Impulse/force in pounds for the time frame


by waynexk8
Tags: frame, impulse or force, pounds, time
DaleSpam
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#181
Feb29-12, 05:27 AM
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Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.
douglis
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#182
Feb29-12, 05:49 AM
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Quote Quote by DaleSpam View Post
Wayne, do these people know that their private correspondence with you is being posted on the Internet including their name? This seems like a very bad practice unless you have their explicit permission.
Those discussion are from many years ago and have been posted in more than 10 different forums around the internet(without any permission of course).The same is also true for the discussions here!

Twelve pages and so many threads for a question that can be answered in just one line!
The average force is always the weight regardless the lifting speed.
waynexk8
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#183
Feb29-12, 07:58 AM
P: 399
What forces have you the slow got that can make up the propulsive forces in the studys, there are no extra force in the slow reps, are there ???

Back later to catch up on all posts.

Wayne
jmmccain
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#184
Mar1-12, 12:07 AM
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Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.
sophiecentaur
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#185
Mar1-12, 04:24 AM
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Quote Quote by jmmccain View Post
Wayne,

May I suggest an experiment? Do you have access to a bathroom scale and two video cameras?

Set up one camera to record the scale reading and perform your various fast and slow exercises. Afterwards, view the recording frame by frame and plot the scale reading for each frame (like douglis' graph, but it won't look the same). Don't forget to subtract your own weight from all the readings. This will rather conclusively demonstrate the difference in forces.

Add up all the readings for a repetition (again, adjusted for your weight). Then divide by the number of frames over which the readings were observed. This will be the mathematical average and should be very close to the weight you are lifting. How close depends upon how well the experiment is performed.

The second camera is used to record the movement of the weight with a tape measure, or other suitably visible scale, in the background. Again, go through frame by frame and plot the position of the weight. This should look something like douglis' graph. Next, plot the change in position from one frame to the next (velocity). Next, plot the change of this change (acceleration) from one frame to the next. This third plot should look just like the first plot of the scale readings.

F = ma

The first plot of the scale readings is force. The last plot is acceration. Since the mass doesn't change, the two remain proportional to each other and the plots should be very similar.

The scale and cameras won't lie. If you need help decipering the results and making the appropriate plots, well, I may be busy, but there are others around who will help.
I'd advise not getting too involved with this. All your suggestions have been made many times before, in this thread and earlier threads. Wayne does not believe in the accepted ideas of Physics. He has his own models and vocabulary of Physics.

Also, your thought experiment on the bathroom scales would not show the sampled forces accurately or frequently enough to convince Wayne. There would be errors which he would jump on and claim that the experiment showed him to be right. The scales and camera would lie in practice unless a much more sophisticated system were used.

In any case, what you say would only apply to free lifts and not to exercises on machines that introduce friction. That also confuses Wayne and strengthens him in his misconceptions.
waynexk8
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#186
Mar1-12, 08:36 PM
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I wish you and D. would do the same, and answer some of my questions ???

Quote Quote by sophiecentaur View Post
OK. You go from London to Bristol at an average speed of 62miles per hour. You could work that out easily enough. But how would you, Wayne, work out what your 'total speed' was? That's how daft your idea of total force is.
You got a point on that one, see what you mean, WILL think more on this one, BUT why then are you, and D. saying, the average force is the same ??? I think that you have taken my maximum force from the acceleration, and then took away my minimum force from the deceleration, but ďwhyĒ did/do you do this ??? However ever, speed is a little different to force, take a look on this video, they measure the maximum force and minimum force, and then they calculate the average force, which will be the same, but we do NOT want this, this is what I keep ON telling D. over and over, we do not want the average force, we want all the three average force, the forces from the negative, the force when lowing the weight, the peak force, the huge peak force from the transition from negative to positive, and the positive force, the forces when lifting the weight.

NOW, if the weight was 80 pounds and my 1RM was a 100 pounds, if I start the lift from the top, lower in .5 of a second and lift in .5 of a second for the fast, and lower in 3 seconds and lift in 3 seconds.

1,
The fast,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ??? 140 ??? Then the tensions on the muscles to accelerate the weight up would be close to 100. 79 + 140 + 100 = 319/3 = 106 over 1 second.

2,
The slow,
The tension on my muscles to lower it would be just under 80, call it 79, then at the transition from eccentric to concentric, would have the maximum tensions on the muscle, say ??? 85 ??? Then the tensions on the muscles to accelerate the weight up world be close to 80. 79 + 85 + 80 = 244/3 = 81 over 6 seconds.

3,
Divide the slow of 6 seconds by the fast of 1 second, 81/6 = 13 over 1 second.

