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  • 06/22/2011, 03:00PM ET

I will now defend the undefendable

Per Sempre (5-22-1) vs Sabih (122-54-26)
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Twenty bodies sprint towards one and other, destined to collide. Ten on each side if you don't count the guy with the ball and kicker.

According to sports-science {commonly seen on ESPN} getting hit by an "average" linebacker in the NFL at normal game speed is equivalent to getting hit by a volkswagen beetle at about 27-28 miles per hour. Enough to injure any normal human being in a car accident. Now imagine that same "average" linebacker getting a 50-60 yard running head start before impact. Sounds like he could do some serious damage.

I understand that football is a physical game and that is part of the enjoyment. But what we forget sometimes is that "football" is their JOB, not their LIFE.

There is a life after sports for the guys we cheer for every sunday and if eliminating the kickoff is small way we can take another measure to protect them then I dont see why it shouldnt be done.

Whichever team wins the coin toss should start with the ball at their own 25 yard line. If the team wants to try for an "onside kick" i heard an idea that i liked where the team would have to pick up a 4th and 15 or 20 to determine if they could keep possesion.

GL and MTC


I am in-favor of player safety but eliminating kick offs is the most ridiculous thing I have ever heard.

Injuries are a part of all sports. The players are aware of the risks when they take the field. When I worked in a warehouse and as an assembly person, I was making twice my friends flipping burgers partially because of the physical risk of the job. And these were inherent risks and part of the job.

Unlike blows to the head which the NFL tried to eliminate last year, kickoffs actually serve a purpose in the game. By reducing the running starts, the NFL has already reduced some of the risk of injury on kickoffs.

I also have a hard time believing your comparison of a beetle to an NFL LB. A beetle weighs over a ton and its made of metal.

A beetle traveling at 28 mph generates over 190 000 newtons of force. (FORCE = MASS * Velocity * Velocity) A LB would have to run at 84 mph to generate the same amount of force.

Also, if you wanna eliminate kickoffs, why not remove punts as well. Its nearly the same thing.


Forfeited Turn


Correction to the initial argument, the LB would have to travel at 213 mph to generate same force as a Beetle. Thank you Willy. (see comment 6 below for details).

Add to the fact that pads reduce the force of the hit by 50%.

The NFL should focus on eliminating hits to the head. You should not be allowed to lead with your head or target the head of a ball carrier doesn't matter if he is a ball carrier, receiver or returner.


Forfeited Turn


Forfeited Turn

Comments (18) Add A Comment

June 22, 2011  08:04 PM ET
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Seriously, can you actually consider this to be a real game simulation. Who is returning kicks here, Peyton Manning.

Comment #2 has been removed
June 22, 2011  09:34 PM ET
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You can complain about the compounding injuries but part of it is actually the players fault.

Yes, I just said that.

The players need to know when to retire.

June 22, 2011  10:25 PM ET
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"football" is their JOB, not their LIFE.

All of my hate and rage. You can't be serious?

June 22, 2011  10:36 PM ET
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If they get a 50-60 yard head start they will eventually reach terminal velocity. They probably will reach it in the first 10 yards. With that being said I think the impact will be about the same for a kick off player with a 60 yard head start and a normal lineback.

June 22, 2011  11:03 PM ET
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Sabeeh - Your point is well taken however your units are all sixes and sevens.

A volkswagen beetle weighs about 1900 lbs. That is the mass equivalent of just under 60 slugs. (The slug is the unit of mass in the English System of Units.) Therefore the stopping force for a 1900 lb beetle traveling at 28 mph (41 ft/sec) is roughly 2430 lbs. In the International System of Units this corresponds to a force close to 10.8 kN (not 190 kN).

A 250 lb linebacker traveling at world record speed (23.4 mph) would require approximately 1.29 kN of stopping force. ( Roughly 9.1 times less.)

For him (or her) to generate 2460 lbs of force he (or she) would have to be traveling at 213 mph. (not 84 mph).

June 22, 2011  11:16 PM ET
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Sabeeh states:

(FORCE = MASS * Velocity * Velocity)

Sorry, Sabeeh, but this is NOT true....
Newton's 3rd Law of Motion is FORCE = MASS * ACCELERATION

and acceleration does not equal velocity*velocity

June 22, 2011  11:21 PM ET
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QUOTE(#6):
Originally posted 11:03 PM ET  06.22 by Box Car Willy

Sabeeh - Your point is well taken however your units are all sixes and sevens.A volkswagen beetle weighs about 1900 lbs. That is the mass equivalent of just under 60 slugs. (The slug is the unit of mass in the English System of Units.) Therefore the stopping force for a 1900 lb beetle traveling at 28 mph (41 ft/sec) is roughly 2430 lbs. In the International System of Units this corresponds to a force close to 10.8 kN (not 190 kN).A 250 lb linebacker traveling at world record speed (23.4 mph) would require approximately 1.29 kN of stopping force. ( Roughly 9.1 times less.)For him (or her) to generate 2460 lbs of force he (or she) would have to be traveling at 213 mph. (not 84 mph).

You lost me. I last took physics 6 years ago in HS.

