Is Average Force the Same in Fast vs. Slow Weightlifting Reps?

[douglis]

As sophiecentaur said,the "clay" experiment will tend to give you the maximum force value rather than the average force.

As for the average force,that seems to bother Wayne,the things are really simple.
When you try to lift a weight,regardless the lifting speed,the weight starts and ends at zero velocity.Therefore the average acceleration is zero too(a=(V2-V1)/t=(0-0)/t=0).
So the average applied force is always equal with the weight(F=mg+ma=mg) regardless the lifting speed.

 

[douglis]

Let me just chuck this stone into the pond an show you just how impossible it is to apply basic Physics to your question. You do not use correct terms all the time and that makes things even harder to square with the Science. Consider this:
You have already dismissed the fact that any force without movement involves no useful work but, if you lower a weight, then Negative work is done. A cycle of lift - rest - lower - rest actually does NO WORK, by the strict definition of Work. However, you don't have regenerative braking or any 'mechanical energy storage (resilience)' in your muscles so they are working away, internally, when stationary or when going up or down. How does that square with any 'average force'? You would have to admit that an exercise machine that pushed towards you with a lot of force and the retreated very easily would still tire you out and develop your muscles. Such a machine would be supplying Work / Energy rather than having work done on it.

This "average force", of which you are so fond, may well be more or less constant throughout, if the sequence is carried out slowly. What acceleration can you achieve with a heavy weight? Not much more than g I'll bet.

Simple Physics cannot (and would not try to) give you your answer because it's what goes on Inside your Muscle Tissue that is the purpose of your exercises and it's what you are trying to calculate.

The only way to get a meaningful answer as to how much energy your muscles are expending would be to analyse the blood that flows through them - or, as a second best, your rate of Oxygen use.

You're absolutely right that the work done can not represent the energy expenditure.A practical method that is used to estimate the energy expenditure is the time-tension integral.Practically is calculated as: mean tension X total contraction time.

 

[sophiecentaur]

You're absolutely right that the work done can not represent the energy expenditure.A practical method that is used to estimate the energy expenditure is the time-tension integral.Practically is calculated as: mean tension X total contraction time.

That's an interesting concept but that would kind of imply that just standing and holding something up would be good exercise?

 

[douglis]

That's an interesting concept but that would kind of imply that just standing and holding something up would be good exercise?

Of course it is!
It's called static holds.It's how the gymnastic athletes train almost exclusively.

 

[sophiecentaur]

So who bothers to pay hefty Gymn fees when they could easily do it for free with a bagfull of bean cans?!

 

[douglis]

So who bothers to pay hefty Gymn fees when they could easily do it for free with a bagfull of bean cans?!

Well...the problem is that as you get stronger you might need the whole shelf of bean cans from the super market so...the gym might be less expensive.

 

[sophiecentaur]

Ha.
You could always make use of a corpulent friend, perhaps. Pressups with him/her sitting on your shoulders. There's no end to my ingenuity in avoiding Gymn fees.

 

[waynexk8]

Let me just chuck this stone into the pond an show you just how impossible it is to apply basic Physics to your question. You do not use correct terms all the time and that makes things even harder to square with the Science. Consider this:

You may be right there, but I am trying. And trying to put it as straightforward as possible, all we are debating, or want to know, that if you had a motor/muscle/pulley, which rep speed done with the same weight and for the same time, would put the most tension on the motor/muscle/pulley.

With the pulley would not it be possible to measure the tension of/on the rope/cable, the rope would represent the muscle, tendons and other things pulling the weight on the pulley.
Could not we use Newton's third law of motion ? So the tension would be the force or the same as the force.

You have already dismissed the fact that any force without movement involves no useful work

I am not sure what you mean there ? I have not said any work done static is not useful ?

You have already dismissed the fact that any force without movement involves no useful work but, if you lower a weight, then Negative work is done.

Yes negative work is done. And when negative work is done by the muscles, they are roughly 40% stronger, meaning you can lower 40% under control, more that your positive RM strength.

A cycle of lift - rest - lower - rest actually does NO WORK, by the strict definition of Work. However, you don't have regenerative braking or any 'mechanical energy storage (resilience)' in your muscles so they are working away, internally, when stationary or when going up or down. How does that square with any 'average force'?

