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

[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 totaly 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 hight 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.

 

[Dale]

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

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

 

[waynexk8]

Sorry about my long posts, I have been told before, just can't do it any other way.

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

Yes right. However, when I lift a weight up and lower it down, I have to use a new or exert physical force once again that needs energy to do this. I mean the same weight is not going to keep being lifted up and down on its own after I just lift it once, as each and every time I need to use the same amount of force to lift and lower the weight, but every time I use this force that needs energy it’s not the first force I used.

Or maybe you call this more work ? However work will be the amount of energy transferred by a force acting through a distance. So if I lift the weight up and down a second time, I am using more energy transferred by more of the same force over a greater distance than the first lift, or you may call that more work, I call it using more of the same force, if you sort of get me.

DaleSpam, have to say to you and the rest for your time and helping me, and I know the above sounds a bit complicated to you, but you have been talking physics for many years, however I have not, so please bear with me on my layman’s way of putting thing.

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

I sort of get that; you mean it’s like dISPLACEMENT?AS IF I WALK SO AND SO MILES AND COME BACK TO THE SAME PLACE THERE IS NO dISPLACEMENT.

However if work will be the amount of energy transferred by a force acting through a distance. I have coved a distance up and down 2m in this example, thus work has been done, force has been used and energy ?

As to move a weight up and down I have to use physical force and energy in the up lift and the down lower ?

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.

Right we agree on more muscle activity.

But surely there must be more average force used, or should I say more of the same average force e used, as I said above, each and every time I lift the weight I have to use a new or separate force as the first effort of force will not lift the weight.

Then you say the same for work !

Ok what in physics will be a good indicators of muscle activity ? Power ? As power is the rate at which work is performed and energy is converted. But then you say no to work.

Confusing.

What we really need is with a muscle or machine, which puts the most tension on the muscles, to me and most it has to be the faster reps as of the higher high and higher peak higher done 6 times in the same time frame as the slow reps.

To me in this video, I have failed roughly 50% faster because I have used more overall or total strength up faster, but you say this is wrong ! How else would I fail if I was not using my force/strength up faster ? Or maybe that’s basically it; I am just using my force/strength up faster ?

http://www.youtube.com/user/waynerock999?feature=mhum#p/a/u/0/sbRVQ_nmhpw

Wayne

 

[JaredJames]

However if work will be the amount of energy transferred by a force acting through a distance. I have coved a distance up and down 2m in this example, thus work has been done, force has been used and energy ?

Force is a vector. It has magnitude and direction.

In the raising stage force is upwards, in the lowering stage force is downwards. The directions 'cancel out'.

Because work = force x distance, work also has this direction. So raising gives you work upwards and lowering gives you work downwards - both equal - they cancel out.

 

[waynexk8]

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

Well said, I cannot understand why D. thinks this way, I did try to explain this to him from the start, and it’s something small I think he’s got wrong but just cannot see it, happens to us all.

DaleSpam, I have been asked to keep repeating and posting my; “Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry.” All the time, too which you find on the page before this.

However why do you think that D. is ignoring science/physics and the way energy is and has been calculated for many years ? As every nutrition site or book states that if you do any activity twice as fast in the same time frame, as you have covered twice the distance you used twice the energy ?

Wayne

 

[waynexk8]

Force is a vector. It has magnitude and direction.

In the raising stage force is upwards, in the lowering stage force is downwards. The directions 'cancel out'.

Because work = force x distance, work also has this direction. So raising gives you work upwards and lowering gives you work downwards - both equal - they cancel out.

Yes I sort of get that, but all I am saying is you need physical force/strength and energy in both direction, to lift the weight up, and to lower it under control.

Wayne

 

[JaredJames]

Yes I sort of get that, but all I am saying is you need physical force/strength and energy in both direction, to lift the weight up, and to lower it under control.

You add the energy use as you would add distance travelled.

This is independent of work.

When you calculate work you use force and distance. You make the distance into displacement.

 

[douglis]

Careful here, you are using the wrong terminology. The energy expenditure is different, you mean that the work done is identical.

The work done is the integral of the force times the differential displacement. That is always 0 over a closed loop in a conservative field like gravity.

Energy expenditure is more complicated than the work done and includes all of the "messy" biological inefficiencies. So the energy expenditure is higher for the fast reps, but not easy to calculate.

Yes...I examine the case from physics POV and I ignore the biological inefficiencies.

Do you agree with this 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.''

 

[JaredJames]

Do you agree with this rocket example?

I've just responded to this.

You are ignoring basic things and confusing the difference between acceleration and constant velocity.

Energy use is not the same when you have two different accelerations.

 

[Dale]

Do you agree with this rocket example?

I think that the rocket example is irrelevant for this thread. Muscles and rockets operate on very different principles and I don't see the value of the analogy in this context.

