most force/strength used in the same time frame

douglis

I see the problem is a little deeper than I thought...

sophiecentaur

=41 if you follow the bodmas convention. Look it up. Anything X zero is zero.
To make it clear, you should use brackets.

Zula110100100

Yeah...I would say 41..and read it 40+(40*0)+1 = 40+0+1 = 41

I want you to do that expiriment because I think it will show that activity does not equal force, I think if you hold 20lbs for 1 second, it will be some amount that is less than double activity to hold 40lbs for 1 seconds.

In my opinion you need to change your angle of attack, the answer given for applied force obviously doesn't give you what you were looking for, rather than try to figure out how that answer is wrong, perhaps you should look for how that answer relates to what you're looking for.

What are you trying to find out about PS resistance training overall?

Zula110100100

bodmas? I learned please excuse my dear aunt sally (Parenthesis, exponents, multiplication, division, addition, subtraction)

douglis

The answer for the applied force gives exactly what he was looking for.He just don't like the answer!

sophiecentaur

Brackets 'of' division, multiplication, addition & subtraction was what they gave me. Then, later, it became BIDMAS, with the I for Indices.

waynexk8

Hi,

Just did the static test, and some others, they were all with about 80% and for 15 seconds on the leg extension.

Static hold half way up. Average muscle activation = 92.1.

Slow. Average muscle activation = 159.1.

Fast. Average muscle activation = 191.7.

Wayne

waynexk8

Hmm, see you point, but this time when there will be no force, will be very small, about ??? Say we semented the whole .5/.5 rep of 1 second in a 100 parts, the rest time maybe just 1 part, so how and why do we need to average in something so small, as this will also be in the slower rep, but for a little longer.

Wayne

waynexk8

Yet again you fail to answer.

After thinking about this, I make it 1, as what I did was x 80 x 0 the other way around, like this, 0 x 80, so if you x 80 by 0 it says at 80, but if you do it like this, 80 x 0, your timesing 0 by 80, but if you x anything by 0, its always going to be 0.

Not sure how you get 41 ??? But no matter, as Douglas did not even reply, it seems like he did not then understand the problem I was trying to get round.

Wayne

Zula110100100

If the weight and time were the same, and they started and ended at v=0, then the force-applied impulses are the same in each. This shows that Average muscle activation does not directly correlate to fore-applied impulse. Do the same thing with different amounts of weight, each held for the same time, in the same position.

Lets say we have a force of 1N, we apply that force at two velocities, one is 12m in 6s = 2m/s the other is 2m in 6 seconds = .33m/s, the power is 2(1) = 2watts for the fast one, and .33(1) .33watts for the slow one. The same force, different velocities.

waynexk8

Don’t get that, but no prob.

Ok, get you a bit more, will do the tests tomorrow with a few different weights, to late here now,

Great, like you attack, yes I think it’s great to change our angle of attack, and try and look at it from all different angles, and great you’re here, as someone very unbiased, and new to look at the debate.

PS ???

wayne

Zula110100100

Because it is a well accepted convention that you do the multiplication before you do addition
so 40 + 40 x 0 + 1 you multiply 40 by 0 and have 40+0+1=41

waynexk8

Could not this be not true ??? As in my theory, when the fast reps are on their high peak forces, their accelerations, then on the low forces the decelerations, when they are on the low force decelerations, the slow force is still on its medium forces, and the medium forces of the slow reps, cannot make up or balance out the high peak forces, with their medium forces. The only way they could make up these forces, if the slow reps went on longer, but they do not.

Best I say this, as the faster reps will have what 12 v=0 starts and stops, too the slows just 2 v=0.
As the full debate is this; which rep/s use the most overall or total force or strength, using roughly 80% of the persons 1RM, and moving the weight 1m up and 1m down, 6 reps at .5/.5 = 6 seconds, moving the weight 12m in all, or 1 rep at 3/3 = 6 seconds, moving the weight 2m in all

As you know, an is the integral of a force with respect to time. When a force is applied to a rigid body it changes the movement of that body. A small force {slow rep} applied for a LONG TIME can produce the same movement change as a large force {fast reps} applied briefly, because it is the product of the force and the time for which it is applied that is important. The impulse is equal to the change of momentum. However both reps are done in the same time frame.


Will do.


It’s getting late here so not thinking straight, as its 1.30. But to accelerate a weight up, and thus to reach different heights in the same time, you would have to use more Newton’s of force to reach the higher high, or move it at a faster m/s ???

Wrote this earlier.
I can work out the power, learnt this from a site, {hope its right} as seen below and to be honest, I thought it would be basically as simple as this, was I wrong.
Calculate how much power/strength I would be used on both rep speeds. Distance weight of 91 kg moved 1.85 M.

To determine the force we will need to figure out what the weight of the barbell is (W = mg = 91 kg x 9.81 m/s = 892 kg.m/s or 892 N). Now, if work is equal to Force x distance then, U = 892 N x 1.85 m = 1650 Nm.

We can calculate that lifting a 91kg barbell overhead a distance of 1.85 m required 1650 J of work. You will notice that the time it took to lift the barbell was not taken into account.

Let us only add up the positive part of the lift.

Concept of power however, takes time into consideration. If for example, it took .5 seconds to complete the lift, then the power generated is 1650 J divided .5 s = 3300 J/s.

If it took 2 seconds to complete the lift, then the power generated is 1650 J divided 2 s = 825 J/s.

Wayne

waynexk8

Ok thx, will read that a few more times tomoorow, to late here now, half aslepp.

Wayne

waynexk8

Why cant anyone asnswerr this please ???

1,
We both have a 100 pounds of maximum strength {or a machine is lifting} that we can bench press. We use 80% 80 pounds, I move the weight 1000mm in .5 of a second, I will be accelerating this weight for 60% for the ROM, {range of motion} of for the concentric rep, you will be moving the same weight in .5 of a second for only 166mm.

Question,
If I have moved the weight more distance in the same time frame, does that not that mean that I have used more force ??? {force and strength are the same I would have thought, but let’s just go for force now} As even if you just count my acceleration its 600mm, that far moré distance than the slow.

2,
I bench press the same weight same distance of 80% {and this weight of 80% is important, as my maximum is only 20% more than my maximum} I lift at .5/.5 for 6 reps = 6 seconds covering a distance of 12m. The slow lifts at 3/3 for 1 rep = 6 seconds, covering a distance of only 2m.

Question, basically the same as the above.

As a force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1Newton, if you move it futher in the same time from you have to use more force, N’s.

Wayne

douglis

Wayne...you got your answer.What's clear is that you don't understand the answer or you don't like it.

In both cases the applied force is equal with gravity's impulse over 6 seconds....so it's the same.

Yes...I failed to answer so I asked my 6 years old son and told me it's 41.

sophiecentaur

You seem to want answers about this without having to make any effort to know the basics of Science or even primary School Arithmetic. We just can't help you on your terms. You have had many answers (the same answer put in different ways, actually) in a attempt to get the message over to you. Yet you refuse to give an inch (or even 2.54cm) so I am unsubscribing from this thread. I suggest you return the the pub / club / locker room and have your conversation with people who have no use for real Science but who just want an hour's worth of idle, anecdotal chit chat.