Heavy bike riders descend faster?

[Trunk]

There is a very ingrained attitude in the bike world that heavier riders descend faster.

Of course I know that all objects fall at the same rate in a vaccuum.

Is there something else going on when you put heavy people on a bike that will make a heavy rider descend faster? Do 2 people with the same cross-section go down a hill at the same rate?

The notion seems wack to me, but there's a lot of people out there firmly convinced of it, like this guy (http://www.mayq.com/Best_european_trips/Cycling_speed_math.htm).

You can find mention of it on any bicycle bulletin board.

Related question. . .

Does a cannonball and a glass sphere of the same size fall at the same rate in air?

Do they roll down a plank at the same rate?

Do two cyclinders of the same size/different weight roll down a plank at the same rate if their coefficient of friction is the same?

[VunderBob]

There is a very ingrained attitude in the bike world that heavier riders descend faster.

Of course I know that all objects fall at the same rate in a vacuum.

Is there something else going on when you put heavy people on a bike that will make a heavy rider descend faster? Do 2 people with the same cross-section go down a hill at the same rate?

The notion seems wack to me, but there's a lot of people out there firmly convinced of it, like this guy (http://www.mayq.com/Best_european_trips/Cycling_speed_math.htm).

I looked over the page, and with my cursory glances I did not see where he said heavier cyclists go downhill faster. The math he did looked OK, but I didn't try to critique it.

All things being equal (bearing resistance, tire pressure, etc.) two cyclists of unequal mass should descend a slope at the same rate. However, it will take more braking force for the heavier cyclist to stop, and if the the heavy rider is significantly heavier, the effect will be noticable in the stopping distance. My guess is that is where the perception of heavier riders going faster really comes from.


Related question. . .

Does a cannonball and a glass sphere of the same size fall at the same rate in air?

Do they roll down a plank at the same rate?

Do two cyclinders of the same size/different weight roll down a plank at the same rate if their coefficient of friction is the same?
1) Yes

2) Yes

3) Given a surface that doesn't compress or crush under a rolling object, friction plays very little into the dynamics of an object rolling down a slope. It's there if you analyze the forces, but doesn't have much of an effect except to keep the rolling object rolling instead of sliding.

[Hoodoo Ulove]

Gravity powered racers. such as soapbox derby cars, have a maximum weight rule, and the cars are ballasted to be just under it. A heavier guy on a bike, unless his drag is increased proportionately to his weight (unlikely, since mass is roughly proportional to the cube of linear dimensions and cross-sectional area proportionate to the square), will have a higher maximum speed downhill. We're talking here about a situation where he's maxed out his pedaling and is coasting.

[Hoodoo Ulove]

We're assuming here that the light and the heavy guy are both tucked in good aero positions. Actually, if you are insanely brave, you could go faster by putting your legs straight back, laying on your stomach on the seat. Cornering would be a problem, though.

[Trunk]

So, Hoodo are you stating that a heavier guy and a lighter guy, tucked into the same aero position will descend at different rates?

Forget about pedalling. . .assume they just crested the hill at the same rate and now are just tucked.

If so, I'll let you and Vunderbob hash it out. I'm just the instigator.

[Quercus]

Let's start from the bottom:
Does a cannonball and a glass sphere of the same size fall at the same rate in air?

(Assuming you mean a cannonball that is much denser, therefore heavier when the same size)
Initially, when they're both moving slowly, so air resistance doesn't matter, yes.
But once air resistance becomes important, the cannonball will go faster, so no.

There are three things going on: force of gravity (which is proportional to the mass), inertia (how much it resists acceleration, which is also proportional to mass) and the force of air resistance (which is more or less proportional to surface area, not mass). In a vacuum the cannonball has a larger gravity force, but also a larger inertia by exactly the same amount, so they fall at the same speed. But the air resistance force is the same for both balls; and because the glass sphere has less inertia it is slowed down more.

For rolling down a plank, it's the same thing, except usually speeds are low enough that air resistance doesn't matter much, so the cannonball will only be a tiny bit faster, probably undetectably so.


Now on to bicycles:
You've still got gravity, inertia and air resistance, plus some rolling resistance (friction on the wheels and bearings). Now rolling resistance is partly proportional to weight, and partly constant, so the lighter rider is affected a little more. And air resistance is important at hill-descending speeds, so the lighter rider is significantly more affected by air resistance.
Therefore the heavier rider has a speed advantage.


Plus, there's the idea that a heavier rider might not only have a slight advantage descending but also a disadvantage climbing, so relatively speaking, descending is even better. Consider a rider that's, well, let's not mince mince words, he's fat. He's got the same amount of muscle for power as the skinny guy, but has to lift a lot more mass, so he's much slower going uphill. Going downhill, though, gravity is more important that muscle power, so the fat guy has no disadvantage. And if fact a slight advantage, so he's a lot better off descending. This works even if he's not blubbery fat and out of shape, but just broad-shouldered with big arm muscles that don't push the bike forward.

So putting these together, I think it's not ridiculous to think that a smaller, lighter rider should plan to try and gain ground on uphill sections, while assuming he'll lose some on downhill.

[Fromage A Trois]

In terms of rolling cylinders down a slope, I remember at university covering the angular moment of inertia of an object.

The moment of inertia of an object, say a cylinder, is a measure of the mass distribution of the object. It's to do with angular momentum, and to simplify matters, the greater an object's moment of inertia, the harder it is for it to speed up or slow down rotation.

If there were two cylinders of the same weight, one hollow so most of the mass was concentrated at it's edge, and one uniform in density, the unform one would roll down a slope quicker. This is because it would have a lower moment of inertia, and hence would need more energy to get it spinning and therefore rolling down the slope. Similarly it would need more energy to stop.

Not related to the two bike riders (unless one of them has hydraulic tyres instead of pneumatic ones for some reason). ;)

[Tapioca Dextrin]

There is a very ingrained attitude in the bike world that heavier riders descend faster.

Of course I know that all objects fall at the same rate in a vaccuum.

On an anecdotal level, it's certainly true that bigger riders descend better than little guys, but it's not due the force of gravity, it's the force of competition.

Little guys generally have the highest power/weight ratio and therefore are generally better at uphill. No problem there.

From the quoted aticle.

On curvy mountain roads, total speed will be limited to what is safe. Power and gravitational acceleration make no difference.

In straight line descents without pedaling, on wide roads (ignoring the effects of wind), bikes will accelerate until wind resistance plus frictional resistance equals the acting force of gravity.

In the real world of competitive cycling, there are no straight line descents. There are always turns, switchbacks, sand, water and motor oil on the way down. That's the conditions that allow the bigger guys to catch up. The little guys (generally speaking) have little to gain by going really fast downhill, but the people following might have a lot to lose.

