How can this be average force ?

[waynexk8]

The cars are not expending any energy to create tension on the cables.

Gravity on acting on the car creates a downwards force (the weight) which puts tension on the cable. No energy is expended in this process and the cables don't expend energy supporting the cars.

The cars are not expending any energy to create tension on the cables.

Gravity on acting on the car creates a downwards force (the weight) which puts tension on the cable. No energy is expended in this process and the cables don't expend energy supporting the cars.

Would not the cables {the muscles} not have to use energy to just hold the cars {weight} and use a force to keep that at a position ? However that was just a simple scenario to try and make more/different sense of this problem.

However all, could we go back to the muscles are using more energy when lifting the weight up and down faster in the same time frame, thus there must be a scientific reason for this, I say it’s because the muscles have more activity, and are using more overall or total force, otherwise, why are the muscles use more energy ? We could put this debate to a machine if its easier !

Wayne

 

[JaredJames]

Would not the cables {the muscles} not have to use energy to just hold the cars {weight} and use a force to keep that at a position ?

Force yes, energy no.

However all, could we go back to the muscles are using more energy when lifting the weight up and down faster in the same time frame, thus there must be a scientific reason for this, I say it’s because the muscles have more activity, and are using more overall or total force, otherwise, why are the muscles use more energy ? We could put this debate to a machine if its easier !

The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

 

[waynexk8]

Here is what someone wrote from this forum.

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

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

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

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

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

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

So what some of you are saying is, that force must be only expressed as either the work done (force x distance) or the impulse applied (force x time).}

So as we all know, the higher high forces and higher high forces are higher, thus these would be why the energy used was more, and thus put a greater ammout of tension on the muscles




So let's see if I get this right, force can = work and impulse.

So I think we all can agree, that the faster rep will have a greater quantity of work, and if work = force, then that must mean more force ? As the faster reps have moved the weight far far far further then the slow rep in the same time frame, work = force x distance.

And

As impulse = force x time. The small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. However as I have applied a higher force for the same time, would not the impulse be higher ?

Wayne

 

[JaredJames]

So as we all know, the higher high forces and higher high forces are higher, thus these would be why the energy used was more, and thus put a greater ammout of tension on the muscles

I have no idea what you are talking about. Complete gibberish.

I explained in my previous post why more energy was used. Period. End of story.

To reiterate for you, you are imparting KE on the weight from CE in your body. The more KE you impart the more CE your body must use. Therefore, the faster you move the weight the you get significant energy increases (as I showed before with the difference in KE from 1m/s to 2m/s goes from 0.5 Joules to 2 Joules, so you need more chemical energy to gain a faster rep speed).

 

[waynexk8]

Force yes, energy no.


The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

Great, that’s basically what I think.

What some other people think, and they are in the very very very lower of the training World, they are the one grain of sand in the World, the rest of us are the rest of the grains of sand, not sure how low that is of a percentage, but as you can see its low, ROL.

Here are how the top people train, trying to move a heavy weight as fast as possible.
Derek Poundstone,

Mariusz Pudzianowski,

Ronnie Coleman


They think that when we are deceleration in our faster reps, that then is where their more constant lower force make up the force, but as I sort of proved with my percentage theory, it’s not so. fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and the lower forces of the slower reps can not make up the force in the same time frame.

Wayne

 

[sophiecentaur]

All this about bridges is irrelevant nonsense.
Also, as has been said many times, if you don't use the right definitions and the right terms there is NO Physics that can apply.

No one has disagreed about the fact that you expend more Power doing faster reps. But there is no simple relationship between this and the Work Done (defined, as it is, within Physics). You can keep on about your numbers and stats but that won't alter this fundamental problem. Why do you keep trying? It just gets up people's noses. Discuss it as much as you like in a non-technical group but don't expect to get anything out of this forum if you're no prepared to go along with the basic rules of Physics.

