The question no one seems to know average force

[Dale]

But when I am repping, I do not think I come to a stop at all ? See below video.

You are going up at one point in time and then you are going down at another point in time. Therefore at some point you have to come to v=0 in order to go the other way, even if it is just for an instant. Of course, since you are using free weights there is some horizontal movement, but that is not doing any work against gravity. In any case, even including the horizontal velocity, over one full repetition your initial and final velocity are the same so the average net force is certainly 0 over any full repetition and therefore the average force over any full repetition is equal to the weight (but pointed upwards).

QUESTION 1,
If I am only accelerating for 80% of the ROM {range Of Motion} or the rep, why do we count the other 20% ?

I think this is the key question. The time averaged force over an entire rep is obviously not the quantity you really want to know. I think you need to choose a better metric, one that really represents what you are interested in. You certainly could arbitarily pick some smaller part of a rep over which to average the force, but I just think that average force (over any interval) is simply not the quantity you really want to know. I think something like average power would be much more meaningful.

Don't forget, that force is a vector quantity, so if you lift something upwards with 2000 lb force and then pull it downwards the same amount of time with 1840 lb force then your average force is 160 lb upwards. I don't think that is what you want. Your muscles don't care if they are pulling up or down, left or right, they get tired either way. I think you really would be better off with a scalar quantity like power or work.

 

[waynexk8]

You are going up at one point in time and then you are going down at another point in time. Therefore at some point you have to come to v=0 in order to go the other way, even if it is just for an instant. Of course, since you are using free weights there is some horizontal movement, but that is not doing any work against gravity. In any case, even including the horizontal velocity, over one full repetition your initial and final velocity are the same so the average net force is certainly 0 over any full repetition and therefore the average force over any full repetition is equal to the weight (but pointed upwards).I think this is the key question. The time averaged force over an entire rep is obviously not the quantity you really want to know. I think you need to choose a better metric, one that really represents what you are interested in. You certainly could arbitarily pick some smaller part of a rep over which to average the force, but I just think that average force (over any interval) is simply not the quantity you really want to know. I think something like average power would be much more meaningful.

Don't forget, that force is a vector quantity, so if you lift something upwards with 2000 lb force and then pull it downwards the same amount of time with 1840 lb force then your average force is 160 lb upwards. I don't think that is what you want. Your muscles don't care if they are pulling up or down, left or right, they get tired either way. I think you really would be better off with a scalar quantity like power or work.

Hi and thx, learing a lot here.

Right I will try and work out the power later today, please do not do it for me, as I would like if you or anyone else has time to check and comment on the way I do it.

But in the meantime if anyone has time, could you please tell me what was wrong with the below please ? As I thought this workded out the extra force {strength} I was using as I was accelerating the weight, but the other person was not.


Too anyone, I still do not know what is wrong with the below for working out the extra force I used to accererate the same weight faster ?


F=MA. One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. 1 kg exerts a force of approximately 9.8 N.

MANY Ns WOULD YOU NEED TO ACCELERATE 50kg MASS 1 M/S/S,
AND HOW MANY Ns WOULD YOU NEED TO ACCELERATE 50KG MASS 6 M/S/S,

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

As 2 seconds is 4 times slower than .5 I divided .5 by 4.

Slow rep, moving at 2s per 1m
F=50kg x .125 = 6.25N

Fast rep, moving at .5s per 1m
F=50 kg x .5m = 25N.

Thus I need and have used {lets take off the 20% for the deceleration phase on the faster reps} N18.75 = 1.875kg MORE force {strength} on the fast reps to accelerate if faster in the same time frame !

So for doing 12 reps = 22.5kg

SO ON MY FAST REPS ACCELERATING AT .5/.5 {THAT’S .5 OF A SECOND FOR THE CONCENTRIC AND .5 SECOND FOR THE ECCENTRIC} AND SLOW REPS AT 2/4 USING A 100 POUNDS FOR 12 REPS, I HAVE USED 49.5 POUNDS MORE FORCE {STRENGTH} NOT COUNTING THE ECCENTRIC.

Wayne

 

[waynexk8]

Here is my go at working out the power {work energy} of both reps for just the concentric then for the concentric and eccentric, comments and advice please. NOT SURE I AM DOING RIGHT IN CONVERTING THE JOULES NMS TO POUNDS AS, One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. 1 kg exerts a force of approximately 9.8 N.

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

To determine the force we will need to figure out what the weight of the barbell is (W = mg = 50 kg x 9.81 m/s² = 490.5 kg.m/s² or 490.5 N).
Now, if work is equal to Force x distance then, U = 490.5 N x 1m = 490.5 Nm. 1 Newton meter “Nm” is equal to 1 joule, so we can say 490.5 joules of work was done. However, the concept of power takes time into consideration.

