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

[sophiecentaur]

This thread never seems to get anywhere. The only bit of useful sense to be injected lately has been the introduction of the notion of efficiency. It 'explains' the apparent paradox of the rockets and it is the nub of the weightlifter's muscles problem. Weightlifting has zero efficiency because there is no gain in GPE when the guy has finished. So much of the chat seems to have ignored the non-connection between work done and energy put in and, instead, tried to coax some sort of Law out of the process by relating non related terms and even non-terms.
Muscles are not springs or bridges. They are actually more like rockets because their efficiency depends upon movement. A sagging muscle is very much like using retro rockets to land. A lot of energy is used up despite a decrease in GPE.

 

[douglis]

douglis, I fully understand that two rockets moving at different speeds will stop in different distances when subjected to the same deceleration (gravity).

However, if you have two rockets moving at different speeds and constrain the stopping area to the distance the slowest takes to stop, you must apply additional deceleration to the faster one in order to achieve the required stopping distance.

In this case with the reps, the constraint placed earlier was that they are both the same distance traveled for the weight (my example was 1m).

In which case applying equal deceleration won't stop them in the same time.

A basic example is a car at 30mph will coast to a stop in Xm. To get a car going 60mph to stop in Xm you need to apply the brakes.

That's not the case we're discussing.Wayne compares 6 fast reps vs 1 slow so we're not talking about the same distance travelled.
So do you agree with my above rocket example(the last one)?

Gravity only applies one acceleration. So whether you are traveling at 1m/s or 100m/s it is always the same acceleration due to gravity.

The only way to increase the final deceleration is to apply additional force. If you don't, the faster rep will continue until it stops on its own - not within the described limit.

I was referring to the length of the deceleration phase as 'greater' and not the magnitude.

 

[douglis]

This thread never seems to get anywhere. The only bit of useful sense to be injected lately has been the introduction of the notion of efficiency. It 'explains' the apparent paradox of the rockets and it is the nub of the weightlifter's muscles problem. Weightlifting has zero efficiency because there is no gain in GPE when the guy has finished. So much of the chat seems to have ignored the non-connection between work done and energy put in and, instead, tried to coax some sort of Law out of the process by relating non related terms and even non-terms.
Muscles are not springs or bridges. They are actually more like rockets because their efficiency depends upon movement. A sagging muscle is very much like using retro rockets to land. A lot of energy is used up despite a decrease in GPE.

Exactly!
At last someone understood my point.Maybe it's my fault because my English is not the best.

 

[JaredJames]

That's not the case we're discussing.Wayne compares 6 fast reps vs 1 slow so we're not talking about the same distance travelled.

From the OP:

Some people say, that if you lift a weight, as in weightlifting/bodybuilding. Slow repetition, up 1m and down 1m, one time at 3 seconds up and 3 seconds down, and let’s call it a 100 pounds. And then with the same weight, fast repetition, for same distance, but up in .5 of a second, and down in .5 of a second, 6 times = 6 seconds as well, that the average forces thus tensions on the muscles are the same in the long run.

We're very much are talking about the same distance travelled.

You cannot have the same deceleration on two objects at different velocities and expect them to stop in the same distance.

 

[douglis]

A basic example is a car at 30mph will coast to a stop in Xm. To get a car going 60mph to stop in Xm you need to apply the brakes.

After a second thought what you say here doesn't make sense.
You don't have to apply the brakes.You just have to pull your foot from the pedal much earlier when a car going 60mph and friction will do the job(stop in Xm).
Just like in the fast rep you'll stop applying force much earlier and gravity does the job(stop at the end of rep).