And how can you say (above) "if work = force" in the same sentence as "work = force times distance"? Can you not see the contradiction? Also, Impulse has nothing to do with Work.

Just read a basic Mechanics / Dynamics book and get some discipline in your Science and you may have a chance of getting this.

 

[JaredJames]

They think that when we are deceleration in our faster reps, that then is where their more constant lower force make up the force, but as I sort of proved with my percentage theory, it’s not so. fast rep, {split up into 5 segments} 100, 100, 100, 80, 20. {second fast rep 140, 100, 100, 60, zero} Slow rep, 80, 80, 80, 80, 80. The averages are/seem the same, but the peak forces are higher.


100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and again, 100 = 20 or 25% more than 80. Or second rep, 140 = 60 or 75% more than 100, then 100 = 20, or 25% more than 80, and again, 100 = 20 or 25% more than 80, and the lower forces of the slower reps can not make up the force in the same time frame.

Wayne

Please stop using these numbers, I don't think they mean anything and I don't think anyone else knows what they mean either.

Frankly, from the explanations you have given it is clear that not even you are that sure of what you're talking about.

Sorry to sound so blunt, but I don't think dragging this out any further is going to be productive.

 

[waynexk8]

I have no idea what you are talking about. Complete gibberish.

Sorry, I meant the higher forces for the first part of the faster reps, please see.

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


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

I explained in my previous post why more energy was used. Period. End of story.

Yes ok thank you.

To reiterate for you, you are imparting KE on the weight from CE in your body. The more KE you impart the more CE your body must use. Therefore, the faster you move the weight the you get significant energy increases (as I showed before with the difference in KE from 1m/s to 2m/s goes from 0.5 Joules to 2 Joules, so you need more chemical energy to gain a faster rep speed).

Yes get you, to use more KE, you have to have more muscle activity and to do this you have to use a higher force/strength to move an object from a to b faster, as moving the object 1m in 1 second, but to move the object 6m in 1 second takes more higher high force/strength and higher peak force/strength. As if you do not use a higher force, the object will not move so fat in the same time frame. Thus because the more KE, from the higher high and peak forces the more energy is used.

Big thank you for your help and time, but maybe more for your patience.

Ho and for the person that thought I maybe having you on, I was definitely not, as I have made many videos proving you fail far far far faster in the faster reps, here one of me showing this.

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

Wayne

 

[JaredJames]

I have made many videos proving you fail far far far faster in the faster reps, here one of me showing this.

Why would you need a video showing this?

It is clear from even basic energy calcs that you use significantly more energy to go faster. So anyone who questions this clearly doesn't understand the physics and frankly, doesn't have a say. Especially not to dismiss it.

 

[waynexk8]

Please stop using these numbers, I don't think they mean anything and I don't think anyone else knows what they mean either.

These numbers were just meant as a basic guideline, nothing more, just somthing to get people to see more into the debate. As we were using 80 pounds as 80% of or RM {repetition maxumun} the most we could lift for once. So we just said for the first part of the rep we were useing a 100 as in pounds, then still a 100, then still a 100, then down to 80 for the start of the deceleration, then down to 20.

Frankly, from the explanations you have given it is clear that not even you are that sure of what you're talking about.

Well I do know what I am talking about, as all the below agree with me, lots are biomacanics, kinsologists and have master in physics, but I find it hard to put into words. Momentum, or as I like to say, off loading may be an issue, if the weight is to light, and you are not accelerating the weight enough.

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

Roger Enoka,
The number of muscle fibers activated to lift a weight depends on two factors: (1) the amount of weight; and (2) the speed of the lift. Although more muscle fibers are activated during fast lifts, they are each generating MORE force. We know this because the rate at which the muscle fibers are activated by the nervous system increases with contraction speed.

The force that a muscle must exert to move a load depends on two factors: the mass of the load and the amount of acceleration imparted to the load. The number of muscle fibers recruited during the lift increases with the speed the lift.


