# How does weight influence your speed when descending?

I have recently bought a road bike and went on a little trip with a friend who is also a novice.

We have approximately the same height but he weighs a lot more (I weigh 67-68kg for 1m81 and he weighs around 80-85kg).

While descending a road, he outclassed me easily. It made me wonder :

Suppose that two people have exactly the same characteristics (same bike, same height, same equipment,...) but a different weight and a corresponding different shape (one is fit and the other overweight or more muscular). If they both ride perfectly (i.e. in an optimal manner), who is going to go faster ?

If the road and tires were perfectly smooth and there was no air, physics tells us that these two people would go at the exact same speed.

Theoretically, the heavier person has a less aerodynamic shape if his/her additional weight is the result of fat and not muscle, so if the road and tires are still perfectly smooth and if there is air, the lighter person should be faster (assuming that the "aerodynamic theory" is right).

Now, add the fact that the road and tires are not perfectly smooth and that I have probably forgotten important factors, how to know which one will be faster ?

I could have asked this question in the Physics community but I bet it is something known in the bicycles one.

• See also [bicycles.stackexchange.com/questions/10531/… do I descend faster on the straightaway?) and the search weight descent speed – Móż Sep 5 '16 at 4:51
• Thanks @Móż . My major question was about the weight and corresponding shape influences and this is addressed in the first link you've given, although the argument based on gravity is not really developed. – MoebiusCorzer Sep 5 '16 at 5:00
• Weight rises linearly with volume, which rises as the cube of linear dimensions, while the frontal area (the main factor in aerodynamic drag) rises as the square. Hence the heavier person accelerates faster and had a higher terminal velocity. – andy256 Sep 5 '16 at 5:57
• The fastest rider downhill is one who has never yet washed out on a corner. Personally I brake way too early now, compared to what I used to do. My former PRs are 5-10% less than what I am ballsy enough to try. – Criggie Sep 5 '16 at 10:05

The heavier person will present more area to the wind, but this is mitigated by two factors: The bicycle presents a fixed area to the wind and the area presented by the heavier person is not proportional because of the 2/3 power law. If you just scale up a rider by a factor in mass, the volume increases in proportion, but the frontal area scales up as the 2/3 power of the weight ratio because the dimension along the direction of travel does not contribute. Both of these mean a heavy rider on a bike with a constant grade will descend faster with no power input besides the hill.

• Yep. Typed my comment but didn't press send ... – andy256 Sep 5 '16 at 6:15
• What do you mean by "2/3 power of the weight ratio" (it's the term "power" I don't get. Is it (2/3)^(weight ratio) ?). Your answer seems not to address gravity forces while other answers do. Why is it so ? – MoebiusCorzer Sep 5 '16 at 11:10
• @MoebiusCorzer No, (weight ratio)^(2/3). The effect of gravity is implicit in this answer but it could be made more explicit. – David Richerby Sep 5 '16 at 12:57
• The 2/3 power law is a widely recognised scaling effect. Whenever a solid form is scaled up, areas increase in proportion to the length to the power two, while volumes increase in proportion to the length to the power three, and so the ratio between area and volume increases by length to the power 2/3 – bdsl Sep 5 '16 at 14:52
• @MoebiusCorzer It's also known as the square-cube law: volume and mass go up as the cube of the scaling factor, but area goes up by the square. Normally that's used for surface area but it also works for frontal area in this case. In other words, if you double the size (height) of the rider their mass goes up by 2^3=8x, but their frontal area only goes up by 2^2=4x, and they have twice as much weight for each unit of frontal area and they will roll down the hill faster. – Móż Sep 5 '16 at 22:46

If it is harder to get up the hill it has to be easier to get down.

Assume you are two rocks of same shape and density dropped from mile up. What the is the relative terminal velocity?

Two forces at work that are equal at terminal velocity

• gravity = c1 * r^3

• wind resistance = c2 * r^2

gravity / wind resistance = c3 * r

velocity1 / velocity2 = r1 / r2

If one weighs twice as much

r1^3 / r2^3 = 2

r1 / r2 = 2^1/3 = 1.26 = velocity1 / velocity2

OK you are not a rock and you are on a bicycle. Same forces at work.

Going up you pay full price for weight and coming down you only get paid pack the cube root.

• That first line really says it all - well put. – Criggie Apr 27 '17 at 0:53
• First time the maths has made sense to me – Kilisi Apr 28 '17 at 10:52

If you drop a styrofoam ball and the same size rock ball in a vacuum they will fall exactly the same. It's because they accelerate with the same gravitational acceleration.

While falling both transform their potential energies into kinetic energies, so:

Mass x Grav_accel x Height = 1/2 x Mass x Velocity^2

We can see it does not matter how much weight the object has, because the Mass is on both sides of equation. The Velocity is only proportional to Height so both objects fall the same.

Now if you drop them in air environment - both objects will have to overcome air drag.

The air drag is not dependent on the Mass of object but only on it's shape, velocity, and the environment. If both objects would fall the same, they would both need the same energy to overcome the air drag. This energy is taken from the kinetic energy of the object to push the air molecules out of the way.

But because the heavier object has bigger potential energy from the start (and bigger kinetic energy in the end) the air drag takes relatively smaller part away from the kinetic energy.