Ok thatís wrong, but how would you do this please ??? What if we added them up ??? As in this debate/test/study, average means nothing, so itís fast, 79 + 140 + 100 = 319 over 1 second. Slow 79 + 85 + 80 = 81 over 6 seconds, 13 over 1 second.

Take a look at this video, they say how they work out the average force on it, and they show the maximum force and minimum force, click on click to preview. What you and D. are doing, is taking to points, and taking the maximum and minimum forces and averaging them up, well of course the average force is going to be the same, but it means nothing here, you are just taking the maximum and minimum, and adding them together and averaging them out, thatís what we donít want here, we NEED to add in the maximum acceleration force, the maximum peak force, {of the transition from negative to positive, this maximum peak force, can put up to 140 pounds of force on the muscles} and the maximum deceleration force, what you are doing is leaving out the most important, the maximum peak force from the transition, why are you leaving this force out, which put the huge tensions on the muscles, and the debate is, which puts the most over the same time frame, the most overall or total tension on the muscles.

http://webcache.googleusercontent.co...&ct=clnk&gl=uk

Wayne
waynexk8
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#187
Mar1-12, 08:53 PM
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Quote Quote by sophiecentaur View Post
I have read the paper an I think I can see your problem. I find little to disagree with what is written.

They are not considering the whole cycle (lift/lower) when they refer to mean force.
Right could we get to this point, as D. would not explain, could you explain on what you think they are leaving out ??? as I do not get it.

Quote Quote by sophiecentaur View Post
They are not only discussing free lifts but machines that present RESISISTIVE loads. Everything changes in that case because you are not just changing gravitational potential energy in that case but work is being done in overcoming friction.
Please forget the machines, this at this moment is just about free weights.

Quote Quote by sophiecentaur View Post
If you had read what they say then you would not think they are disagreeing with established physics at all. You did not understand what you were arguing about because you insisted on giving details instead of condensing your questions into something meaningful.
I did not think they are disagreeing, or never said they were disagreeing with established physics.

What I said, was what forces do you think you have that can make up of balance out the higher propulsive forces of the fast in the studies ???

Letís take the mean propulsive forces, slow 6.2mean in 10.9 seconds. Fast 45.3mean 2.8 seconds, now letís divided the mean slow of 10.9 seconds by the fast 2.8 = 3.8, so now letís divide the slow mean by 3.8 = 1.6.

Fast mean for 2.8 seconds = 45.3.

Slow mean for 2.8 seconds = 1.6.


Please what forces have I left out that the slow has to make up or balance out these ???

http://www.jssm.org/vol7/n2/16/v7n2-16pdf.pdf

Wayne
waynexk8
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#188
Mar1-12, 08:55 PM
P: 399
Will get back to jmmccain and Dalespam tomorrow, funny I have two very good high definition cameras.

Wayne
waynexk8
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#189
Mar2-12, 07:16 AM
P: 399
No time just now, to read the posts and answer.

Just wanted the state why average force means nothing in this debate

I lift 80 pounds at ďanyĒ speed, letís go for 1/1, {1 second positive and 1 second negative} you all are saying that the average force for 1 repetition, and the average force for a 100 repetitions are the exact same ??? However, we all know that you will exert far more total or overall force doing the 100 repetitions. You will exert a force the same as you do for 1 repetition {forgetting fatigue here} = 2 seconds, for 200 seconds, so its overall or total force output x 200 seconds.

So the question is, why does everyone keep commenting on the average force is the same, when it has no reverence in this debate ??? What we are looking for, is in which repetition speed, in the same time frame, puts the most overall or/and total tensions on the muscles ???

As I have said, as you fail 50% faster using the .5/.5 to the 3/3 using 80% this does mean thet you have put more tension on the muscles faster, right ??? And to put more tension on the muscles, and faster, you will have to use more force, if not please say why you think not ??? PLEASE will someone try to answer this.

Are there anyone out there that thinks if you fail to lift a weight up after say 10 seconds, as you are lifting it faster, that it is not because you have put more tension on the muscles than the slow lifting, if so say why please.

Wayne
sophiecentaur
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#190
Mar2-12, 10:57 AM
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You were the one who asked what average force was. You also asked about Total Force.
Do keep up!
sophiecentaur
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#191
Mar2-12, 01:55 PM
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Wayne
Can you quote me a single statement, from anyone else, on this forum or in the communications with others that you have shown us, that indicates what you are talking about makes any sense in terms of the Physics? Everything I have read looks like non-committal and polite put-downs. When will you get the message?
Have you not realised that no one wants to discuss your endless, blow-by-blow descriptions of a lifting session? If you cannot be bothered to condense your questions into a digestible form then I, for one, can't be bothered to read them. I (/we all) have given you all the facts about the very straightforward Physics that relates to the process.
You haven't been prepared to accept it because it "doesn't feel right to you", somehow and you haven't been prepared even to use 'our' vocabulary. Can you be surprised that you aren't getting satisfaction?
waynexk8
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#192
Mar2-12, 07:26 PM
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Quote Quote by DaleSpam View Post
Hi waynekx8, I didn't notice that you were back.