The rate at which any motor unit, low or high threshold, can discharge action potentials is not maximal during slow contractions. As contraction speed increases, so does discharge rate for all motor units.

The most common finding is that it is the intermediate fiber type, the fast muscle fiber (type IIa) that experiences the biggest increase in size (strength) in individuals who perform conventional weight lifting (heavy loads,) and body building (lighter loads, fast/explosive reps) training. Neither type of training appears to have a significant effect on the size of types I and IIx fibers.

William Kraemer,
http://www.education.uconn.edu/direc...ails.cfm?id=44

Steven Plisk,
http://www.excelsiorsports.com/files...e_Training.pdf

It’s generally understood that a certain threshold of training intensity is needed to effect positive adaptation, but many athletes and coaches still believe that resistance must be sufficient that the weight can’t — or shouldn’t — be moved very fast. I intend to challenge this proposition, and to make a case for the fact that acceleration is the name of the game even when executing basic structural movements (e.g. the squat and deadlift). It’s really just a simple matter of understanding the fundamental nature of force, and of putting this concept into practice regardless of task or workload.

F=m•a Revisited

At first glance, “force is the product of mass and acceleration” appears to imply that there is no force without motion (or vice-versa), but that’s not necessarily the case. For example, since gravity is expressed as an acceleration constant [~9.8 m/sec2], a vertical force of ~980 kg•m/sec2 (or Newtons) would be required to hold a 100 kg barbell in place statically.

Despite the apparent simplicity of F=m•a, the inability or unwillingness to grasp its functional
significance is an underlying cause of the nonsense taking place in many weightrooms. This concept is neither contrived nor trivial, and shouldn’t be tucked away in a physics textbook until needed to support some abstract opinion. In fact, it’s a foundational principle upon which all motion is based (with strength training being no exception). When you consider that any movement is essentially an act of defying gravity — which itself is an accelerative force — the central issue becomes: What is being moved, and how fast?

Vladimir M. Zatsiorsky,
http://www.hhdev.psu.edu/kines/facul...orsky%20CV.pdf

Westside Barbell,
http://www.westside-barbell.com/articles/

Dr. Yuri Verkhoshansky,
http://www.verkhoshansky.com/

Dr. Hatfield, (aka Dr. Squat)
http://drsquat.com/content/knowledge...-look-strength

Per Aagaard Professor, PhD
Institute of Sports Science and Clinical Biomechanics
University of Southern Denmark

When a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

For instance, a 120 kg squat can easily produce peak vertical ground reaction forces (beyond the body mass of the lifter) of 160-220 kg's when executed in a very fast manner! Same goes for all other resisted movements with unrestricted acceleration (i.e. isokinetic dynamometers (and in part also hydraulic loading devices) do not have this effect).

This means that higher forces will be exerted by MORE muscles fiber when a given load is moved at maximal high acceleration and speed - i.e. contractile stress (F/CSA) will be greater for the activated muscle fibers than when the load is lifted slowly...
best wishes
Per

Sorry to sound so blunt, but I don't think dragging this out any further is going to be productive.

That’s ok; you are a gentleman and a scholar, as I said before, and to all that have helped, thank you for your time and help.

Wayne

 

[JaredJames]

Well I do know what I am talking about,

Really? Then you follow with:

Acceleration is not momentum,

Good

its the opposite of momentum,

Bad

acceleration requires more force/strength, where momentum requires less force/strength.

Ugly

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

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

That was enough for me to not read on.

 

[douglis]

The muscles use more energy going faster because they are required to in order to gain the higher velocities. (KE = 0.5mv² and m is constant, you're increasing v which gives a higher KE.)

They are required to generate significantly more force and as such expend more energy doing so.

You want a bullet to travel faster you have to give it a lot more kinetic energy, which is done by using a more powerful explosion.

