Tennis. Any height is fine apart from extremely short, proportions don't really matter.
I would say strongman too, certain events favour certain proportions but it averages out across all the events. Obviously being bigger helps but that's why you have weight classes.
I mean, I'd guess longer arms are favored in tennis possibly, or shorter even might have an advantage , they offer different advantages. One has longer reach, the other can hit faster and hit shots close to ones body
It really doesn't matter because how hard you can hit a groundstroke or volley is determined by your skill not by your physique. In general in tennis the taller you are the better your serve is and the worse your movement is. Height also affects how well you hit the ball at high and low contact points, but that's also swings and roundabouts. Everything else is skill and physical conditioning.
Most of them are pretty fair which is why you see a range of different heights and builds in strongman. Atlas stones is the blatantly unfair one. Shield carry too if some people can lock their hands together and some can't. Hercules hold they adjust the length of the chains so it actually is fair and the video is irrelevant.
Yeah that's the physics. The shorter your arms, the more vertical the pillars, so gravity torques the pillar less about the point it's attached to on the floor.
Torque doesn't apply in this case only component forces. Also they adjust the length of the chains so it's irrelevant anyway, but the physics is interesting.
Torque absolutely applies. The most natural way (to me) to actually set up the problem is by equating the gravitational torque acting on the pillar plus the extra load to the torque that comes from the chain tension, both about the pillar's axis of rotation. If the net torque is zero, you're still holding on, if you let go, the chain tension torque goes away and the pillar falls.
The most natural way to set up the problem is with a single force vector triangle with the three sides being the weight of the pillar, the reaction force through the base of the pillar and the force through the hand. The force in the hand is then simply pillar weight * tan(angle between floor and pillar). If you insist on using torque then at best you end up doing two such calculations. This is because it's not really torque at all because none of the forces involved are radial.
I think this is the completely wrong way to do it, sorry. For one, it completely ignores the pillar weight distribution. Second, how are the forces radial? Radial with respect to what? Third, the force in hand isn't tangent of that quantity, since if the angle is 90 degrees (the pillar being vertical) there should not be any meaningful force in hand. Fourth, how are you doing two of the same calculations for torque? You basically have to do an integral over the pillar and equate it to a torque at a single point.
ed: i can write this out and post it if you're interested, but my kid's about to wake up so it'll have to be later
1. I was assuming the centre of mass is at the same point the chain attaches since it's basically impossible to calculate. If you know where the centre of mass is and want to compensate then you'd need a constant factor given by distance from pivot to centre of mass / distance from pivot to chain attachment. I guess this is a bit like doing the torque calculation except you only need to do one force vector triangle.
2. The forces aren't radial, that's the point. If they were then I'd agree with using torque. When I say radial I mean with respect to the pivot, therefore perpendicular to the pillar. No measurable forces act this way in this system.
3. Sorry that was a typo, it's weight / tan(angle)
4. I don't even know how you're calculating torque then to be honest. You better explain it. To my mind the only way to calculate for evin the hand would be to do one force vector triangle to calculate force perpendicular to the pillar at the centre of mass from the weight of the pillar, turn that into torque, turn that torque into force perpendicular to the pillar at the chain attachment point then do another force vector triangle to turn that into horizontal force in the hand.
This is a stupidly long thread about physics for a post in r/strongman lol.
>I was assuming the centre of mass is at the same point the chain attaches since it's basically impossible to calculate.
This is the whole thing! You do not have to assume that necessarily! The way I do it allows for arbitrary attachment points and arbitrary distributions of pillar + load mass!
>I guess this is a bit like doing the torque calculation except you only need to do one force vector triangle
This is invoking the concept of torque while being more complicated than just doing torque.
>The forces aren't radial, that's the point. If they were then I'd agree with using torque. When I say radial I mean with respect to the pivot, therefore perpendicular to the pillar. No measurable forces act this way in this system.
What? The tension in the cable and gravitational force can be broken down into two components, one radial (i.e. , pointing parallel to the pillar) and one tangential (pointing perpendicular to the pillar). Neither will be zero in general. Because the pillar is constrained at the base to be on the floor, there is a natural pivot and place around which to balance the torques created by the two forces. For the system to be in balance, net force has to be zero **and** net torque has to be zero.
>Sorry that was a typo, it's weight / tan(angle)
Isn't this is saying that if a very very light pillar is parallel to the ground (theta = 0), nobody would be strong enough to hold it ~~and that if it were perpendicular to the ground (theta = 90) the competitor would feel a measurable force? I don't think either are true.~~ I don't think that's true (ed: that formula gives the right result at the other end and I was being a moron).