 

[waynexk8]

What I attempt to do is to move/accelerate a load/weight as fast as possible, the more force I will need to do it.

waynexk8 wrote;
Well I do know what I am talking about,
jarednjames wrote;
Really? Then you follow with:
waynexk8 wrote;
Acceleration is not momentum
jarednjames wrote;
Good
waynexk8 wrote;
Its the opposite of momentum
jarednjames wrote;
Bad

Hi jarednjames,
First, I best say, that when I say momentum, I think I used the wrong word for a physics forum, momentum just means movement in physics ? However in everyday life, momentum , means the kinetic energy an object has once you apply a great enough force, eg, I lift a weight as fast as possible, and if I stop applying force this weight still might travel ? 100mm. Thus a lighter weight moved fast will move easer and faster as of the momentum, {kinetic energy} but the heaver weight going closer to my RM {repetition maximum} will have has as much momentum, {kinetic energy}as if I try to accelerate it as much as I can with my highest force, I stay in front of any momentum, {kinetic energy} then when I need to slow the weight down {which is done at high speeds} for the deceleration for the transition from the concentric to the eccentric, I absorbed any momentum, {kinetic energy} when reversing the direction.

waynexk8 wrote;
acceleration requires more force/strength, where momentum requires less force/strength
jarednjames wrote;
Ugly

So what I was trying to say there, was that if I try to accelerate a weight with as much force as possible, {and remember the weight is 80% of my RM, and only traveling arms length} I am in front of the build up of momentum, {kinetic energy} and in charge of the situation/lift and it’s my force and my force only that’s moving the weight until the transition from the concentric {up part of the lift} to the eccentric. {down part of the lift}

As if I was using a lighter weight, momentum, {kinetic energy} would require less force/acceleration, as the lighter weight could fly out of my arms with far less force/acceleration than the heaver weight.

jarednjames wrote;
Momentum does not require force in any form. Any moving object has momentum but it does not need to be subject to any force.

Not sure what you mean there, as if momentum means movement or kinetic energy, first you have to use a force of some kind to get the object moving ? Unless you mean all objects are moving as of this World and all objects on it are moving though the Universe at a hell of a speed.

jarednjames wrote;
Any moving object has momentum but it does not need to be subject to any force.

Yes.

jarednjames wrote;
The more you accelerate a mass, the more momentum it gains. That's the only link there is between them.

Yes.

jarednjames wrote;
That was enough for me to not read on.

The rest were from World renowned people who have Masters PhD’s in Physics, Biomechanics and Kinology, thus I would imagine you would agree with what they say.

Wayne

 

[sophiecentaur]

"momentum just means movement in physics ? However in everyday life, momentum , means the kinetic energy an object has once you apply a great enough force. . . .etc"

How can you think you are making a useful contribution when you make statements like this? I suggest you get an elementary textbook and paste the relevant pages on your shaving mirror so that you can see them every morning. Those "?" marks presumably mean you are not quite sure about these terms so why do you still feel qualified to argue about all this?

 

[JaredJames]

However in everyday life, momentum , means the kinetic energy an object has once you apply a great enough force

Please just stop. This is complete and utter non-sense. Momentum means no such thing.

The rest were from World renowned people who have Masters PhD’s in Physics, Biomechanics and Kinology, thus I would imagine you would agree with what they say.

It is you who does not understand the basics and as such aren't using them correctly. As such, what you say is based on your incorrect understanding of what these people say, not what they are actually publishing.

 

[JaredJames]

After a second thought what you say here doesn't make sense.
You don't have to apply the brakes.You just have to pull your foot from the pedal much earlier when a car going 60mph and friction will do the job(stop in Xm).
Just like in the fast rep you'll stop applying force much earlier and gravity does the job(stop at the end of rep).

Certainly. But you're missing the point.

Let's say g=1m/s² and you have a 2m track vertically.

You accelerate an object at 1m/s² vertically and then let gravity decelerate it. Total travel time is 2 seconds.

Now, let's say you want it to travel it quicker.

That would mean, simply, an acceleration of 2m/s² and a deceleration of 2m/s². The problem is we only have g to decelerate.

So you accelerate at 4m/s² for 0.25 seconds - this leaves your final velocity at 1m/s and g can decelerate you to a stop on the 2m mark. Giving you a total travel time of 1.25 seconds instead of 2.

This means that you have a minimum transit time of > 1 second and must stop acceleration at 1m/s in order to only require g to stop you by the end of the course.

So now we have a rep time for your weight of 2 seconds and 1.25 seconds.