It looks like you are still confusing "work done" and "energy expended" as we discussed last year: http://www.physicsforums.com/showthr...15#post3190515
Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ??? And we know for a fact that the fast does more work and uses more energy.

Average means nothing in this debate. What I am looking for is the higher overall or total forces used, thus more overall or total force on the muscles, just take this example.

Lift a weight up and down for 10 seconds, lift a weight up and down for 6o seconds, the average force is the same, but lifting the weight for 60 seconds WILL and DOES put more total or overall tension on the muscles, this means the total or overall force was more.

What we WANT to know is the same for lifting the weight up and down 6 times in 6 seconds, and lifting the weight up and down 1 time in 6 seconds.

Now I say without a shadow of a doubt, itís the fast, and here is why. If you took both repetitions speeds to momentary muscular failure, meaning if you lifted the weights at the two given speeds until you could not move the weight, you would fail about 50% faster lifting the weight the fast way. This can ONLY mean one thing, you put more tension on the muscles faster, thus the muscles failed faster, thus more force must have been used in not only the same time frame, but in less time frame, you also use more energy in the fast, why ??? Itís because your putting out more force and tension on the muscles, if not, why please ??? As no one here is giving me a direct answer, they seem to not want to, or cant ???

Wayne
sophiecentaur
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#193
Mar3-12, 03:45 AM
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Quote Quote by waynexk8 View Post
Hi again Dalespam,

Actually I am not confusing work done and energy expended, "unless" work done means the overall or total force used ??? And we know for a fact that the fast does more work and uses more energy.

Wayne
That statement is meaningless. There is no "debate" possible on that basis. Why not do us all a favour and use PF language?

Total force is as daft as total speed. Come to terms with that.
DaleSpam
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#194
Mar3-12, 07:15 AM
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Quote Quote by waynexk8 View Post
Actually I am not confusing work done and energy expended ... And we know for a fact that the fast does more work and uses more energy.
Yes, you are still confusing them. In fact, this quote proves that you are confusing them since you finish with the confused statement that the fast does more work and uses more energy.

The correct statement is that the fast does the same amount of work and uses more energy (i.e. the human machine is less efficient at doing fast work). If you were not still confusing the concepts then you would not still be making the same incorrect statements.

Quote Quote by waynexk8 View Post
What I am looking for is the higher overall or total forces used,
Let f(t) be the force exerted by the human on the weight at time t. Please define overall or total force in terms of f(t).

For example, average force from time [itex]t_i[/itex] to time [itex]t_f[/itex] is:
[tex]\overline{\mathbf{f}}=\frac{\int_{t_i}^{t_f} \mathbf{f}(t) \, dt}{t_f-t_i}[/tex]

Please provide a similar rigorous definition for total or overall force.
sophiecentaur
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#195
Mar3-12, 07:21 AM
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@Dalespam
I see you are being drawn into this. Beware - madness lies there.
DaleSpam
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#196
Mar3-12, 07:27 AM
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Quote Quote by sophiecentaur View Post
@Dalespam
I see you are being drawn into this. Beware - madness lies there.
Thanks for the warning sophiecentaur, but I have worked with wayne before, same topic different thread. I will only stick around while it is fun.
sophiecentaur
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#197
Mar3-12, 07:32 AM
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It's quite incredible how he bounces back with the same question, which, in itself, has no meaning.
DaleSpam
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#198
Mar3-12, 07:51 AM
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I was trying to think of weight lifting analogy to describe his behavior in terms that he could understand, but my weight lifting vocabulary is fairly limited.

Basically, I want to get across the point about how stupid I would be if I were to go to him and a bunch of other weight lifting experts and ask for their advice on a weightlifting problem and when all of them tell me the same thing I then ignore their expertise and continue doing the opposite. Particularly if they are united in their opinion and persist in their opinion after detailed questioning.

They are experts in weight lifting, I know that I am ignorant on the subject, and since they all agree with each other, and my opinion disagrees with them, the only reasonable thing to do is to recognize that my uninformed opinion is almost certainly wrong and try to understand their advice.

Also, it wouldn't be helpful for me to call a weight machine a dumb bell, nor would it be helpful for me to talk about things they have never heard of like the supercalifragilistic rep without describing in detail what I mean by that. Furthermore, if they corrected me on my incorrect use of the term weight machine and dumb bell and I continued to use it incorrectly (for years) they would reasonably become frustrated.


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