I had to go convert everything to metric. I first converted velocity to meter/second and weight to KG. Weight for a beetle was listed @ 1230 kg. 28 MPH is about 12.52 meters/second. So 1230*12.96*12.96 got me 194k of force in newtons.

Then reverse engineer to get LB speed.

You know more than I do, where did I go wrong.

Comment #9 has been removed
June 23, 2011  12:22 AM ET
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QUOTE(#7):
Originally posted 11:16 PM ET  06.22 by Really??

Sorry, Sabeeh, but this is NOT true....
Newton's 3rd Law of Motion is FORCE = MASS * ACCELERATION

and acceleration does not equal velocity*velocity

That's Newton's 2nd Law.

But yeah, Sabeeh's formula is off.

June 23, 2011  12:34 AM ET
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QUOTE(#8):
Originally posted 11:21 PM ET  06.22 by Sabih

You lost me. I last took physics 6 years ago in HS.I had to go convert everything to metric. I first converted velocity to meter/second and weight to KG. Weight for a beetle was listed @ 1230 kg. 28 MPH is about 12.52 meters/second. So 1230*12.96*12.96 got me 194k of force in newtons. Then reverse engineer to get LB speed. You know more than I do, where did I go wrong.

First of all THIS AGAIN is correct. The unit of force is, mass X acceleration, not mass X velocity squared, as you used. However the terms, stopping force, or impulse or change in momentum (which is what I used) are dimensioned in units of mass X velocity. I assumed that is what you meant to calculate. Quite literally this quantity is related to force via the time that the force acts on the body. I just assumed you were taking the time of interaction for the beetle and the linebacker with the ball carrier to be the same and thus swallowing the units of seconds.

Where you went wrong:
First, 28 mph converts to 41.07 ft/sec (28 [mile/hour] X 5280 [ft/mile]/3600 [sec/hour]), not 12.52 ft/sec.

Second, what you calculated is a quantity that scales with the kinetic energy of beetle. That is, energy = mass X velocity X velocity, not force. Hence what you were discussing (again provided the interaction times are the same and off by the conversion errors and a factor of 1 half) is the power needed to stop the ball carrier.

Nevertheless, as I said before your point is correct.

June 23, 2011  12:39 AM ET
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QUOTE(#11):
Originally posted 12:34 AM ET  06.23 by Box Car Willy

First of all THIS AGAIN is correct. The unit of force is, mass X acceleration, not mass X velocity squared, as you used. However the terms, stopping force, or impulse or change in momentum (which is what I used) are dimensioned in units of mass X velocity. I assumed that is what you meant to calculate. Quite literally this quantity is related to force via the time that the force acts on the body. I just assumed you were taking the time of interaction for the beetle and the linebacker with the ball carrier to be the same and thus swallowing the units of seconds.Where you went wrong:First, 28 mph converts to 41.07 ft/sec (28 [mile/hour] X 5280 [ft/mile]/3600 [sec/hour]), not 12.52 ft/sec.Second, what you calculated is a quantity that scales with the kinetic energy of beetle. That is, energy = mass X velocity X velocity, not force. Hence what you were discussing (again provided the interaction times are the same and off by the conversion errors and a factor of 1 half) is the power needed to stop the ball carrier.Nevertheless, as I said before your point is correct.

And this is why I failed at math.

June 23, 2011  12:13 PM ET
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I think I calculated energy instead of force.

Anyways. Here is another sports science looking at the force of a Ray Lewis Bull rush.



Says a Ray Lewis bull rush has a force of 1000 lbs.

June 23, 2011  10:02 PM ET
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First vote for left comes after the Forfeited Turn.

June 23, 2011  10:03 PM ET
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QUOTE(#14):
Originally posted 10:02 PM ET  06.23 by (Un) i like cheese

First vote for left comes after the Forfeited Turn.

It was more convincing than the first argument....

June 25, 2011  12:17 AM ET
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------------------------------------------------------------->

June 25, 2011  01:00 AM ET
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QUOTE(#6):
Originally posted 11:03 PM ET  06.22 by Box Car Willy

Sabeeh - Your point is well taken however your units are all sixes and sevens.A volkswagen beetle weighs about 1900 lbs. That is the mass equivalent of just under 60 slugs. (The slug is the unit of mass in the English System of Units.) Therefore the stopping force for a 1900 lb beetle traveling at 28 mph (41 ft/sec) is roughly 2430 lbs. In the International System of Units this corresponds to a force close to 10.8 kN (not 190 kN).A 250 lb linebacker traveling at world record speed (23.4 mph) would require approximately 1.29 kN of stopping force. ( Roughly 9.1 times less.)For him (or her) to generate 2460 lbs of force he (or she) would have to be traveling at 213 mph. (not 84 mph).

You don't need to put the "Or she" in there, because there are no women playing in the NFL.

 
June 25, 2011  01:05 AM ET
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QUOTE(#17):
Originally posted 01:00 AM ET  06.25 by Confucius. Trout

You don't need to put the "Or she" in there, because there are no women playing in the NFL.

Unless you count Jay Cutler.

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