The problem is, as per my previous example, the energy required to generate the latter is significantly greater than the former. Which means you need more CE to generate the latter acceleration.

Given that no energy is used to stop the weight, that means you only have the energy use to generate the initial acceleration to take into account. Which are not equal in both cases.

 

[waynexk8]

Please just stop. This is complete and utter non-sense. Momentum means no such thing.

In everyday life with everyday people, momentum means just that. Let’s say they are pushing a car, they will first say I was pushing the car and I got it moving, it had movement, than when they got it going faster with would say as I pushed harder the car moved more easy, as it had momentum.

As I said in physics momentum means movement, but to most people it means kinetic energy.

It is you who does not understand the basics and as such aren't using them correctly.

Do you not mean incorectely ? However as I said, most people think momentum is kinetic energy.

But sorry for using the wrong terms, at least I explained and said I did.

As such, what you say is based on your incorrect understanding of what these people say, not what they are actually publishing.

I am understanding what the people say here, if I did not I would ask, however remember, you are all top physics people, I am just learning, so you will have to excuse me on some things and maybe explain a little more in layman’s terms. As if you came a started at my Factory fabricating wrought iron, drive gates and so forth, I am sure you would make many mistakes in the first few years, where I would help you out.

Look sorry I used the wrong terms, but it’s not easy learning all physics at my age, and I am trying.

Wayne

 

[JaredJames]

In everyday life with everyday people, momentum means just that. Let’s say they are pushing a car, they will first say I was pushing the car and I got it moving, it had movement, than when they got it going faster with would say as I pushed harder the car moved more easy, as it had momentum.

As I said in physics momentum means movement, but to most people it means kinetic energy.

1. That is a good indicator not to use your own meanings here.
2. I have never heard that meaning. From what I've heard, people understand that momentum simply means a big object moving has more than a small object moving at the same speed
3. What the public think is irrelevant. It is the officially accepted terms - as per the dictionary - that matter. If the public use a word without knowing its definition, it doesn't mean anything.

Do you not mean incorectely ?

No, I said you aren't using them correctly.

Aren't = are not.

I am understanding what the people say here, if I did not I would ask, however remember, you are all top physics people, I am just learning, so you will have to excuse me on some things and maybe explain a little more in layman’s terms.

Layman's terms does not mean using incorrect terms (your momentum example).

 

[waynexk8]

Certainly. But you're missing the point.

Let's say g=1m/s² and you have a 2m track vertically.

You accelerate an object at 1m/s² vertically and then let gravity decelerate it. Total travel time is 2 seconds.

Now, let's say you want it to travel it quicker.

That would mean, simply, an acceleration of 2m/s² and a deceleration of 2m/s². The problem is we only have g to decelerate.

So you accelerate at 4m/s² for 0.25 seconds - this leaves your final velocity at 1m/s and g can decelerate you to a stop on the 2m mark. Giving you a total travel time of 1.25 seconds instead of 2.

This means that you have a minimum transit time of > 1 second and must stop acceleration at 1m/s in order to only require g to stop you by the end of the course.

So now we have a rep time for your weight of 2 seconds and 1.25 seconds.

The problem is, as per my previous example, the energy required to generate the latter is significantly greater than the former. Which means you need more CE to generate the latter acceleration.

Given that no energy is used to stop the weight, that means you only have the energy use to generate the initial acceleration to take into account. Which are not equal in both cases.

Well I do understand and total agree with you on the energy, and I cannot see douglis does not and has got this wrong. He’s a great guy and one of the people I have been debating with, he’s an Engineer and knows Physics quite well, but has got this very wrong.

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. The greater the force exerted by a task, the more rapidly the muscles fatigue.

So to review, more power, work will equal more energy used, equals more force used, equal more acceleration, speed, velocity.

Wayne

 

[waynexk8]

1. That is a good indicator not to use your own meanings here.
2. I have never heard that meaning. From what I've heard, people understand that momentum simply means a big object moving has more than a small object moving at the same speed

Not in all circles. If you came over to some of the training forums, not do they only use the wrong words, they actually make up words that have no real meaning or science back up, they do this just to try and sound right.

