I read that sprinters produce a force of 4 times their body weight into the ground. My question is- shouldn’t they produce the same force all the time however faster and faster? Why does the faster you go the more force you need to put into the ground? Why isn’t the same force you need to put only in less time?
One way to simplify this is using the impulse-momentum theory.
- you're standing still on one leg. So in any given second, you are on the ground for 1 secons. What force is that leg putting into the ground? 1x your body weight, lets call it 100 pounds for simplicity. So you are putting in 100lb-seconds. We therefore know 100 lb-second is required to keep you from accelerating up or falling downward.
- imagine now, you are on the ground for exactly one half of any given second (ie, a slow jog). We need 100lb-s to maintain steady state, so simple algebra tells us we need to put out 200 pounds of force for that half second (200 pounds x .5s equals 100lb-s)
- now imagine you are only on the ground for 1/4 of any given second (typical of a fast sprint). We still need that same 100 lb-s. So we know we need 400 pounds of force (400 lb x .25s equals 100 lb-s).
Due to the mechanics of how people sprint, the faster you are on the ground, the less time you spend on the ground and the more air time you have. So as your speed increases, your vertical force increases proportionally.
The force imparted by a sprinter determines their ground contact time. The quantities are related. You can’t hold force constant since it’s the very mechanism by which ground contact time is reduced.
Even in theory, decreasing ground contact time without changing force would actually inhibit your speed, not increase it. Reduced contact time for the same force means reduced impulse, and reduced impulse means reduced change in velocity.
Sdfsdfsdfsdfsdfsdf wrote:
One way to simplify this is using the impulse-momentum theory.
- you're standing still on one leg. So in any given second, you are on the ground for 1 secons. What force is that leg putting into the ground? 1x your body weight, lets call it 100 pounds for simplicity. So you are putting in 100lb-seconds. We therefore know 100 lb-second is required to keep you from accelerating up or falling downward.
- imagine now, you are on the ground for exactly one half of any given second (ie, a slow jog). We need 100lb-s to maintain steady state, so simple algebra tells us we need to put out 200 pounds of force for that half second (200 pounds x .5s equals 100lb-s)
- now imagine you are only on the ground for 1/4 of any given second (typical of a fast sprint). We still need that same 100 lb-s. So we know we need 400 pounds of force (400 lb x .25s equals 100 lb-s).
Due to the mechanics of how people sprint, the faster you are on the ground, the less time you spend on the ground and the more air time you have. So as your speed increases, your vertical force increases proportionally.
I think this explanation is very wrong. In order to keep 100 labs on the ground from collapsing you need to produce 100 labs, be it 1/2 a second on the ground or a second. Elite sprinters ground contact time is about 0.08 seconds. According to your logic a 200 pound sprinter would have to produce 2500 pounds into the ground. They don’t. They produce around 4 times their body weight- around 800-900 pounds force
Ithinkitswrong wrote:
Sdfsdfsdfsdfsdfsdf wrote:
One way to simplify this is using the impulse-momentum theory.
- you're standing still on one leg. So in any given second, you are on the ground for 1 secons. What force is that leg putting into the ground? 1x your body weight, lets call it 100 pounds for simplicity. So you are putting in 100lb-seconds. We therefore know 100 lb-second is required to keep you from accelerating up or falling downward.
- imagine now, you are on the ground for exactly one half of any given second (ie, a slow jog). We need 100lb-s to maintain steady state, so simple algebra tells us we need to put out 200 pounds of force for that half second (200 pounds x .5s equals 100lb-s)
- now imagine you are only on the ground for 1/4 of any given second (typical of a fast sprint). We still need that same 100 lb-s. So we know we need 400 pounds of force (400 lb x .25s equals 100 lb-s).
Due to the mechanics of how people sprint, the faster you are on the ground, the less time you spend on the ground and the more air time you have. So as your speed increases, your vertical force increases proportionally.
I think this explanation is very wrong. In order to keep 100 labs on the ground from collapsing you need to produce 100 labs, be it 1/2 a second on the ground or a second. Elite sprinters ground contact time is about 0.08 seconds. According to your logic a 200 pound sprinter would have to produce 2500 pounds into the ground. They don’t. They produce around 4 times their body weight- around 800-900 pounds force
My explanation is correct.
Lets go with your 200 pound sprinter. 0.08s on the ground at a time. 4 steps per second is .32s on the ground per second. This would therefore require them to produce about 3.1x their body weight in force. So for a 200 pound sprinter, they need 200 lb-s for every second, so they would generate about 620 pounds of force, because 620 pounds x .32s is about 200 lb-s.
Where you went wrong is not accounting for the fact that they are taking multiple steps per second.
Impact-momentum is a very simple way of thinking of these sorts of problems, which is why I used it to illustrate the idea.
Sdfsdfsdfsdfsdfsdf wrote:
Ithinkitswrong wrote:
I think this explanation is very wrong. In order to keep 100 labs on the ground from collapsing you need to produce 100 labs, be it 1/2 a second on the ground or a second. Elite sprinters ground contact time is about 0.08 seconds. According to your logic a 200 pound sprinter would have to produce 2500 pounds into the ground. They don’t. They produce around 4 times their body weight- around 800-900 pounds force
My explanation is correct.
