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I have had this question for a while. I want to know how I can get down a hill as fast as I can. So, I was wondering if increasing the weight of a bike increases the speed on downhill stretches, or will it just slow me down because of rolling resistance, drag or other factors?

I am talking long stretches here, and long, 10-30% hills. I have also checked the road surface, and there are no bends, and I can easily see if traffic comes along. It is very rarely used for traffic anyway.


EDIT: I would also like to factor in things like rolling resistance in this question.

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Consult Galileo. The answer is no. –  Carey Gregory Apr 25 '14 at 15:51
@CareyGregory, excluding drag of course: F=ma=mg-Fd, where Fd is the drag force which doesn't scale with mass. Try dropping a balloon and a (soccer) football of the same size. Weight will have a small effect on rolling resistance and (assuming it's streamlined) none on drag so it will make you accelerate faster but not by much - better to consider the coefficient of drag. –  Chris H Apr 25 '14 at 16:18
My stats show that 4 years ago, when I was 25kg heavier than I am now, I used to descend faster than I do now. They also show that I was a slower overall cyclist then than I am now. So I suspect with a heavier bike the same will apply (but to a lesser effect as the difference between a light and a heavy bike would be less). –  PeteH Apr 25 '14 at 16:24
Increased weight, all things being equal, will increase downhill speed. However, the increase will not be all that significant for relatively shallow slopes. (Galileo was considering situations where air resistance wasn't a major factor. But with bikes it is THE major factor.) –  Daniel R Hicks Apr 25 '14 at 20:25
(If tires are sufficiently inflated, increased weight will only negligibly affect rolling resistance.) –  Daniel R Hicks Apr 25 '14 at 20:26

8 Answers 8

up vote 10 down vote accepted

The main thing you have to consider at speed is drag: The force F on you+bike (mass m) is:

F = ma = mg sin Q - F_d - F_rr

where a is your acceleration, g is the acceleration due to gravity and Q is the hill angle to the horizontal. F_d is the drag force which doesn't scale with mass. Try dropping a balloon and a (soccer) football of the same size and you'll see this in action. F_rr is the rolling resistance.

Drag dominates rolling resistance at any decent speed for a well-set-up bike: More detail than you probably want or a nice graph so you should be able to neglect rolling resistance.

Weight will have a small effect on rolling resistance and (assuming it doesn't affect the cross-section) none on drag so it will make you accelerate faster but not by much.

You might want to experiment: start with some measured speeds with no extra load, then test the extra mass, and a dummy load of the same volume/shape/mounting made of polystyrene or cardboard.

The biggest effect you'll have is on the drag - so the first thing to try is a better tuck, not wearing flappy clothes, etc.

For really high speeds a fairing is the way to go - compare the hour record for a fully faired recumbant (~92km) with a UCI racing bike (~52km). There are no stats I can find specifically for an upright with a fairing.

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+1 Tuck, not wearing flappy clothes! –  andy256 Apr 25 '14 at 23:37
One other thing occurred to me - if you're measuring average speed (i.e. time between top and bottom) you might gain more by accelerating faster at the beginning (less weight) than by increasing your terminal velocity. Oh, and pump up the tyres as hard as they'll permit. –  Chris H Apr 28 '14 at 10:53

If there were friction it'd be the exact same (remember the hammer and feather on the moon deal). However - your net drag isn't really going to be proportional to your mass. The air resistance isn't proportional at all to your mass (although the rolling drag is). Due to this, you'll have less drag per kilogram with more weight which will cause you to go faster, yes.

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The short answer is yes. See Chris's post for the long answer.

The main reason for this "answer" is to encourage great caution.

As a teenager (in the previous millennium), I grew up in hilly area. There were two descents we used to do regularly - a two mile run from my friends farm to home, and a steep hill of one mile. My best time to get home was 2:11, with lots of pedaling (I wont mention passing the cars). On the steep hill we would crack 1 minute with no pedaling. The point is we were young and stupid.

Recently, my brother met some similar aged kids, who were keen on descending a serious hill we know, as fast as possible. Although he advised against it, they did it anyway. One crashed and smashed his skull. From our own experience on that hill, he would have been doing about 60kph. Only.

You mention 30%. That's steep:

The sign says "3 drains"

The steep part is where the road drops out of sight.

The final cautionary point is bike stability and brakes. Bike brakes are not designed for those speeds. And your frame geometry is an unknown factor - will it "get the wobbles"? The only way of you knowing is to do it. Problem is, if it does you'll most likely crash. Hard.