3. What the public think is irrelevant. It is the officially accepted terms - as per the dictionary - that matter. If the public use a word without knowing its definition, it doesn't mean anything.

Yes agree, sorry it was my fault.

No, I said you aren't using them correctly.

Aren't = are not.

Yes get you.

Layman's terms does not mean using incorrect terms (your momentum example).

Suppose your right there again. I do admit I was in the wrong and used the word wrong, and do not want to spoil this debate as you all helping me and giving advice, so sorry again and I will remember where I am in future.

Wayne

 

[waynexk8]

"momentum just means movement in physics ? However in everyday life, momentum , means the kinetic energy an object has once you apply a great enough force. . . .etc"

How can you think you are making a useful contribution when you make statements like this? I suggest you get an elementary textbook and paste the relevant pages on your shaving mirror so that you can see them every morning. Those "?" marks presumably mean you are not quite sure about these terms so why do you still feel qualified to argue about all this?

Ok, I did say sorry above.

Wayne

 

[douglis]

Certainly. But you're missing the point.

Let's say g=1m/s² and you have a 2m track vertically.

You accelerate an object at 1m/s² vertically and then let gravity decelerate it. Total travel time is 2 seconds.

Now, let's say you want it to travel it quicker.

That would mean, simply, an acceleration of 2m/s² and a deceleration of 2m/s². The problem is we only have g to decelerate.

So you accelerate at 4m/s² for 0.25 seconds - this leaves your final velocity at 1m/s and g can decelerate you to a stop on the 2m mark. Giving you a total travel time of 1.25 seconds instead of 2.

This means that you have a minimum transit time of > 1 second and must stop acceleration at 1m/s in order to only require g to stop you by the end of the course.

So now we have a rep time for your weight of 2 seconds and 1.25 seconds.

The problem is, as per my previous example, the energy required to generate the latter is significantly greater than the former. Which means you need more CE to generate the latter acceleration.

Given that no energy is used to stop the weight, that means you only have the energy use to generate the initial acceleration to take into account. Which are not equal in both cases.

You overcomplicate things.Forget for a moment that you totally stop applying force after the acceleration phase(which is not realistic after all) and think of the following example.

In a fast rep you accelerate the weight by using g+2 m/s^2 and you decelerate with g-2 m/s^2.On average you use acceleration equal with g and force equal with mg.
On a slower rep you accelerate the weight by using g+1 m/s^2 and you decelerate with g-1 m/s^2.Again on average you use acceleration equal with g and force equal with mg.

Of course the duration of the slower rep is longer but the average force per unit of time is in both cases mg hence the energy expenditure per unit of time is identical.

But you didn't answer me.Do you agree with my rocket example(see below)?I believe it represents perfectly what we discuss here.

''Let's say a rocket starts accelerating from the Earth upwards and after a while the engine shuts down.The rocket decelerates and reaches a maximum height before it will start falling.
Let's say it reached the max H in exactly one minute.The average acceleration for that trip is zero(starting and ending velocity zero) hence the average force is exactly equal with the weight of the rocket.
Now if the rocket was just standing still in the air for that minute again its engine would have used force equal with the rocket's weight for 1min.

In both cases the rocket will use exactly identical fuels since it used the same average thrust for 1 min.''

 

[waynexk8]

D. wrote;
After a second thought what you say here doesn't make sense.
You don't have to apply the brakes.You just have to pull your foot from the pedal much earlier when a car going 60mph and friction will do the job(stop in Xm).
Just like in the fast rep you'll stop applying force much earlier and gravity does the job(stop at the end of rep).

You never stop applying force, to slow the weight down you “still” have to use “a” force. IN ADDITION, you “HAVE” to use a force to lower the weight.

D. why do you not acknowledge the way that energy is used a calculated from the human body ? This how its done all over the World.

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. The greater the force exerted by a task, the more rapidly the muscles fatigue.

Wayne

 

[JaredJames]

You never stop applying force, to slow the weight down you “still” have to use “a” force. IN ADDITION, you “HAVE” to use a force to lower the weight.

You don't have to keep applying a force.

