you have so far posted 7 times on this topic with only 1 post offering some numbers
that was something about calories from runnersworld
why can't you actually offer some theoretical clockings ???
the OP wants to know if the fat boy made as much effort as skinny one ?
if not, what clocking woud the fat boy have had to do to do so ???
offer a damn number...
fisky wrote:Thanks, but I'm still confused. Under the above formula, once a runner reaches a constant speed, no further work is done since horizontal acceleration is zero.
Example: If mass = 60kg and displacement = 1 mile, but acceleration = zero, then 60x1x0=0
it is not acceleration we are concerned about
it is kinetic energy as starter
that is given as 0.5mv^2
that has to be modified because of drag
at equilibrium pace = flat-out from gun-to-tape, you are limited from physics only by drag
( & only by actual physical leg-turnover speed if running on moon with 0kg scuba in 0kg pressurised suit ) :
0.5mv^3
http://scienceworld.wolfram.com/physics/DragPower.htmlventolin^3 wrote:
fisky wrote:Thanks, but I'm still confused. Under the above formula, once a runner reaches a constant speed, no further work is done since horizontal acceleration is zero.Example: If mass = 60kg and displacement = 1 mile, but acceleration = zero, then 60x1x0=0
it is not acceleration we are concerned about
it is kinetic energy as starter
that is given as 0.5mv^2
that has to be modified because of drag
at equilibrium pace = flat-out from gun-to-tape, you are limited from physics only by drag
( & only by actual physical leg-turnover speed if running on moon with 0kg scuba in 0kg pressurised suit ) :
0.5mv^3
http://scienceworld.wolfram.com/physics/DragPower.html
Energy is not power, and you can't just multiply KE by v "because of drag".
Also, your assumption that a runner is only working to overcome drag at a steady pace is bogus. If you've run in a tailwind, you probably know this. If not, work through the drag power equation you linked to with some reasonable assumptions:
P = .5*rho*Cd*A*v^3 with rho = 1.2 kg/m^3, Cd = .6, A = .5 m^2, v = 5 m/s
so power to overcome drag for at least one human is 23 W at 5:22 pace. 23 watts is nothing, but most people have trouble maintaining 5:22 pace for extended periods of time. so, there's more to the story than drag.
What's missing? In every step of a run:
-We decelerate a bit when our foot touches down.
-We catch ourselves from the brief moment of airtime (freefall) by exerting a large force on the ground.
-We reaccelerate at toe off.
-We launch ourselves back up into the air.
-We soar through the air (100-150 milliseconds or so)
This cycle repeats 2.5 to 3 times per second. This is where the energy goes.
fisky wrote:
Thanks, but I'm still confused. Under the above formula, once a runner reaches a constant speed, no further work is done since horizontal acceleration is zero.
Example: If mass = 60kg and displacement = 1 mile, but acceleration = zero, then 60x1x0=0
Runners are constantly accelerating, and the a vector is constantly changing. In the air, we're acclerating down (gravity) and a bit backwards (drag). On the ground, we add another brief backwards component (braking), replace the gravitation accel down with an upward accel (by pushing against the ground), and add a forward component to regain the speed we lost (push off).
If we take a macro view and say a = 0 and displacement = distance run, then W = m*a*d doesn't help because all real work gets tossed in the "internal losses" bin.
I'm not sure it is worth arguing with him. He has deleted all the posts where I have been critical of his drag-only idea.
knox harrington wrote:Energy is not power, and you can't just multiply KE by v "because of drag"
you are utterly clueless about physics
that is formula from one of best physics sites on web
you have better physics ???
then there is working example from "oxford uni" :
http://www.atm.ox.ac.uk/rowing/physics/basics.html#section5anyone with a clue about physics & maths woud understand it is all about v^3 & mass is easily a workable parameter to answer OP's question
Also, your assumption that a runner is only working to overcome drag at a steady pace is bogus
nonsense
it is hugely main factor at equilibrium
it determines equilibrium
If you've run in a tailwind, you probably know this
nonsense
any +ve wind just etches up the equilibrium speed, nothing more within limitations of anatomy & physical leg-turnover
If not, work through the drag power equation you linked to with some reasonable assumptions:
P = .5*rho*Cd*A*v^3 with rho = 1.2 kg/m^3, Cd = .6, A = .5 m^2, v = 5 m/s
so power to overcome drag for at least one human is 23 W at 5:22 pace
no one asked for wattage
OP didn't
it's required to answer : are the 2 efforts =, if not, who is superior & by implication what are = efforts
you answered none of these
23 watts is nothing, but most people have trouble maintaining 5:22 pace for extended periods of time
no one asked
all you have done is insert numbers into a formula i linked to & offered nothing to OP question
so, there's more to the story than drag
nonsense
with very similar surface areas which are assumed, the only significant factor is drag
What's missing? In every step of a run:
-We decelerate a bit when our foot touches down.
-We catch ourselves from the brief moment of airtime (freefall) by exerting a large force on the ground.
-We reaccelerate at toe off.
-We launch ourselves back up into the air.
-We soar through the air (100-150 milliseconds or so)
This cycle repeats 2.5 to 3 times per second. This is where the energy goes
i'm surprised to see in this extraneous drivel no mention of the relative swing of balls in their sack sideways/up-down/anterior-posterior vector for the 2 guys affecting centre of mass/momentum considerations ???
knox harrington wrote:Runners are constantly accelerating, and the a vector is constantly changing. In the air, we're acclerating down (gravity) and a bit backwards (drag). On the ground, we add another brief backwards component (braking), replace the gravitation accel down with an upward accel (by pushing against the ground), and add a forward component to regain the speed we lost (push off).
If we take a macro view and say a = 0 and displacement = distance run, then W = m*a*d doesn't help because all real work gets tossed in the "internal losses" bin.
where is the relative swing of balls in their sack sideways/up-down/anterior-posterior vector for the 2 guys affecting centre of mass/momentum considerations in your grand scheme to determine OP's answer ???