 

[douglis]

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.

O.K...let's start from here.
We agree that acceleration requires more energy and also the greater the acceleration the greater the energy.
So in this rocket example let's say that the rocket used force that caused initial acceleration equal with 2g(if we assume that was accelerated for 30sec) and then since the engine shut down used zero force for the final deceleretion(the last 30sec).
On average the force that was used produced average acceleration equal with g or else the average force was mg for 1 minute.
Exactly like the force that was used by the rocket that was standing still in the air.

Since in both cases the average acceleretion that the engined produced was equal I can't see how the fuels can be different.The fuels that the moving rocket spent when using 2g for 30 sec are exactly equal with the fuels that the standing rocket spent by using g for 1min.

 

[JaredJames]

Since in both cases the average acceleretion that the engined produced was equal I can't see how the fuels can be different.The fuels that the moving rocket spent when using 2g for 30 sec are exactly equal with the fuels that the standing rocket spent by using g for 1min.

They most certainly are not.

Average acceleration doesn't apply to the fuel use because it isn't linear.

As previously, fuel use is exponential. If you double your speed, the fuel use more than doubles.

So to double your speed may half your travel time but it more than doubles fuel use.

1g for a minute may use 1L of fuel. 2g for 30 seconds will use 4L. 4g for 15 seconds will use 15L etc.

Obviously, it's not as straight forward as that but that's the gist of it. 1g for a minute does not equal 2g for 30 seconds.

 

[douglis]

They most certainly are not.

Average acceleration doesn't apply to the fuel use because it isn't linear.

As previously, fuel use is exponential. If you double your speed, the fuel use more than doubles.

So to double your speed may half your travel time but it more than doubles fuel use.

1g for a minute may use 1L of fuel. 2g for 30 seconds will use 4L. 4g for 15 seconds will use 15L etc.

Obviously, it's not as straight forward as that but that's the gist of it. 1g for a minute does not equal 2g for 30 seconds.

Look...I tried to make an equivalent with weight lifting and I assumed that fuel expenditure is linear.If we don't make that assumption the discussion goes too far.

 

[JaredJames]

Look...I tried to make an equivalent with weight lifting and I assumed that fuel expenditure is linear.If we don't make that assumption the discussion goes too far.

In no case, ever, is fuel consumption linear. Weight lifting and rockets alike.

To make that assumption would completely change the physics and make it so that twice the speed with double the distance would give you the same energy use. Complete non-sense.

 

[douglis]

In no case, ever, is fuel consumption linear. Weight lifting and rockets alike.

To make that assumption would completely change the physics and make it so that twice the speed with double the distance would give you the same energy use. Complete non-sense.

I insist that speed and distance are irrelevant and keep the nonsense comments for youself.
Forget for a moment the acceleration phases.Two rockets with different constant speeds will spend identical fuels for the same duration if you exclude the air resistance.

Again at the acceleration part now...do you believe that double force will result in more than double fuel expenditure?

 

[JaredJames]

I insist that speed and distance are irrelevant and keep the nonsense comments for youself.
Forget for a moment the acceleration phases.Two rockets with different constant speeds will spend identical fuels for the same duration if you exclude the air resistance.

Constant speed yes, but it has to get to that speed which involves acceleration which, depending on the required velocity is different (as you are aware). To accelerate to 50mph does not take half the fuel it takes to accelerate to 100mph.

For simplicity you can ignore constant speed and simply have acceleration and deceleration.

If deceleration is due to gravity the rocket expends no fuel.

Leaving you only acceleration to consider for fuel use and the values for different accelerations (1m/s², 2m/s², 4m/s² etc) do not have the same fuel use and it is not simply double fuel as you double acceleration.

Again at the acceleration part now...do you believe that double force will result in more than double fuel expenditure?

There's nothing to believe. It is fact.

 

[douglis]

Constant speed yes, but it has to get to that speed which involves acceleration which, depending on the required velocity is different (as you are aware). To accelerate to 50mph does not take half the fuel it takes to accelerate to 100mph.

Yes...but the constant speed case proves that it's not about speed and distance but only about acceleration.
Anyway...I find more interesting the acceleration/force part right now.

There's nothing to believe. It is fact.

I have no reason to doubt you.
So...let's say I hold a 50 pounds weight for a minute and then a 100 pounds weight for another minute.Since the force/energy relation is not linear...in the second case I'll spend more than double energy.I guess the magnitude of the exponent is different in any case of engine/muscle.
That's really interesting and changes everything.Not that I don't believe you but can you give me a link or something I can search about it?

 

[waynexk8]

D. I did tell you distance was important, but you did not want to listen to me for some reason. Time and distance.

What D. is saying is that if you climb 10m up a robe in 100 seconds, and then climb up a rope 5m in 100 seconds that you will use the same energy.

