thats why I put this in.

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. The impulse is equal to the change of momentum.

Some of the others seem to think, that because their rep is longer they have more time under tension, yes they have more time under tension, but their longer time under tension is a FAR lower tension that I am creating in my large force applied briefly, thats what they do not seem to understand. Then add in my other 5 reps with large force applied briefly, and in my opinion you have far far far more tension in the faster reps = more overall force.

And here is my laymans terms is what is happening.

BUT I CAN HIT THE PIECE OF STEEL SAY AT LEAST 1 MORE TIME IN THE SAME TIME FRAME AS YOU, bending the steel the exact same distance again, making it touch, THUS PUTTING TWICE THE AMOUNT OF TENSION ON THE STEEL IN THE SAME AMOUNT OF TIME AS YOU, BUT IN OUR CASE ITS THE MUSCLES WE PUT THE TENSION ON, AND I PUT THE EXACT SAME AMOUNT OF TENSION ON THE MUSCLES THAT YOU DO ONCE IN 2 SECONDS CONCENTRIC, 2 TIMES, OR ONE MORE TIME THAT YOU.

Sorry for my long posts, I have m=been told before.

LOL. Your Threads...

What big ears you have when you eat Atavis ROL, just frendeliy leg pulling.

Wayne

[waynelucky2]

Because that's what the basic physics tell us.

Look D. I am training in the real World. You once told me that if you go up then down that cancels each other out, and not work has been done, but every one that reps up and then down actually knows that work has been done in both directions.

So lets stop talking like this please, its not helping the debate.

If I jump in my can, and hit the gas and reach 60mph in 4.5 seconds, and you jump in your car and you reach 60mph in 10 seconds, I have accelerated faster than you, so lets keep to real World applications, meaning that when you do a concentric at 2 seconds you are only accelerating for say the first 20% then you move at a basically constant speed, then use the last 190% to decelerate. I on the other hand most probably accelerate for 80% of the rep and use the last 20% to decelerate.

So lets keep the debate as what happens in the real World, as most probably this is why you and Ron and some others are having difficulty in that I am saying.

Sometimes I believe that you can't even read.
The physicist(not me) said:
"Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0."

Got it?Even though the speed is not constant the average acceleration is zero(even if you accelerate for 80%) and the force equal with the weight regardless the speed and the distance.


I'm saying exactly that and it's been proven to you many times by the above.
But what am I doing here?Clearly there's no hope for you.Ignored from now.

Look, I can go from a standing start to an ending stop, both ends at zero. I first hit the gas in my car, and see with my own eyes that I can accelerate up to 178mph if I wanted to, and thats sort of the same with planes, people who can climb ropes very fast, 100m sprinters for the first half of the race, and in reps when I move multiple times in my .5/.5 reps, just after the concentric, I am accelerating for as much as I can.

In physics, and more specifically kinematics, acceleration is the change in velocity over time. Acceleration is the rate at which something speeds up or slows down.

And note what Jeff says, and I am, am I not generating a force far greater than my 80% I am trying to use 100% thats 20% more than 80%

Jeff Pinter wrote;
Is there offloading during small portions of the ROM? Yes. Are there forces greater than the resistance during major portions of the ROM? Yes. we are accelerating the load (typically the first 80 - 90 percent of the lift in most cases). That is, as long as we are accelerating the load, we MUST be generating a force greater than the load itself. To say otherwise would violate Newton's Laws. And it does not matter what the speed is...as long as the load is accelerating the force must be greater than the load.

The only possible caveat here is the issue of "terminal velocity". That is, if the load is light enough then the speed can become so great that it exceeds terminal velocity, and the force generated is reduced due to physiological reasons. I'm not sure what this speed is, or what %RM the load would be in this case, but BIO probably has a good idea of the details.

It should be noted that the same type of reasoning applies to the case where the speed is constant. In this case the force generated must exactly match the load...and again it does not matter what the speed is (in either the concentric or eccentric for that matter).

Jeff

As John, and I did say something very similar from the very start of the thread in 1901, stop playing games please.

Wayne

[waynelucky2]

its boring without you brah ROL, thx for that, made my day.

Im back, bigger, meaner and duller, sorry more intelligent.

Right back in, had my password reset.

Hope this is ok Ron, to carry on the debate here, as the more input the better.

Hi Ron, and thx for the explanation. You should ask John on the machines he has invented, I really mean that.



Right, if your right, and I say you are, this gets me back to what I thought before, as when HIT83 put down those equations, I thought we/I was wrong from the last debate, before you got here, yes believe it or not this has been going on, on the average force thing for about 6 months earlier, and the rep speed thing about 4 years now, so no need to worry if your going mad, as you have plenty of time rest, and the rest of us are now mad ROL.



So as Thought before and worked out before with the help of Jeff Pinter the physicist, the average force are the same, but read on.

Nwlifter wrote:

You still don't get how average force works with accelerations. Why don't you just read this one thing,just once. You keep complaining I don't show you why, but everytime I do, you skip my posts. My bet is that you'll just skip this and write a bunch of equations again, forgetting the basic laws of physics I'm about to put here. But what the heck, I'm just waking up, drinking my coffee, might as well for the fun of it.



