overweight riders have cellulite, these dimples cause a boundary layer of air and reduce drag..blah..blah.
The fatter you are, the more gravity affects you, meaning more watts for the coasting descent.
Now, the fatties also have a bigger surface area for wind to smash against.
These two forces work against eachother, and the real question is, do the forces balance perfectly (no change in descent speeds coasting) or does one have more effect than the other?
Because people are like most animals that fit the cube-square law, as you get bigger, your surface area grows as a square function, but your volume (and mass) grow as a cube. So heavier people will have a more efficient descending body (better power output from gravity when compared to surface area).
How the F should I know? Are you saying I'm fat?
For example, consider a sphere:
projected area = pi*r^2
mass = ρ*(4/3)*pi*r^3
So, the ratio m/A (which from your equation above is proportional to Vt) is greater for a larger object than for a smaller one, given constant proportions and composition.
I agree with Rankin.
The answer, of course, is that wind resistance exists for any non-zero velocity and there is no abrupt change. The implication of that is the short time of which you write is, in fact, also zero.
I believe this chart says it all:
Ruthlessly stolen from another poster.
Pulling the trigger as often as possible.
a feather and a hammer will both fall at the same speed on the moon.
but a hammer and a piece of neutron star matter the size of a bowling ball will not.
the only reason why it appears that the feather and hammer fall at the same rate is that relative to the moon's mass, their mass is too close to one-another.
Now if I can just find a rider with enough mass to reach relativistic velocities to draft. I hear that the podium girls at the event horizon are really cool.
Answer the question you weenies.
Um...I think the answer is:
Drag force can be approximated by D(v) = kv, where v is velocity.
on an incline, the force pushing down is mg* cos(theta), where theta is an incline
at terminal velocity (on an incline - not falling directly), kv = mg*cos(theta), so, basically,
v is an increasing function of m. That's it.
Actually, it doesn't even matter what the functions are. As long as D(v), the drag, is a monotonically increasing function of v, and force of gravity is G(m), an increasing function of m, we have that, at coasting speed,
D(v) = G(m) => v = D^-1(G(m)), which is increasing in m
Last edited by scr660; 12-04-08 at 02:28 PM.
Spoken like a true physicist. =]
Of course riders are not spheres, but I suspect the following research would be helpful in determining the scaling relationship:
That allows us to relate m to A and then find that Vt scales by m^.32 which means the Vt is a very weak function of m. Not even linear.
Last edited by uberclkgtr; 12-04-08 at 02:36 PM.
Short answer: frontal area of the larger rider is basically the same as the frontal area of the smaller rider, hence, air resistance is basically equal for both. Gravity acts on the big rider with more force than on the smaller rider. Big force minus air drag is greater than little force minus air drag, so big rider falls faster.
General long answer:
Cat 2 Track, Cat 3 Road.
"If you’re new enough [to racing] that you would ask such question, then i would hazard a guess that if you just made up a workout that sounded hard to do, and did it, you’d probably get faster." --the tiniest sprinter
D(v) = G(m) => v = D^-1(G(m)), which is increasing in m, so w/e, the point is still that fatter people descend more quickly.
Of course, if you don't believe that D(v) and G(m) are increasing in v and m, you might be an idiot