Why do heavier cyclists descend faster?
12-04-08, 02:32 PM
#26
out walking the earth
For today's lesson in physics:
"A free falling object achieves its terminal velocity when the downward force of gravity (Fg)equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.
As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight (mg). Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.
Mathematically, terminal velocity is given by
Vt = sq rt (2mg/ρACd)
where
Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling, and
A = projected area of the object. "
I hope that answers your question for you.
"A free falling object achieves its terminal velocity when the downward force of gravity (Fg)equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.
As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight (mg). Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.
Mathematically, terminal velocity is given by
Vt = sq rt (2mg/ρACd)
where
Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling, and
A = projected area of the object. "
I hope that answers your question for you.
12-04-08, 02:35 PM
#28
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Ahh, interesting. So it's only relevant when pedaling, not coasting (except for the conservation of momentum part, which is also a good answer). So say a skinny dude putting out 150 watts will go 15 mph on the flat. The big guy puts out 200 watts to go 15 mph on the flat. They are both working at 60% intensity to go that 15 mph. When you get to the downhill, the 200 watts of the big guy suddenly make him go faster than the 150 watts of the skinny dude because his heavier weight is no longer working against him. Something like that?
12-04-08, 02:35 PM
#29
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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).
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).
12-04-08, 02:42 PM
#30
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12-04-08, 02:51 PM
#31
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And if two riders of different weight start coasting downhill from rest so they both have 0 momentum? Are you saying that then they will descend at the same speed?
12-04-08, 02:59 PM
#33
Peloton Shelter Dog
How the F should I know? Are you saying I'm fat?
12-04-08, 03:01 PM
#34
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12-04-08, 03:04 PM
#35
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All variables being equal, and for a short period of time, yes. Until the wind resistance begins to play a role, they will be at the same speed.
12-04-08, 03:09 PM
#36
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For today's lesson in physics:
"A free falling object achieves its terminal velocity when the downward force of gravity (Fg)equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.
As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight (mg). Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.
Mathematically, terminal velocity is given by
Vt = sq rt (2mg/ρACd)
where
Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling, and
A = projected area of the object. "
I hope that answers your question for you.
"A free falling object achieves its terminal velocity when the downward force of gravity (Fg)equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.
As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight (mg). Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.
Mathematically, terminal velocity is given by
Vt = sq rt (2mg/ρACd)
where
Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling, and
A = projected area of the object. "
I hope that answers your question for you.
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.
12-04-08, 03:10 PM
#38
Senior Member
At what speed do you believe wind resistance begins? What changes suddenly as that threshold is crossed?
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.
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.
12-04-08, 03:15 PM
#40
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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.
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.
12-04-08, 03:18 PM
#41
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At what speed do you believe wind resistance begins? What changes suddenly as that threshold is crossed?
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.
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.
12-04-08, 03:19 PM
#42
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Answer the question you weenies.
12-04-08, 03:23 PM
#43
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12-04-08, 03:23 PM
#44
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At what speed do you believe wind resistance begins? What changes suddenly as that threshold is crossed?
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.
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.
12-04-08, 03:24 PM
#45
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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
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 03:28 PM.
12-04-08, 03:25 PM
#46
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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:
https://www.springerlink.com/content/t617t89lj948gw96/
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 03:36 PM.
12-04-08, 03:26 PM
#47
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12-04-08, 03:26 PM
#48
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12-04-08, 03:27 PM
#49
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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.
Long answer:
General long answer:
Long answer:
General long answer:
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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
12-04-08, 03:28 PM
#50
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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, 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
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