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200 Max and 100 Max, Twice as Strong?

i got a stupid question:

If someone maxes 200 kg

does he have double the strength of someone that maxes 100 kg?

this is a tricky one because the one that maxes 200 kg could probably do 100 for like 20-30 reps or something, so is he 30 times stronger or 2 times stronger?

Hmmmm… I’m strangely intrigued by this.

Both actually make sense. I think the problem is that the strength is being measured two different ways.

It’s like saying “how big is the earth compared to the sun?” One compares the volume and other measures the circumference. Maybe earthquakes are a better example: adding one on the richter scale increases it exponentially.

So I would have to say that until a person unifies the comparison of strength, there really is no right or wrong.

You need to define ‘strong’ before you can answer that question.

Their lift also depends on their leverages, which vary depending on what you are doing.

If you can lift twice as much weight as someone, then you are twice as strong.

[quote]csulli wrote:
If you can lift twice as much weight as someone, then you are twice as strong.[/quote]

Reading the entire posts helps.

He can also lift 30 times as much weight as the guy (100x30 vs 100x1)…sooooooo

Really, does it matter?

they are simply alot stronger yeah?

Absolute strength or Relative Strength. Does the guy who can only lift 100kg WEIGH 50kg where the 200kg lifting

guy weighs 300KG ? ? ?

This has got to be the most interesting pointless thread…

IMO:
Ratio of wilks score.

Hmmm sounds like powerlifting philosophy…

[quote]VTBalla34 wrote:
Reading the entire posts helps.
[/quote]

[quote]apple12345 wrote:
is he 30 times stronger or 2 times stronger?[/quote]

[quote]csulli wrote:
If you can lift twice as much weight as someone, then you are twice as strong.[/quote]

So… I read the whole post and said 2 times, not 30 times.

Reading at all helps? Lol

It’s just a silly question though.

I can’t believe how some people are arguing about this.

Listen up folks, this will delineate things very clearly for you :smiley:

Subject A: Bench 225x1
Subject B: Bench 225x2

Estimated 1 rep Max for A: 225; Estimated 1 rep Max for B: 235

B is not “2x” stronger than A, despite the fact he can push the same quantifiable mass 2x instead of 1x.

The difference is only ~10 pounds in their overall productive force & work output.
You will have to look at the ratio between 10 pounds versus the overall mass being pushed, since that is the difference. That is the % increase in strength,which is very low.

this is when the mind of brute comes in handy …

[quote]Ethan7X wrote:
I can’t believe how some people are arguing about this.

Listen up folks, this will delineate things very clearly for you :smiley:

Subject A: Bench 225x1
Subject B: Bench 225x2

Estimated 1 rep Max for A: 225; Estimated 1 rep Max for B: 235

B is not “2x” stronger than A, despite the fact he can push the same quantifiable mass 2x instead of 1x.

The difference is only ~10 pounds in their overall productive force & work output.
You will have to look at the ratio between 10 pounds versus the overall mass being pushed, since that is the difference. That is the % increase in strength,which is very low.[/quote]

So you mean that if x energy is needed to move a weight from point a to point b, x2 energy is not needed to move it twice? strong logic

Pff everyone knows that fran time only matters.

[quote]apple12345 wrote:

[quote]Ethan7X wrote:
I can’t believe how some people are arguing about this.

Listen up folks, this will delineate things very clearly for you :smiley:

Subject A: Bench 225x1
Subject B: Bench 225x2

Estimated 1 rep Max for A: 225; Estimated 1 rep Max for B: 235

B is not “2x” stronger than A, despite the fact he can push the same quantifiable mass 2x instead of 1x.

The difference is only ~10 pounds in their overall productive force & work output.
You will have to look at the ratio between 10 pounds versus the overall mass being pushed, since that is the difference. That is the % increase in strength,which is very low.[/quote]

So you mean that if x energy is needed to move a weight from point a to point b, x2 energy is not needed to move it twice? strong logic
[/quote]

Technically speaking, since the barbell is returning to its original location in both instances, by definition the work accomplished in both is ZERO.

[quote]Johnny T Frisk wrote:
Pff everyone knows that fran time only matters.[/quote]

Truth

[quote]VTBalla34 wrote:

[quote]apple12345 wrote:

[quote]Ethan7X wrote:
I can’t believe how some people are arguing about this.

Listen up folks, this will delineate things very clearly for you :smiley:

Subject A: Bench 225x1
Subject B: Bench 225x2

Estimated 1 rep Max for A: 225; Estimated 1 rep Max for B: 235

B is not “2x” stronger than A, despite the fact he can push the same quantifiable mass 2x instead of 1x.