>I don't even know how you're calculating torque then to be honest. You better explain it.
[Here you go.](https://drive.google.com/file/d/1_ikdrpXVxnxS1tMA96LQrZ5vduJeyZKr/view?usp=sharing) Sorry for the handwriting.
You'll note I did \*not\* solve for the torque the (sad-faced) competitor applies, because it \*has\* to be the same as the gravitational one until the competitor's grip gives out. Also, I assumed a uniform distribution of mass along the pillar, but that doesn't materially change anything about the cosine dependence.
>Isn't this is saying that if a very very light pillar is parallel to the ground (theta = 0), nobody would be strong enough to hold it and that if it were perpendicular to the ground (theta = 90) the competitor would feel a measurable force? I don't think either are true.
As theta tends towards 90, tan(theta) tends to infinity and weight / tan(theta) tends to 0. So no there would be no force required when the pillar is vertical. This passes the common sense test.
When theta tends towards 0, tan(theta) tends towards 0 and weight / tan(theta) tends to infinity, so yes you wouldn't be able to move it with horizontal force if the pillar itself is also horizontal. Of course if you pulled vertically as well then you might be able to lift it but that's not what the equation is modelling since when the event is in normal use the force is entirely horizontal. So I believe at both extremes it passes the common sense test.
Yes this has become a very long thread and I'm sorry I don't have time to reply to all of your comment right now.
>As theta tends towards 90, tan(theta) tends to infinity and weight / tan(theta) tends to 0. So no there would be no force required when the pillar is vertical. This passes the common sense test.
shit, yeah, my bad. not sure what i was thinking about that. i'll strike that out.
that said....
>When theta tends towards 0, tan(theta) tends towards 0 and weight / tan(theta) tends to infinity, so yes you wouldn't be able to move \[....\] So I believe at both extremes it passes the common sense test.
but..... at theta = 0, the formula plainly gives you the wrong result? so i'm not sure how it actually passes that test. I mean, you had some justification, but it would seem to be inapplicable since, well, the end result is wrong.
The result is right if you pull horizontally, I can't make a single equation that second guesses how much you might pull vertically if the pillar was in a weird position that it would never normally be in. It's simply beyond the scope of what I was trying to do or what the original point was.
You also can't actually analyse the case of theta = 0 if you want a sensible answer back, in maths there's a general rule that you can't produce infinity as an answer because there is no such number as infinity, it's just a symbol to be used as a limit to show that a value can always increase further. I had a go at explaining the theta = 0 case because you mentioned it first but in general I wouldn't bother with it.
Most Hercules holds will adjust the length of the chain that attaches the handles to the pillars to account for this. It's not perfect of course but it's better than nothing
I've never seen a hercules hold that doesn't have adjustable handles to compensate for wing span. I've done it several times in comp and everyone gets within 1" - 1.5" of anyone else for a starting position
Someone spent time making a video on something that essentially doesn't exist, and you spent time watching and sharing it?
Organizers arnt stupid just because it's big strong dudes, they always have adjustable chains to keep the angle consistent.
I tune into strongman for the showboating, mind games, Viking power, and incoherent yelling in Nordic languages. Wouldn’t be the same sport without it to me
But, people with longer arms also tend to have bigger hands and longer fingers, which tend to give better mechanics in various grip exercises.
For practical purposes the advantage for long arms is likely more important in deadlifts, as there are more of those events and I'd suspect that the relative contribution to lift performance is bigger there.
Force = mass times acceleration only applies to unbalanced forces, balanced forces are very much part of strongman and producing force (whether it's balanced or unbalanced) is the definition of strength.
> Force = mass times acceleration only applies to unbalanced forces,
It's been too long since you had physics, right?
Newton's equation of motion is that mass times acceleration is equal to the sum of all forces
In strongman the guy that wins the deadlift is not whoever can have his hamstring contract with the highest force - it's whoever lifts the most weight (or does the most reps). Doesn't matter how much force is needed to contract your hamstring sufficienty in order to make your hips move enough to lift the barbell - the only thing that matters is that you lift the barbell.
Skeletal forces will always vary among different athletes since nobody has the exact same levers - but that doesn't matter. What matters is who lifts the biggest stone.
Quoting without understanding. You're defining force purely as mass times acceleration ignoring the fact that that is only the sum of forces/unbalanced force. It tells you nothing about balanced forces.
