# How can this be average force ?

waynexk8
how can this be average force ???

The short of the question, is which puts the most tension on the muscles ???

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.

HOWEVER, I cannot see this.

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

But let’s now do a thought experiment.

You position a piece of clay in-between the fast and slow lifters hands and the barbell, and if you than take the clay out of both lifters, I would say the faster repetition would have flattened the clay out as much as 3 times that of the slow lifter.

But how can that be if the average forces are the same, as when the bar is decelerating on the faster repetition, and not using much force, the slow rep are still using their medium force, thus they should catch up and flatten out the clay as much as the fast reps, but they do not do they ???

THEREFORE, does that mean the average forces are NOT the same.

I say the average forces cannot be the same, as the high higher forces, and the higher peak forces, {peak forces, the forces on the second and conceding repetitions, as of the transition from eccentric to concentric} are far greater total force.

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

Wayne

JaredJames

Why does the force change for the fast repetitions?

If the fast repetitions always have the same duration then the acceleration is the same and as such the force is constant. It doesn't drop off as per your figures.

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

Let's say it takes 10N to lift the weight in 1 second (a = f/m), if you apply 20N it would only take 0.5 seconds and so on. But the force applied per unit time is the same.

The difference comes when you look at energy. The kinetic energy of moving the weight (100lbs) at an average velocity of 1m/s is 22 joules, but the KE of moving the weight at 2m/s is 90 joules. So the higher the average velocity of the weight, the more the energy use increases - which is why you get tired quicker.

* All figures are for guidance only.

Last edited:
waynexk8

EDIT: Hang on, I'll try and put this better.

K thx.

Wow I thought that was fast.

Wayne

Gold Member

The "clay" experiment will tend to give you the maximum force value rather than the average force because clay is a non-linear medium.

Also, it's pretty well impossible to relate what goes on in your muscle tissue to simple mechanical work and force. Remember, if you merely maintain a load at a fixed height, the work done on the load is zero. But your muscle fibres are constantly contracting and relaxing, passing the load to one another. This corresponds to a constant amount of energy being expended in flexing the inelastic bits of the tissue.
I wouldn't mind betting that the nearest answer you can get is to say that the exercise that hurts the most is probably actually doing the most 'work' towards developing the muscles.

waynexk8

Why does the force change for the fast repetitions?

If the fast repetitions always have the same duration then the acceleration is the same and as such the force is constant. It doesn't drop off as per your figures.

Where are you getting your figures from:

Very late here not much time.

I got the figures from a rough estimate, as just say you are bench pressing this weight, and its 80% or your RM {repetition maxumum} So you would have say a deceleration after say ??? 70% of the concentric, for the transistion from the concentric to the ecentric.

Let's say it takes 10N to lift the weight in 1 second (a = f/m), if you apply 20N it would only take 0.5 seconds and so on. But the force applied per unit time is the same.

The difference comes when you look at energy. The kinetic energy of moving the weight (100lbs) at an average velocity of 1m/s is 22 joules, but the KE of moving the weight at 2m/s is 90 joules. So the higher the average velocity of the weight, the more the energy use increases - which is why you get tired quicker.

* All figures are for guidance only.

Will have to get back to you tommorow sophiecentaur, bed calls.

Wayne

JaredJames

Very late here not much time.

I got the figures from a rough estimate, as just say you are bench pressing this weight, and its 80% or your RM {repetition maxumum} So you would have say a deceleration after say ??? 70% of the concentric, for the transistion from the concentric to the ecentric.

If you move a 100lb weight, 1m in 0.5 seconds each time, it will always take the same amount of force. If you reduce the force, you increase the time. But, the applied force per time will always be the same.

As you can see from the rough numbers I posted, the force per time applied is always the same, but the energy requirements are what change. So it takes significantly more energy to move the weight twice as fast.

Gold Member

If you move a 100lb weight, 1m in 0.5 seconds each time, it will always take the same amount of force. If you reduce the force, you increase the time. But, the applied force per time will always be the same.

As you can see from the rough numbers I posted, the force per time applied is always the same, but the energy requirements are what change. So it takes significantly more energy to move the weight twice as fast.

