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Is the Olympic running race fair?

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7

1

$begingroup$

I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)

Isn't that weird? I mean, just look at the curvatures of each lane!

---

---

(Source)

Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.

(The extra centrifugal force is roughly 2 Newtons from my calculation).

I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?

newtonian-mechanics friction work everyday-life centrifugal-force

share|cite|improve this question

edited 5 hours ago

Krishnanand J

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
  $endgroup$
  – Yi Jiang
  9 hours ago
  
  
  

  
  
  
  


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  $begingroup$
  Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
  $endgroup$
  – Jon Custer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
  $endgroup$
  – Krishnanand J
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
  $endgroup$
  – Jon Custer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
  $endgroup$
  – BowlOfRed
  6 hours ago
  

  
  
  
  

 | 
show 2 more comments

7

1

$begingroup$

I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)

Isn't that weird? I mean, just look at the curvatures of each lane!

---

---

(Source)

Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.

(The extra centrifugal force is roughly 2 Newtons from my calculation).

I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?

newtonian-mechanics friction work everyday-life centrifugal-force

share|cite|improve this question

edited 5 hours ago

Krishnanand J

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either.
  $endgroup$
  – Yi Jiang
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
  $endgroup$
  – Jon Custer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right?
  $endgroup$
  – Krishnanand J
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
  $endgroup$
  – Jon Custer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
  $endgroup$
  – BowlOfRed
  6 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)

Isn't that weird? I mean, just look at the curvatures of each lane!

---

---

(Source)

Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.

(The extra centrifugal force is roughly 2 Newtons from my calculation).

I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?

newtonian-mechanics friction work everyday-life centrifugal-force

share|cite|improve this question

edited 5 hours ago

Krishnanand J

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

$endgroup$

share|cite|improve this question

edited 5 hours ago

Krishnanand J

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

share|cite|improve this question

edited 5 hours ago

Krishnanand J

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

asked 10 hours ago

Krishnanand JKrishnanand J

1,64833 gold badges99 silver badges2929 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).

share|cite|improve this answer

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
  $endgroup$
  – puppetsock
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
  $endgroup$
  – Mark
  1 hour ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.

Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.

All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.

share|cite|improve this answer

edited 58 mins ago

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant "scalar" and not "cross" product.
  $endgroup$
  – hyportnex
  1 hour ago
  

  
  
  
  

add a comment
 | 

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6

$begingroup$

Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).

share|cite|improve this answer

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
  $endgroup$
  – puppetsock
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
  $endgroup$
  – Mark
  1 hour ago
  

  
  
  
  

add a comment
 | 

6

$begingroup$

Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).

share|cite|improve this answer

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded.
  $endgroup$
  – puppetsock
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @puppetsock, my general recollection is that the innermost lane simply isn't used during sprints.
  $endgroup$
  – Mark
  1 hour ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).

share|cite|improve this answer

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

answered 9 hours ago

Ben51Ben51

4,8321010 silver badges3131 bronze badges

1

$begingroup$

From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.

Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.

All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.

share|cite|improve this answer

edited 58 mins ago

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant "scalar" and not "cross" product.
  $endgroup$
  – hyportnex
  1 hour ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.

Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.

All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.

share|cite|improve this answer

edited 58 mins ago

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant "scalar" and not "cross" product.
  $endgroup$
  – hyportnex
  1 hour ago
  

  
  
  
  

add a comment
 | 

$begingroup$

From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F cdot vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.

Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.

All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.

share|cite|improve this answer

edited 58 mins ago

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

$endgroup$

share|cite|improve this answer

edited 58 mins ago

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges

answered 2 hours ago

BowlOfRedBowlOfRed

20.4k22 gold badges3535 silver badges5454 bronze badges