The muscles use chemical energy and convert it into kinetic energy. When moving faster there is more kinetic energy required (as above) and so more chemical energy required to be converted. So your muscles use more energy going faster than slower. (From above, moving the weight faster = higher KE and so more CE needs to be converted to KE.)

jarednjames,
If we want to calculate the total energy we must take into account the whole lifting of the weight(let's say it's been lifted for a height h) which ends with zero velocity.So the whole KE,that was gained from the conversion of CE,is finaly converted to PE equal with mgh.
So the theoretical minimum of the energy expenditure when you lift a weight is mgh assuming 100% efficiency.Here's the tricky part and I'll try to make my point with a more simple example.

In the lifting weight example the starting and ending velocity is zero hence the average acceleration is zero so,for simplicity's sake,I'll use constant speeds in my example(average acceleration is zero too).
Two rockets are moving upwards with constant speed.The first with 50mph and the second with 100 mph(the air resistance is excluded).Since the speed is constant the engines of both rockets are using force equal with the rocket weight.
After an hour the engine of the the fast rocket has produced exactly double work than the slow rocket.But the total fuel expenditure will be exactly the same for both rockets.It has to be this way since the rockets use the same force for the same duration.Both the engines were just balancing the weight of the rocket for an hour.The only difference is that the engine of the fast rocket worked with double efficiency.

So I believe that in all cases where the average acceleration is zero(like in weight lifting and in the rockets example) and the same average force is being used,the rate of energy expenditure does not depend on speed and distance.

 

[JaredJames]

Two rockets are moving upwards with constant speed.The first with 50mph and the second with 100 mph(the air resistance is excluded).Since the speed is constant the engines of both rockets are using force equal with the rocket weight.
After an hour the engine of the the fast rocket has produced exactly double work than the slow rocket.But the total fuel expenditure will be exactly the same for both rockets.It has to be this way since the rockets use the same force for the same duration.Both the engines were just balancing the weight of the rocket for an hour.The only difference is that the engine of the fast rocket worked with double efficiency.

So I believe that in all cases where the average acceleration is zero(like in weight lifting and in the rockets example) and the same average force is being used,the rate of energy expenditure does not depend on speed and distance.

This is non-sense.

What you have just said is that two vehicles traveling at constant velocities (different velocities) use the same fuel. Don't be ridiculous.

You can't just exclude air resistance - obviously if you throw out the variables you get rubbish like this because your resisting force only comes from gravity which is constant. Even then you need to realize that it takes more fuel to produce the thrust in the high speed scenario.

Fuel use is not linear and so a vehicles traveling at twice the speed of the first will use significantly more fuel - even if it is at constant speed.

 

[douglis]

This is non-sense.

What you have just said is that two vehicles traveling at constant velocities (different velocities) use the same fuel. Don't be ridiculous.

You can't just exclude air resistance - obviously if you throw out the variables you get rubbish like this because your resisting force only comes from gravity which is constant. Even then you need to realize that it takes more fuel to produce the thrust in the high speed scenario.

I excluded the air resistance to create an equivalent example with weight lifting where obviously the air resistance is negligible.I thought there was no need to explain that.
So the thrust is always equal with the weight in any constant speed scenario and so is the fuels expenditure.

Fuel use is not linear and so a vehicles traveling at twice the speed of the first will use significantly more fuel - even if it is at constant speed.

Only if you take into account the air resistance.

 

[JaredJames]

I excluded the air resistance to create an equivalent example with weight lifting where obviously the air resistance is negligible.I thought there was no need to explain that.
So the thrust is always equal with the weight in any constant speed scenario and so is the fuels expenditure.

You're ignoring initial energy expenditure. The initial acceleration (as per my previous example) uses significantly more energy to achieve the required rocket velocity (or rep time) and so already you have used more energy than the slower rocket (or rep). Even if they use the same energy at constant velocity it takes significantly more to achieve (and stop) the faster one.

You can't just wave this off.

It doesn't matter what happens to the KE after the weight stops. You must provide significantly more KE in the faster rep and so that means more CE. As none of that KE gets translated back into CE it's irrelevant and you have the energy expenditure being higher for the faster rep.