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Woah. Did he survive –  George Apr 26 '14 at 10:10
Yes. I haven't heard if there's any permanent damage yet. Many of us like to go fast. Just be real careful. Wear a helmet and build up speeds slowly to get experience. Consider knee and elbow pads. Keep your tires hard. That's another easy way die. –  andy256 Apr 26 '14 at 14:47
60kph is only 37mph, and not particularly fast on a good road with a good bike. On a road (not off-road) the main danger from speed is "shudder" due to the geometry of the bike and the weight distribution, often combined with the roughness of the road. A cyclist needs to understand his own limitations in such a situation. –  Daniel R Hicks Apr 27 '14 at 14:33
(And you will not likely crash if your bike "gets the wobbles" so long as you don't panic and gradually slow the bike while firmly bracing the bars. But it scares the &excrement out of you enough that you're not likely to want to do it again.) –  Daniel R Hicks Apr 27 '14 at 15:57

This may be half-remembered physics coming into play here, but I've raced (substantially heavier) friends down hills where they've been on bikes which should - be all accounts - be slower than my own, yet they've won. Could momentum come into play over rougher terrain? By that, I mean if a rider carrying a heavier weight were to hit a bump, or hole etc, they would experience less deceleration due to their increased momentum compared to a rider with a lighter weight.

In a perfect vacuum, of course, acceleration remains at 9.81 m/s/s and, as such, mass would have no impact on velocity.

P.S. I tried to post this in a comment, but I don't have enough reputation yet.

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In a practical sense, increased weight will make you go downhill faster, for the same reason decreasing weight will make you go uphill faster

See this interactive calculator. If you set the gradient to "-10", set "Power P (watts)" to 0.001 (i.e almost zero):

  • With rider weight set to 50kg you will go about 59.54km/h
  • With rider weight set to 75kg you will go about 71.22km/h
  • With rider weight set to 100kg you will go about 81.25km/h

However that is a fairly massive increase in weight. Increasing your body weight that much would be "unhealthy", and increasing the bike weight would very likely negatively influence handling..

Smaller increases in weight (a few kilograms etc) will increase your speed downhill somewhat.. but it would probably be far more productive to work on reducing aerodynamic drag.

An extreme example would be a recumbent bicycle (much smaller frontal area than a typical road bike, and can have an outer shell to further reduce drag), however there are plenty of other ways to do this, like adopting rather unsafe positions on the bike, or wearing funny shaped helmets and more aerodynamic clothing (triathlon/timetrial skinsuits)

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+1 for the useful calculator and vid –  George Apr 28 '14 at 2:42

Yes in some sense. It's like some skiers try to get more weight in order to get extra velocity. But it would hard to have a good control compare to when you were lighter before :)

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The accepted answer states drag does not scale with mass. Which is true. But drag does scale with frontal area. It is fair to assume the two frames are made of the same material therefore the larger frame has a larger frontal area.

Rather than compare balloon and soccer ball of the same size a more appropriate comparison is two rocks of the same density but different sizes.

Terminal velocity is when the force of gravity equals the drag
See this link for the calculation of terminal velocity
Terminal velocity

In the equation once you take out the constants
vterminal is proportional to square root (mass / area)
vterminal is proportional to square root (r cubed / r squared)
vterminal is proportional to square root (r)

So at constant density if you double r the terminal velocity increases by 1.414
Double r is eight times the mass for only 1.414 the terminal velocity.

Drag proportional to v squared is the real drag (pun intended)

Now lets pretend you could double your mass and keep the same area
vterminal is proportional to square root (mass / area)
vterminal is proportional to square root (mass / constant)
vterminal is proportional to square root (mass)
If everything was constant (including your area and rolling resistance)
If you increased mass by 2 you would increase terminal velocity by 1.414
If you increased mass by 4 you increase terminal velocity by 2
Rolling resistance is not constant so it would be less than that 1.414 and 2

Let say 180 lb rider and add 20 lbs of lead to the frame - that is only 5% straight down hill
On a 10 grade that is only 0.846% - 40 mph versus 40.43 mph (without accounting for rolling resistance).

Even climbing bikes are designed to be light.
Basically up the hill you pay for all the weight and down the hill you only get credit for the square root of the weight.
Drag proportional to v squared is the real drag

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As usual, I think the best way to consider this is through energy;

in moving from rest at the top of a hill (height h), the conservation of energy applies between potential energy at the top and kinetic at the bottom:

Mgh = MV^2 + losses(due to aero and rolling resistance)


V = sqrt( gh - losses/M )

as losses are not proportional to mass, then factoring them down by mass reduces their influence on speed for the heavier rider, regardless of whether terminal velocity has been reached or not

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