You stop pushing upwards and the weight will stop due to gravity - no input from you required.

Please stop quoting that same section in bold over and over.

 

[waynexk8]

Certainly. But you're missing the point.

Let's say g=1m/s² and you have a 2m track vertically.

You accelerate an object at 1m/s² vertically and then let gravity decelerate it. Total travel time is 2 seconds.

I am not going to pretend I understand your calculations. However, gravity does NOT let the weight down, this is not the debate, the muscles control it down.

As this is my debate could we have a small recap please.

Fast reps moves the weight up 1m and down 1m for 6 times in 6 seconds = 12m in total.

Slow rep moves the weight up 1m and down 1m for 1 time in 6 seconds = 2m in total.

The muscle controls and moves the weight up and down.

Now, let's say you want it to travel it quicker.

That would mean, simply, an acceleration of 2m/s² and a deceleration of 2m/s². The problem is we only have g to decelerate.

So you accelerate at 4m/s² for 0.25 seconds - this leaves your final velocity at 1m/s and g can decelerate you to a stop on the 2m mark. Giving you a total travel time of 1.25 seconds instead of 2.

This means that you have a minimum transit time of > 1 second and must stop acceleration at 1m/s in order to only require g to stop you by the end of the course.

So now we have a rep time for your weight of 2 seconds and 1.25 seconds.

The problem is, as per my previous example, the energy required to generate the latter is significantly greater than the former. Which means you need more CE to generate the latter acceleration.

Yes that what we all have been saying; the energy required to generate the latter is significantly greater than the former.

Here is a small part of the debate on energy you might like to read.

BIO-FORCE wrote:
While it is perfectly simple to see that walking a flight of stairs and running a flight of stairs has:

1) The same average force
2) The same amount of work
3) The same energy use
4) Speed is different
5) Duration is different


douglis wrote:
The numbers 1,2,3 are wrong.
Running a flight of stairs is typical case where kinetic energy(speed) is preserved after each step. So the average force is greater and the energy expenditure is greater too.

Wayne wrote;
D. you will find Johns 1 to 5 right.

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

douglis wrote:
Do you understand to what you just agreed?
At number 2 Jeff said:"the rate of energy usage is independent of rep speed".
Isn't that exactly what I say the last 8 months and you disagree?

BIO-FORCE wrote:
Glad you caught that since the "rate" of energy use IS greater. Rate has a time component. If an action is performed in 1 second and the same work is performed in 5 seconds, then the "rate" of energy use is greater for the faster (shorter duration) rep.[/quote]

Wayne wrote;
D. you said that you use the same energy {calories} was used doing the same work for the same speeds, this is NOT true,

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

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

Note the 10mph used MORE {double} the {energy calories} That the 5mph did the same time frame.

Given that no energy is used to stop the weight, that means you only have the energy use to generate the initial acceleration to take into account. Which are not equal in both cases.

Not sure what you mean there, as the muscles have to use a force thus energy constantly doing the reps. They use force and energy when acceleration and when decelerating, and when moving the weight down, there will and cannot be any constant speed. Also they will use a huge weight as of the peak forces the transition from negative to positive.

Wayne

 

[waynexk8]

You don't have to keep applying a force.

You stop pushing upwards and the weight will stop due to gravity - no input from you required.

Yes, I understand what you mean there. However, this is not the case in Weightlifting, as I constantly try to push up with as much force as possible, and then immediately start to lower the weight under control.

One more thing that might interest you, we worked out that so and so a weight pushed at full force for a certain distance, if the person immediately stopped the force the weight would keep on moving 3 inches, “however” with the human body this does not happen, it was BIO-FORCE above who said why. It’s because of the biomechanical advantages and disadvantages thought the ROM, {range of motion} meaning the muscles cannot push with a constant force like a machine and how the calculations are worked out.

Please stop quoting that same section in bold over and over.

Sorry about that, but D. try’s to ignore science and how people have been working these things out for years, and try’s to ignore what it right.

Wayne

 

[JaredJames]

Sorry about that, but D. try’s to ignore science and how people have been working these things out for years, and try’s to ignore what it right.