If something starts at zero and ends at zero, that means it had to slow down exactly as much as it sped up.



This means, if you take a 100lb barbell and apply 150 to accelerate it, then the deceleration HAS to be some number lower then 100 for the correct time period for it to end up at zero. Since you accelerate for longer, the deceleration force must be very low to get the barbell to stop so it doesn't smack you in the face at the end. Knowing this law of physics means, we don't even have to bother calculating all that, as we KNOW the lower force/time during deceleration is the same amount as the acceleration force/time during acceleration.

But I do not think the time for the deceleration is the same as for the acceleration ??? I would say on the second to last reps, thats coming out of the eccentric to the concentric, that the acceleration, or trying to accelerate {lets say for now I can accelerate the weight 80% of the way, unless you do not think so} it, will be about 80% {actually is seems when I am repping I am accelerating the weight up to 99% then immediately change direction} and then use the 20% for the deceleration and transition.

Nwlifter wrote:

This leaves us with the load as the average force for concentric. Same with eccentric. If you drop it fast, your applying almost zero force, but a high peak force for a quick split second at the end to stop it. This equals out to the same average force as a slower eccentric with more force for longer. This also leaves both fast and slow with an average force equal to the load.

This is where it gets tricky. As you just seemed to say that both the fast and slow rep have the same average forces, thus tensions to the muscles equals the same, but not the same per unit of time, thus it cannot be the same per unit of time can it ??? As your rep = 3/3 and my rep = .5/.5



So are you actually saying both reps have the same average force, but not per unit of time ??? As your rep on the concentric must be producing less for per unit of time, as yours lasts 3 seconds and mine .5 of a second, thus if you was using the same force per unit of time, more force per unit of time, the average force for both the concentric would not be the same would it ??? Lets just chat concentric for now.



This example I gave above might help you see what I am trying to get at.



BUT I CAN HIT THE PIECE OF STEEL SAY AT LEAST 1 MORE TIME IN THE SAME TIME FRAME AS YOU, bending the steel the exact same distance again, making it touch, THUS PUTTING TWICE THE AMOUNT OF TENSION ON THE STEEL IN THE SAME AMOUNT OF TIME AS YOU, BUT IN OUR CASE ITS THE MUSCLES WE PUT THE TENSION ON, AND I PUT THE EXACT SAME AMOUNT OF TENSION ON THE MUSCLES THAT YOU DO ONCE IN 2 SECONDS CONCENTRIC, 2 TIMES, OR ONE MORE TIME THAT YOU.



[b]Therefore, my next questions are,



1,

Is your rep using less, the same or more force per unit of time ???



2,

Do you think that as my 1 rep is .5/.5 and covers a distance of 1m up and 1m down in 1 second =2m, that as your rep is say 3/3 and you move the same weight only .16m up and .16 down in 1 second = .32m needs the same force output and thus tension from the muscles, and if so why. Why do you think your rep puts out more force per unit of time, when I have moved the weight over 500% more in the same time frame ???



3,

Do you think that my 6 reps at .5/.5 that covers a distance of 6m up and 6m down in 6 seconds = 12m, and that as your rep is say 3/3 and you move the same weight only 1m up and 1m down in 3 seconds = 2m needs the same force output and thus tension from the muscles, and if so why. Why do you think your rep puts out more force per unit of time, when I have moved the weight over 500% more in the same time frame ???

Nwlifter wrote:

Don't say 'why do you say this' or 'how can you say this', honestly, it doesn't matter if you agree or if you like this, it's the way it works. I wish I could change some of the laws of physics too, but just wanting them changed, doesn't change them. The reason I don't bother with calculations, are they are not needed once you understand these laws. It's the reason even Enoka and all the physics geniuses told you the same, and why we end up with average force and time.

But the times of a .5/.5 and 3/3 are different.

Nwlifter wrote:

You burn a lot of energy for a shot time and that makes the bar go faster and farther, but you burn a lot less while it's traveling.

Best get something straight here. When I am pushing up on a rep, there is NO offloading at all, there is nothing helping me lift the weight lift offloading, and even if I stopped with open arms at the end of the rep, the weight would not travel out of my open hand, as of the biomechanical advantages, and disadvantages. I mean I do not just use the first 20% of the peak forces, and let these forces keep on moving it, I MUST push constantly hard all of the time to keep it moving this fast..



As Jeff said,

we are accelerating the load (typically the first 80 - 90 percent of the lift in most cases). That is, as long as we are accelerating the load, we MUST be generating a force greater than the load itself. To say otherwise would violate Newton's Laws. And it does not matter what the speed is...as long as the load is accelerating the force must be greater than the load.

Nwlifter wrote:

Just like a bullet, all the energy is spent immediately to send it on it's way, then it goes very fast and very far with no more help.

But as I said, this does not happen with 80% I have to use the most highest of effort I can to keep the weight moving that fast, its very hard to try to keep doing this rep after rep.



Momentum, or as I like to say, off loading may be an issue, if the weight is to light, and you are not accelerating the weight enough.

Acceleration is not momentum, its the opposite of momentum, acceleration requires more force/strength, where momentum requires less force/strength.

Nwlifter wrote:

If you could just slowly use the same energy, the bullet would go slower and not as far from that same amount of energy. But it would still be the same amount.