[Padeye]

Quercus, heavy riders are at a profound disadvantage in hills. I only weighed about 210 in my road bike riding days and it was always a struggle to keep up with my friends in the hills around San Diego. No one goes uphill fast so the amount of weight one has to lift if the biggest factor. It might have been an advantage going downhill but drag is the biggest factor at higher speed which marginalized any advantage I might have. In our suicide runs down the hill on Torrey Pines near Del Mar I might have slight edge but only by spinning out a 108" top gear.

Oh, bulky guys never win the polka dot jersey but upper body strength is a factor when standing in the pedals to climb hills.

[matt]

Does a cannonball and a glass sphere of the same size fall at the same rate in air?

Do they roll down a plank at the same rate?

Do two cyclinders of the same size/different weight roll down a plank at the same rate if their coefficient of friction is the same?

I beg to differ...

1) They shouldn't do. If we assume the same aerodynamic drag, the cannonball will be heavier and should fall faster. And since air was specified as the medium, neglecting drag would be naughty...

2) Not sure. Even if we can neglect aerodynamic drag, the situation is a little complicated. As the balls roll down the ramp, they are converting their gravitational potential energy into linear kinetic energy and also rotational kinetic energy. The cannonball will convert more energy per metre of descent because it is heavier, but it also has a greater rotational moment of inertia. Hmm. Going to bug me, that one. (Actually, think I've nailed it, see below.)

3) Similar considerations to (2). But wait! As far as rolling is concerned, there's no difference between a cylinder of fixed dimensions but double the density, and one that is twice as long and the original density. The twice-as-long cylinder has to roll at the same rate as the original cylinder - it's exactly the same as two of the original cylinders rolling side by side. So the double-density cylinder will also roll at the same rate. By extension, if we can leave aerodynamic drag out of it, all uniform cylinders of a particular radius will roll down a ramp at the same rate, regardless of their weight. The same argument applies to the balls as well - the glass ball and the cannonball will roll at the same rates.

Re the bicycle, bigger things of the same density fall faster than smaller things, just because of their smaller area-weight ratio. E.g. sawdust falls a lot slower than chunks of wood, even though they have the same density. So the bigger guy should descend faster. A guy exactly twice as big is going to be eight times as heavy (twice as tall, twice as wide, twice as thick) but only four times the cross section.

[Trunk]

I beg to differ...
There was nothing to differ with. Those were questions.


Re the bicycle, bigger things of the same density fall faster than smaller things, just because of their smaller area-weight ratio. E.g. sawdust falls a lot slower than chunks of wood, even though they have the same density. So the bigger guy should descend faster. A guy exactly twice as big is going to be eight times as heavy (twice as tall, twice as wide, twice as thick) but only four times the cross section.
Except you're not standing on the bike with your arms spread out to the sides.

Imagine if both riders were in the same aerodynamic shell, if you will. . .heavier guy still go faster?

On an anecdotal level, it's certainly true that bigger riders descend better than little guys, but it's not due the force of gravity, it's the force of competition.
That is decidedly NOT the meaning of the people who claim heavier riders descend faster.

I do understand what you're getting at though, cycling-wise.

Too many dissenting opinions in this thread so far. . .can someone assert their authori-TAY?

[VunderBob]

Too many dissenting opinions in this thread so far. . .can someone assert their authori-TAY?

Two riders, on bikes as close to identical as possible, in a shell will descend a slope at close to identical speeds provided they are passively riding.

I failed to take into account active riding. There, the heavier rider might have a slight advantage.

[Waverly]

Gravity powered racers. such as soapbox derby cars, have a maximum weight rule, and the cars are ballasted to be just under it. A heavier guy on a bike, unless his drag is increased proportionately to his weight (unlikely, since mass is roughly proportional to the cube of linear dimensions and cross-sectional area proportionate to the square), will have a higher maximum speed downhill. We're talking here about a situation where he's maxed out his pedaling and is coasting.I think you might be confusing the reason for heavier soap box racers. It’s the same reason that, all things being equal, a heavier biker may appear to go down hill faster.

Let’s first establish that acceleration due to gravity is the same for all of these objects. It is; and we aren’t going to cram mass into a meter per second per second unit of measure.

However, gravity is not the only force at work. Friction and air resistance both will impart some force against the forward movement. Here’s where the difference is observed.

Depending upon the slope and where you are in the decent, it’s possible for a heavier rider to begin accelerating more slowly, since the weight can increase the friction at the axel. The steeper the slope, the less you will see this particular effect, and at some point during the decent, mass as it applies to momentum (mass is a factor in momentum) may overtake the friction at the axel and overcome a portion of the decelerating forces.

Traction is also higher for the heavier rider, but I’m not sure it is a major effect here.

[Colophon]

The same thing is true with downhill skiing - heavier guys have the advantage.

As others have said, weight (and therefore the gravitational force on the person) goes up roughly as the cube of linear dimensions, whereas surface area (and therefore the drag force acting against movement) goes up as the square of linear dimensions.

So, the bigger a person is, the faster they are going to go, in general, as the ratio of drag to gravitational force is smaller. The famous "feather and hammer" experiment is an extreme demonstration of the same effect.

Of course, on skis you have the added factor that more weight means more pressure on the snow, which makes more of a melting effect and better lubricates the passage of the ski over the snow surface.

[Hoodoo Ulove]

Two riders, on bikes as close to identical as possible, in a shell will descend a slope at close to identical speeds provided they are passively riding.

I failed to take into account active riding. There, the heavier rider might have a slight advantage.
I take it you are referring here to HPVs with aero fairings. If so, you are even wronger than before, since th bigger guy now has no additional aerodynamic drag. Same force slowing it, greater force pushing it - the big guy wins.

Waverly, with bikes at high speed the aero forces are so much greater than the rolling friction that the latter can be ignored.

Tapi's point that in the real world of competitive cycling, downhill speeds are limited by cornering speed is of course correct.

[scr4]

Too many dissenting opinions in this thread so far. . .can someone assert their authori-TAY?
There is no dissent. People are just getting lost in the minutiae.

With varying degrees of approximation:


Two balls of different mass, dropped in vacuum (i.e. no air resistance) - this is the classic high school thought expeirment. The two balls fall at the same rate.
Two balls of identical size and different mass, dropped in air (air resistance not ignored): the heavier ball falls faster. Air resistance is the same for both balls, but the heavier ball is pulled downward with a greater force (of gravity).
Two objects of identical size and shape, and different mass, rolling down a hill on identical wheels: The heavier object rolls down faster, for the same reason.
Two objects of identical density and different mass, rolling down a hill on identical wheels: the heavier object will roll faster. Since the density is the same, the heavier object has a larger mass-to-surface-area reatio. That means if you make the object larger, mass (and therefore grativtational pull) increases faster than air resistance does.