The difference is only ~10 pounds in their overall productive force & work output.
You will have to look at the ratio between 10 pounds versus the overall mass being pushed, since that is the difference. That is the % increase in strength,which is very low.[/quote]

So you mean that if x energy is needed to move a weight from point a to point b, x2 energy is not needed to move it twice? strong logic
[/quote]

Technically speaking, since the barbell is returning to its original location in both instances, by definition the work accomplished in both is ZERO.
[/quote]

No, you need to use a line or path integral to evaluate work. That means it matters what happens as the force vector moves through space, even though it ends up back at point a.

http://hyperphysics.phy-astr.gsu.edu/hbase/intare.html#c2

Also, to apple, we’re not talking about if someone has twice the energy (work) capacity, we’re debating if they are twice as strong. And so we’re back at debating the definition of strong. I don’t think it should be defined in terms of fundamental physical concepts. I think it is the bar weight used on a successful lift, period.

Also I don’t think the concept of “twice as strong” has any meaning. A 200lb bench is nothing compared to a 400lb bench. There needs to be some sort of nonlinear scaling (like logarithmic but not actually a log scale) between 0 and the maximum bench (Mendy’s 715?).

[quote]grettiron wrote:

[quote]VTBalla34 wrote:
Technically speaking, since the barbell is returning to its original location in both instances, by definition the work accomplished in both is ZERO.
[/quote]

No, you need to use a line or path integral to evaluate work. That means it matters what happens as the force vector moves through space, even though it ends up back at point a.

http://hyperphysics.phy-astr.gsu.edu/hbase/intare.html#c2

[/quote]

A and B are the same position, so any integral will result in 0.

[quote]VTBalla34 wrote:

[quote]grettiron wrote:

[quote]VTBalla34 wrote:
Technically speaking, since the barbell is returning to its original location in both instances, by definition the work accomplished in both is ZERO.
[/quote]

No, you need to use a line or path integral to evaluate work. That means it matters what happens as the force vector moves through space, even though it ends up back at point a.

http://hyperphysics.phy-astr.gsu.edu/hbase/intare.html#c2

[/quote]

A and B are the same position, so any integral will result in 0.[/quote]

Yes, for the regular integral you learned in high school calculus. A path integral is different.

Best way to illustrate this is to think of a pressure-volume plot of a heat engine cycle. By virtue of being a cycle, your start and end points are the same. So called P-V work is not zero though, and that is because work is evaluated using a path/line integral. Numerically, if you have a vector containing your P and V points on the curve, total work is evaluated by sum(P.diff(V)), where "." is an element-wise vector multiplication and diff(V) is the numerical equivalent to dV (differential volume change).

You would use the same path integral to evaluate the work done by a force translating in space, though with 2 vector quantities F and x you probably need to use a dot product to integrate the component of F in the x direction.

[quote]grettiron wrote:

[quote]VTBalla34 wrote:

[quote]apple12345 wrote:

[quote]Ethan7X wrote:
I can’t believe how some people are arguing about this.

Listen up folks, this will delineate things very clearly for you :smiley:

Subject A: Bench 225x1
Subject B: Bench 225x2

Estimated 1 rep Max for A: 225; Estimated 1 rep Max for B: 235

B is not “2x” stronger than A, despite the fact he can push the same quantifiable mass 2x instead of 1x.

The difference is only ~10 pounds in their overall productive force & work output.
You will have to look at the ratio between 10 pounds versus the overall mass being pushed, since that is the difference. That is the % increase in strength,which is very low.[/quote]

So you mean that if x energy is needed to move a weight from point a to point b, x2 energy is not needed to move it twice? strong logic
[/quote]

Technically speaking, since the barbell is returning to its original location in both instances, by definition the work accomplished in both is ZERO.
[/quote]

No, you need to use a line or path integral to evaluate work. That means it matters what happens as the force vector moves through space, even though it ends up back at point a.

http://hyperphysics.phy-astr.gsu.edu/hbase/intare.html#c2

Also, to apple, we’re not talking about if someone has twice the energy (work) capacity, we’re debating if they are twice as strong. And so we’re back at debating the definition of strong. I don’t think it should be defined in terms of fundamental physical concepts. I think it is the bar weight used on a successful lift, period.

Also I don’t think the concept of “twice as strong” has any meaning. A 200lb bench is nothing compared to a 400lb bench. There needs to be some sort of nonlinear scaling (like logarithmic but not actually a log scale) between 0 and the maximum bench (Mendy’s 715?). [/quote]

The reason i started wondering about this it seems so weird that average people benching 200 lbs got half the strength as someone serious and big enough to bench 400 lbs. There has to be a nonlinear scale of strength, but how does that make sense? obviously if x muscles is needed to move x weight from exact point a to point b the double would be needed to move double the weight or am i missing something here?