YES, thats what I am talking about, as I am actually using more force/strength per unit of time, and more energy per unit of time, thats why I keep saying before, every activity you do faster you use more energy, twice as fast twice as much energy, six times as fast 6 times more energy.



The point is what as you said the bullet from the gun, my guns energy or force/strength is not used straight always, its being used over and over in my 5 more reps that yours.



1,

Normal gun uses a force, force makes bullet move a certain distance, then carries on, on its own.



2,

Slow gun, you use low force, but for a longer time, say 2 seconds, {that sounds like a good number ROL,} bullet goes 100m.



3,

Fast gun, you use high force for a short time, .5 of a second, bullet goes 100m.

As both bullets go the same distance, they use the same average force, thus tension on the muscles.



4,

Automatic machine fast gun.

Again, you use high force for a short time, but lots of times, .5 of a second, bullet goes 100m, .5 of a second, bullet goes 100m, .5 of a second, bullet goes 100m, .5 of a second, bullet goes 100m, .5 of a second, bullet goes 100m, .5 of a second, bullet goes 100m.



So this time you use high force for a short time, but 6 times consecutively, thus if it was the same bullet, and it is really, it would have gone 600m and put 6 times more tension on the muscles than 1 fast rep.



5,

Actually it might not work exactly like that, as your 2 seconds going up is still quite fast, so my 1 fast rep at .5, might not be putting the exact same tension on the muscles, but then again it should, as you yourself said, they are the same average forces, and even thou yours is longer, yours has a low force output per time, but mine has a shorter time, but the highest force output possible.



And if you stretched your rep to say 4 seconds up, you then would be using even less force, but over a longer period



As you yourself said, If you could just slowly use the same energy, the bullet would go slower and not as far from that same amount of energy. But it would still be the same amount.



YES, the same amount of energy used for different times, = less force used for more time, its the same as the running examples, you run slower using less force, but run like this over a longer time, and you will in the end cover the same distance, as the fast runner using more force but in a shorter time, but the fast runners, runs on and on and on, THIS MUST FROM THE EXAMPLES BELOW USE MORE ENERGY, WORK {POWER} AND MORE FORCE/STRENGTH.



AS MORE {ENERGY CALORIES} ARE USED IN THE FASTER {SHORTER DURATION} RUN {REP}





1,

Time ran 1 hour, Bodyweight 130 pounds Running, 10 mph (6 min mile) 944 {calories} 10 mile ran 944 {calories}

Work done 10 miles = 10mph = 10 miles in 1 hour = 944 {energy calories}





2,

Time ran 1 hour, Bodyweight 130 Running, 5 mph (12 min mile) 472 {calories}



Work done 10 miles = 5mph = 10 miles in 2 hours = 944 {energy calories}



http://www.nutristrategy.com/activitylist3.htm



Its like what Kelei said over at BB.com, Wayne you would be right if you was just talking the concentric, as you move the weight 6m to their 1m, just you are generating more force per unit of time, but then again you lose it on the eccentric. But he did not realise that the eccentric is so under loaded, you can NOT in any way make up the lost forces.



Wayne

[waynelucky2]

Nwlifter wrote:

OK, there I gave you again, a long explanation. You either won't read it,

I love to read all you and the rest write.

Nwlifter wrote:

or you will just argue with it.

Your right there, but this is a debate after all, thus I am debating the facts.

Nwlifter wrote:

But I am just writing the facts from the laws of physics.

Yes you are writing facts of the laws of physics, and I agree with them, but think you and D. as explained above forgot to add in that are the average forces are the same for 1 rep at .5/.5 and for 1 rep at 2/4 the tensions MUST be the same on the muscles right ??? So if they are the same, this means that I put another 5 more exact same tensions on the muscles in the same time frame. And its NOT like you were thinking that your more under tension, you more under LESS tension are you not ???



You must be, BECAUSE IN THE SAME TIME FRAME, AS MY .5 OF A SECOND, WHERE I MOVE THE WEIGHT 1M, YOU HAVE ONLY MOVED IT ONLY .16 ON A METER, THAT MEANS I HAVE MOVED MY WEIGHT 500% MORE IN THE SAME TIME FRAME, AND THAT TAKES MORE FORCE/STRENGTH.



This is why I keep saying, 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. The impulse is equal to the change of momentum.

Nwlifter wrote:

I didn't invent these facts, the universe did. People like Newton revealed them, so you must take this up with either him or the universe

Fuuny old things these laws, I think its sort of all or nothing, meaning basically these laws have to work like this, or we and 99.9% of things would just not be here.

Nwlifter wrote:

You really really need to just research these basic concepts. Seriously, before you try to find why they don't fit your hypothesis, just read about accelerations and average force. Get these basic laws ingrained. Before you know it, you will see the errors in your force ideas.

Yes your right there, but as you will see, I do not find I am in errors, unless you think you need the same force to move a 100kg .16 of a meter to 1m in the same time frame ??? As a force is a push or pull that can cause an object with mass to change its velocity to accelerate, and in the faster rep that is what is happening, a larger force is changing the velocity of the weight, and making it accelerate more the other weight, which is using less force per unit of time.


