# Descending without braking: where's the sweet spot?

This may get too math-heavy, but it seems that only a few variables are needed to estimate the optimum slope to descend without gaining/losing speed.

I looked around, but wasn't able to find anything directly relevant- just discussions about drag at high speed.

You are going at constant speed when the driving force from gravity Fg is equal to the drag (air resistance) Fd plus friction (rolling resistance) Ff:

Fg = Fd + Ff

When coasting down a hill, the driving force is the component of the gravitational force parallel to the road:

Fg = m g sin(a)

Here, a is the slope (angle with the horizontal), m is the mass of bike and rider, and g is the gravitational constant. Slopes on roads are often measured in percent, and for small angles sin(a) is equal to the slope in percent. e.g. for a slope of 10% (100m descent on 1km of road) sin(a) = 0.1.

Drag is proportional to the square of the velocity v, the projected frontal area A and a drag coefficient cd, and the density of air rho:

Fd = 1/2 rho v^2 cd A

The drag coefficient cd depends on the shape of the object - streamlined bodies have a low cd, but for a bicycle with rider I find approximately cd = 1 in Wikipedia. I'd estimate the frontal area of a cyclist to about 0.4m^2.

Friction (rolling resistance) is proportional to a coefficient cr and the normal force (the weight component perpendicular to the road) Fn = m g cos(a):

Ff = cr m g cos(a)

The coefficient of rolling resistance cr will depend on the bike (lubrication, tire pressure, surface roughness etc.). For car tires on concrete, Wikipedia gives cr=0.01, so the Ff would be about 10N. That's about a tenth of the driving force on a 10% slope (see above), but on a rough road surface it might well be much more.

So we end up with

m g sin(a) = Fg = Fd + Ff = 1/2 rho v^2 cd A + cr m g cos(a)

Solving this for v gives:

v^2 = 2 m g (sin(a) - cr cos(a) ) / (rho cd A)

Putting in g=10m/s^2, rho=1.2kg/m^3, cd=1, A=0.4m^2, and m=100kg (nice round number for rider and bike), I get

v = sqrt( sin(a)-cr cos(a) ) * 64 m/s

The rolling resistance cr essentially is like reducing the slope somewhat, so a slope of 10%=0.1 with cr=0.01 is the same as a slope of 9% on a perfectly rolling bike. So, for a slope of 10%, the sqrt becomes 0.3, and I get a speed of 19m/s = 70km/h

It's not unrealistic, but perhaps a bit faster than I would have expected. However, there are simplified assumptions, friction and rolling resistance is probably higher and cd and A have a big effect, they all depend on the bike and the rider.

From the drag equation you can see why the racing position is with the head down. This position decreases the frontal area A considerable, compared to an upright position, and it makes the body somewhat more streamlined (reduces cd). As the drag goes with the square of the velocity, this gets more important at higher speed.

• 70km/h seems high but math seems correct. I would not have thought of the cos on the rolling resistance. I would not use "but" on frontal area. Make it two statements and start with area. – paparazzo Jul 19 '15 at 19:07
• @Blam: Thanks. There are many uncertainties in the parameters cd, A and cr, so I don't think we can get precise numbers. For example, Wikipedia gives cd=1 for a road bike, but 1.1-1.3 for a ski jumper (which I'd thought more streamlined), so this number is quite uncertain. If cd=1.3 instead of 1.0, then the numbers have to be divided by sqrt(1.3)=1.14, i.e. 61km/h for 10% slope. – Stephan Matthiesen Jul 20 '15 at 6:50
• @Blam: I don't understand what you mean by 'I would not use "but" on frontal area.' Can you explain that please? – Stephan Matthiesen Jul 20 '15 at 6:58
• But makes area seem secondary and a product of the CD. Tuck reduces frontal area considerably and improves the CD. – paparazzo Jul 20 '15 at 10:23
• @Blam Oh, I see what you mean now, I've changed the sentence. Area is probably the bigger factor. Thanks. – Stephan Matthiesen Jul 20 '15 at 17:12

Drag increases with speed. Drag is both rolling resistance and wind resistance. A steeper slope is more gravitational pull. Terminal velocity is when the gravitational pull equals the drag. On a mild slope it will be only a few miles an hour. On a very steep slope it might be over 50 mph. A 7% slope is around 20 mph on road bike on a road.

• ...depending on the road surface, the tyres (knobbly/slick), the air pressure in said tyres, the direction of the wind and your posture on the bike. Just to name a few more factors. – Carel Jul 19 '15 at 8:13
• @Carel Yes there are many factors. But with any set of factors drag increases with speed. There will be a speed (terminal velocity) where drag equals gravitational pull. I did not mean to imply that any bike on a given slope would have the same terminal velocity. Made an edit. – paparazzo Jul 19 '15 at 8:55
• Are you sure about a 7% slope giving a terminal velocity of 20mph? 7% isn't super-steep but it's steep enough for warning signs in many places; 20mph isn't very fast. – David Richerby Jul 19 '15 at 21:33
• @DavidRicherby I said around. For A set of factors at 7% that is what I got. What do you think it is? Are you sure that is wrong? The warning sign is probably there for trucks and that is whole different set of factors. – paparazzo Jul 19 '15 at 22:09
• @Blam I'm not sure it's wrong: that's why I asked. And I agree that trucks are a whole 'nother thing. However, a few kilometers of 7% grade is a categorized climb in the Tour de France and, for example, the famous Col du Tourmalet is 17km at an average of 7.3%, and is in the highest category. It seems surprising to me that terminal velocity down such a mountain would be only 20mph. – David Richerby Jul 19 '15 at 22:42

In almost every slope you will hit a speed "sweet spot" where the gravity pull and drag from the wind will cancel each other out, this can be at 70 km/h for a 8% slope in aero position or 15 km/h for a 2% slope on an hybrid bike.

• This is a good point. My fastest regular downhill is closed to cars for a couple of weeks. I detect an experiment coming on (posture only, I ride the same hybrid every day) – Chris H Jul 20 '15 at 7:44

Newtons second Law:

F=ma

where m is your mass (including everything, bicycle, luggage), a is acceleration. you look for solutions with a=0 here force F consists of:

gravity: m* g* cos(α), g is gravitational acceleration, earth average: g=9.81m/s², α is the angle between path and vertical.

air drag: ρ/2*c *A * v² where ρ is air density, c drag coefficient, A reference area, v your velocity

friction: d*v, where d is your coefficient of friction your input: Fh. If you have a power meter Fh=P/v where P is your power.

than you solve for α:

α = acos((Fh - ρ/2 * c * A * v²- d * v) / (m*g))

now you're somewhat stuck, because some of the parameters are hard to measure: c *A and d. There are numbers around for that ( c *A= 0.4, d=1Wh/km ) but the air drag part depends on your clothes, how you are actually sitting on the bike etc. Friction coefficient depends not only on your tires and wheel ball bearing but also on the road surface.

However as a cyclist you rarely get too choose your slope of descent, and there is always some equilibrium velocity where you neither lose or gain speed, since air drag grows with velocity squared - at high velocity it will dominate. At low velocity, without input of you, either gravity or friction will dominate, since they grow linearly in velocity. Of course that equilibrium velocity may be beyond what you can go through curves or what you're comfortable with, or what cars are going.