I have a feeling that you are not using the right quantities in your attempted analysis.
What soes "force per time" tell you (Dimentionally, that's MLT-3)? Force TIMES time is Impulse, or rate of change of momentum. This could be relevant, possibly.
Force times height would give you the work done. Force times speed gives the Power being developed. Or just Force could be of interest.
These are the quantities that, imho, you should be considering if you want to discuss the Physics of what's happening. If you have an ipod you can strap it to your weights and use a downloaded accelerometer app to tell you something about the motion of your weights.

Your weights will know about Force, Power and Energy delivered and you can measure all of them. But the situation in your muscles is much more complicated than that. As well as just increasing muscle power / strength, good training will optimise how you actually apply your strength to the physical task; the way the muscle fibres are behaving. As I said, earlier, the only way to assess what a particular exercise is doing for you may well be how much discomfort it's causing - as well as how fast you are improving, of course.

It's interesting that the only official, direct 'Gymn' related sport is weightlifting - one weight lifted once bevause it is nicely objective and measurable. They have unofficial competitions such as pressups and dragging weights about but I don't know of anything other than weightlifting in Olympics and the like. I think this actually reflects what I have been saying.

JaredJames

I have a feeling that you are not using the right quantities in your attempted analysis.

I was trying to give a brief example, that just ended up being the best way I could come up with at the time.

The OP speaks of doing repetitions (3 second and 0.5 second) and somehow claims the force used reduces each time. This is incorrect. The force to lift the weight 1m in 3 seconds is constant, as is the force to lift it 1m in 0.5 seconds. Each time you lift it, the mass is constant and therefore to gain the required acceleration to achieve the task in the required repetition time you must impart the same force each time. It doesn't drop off.

I also tried to demonstrate that the energy required during shorter reps is significantly more than during longer ones.

Gold Member

I suppose it's always worth trying to get a simple 'mechanical' model but I think there are far too many imponderables. For instance, what is the velocity profile during the lift in a fast rep compared with that in a slow rep? After a brief pause (in the slow reps) I should have expected there to be more muscle effort than without a pause (as in the short rep) available due to recovery. So you would HAVE to do some objective measurements before coming to any serious conclusions. Does modern exercise equipment provide you with such info, perhaps?
And the Energy requirement would depend on the total number of lifts, of course - if you just consider the Work Done. But if there is any elasticity involved (as with energy storage in the tendons when running) the fast reps could benefit from the 'bounce' effect.
This is just tooooop hard.

waynexk8

On lunch break, so no time for a full reply until later, but thank you for all the replies and your time.

Its seem some are saying the forces are constant, however I cannot see/understand this, as if I am at first accelerating the weight up on the concentric, do not I need to then use less force for the deceleration phase for the transition from the concentric to the eccentric ???

Here is a video I made, and you can see I fail roughly 50% faster in the faster reps. However I am debating with an Engineer, who is very good at Physics, and stats because he says the average forces are the same, then the overall/total force thus tension on the muscles is the same, I say a very BIG no to this, as of the % I put down above on the higher high forces, and the higher peak forces, and the clay test.

In addition, in the faster rep I have moved the same weight 12m to the slow rep of only 2m. Also as we know and I have worked out, Power {work energy} is also far higher with the faster reps.

Best say, this debate is not about which makes for the best size or strength gains, it is just about at this moment in time, what rep/s put the most overall/total force output, thus MOST, overall/total tension on the muscles. 1 rep at 3/3 = 6 seconds or 6 reps at .5/.5 or if we did more reps of both, 4 reps at 3/3 = 24 seconds or 24 reps at .5/.5 = 24 seconds.

Wayne

JaredJames

In addition, in the faster rep I have moved the same weight 12m to the slow rep of only 2m. Also as we know and I have worked out, Power {work energy} is also far higher with the faster reps.

And that is why you fail faster in the fast reps - nothing to do with the force. Your energy consumption / use is far higher.

waynexk8

The "clay" experiment will tend to give you the maximum force value rather than the average force because clay is a non-linear medium.

Hmm, that’s what someone else just said on another forum. However I can not see that, as say you are bench pressing with the clay between your hands and the barbell, the clay will now act/flatten throughout the range of motion, if there is a low force, median force, high force or peak force, thus the clay will take all forces on it like the muscles are, thus the clay is showing all the exact same tensions the muscles will, will it not ???

So if I am correct, and please state if and where you think I am not, as I know you will. If the average forces were the same, the clay should flatten the same, as when the weight is decelerating on the faster reps, but it does not flatten out the clay as much, the slow reps do not make up for what they lost to the faster reps in the first half or more of the range of motion, from the peak and high forces.