Yes. However, we all know that physical work has been done, and has to be done to move the weight up and done as you say. I myself do not think average force can apply here, and here are my reasons again, reworded.

1,
As with the clay test, ”ALL” forces/tensions will HAVE to be shown on the clay, thus if as some think the forces/tensions were the same, then when the SO CALLED DECELERATION phase of the fast rep take place, “WHY” does not the slow rep make up for it here, and flatten the clay as much ? NO ONE SEEMS TO WANT TO COMMENT ON THAT ? It’s because I would think that it’s the peak forces, at the very start of the rep, and the high, or higher force than the slow for most of the rep, that’s why the clay will flatten more, as overall the slow rep must have lower forces/tensions.

2,
Fast and slow repetition/s, {split up into 5 segments, concentric only} 100, 100, 100, 80, 20. {Second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher. That is a rough estimate.

I say the average forces cannot be the same, as the high higher forces, and the higher peak forces, {peak forces, the forces on the second and conceding repetitions, as of the transition from eccentric to concentric} are far greater total force. I would think that some are people are calculating the average forces wrong here, as, it would be the percentages you need to add up below like I have done.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, thus I cannot see how the forces are the same.

1,
Therefore, we have on the first fast rep 60 or 75% MORE force/tension with the faster rep.

2,
Then on the second and conceding reps, we have on the first fast rep 100 or 125% MORE force/tension with the faster rep.


Also, impulse is higher as show from another member of this forum, no one has commented on this ?.

You would have to admit that an exercise machine that pushed towards you with a lot of force and the retreated very easily would still tire you out and develop your muscles. Such a machine would be supplying Work / Energy rather than having work done on it.

Yes, you’re right. However I would be working against it ? Not sure what you’re getting at there, will have to think about that one more.

This "average force", of which you are so fond, may well be more or less constant throughout, if the sequence is carried out slowly. What acceleration can you achieve with a heavy weight? Not much more than g I'll bet.

Please see above, I am not found of the average theory at all. All weights are the same in both reps.

However, I am now thinking that, that should, or may should not be the case, more on that later.

Simple Physics cannot (and would not try to) give you your answer because it's what goes on Inside your Muscle Tissue that is the purpose of your exercises and it's what you are trying to calculate.

Yes I see that, but could we not use the force/tension on the cable/rope above to work something out ?

The only way to get a meaningful answer as to how much energy your muscles are expending would be to analyse the blood that flows through them - or, as a second best, your rate of Oxygen use.

Yes, totally agree there, analyse the blood that flows through them and your rate of Oxygen use. And as you and many have said, the faster reps must use more and far more energy, as they are using MORE power {work energy} actually its quite contradictory to say that the fast reps do not use far far far more energy, as they are as we all know and agree using more power, {work energy} And many proved that before with many calculations, which agree with me and every nutrition book and site.

SO as we are using more power, {work energy} that must, or can only mean we are using more force thus tension on the muscles, right ? If not why would we be using more power {energy work} ?

Thank you for your answers and your time.

Wayne

 

[sophiecentaur]

Look. You are trying to mix up the Physics and the Physiology.
The Physics is totally trivial and can be calculated from any elementary Physics knowledge. You can work out the Potential and Kinetic Energies at any stage. You seem to want to relate all this directly to the Physiology of the situation. I don't think it can be done (and neither, it appears, does anyone else) so this can't go any further.

Faster Reps (simple lifts) do no more work per lift than slow reps but the Power (you have to go along with the strict definitions or it's not worth being on this Forum) of course, is proportional to the rate of doing the Reps. That's so blindingly obvious that it's not worth discussion.

"Yes I see that, but could we not use the force/tension on the cable/rope above to work something out ? "
What particular "something" did you have in mind? I could suggest many possible Sumns that one could do but would they be worth a candle? I think not.

 

[waynexk8]

It’s a bit late here to answer the above, but I find it strange no one is yet answering what I write. I would like to say, that Douglas has a history of not answering direct questions, as he feels that if he does they will prove him wrong.