Wayne

[waynelucky2]

Nwlifter wrote:
Force is not 'per unit of time' that's what's throwing you. Force happens 'over time', like sunlight, it doesn't require time to 'mount up'.

You and D. keep saying this, but I know that, I have always known this, and I am wondering why you keep saying it ???

Maybe because its me saying per unit of time ??? But what I mean by this is your concentric is producing less force per unit of time, not average, but per unit of time.

Here is where I think your slipping up.

Do you think you are producing the same amount of force with your 3 second concentric to my .5 concentric ??? And if so why ??? As you say we both produce the same average force, so the force/strength thus tension on the muscles must be the exact same on the muscles yes ??? But then before you have said no, because your rep is longer, but thats where you slip up, if the average forces are the same, TIME has all ready been added in has it not, as if it was not, the average forces would be different, but they are not, they are the same. Thus, you cannot turn around and say your rep are under more time, as my rep was of a HIGHER force.

This is why I keep saying this;

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. The impulse is equal to the change of momentum.

Your slow rep is for a long time and only produces a low force, but mine is for a short time, but produces a high force, so in the end both forces are the same. This is what you and D. keep missing, as it does not matter that your rep was for longer, IT WAS LONGER BUT OF A LOWER FORCE, mine was shorter but for a higher force, thus I have to say again, with time added in, both forces after time were the same.

Is your rep using less, the same or more force per unit of time ???

Nwlifter wrote:
Absolutely the same AVERAGE per unit of time. Less at first, more later.

THIS, is where you are wrong. How can your rep be using the same force per unit of time ??? As if it was, you would pick up the weight in .5 of a second the same as I have, as explained above; you are NOT using the same amount of force per unit of time. Your force is lower, thats why it takes you 2 seconds to lift the weight.

Or lets put it another way, and I said this before, I do my high force rep for .5 of a second, and I move the 100 pounds 1m, you do your low force rep for the same, .5 of a second, but you only move the 100kg .16 of a meter.

SO THIS MEANS THAT PER UNIT OF TIME, YOUR LOW FORCE REP IS FAR LOWER OUTPUT THAN MINE. Or just compeered these 2 reps and tell me which has the highest force output thus tension on the muscles, and why ??? My high force rep for .5 of a second, and I move the 100 pounds 1m, you do your low force rep for the same, .5 of a second, but you only move the 100kg .16 of a meter ???

But when and when only you do your low force rep for 3 seconds does it add up to the same force as mine. You and D. seem to think at times, that because your 1 concentric = 3 seconds and mine .5 of a second, you muscles are producing the same force as me and for longer, but you are actually producing less force for longer, thus it makes no difference your under tension longer, as its LOWER tension for longer.

Ha that why you think I think force adds up over time, no I do not mean it that way its like this, I have said this before, and its the same in what I am on about in the reps, yours is a small force applied for a long time, which 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. The impulse is equal to the change of momentum.

I CAN HIT THE PIECE OF STEEL SAY AT LEAST 1 MORE TIME IN THE SAME TIME FRAME AS YOU, bending the steel the exact same distance again, making it touch, THUS PUTTING TWICE THE AMOUNT OF TENSION ON THE STEEL IN THE SAME AMOUNT OF TIME AS YOU, BUT IN OUR CASE ITS THE MUSCLES WE PUT THE TENSION ON, AND I PUT THE EXACT SAME AMOUNT OF TENSION ON THE MUSCLES THAT YOU DO ONCE IN 2 SECONDS CONCENTRIC, 2 TIMES, OR ONE MORE TIME THAT YOU.

You and D. are not getting that bit, I know force does not need time to mount up, but because your slow rep is of a lower force than mine, IT DOES NEED MORE TIME TO MOVE THE SAME 100KG WEIGHT THE SAME DISTANCE, THUS IT SORT OF DOES NEED TIME TO MOUNT UP.
Its like a fast car producing high force, it can pull a 100kg for 1 mile in 1 minute, the other car thats using a low force, can also pull the 100kg 1 mile, but as its only pushing out a low force, its going to take 6 minutes to do it.

Nwlifter wrote:
That's where average force comes in, you use more at first, less later and much less on the eccentric. The slow rep uses less at first, but more later and more on the eccentric. That's why it's the same average.

You did not really answer the question. It was, do you think that as my 1 rep is .5/.5 and covers a distance of 1m up and 1m down in 1 second =2m, and that as your rep is say 3/3 and you move the same weight only .16m up and .16 down in 1 second = .32m.

Nwlifter wrote:
You have lots of peak force at first which makes the load move fast,

Yes.

Nwlifter wrote:
then it coasts for the rest of the rep. Like shooting a bullet from a gun.

What do you mean by coast ??? As there is NO way that I just push from the start and the weight moves on its own, it does NOT happen, I am pushing with all my might, all my force from the start unto the end, and at the end I make my muscles work in reverse to do the eccentric. Its NOT like I throw the weight from the first 20% of the rep, then it moves on its own, I mean thats impossible, as if at any TIME I stop pushing and do not use and force, the weight will immediately stop and fall back down. Yes when you give an object constant acceleration, it will start to speed/move on its own, but in your examples it would be so small, its not worth taking into account.