Fuel use is not the same.

Only if you take into account the air resistance.

As above re energy expenditure.

Douglis, I'm slightly concerned your post came almost perfectly from here:

http://lofi.forum.physorg.com/Energy-In-Bench-Press_28867.html

Same debate by the look of it. Certain posts / responses are identical to here. Only we seem to be on a one month delay.

 

[douglis]

You're ignoring initial energy expenditure. The initial acceleration (as per my previous example) uses significantly more energy to achieve the required rocket velocity (or rep time) and so already you have used more energy than the slower rocket (or rep). Even if they use the same energy at constant velocity it takes significantly more to achieve (and stop) the faster one.

You can't just wave this off.

Again to create an equivalent example with weight lifting we must assume that the rocket stops at the end only due to gravity and not with opposite thrust.
So...the initial acceleration(which requires more fuels) is exactly balanced by the final deceleration(requires less fuels).On average the fuels are identcal as if the motion would be constant.After all that's what practically means zero average acceleration.

It doesn't matter what happens to the KE after the weight stops. You must provide significantly more KE in the faster rep and so that means more CE. As none of that KE gets translated back into CE it's irrelevant and you have the energy expenditure being higher for the faster rep.

Fuel use is not the same.

Again the same thing with other words.
The KE is gained during the acceleration phase.The more CE that is spent in a fast rep(and is converted to KE) during the initial acceleration,compared with the slow rep,is exactly balanced by the less CE that is spent during the final deceleration(where the gained KE is converted to PE).
The less mechanical energy(PE+KE) that is produced per unit of time in a slower rep is balanced by more heat production.It MUST be this way since the same average force is used per unit of time.
That's also the case with the constant speed rockets.The more mechanical energy that the faster rocket uses is exactly balanced by the more heat that the slow rocket produces.They spend exactly the same fuels with the difference that the engine of the slow rocket works less efficiently.

As above re energy expenditure.

Douglis, I'm slightly concerned your post came almost perfectly from here:

http://lofi.forum.physorg.com/Energy-In-Bench-Press_28867.html

Same debate by the look of it. Certain posts / responses are identical to here. Only we seem to be on a one month delay.

Yes...an old debate.I wasn't the one who brought it back.

 

[Dale]

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.

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

 

[JaredJames]

Again to create an equivalent example with weight lifting we must assume that the rocket stops at the end only due to gravity and not with opposite thrust.
So...the initial acceleration(which requires more fuels) is exactly balanced by the final deceleration(requires less fuels).On average the fuels are identcal as if the motion would be constant.After all that's what practically means zero average acceleration.

What are you talking about? In both cases (fast and slow rocket) if they use gravity only to stop they don't use any fuel. Meaning the only important parts are initial acceleration and energy requirement for constant velocity. Hence the faster rocket requiring more fuel.

If you have a fast rep and a slow rep, both covering the same distance, you will need to use energy to stop the faster one if the slower one is only using gravity to stop. Meaning the fast rep requires additional energy to stop.

Why are you trying to say the fast rep uses less energy to stop than the slow rep? That's the only way the energy values would balance out.

Slow rep: CE to KE to get the weight moving. KE used to overcome resistance. Gravity decelerates the weight to a stop, but CE is still used to support it at the peak.

Fast rep: More KE required so more CE to get the weight moving. KE used to overcome resistance. Gravity plus additional CE required to decelerate the weight over the same distance to peak, again CE is still used to support it at the peak.

The fast rep requires energy to slow it. More KE is dissipated as heat in the fast reps.

Either way, the fast rep requires more CE (significantly more) and the two don't balance out at the end. Don't confuse the final GPE values - which are equal with both weights - with the fact the fast rep has more energy during travel and as such has to dissipate it as heat to bring it to a halt.

If allowed to stop solely using gravity, the faster rocket / rep will travel further. You must introduce resistance (from thrusters / muscles) in order to stop them in the same distance.