However Douglas, if you look back on some of the posts here, you will find all say that there is more power {work energy} used, you yourself do Not agree to the energy part do you ? But you do agree that there was more work done, as the fast rep moved the weight up and down 6 times, and the slow rep just up and down one time. Please after reading what was written, tell me why you think that there is the same energy/calories used in both rep/s.

When both reps have the same time frame, the same weight moved, however the fast reps will move the weight 12m to the slow rep 2m. Why do you not agree on this when EVERY nutrition book and site agrees that you do use far far far more energy ?

Also, how can you {not sure if you know this} say that the same amount of energy is used in both reps, when basically energy is worked out by the more you use, by the more power {work} has been done ? Basically they put you in a chamber with a mask on, and the more oxygen you use, and several other things, works out that you use more energy, and everybody knows that if you climbed a 10m ladder up and down 6 times, in 1 minute, to 1 time in 1 minute, you would be much more out of breath on the faster climb.

1,
Please could you tell me, that on my clay test, you seem to think that when I am decelerating, that’s when your constant medium force reps will catch up, and make up for the force/tension that they lost, PLEASE tell me why on the clay experiment they do NOT catch up ? As ALL the force/tensions will HAVE to be put on the clay will they not ?

2,
Fast and slow repetition/s, {split up into 5 segments, concentric only} 100, 100, 100, 80, 20. {Second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher. That is a rough estimate.

100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, thus I cannot see how the forces are the same. And impulse is higher.

, So please could anyone explain to me how the first fast rep it has 60 or 75% force/tension, and on the second and conceding reps, the fast reps have 100 or 125% more force tension, that 60 or 75% or 100 or 125% is NOT more than the force/tension of the slow rep ? And I am NOT taking about average force/tension, I am talking of the total/overall higher high and peak force/tensions, just tell me how 60 or 75% or 100 or 125% is NOT more than the force/tension of the slow rep ?

I mean it’s like saying a 100 is not 20 or 25% more than 80, but how can you say or work that out ?
3,
Why has no one commented on the below ? Slow rep = 1.06 and the fast rep with its peak and higher high = 3, I mean that’s what I have been saying all along, see my percentages above, the faster reps have a higher, and far higher 200% higher force/tension.

Make some assumptions and see what the numbers say. You are moving with peak to peak amplitude of 1m, in either 6 sec (3 up 3 down) or 1 sec (0.5 up 0.5 down).

Assume the weight is doing simple harmonic motion up and down. That is probably not a very good approximation but it's easy to calculate

Amplitude = 0.5m, frequency = 0.167 Hz or 1 Hz.

Maximum acceleration = a times (2 pi f)^2
= 0.55 m/s or 19.7 m/s
= about 0.06G, or 2G
So the maximum force would be 1.06 times the weight lifted for the slow case, and 3 times for the fast case.

Don't take those numbers as "accurate" but they do suggest there would be an effect.

The lifter might apply a large force for a short time and then ease off, rather than a smaller force for the whole 3 seconds. That would reduce the difference in the peak force.

Both reps only have the same average force if you average the up and down forces together. The average force up will be higher for faster reps and the average force down will be lower. (That statement includes some more hidden assumptions about how the lift is done, but the basic idea is probably correct).

4,
How much force is used moving 80 pounds at .5/.5 up and down, 1m for 6 times in 6 seconds, and how much force is used when moving 60 pounds at 3/3 up and down, 1m for 1 time in 6 seconds ?

Wayne

 

[sophiecentaur]

Wayne.
Acres of writing achieves nothing on its own. I'm sorry but I just can't stagger through it all and I bet the same goes for most followers of this thread.
I am not even sure what you want out of this any more. There are no "answers" as you describe them because you haven't posted a proper question.
You reject physics and you ignore basic engineering facts. Where is there to go with this?

 

[JaredJames]

I've given up reading through it as well. It's just non-sense for the most part, given the lack of physics knowledge going into it.

This thread is dead. We've all given you the same thing and each time you argue back at us.

From a physics perspective there's little more we can do - as sophie points out it's all about physiology here. And I don't think anyone here (immediate responders) are capable of answering that.