Slow reps power,
It took 2 seconds to complete the lift, so the power generated is 490.5 J divided by 2 s = 245.25 J/s. {55 pounds}

Fast reps,
It took .5 of second to complete the lift, so the power generated is 490.5 J divided by .5s = 981 J/s. {220 pounds – off 20% for the deceleration = 178 pounds}

Therefore, for the concentric only I use roughly 300% more power {work energy}

Concentric and eccentric,
Slow rep is moving 50kg at a constant 2 seconds concentric for 1 metre, and at a constant 4 seconds for the eccentric for 1 metre. It takes 6 seconds to travel 2m. Fast rep is accelerating 50kg 1 metre concentric and eccentric every .5 of a second. In addition, does this 6 times in 6 seconds, thus it has moved 50kg 5 times the distance in the same time frame as the 2/4 second slow rep.

Slow reps power,
It took 6 seconds to complete the lift, so the power generated is 490.5 J divided by 6s = 82 J/s. {18 pounds}

Fast reps,
It took 1 second to complete the lift, so the power generated is 490.5 J divided by 2 s = 490.5 J/s. x 490.5 by 6 {as I am doing 6 reps in the same time frame as the 1 rep at 2/4} = 2943 J {660 pounds ?}

And for the concentric and the eccentric, I use roughly 3500% WOW, but that seems awful high, and when I turned the joules NMs in pounds, I must have done something wrong ?

In biomechanics then, power is the rate of doing work and can be expressed algebraically as: (Power = Force x velocity) (Work = Force x distance) and (Power = Work divided by time)

Wayne

 

[Dale]

But in the meantime if anyone has time, could you please tell me what was wrong with the below please ?

Sure, I will go over it.

F=MA. One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. 1 kg exerts a force of approximately 9.8 N.

Fine.

MANY Ns WOULD YOU NEED TO ACCELERATE 50kg MASS 1 M/S/S,
AND HOW MANY Ns WOULD YOU NEED TO ACCELERATE 50KG MASS 6 M/S/S,

As you said above f=ma so
50 kg 1 m/s² = 50 N
50 kg 6 m/s² = 300 N

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

As 2 seconds is 4 times slower than .5 I divided .5 by 4.

Slow rep, moving at 2s per 1m
F=50kg x .125 = 6.25N

Fast rep, moving at .5s per 1m
F=50 kg x .5m = 25N.

OK here is the first mistake. Your units are not correct, remember what you said above, force (N) is mass times acceleration (kg m/s²), not mass times distance (kg m) as you have here.

What you need to do is to calculate the acceleration. Acceleration is the change in velocity divided by the change in time, and the velocity is the change in distance divided by the change in time. So let's go back up to the top. Your mass is moving at a constant 2 s for 1 m, so v = Δx/Δt = 1m/2s = 0.5 m/s. Now, you say that the speed is constant so the change in v is Δv = 0, so the acceleration is also 0.

Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0.

Thus I need and have used {lets take off the 20% for the deceleration phase on the faster reps} N18.75 = 1.875kg MORE force {strength} on the fast reps to accelerate if faster in the same time frame !

So for doing 12 reps = 22.5kg

Again, you have similar problems with units, you cannot equate Newtons and kilograms.

 

[Dale]

Here is my go at working out the power {work energy} of both reps for just the concentric then for the concentric and eccentric, comments and advice please. NOT SURE I AM DOING RIGHT IN CONVERTING THE JOULES NMS TO POUNDS AS, One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. 1 kg exerts a force of approximately 9.8 N.

This is correct except that you cannot convert joules into pounds. You could convert it into calories (1 kcal = 4187 J)

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

If you are doing 4 fast reps in the same time as 1 slow rep then it has moved 4 times the distance in the same time.

To determine the force we will need to figure out what the weight of the barbell is (W = mg = 50 kg x 9.81 m/s² = 490.5 kg.m/s² or 490.5 N).
Now, if work is equal to Force x distance then, U = 490.5 N x 1m = 490.5 Nm. 1 Newton meter “Nm” is equal to 1 joule, so we can say 490.5 joules of work was done. However, the concept of power takes time into consideration.

All correct.

Slow reps power,
It took 2 seconds to complete the lift, so the power generated is 490.5 J divided by 2 s = 245.25 J/s. {55 pounds}

All correct except for the conversion to pounds.

Fast reps,
It took 1 seconds to complete the lift, so the power generated is 490.5 J divided by 1s = 490.5 J/s. {110 pounds – off 20% for the deceleration = 88 pounds}

I wouldn't take off for the deceleration, you are still doing work. And the time is 0.5 s for a fast rep. So you for the power you should have 490.5J/0.5s = 981 J/s.

Sorry, that is all I have time for now, but I am sure you get the idea. You are going 4 times as fast so you are doing the same amount of work in 1/4 the time which implies 4 times the power, despite the fact that the average force is the same.

 

[waynexk8]

Sure, I will go over it.

Fine.

As you said above f=ma so
50 kg 1 m/s² = 50 N
50 kg 6 m/s² = 300 N

OK here is the first mistake. Your units are not correct, remember what you said above, force (N) is mass times acceleration (kg m/s²), not mass times distance (kg m) as you have here.