So no matter how you think about it, the heavier rider wins.

[Trunk]

There is no dissent. People are just getting lost in the minutiae.

With varying degrees of approximation:


Two balls of different mass, dropped in vacuum (i.e. no air resistance) - this is the classic high school thought expeirment. The two balls fall at the same rate.
Two balls of identical size and different mass, dropped in air (air resistance not ignored): the heavier ball falls faster. Air resistance is the same for both balls, but the heavier ball is pulled downward with a greater force (of gravity).
Two objects of identical size and shape, and different mass, rolling down a hill on identical wheels: The heavier object rolls down faster, for the same reason.
Two objects of identical density and different mass, rolling down a hill on identical wheels: the heavier object will roll faster. Since the density is the same, the heavier object has a larger mass-to-surface-area reatio. That means if you make the object larger, mass (and therefore grativtational pull) increases faster than air resistance does.


So no matter how you think about it, the heavier rider wins.

So, you're telling me that in a VACCUUM, the objects fall at the same rate because there's no air resistance. Fine.

But your list item 2. . .in AIR, two identical balls fall at different rates EVEN THOUGH THE AIR RESISTANCE IS THE SAME because the force of gravity was higher on the second ball?

Isn't that the same gravity that was acting on the balls in the vaccuum?

You're saying air resistance is NOT ignored, but then you're saying the air resistance is the same for both objects. FINE. But the claim you're making (that the force of gravity is stronger on the denser ball) applies just as much to the experiment in a vaccuum.

Yes, there is dissent. You're telling me the heavier rider goes faster downhill. Vunderbob is saying the opposite. That's what I'm calling "dissent".

[Hoodoo Ulove]

So, you're telling me that in a VACCUUM, the objects fall at the same rate because there's no air resistance. Fine.

But your list item 2. . .in AIR, two identical balls fall at different rates EVEN THOUGH THE AIR RESISTANCE IS THE SAME because the force of gravity was higher on the second ball?

Isn't that the same gravity that was acting on the balls in the vaccuum?

You're saying air resistance is NOT ignored, but then you're saying the air resistance is the same for both objects. FINE. But the claim you're making (that the force of gravity is stronger on the denser ball) applies just as much to the experiment in a vaccuum.

Yes, there is dissent. You're telling me the heavier rider goes faster downhill. Vunderbob is saying the opposite. That's what I'm calling "dissent".
We're talking about terminal velocity here. No acceleration. Inertia not a factor. Just drag vs force of gravity.

[scr4]

Imagine two motor vehicles, a souped-up truck and a small Japanese car. The truck obviously has a larger engine, but also more inertia to overcome - in other words, the power-to-weight ratio is the same. So they both do, say, 0-60mph in 6 seconds. That's roughly analogous to two balls falling under vacuum: one is heavier than the other (i.e. more power), but it has more inertia so it cancels out, and they both accelerate at the same rate.

Now put identical drag chutes on both cars. These act as brakes and provide identical braking force to each car. Will the two cars still have the same 0-60 time? Of coures not, the same braking force will have a larger effect on the smaller car. The bigger truck wins hands down.

[Trunk]

We're talking about terminal velocity here. No acceleration. Inertia not a factor. Just drag vs force of gravity.

Who's talking about terminal velocity?

When did we start talking about terminal velocity?

I thought we were talking about a bike rider going down a hill.

Besides, not that you don't know this, but there is no terminal velocity in a vaccuum.

Saying that the air-resistance is the same on two spheres (as SCR4 said) is pretty much the same thing as negating the effect air resistance which is the main point of talking about what a vaccuum does.

If two spheres have the same air-resistance, then how does one get going faster than the other?

[scr4]

I thought we were talking about a bike rider going down a hill.
A bike going down a hill would usually reach terminal velocity quickly, and therefore it's a useful illustration of why a heavier bike rider descends faster. But this approximation is not strictly necessary.

Besides, not that you don't know this, but there is no terminal velocity in a vaccuum.
We're not talking about riders in vacuum.

Saying that the air-resistance is the same on two spheres (as SCR4 said) is pretty much the same thing as negating the effect air resistance which is the main point of talking about what a vaccuum does.
No. Just because two effects are numerically identical does not mean you can ignore both. The identical force has a more significant effect on the smaller object.

If two spheres have the same air-resistance, then how does one get going faster than the other?
Because one is pulled downward more strongly than the other.

[Waverly]

There is no dissent.I think you are confusing this greater force with greater acceleration. Acceleration here is due to gravity and is constant. Therefore velocity can be calculated given distance and time, and is likewise unaffected by mass. Mass is playing a part, but you are miss-applying Newton’s second law.

Let’s look at it: F = m * a, where (a) is going to be the gravitational constant (g).
F = M * g
Pick any mass you like, as large as you like, and you will see F increase and balance the equation, but (g) remains constant. A bigger (F) only equals a bigger (a) if 1) you hold mass constant, and 2) we aren’t talking about gravity, where (g) is constant by definition. 9.8m/s/s

Mass factors in as more of a first law consideration. A [heavy] body in motion tends to stay in motion [more so than a lighter body.]

So yea. I do dissent. I’m am correct, and you are wrong.

[Trunk]

Because one is pulled downward more strongly than the other.

Is not that same sphere pulled downward more strongly in a vaccuum?

[CurtC]

Tuck, for the case of two unequal weight spheres falling through air: the air resistance will be the same, but the force of gravity will be higher for the heavier sphere because it's, well, heavier. Gravity is also higher in the case of the vaccuum, but the difference with air is that the force due to air resistance will be smaller in proportion, for the heavier ball vs. the smaller ball.

For the question of heavier bikes, it's the same thing, and I've seen it many times. On organized distance bike rides, the riders tend to naturally sort themselves out by average speed over the first few miles (the people riding around you are the ones who go about the same speed as you, since you all started about the same time). When I'm near a tandem bike, they go much faster on the downhill slopes because of their weight. Of course, I go faster on the uphills (since our average is the same and they're faster downhill, it follows that I'm faster uphill). I've also seen it simply with heavier riders, but the effect is less noticeable.

[Trunk]

Because one is pulled downward more strongly than the other.
Is it not also pulled downward more strongly than the other in a vaccuum?

CurtC -- if you put two of you on a tandem, you should go up hill faster.

The people weigh twice as much.

The bike weighs less than twice as much.

Your power output should be doubled. Or at least close to doubled, withing the limitations of the tandem drive train.

Going UP hills is TOTALLY about power to weight ratio. And the amount of power that an "average" heavy rider generates doesn't make up for the extra amount of power he generates over the "average" light rider.

Exceptions abound.

[UncaStuart]

Totally anecdotal info to follow:

I ride a road bike and a tandem. Combined weight on the road bike is around 220 pounds; on the tandem, 360.