 

[waynexk8]

Wayne.
Acres of writing achieves nothing on its own. I'm sorry but I just can't stagger through it all and I bet the same goes for most followers of this thread.

I am unsure why you say that, as basically the questions are quite straightforward, let me shorten them for you.
1,
Basically, “if” the average forces are the same, why dose not the slow rep fatten the clay as much when it’s still producing is medium force when the fast rep is supposed to be decelerating and using less force than the load.

2,
How is 60 or 75% or 100 or 125% from the fast reps, NOT more than the force/tension of the slow rep ?

3,
Just wondered why no one commented on the impulse, when the Slow rep = 1.06 and the fast rep with its peak and higher high = 3. As impulse is the integral of a force with respect to time, the small force {slow rep} applied for a long time can produce the same momentum change as a large force {fast rep} applied briefly, because it is the product of the force and the time for which it is applied that is important. BUT in this case the force is LOWER in the faster rep because the time is the same.

I am not even sure what you want out of this anymore. There are no "answers" as you describe them because you haven't posted a proper question.

Not sure what you mean there ? As I put all the variables in. Which rep puts out the most overall/total force, or puts the most tension on the muscles. Both people use 80% call that 80 pounds, they move the weight up 1m and down 1m. One rep is at a speed of .5/.5 and they do 6 reps in 6 seconds. The other rep is at a speed of 3/3 and they do 1 rep in 6 seconds. We can take all the biomechanical advantages and disadvantages of the muscles out of this, and say it’s a machine moving up and down if you like.

You reject physics and you ignore basic engineering facts. Where is there to go with this?

I not sure why you say I reject physics ? As I do not, as if we know all the variables that I thought physics could would it out. And I do not ignore ignore basic engineering facts, I used physics, and engineering to design, fabricate, all on my own, only help needed was to install.

Only an answer.

Please I am not trying to be awkward, if I am not supplying all the facts, and it’s not possible to work out then please say, however I myself can easy work out the power of the lifts, thus thought the force/tension would be able to be done.

http://www.youtube.com/user/waynerock999?feature=mhum#p/u/8/4s73iACXSlE

Wayne

 

[waynexk8]

I've given up reading through it as well. It's just non-sense for the most part, given the lack of physics knowledge going into it.

I did not say I had a huge physics knowledge, that’s why I am posting and asking the question here, I thought I put the question quite plain and put all the variables in, if I missed something please say.

This thread is dead. We've all given you the same thing and each time you argue back at us.

From a physics perspective there's little more we can do - as sophie points out it's all about physiology here. And I don't think anyone here (immediate responders) are capable of answering that.

Ok, but a big thank you for your help and time, I thought there would be a total/overall force, and the same for tension on the muscles, looks like physics cannot answer this question. I thought it would, as I found it easy to work out the power involved, thought we could find the force and tension.

Wayne

 

[waynexk8]

Of course it is!
It's called static holds.It's how the gymnastic athletes train almost exclusively.

No they do not D.

I knew you "you" could NOT answer my questions. Here is my proof and evidence that you D. were very wrong, basically I have worked the answer out for you, with the most basic of physics, and my clay test.

If anyone things the below is wrong, please say why, as I do not like to say things that are wrong.





Fast and slow repetition/s, {split up into 5 segments, concentric only} 100, 100, 100, 80, 20. {Second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher. That is a rough estimate.

100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, thus I cannot see how the forces are the same. And impulse is higher.

So please could anyone explain to me how the first fast rep it has 60 or 75% force/tension, and on the second and conceding reps, the fast reps have 100 or 125% more force tension, that 60 or 75% or 100 or 125% is NOT more than the force/tension of the slow rep ? And I am NOT taking about average force/tension, I am talking of the total/overall higher high and peak force/tensions, just tell me how 60 or 75% or 100 or 125% is NOT more than the force/tension of the slow rep ?

I mean it’s like saying a 100 is not 20 or 25% more than 80, but how can you say or work that out ?

Wayne

 

[sophiecentaur]

If Scientists approached every question in the way you seem to be doing, Wayne, they would never have got anywhere. The whole point of Science is to reduce a problem to it's simple basics. Every post of yours just consists of reams and reams of instances and you refuse to apply what we have been saying to the problem - possibly because our replies are not in the form you want. Well. I'm sorry. My training has always been feet on the ground Science and I can't help wanting explanations to be reductionist.