What you need to do is to calculate the acceleration. Acceleration is the change in velocity divided by the change in time, and the velocity is the change in distance divided by the change in time. So let's go back up to the top. Your mass is moving at a constant 2 s for 1 m, so v = Δx/Δt = 1m/2s = 0.5 m/s. Now, you say that the speed is constant so the change in v is Δv = 0, so the acceleration is also 0.

Now, in reality the speed is not constant, but it starts and stops at 0 so the average acceleration is 0.

Again, you have similar problems with units, you cannot equate Newtons and kilograms.

Hi, and thanks for your time and helping me.

So you are saying the below is wrong ?

Slow rep is moving 50kg at a constant 2 seconds for 1 metre. It takes 2 seconds to travel 1m. Fast rep is accelerating 50kg 1 metre every .5 of a second. And does this 4 times in the same time frame as the slow rep. Thus, it has moved 50kg 3 times the distance in the same time frame as the 2-second slow rep.

As 2 seconds is 4 times slower than .5 I divided .5 by 4.

Slow rep, moving at 2s per 1m
F=50kg x .125 = 6.25N

Fast rep, moving at .5s per 1m
F=50 kg x .5m = 25N.

However, I found the below and checked on there and they seem right ? Unless I am calculating the wrong thing ? I will read what you wrote again, but getting a little sleepy, and as you know its best not to go over things then, as its getting late here, its gone 01.00

http://www.ajdesigner.com/phpforce/force_equation.php

Wayne

 

[waynexk8]

This is correct except that you cannot convert joules into pounds. You could convert it into calories (1 kcal = 4187 J)

If you are doing 4 fast reps in the same time as 1 slow rep then it has moved 4 times the distance in the same time.


All correct.

All correct except for the conversion to pounds.

I wouldn't take off for the deceleration, you are still doing work. And the time is 0.5 s for a fast rep. So you for the power you should have 490.5J/0.5s = 981 J/s.

Sorry, that is all I have time for now, but I am sure you get the idea. You are going 4 times as fast so you are doing the same amount of work in 1/4 the time which implies 4 times the power, despite the fact that the average force is the same.

Thx get what you say there, and see how silly I was trying to change Js to pounds.

Wayne

 

[waynexk8]

I WAS JUST THINKING, Maybe that we all are forgetting that the slow reppers are going at a constant speed, and NOT accelerating, BUT I am NOT going at a constant speed, as I AM accelerating all the time, and there is a BIG difference between going at a Slow constant speed, and accelerating at a VERY FAST ACCELERATING SPEED.

THUS IF THE BELOW IS RIGHT, HOW CAN EVEN THE AVERAGE FORCE BE THE SAME ? IF I AM ACCELERATING AND THE OTHER AT A CONSTANT SPEED, ARE NOT THINGS VERY DIFFRENT ?

They push a 100 pounds at 2/4 {we are just counting the concentric} for 1m with a constant extra force {strength} of 6.25N. That works out they are using 1.4 pounds extra for the whole duration of 2 seconds.

I can push that same 100 pounds at .5/.5 for 2m, accelerating with an extra force {strength} of 25N. That works out I am using 5.6 pounds extra for the whole duration of 2 seconds. Take off 20% for the deceleration = 4.4 pounds. But I am just having mini breaks, the eccentrics which you are not having.

Therefore, in 2 seconds of lifting for the slow 1 rep, and me lifting for 2 seconds fast, I have used OVER 200% more force {strength}

Thus if we did 12 reps you would use an extra 16.8 pounds, and I would use an extra 52.8 pounds, or 36 pounds MORE force {strength} that they in one set, and that's a VERY lot that's as I said over 200% more in each rep and in each set.

Wayne

 

[Dale]

THUS IF THE BELOW IS RIGHT, HOW CAN EVEN THE AVERAGE FORCE BE THE SAME ? IF I AM ACCELERATING AND THE OTHER AT A CONSTANT SPEED, ARE NOT THINGS VERY DIFFRENT ?

Don't forget that acceleration and force are vector quantities. So if you use 2000 lb force to accelerate it really fast to the left and then 2000 lb force to accelerate it really fast to the right (which you have to do) then the average force is 0. Similarly with up and down forces and accelerations. No matter what you do, over any cycle the average force is equal to the weight (but pointed up).

 

[waynexk8]

Thx for the above, not much time now DaleSpam.

If you move a load faster over a given distance you are producing more force/power, it is impossible for this not to be the case. Moving 100 pounds a distance of 2 feet in 1 second requires more force/power than moving 100 pounds a distance of 2 feet in 2 seconds.

Agreed.

In my example, the rep was 1m concentric and 1m eccentric.

However, something is still bugging me on these average forces, but think I have worked out where we are all going wrong, no time now. But why are we counting the decelerating phase of the fast concentric, thus why are we talking 20% off ? I know saying the following may seem a little silly, but in a drag race, they only count the force or in their case speed/time that was done in the ? Mile they raced, they did not count the decelerating part of it. Got to go later.

Wayne