Fun site, looks like both our reps, your using a low for a longer time, mine using a high force for a shorter time.

http://id.mind.net/~zona/mstm/physic...work/work.html

Nwlifter wrote:
A slow rep is like a tractor moving slow with medium force the whole time.

Right.

Nwlifter wrote:
I kept trying to use isometrics to get you to see. Please think about this. Lets say your reps have an average force of 100lbs and you do 6 of them. That's 100lbs for 6 seconds right?

Here is where we have to go outside physics I think, as if I use a 100 pounds of force on my first rep, I also use it on the next 5 right ??? However, this is not telling us what we want to know, we are not looking at how much power, is used, as power is the rate at which work is performed or energy is converted. And we are not looking at how much work, as work is the amount of energy transferred by a the force acting through a distance.

We want the actual overall force/strength that was used. You are taking the full weight of the 100kg just like me, however I am putting more concentric force = faster movement into the weight for the same amount of time, so this is actually saying it, I am putting more concentric force, more force, I am putting out more force. And I am moving it 6 times faster than you. Yes the in the eccentrics you are doing this to me, but its so under loaded, it cannot in anyway make it up.

Nwlifter wrote:
What if you just hold a 100lb barbell for 6 seconds. There is zero movement right? Zero distance right? But you still had 100lbs of force for 6 seconds. See? Or what about if you held 120lbs for 6 seconds, that's more force then if you did 6 of your reps at 100lbs force. And it had ZERO distance or movement. Think about that, really think about that .. please

Yes get what your saying there, but lowering a weight down use the least force, then a little more holing it, then a little move moving it, then and little more moving it faster, and the faster you move it the more force is needed.

Wayne

[waynelucky2]

Nwlifter wrote:
That is the very reason we use average forces. Yes, with fast reps the eccentric is veryunloaded, but with slow reps it's NOT unloaded. That's why your training has higher PEAK forces. PEAK force means you have short periods of high force and other periods of low force.

Hi Ron,
With the eccentric, you do not have the sliding filament system; you have the opposite, more like a braking system right ??? And this braking system is FAR more efficient, meaning you use less muscle fibers and can lower a lager load under control.

So this means that when, and always when your using a weight for normal training, which is basically for the concentric strength and portion of the lift, when EVER you are in the eccentric portion of the lift, where you do it fast or slow, its totally under loaded.

Yes the faster rep makes it even more under loaded, but as I said, as SOON as you start lowering this weight its under loaded for the eccentric. Its like using 40 to 50% on the concentric.

Ron you should know this, unless you have not studied this, its under loaded and the little you make back can in no way at all make up for what you lose in the concentric. Why do you think that when using 80% for the concentric, then when you move into the eccentric which is like using 40 to 50% that 40 to 50% is not under loaded ??? As using 40 to 50% for any exercise, is VERY under loaded.

Wayne wrote
Nwlifter wrote:
But I do not think the time for the deceleration is the same as for the acceleration ??? I would say on the second to last reps, thats coming out of the eccentric to the concentric, that the acceleration, or trying to accelerate {lets say for now I can accelerate the weight 80% of the way, unless you do not think so} it, will be about 80% {actually is seems when I am repping I am accelerating the weight up to 99% then immediately change direction} and then use the 20% for the deceleration and transition.

The time isn't the same, it doesn't have to be. Way less force takes less time to slow it down. So you have .3 seconds of very high force and .2 seconds of almost zero on the concentric.

Not sure what you mean by, the time is not the same, it does not have to be ???
Yes way less force takes less time to slow down.
If I am accelerating the weight up 80% I have .4 seconds of very high force, and .1 of deceleration to zero force. But your rep must have this as well, it will just be for a shorter distance and time.

Nwlifter wrote:
I know what your getting at, but the thing is, if the load stops or reverses direction it means it HAD to decelerate the same as it accelerated. If it decelerated less, it would still be going forward.

You and D. keep saying this, but I know that, I have always known this, and I am wondering why you keep saying it ???

Nwlifter wrote:
Because you keep talking about total force. It's like saying 'total sunlight' outside. Or total heat. If something is increases to 100 degrees 5 times, you don't have a 'total temperature of 500'. same with force.

No I know you will not have a total temperature of 500, but with your slow rep you are saying outside for say 3 minutes at 80c but I am going outside at 100c but for 6 times at 30 seconds each, so each time I take more total temperature, as I am taking 100c every time, and for the same length of time as you taking the lower temperature of 80c.

You and D. are not getting that bit, I know force does not need time to mount up, but because your slow rep is of a lower force than mine, IT DOES NEED MORE TIME TO MOVE THE SAME 100KG WEIGHT THE SAME DISTANCE, THUS IT SORT OF DOES NEED TIME TO MOUNT UP.

Nwlifter wrote:
No, remember I explained that with the bullet example?

No, its NOT like your bullet example, your low force needs a longer time to move the weight does it not ???

Its like a fast car producing high force, it can pull a 100kg for 1 mile in 1 minute, the other car thats using a low force, can also pull the 100kg 1 mile, but as its only pushing out a low force, its going to take 6 minutes to do it.

Nwlifter wrote:
But if both go a mile, and stop at the end, the fast car accelerates for a while, then has to let off the gas for a while. The slow car and keep the force even for a lot longer.