There are no answers to your rambling questions because you have not defined the questions properly. Just apply the basic definitions of Force, Work and Power and you can make your own conclusions.

If you still don't know what I mean, then read some other long threads and you'll see the satisfactory ones are never like this one.

 

[waynexk8]

If Scientists approached every question in the way you seem to be doing, Wayne, they would never have got anywhere. The whole point of Science is to reduce a problem to it's simple basics. Every post of yours just consists of reams and reams of instances and you refuse to apply what we have been saying to the problem - possibly because our replies are not in the form you want. Well. I'm sorry. My training has always been feet on the ground Science and I can't help wanting explanations to be reductionist.

Hi there all and sophiecentaur, I have been thinking what you have said above, and would like to try and make it more simpler. Also what you said here;

“Faster Reps (simple lifts) do no more work per lift than slow reps but the Power (you have to go along with the strict definitions or it's not worth being on this Forum) of course, is proportional to the rate of doing the Reps. That's so blindingly obvious that it's not worth discussion.”

So if I do 6 reps at .5/.5 to 1 rep at 3/3. We can all agree that the .5/.5 have used more power. {work energy} HOWEVER, here we get to the root of the problem, “why/how” is more work energy “power” used, what muscular need is making the muscles use more ? power. {work energy}

I would like to hear the member answer to this before mine please, or if you like I will put mine in

There are no answers to your rambling questions because you have not defined the questions properly. Just apply the basic definitions of Force, Work and Power and you can make your own conclusions.

If you still don't know what I mean, then read some other long threads and you'll see the satisfactory ones are never like this one.

Hope I have done that now, and thanks for your time and help.

Wayne

 

[JaredJames]

I'm curious what is "work energy" and why are you calling it power?

I may have missed something over the last few years of learning.

 

[waynexk8]

I'm curious what is "work energy" and why are you calling it power?

I may have missed something over the last few years of learning.

Power is the rate at which work is performed or/and the energy converted. So when I lift a barbell faster up and down say 1m each way, I am doing more work and using more energy {power “work energy”} in the same time frame, and covering more distance. As I can lift the barbell 12 times up and down at .5/.5 in 6 seconds, but only 2 times at 3/3.

If you lift a 91kg barbell 1.85 m overhead. How much work is done ?

To determine the force we need to figure out what the weight of the barbell is (W = mg = 91kg x 9.81 m/s² = 892 kg.m/s² or 892 N).

Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.
Unit of measure for Work is the units of force times units of length (Nm).1 Newton meter “Nm” is equal to 1 joule, hence 1 joule is the work done when a force of 1 N moves through a distance of 1 m in the same direction as the force.

Power refers to the amount of mechanical work done in a time frame. The main component to note here is time.

Earlier, we calculated that lifting a 91kg barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account. It could have taken two seconds or four or more. No matter. The same amount of work is done regardless of how long it takes to complete the movement. The concept of power however, takes time into consideration. If for example, it took four seconds to complete the lift, then the power generated is 1650 J ÷ 4 s = 412.5 J/s. If on the other hand it only took 2 seconds to complete the lift, then the power generated is 1650 J ÷ 2 s = 825 J/s. Hence, the person who can lift the barbell in two seconds is more powerful than the one who lifted the barbell in four seconds. So while the work done remains constant, more power is developed when mechanical work is done more quickly. Indeed, power can be thought of as how quickly or slowly work is done.


Wayne

 

[JaredJames]

Power is the rate at which work is performed or/and the energy converted.

Power refers to the amount of mechanical work done in a time frame. The main component to note here is time.

I know precisely what power, work and energy are and how they relate to each other. What I asked was what is "work energy" and why do you call this power?

 

[waynexk8]

I know precisely what power, work and energy are and how they relate to each other. What I asked was what is "work energy" and why do you call this power?

Hi there jarednjames, yes I knew you knew what power, work and energy are, and far far far more about physics than me, however I thought I put that in my first sentence, and paragraph ?