Lets go with your 200 pound sprinter. 0.08s on the ground at a time. 4 steps per second is .32s on the ground per second. This would therefore require them to produce about 3.1x their body weight in force. So for a 200 pound sprinter, they need 200 lb-s for every second, so they would generate about 620 pounds of force, because 620 pounds x .32s is about 200 lb-s.
Where you went wrong is not accounting for the fact that they are taking multiple steps per second.
Impact-momentum is a very simple way of thinking of these sorts of problems, which is why I used it to illustrate the idea.
You are wrong. Wrong calculation. The sprinter does not have to keep 200 pounds from collapsing in the total time he spends on the ground relative to a second or a taken time. He has to produce a resistance to his body weight per stride/ per impact on the ground, and not per time on the ground out of another time.
Hghh wrote:
Sdfsdfsdfsdfsdfsdf wrote:
My explanation is correct.
Lets go with your 200 pound sprinter. 0.08s on the ground at a time. 4 steps per second is .32s on the ground per second. This would therefore require them to produce about 3.1x their body weight in force. So for a 200 pound sprinter, they need 200 lb-s for every second, so they would generate about 620 pounds of force, because 620 pounds x .32s is about 200 lb-s.
Where you went wrong is not accounting for the fact that they are taking multiple steps per second.
Impact-momentum is a very simple way of thinking of these sorts of problems, which is why I used it to illustrate the idea.
You are wrong. Wrong calculation. The sprinter does not have to keep 200 pounds from collapsing in the total time he spends on the ground relative to a second or a taken time. He has to produce a resistance to his body weight per stride/ per impact on the ground, and not per time on the ground out of another time.
What you just wrote makes no sense, maybe you could re write what you mean?
Impulse momentum theorem is very basic physics... I'm not wrong.
It can be summed up with:
On the ground 100% of the time? Average force is 100% of body weight.
On the ground 50% of the time? Average force is 200% of body weight.
On ground 25% of the time? Average force is 400% of body weight.
Etc.
Sprinter question wrote:
I read that sprinters produce a force of 4 times their body weight into the ground. My question is- shouldn’t they produce the same force all the time however faster and faster? Why does the faster you go the more force you need to put into the ground? Why isn’t the same force you need to put only in less time?
Simple, speed +mass = force.
The more speed that they produce with their turnover, plus the weight of their body will produce a lot of force. Also, they strike with a dorsiflexed foot, which generates a lot of ground force.
Sdfsdfsdfsdfsdfsdf wrote:
Hghh wrote:
You are wrong. Wrong calculation. The sprinter does not have to keep 200 pounds from collapsing in the total time he spends on the ground relative to a second or a taken time. He has to produce a resistance to his body weight per stride/ per impact on the ground, and not per time on the ground out of another time.
What you just wrote makes no sense, maybe you could re write what you mean?
Impulse momentum theorem is very basic physics... I'm not wrong.
It can be summed up with:
On the ground 100% of the time? Average force is 100% of body weight.
On the ground 50% of the time? Average force is 200% of body weight.
On ground 25% of the time? Average force is 400% of body weight.
Etc.
Wrong. Percentage of time spent on the ground has no room in this equation. Every impact of the body to the ground- you will have to produce a resistance force in order not to collapse. My question is why this force has to be big at high speeds and doesn’t the same only produced faster? Shouldn’t what need to grow is the speed of the production of the force?
Jffhhh wrote:
Sdfsdfsdfsdfsdfsdf wrote:
What you just wrote makes no sense, maybe you could re write what you mean?
Impulse momentum theorem is very basic physics... I'm not wrong.
It can be summed up with:
On the ground 100% of the time? Average force is 100% of body weight.
On the ground 50% of the time? Average force is 200% of body weight.
On ground 25% of the time? Average force is 400% of body weight.
Etc.
Wrong. Percentage of time spent on the ground has no room in this equation. Every impact of the body to the ground- you will have to produce a resistance force in order not to collapse. My question is why this force has to be big at high speeds and doesn’t the same only produced faster? Shouldn’t what need to grow is the speed of the production of the force?
It took me until now to realize that you're trolling. So i give you 7/10.
What took you long is to find out you were wrong and don’t have an answer
because sprinters are very muscular & strong, and long distance runners are thin & emaciated
/thread
Hfffh wrote:
What took you long is to find out you were wrong and don’t have an answer
Not taking the bait. I already figured out you were trolling and gave you some solid points.
You are full of crap. Let’s go with your theory. Elite sprinters are 0.08 seconds on the ground and 0.12 seconds in the air. That means 40% of the time on the ground. They produce almost 1000 pounds force to the ground (see video below). 40% of 1000 pounds is 400 pounds. According to you they weight 400 pounds.
My question is - shouldn’t they produce the same force all the time however faster and faster?
I have no idea whether they "should" or not, but the fact of the matter is they do not.
if you can't be bothered to read the paper I link to below, "Vertical ground reaction forces ... and periods of ground force application ... were measured using a custom, high-speed force treadmill."
they measured them because they are not the same. doh!
they found that, "the stance phase limit to running speed is imposed not by the maximum forces that the limbs can apply to the ground but rather by the minimum time needed to apply the large, mass-specific forces necessary."
in other words, running speed is not determined by changes in ground forces, but by changes in ground contact time.
cheers.
https://www.ncbi.nlm.nih.gov/pubmed/20093666