When on the tandem, we engage in "tandem leap frog," where we are passed by single bikes going up, and we pass them going down.

While on my single bike, I have to spin out to reach the low 40 mph range while descending a smooth, straight 7% hill. However, on the tandem we coast into the high 40s, when the "automatic speed govenor" (my wife) kicks in and we start to brake--and are passed by other tandems that get into the mid 50s.

On long gradual descents, where we can tick along easily at 38mph, we seem to pick up a long train of single bikes who tuck in to draft us after we scoot by them.

As I said, anecdotal. I've always assumed that the greater mass was better able to overcome the air resistance, but I was a psych major, y'know.

[zut]

I think you are confusing this greater force with greater acceleration. Acceleration here is due to gravity and is constant. Therefore velocity can be calculated given distance and time, and is likewise unaffected by mass. Mass is playing a part, but you are miss-applying Newton’s second law.

Let’s look at it: F = m * a, where (a) is going to be the gravitational constant (g).
F = M * g
Pick any mass you like, as large as you like, and you will see F increase and balance the equation, but (g) remains constant. A bigger (F) only equals a bigger (a) if 1) you hold mass constant, and 2) we aren’t talking about gravity, where (g) is constant by definition. 9.8m/s/s

Mass factors in as more of a first law consideration. A [heavy] body in motion tends to stay in motion [more so than a lighter body.]

So yea. I do dissent. I’m am correct, and you are wrong.Actually, no. scr4 is correct. He's talking about forces due to air resistance.

To simplify the problem, think about two blobs that are falling through the air. The blobs are the same size and shape, but different masses -- call the masses m and M. Each blob has two forces on it: the drag force (FD) and the body force due to gravity. If the blobs have the same velocity, the drag force is the same for each. The body force, however, is different. In one case, the body force is (mg), and in the other it's (Mg).

So the total force on each blob is different also: In one case it's (mg-FD), and in the other it's (Mg-FD).

Since acceleration is equal to force divided by mass (F=ma, right?), the acceleration of the two blobs is different. am = (g-FD/m) and aM = (g-FD/M). In the absence of air, the drap force is zero, and the acceleration of each simplifies to just g, as you would expect. But with air resistance, the acceleration is not constant, and it's not the same for the two blobs.

[Waverly]

Actually, no. scr4 is correct. He's talking about forces due to air resistance.Perhaps I did misundertand the post, but even as I read it again, I see someone correctly identifying that force increases with mass, but extending this to mean that this same increase in force has caused the higher velocity. Have a look:scr4: Two objects of identical density and different mass, rolling down a hill on identical wheels: the heavier object will roll faster. Since the density is the same, the heavier object has a larger mass-to-surface-area reatio. That means if you make the object larger, mass (and therefore grativtational pull) increases faster than air resistance does.Air resistance is mentioned, but only but only in terms of it not increasing as quickly as what seems to me the erroneus attribution of accelaration due to mass.

I agree with the remainder of your post, since it is clear your F=ma is in the opposite direction to your F=mg. My apologies to scr4 if I have continue to misundersand his/her meaning.

[zut]

Perhaps I did misundertand the post, but even as I read it again, I see someone correctly identifying that force increases with mass, but extending this to mean that this same increase in force has caused the higher velocity. Have a lookI'm 99% certain that scr4 is saying exactly the same thing that I am. The key is this little bit, from what you quoted: "the heavier object has a larger mass-to-surface-area ratio." His post is cast as a description of the scale effect. So he's not saying that the higher body force causes a higher velocity per se, but the rather that the body force is proportionately larger than the drag force, and thus results in a higher velocity.

Or, to cast it ina different light, remember that I was claim above that the total acceleration on a body is aM = (g-FD/M). Well, FD is proportional to area, so when scr4 talks about the "mass-to-surface-area ratio," he's talking about FD/M. In other words, as you scale a blob up, the mass-to-surface-area ratio changes, which means that FD/M changes, which means that aM changes, which means that the maximum velocity changes.

[treis]

scr4 is indeed correct. I think at this point it is useful to get some forumlas on the table.

I am assuming that the wheels roll without slipping, the bearings in the bike are frictionless and the biker and bikes are identical except for mass.

For you playing along at home draw your best representation of a biker riding down a hill (don't draw the hill though) and a similair one next to it.

Got that?

Ok good now lets start adding forces to the one on the left. First lets add the force due to gravity which is mass*acceleration due to gravity pointed straight down. Next add a an air resistance force parallel to the hill pointing up the hill. There is also a friction force but assuming we have the same bike it is inconsequential so we can ignore it. Same goes for the Normal force from the hill onto the wheels.*

Now draw an = in between the two pictures and draw a force parallel with the hill pointing down the hill.

(waits patiently)

Ok good we have now drawn our free-body diagram with all forces. Let us write the equation for the force parallel to the hill with down positive. From our left picture we have mass*acceleration due to gravity*sin(theta) - Force due to air resistance and from our right we have mass*acceleration of the rider. Putting these together we have mg*sin(theta)-Far=ma where m=mass, g=gravitational acceleration Far=force due to air resistance, theta=angle of the hill and a=acceleration of the body

Now Far is governed by the equation Far= 1/2CpAv^2 where C=a coefficient dependedant on the material of the object i.e. rough objects fall slower than smooth objects, p=density of air, A=crossectional area and v=velocity. Now in our situation here we can ignore p,C and A becuase they are the same.

Our equation now looks like mg*sin(theta)-Zv^2=ma where Z is some constant that equals 1/2CpA. solving for a we get g*sin(theta)-Zv^2/m=a. The greater the mass of the object the smaller the term Zv^2/m is and consequentially the higher the acceleration of the biker. In this case the acceleration for both bikes is equal (g*sin(theta)) but the acceleration due to air resistance depends on the mass.

* We can ignore the normal force becuase it is canceled out by mgcos(theta). We can also ignore friction becuase it is equal for both bikes at the same velocity becuase we assumed the wheel does not slip.

Is not that same sphere pulled downward more strongly in a vaccuum?

Lets go ahead and draw another free body diagram for this situation. Draw two balls and the one on the left draw an arrow straight down and label it mg and on the right ball draw an arrow straight down and label it ma. Writing the equation down we get mg=ma. The masses cancel out so the acceleration of the object is equal to g. However the value of the force is equal to mg so it depents on mass. If you apply the same force to say a couch and a empty cardboard box you of course know that the cardboard box will accelerate much faster than the couch.

Now in a case with air resistance you need to add a Far to the left ball. Using the same reasoning as we did with the bike this force is equal to Zv^2. Writing Newtons 2nd law we get mg-Zv^2=ma. Solving for a we get a=g-Zv^2/m. Again the Zv^2/m term depends on mass so the two objects accelerate at a different rate. As before the acceleration due to gravity is the same for both balls but the acceleration due to air resistance is different.