Power is the rate at which work is performed or/and the energy converted. So when I lift a barbell faster up and down say 1m each way, I am doing more work and using more energy {power “work energy”} in the same time frame, and covering more distance. As I can lift the barbell 12 times up and down at .5/.5 in 6 seconds, but only 2 times at 3/3.

Basically if I lift a barbell 12 times up and down at .5/.5 in 6 seconds, but only 2 times at 3/3. I have used more power {work energy}

Wayne

 

[waynexk8]

I know precisely what power, work and energy are and how they relate to each other. What I asked was what is "work energy" and why do you call this power?

Hi there jarednjames,

Your last post seems to have gone ? You said somthing like; You can have the same work if its done fast or slow.

Yes.

However, so if I do 6 reps at .5/.5 to 1 rep at 3/3. We can all agree that the .5/.5 have used more power. {work energy} HOWEVER, here we get to the root of the problem, “why/how” is more work energy “power” used, what muscular need is making the muscles use more ? power. {work energy}

I would like to hear the member {as many as have to time to answer please, if it’s not too difficult} answers to this before mine please, or if you like I will put mine in, as all here are well versed in physics.

Wayne

 

[JaredJames]

Hi there jarednjames,

Your last post seems to have gone ? You said somthing like; You can have the same work if its done fast or slow.

Yes.

Although it is correct in certain areas, I don't believe it reflects correctly for your example.

However, so if I do 6 reps at .5/.5 to 1 rep at 3/3. We can all agree that the .5/.5 have used more power. {work energy} HOWEVER, here we get to the root of the problem, “why/how” is more work energy “power” used, what muscular need is making the muscles use more ? power. {work energy}

I would like to hear the member {as many as have to time to answer please, if it’s not too difficult} answers to this before mine please, or if you like I will put mine in, as all here are well versed in physics.

1. Stop using "{work energy}" or you'll be shot down for incorrect terminology again.
2. I'll answer it but please clarify what 0.5/0.5 and 3/3 means before I do.

 

[waynexk8]

Although it is correct in certain areas, I don't believe it reflects correctly for your example.


1. Stop using "{work energy}" or you'll be shot down for incorrect terminology again.
2. I'll answer it but please clarify what 0.5/0.5 and 3/3 means before I do.

Hi jarednjames, sorry been away for a few days.

Ok, sorry, I thought that was right.
Most know the bench press so we will use that. Also shall we say that the concentric = 1m and thus the eccentric = 1m. 1RM means your repetition maximum, or the most you can lift for one time, and on this example I have used a 100 pounds, and 80% too which is quite easy 80 pounds.
.5/.5 means that you lift the weight for the concentric in .5 or half a second, and lower in .5 or half a second, the 3/3 is you lift the weight for the concentric in 3 seconds and lower in 3 seconds.
Thx for your time and help.

Wayne

 

[sophiecentaur]

Most know the bench press so we will use that. Also shall we say that the concentric = 1m and thus the eccentric = 1m. 1RM means your repetition maximum, or the most you can lift for one time, and on this example I have used a 100 pounds, and 80% too which is quite easy 80 pounds.
.5/.5 means that you lift the weight for the concentric in .5 or half a second, and lower in .5 or half a second, the 3/3 is you lift the weight for the concentric in 3 seconds and lower in 3 seconds.
Thx for your time and help.

Wayne

"Most" probably do not know the bench press except in so far as many people might recognise one if they saw one. The terms eccentric and concentric, in this context, have absolutely no meaning to non-gym members. It would be much better if you were to state accurately the physics and the mechanics of a described situation and not launch into foreign terminology. The whole reason for viewing the world through 'Physics" eyes is to eliminate all the specific stuff and to try to arrive at as simple a description of a system to which Physics theory can be applied..
Your arms know nothing of what they are pushing or pulling. They are only aware of forces, distances and times. If you could state your problem in just those terms then you could probably answer it yourself. I don't think anyone can give you a proper answer because you haven't yet defined the question.

 

[douglis]

I will try to help Wayne by simplifying his question.
Let's say two weightlifters lift 100 pounds.The first one lifts them in 1sec and the second in 10sec.Obviously the first weightlifter used 10 times greater power than the second.
For that reason Wayne somehow believes that the first weightlifter also used greater force.