But thats wrong, I do NOT let of the gas, yours is a wrong and not right example. The slow rep uses LESS force for the same amount of time, well just about the same time, as I might need 20% to decelerate, you might need 10% But then my peak forces make up and overtake that by a huge margin.

As the eccentric is so under loaded, its like using 40 to 50% for the concentric. Why do you say you push on the brick for 6 seconds ???

Nwlifter wrote:
No, the eccentric with slow is very close to the same loading as the concentric. Whereever that info. came from, they are flat out wrong and do NOT understand physics at all. You have read physics, you know then that if you do an eccentric at a constant speed, the force is equal to the load to maintain that speed.

Its NOT close to the same loading, as, as soon as you hold or lower a weight, the braking system starts, which as I explained is far more efficient, thus the load is under loaded. I am surprised you do not know this ???

Physics does not work here now does it, how can you even say you can you physics in its right way here in this example ??? you have to turn to biomechanics and muscle physiology With the eccentric, you do not have the sliding filament system; you have the opposite, more like a braking system right ??? And this braking system is FAR more efficient, meaning you use less muscle fibers and can lower a lager load under control.

Wayne

[all pro]

Date: Thu, 6 Nov 1997 01:15:56 -0700
From: _MIKE_ <mremple@EXCAL.NET>
Subject: Re: Louis Simmons methods


From: Denilson P. da Costa
>Dear fiends, Im writing again , as Ive got no answers the first time

Well I am not surprised you got no answers, because explaining Louie
Simmon's program is a challenge at the best of times - but it sucks getting
zero replies, so I will try. When I read on Deepsquatters site about his
Louie experience, I thought it couldn't be that complicated. I got the
video's thinking that I would totally understand the program afterwards. Not
quite. It raised as many questions as it answered.

Part of the problem in (re)constucting his philosphy is that he doesn't
fully explain why he has you doing certain things. One thing I have learned
is that "a strong man is strong in the BACK of his body" (he says this
alot). Another thing is that his system evovles all of the time. So you
might hear him talk about squating at a certain percentage on his video, a
different % in his PLUSA articles and, yet a different one again from
someone who just trained with him. So I will try to give you a brief
overview of the tennets of his philosophy - as I understand them [anyone
else can correct me]

First of all he uses lighter weights for 1 day each of squatting and
benching to develop explosive speed (aka compensatory acceleration)- all
squats are done off of a box using suit bottoms only. On the other days he
uses similar excercises to main lifts with really heavy weight (many to max
singles). You alternate the heavy excercises every 3-4 weeks to prevent
getting stale. He specializes in working bodyparts and ranges of motion
where a Squat suit and bench shirt don't help (ie lockout)- good unless you
lift RAW.

On your Speed bench day you use 55% of your 1RM, 60% of your shirtless max.
You do 8 sets of 3 reps, trying to explode from the bottom taking 45 seconds
of rest - no shirt. Then throw a couple heavy singles at the end - I *think*
to asess how heavy weight feels (around 70-85%).You also do lot's of tricep
work, with a couple of sets for Lats, rotator cuffs and shoulders. You test
your bench every 4th week and adjust to your near weight at 55 or 60%, he
tries to enter a competion every 8th week - so you would probably test if
you weren't.

On the Speed squat day you use 50-60%*, progressing by about 2.5% per week
(only if you maintain speed till the last set) doing 10-12 sets of 2 with
45-60 second rests. You do these off of 4 box heights, rotating box height
after 3-4 weeks. Using suit bottoms, you sit down on box for a moment (matt
on top for slight support) and explode up. You use 50 % of your 1RM at that
height and go up 2.5% per week, until the last week were you take a weight
(test for 1RM). This is the 1RM you will use to figure out your percentages
next time you use that height. Then you change heights and do 3-4 weeks
again. Test your normal Squat every 8th week. The heights are 2" above,
paralel, 1 " below, and far below paralel. You would usually do a couple
sets of HEAVY good morning, wit heavy ab work, heavy sidebends, and calf and
or neck work.

On the Heavy Bench day, you do similar excercises to the bench, like decline
bench, close grip incline, Floor presses, close grip bench off rack
(starting at 2" adding an inch each week) - All of these are done to a max
single for 3-4 weeks, then you change to a new excercise. With some heavy
tricep work (some partials -1/2 and 1/4 dips), Shoulder work, lat work and
some rotator cuff work.

Heavy Squat day - On this day you do your Deadlifts ( mostly off of racks
4-6" from lock out, or off of boards -depending upon your weakness (start or
finish))
After this you would do heavy Squat type excercises like Zercher squats,
orkneeling quats (6-10), or Belt squats. Followed by heavy ab work, lighter
errector work, optional biceps. When you do have a contest coming up try
15x1 @65%, next week 15x1 @ 70%, 12x1 @ 75%, 8x1 @ 80%, 6x1 @ 85%, do
contest - all of these have 50 sec rests!

-Phew! that is the short version!