Putting it in bold and then things in bold and blue after my request not to use bold does not improve the quality of your posts and stops me reading them!

There is no need for it. Quoting others when you clearly don't understand the material yourself doesn't help.

Gravity is constantly acting on the weight, whether you like it or not.

It constantly works against you when raising it and helps you when lowering it - it is easier to lower a weight than raise one.

On the way down, there is nothing outside of air resistance acting against your motion.

As this is my debate could we have a small recap please.

We have, multiple times, but you just keep adding non-sense and using incorrect terminology.

Fast reps moves the weight up 1m and down 1m for 6 times in 6 seconds = 12m in total.

Slow rep moves the weight up 1m and down 1m for 1 time in 6 seconds = 2m in total.

The muscle controls and moves the weight up and down.

You keep saying these things, but the fact you are waiving off gravity is a clear problem on your part. You really need to look of a physics 101 book and get to grips with it (as sophie keeps asking you to).

 

[waynexk8]

Hi Wayne,

I just noticed this thread despite that it has been ongoing for a while. Some time back you and I discussed the concept of "average force" in quite some detail. If you will recall, as long as you start and stop in the same location each rep the average force is always equal and opposite to the gravitational force (remember that force is a vector quantity). The speed at which you do a rep does not influence the average force at all. I think that the conclusion from that previous thread was essentially that average force is not a useful measure for what you are really interested in.

Hi DaleSpam,

Well the speed does differ in that I use a far far far higher high force and a high peak force, otherwise the weight would not move so fast and farther than the slower rep.

Yes your right, and I did actually say this from the start, and even last year started a thread call average force does not apply here, as that’s all, it is average force.

One,
Two runners, one could run 100m at, 10s, 10s, 10s, 10s, 15s the other could lift run in 12s, 12s, 12s, 12s, 7s. Averages are the same, however runner 1 uses more energy, does more work faster, uses more force and speed/velocity for the biggest part.

Two,
That all it is, is average, the thing is I am using that average force far far far more times, that’s why I am moving the weight 500% MORE in distance using my faster reps than the slower reps.

You may be interested in a more realistic model of muscle contraction:
http://en.wikipedia.org/wiki/Hill's_muscle_model

Thank you for the link. Yes, I am aware of Hills work, thank you very much.

Wayne

 

[waynexk8]

Putting it in bold and then things in bold and blue after my request not to use bold does not improve the quality of your posts and stops me reading them!

There is no need for it. Quoting others when you clearly don't understand the material yourself doesn't help.

Ok sorry about the quoting in bold, will not do it again, only wanted to catch your and other attention, but as you say it could have the opposite effect.

Gravity is constantly acting on the weight, whether you like it or not.

Yes, I never said other.

It constantly works against you when raising it and helps you when lowering it - it is easier to lower a weight than raise one.

Yes, again I never said or thought other.

However as you know, the force to push a weight up at full force, from a standing start, would be far less force thus tension on the muscles than if you let the weight drop from a height of say 1m, then you had to stop it and push up, as of the downward acceleration components.

On the way down, there is nothing outside of air resistance acting against your motion.

Not sure if I get you there. As say we let the weight drop from the top height of the rep, it would fall to the floor faster than I am lowering it, thus I am and have to use force/enegy to lower it, even thou as you said its easier to lower it.

Also, you might find of interest, that the muscles are roughly 40% stronger at lowering a weight, as if you took someone’s RM, you could add 40% to this and they could lower it under control. This is because the muscles fibers are sort of like the fins on a fish’s back, and slide when pushing up, but when gown down they do not slide as good they catch more on the fins. Also less of the muscle itself needs to be used when lower as of this action.

We have, multiple times, but you just keep adding non-sense and using incorrect terminology.

Its hard me putting in all the correct terms, as I am only getting back into physics, more here are Master, but I am trying to learn.

You keep saying these things, but the fact you are waiving off gravity is a clear problem on your part. You really need to look of a physics 101 book and get to grips with it (as sophie keeps asking you to).

I do not understand when you say I am waving of gravity, as I am or definitely would not, as gravity is the main resistance.