My point is that in the faster reps, the peak forces, and not just the peak force, but also the peak force and the higher high force in the first 10 to ??? 70% of the range of motion, are FAR higher than the forces that are in the slow rep when the fast rep is decelerating. And this would also apply to reps like 2 reps at 5/5 and 1 at 10/10. As the faster you move the weight up against gravity and air resistance, the more you will flatten the clay like with G-force.

As I said, THEREFORE, does that mean the average forces are NOT the same.

I say the average forces cannot be the same, as the high higher forces, and the higher peak forces, {peak forces, the forces on the second and conceding repetitions, as of the transition from eccentric to concentric} are far greater total force.

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

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

The below is sort of like the above. I am looking at the reps split into 5 segments, but looking at the second rep of the faster reps, as it has the peak forces.

Slow rep 80, fast rep 140, fast = 60 or 75% more for the fast rep.
Slow rep 80, fast rep 100 = 20 or 25% more for the fast rep.
Slow rep 80, fast rep 100 = 20 or 25% more for the fast rep.

So the high and peak forces are 100 or 125% more on the fast reps. As I do not think we need to count up the last two segments, as the slow reps are just using the force of the weight. And the fast reps I think are using less force than the weight. Or maybe we have to count them, but not thought enough about this part yet.

Also, it's pretty well impossible to relate what goes on in your muscle tissue to simple mechanical work and force. Remember, if you merely maintain a load at a fixed height, the work done on the load is zero. But your muscle fibres are constantly contracting and relaxing, passing the load to one another. This corresponds to a constant amount of energy being expended in flexing the inelastic bits of the tissue.
I wouldn't mind betting that the nearest answer you can get is to say that the exercise that hurts the most is probably actually doing the most 'work' towards developing the muscles.

I find it odd that we can easy measure the Power, but not the force.

Wayne

waynexk8

Get back to the rest later.

And that is why you fail faster in the fast reps - nothing to do with the force. Your energy consumption / use is far higher.

Now this is what someone else said, “but” why am I using more energy ??? Or more power {work energy} I say its because I must be using more total or more overall force thus tension on the muscles.

As how can I be using more energy or more power {work energy} without using MORE force ??? As soon as I move a muscle, the first things I use are force and energy, you can not have one without the other, so if I am using more energy, I must be using more force. As in the reps, 1 at 3/3 and 6 at .5/.5 on the .5/.5 I have moved the weight 12m to the slow reps 2m, that’s 8m more, so how can you move the same weight 8m more in the same time frame without using more force ??? Or how can I use more energy without using more force ???

Wayne

Gold Member

Wayne. You just need to sort out your basic definitions of work and power etc. before you try to apply the terms in a physics discussion. Force is only one part of it. If you are delivering the same number of actions quicker then you are developing more power. That makes a huge difference to your system and how you feel or when you fail.

JaredJames

As sophiecentaur said, I'd just add the following clarification:

When an object is moving at velocity v with mass m, it has a certain KE dictated by KE=0.5mv2. That is, to move it at the velocity you must give it that amount of kinetic energy.

So in your case, m is constant which means the deciding factor is velocity.

Let's say that in the slow reps the average velocity is 1m/s, that means your muscles must provide 22 joules of energy. Now let's assume the average velocity for the fast reps is 2m/s, that means your muscles must provide 90 joules of energy. This is all because of the v2 term in the KE equation.

The faster you move something, the more energy it takes. As you can see, simply doubling the velocity (which would halve the rep speed) requires over four times more energy. Each time you double speed of the reps, and as such halve the time for the reps, you are increasing the energy requirement in this manner.

That is why you're using more energy.

ajthrax1

this discussion is closed quite simply through an application of calculus.

We begin with force, better put as F=ma
Integrating with respect to acceleration, we are left with $$\int$$F=mv.
Momentum is the integral (wrt a) of force. In order to view the average force applied, we must take the integral of ma (wrt a) with a lower limit of t0=0 and an upper limit of t1=x and divide it by (t1-t0). By looking solely at the "peaks" of the forces, we do not get a clear picture.
Those who mentioned the kinetic energy output are even more correct, as the integral with respect to velocity of momentum is kinetic energy( $$\int$$mv= .5mv$$^{2}$$).That makes KE the double integral of force. Applying basic integral rules, we see that if the integral of one force x is > force y's, than the double integral of force x is > force y's.
Assuming the possibility of reimann sums for the summation of the two graphs necessary for the examples given by OP, we can give an uncertain of +- 10 units. Would anyone like to see the calculations made? lol

waynexk8

As sophiecentaur said, I'd just add the following clarification:

When an object is moving at velocity v with mass m, it has a certain KE dictated by KE=0.5mv2. That is, to move it at the velocity you must give it that amount of kinetic energy.