This is clearly not true in my opinion.When you lift a weight(regardless if it's in 1 or 10sec) the starting and ending velocity is zero hence the average acceleration is zero too[a=(V2-V1)/t=(0-0)/t=0].So the average force that you use is (F=mg+ma=mg) equal with the weight regardless the lifting speed.

As for the energy that was used...from physics POV the work is identical in both cases.But since the second weightlifter(10sec lifting) used the same average force for 10 times longer I believe it's obvious that he spent more energy.

Now specifically in Wayne's example.
He tries to compare the force and energy in two cases.Either you lift and lower a weight 6 times in 6 seconds or once in 6 seconds again.
As I believe I proved above the average force is the same in both cases.As for the energy that was used...the work done in both cases is zero from physics POV.But since the same average force was used for 6 seconds in both cases my guess is that also the energy expenditure is the same or with biomechanics terms the time-force integral is identical.

 

[waynexk8]

Sorry for the long post, I have been told about this before.

"Most" probably do not know the bench press except in so far as many people might recognise one if they saw one. The terms eccentric and concentric, in this context, have absolutely no meaning to non-gym members. It would be much better if you were to state accurately the physics and the mechanics of a described situation and not launch into foreign terminology. The whole reason for viewing the world through 'Physics" eyes is to eliminate all the specific stuff and to try to arrive at as simple a description of a system to which Physics theory can be applied..
Your arms know nothing of what they are pushing or pulling. They are only aware of forces, distances and times. If you could state your problem in just those terms then you could probably answer it yourself. I don't think anyone can give you a proper answer because you haven't yet defined the question.

Ok, sorry, will try it this way.

It does not matter what lift we use, however the bench press is when you lay on a bench, and press up from the chest and back down. We shall we say that the press up = 1m and thus the lowering down = 1m. 1RM means your repetition maximum, or the most you can lift for one time, and on this example I have used a 100 pounds, and 80% too which is quite easy 80 pounds.

.5/.5 means that you lift the weight for the press up in .5 {half a second} and lower in .5 {half a second} the 3/3 is you lift the weight for the press up in 3 seconds and lower in 3 seconds.

Thx for your time and help.

I will try to help Wayne by simplifying his question.

Hi D. and a big thank you for reengaging me in the debate. I thought you had realized you were wrong after debating with John, however it seems that you still to not understand where or how you are wrong..

Let me answer you two questions, and you can scroll back in this thread to find what I say is true, or look on every nutrition site or book. Just borrowed this from your other friend.
Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry. Many Statistics will be measured eg; Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, to determine the effects of exercise at different intensities. On every occasion when an activity has been practiced at a faster rate, the heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, has been higher, therefore energy expenditure, nutrient oxidation will be higher.
The increase in the rate of energy expenditure/cost will be due the increase in the intensity/force of muscle activity. When a muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy.


1,
More Energy/calories are used in the faster rep in the same time frame, yes or no ? And please if no; give a complete answer, with physics and in layman’s terms. Please do not say because you think it is, or as you think the average forces are the same, as if you think the average forces are the same, you still have to explain why you also think the Energy/Calories used are the same.

As if Power, and as we all know, Power is the rate at which work is performed Energy is converted. In the faster rep, work is performed faster, thus why do you think energy is not needed nor converted and used faster ? What you say, as you do say Power is more in the faster reps, then you say Energy is the same, is totally contradictory.

2,
If and when you say that more Energy is used in the faster reps, could you please say why the muscle has to use more Energy ?

a,
Is it because the muscle uses the same intensity, force, average force, total highest force, muscle activity. The muscle is subjected to a greater acceleration, speed or velocity, EMG reading will stay the same.

Or,

B,
Is it because the muscle uses the MORE acceleration, speed, velocity, energy, work, power, intensity, force, average force, total highest force, and muscle activity, to state a few. As all tests/studies show when the muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy, work power, intensity, muscle activity.


For all, D. is my friend from Grease, whom I am having THE DEBATE. However, he as I said as a history of not answering questions when they are either too hard, he does not understand them, and does not like to ask, or if is shows his theory is wrong. He will and has never answered question one.