[all pro]

Louie Simmons
Have you ever thought that ideas should be reversed? What if we
were born with the wisdom and the reasoning of a 65-year-old? We
would make more right decisions and possibly stay out of trouble and
make the most of our time while we’re young. Then as we get older,
we could start thrill-chasing and taking chances that instinctively we
would never consider. This of course would lead us to live by the
code of the poet Dylan Thomas, “Do not go gentle into that good
night, old age should burn and rage at close of day; rage, rage
against the dying of the light” (1940). I try to live as Dylan Thomas put
into words, and yes, I have the scars to prove it. But of course we can
never live our lives in reverse.
There are a few that have read the exploits of a person who has been
said to have made great progress doing let’s say eccentrics. But were
there other factors involved in their training? I have read several
articles by sports experts around the globe, yet none have conclusive
evidence that eccentrics work. Mel Siff in Supertraining explains
eccentrics as action in which the proximal and distal muscle
attachments move away from one another. Eccentric work uses
significantly less energy than concentric work. When doing slow
eccentrics with large loads, there is no reason to associate these
advantages with the possibility of developing the ability to move
quickly and powerfully in concentric work. Lowering weights slowly
builds larger muscles for body building but will not assist concentric
actions.
Let’s look at depth jumps. One is accelerating close to 9.8
meters/second/second when one lands on the floor. Everyone knows
they work, so why would you lower a weight at 0.1 or 0.2
meters/second and destroy the stretch reflex? In the book Science of
Sports Training, T. Kurz states that some athletes can lower 10% to
60% more than they can overcome concentrically.
All this said, slow eccentrics have no place in powerlifting. To build
larger muscles, yes. If you want to become very sore, yes. What does
it matter if you can lower 60% more than you can raise? If I recall
correctly, you must raise the bar from the floor in a clean, snatch, and
deadlift. In the bench press after the bar is lowered to the chest, after
the pause, you must raise the bar to completion. The squat is similar.
After lowering to parallel, you must, for a fraction of a second, hold
the weight statically and then raise to completion. If you load the bar
with 60% more than your best squat of, say, 1000 pounds, the total
bar weight would be 1600 pounds. Does this sound like a good idea
to you?
A weight that feels fine at the top becomes much too heavy in the
bottom. How can a bar weight be perfect at the top and also at the
bottom? It can’t. That’s precisely why we use a combination of bar
weight and bands or chains to accommodate resistance, causing a
reactive method. But that’s another story.
The key to eccentric success is overspeed eccentrics. Hopefully, you
already know that force equals mass times acceleration. But the force
is almost always connected to concentric movements. What about
eccentric work? Light weights can be lowered with greater
acceleration than heavy weights, just as in concentric movements.
You must understand that the largest force may not always be
associated with the heaviest loads. Jump-Stretch bands can produce
much greater acceleration properties by pulling the bar down by
means of great tension. This causes overspeed eccentrics, adding to
kinetic energy. If one could triple the bar speed on the eccentric
phase, it would produce 9 times the kinetic energy. In a simple action,
such as a depth jump, a 200-pound man jumping off a platform of 10
feet will produce many times his body weight.
Force will decrease on concentric movements due to deceleration,
while force will increase on eccentric movements due to acceleration.
This leads us to a problem that many never consider: optimal
eccentrics. How much muscle tension should one use on eccentric
movements? If it is true that an individual can lower 60% more than
they can raise, then applying 100% of his eccentric strength, he could
hold the bar at any position. However, he would destroy the stretch
reflex with heavy weights. Overcoming inertia is done with light
weights with a fast eccentric stop to build reversal strength. This is
ballistic training. The bar never touches the chest. One to three
inches off the chest is recommended. If you drop a 10-pound rock
and a 1-pound rock from the same distance, they hit the ground at
the same time. But as one lowers a heavy weight, as the weight
increases, the eccentric phase slows down. This is due to too much
eccentric muscle action.
There are optimal bar speeds for velocity training, where the objective
is fast movements with light loads, and for force training, where the
bar speed may be zero or very low, to produce maximum force.
There are optimal bar speeds for these efforts, and they are always
measured in concentric movements. So should there not be an
optimal eccentric speed? Of course, it is just that no one has ever
considered it. Until now.
How can you learn to optimally lower heavy loads in the pressing and
squatting or good morning exercises? You can use a foam block for
assisted eccentrics training. In my experiments with a Tendo unit,
comparing the eccentric and concentric bar speed with speed
strength benching and squatting, the difference was one-tenth of a
meter per second. The same was true for a circa-max phase in the
squat. A combination of band tension and weight was used. The
bands cause an accommodating resistance effect on the concentric
phase and an overspeed eccentric effect on the eccentric phase,
increasing kinetic energy in the stretch reflex.
The eccentric/concentric phase on speed strength should be 0.7 to
0.8 meters/second. On near-max weights, this should be 0.5
meters/second. Remember, this was done with adding bands to the
bar. But how do you lower just weight in a fast manner? This is done
by using roughly 60% of your eccentric potential. I have done ballistic
benching with 200 pounds when my raw bench was 500. As you can
see, I am dropping almost 40% of my best raw bench. I was lowering
the bar at around 0.5 meters/second.
Watching Elite benchers at Westside perform at similar eccentric
speed, I noticed that this has led to a faster concentric phase. How
did we learn to lower heavier loads in the bench and squat? We use
foam blocks to bench off of, lowering the plates onto the foam. We sit
on the foam for box squatting. Here are some of the methods we use.

http://westside-barbell.com/westside...april_2008.pdf

[waynelucky2]

Nwlifter wrote:
Wayne, that is a lot to sort through and only takes us off track.