So in your case, m is constant which means the deciding factor is velocity.

Let's say that in the slow reps the average velocity is 1m/s, that means your muscles must provide 22 joules of energy. Now let's assume the average velocity for the fast reps is 2m/s, that means your muscles must provide 90 joules of energy. This is all because of the v2 term in the KE equation.

Not sure if I am not seeing something here, so sorry in advance.

Why am I using more energy ??? To move the weight any distance you have to use a force, and to move it further you need to use a lager force, or the same force for longer. As you can not use more energy without using more force, well in this case anyway, as energy on its own will not more the weight, you need a force with is powered by energy.

However, why am I using more energy, you seem to be saying that I use more energy but not more force/strength, how can I move something faster, 6 times the distance in the same time frame and not use more force ??? You seem to be saying that a 100 force can move an object 100m in 1 second and 1m in 1 second ???

The faster you move something, the more energy it takes.

Right. And more force, a lager force, or the same force for longer.

The faster you move something, the more energy it takes. As you can see, simply doubling the velocity (which would halve the rep speed) requires over four times more energy. Each time you double speed of the reps, and as such halve the time for the reps, you are increasing the energy requirement in this manner.

That is why you're using more energy.

Sorry its late here, will have to get back to the rest.

Wayne

JaredJames

You really need to check your definitions and understanding of the various words being thrown around here.

You seem to be saying that a 100 force can move an object 100m in 1 second and 1m in 1 second ???

I've never said that - to reduce the time to move an object a distance of 1m, you have to increase the force. By increasing the force, the acceleration becomes larger:

a = f/m therefore if you double to force to 2f you get 2a = 2f/m.

Let's say we have an object that is 1kg. To move that 1m in 1s (1m/s) requires a KE of 0.5mv2 = 0.5*1*1 = 0.5 Joules of energy.
Now, to move it 100m in 1s (100m/s) requires a KE of 0.5*1*10000 = 5000 Joules. So in the first case I need a tiny amount of energy, in the second I need a huge amount in comparison.

For your case to move that 1m in 1s (1m/s) requires a KE as above (0.5 Joules).
Now, to move it 1m in 0.5s (2m/s) requires a KE of 0.5*1*4 = 2 Joules. So again you can see how simply halving the time of the repetition requires you to use more energy to complete it. The time applied is considered in the velocity figure.

For you to move the weight 1 rep in 1s requires 0.5 Joules - that is the energy you must provide to do it.
For you to move the weight 1 rep in 0.5s requires 2 Joules - again, that is the energy you must provide to do it.

If you do not provide that energy, you can't complete the rep in the required time.

waynexk8

Wayne. You just need to sort out your basic definitions of work and power etc. before you try to apply the terms in a physics discussion. Force is only one part of it. If you are delivering the same number of actions quicker then you are developing more power. That makes a huge difference to your system and how you feel or when you fail.

Hi there, and thx for helping me.

Yes I see force is only part of it, however, to get to the root of the debate, it would be which rep speed with the same weight done for the same time frame puts the most tension on the muscle, thus which ever one will also tell us which one puts out the most overall/total force. As if one puts more force out it must put most force out ???

Yes I see and can work out how much power I am using, working that out about 3 years ago got me back into physics, after many years. However of I use more power {work energy} I am moving faster, and to move faster I must be using more, total/overall force/strength ??? Or how else could I be using more power {work energy} if its not by using more total/overall force/strength ???

As power is the rate at which work is performed and energy converted, and work is the amount of the energy transferred the force going through a distance. THUS, more power would have to mean more force going through a distance, {or more distance in this instance} in the same time frame. As work is a force acting through a distance, is not energy always equivalent to the amount of force used ???

If you are delivering the same number of actions quicker then you are developing more power. That makes a huge difference to your system and how you feel or when you fail.

Yes, I definitely agree with that.