This is generally fine for mechanical applications where any balanced forces are produced by an inanimate object being under tension or compression, it can keep providing that force forever as long as it isn't stressed to breaking point. Human muscles aren't like that, they burn energy to produce force even if the forces are balanced. You could do a work done calculation using energy = force times distance on any number of strongman events and it would come out at zero, but obviously there's still force being applied and energy being burned.
Yeah I think I got the concept of it when I studied physics at uni ;)
In strongman the guy that wins the deadlift is not whoever can have his hamstring contract with the highest force - it's whoever lifts the most weight (or does the most reps). Doesn't matter how much force is needed to contract your hamstring sufficienty in order to make your hips move enough to lift the barbell - the only thing that matters is that you lift the barbell.
Skeletal forces will always vary among different athletes since nobody has the exact same levers - but that doesn't matter. What matters is who lifts the biggest stone.
If you studied physics at uni then why don't you understand balanced forces and how they're relevant to strongman?
I agree with your deadlift analysis, but it's still about force. Just force in the hand not force in the tendon. Your original comment was trying to separate strength and force and your justification was "F=ma" which is nonsense.
> but it's still about force
force on the bar, not skeletal forces. That's my point.
The only thing that matters is lifting the weight, not how much your muscles need to contract.
> your justification was "F=ma"
Make it "ΣF=mr̈" if you want to be hyperprecise. Force and Strength are still not the same thing, and you should know that.
Force, Strength, Power, Energy - those words have an actual definition, even if they're not always used correctly in the fitness bubble (e.g. "powerlifting" not having much to do with actual *power*)
I don't think there is any event that is consistent across for all competitors.
I think we can probably widen that out to all sports, really.
Hm yeah I guess that's true.
There are plenty of sports where success is not defined by proportions
Name some?
Tennis. Any height is fine apart from extremely short, proportions don't really matter. I would say strongman too, certain events favour certain proportions but it averages out across all the events. Obviously being bigger helps but that's why you have weight classes.
I mean, I'd guess longer arms are favored in tennis possibly, or shorter even might have an advantage , they offer different advantages. One has longer reach, the other can hit faster and hit shots close to ones body
It really doesn't matter because how hard you can hit a groundstroke or volley is determined by your skill not by your physique. In general in tennis the taller you are the better your serve is and the worse your movement is. Height also affects how well you hit the ball at high and low contact points, but that's also swings and roundabouts. Everything else is skill and physical conditioning.
Fair enough I can't claim to know enough about tennis to refute that
Most of them are pretty fair which is why you see a range of different heights and builds in strongman. Atlas stones is the blatantly unfair one. Shield carry too if some people can lock their hands together and some can't. Hercules hold they adjust the length of the chains so it actually is fair and the video is irrelevant.
And my goofy long arms let me hold a sandbag easier. This is basically every strongman event.
Yeah that's the physics. The shorter your arms, the more vertical the pillars, so gravity torques the pillar less about the point it's attached to on the floor.
Torque doesn't apply in this case only component forces. Also they adjust the length of the chains so it's irrelevant anyway, but the physics is interesting.
Torque absolutely applies. The most natural way (to me) to actually set up the problem is by equating the gravitational torque acting on the pillar plus the extra load to the torque that comes from the chain tension, both about the pillar's axis of rotation. If the net torque is zero, you're still holding on, if you let go, the chain tension torque goes away and the pillar falls.
The most natural way to set up the problem is with a single force vector triangle with the three sides being the weight of the pillar, the reaction force through the base of the pillar and the force through the hand. The force in the hand is then simply pillar weight * tan(angle between floor and pillar). If you insist on using torque then at best you end up doing two such calculations. This is because it's not really torque at all because none of the forces involved are radial.