Mass x Grav_accel x Height = 1/2 x Mass x Velocity^2 + 1/2 x Velocity^2 x Some_constant

This is why the heavier object falls faster in drag environment.

Now if the objects have same density and one is bigger heavier and the other is smaller and lighter:

Air drag depends on the drag_coefficient which largely depends on the Cross section. Mass (when the density is constant) depends on the Volume.

Volume of sphere is: 4/3 x π x r^3, Cross section of sphere is π x r^2

This means the Mass increases 1.33 x radius times faster than Cross section for bigger objects, giving them falling advantage.

Thats why dust of the same material falls very very slowly and chunks of the same material fall fast.

• Your energy explanation doesn't work. Being in a vacuum or not doesn't change the potential energy. Yet, in the vacuum, the rock and styrofoam reach the same speed, whereas in air, the rock is faster. So it can't be about potential energy. – David Richerby Sep 6 '16 at 7:52
• @DavidRicherby In vacuum I said nothing about potential energy. Vacuum does not change potential energy and I don't know where you concluded I think it does. I said the heavier object does overcome air drag better because of it's potential energy which is totally true. I can show the physics equations if you like. I'll edit the answer to make it more clear and better, because you are not the only one that didn't understood it. – Jerryno Sep 6 '16 at 8:14
• I didn't conclude that you think it does. I pointed out that your argument about why the rock falls faster in air uses no properties of being in air, so it also argues that the rock falls faster in a vacuum. An argument that reaches false conclusions must be incorrect. – David Richerby Sep 6 '16 at 8:29
• @DavidRicherby well I stated in air is air drag - that was the property of being in air. And that there is none in vacuum - so the cases are not the same. I made the answer more rigorous with better reasoning. – Jerryno Sep 6 '16 at 8:49
• I find the above explanation technically accurate and reasonably complete and comprehensive. What I can't comprehend is the down votes. – Daniel R Hicks Sep 6 '16 at 11:45

If the heavy person and the light person were identical in all ways except for their weight (e.g., – warning, thought experiment only; do not do this – you, vs you after drinking a litre of mercury), then the heavy person will be faster downhill in a straight line.

The reason for this is that there is a greater gravitational force pulling them down the hill, whereas by far the most significant resistive force is air resistance, which depends on speed and shape (which we assumed to be identical) but not mass. This means that, when freewheeling down a hill, the heavy cyclist will be able to travel faster before air resistance balances out the gravitational force. The same is true when you add the force of pedalling to the equation, since we're assuming that both cyclists can put out exactly the same power.

However, this picture isn't entirely realistic as I've made a huge bunch of simplifying assumptions. In reality, the heavy cyclist will be larger, so will have more air resistance. I'm not sure what the trade-off would be, there. I've also assumed that the heavier cyclist will have the same rolling resistance as the lighter one. That's not going to be true but air resistance is much more significant so this shouldn't make a big difference. Also, I've only looked at straight-line speed. In real cycling, you have to turn corners, which usually requires slowing down. A heavier cyclist will need to brake earlier because, for a given speed, they have more kinetic energy to bleed off into their brakes. I'm not sure how much of the gain that would cancel out.

• I used to have a flask of mercury ... it is very heavy :-) – andy256 Sep 5 '16 at 9:12
• @andy256 Yeah, a litre of mercury is 13.5kg. It's a really surprising substance: you just don't expect a liquid to be dense enough that lead can float in it... – David Richerby Sep 5 '16 at 9:17
• "I'm not sure what the trade-off would be" -- for what it's worth, elite cyclists show a bit more variation in body shape than many "pure performance" sports. So it really isn't entirely clear how the power-aerodynamics trade-off should work, there appears to be more than one correct answer. – Steve Jessop Sep 5 '16 at 18:33

Assuming you both have the same shape (but he has more density, so he weights more):

If there were no air, both of you would drive at the same speed, due to the gravity acceleration (the same for both).

If there were usual atmosphere, you both would be accelerated downwards due to gravity (same acceleration), and your aerodynamic drag force would be the same (you have the same shape, and - in the beginning, at the comparison moment - at the same speed). As force accelerates you proportional to mass, drag would decelerate less than your friend, so he would reach greater speed.

• This misunderstands even the simplified physics you're assuming, and isn't relevant to the question at all sine both riders require atmosphere to survive. Rolling resistance is affected by weight, the heavier (or more dense, in your weird setup) rider will have higher rolling resistance and thus be slower than the lighter one ... in a vacuum. So insofar as it's relevant, your answer is also wrong. – Móż Sep 5 '16 at 21:44
• @Móż - On any sort of "real" downhill, with a decent bike and tires, rolling resistance is trivial. And rolling resistance will not increase in proportion to weight, until the tire is severely deformed. – Daniel R Hicks Sep 6 '16 at 11:49
• @DanielRHicks you might consider the M term in the RR calculations trivial, I couldn't possibly comment. – Móż Sep 6 '16 at 20:44
• @DanielRHicks see for example this long answer by R. Chung, who does appear to know a bit about such things. He thinks mass affects rolling resistance... and that it's important. Try telling the Battle Mountain kids that it's not, even with the slope (which, BTW, R.Chung thinks is unimportant and I disagree). Also, FWIW I have not upvoted the accepted answer because I think it's also useless, but not so useless as to deserve a downvote. – Móż Sep 7 '16 at 1:57