Wayne

 

[Dale]

Well the speed does differ in that I use a far far far higher high force and a high peak force, otherwise the weight would not move so fast and farther than the slower rep.

Yes, the peak force is higher for the faster reps. Not the average force. I think peak force would be a much more relevant measure in this context.

That all it is, is average, the thing is I am using that average force far far far more times

Force isn't something that gets used up, so I don't know in what sense you mean this. If you are exercising with a given weight for 10 s then you are exerting the same average force over the same amount of time regardless of whether or not you do 2 or 5 reps during that time. The peak force will be very different.

Also, the external work done (on the weight) will be 0 in all cases that you start and stop at the same point. So I think that work done will not be a useful measure either. Energy consumed will be higher for the faster reps since humans are less efficient when we go faster, but that is not a number that can easily be calculated and would require some Hill-style modeling.

 

[waynexk8]

Yes, the peak force is higher for the faster reps. Not the average force. I think peak force would be a much more relevant measure in this context.

Yes me too. But not just peak, the highs are higher also; just look at this rough example, the others here have seen this so sorry.

The persons max force is a 100 pounds, he uses 80 pounds weight for the rep/s, meaning he’s lifting a 80 pound weight, but will use on the faster reps 100 pounds of force. The concentric of the rep here is split up into five parts. This is the second rep, as it has the huge peak forces from the transition from eccentric to concentric.

Fast rep,
140, 100, 100, 40, 20.

Slow rep,
80, 80, 80, 80, 80.

The peaks are higher and the highs. What some people think is when the fast rep is on its lows, as for the deceleration phase that’s where the constant median forces of the slow rep catch up, but this is not so. As 140 is 60 or 75% more than 80, and 100 is 20 or 25% more than 80, and again 100 is 20 or 25% more than 80. And the medium forces of the slow rep cannot compete with this.

Also, the faster reps have moved the weight 12m to the slow rep 2m, in the same time frame. That must mean a huge force must be used to move the weight 5005 more in distance in the same time frame.

Force isn't something that gets used up, so I don't know in what sense you mean this. If you are exercising with a given weight for 10 s then you are exerting the same average force over the same amount of time regardless of whether or not you do 2 or 5 reps during that time. The peak force will be very different.

Yes I get what you’re saying, maybe I am saying it wrong, however what I mean is that each and every time I lift the weight, I am yes using that same force, but I am using it again and again, not just once. Maybe I should say work done (force x distance) or the impulse applied (force x time). But to me each and every time I lift the weight I am using another force/strength, as once I lift the weight up and down, I have to exert a new force/strength yet again, each and every time, as the first force/strength has gone and been used up, and will not lift the weight a second time. I know you know all that, but if I explain my way of thinking it might be better for you to help me.

Also, the external work done (on the weight) will be 0 in all cases that you start and stop at the same point. So I think that work done will not be a useful measure either. Energy consumed will be higher for the faster reps since humans are less efficient when we go faster, but that is not a number that can easily be calculated and would require some Hill-style modeling.

Not sure what you say there, as if I lift a weight up 1m and lower it down 1m I have done work, as work will be the amount of energy transferred by a force acting through a distance, I have coved a distance on 2m.

Now this is one of my questions, as you say more energy will be used, buy “why” is this ? I say because there is more muscle activity, eg more force/strength used, otherwise why else is there more energy used ?

Late here thanks all for you help and time, zzz.

Wayne

 

[Dale]

to me each and every time I lift the I am using another force/strength, as once I lift the weight up and down, I have to exert a new force/strength yet again, each and every time, as the first force/strength has gone and been used up, and will not lift the weight a second time.

Force is not a conserved quantity. It doesn't get used up.

if I lift a weight up 1m and lower it down 1m I have done work

No, the work done over any closed path is 0 in a conservative field like gravity.

Now this is one of my questions, as you say more energy will be used, buy “why” is this ? I say because there is more muscle activity

Yes, there is more muscle "activity", but not more average force nor more work done. Therefore neither average force nor work are good indicators of muscle activity.

 

[douglis]

Yes, the peak force is higher for the faster reps. Not the average force. I think peak force would be a much more relevant measure in this context.