Wayne

Gold Member

Energy / work is the product of force and distance - not just force.
You seem almost obsessed with making Force the only quantity of any interest. I (and others) keep pointing out that it isn't.
A railway locomotive sitting on its track, is exerting a huge force on the track yet NO work is being done.

waynexk8

Energy / work is the product of force and distance - not just force.

Yes right, the more force you use the more distance you cover in the same time frame ???

You seem almost obsessed with making Force the only quantity of any interest. I (and others) keep pointing out that it isn't.

“Sorry” If I sound like that, however all the other things power {work energy} can only be done if your exert/use force, if you see what I mean. I am only trying to find out how much force that both rep/s have used overall/total, as then the one with the most force will have the most tension on the muscles.

I myself {and sorry here I go again} would have to say the faster reps, as they cover 6 times the distance with the same weight in the same time frame.

Also some posted the below, could anyone explain or expand on this more please ??? As I have asked this question before, but no one here seems to be able to work it out with physics, and that surprises me so much.

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

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

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

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

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

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

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

I thought both the up and down were both the same average ??? However if you look at the clay test, they cannot be. I JUST THINK THAT WE ARE MISSING SOMETHING IN THESE AVERAGE FORCE, could it be the percentages I put up ???

A railway locomotive sitting on its track, is exerting a huge force on the track yet NO work is being done.[/QUOTE]

Yes.

Wayne

JaredJames

“Sorry”

There's no need for attitude. What sophiecentaur said about you is correct.

I'd recommend you do some basic reading on the materials you are trying to discuss. Your posts (to me) appear very cluttered and incoherent. I think you know what you want to say, but your lack of knowledge of the specifics is preventing you describing it.

waynexk8

You really need to check your definitions and understanding of the various words being thrown around here.

I've never said that - to reduce the time to move an object a distance of 1m, you have to increase the force. By increasing the force, the acceleration becomes larger:

a = f/m therefore if you double to force to 2f you get 2a = 2f/m.

Let's say we have an object that is 1kg. To move that 1m in 1s (1m/s) requires a KE of 0.5mv2 = 0.5*1*1 = 0.5 Joules of energy.
Now, to move it 100m in 1s (100m/s) requires a KE of 0.5*1*10000 = 5000 Joules. So in the first case I need a tiny amount of energy, in the second I need a huge amount in comparison.

For your case to move that 1m in 1s (1m/s) requires a KE as above (0.5 Joules).
Now, to move it 1m in 0.5s (2m/s) requires a KE of 0.5*1*4 = 2 Joules. So again you can see how simply halving the time of the repetition requires you to use more energy to complete it. The time applied is considered in the velocity figure.

For you to move the weight 1 rep in 1s requires 0.5 Joules - that is the energy you must provide to do it.
For you to move the weight 1 rep in 0.5s requires 2 Joules - again, that is the energy you must provide to do it.

If you do not provide that energy, you can't complete the rep in the required time.

Yes I do agree with that and have always said that, but here I go again and sorry, however energy cannot move the weight fast or slow on its own, first you have to have a force/strength, then that will use energy, the force will do work over a distance, and as you say the faster is moves the same weight in the same time frame the more energy you will use, so if you are using more energy you “must” be using more force, if not, why are you using more energy.

That’s why I said, if the averages forces are the same, what else are we missing, the percentages I put down ??? Or are the forces really average, and are they average on which reps ???

Could we look at it this way.

1,
1 rep at .5/.5 = 1 second.
2,
6 reps at .5/.5 = 6 seconds.
3,
1 rep at 3/3 = 6 seconds.

Which of reps 1 and 2 have the same average force as 3 ??? Please, as we know that 2 will have more force than 1.

Late here, will have to get back to any I missed tomorrow.

And thank you all for your time and help.

Wayne

waynexk8

There's no need for attitude. What sophiecentaur said about you is correct.

I'd recommend you do some basic reading on the materials you are trying to discuss. Your posts (to me) appear very cluttered and incoherent. I think you know what you want to say, but your lack of knowledge of the specifics is preventing you describing it.

Yes you are right, but I am trying my best, as to me it all seems quite straight forward the way I am saying it, and putting it down.

I did not mean to have an attitude or anything, and do very much thank you all for helping.