Hope that helps.

[scr4]

Waverly, you are ignoring air resistance. This is not a valid approximation for this problem, and therefore your answer is incorrect.

Yes, I undersatnd F=ma and F=mg. Acceleration is a=f/m=mg/m=g, which is independent of mass. Which is why everything falls at the same rate - in vacuum!

Now add air resistance, so that net force is F=mg-r(v) where r(v) is the air resistance of the object at speed v. Then acceleration is a=f/m=(mg-r(v))/m=g-r(v)/m. Do you see that if two objects had the same r(v) but different m, they would accelerate at different rates?

Or to put it in English: the same force acting on a more massive object has a less significant effect.

[Waverly]

Waverly, you are ignoring air resistance.No, no I am not. from above: However, gravity is not the only force at work. Friction and air resistance both will impart some force against the forward movement. Here’s where the difference is observed.I can see now we are saying the same thing. Your earlier post just wasn't clear to me.

I can also see that my spellcheck isn't working, so maybe my posts aren't much clearer.

[Quercus]

Trunk, You want authoritay? The authoritay of this Ivy-league physics degree (and not from some afraid-of-numbers-school like Brown, either) says scr4 is right. You want, I'll get my father's PhD in physics to weigh in. So respect!

And the tandem slower-uphill faster-downhill phenomenon is most likely explained by the fact that a tandem is not that much lighter than two single bicycles, but has only a tiny bit more air resistance than one single bicycle.

Therefore going uphill, where it's slow enough that air resistance doesn't matter much, and it's all about pushing mass against gravity, the tandem has little advantage. But going down, where it's mostly about air resistance, the tandem has a huge advantage (for the same reason that the cannonball falls faster than the nerf ball).

[Trunk]

Trunk, You want authoritay? The authoritay of this Ivy-league physics degree (and not from some afraid-of-numbers-school like Brown, either) says scr4 is right. You want, I'll get my father's PhD in physics to weigh in. So respect!

That's exactly what I wanted.

Someone dropping PHAT IVY on my ass and jamming the physics degree up in my grill!

[Hoodoo Ulove]

Argument by authority sucks! And the tandem slower-uphill faster-downhill phenomenon is most likely explained by the fact that a tandem is not that much lighter than two single bicycles, but has only a tiny bit more air resistance than one single bicycle.
If you read CurtC's post a bit more closely, you'll see that he's comparing tandems and singles with the same average speed, so no explanation is necessary. Probably the single riders are just stronger climbers.

[scr4]

No, no I am not [ignoring air resistance].
I'm confused as to what your position is. Are you still claiming that even if you take air resistance into account, two objects with identical size and different mass will fall at the same rate? Or are you retracting your earlier accusation that I am incorrect?

[Waverly]

I'm confused as to what your position is. Are you still claiming that even if you take air resistance into account, two objects with identical size and different mass will fall at the same rate? Or are you retracting your earlier accusation that I am incorrect?Holy shit. I’m beginning to wonder, typos aside, if I’m conversing in English. I said, and I believe a quick perusal of my posts will bear this out, that:

1) Acceleration due to gravity will be the same for both bikers.
2) Decelerative forces such as friction and air resistance will act in the opposite direction.
3) The net result of these forces will have a greater impact on the lighter biker.
4) The heavier biker will move faster once past the point where air resistance overrides friction. It’s a fine point, but if the slope is low enough and the biker heavy enough, it’s possible that this never happens.
5) Acceleration due to gravity will be the same for both bikers. Right or wrong, I thought you were suggesting that this increases because the force of gravity increases.

Sorry. I’m not capable of stating it more clearly.

[Hoodoo Ulove]

That's clear enough. And your point 4 is a good one. On the other hand, your earlier assertion that "traction is higher for the heavier rider" is questionable.

[Princhester]

This topic, or at least some variation on the "do heavier things fall faster" argument comes up about once a fortnight around here. The suggestion that in air things of the same size fall at the same speed regardless of differing weights is one that an amazing number of people will fight to the death to defend. Die they always do eventually, but it's a remarkably persistent meme.

What I find interesting is that it tends to be people who have at least 8/10ths of a clue who are dead wrong about it (see below). They have enough physics to have heard the "all things fall at the same speed in a vacuum" rule and they apply that willy nilly. Another interesting thing is that while it is common for uneducated people to apply common sense and ignore counter-intuitive science, it is so much rarer to see educated people insisting on a wrong answer as a result of misapplication of counter-intuitive science in defiance of common sense.

That's exactly what I wanted.

Someone dropping PHAT IVY on my ass and jamming the physics degree up in my grill!

You can't be too sure. I once had dinner the night before a trial with an expert witness (a Phd in fluid dynamics, fer chrissakes) and one of my hobbies (water rockets) came up and we got onto this whole topic and he insisted that all else being equal, heavier things don't fall faster in air. Eventually I managed to convince him but not before it got down to writing formulae on napkins.

He was not at all happy about having his ass handed to him in something so close to his own field by a lawyer.

[flight]

Physics degree? Pishaw. I am going to throw my PhD in Aerospace Engineering in here to say that you are.. umm... absolutely right Zut. Your equations are also the best mathematical explanation in this thread.

I am also going to expand this to say that exactly the same thing applies to two uniform cylinders of different density (and therefore mass). The fact we are creating both rotational and translation motion comes out in a wash if the cylinders have the same dimensions.

One last consdieration: The heavier rider, though still at an advantage downhill, will be at less of an advantage than this initial glance would suggest. You see, most people become larger with increasing mass, and have correspondingly larger drags. Drag is extraordinarily difficult to determine mathematically, but suffice to say that the big guy's edge will be slightly reduced.

[flight]

Oh, Waverly, I would avoid the concept of adding accelerations as it often tends to confuse people and lead to just these kinds of problems. In fact, I think that was why some of your posts were misunderstood. It is better to just calculate the force due to gravity where g is a conversion factor and only convert back to acceleration when you have a resultant force. My 2 cents from common mistakes I see.

[galt]

Trunk, I think I can help. I see a point you may have overlooked:
Is not that same [heavier] sphere pulled downward more strongly in a vaccuum?
Yes, in a vaccuum, the heavier sphere is pulled downward more strongly than the lighter sphere. However, this doesn't mean it accelerates more. It has a greater resistance to acceleration due to its inertia. The net result is that the heavy sphere and the light sphere fall at the same rate in a vaccuum, even though the heavier sphere is being pulled more strongly.

heavy sphere in vaccuum:
a) large downward gravitational force
b) large inertia opposing acceleration

light sphere in vacuum:
a) small downward gravitational force
b) small inertia opposing acceleration

In both of these, a+b works out to the same acceleration.