As we know, and as science has told us for many years, when you do something slower, the only time you will use the same energy is when you travel the same distance as in the person who has done the activity faster.

Wayne

 

[waynexk8]

Also, this debate dose “NOT” only depend on ACCELERATION, as if I ran uphill at a speed of 1ms for 5 hours and also ran uphill at a speed of 5ms for 5 hours, the ONLY time that the speed of 1ms would use the same energy is when it covered the same distance.

As you can see by the example, acceleration energy would be negotiable. This is because we HAVE to take into consideration, Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, will ALL be more, higher when moving faster for the same time frame, however the faster moving will cover more distance.

Wayne

 

[Dale]

I mean the same weight is not going to keep being lifted up and down on its own after I just lift it once

It would if our arms were elastic, like springs. The weight certainly could continue to be lifted up and down on its own, precisely because force is not conserved and work is not done. You are confusing biology with physics.

you may call that more work, I call it using more of the same force, if you sort of get me.

When you are discussing physics it is very important to use the correct terminology. These are not just words with ambiguous meanings that depend on context to "sort of get". These are precisely defined technical terms with specific and exact meanings. I think that more than half of the problem in this thread and the other is that you persistently continue to use the same incorrect terminology even after you have been corrected. The frustration that you occasionally hear directed at you stems from that consistent behavior. It is OK to not know the right word in the beginning, that is the purpose of an educational site like this, but once it has been explained to you it is not OK to continue misusing precisely defined physics terminology.

I have coved a distance up and down 2m in this example, thus work has been done

No work has been done on the weight if you lift it up 1 m and then down 1 m. As you lift it up, the force is directed up and the displacement is also up, therefore you are doing work on the weight. As you let it down, the force is still directed up but the displacement is down, therefore you are doing negative work on the weight (the weight is doing work on you). A concentric contraction does positive work, an eccentric contraction does negative work, and over one full rep the work is 0.

But surely there must be more average force used

There is not. I have proved that in our previous conversation.

Ok what in physics will be a good indicators of muscle activity ?

The energy consumed would be a good indicator of muscle activity. This is different from the work done, but it is also not a simple quantity to calculate and would require some complicated modeling similar to what Hill did, but including metabolism. It could be measured with the room calorimetry as you mention, but not easily calculated in advance.

 

[waynexk8]

You add the energy use as you would add distance travelled.

Yes agree.

This is independent of work.

So you’re saying physical force/strength in both directions, to lift the weight up, and to lower it under control, is independent of work ? If that’s true I am glad we got that out of the way.

When you calculate work you use force and distance. You make the distance into displacement.

I find this a little hard to understand. So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ? Or have I got the wrong end of the stick.

In my way of thinking, like in the weighting lifting repping example, in .5 of a second, I have moved the weight 1m, however the slower moving rep has only moved the weight .16 of a meter, or one sixth of what I have moved in the same time frame. To me that means I have used far far far more force/strength.

Wayne

 

[waynexk8]

It would if our arms were elastic, like springs. The weight certainly could continue to be lifted up and down on its own, precisely because force is not conserved and work is not done. You are confusing biology with physics.

When you are discussing physics it is very important to use the correct terminology. These are not just words with ambiguous meanings that depend on context to "sort of get". These are precisely defined technical terms with specific and exact meanings. I think that more than half of the problem in this thread and the other is that you persistently continue to use the same incorrect terminology even after you have been corrected. The frustration that you occasionally hear directed at you stems from that consistent behavior. It is OK to not know the right word in the beginning, that is the purpose of an educational site like this, but once it has been explained to you it is not OK to continue misusing precisely defined physics terminology.

No work has been done on the weight if you lift it up 1 m and then down 1 m. As you lift it up, the force is directed up and the displacement is also up, therefore you are doing work on the weight. As you let it down, the force is still directed up but the displacement is down, therefore you are doing negative work on the weight (the weight is doing work on you). A concentric contraction does positive work, an eccentric contraction does negative work, and over one full rep the work is 0.

There is not. I have proved that in our previous conversation.

The energy consumed would be a good indicator of muscle activity. This is different from the work done, but it is also not a simple quantity to calculate and would require some complicated modeling similar to what Hill did, but including metabolism. It could be measured with the room calorimetry as you mention, but not easily calculated in advance.

Hi DaleSpam,

Just going out for the night, so will have to get back to this one, however as I have said before, a big thanks to all for your help and time. Hey, with a few more weeks of this, I will have Stephen Hawking begging me to help him with some of his harder equations ROL.

Wayne

 

[douglis]

Wayne...check BB.com.

 

[JaredJames]

So you’re saying physical force/strength in both directions, to lift the weight up, and to lower it under control, is independent of work ? If that’s true I am glad we got that out of the way.