Wayne

Gold Member

Wayne, your first and last responses to my last post say it all. You are wanting a 'Physics' explanation for this yet you are rejecting it all the way along. You do not seem to realise where you are going wrong in this, either, As I have said before, read the basics of proper Physics and use them to see the actual meaning of my statements. The word "Product" has a specific mathematical meaning, for a start.

waynexk8

Wayne, your first and last responses to my last post say it all. You are wanting a 'Physics' explanation for this yet you are rejecting it all the way along. You do not seem to realise where you are going wrong in this, either, As I have said before, read the basics of proper Physics and use them to see the actual meaning of my statements. The word "Product" has a specific mathematical meaning, for a start.

Hi sophiecentaur, and thx for helping me. As you say, I may not seem to realise where I are going wrong, so could we clear the below up first

I know my physics is not very good, however I do not see how I am rejecting it. As some of you are telling me that I am using more energy in the faster reps, and this is what I have also said and agreed with all along, and several months before I posted here. However, we are talking of the same energy are we ??? As when I think of energy in this debate, energy means human energy used, so the energy expenditure needed for human muscles to do 6 reps at .5/.5 to 1 rep at 3/3 in the same time frame with the same weight will be far high, far higher for the faster reps. Could we clear this bit up first please. Do we all agree there ??? And this goes for all horizontal and all vertical movements when using the human body, all nutrition and calories books state this.

Wayne

waynexk8

After clearing the above up, could we also clear this one up please.

1,
1 rep at .5/.5 = 1 second.
2,
6 reps at .5/.5 = 6 seconds.
3,
1 rep at 3/3 = 6 seconds.

Which of reps 1 and 2 have the same average force as 3 ??? Please, as we know that 2 will have to use more/longer force/strength than 1. And which has the most overall force/strength, again we know that 2 will have to use more/longer force/strength than 1, as it uses more power.{work energy}

And thank you all for your time and help.

Wayne

waynexk8

this discussion is closed quite simply through an application of calculus.

We begin with force, better put as F=ma
Integrating with respect to acceleration, we are left with $$\int$$F=mv.
Momentum is the integral (wrt a) of force. In order to view the average force applied, we must take the integral of ma (wrt a) with a lower limit of t0=0 and an upper limit of t1=x and divide it by (t1-t0). By looking solely at the "peaks" of the forces, we do not get a clear picture.

However, I do not want to look at just the peaks, I want to look at each and every one of the 6 reps at .5/.5 = 6 seconds, moving a weight 12m. And to look at the one rep at 3/3 = 6 seconds moving a weight 2m. And see which uses the most overall/total force, or if its easier for you all, to see which in the 6 seconds puts the most tension on the muscles, as force in this case = tension.

Those who mentioned the kinetic energy output are even more correct, as the integral with respect to velocity of momentum is kinetic energy( $$\int$$mv= .5mv$$^{2}$$).That makes KE the double integral of force. Applying basic integral rules, we see that if the integral of one force x is > force y's, than the double integral of force x is > force y's.
Assuming the possibility of reimann sums for the summation of the two graphs necessary for the examples given by OP, we can give an uncertain of +- 10 units. Would anyone like to see the calculations made? lol

Sorry you have lost me there, or are you teasing ??? If not, which do you work out has the most tension on the muscles, and by how much.

Wayne

waynexk8

After clearing the above up, could we also clear this one up please.

1,
1 rep at .5/.5 = 1 second.
2,
6 reps at .5/.5 = 6 seconds.
3,
1 rep at 3/3 = 6 seconds.

Which of reps 1 and 2 have the same average force as 3 ??? Please, as we know that 2 will have to use more/longer force/strength than 1. And which has the most overall force/strength, again we know that 2 will have to use more/longer force/strength than 1, as it uses more power.{work energy}

And thank you all for your time and help.

Wayne

As I image all with say, if one rep has the same average force, then several reps will have the same average force, however, there will and must be more force/strength used to do two or six reps rather than one, so this also means that there will be more tension used to do two or six reps rather than one, so does average force in this debate actualy mean anything ???

Wayne

Gold Member

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

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

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

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

douglis

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

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

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douglis

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

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

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

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

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

Gold Member

You're absolutely right that the work done can not represent the energy expenditure.A practical method that is used to estimate the energy expenditure is the time-tension integral.Practically is calculated as: mean tension X total contraction time.
That's an interesting concept but that would kind of imply that just standing and holding something up would be good exercise????

douglis

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

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