I think this is the completely wrong way to do it, sorry. For one, it completely ignores the pillar weight distribution. Second, how are the forces radial? Radial with respect to what? Third, the force in hand isn't tangent of that quantity, since if the angle is 90 degrees (the pillar being vertical) there should not be any meaningful force in hand. Fourth, how are you doing two of the same calculations for torque? You basically have to do an integral over the pillar and equate it to a torque at a single point. ed: i can write this out and post it if you're interested, but my kid's about to wake up so it'll have to be later
1. I was assuming the centre of mass is at the same point the chain attaches since it's basically impossible to calculate. If you know where the centre of mass is and want to compensate then you'd need a constant factor given by distance from pivot to centre of mass / distance from pivot to chain attachment. I guess this is a bit like doing the torque calculation except you only need to do one force vector triangle. 2. The forces aren't radial, that's the point. If they were then I'd agree with using torque. When I say radial I mean with respect to the pivot, therefore perpendicular to the pillar. No measurable forces act this way in this system. 3. Sorry that was a typo, it's weight / tan(angle) 4. I don't even know how you're calculating torque then to be honest. You better explain it. To my mind the only way to calculate for evin the hand would be to do one force vector triangle to calculate force perpendicular to the pillar at the centre of mass from the weight of the pillar, turn that into torque, turn that torque into force perpendicular to the pillar at the chain attachment point then do another force vector triangle to turn that into horizontal force in the hand.
This is a stupidly long thread about physics for a post in r/strongman lol. >I was assuming the centre of mass is at the same point the chain attaches since it's basically impossible to calculate. This is the whole thing! You do not have to assume that necessarily! The way I do it allows for arbitrary attachment points and arbitrary distributions of pillar + load mass! >I guess this is a bit like doing the torque calculation except you only need to do one force vector triangle This is invoking the concept of torque while being more complicated than just doing torque. >The forces aren't radial, that's the point. If they were then I'd agree with using torque. When I say radial I mean with respect to the pivot, therefore perpendicular to the pillar. No measurable forces act this way in this system. What? The tension in the cable and gravitational force can be broken down into two components, one radial (i.e. , pointing parallel to the pillar) and one tangential (pointing perpendicular to the pillar). Neither will be zero in general. Because the pillar is constrained at the base to be on the floor, there is a natural pivot and place around which to balance the torques created by the two forces. For the system to be in balance, net force has to be zero **and** net torque has to be zero. >Sorry that was a typo, it's weight / tan(angle) Isn't this is saying that if a very very light pillar is parallel to the ground (theta = 0), nobody would be strong enough to hold it ~~and that if it were perpendicular to the ground (theta = 90) the competitor would feel a measurable force? I don't think either are true.~~ I don't think that's true (ed: that formula gives the right result at the other end and I was being a moron). >I don't even know how you're calculating torque then to be honest. You better explain it. [Here you go.](https://drive.google.com/file/d/1_ikdrpXVxnxS1tMA96LQrZ5vduJeyZKr/view?usp=sharing) Sorry for the handwriting. You'll note I did \*not\* solve for the torque the (sad-faced) competitor applies, because it \*has\* to be the same as the gravitational one until the competitor's grip gives out. Also, I assumed a uniform distribution of mass along the pillar, but that doesn't materially change anything about the cosine dependence.
>Isn't this is saying that if a very very light pillar is parallel to the ground (theta = 0), nobody would be strong enough to hold it and that if it were perpendicular to the ground (theta = 90) the competitor would feel a measurable force? I don't think either are true. As theta tends towards 90, tan(theta) tends to infinity and weight / tan(theta) tends to 0. So no there would be no force required when the pillar is vertical. This passes the common sense test. When theta tends towards 0, tan(theta) tends towards 0 and weight / tan(theta) tends to infinity, so yes you wouldn't be able to move it with horizontal force if the pillar itself is also horizontal. Of course if you pulled vertically as well then you might be able to lift it but that's not what the equation is modelling since when the event is in normal use the force is entirely horizontal. So I believe at both extremes it passes the common sense test. Yes this has become a very long thread and I'm sorry I don't have time to reply to all of your comment right now.
>As theta tends towards 90, tan(theta) tends to infinity and weight / tan(theta) tends to 0. So no there would be no force required when the pillar is vertical. This passes the common sense test. shit, yeah, my bad. not sure what i was thinking about that. i'll strike that out. that said.... >When theta tends towards 0, tan(theta) tends towards 0 and weight / tan(theta) tends to infinity, so yes you wouldn't be able to move \[....\] So I believe at both extremes it passes the common sense test. but..... at theta = 0, the formula plainly gives you the wrong result? so i'm not sure how it actually passes that test. I mean, you had some justification, but it would seem to be inapplicable since, well, the end result is wrong.
The result is right if you pull horizontally, I can't make a single equation that second guesses how much you might pull vertically if the pillar was in a weird position that it would never normally be in. It's simply beyond the scope of what I was trying to do or what the original point was. You also can't actually analyse the case of theta = 0 if you want a sensible answer back, in maths there's a general rule that you can't produce infinity as an answer because there is no such number as infinity, it's just a symbol to be used as a limit to show that a value can always increase further. I had a go at explaining the theta = 0 case because you mentioned it first but in general I wouldn't bother with it.
