Bike Forums > Why do heavier cyclists descend faster?
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 Road Cycling “It is by riding a bicycle that you learn the contours of a country best, since you have to sweat up the hills and coast down them. Thus you remember them as they actually are, while in a motor car only a high hill impresses you, and you have no such accurate remembrance of country you have driven through as you gain by riding a bicycle.” -- Ernest Hemingway

12-04-08, 02:32 PM   #26
gsteinb
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 Originally Posted by <3 2 Ride 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.
so what happens when, say, a plane on it's way to NY falls from the sky? will it fall at the same rate as PCad down Gate Hill Road?

 12-04-08, 02:34 PM #27 BikeIndustryGuy Guest   Bikes: Posts: n/a Mentioned: Post(s) Tagged: Thread(s) Quoted: Post(s) overweight riders have cellulite, these dimples cause a boundary layer of air and reduce drag..blah..blah.
12-04-08, 02:35 PM   #28
sfwhiteman
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 Originally Posted by ROJA 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?
No, sorry... I didn't mean the heavier rider was "pushing" against the air resistance with his leg muscles (pedaling). I meant that his greater weight (mass) itself generates more force against the air resistance.

 12-04-08, 02:35 PM #29 the_mac Senior Member   Join Date: Apr 2008 Bikes: Posts: 93 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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).
12-04-08, 02:42 PM   #30
rankin116
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 Originally Posted by the_mac The fatter you are, the more gravity affects you, meaning more watts for the coasting descent.
That's not true. It affects us all the same. Some just have more momentum than others to overcome the air resistance.

12-04-08, 02:51 PM   #31
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 Originally Posted by rankin116 That's not true. It affects us all the same. Some just have more momentum than others to overcome the air resistance.
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   #32
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 Originally Posted by asgelle 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?
No, but not for the reason you listed.
__________________
Burn the incline - V5

 12-04-08, 02:59 PM #33 patentcad Peloton Shelter Dog     Join Date: Nov 2005 Location: Chester, NY Bikes: 2013 Scott Foil, 2016 Scott Addict, 2015 Scott Scale 700SL MTB Posts: 60,053 Mentioned: 44 Post(s) Tagged: 0 Thread(s) Quoted: 711 Post(s) How the F should I know? Are you saying I'm fat?
12-04-08, 03:01 PM   #34
asgelle
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 Originally Posted by cuski No, but not for the reason you listed.
Where in the two questions I posted can you find any semblence of a reason listed?

12-04-08, 03:04 PM   #35
rankin116
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 Originally Posted by asgelle 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?
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
nafun
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 Originally Posted by <3 2 Ride 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.
That is one half of the explanation. The other half is how projected area scales with an objects size verses how mass scales. For a given dimension d, projected area scales with d^2, while mass scales with d^3.

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, 03:09 PM #37 rydaddy Type 1 Racer     Join Date: Apr 2006 Location: Davis, CA Bikes: A dozen or so. Posts: 2,579 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) I agree with Rankin.
12-04-08, 03:10 PM   #38
asgelle
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 Originally Posted by rankin116 . Until the wind resistance begins to play a role, ...
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.

 12-04-08, 03:14 PM #39 IAMTB Senior Member     Join Date: Dec 2005 Location: Iowa Bikes: Posts: 349 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) I believe this chart says it all: Ruthlessly stolen from another poster.
 12-04-08, 03:15 PM #40 Grumpy McTrumpy gmt     Join Date: Sep 2008 Location: Binghamton, NY Bikes: Posts: 12,504 Mentioned: 2 Post(s) Tagged: 0 Thread(s) Quoted: 43 Post(s) 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.
12-04-08, 03:18 PM   #41
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 Originally Posted by asgelle 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 overuse of English, and commas, can, in fact, be an effective tool to appear smarter when one doesn't bother to understand the original response.

 12-04-08, 03:19 PM #42 patentcad Peloton Shelter Dog     Join Date: Nov 2005 Location: Chester, NY Bikes: 2013 Scott Foil, 2016 Scott Addict, 2015 Scott Scale 700SL MTB Posts: 60,053 Mentioned: 44 Post(s) Tagged: 0 Thread(s) Quoted: 711 Post(s) Answer the question you weenies.
12-04-08, 03:23 PM   #43
asgelle
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 Originally Posted by BikeIndustryGuy The overuse of English, and commas, can, in fact, be an effective tool to appear smarter when one doesn't bother to understand the original response.
F=m a

F_g = m g

F_d = 1/2 rho Cd A v^2

Lim [F_d] |V->0 = 1/2 rho Cd A v^2

12-04-08, 03:23 PM   #44
rankin116
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 Originally Posted by asgelle 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.
I see your point. I guess what I was getting at is that it will take a little while for the wind resistance to affect the smaller rider enough for us to notice it.

 12-04-08, 03:24 PM #45 scr660 Senior Member   Join Date: Feb 2008 Bikes: Posts: 527 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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 03:28 PM.
12-04-08, 03:25 PM   #46
uberclkgtr
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 Originally Posted by nafun 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.
Which is the key here. So if riders were spheres, the bigger ones would go faster down a hill.

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 03:36 PM.

12-04-08, 03:26 PM   #47
asgelle
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 Originally Posted by scr660 Drag force can be approximated by D(v) = kv, where v is velocity.
Not anywhere I'd like to ride.

12-04-08, 03:26 PM   #48
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 Originally Posted by patentcad Answer the question you weenies.
I'll answer it myself:

PCD M > Gsteinb M = DNHL PWN

(that means I'm fatter than Gstein and I'll drop him on any descent)

 12-04-08, 03:27 PM #49 Brian Ratliff Senior Member     Join Date: May 2002 Location: Near Portland, OR Bikes: Three road bikes. Two track bikes. Posts: 10,095 Mentioned: 1 Post(s) Tagged: 0 Thread(s) Quoted: 28 Post(s) 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: __________________ 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
scr660
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 Originally Posted by asgelle Not anywhere I'd like to ride.
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