 

[douglis]

What are you talking about? In both cases (fast and slow rocket) if they use gravity only to stop they don't use any fuel. Meaning the only important parts are initial acceleration and energy requirement for constant velocity. Hence the faster rocket requiring more fuel.

The deceleration phase(where the gravity stops the rocket) is much longer at the fast rocket and proportional with the acceleration phase.

If you have a fast rep and a slow rep, both covering the same distance, you will need to use energy to stop the faster one if the slower one is only using gravity to stop. Meaning the fast rep requires additional energy to stop.

Why are you trying to say the fast rep uses less energy to stop than the slow rep? That's the only way the energy values would balance out.

Slow rep: CE to KE to get the weight moving. KE used to overcome resistance. Gravity decelerates the weight to a stop, but CE is still used to support it at the peak.

Fast rep: More KE required so more CE to get the weight moving. KE used to overcome resistance. Gravity plus additional CE required to decelerate the weight over the same distance to peak, again CE is still used to support it at the peak.

The fast rep requires energy to slow it. More KE is dissipated as heat in the fast reps.

Either way, the fast rep requires more CE (significantly more) and the two don't balance out at the end. Don't confuse the final GPE values - which are equal with both weights - with the fact the fast rep has more energy during travel and as such has to dissipate it as heat to bring it to a halt.

jarednjames...have you ever tried to lift weights?You always rely on gravity in order to stop the weight.You never use force from the antagonist muscle(opposite thrust) to do the job.
That means the greater initial acceleration in faster lifting is ALWAYS accompanied by greater final deceleration.The average acceleration is ALWAYS zero or else the weight will fly from your hands.

If allowed to stop solely using gravity, the faster rocket / rep will travel further. You must introduce resistance (from thrusters / muscles) in order to stop them in the same distance.

Here's another example to help you understand my point.

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]

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

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

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

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

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

That means the greater initial acceleration in faster lifting is ALWAYS accompanied by greater final deceleration.

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

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

 

[sophiecentaur]

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

 

[douglis]

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

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

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

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

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

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

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

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

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

 

[douglis]

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

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

 

[JaredJames]

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

From the OP:

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

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

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

 

[douglis]

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

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

 

[waynexk8]

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

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

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

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

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

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

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

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

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

Yes.

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

Yes.

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

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

Wayne

 

[sophiecentaur]

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

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

 

[JaredJames]

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

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

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

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

 

[JaredJames]

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

Certainly. But you're missing the point.

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

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

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

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

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

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

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

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

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

 

[waynexk8]

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

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

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

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

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

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

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

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

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

Wayne

 

[JaredJames]

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

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

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

Do you not mean incorectely ?

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

Aren't = are not.

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

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

 

[waynexk8]

Certainly. But you're missing the point.

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

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

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

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

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

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

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

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

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

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

Energy Expenditure and Nutrient Oxidation will and have been established in a room calorimetry. Many Statistics will be measured eg; Heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, to determine the effects of exercise at different intensities.

On every occasion when an activity has been practiced at a faster rate, the heart rate, oxygen consumption, fat consumption, protein consumption and carbohydrate consumption, has been higher, therefore energy expenditure, nutrient oxidation will be higher.

The increase in the rate of energy expenditure/cost will be due the increase in the intensity/force of muscle activity. When a muscle is subjected to a greater acceleration, speed or velocity, EMG reading will go up, when muscle uses more force, more force will equal more acceleration, speed, velocity, equals more energy. The greater the force exerted by a task, the more rapidly the muscles fatigue.

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

Wayne

 

[waynexk8]

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

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

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

Yes agree, sorry it was my fault.

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

Aren't = are not.

Yes get you.

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

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

Wayne

 

[waynexk8]

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

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

Ok, I did say sorry above.

Wayne

 

[douglis]

Certainly. But you're missing the point.

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

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

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

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

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

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

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

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

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

You overcomplicate things.Forget for a moment that you 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.''