A concentric rep, has a period of acceleration, (force higher then load), a period of constant speed (no matter how short, it's there even if just in transision, equal to the load) and a period of deceleration (with a force less then the load)
An eccentric rep, also has these 3 periods. But less force acclerates and more force slows the load.

Ok yes.

Nwlifter wrote:
Slide down a rope really fast, it takes very little force to go down fast.
But climb down a rope slowly and see it takes much more force.
Let a car coast down a hill fast, it takes very little force from the brakes
Make the car coast slowly down the hill, it takes more force from the brakes

Hmm, yes see your point.

But as the eccentric is so under loaded, and if you use the first half or just over the let the weight gain a little speeds, so it will gain a little more force, thus in turn you will be then trying to slow a heaver weight, well not heavier weight, but you know what you mean, {my English and Grammar and way of putting things is bad, but I am sure you know what I mean} also then you have a larger force going into the transition, and doing it this way will get the highest of the highest forces thus tensions on the muscles. If I did them slow, it would compromise the peak forces by about half, and as these peak forces are one of the main keys to strength and hypertrophy, I would not want to do that, but thats really another debate.

Nwlifter wrote:
Here are some inarguable facts,
• Why time matters: If you have 50lbs of force for 1 second and 100lbs for 1 second the average force is 75. If you have 50 lbs for 2 seconds and 100lbs for 1 second, the average force is 66.66 lbs. This is for figuring average force but..

Right.
•

Nwlifter wrote:
• We know the average force is the same between both styles of reps. This is due to the fast rep having high peaks and low valleys of force changes as the rep proceeds

Right.

However, if the force output is the same for 1 rep at .5/.5 and 6 reps at .5/.5, the 6 reps at .5/.5 can not have the same overall force, lets just think outside physics now, as I am not just talking physics, I am talking everything.

So if I use a high force for a short time = 100 pounds of average force. And if you used a low force for a longer time = 100 pounds of average force, what I do not think you are getting is, TIME has been taken ac****ed for in calculating the force, as when I say I do another 5 reps with another 100 pounds of average force each, you two keep saying, yes, but we are still moving the weight the same time as you.

But as I just said, because your force is low and long, the force has been calculated with the time.

So your average force for your 2/4 rep = 6 seconds average force = 100 pounds. My average force for my 1 rep at .5/.5 = 1 second average force = 100 pounds.

OR, if you say you do not add forces up, lets call force tension, as its force then tension anyway.

So your 1 rep at 2/4 for 6 seconds = 100T on the muscles. My rep at .5/.5 for 1 second = 100T on the muscles.

But then again you will say, you are lifting that weight for 2 seconds, yes but I say as I am lifting it with a far greater force, and I am starting here from the eccentric, to trigger the peck forces remember, and when a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

So its the same tension on the muscle for both reps ???

Nwlifter wrote:
Why distance doesn't matter if you know force. Distance is used to FIND force if you also know mass, time , then you can find force. But if you know force already, and force is what your looking for, there is no reason to have distance in the calculations.

??? But said; Distance is used to FIND force if you also know mass, time , then you can find force.

I still do not get why you two say distance doe not matter ???
My high force rep for .5 of a second, and I move the 100 pounds 1m, you do your low force rep for the same, .5 of a second, but you only move the 100kg .16 of a meter ???
So why are you saying distance does not matter ??? As in .5 of a second, I moved the weight 500% further, and you say this does not mater ??? To me thats like saying a weight lifter pick 500 pounds over his head, but the other one only lifted it 1 inch in the same time frame, but the lifter who lifted it only 1 inch, used the same force as the other lifter ???

Nwlifter wrote:
I am curious now, at first you were trying to compare energy expenditure. What final answer are you looking for between these two methods now?

Hmm, I have always been wanted to compare the total force/strength used, for 6 reps at .5/.5 and 1 rep at 3/3, both with say 80% used, with 2 people the same strength, or in other words which puts the most overall tension on the muscles. And as I failed 55% faster, and all the people failed faster in the study, I would have thought I was right, but I am fair and non biased, so, this failure could be because of the peak forces, too which I did not think they could be from that before, I just did not think of that, but still think that its its because you use more total force/strength, this in turn puts the most total tension on the muscles.

Wayne

[waynelucky2]

Hi All-pro,

Are you trying to say that quite a lot of things work, and variety is good, but you still need a good log, and a good basic routine to go back too ???

So most would do great at first if they did say 10 exercises x 3 sets of each, upped the weight each and every time, and good lots of good clean food down them.

Than after so and so, try the odd different thing to see if it works, as there are some many different combinations of reps, sets weights, and whatever out there.

Yes your right.

Put I only started this for one reason, not to try and tell or change anybodys training, just to see if 1 rep at 3/3 and 6 reps at .5/.5 and which has the most total force/strength/tension from/on the muscles. Or 4 reps at 3/3 and 25 at .5/.5, using the same weight, with 2 people of the same strength.

However, in reality, the person using the .5/.5 rep, would be using a heaver weight.

Wayne