In the atmosphere, those same forces exist: the light sphere is pulled gently by gravity, and the heavy sphere is pulled hard. All other things being equal, they would accelerate the same. However, they also both have an equal drag force pushing upward on them:

heavy sphere in air:
a) large downward gravitational force
b) large inertia opposing acceleration
c) drag force (upward)

light sphere in air:
a) small downward gravitational force
b) small inertia opposing acceleration
c) drag force (upward)

We already know that adding up a+b in both cases results in the same acceleration, so now what happens if we add in the additional force c to each? Well, a=f/m, so the same force produces a different acceleration on bodies of different masses. The more massive body is upwardly accelerated less, so it falls faster.

[flight]

Man, I need to preview to catch posts that have showed up while I am typing. Hoodoo Ulove, traction is indeed better for heavier riders. The tractin can be simply represented as the coefficient of friction (unchanged for different riders) times the normal force. This is simply the weight of the rider and bike (with some reductions when the rider is accelerating downhill), so the heavier rider gets better traction. Remember, however, that he has more inertia to resist as well, so though the traction is better, the actual resistance to sliding will be similar. It gets compllex when you take all these little things into account.

[x-ray vision]

No fancy degree here but I can tell you that when I was a kid my heavier matchbox cars went down ramps faster than my lighter ones. :D

[Chronos]

Quoth Padeye: No one goes uphill fast so the amount of weight one has to lift if the biggest factor.During the Tour de France, some of my office-mates and I looked up some of Lance Armstrong's stats. During the uphill mountain parts of the race, he was putting out nearly half a horsepower just in lifting his mass, without regard even for power wasted on friction or drag.

Quoth Princhester:What I find interesting is that it tends to be people who have at least 8/10ths of a clue who are dead wrong about it (see below). They have enough physics to have heard the "all things fall at the same speed in a vacuum" rule and they apply that willy nilly. Another interesting thing is that while it is common for uneducated people to apply common sense and ignore counter-intuitive science, it is so much rarer to see educated people insisting on a wrong answer as a result of misapplication of counter-intuitive science in defiance of common sense.A constant frustration, in teaching. The first lesson of science, which so many fail to grasp, is that you should observe what actually happens, and what happens is the truth. A balloon falls more slowly than a ball of lead the same size and shape, as anyone who observes knows. If equations and theory tell you otherwise, then either you are misinterpreting the theory, or the theory is wrong.

[Hoodoo Ulove]

though the traction is better, the actual resistance to sliding will be similar.
That was what I meant. Actually, with rubber tires on pavement, the coefficient of friction is a bit better with lower unit load. That's why dragsters run big tires.

[scr4]

Holy shit. I’m beginning to wonder, typos aside, if I’m conversing in English. I said, and I believe a quick perusal of my posts will bear this out, that:

1) Acceleration due to gravity will be the same for both bikers.
2) Decelerative forces such as friction and air resistance will act in the opposite direction.
...
I can't agree with your point 1 there, and this may be the source of our confusion: this statement either ignores air resistance, or treats acceleration as a vector that can be added with other accelerations to find the final acceleration. You can add forces as vectors because an object can have multiple forces acting on it at the same time, but you can't do it with acceleration because an object can't have several different accelerations at the same time.

[Waverly]

scr4,
You may want to double check that. An object may have more than one acceleration. In linear motion it's possible to simplify as per flight's suggestion above. In non linear motion, you have to consider them.

[treis]

I can't agree with your point 1 there, and this may be the source of our confusion: this statement either ignores air resistance, or treats acceleration as a vector that can be added with other accelerations to find the final acceleration. You can add forces as vectors because an object can have multiple forces acting on it at the same time, but you can't do it with acceleration because an object can't have several different accelerations at the same time.

Yeah scr4 you are wrong on this. After all an acceleration vector is nothing more than a force vector divided by a scalar (mass).

The tractin can be simply represented as the coefficient of friction (unchanged for different riders) times the normal force.

You can do this if you want to fail your physics exam. The normal force times the coeffecient of friction is the MAXIMUM force due to friction not necessarily the friction force.

[flight]

Of course treis, but I did not say friction force, I said traction. Though I suppose I could be misunderstood to meant that, I was quantifying the ability of the tire to stick to the road without slippage, which is neatly expressed with the max friction force. This would, course, be using the coefficient of static friction as sliding motion is assumed not to have started yet.

[matt]

There was nothing to differ with. Those were questions.


I was differing with Vunderbob's answers, specifically no. 1. A couple of othe posts appeared while I was writing mine - I really should get into the habit of quoting what I'm replying to!


Except you're not standing on the bike with your arms spread out to the sides.


Doesn't matter. Even if your big guy and little guy are in identical aerodynamic crouches, the bigger guy will have a larger weight:surface-area ratio, so he will go faster freewheeling down a hill.

[Trunk]

Trunk, I think I can help. I see a point you may have overlooked:


Actually, I was just trying to clarify what scr4 was saying.

He made it sound like gravity was accelerating a heavier object faster just because it was heavier.

I think he had done an intermediate step in his head that he was assuming some of use had picked up on.

He might have been applying the mass to what treis called Far, and basically, what you listed as 'c', but he was only mentioning what you called 'a' and 'b'.

Or at least that's how I read it.

treis and zut really clarified it with some equations.

Thanks all.

[zut]

Physics degree? Pishaw. I am going to throw my PhD in Aerospace Engineering in here to say that you are.. umm... absolutely right Zut. Your equations are also the best mathematical explanation in this thread. Why, thanks. Would it be too late to point out that I've got a PhD in mechanical engineering? Not in fluids, but still. I also happen to teach a kinematics class, and I've assigned a project to computationally determine the motion of falling objects, taking into account the air resistance.

It's due tonight. :)

[flight]

Oh yeah? Well I... got nothing really to compare to that. Actually I did write a class paper on the aerodynamics of falling water droplets, but that was way back as an undergrad. Man, you think it is hard to calculate drag on a rigid body? Fluid is a bitch!

[Don Roberto]

Concerning the glass ball and cannon ball rolling down a plank:



2) Not sure. Even if we can neglect aerodynamic drag, the situation is a little complicated. As the balls roll down the ramp, they are converting their gravitational potential energy into linear kinetic energy and also rotational kinetic energy. The cannonball will convert more energy per metre of descent because it is heavier, but it also has a greater rotational moment of inertia. Hmm. Going to bug me, that one. (Actually, think I've nailed it, see below.)

3) Similar considerations to (2). But wait! As far as rolling is concerned, there's no difference between a cylinder of fixed dimensions but double the density, and one that is twice as long and the original density. The twice-as-long cylinder has to roll at the same rate as the original cylinder - it's exactly the same as two of the original cylinders rolling side by side. So the double-density cylinder will also roll at the same rate. By extension, if we can leave aerodynamic drag out of it, all uniform cylinders of a particular radius will roll down a ramp at the same rate, regardless of their weight. The same argument applies to the balls as well - the glass ball and the cannonball will roll at the same rates.



