<Physics police pull up and step out of hovercraft>
arizona.edu (Mark D Williams) writes:
>Strictly speaking, this analogy is misleading. True, the more drag you have,
>the harder it is to get from one end of the pool to the other. However, you
>are suggesting that given equal amounts of power output, for different
>masses, you are doing different amounts of work to go the same distance.
>Work = (Force)*(Distance).
<the ***come out>
This is inadequate to even begin to describe the complex motion of swimming
and the work you do. In ***stroke you accelerate and decelerate quite a
bit. Efficiency of *** flow at differing fat levels is a consideration.
Even the mass of swinging arms and kicking legs comes into play.
>Now, since this is horizontal work (i.e. our
>force acting through a distance is a horizontal distance, not the vertical
>distance which would be relevant if we were considering object mass)
<the rights are read..."You have the right to a calculator...anything you
scribble can and will be used against you on the quiz...">
What about up+down motion of ***stroke and fly? The torsional motion of
freestyle and arm movement?
>the force to consider is not the weight of the objects, but the force
>needed to maintain a constant velocity against the internal friction of
>wheels, bearings, etc. and the force needed to maintain a constant velocity
>against aerodynamic drag. Now, according to fluid mechanics:
>Drag Force = 1/2 * (fluid density) * (Velocity)^2 * (Drag Coefficient) * (X-
<the prisoner is dragged away to the mad scientist's asylum to await trial>
This would be nice if (Velocity)^2 were a nice, constant term. No one has
ever developed a very good mathematical model for such complex and variant
human locomotion, especially immersed in a fluid. Don't try to simplify it
>Since you have stated that aerodynamics, gearing, tires, etc are all the
>this would imply that the drag force for the two vehicles is the same, and
>therefore the work done will be the same, regardless of their mass, if they
>maintain constant velocity over the 1/4 mile, because, according to
>the above equation, with same drag coefficient, and cross sectional area,
>since the two vehicles are the same in all respects except for their mass,
>they will have the same drag force to fight and therefore will do the
>same amount of work (i.e. Drag Force * (1/4 Mile). )
<the death penalty is recommended>
Well, it's about cars but I'll go for it anyway. It still takes less work
to move a lighter car 1/4 mile with the same drag coefficients and such
because of a non-constant velocity (you begin at the starting line and start
at the wall at V=0, to keep the analogy with swimming). The difference in
what the engine can put out will be made up in a faster time between the
markers (less time for the engine to put a chunk of energy, thus a
smaller chunk). The reason is that the lighter car will accelerate to its
maximum velocity faster than the heavy one. There's a certain amount of
energy that has to go into the car itself in the form of kinetic energy
(1/2mV^2), and the rest goes into fighting drag. Car races can be won
and lost in getting out of the pits and accelerating to maximum velocity
right after turns, if they're the same on the straightaways. Kind of like
swimming, where you can beat an equal-speed swimmer on the strength of
starts and turns.
>However, if we look at the drag force equation, for the same velocity,
>and fluid density (obviously we all swim in water), we will presumably
>have a larger drag coefficient, and cross-sectional area if we have
>more body fat--we are presumably rounder! :-) and therefore, the
>drag force should be greater if we have more body fat, and we will do more
>work. Same answer, different method! Phew....
<sentence...20 years to life...>
I still agree that higher body fat (which just means you're out of shape)
means slower swimming. The question of swimmers with a naturally high body
fat level being better swimmers seems answered by statistics, it's not
so. In general, it's OK to say that it's harder to move a large
object through water. It takes more energy to displace the water that the
object moves through. The energy lost in actual drag on the body is poorly
understood when the body is going through such a complicated motion. Even
the drag coefficient term can change in seconds if you break out in a sweat
on the blocks. I think everyone without a few paralell crays should leave
the drag stuff alone (I certainly won't touch it).
Steve-O (if you're a *physicist* and you think I screwed up, then correct
If your ship hasn't come in, swim out to it!