Force applied to an object, moving it a distance gives you work. I have no idea what you're talking about now.

I find this a little hard to understand. So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ? Or have I got the wrong end of the stick.

If you end up in the same place, the weight has been displaced but the final displacement is zero.

Distance is a scalar. This means it simply has a magnitude.
Displacement is a vector. This means it has a magnitude and a direction.

They are not the same thing.

If I move 3m North and 4m East, I have traveled a distance of 7m but my displacement is 5m North East.

Bolded: Ok, now you're just being ridiculous and adding insignificant factors trying to complicate things. For the purpose of simplicity this is irrelevant.

In my way of thinking, like in the weighting lifting repping example, in .5 of a second, I have moved the weight 1m, however the slower moving rep has only moved the weight .16 of a meter, or one sixth of what I have moved in the same time frame. To me that means I have used far far far more force/strength.

I've never argued that. I don't know what you're debating now.

 

[sophiecentaur]

@Wayne
You started this thread 'outside' Physics and your last post is also way outside. You seem to have made no progress with this but see determined, somehow, to prove that you are still 'right'. Why are you bothering still.

 

[Dale]

So is physics saying if I move up and down that as no displacement has been done no work has been done, but what if I move up and then down, but down on a slight angle ? Meaning I am not back in the same place ?

If you do not return to the same place then the path is not a closed loop and the work done may be non-zero. In a gravitational field the work done on the weight depends only on the difference in height between the starting point and the stopping point.

 

[douglis]

I have no reason to doubt you.
So...let's say I hold a 50 pounds weight for a minute and then a 100 pounds weight for another minute.Since the force/energy relation is not linear...in the second case I'll spend more than double energy.I guess the magnitude of the exponent is different in any case of engine/muscle.
That's really interesting and changes everything.Not that I don't believe you but can you give me a link or something I can search about it?

Hi jarednjames,
you said that double force results in more than energy expenditure(expontential relationship) and sounded logical to me.

But I still would like to see some refferences.In the above example,after a little research I did,the relationship seems to be linear.

 

[waynexk8]

Have to say a big thank you to all the people here that have helped.

As looking at the two other forums we are debating on, seems that my friend D. is going more to the side that the faster reps must use more energy. This is what I said.




Therefore, if more energy is used on the faster reps, and in this example I will just say the reps are twice as fast. So we have 100 pounds moved up 1m and down 1m, twice in 2 seconds = 4m. We then have 100 pounds moved up 1m and down 1m, one time in 2 seconds = 2m.

I would say the faster rep uses twice the amount of energy.

I would also say, that because twice the amount of energy was used, was because there was more muscle activity, meaning the higher high forces, and the higher peak forces uses far far far more energy in the same time frame as the slower rep. As it needs to use more energy and higher high forces and higher peak forces to cover twice the distance in the same time frame.

Wayne

 

[sophiecentaur]

Is this still going on?
What is there possibly left to say except to give more and more instances of numbers of "reps" and the fact that the system is not possible to characterise as a piece of mechanics?

 

[douglis]

Therefore, if more energy is used on the faster reps, and in this example I will just say the reps are twice as fast. So we have 100 pounds moved up 1m and down 1m, twice in 2 seconds = 4m. We then have 100 pounds moved up 1m and down 1m, one time in 2 seconds = 2m.

I would say the faster rep uses twice the amount of energy.

Wayne

Wayne...the total displacement is zero so the work done is zero too.End of story.

As for the energy expenditure what only matters is if the higher fluctuations of force in fast reps expend energy in a exponential or linear mode.
That's something that can't be proved or disproved with physics.In the biology forum maybe you can find a better answer.

 

[JaredJames]

Hi jarednjames,
you said that double force results in more than energy expenditure(expontential relationship) and sounded logical to me.

But I still would like to see some refferences.In the above example,after a little research I did,the relationship seems to be linear.

OK douglis, I'm disappointed you haven't done the calcs yourself (seeing as I did them a few pages back).

Let's keep it simple. You have a 1kg object that you accelerate for 1 second.

So we accelerate it at 1m/s², 2m/s² and 4m/s² so the resulting speed after 1s is 1m/s, 2m/s and 4m/s.

This gives each one a final KE of 0.5mv². Which is 0.5J, 2J and 8J respectively.

As you can see it's an exponential increase in energy required per second of acceleration to achieve the required final velocity (and KE).

Now I can see your next question already (and I can see what you did to get a linear relationship).

You're going to point out that in the rep example you accelerate for 1s in the slow reps and only 0.25s in the fast ones (as per my previous example). Which is all well and good, at the end of those times you will have imparted equal KE on the weight (it would be moving at 1m/s in both cases and so would have the same KE). However, the energy required for it to accelerate at each rate is different. Getting to 1m/s in 1s uses significantly less energy than getting to 1m/s in 0.25s.