This reasoning is correct, as far as it goes, but I'd thought I'd illuminate the situation for the TM with a clarifying bit of math:

A uniform disc is simpler than a ball, so we'll use that; all the conclusions will apply to a sphere as well with some slightly different factors.

Consider a uniform disk at the top of an incline. It has mass M and radius R. At the top, before it moves, it has gravitational potential energy, given by PE = Mgh, where h is the height, and g is the acceleration due to gravity (9.8 m/s^2).

The disk rolls down the plank, and its gravitational potential energy is converted to kinetic energy, both linear and rotational. So at the bottom, it has.

KE = 1/2Mv^2 + 1/2Iw^2

Where I is the moment of inertia ( 1/2MR^2 for a uniform disk) and w is the angular velocity.

For a disk that is spinning with angular velocity w, a point on the edge has tangential (i.e., linear) velocity given by v = wr (you'll find this in any freshman physics book). If we consider our disk to roll without slipping, this tangential velocity of a point on the edge is equal to the linear velocity of the center of mass. Thus our KE term is:

KE = 1/2Mv^2 + (1/2)(1/2)MR^2w^2
KE = 1/2M(Rw)^2 + (1/2)(1/2)MR^2w^2 using v= wr
KE = 3/4 M(v)^2.

Now we began with PE, and wound up with KE. So, in classic physics style,

KE = PE
3/4Mv^2 = Mgh
v^2 = 4/3gh
v = sqrt(4/3gh)

Note that the mass of the disk (ball) does not appear in the expression for the velocity.

This treatment, obviously, ignores effects due to air resistance, which, for reasons explained above, will cause the heavier ball to roll faster.

[matt]

Don, you are a scholar!

I on the other hand am a mathematical lightweight and a lazy, lazy man...

[Hoodoo Ulove]

Does a cannonball and a glass sphere of the same size fall at the same rate in air?

Do they roll down a plank at the same rate?
If this is a hollow glass sphere, the concentration of the mass at the surface gives it a higher polar moment of inertia compared to its mass, and thus it will roll more slowly down the plank. This is neglecting aerodynamic drag, which of course would slow it further compared to the cannonball.

[AskNott]

I disagree with Quercus, Treis, and perhaps a few others on the matter of rolling resistance. Treis assumes frictionless bearings and ignores the deformation of the tires. From there on the comparison is interesting but theoretical, because real bearings have friction, and real tires behave differently under heavier load.

Quercus seems to say that the lighter rider would have an advantage in rolling resistance. That cannot be true. Bearings first. Consider that bicycle wheel bearing are quite efficient for their tiny size, but even if they were much bigger, a bearing's friction increases under load. Consider, too, that the heavier rider nears the safe capacity of a bicycle wheel bearing. Surely, when the load approaches the bearing's capacity, the friction (not to mention the possibility of failure) increases.

I'm assuming the lighter rider weighs 160 pounds, and the heavier rider, 260.

Now, tires. We cannot assume that a wheel, with tire, is a disc. The wheel is pretty stable, but the tire deforms at the contact patch (the "footprint" actually in contact with the pavement.) If you roll the empty bike along the street, the contact patch is about 1/4 inch wide. With a rider aboard, it's radically different. At the forward edge of that patch, the sidewalls squish down, and at the trailing edge, they rebound to original shape. That flexing takes energy, and you subtract that from forward-motion energy. (Most of it is expended as heat.)

Your 160 lb. rider has a contact patch that looks, at rest, like a stretched-out football (US). The 260 lb. rider has a longer, fatter contact patch, and the deformation necessary to make it takes more energy than for the lighter rider. The distance between wheel and ground gets perilously close to zero, and the sidewalls get hot from this abuse.

In the overall scheme of things, rolling resistance might not amount to much, but it is different for the two riders, and more for the heavier rider.

[Hoodoo Ulove]

In the overall scheme of things, rolling resistance might not amount to much, but it is different for the two riders, and more for the heavier rider.
But is it proportionately higher? If not, then the benefit is to the big guy.

[zut]

Agreed that the inclusion of rolling resistance makes this a more complicated problem. My gut feel, although I'm willing to be wrong, is that the rolling resistance would scale with (rider + cycle) weight, close enough.

However, from what I'm able to find, aerodynamic drag at reasonable speeds is much greater than rolling resistance. For example, (http://wings.avkids.com/Book/Sports/advanced/bicycle-01.html) "In fact, at 8 mph (3.5 meters/second) the aerodynamic drag of a bicycle and rider is greater than the rolling resistance (wheels on the ground). At 20 mph (11 m/s), the aerodynamic drag is more than 80% of the total drag."

So if you were to, say, graph the speed of two bicycles coasting down a hill from a dead stop, you'd expect the lighter one to pull ahead first (due to rolling resistance) and the heavier one to catch up later (due to aerodynamic drag). Makes for a complicated problem, and depends a lot on your assumptions.

[Quercus]

OK asknott, on to rolling resistance. You understand that, as zut says, rolling resistance is small potatoes compared to wind resistance, so this is an academic debate at best.

But I submit the larger rider has an advantage, going downhill. Reasoning is that some portion of the rolling resistance is proportional to weight (call this proportional rolling resistance, or p.r.r.). Since gravity's force is also proportional to weight, then the p.r.r. affects the speed of both riders equally (The rider that's twice as heavy has twice the p.r.r. force of drag, but has twice the gravitational force pulling him down, too. So the p.r.r. essentially reduces the gravitational force by the same percentage for each rider).
But there is probably some portion of the rolling resistance that is constant and/or depends on speed. For instance, if the rider is pedalling, there's friction from the chain and gears, which won't depend on weight). And this non-weight dependent part of the rolling resistance will slow the heavier rider less.

Now, it's likely (I'll even say probably) true that the biggest part of rolling resistance is in fact proportional to weight, so I'm not saying the effect is large or noticeable. But then again, as we said this is all academic anyway.

[treis]

Of course treis, but I did not say friction force, I said traction. Though I suppose I could be misunderstood to meant that, I was quantifying the ability of the tire to stick to the road without slippage, which is neatly expressed with the max friction force. This would, course, be using the coefficient of static friction as sliding motion is assumed not to have started yet.

Ah I see. I agree with you that the friction force for the heavier biker would be greater. However that does not translate to any practical difference becuase the two bikers will have the same max acceleration correct?

AskNott

I did not include the rolling friction becuase its extremely hard if not impossible to calculate without experimenting.