D. wrote;
Let's say two weightlifters lift 100 pounds.The first one lifts them in 1sec and the second in 10sec.Obviously the first weightlifter used 10 times greater power than the second.
For that reason Wayne somehow believes that the first weightlifter also used greater force.

This is clearly not true in my opinion.When you lift a weight(regardless if it's in 1 or 10sec) the starting and ending velocity is zero hence the average acceleration is zero too[a=(V2-V1)/t=(0-0)/t=0].So the average force that you use is (F=mg+ma=mg) equal with the weight regardless the lifting speed.

As far as for the energy that was used...from physics POV the work is identical in both cases.But since the second weightlifter(10sec lifting) used the same average force for 10 times longer I believe it's obvious that he spent more energy.

HOLD on, so now you are saying that the faster rep uses more energy, great, that’s made my day, so could we go straight to number two please, and please do not say you did all along, as I have numberuss quotes from you stating the opposite.

What D. fails to understand, is that if the average force is the same in one of my reps and one of his, it will be the same for each and every one of my reps, but he does not understand, THAT EACH AND EVERY ONE OF MY REPS THAT HAVE THE SAME AVERAGE FORCE, WILL BE DONE NOT ONCE NOR TWICE, BUT SIX TIMES, AND YOU HAVE TO ADD THESE AVERAGE FORCES UP. That’s why they use more Power, more Power = more muscle activity, and why do you have more muscle activity is it because you use more or the same total force/energy ?




D. you seem to “think” that when the faster rep is decelerating, that is when the constant forces of the slower rep will make up here with the forces they lost in the peak forces, and of the higher high forces in the faster rep. But as the faster rep would flatten the clay far far far more, you assumption on the forces, and average forces, are WRONG, are they not ?

fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, thuis I can not see how the forces are the same. Thus impulse is higher.

Also D. and all, impulse is higher , we worked this out on this forum.

Wayne

 

[sophiecentaur]

Your basic problem here is that you are trying to relate 'Work Done' in Physics terms to Energy Expended by muscles in some simple but exact way.
That just cannot be done because the muscles behave in a very complex way - e.g. when they are standing there just pushing, they are working - unlike the table you rest weights on it.

There are, of course, some simple things like ten lifts in twice the time will use about twice as much energy as five lifts in half the time (obvious). You could possibly say that ten lifts in the same time as five lifts would involve twice the amount of energy BUT, even that could be dodgy because of the different amount of resting time between lifts or the speed of lifting. This is nothing to do with simple Physics but involves how your body works (it wasn't 'designed' in the same way as an electric motor or a steam engine).

We have already discussed the biometric approach - calorimetric rooms and gas analysis and why do you think things are done that way? It's because the method you seem to be wanting is not feasible. All this could easily take you into the realms of Pseudo Science, if you're not careful because there is not shortage of people who have strong opinions, based on very little physical evidence but who can make loads of money as consultants and trainers.

I really think we've done this to death now. My theory is that if the exercise is agony, it may be doing you some good and if it's easy, it may not be. The long term effect may or may not be visible in your improvement in competition. Physics is not capable of predicting the benefit.

 

[douglis]

HOLD on, so now you are saying that the faster rep uses more energy, great, that’s made my day, so could we go straight to number two please, and please do not say you did all along, as I have numberuss quotes from you stating the opposite.

Wayne

I don't know which part of my post you didn't understand to say something like that and to be honest I don't really care.
I won't continue in this discussion.Try to understand what I wrote cause I believe it's really simple.

 

[waynexk8]

Hi all, and thanks for your help and time, also sorry if my physics is not that good, but we all have to learn somewhere.

As we all agree that more power is used, and I will work that out after, {please all I know any of you could work this out easy, but please let me do it, then you can comment on how and if I worked it out right} energy is used in the faster rep, done in the same time frame, and if work is the amount of energy transferred by a force acting through a distance.

This is my way of thinking, if we have used more energy, we must have used more force ? And I think that more force is the higher high forces, and higher peak forces in the first half or three quarters of the faster rep, as I explained above. The high forces are 25% higher, and the peak forces are 75% higher.

Wayne