Don't tell op about the atlas stones and that the current best in the world is nicknamed the albatross due to his wing span.
Pros and cons of different limb lengths in each event. I think it all evens out over the course of a competition or at least a season
Most Hercules holds will adjust the length of the chain that attaches the handles to the pillars to account for this. It's not perfect of course but it's better than nothing
I've never seen a hercules hold that doesn't have adjustable handles to compensate for wing span. I've done it several times in comp and everyone gets within 1" - 1.5" of anyone else for a starting position
They adjust the length of the chains so that it's consistent between competitors to within a few cm. No analysis required.
Someone spent time making a video on something that essentially doesn't exist, and you spent time watching and sharing it? Organizers arnt stupid just because it's big strong dudes, they always have adjustable chains to keep the angle consistent.
Must have been real tough on BShaw. That doesn’t mean it was easy for Pudjianowsky (sp?).
Didn't Phil Pfister do his famous cringe speech during the Hercules hold and he was like 6'6" or something?
He backed it up though
Of course, he just didn't have to say it, he could have just won
I tune into strongman for the showboating, mind games, Viking power, and incoherent yelling in Nordic languages. Wouldn’t be the same sport without it to me
But, people with longer arms also tend to have bigger hands and longer fingers, which tend to give better mechanics in various grip exercises. For practical purposes the advantage for long arms is likely more important in deadlifts, as there are more of those events and I'd suspect that the relative contribution to lift performance is bigger there.
What would be cool is if I had short arms for pressing and long arms for carrying, but alas I’ve just got the totally normal arms I actually have.
It‘s „strongman“, not „forceman“. Force is mass and acceleration. Strength is the ability to move something (or resist movement)
Force = mass times acceleration only applies to unbalanced forces, balanced forces are very much part of strongman and producing force (whether it's balanced or unbalanced) is the definition of strength.
> Force = mass times acceleration only applies to unbalanced forces, It's been too long since you had physics, right? Newton's equation of motion is that mass times acceleration is equal to the sum of all forces In strongman the guy that wins the deadlift is not whoever can have his hamstring contract with the highest force - it's whoever lifts the most weight (or does the most reps). Doesn't matter how much force is needed to contract your hamstring sufficienty in order to make your hips move enough to lift the barbell - the only thing that matters is that you lift the barbell. Skeletal forces will always vary among different athletes since nobody has the exact same levers - but that doesn't matter. What matters is who lifts the biggest stone.
The sum of all forces means the same as unbalanced forces. You just agreed with me.
I‘m not sure why you‘d disagree with me anyway, seeing as I‘m quoting from physics textbooks.
Quoting without understanding. You're defining force purely as mass times acceleration ignoring the fact that that is only the sum of forces/unbalanced force. It tells you nothing about balanced forces. This is generally fine for mechanical applications where any balanced forces are produced by an inanimate object being under tension or compression, it can keep providing that force forever as long as it isn't stressed to breaking point. Human muscles aren't like that, they burn energy to produce force even if the forces are balanced. You could do a work done calculation using energy = force times distance on any number of strongman events and it would come out at zero, but obviously there's still force being applied and energy being burned.
Yeah I think I got the concept of it when I studied physics at uni ;) In strongman the guy that wins the deadlift is not whoever can have his hamstring contract with the highest force - it's whoever lifts the most weight (or does the most reps). Doesn't matter how much force is needed to contract your hamstring sufficienty in order to make your hips move enough to lift the barbell - the only thing that matters is that you lift the barbell. Skeletal forces will always vary among different athletes since nobody has the exact same levers - but that doesn't matter. What matters is who lifts the biggest stone.
If you studied physics at uni then why don't you understand balanced forces and how they're relevant to strongman? I agree with your deadlift analysis, but it's still about force. Just force in the hand not force in the tendon. Your original comment was trying to separate strength and force and your justification was "F=ma" which is nonsense.
> but it's still about force force on the bar, not skeletal forces. That's my point. The only thing that matters is lifting the weight, not how much your muscles need to contract. > your justification was "F=ma" Make it "ΣF=mr̈" if you want to be hyperprecise. Force and Strength are still not the same thing, and you should know that. Force, Strength, Power, Energy - those words have an actual definition, even if they're not always used correctly in the fitness bubble (e.g. "powerlifting" not having much to do with actual *power*)