[Princhester]

IConsider, too, that the heavier rider nears the safe capacity of a bicycle wheel bearing. Surely, when the load approaches the bearing's capacity, the friction (not to mention the possibility of failure) increases.

Ain't no way. A heavy guy on a bike rolling down a hill doesn't come close to the bearing's capacity.

If you have any doubts, consider what the bearing has to cope with under shock loads (hitting a bump, say). Those shock loads woud be orders of magnitude higher than the load the bearing is under on a smooth surface.

Bearings are only very rarely damaged. Even with rough handling.

They are nowhere near their capacity under normal rolling conditions.

The distance between wheel and ground gets perilously close to zero, and the sidewalls get hot from this abuse.

You have a cite for that? I know plenty of guys who are up round the 260 mark who don't have any tire problems, or get noticeably hot tires, or have problems with damaged rims (which they'd have to from bumps if they were riding perilously close to zero margin between rim and ground).

Check out this calculator (http://ida.physik.uni-siegen.de/menn/motion.htm)

Everything else being equal, it shows that going from a 160 to a 260 pound rider will, for the same power output, at say 45-50km/hr result in about 1 km/h speed difference.

45-50 km/h would be a common downhill speed for a modest hill.

Rolling resistance is not worth worrying about.

[CurtC]

So if you were to, say, graph the speed of two bicycles coasting down a hill from a dead stop, you'd expect the lighter one to pull ahead first (due to rolling resistance) and the heavier one to catch up later (due to aerodynamic drag).That doesn't follow. I would expect a heavier bike to be faster, even without a rider, even starting from a dead stop. The rolling resistance will certainly be greater for a heavier bike, but the question is whether it's a greater proportion of the forces on the bike.

What it gets down to is this. Everyone agrees that rolling resistance is greater if the weight of the bike/rider is greater - the rolling resistance is monotonically increasing with weight. But is it a linear increase? For example, with a 2x heavier bike+rider, will the rolling resistance be 3x? Or 2x? Or 1.5x? All this argument that it is monotonically increasing doesn't address the key question of whether it's linear. I suspect that it's slightly less than 2x, giving yet another advantage to the heavier rider in downhill speed.

But in any case, the variation from linearity for rolling resistance is dwarfed by the variation from linearity in wind resistance. A 2x heavier bike+rider will not have anywhere near 2x increased wind resistance, and this is the key that makes the heavier rider go faster downhill.

[Princhester]

The equations are set out in that cite I gave in my last post. Rolling resistance is linear.

To give perspective, I weigh about 210 pounds including bike. According to the calculator given in that cite, by the time I'm doing 11 mph air resistance equals rolling resistance. A quick experiment on the modestly sloping street outside my house showed that I reached that after about 5 seconds and about 15 yards.

[scr4]

The equations are set out in that cite I gave in my last post. Rolling resistance is linear.
No, your cite approximates rolling resistance as a linear function. It doesn't show how accurate this approximation is. I think CurtC is right, it increases slightly slower than linearly. If it were truly linear, then adding extra wheels to a vehicle (while keeping the total vehicle weight constant) would not increase rolling resistance. This seems unlikely to me. (And in my experience, a trike is always slower than a bike with similar weight and similar rider position.)

[Princhester]

Good point.

[S137]

I was puzzled by this same question. I frequently ride bikes with my sons, who are about 50 lb while I am 180 lb. I coast MUCH faster than they do.

They have less weight AND less cross section. More weight should increase friction (rolling and otherwise), so that doesn't help at all.

However, according to Newton:

sum(F) = ma = m*(rate of change of speed, dv/dt)

m * dv/dt = mg (if no air resistance and dv/dt = g, so mass doesn't matter).

HOWEVER, if we model air resistance using the standard constant time velocity squared:

m *dv/dt = mg - K*v^2


If we assume the constant K is the same for both riders (both have the same bike, same cross section, etc etc) then:

dv/dt = g - (K/m) * v^2

We see that the term "K" gets divided by mass, so the effect on the heavy rider is that air resistance is significantly less. He is experiencing the same force, but not the same deceleration.

Good. I can enjoy the rest of my afternoon now!

[fumster]

I am surprised and dismayed that in the 21st century people don't understand such a simple question. Do people really not know that in the absence of a vacuum a Styrofoam ball will fall slower than a ball bearing of the same size? WTF do they teach in school?

I understand that the more subtle questions of rolling friction and the relative cross section of heavier people vs lighter is not obvious.

[Anachronism]

We're assuming here that the light and the heavy guy are both tucked in good aero positions. Actually, if you are insanely brave, you could go faster by putting your legs straight back, laying on your stomach on the seat. Cornering would be a problem, though.

:eek:

http://www.babysitterofthedamned.com/2005/05/old-land-speed-record-pic.html

[Anachronism]

What I find interesting is that it tends to be people who have at least 8/10ths of a clue who are dead wrong about it (see below). They have enough physics to have heard the "all things fall at the same speed in a vacuum" rule and they apply that willy nilly. Another interesting thing is that while it is common for uneducated people to apply common sense and ignore counter-intuitive science, it is so much rarer to see educated people insisting on a wrong answer as a result of misapplication of counter-intuitive science in defiance of common sense.


I couldn't find it but in the 'Jumping with a balloon tied to you' thread, people where still arguing that they where right mathematically even after videos proving they where wrong where posted :rolleyes:

[Reply]

I am surprised and dismayed that in the 21st century people don't understand such a simple question. Do people really not know that in the absence of a vacuum a Styrofoam ball will fall slower than a ball bearing of the same size? WTF do they teach in school?

I understand that the more subtle questions of rolling friction and the relative cross section of heavier people vs lighter is not obvious.

I didn't know that before this thread. It took me a few readings (actually, until S137's final sentence) to finally grasp it.

So much for that fancy college education...

[CookingWithGas]

I was puzzled by this same question. I frequently ride bikes with my sons, who are about 50 lb while I am 180 lb. I coast MUCH faster than they do. Do you all have the same kind of bikes? I used to ride with my girlfriend. I had a 10-speed road bike and she had a hybrid bike and I would overtake her on flat stretches while she was pedaling and I was just coasting. Weight didn't have much to do with it, it was mostly the tires.

Also, zombies coast downhill much faster.

[ZenBeam]

I couldn't find it but in the 'Jumping with a balloon tied to you' thread, people where still arguing that they where right mathematically even after videos proving they where wrong where posted :rolleyes:Can enough helium filled balloons make you leap to great heights? (http://boards.straightdope.com/sdmb/showthread.php?t=201132&highlight=jumping+balloon)

I missed it at the time, but it was an interesting read, since it wasn't obvious what the correct answer would be:

Rosebud.



(go read the thread) :p

[Chronos]

The videos that were posted in that thread were of a subtly different situation than what the OP was asking about.