Force isn't something that gets used up, so I don't know in what sense you mean this. If you are exercising with a given weight for 10 s then you are exerting the same average force over the same amount of time regardless of whether or not you do 2 or 5 reps during that time. The peak force will be very different.

Also, the external work done (on the weight) will be 0 in all cases that you start and stop at the same point. So I think that work done will not be a useful measure either. Energy consumed will be higher for the faster reps since humans are less efficient when we go faster, but that is not a number that can easily be calculated and would require some Hill-style modeling.

Exactly my point.
From physics point of view...the average force and the energy expenditure is identical regardless if in a minute you'll do 100 reps or 1 or even you just hold the weight.The higher fluctuations of force(peak highs but also peak lows) in the 100 reps don't make any difference in the total energy expenditure.

If in reality there's any difference in the energy expenditure it's a matter of muscle's efficiency or physiology and we can't possibly know.Only a physiologist might know the answer.

 

[douglis]

Yes me too. But not just peak, the highs are higher also

Wayne

What are the ''highs''?

 

[JaredJames]

energy expenditure is identical regardless if in a minute you'll do 100 reps or 1 or even you just hold the weight.

Why are you still ignoring initial acceleration?

The energy use in a larger initial acceleration is significantly greater than in slow. As you never gain any of that energy back it is considered lost (moving it faster causes more heat dissipation).

Now you can keep on with this non-sense all you like, but it is a fact that the faster you move the weight the more energy you use.

If I move a 1kg weight a hundred times in a minute I've used a lot more energy than moving it just the once.

What you are saying here is that if I was to pick a bag of sugar off the floor and put it on the kitchen counter taking 1 minute, I'd use the same energy if I picked up 100 bags in the same time. Non-sense.

 

[douglis]

What you are saying here is that if I was to pick a bag of sugar off the floor and put it on the kitchen counter taking 1 minute, I'd use the same energy if I picked up 100 bags in the same time. Non-sense.

That's exactly what I'm saying.The work is 100 times more but the energy is identical since in both cases you use force equal with the bag for a minute.

I'm not ignoring the initial acceleration.You're the one who ignores the final deceleration.

Read my last rocket example and tell me where exactly you disagree.

 

[JaredJames]

That's exactly what I'm saying.The work is 100 times more but the energy is identical.

I'm not ignoring the initial acceleration.You're the one who ignores the final deceleration.

Read my last rocket example and tell me where exactly you disagree.

If gravity is doing the deceleration in both cases, you are expending zero energy during that phase. So when looking at your own energy expenditure you don't need to include the deceleration phase as it has no bearing on the matter.

It takes significantly more energy to accelerate the mass at 2m/s² than it does at 1m/s².

Didn't see your rocket example.

''Let's say a rocket starts accelerating from the Earth upwards and after a while the engine shuts down.The rocket decelerates and reaches a maximum height before it will start falling.
Let's say it reached the max H in exactly one minute.The average acceleration for that trip is zero(starting and ending velocity zero) hence the average force is exactly equal with the weight of the rocket.
Now if the rocket was just standing still in the air for that minute again its engine would have used force equal with the rocket's weight for 1min.

In both cases the rocket will use exactly identical fuels since it used the same average thrust for 1 min.''

During acceleration, thrust does not equal the weight of the rocket.

Fuel use is exponential. The harder you accelerate the fuel use significantly increases - it is not linear.

All this average acceleration stuff is non-sense. You are ignoring the fact that to get a rocket to accelerate at 1m/s² uses significantly less energy than to get it to accelerate at 2m/s². To increase the acceleration by even such a small amount requires a drastic energy increase.

By doubling the acceleration (linear), you halve the time you accelerate (linear) for before allowing gravity to take over and decelerate the rocket. But, the energy to give that double acceleration is significantly more (exponential), not double. So it doesn't give you equal energy use in both cases.

As above, gravity does the deceleration so we can look at the rocket as not using it's own energy (fuel) to stop itself. So only the initial acceleration matters. We can ignore constant velocity for simplicity and just have acceleration and deceleration.