Bike Gremlin

But, this particular example is incorrect. Take a 10 mile long descent. You can reach a terminal velocity in about a mile. You can brake 10 times to zero. Or just once at the bottom. Letting the wind do all the braking.

Similar example is holding the speed below a safe threshold. Whichever that particular speed was - depending on road, skill etc. The faster you let the bike go, the more work is done by the wind, less with the brakes.

gregf83 

Not really clear what you're arguing about here. I gave the definition for terminal velocity. It has nothing to do with pedaling but the terminal velocity of a cyclist is related to the slope and riding position.

The example I provided earlier of a local 800m hill. Rider starts at the top with over 700,000J of potential energy. If he coasts down the whole hill and ends up travelling 70kph at the bottom he will have dissipated over 680,000J of energy in the form of heat on the way down. If he coasts up to terminal velocity and doesn't brake all the potential energy gets dissipated by aero drag. If he want's to go slower than terminal velocity at say 50kph he will have to burn up a lot of energy by heating up his rims. I think that was the original point. By going faster you can brake less.

gregf83 

I didn't read the rest of your post but you are incorrect in the above analysis. Going down the hill at less than terminal velocity will require a much higher amount of heat to be dissipated by your brakes than if you le the wind take care of braking. Do the math.

diamacleod

Rediscovering fundamental physics is fun, though slightly off topic. Most of the recent postings do reflect what I take to be the standard view. At the risk of further illustrating the rule that the most wisdom is to be found in the briefest postings, I will try to summarize that.
The force due to gravity in the downhill direction, per unit mass is g sin(angular gradient). For not very extreme gradients this is approximately (g times (% grade) /100 ).
Thus if air resistance is negligible—as it is if you are going very slowly—you pick up 2 mph per second freewheeling down a 10% grade, since g is about 20 mph/s/s.
The gravitational force is opposed by air resistance that increases as the square of your speed, until at some sustained coasting speed (= terminal velocity in the downhill direction) the forces cancel and you stop accelerating. That coasting speed depends on rider mass and posture but is always proportional to the square root of the gradient. For a normal, moderately aero, posture on the drops, mine is around 40mph on a 6% grade. The approximations suggest twice that (80 mph) for a (sufficiently long) 24% grade.

Applying prathmann’s excellent formula above, my worst speed for overheating would be 46 mph on a 24% descent, or 23 mph on a 6% descent.

GamblerGORD53

I can agree with that speed/ braking chart. I would say the hill needs to be at least 12% to get near limitless speed. On a 6 or 7% hill I have to pedal like mad to get 37 mph. If into a mere 6 mph wind I just can't go more than 30. Same on a short 9% hill where I wizzed up to 47.8 mph with a light tail wind. Into the same wind I get stuck at 35 mph. Air does matter a LOT. I have a video of my top speed there. Maybe these hills are too short to really see, but I doubt it.

I see altitude can make a big difference.

Leebo

OP , just get comfortable going faster? Sit up more and pulse the brakes, start there. I usually top out at 40- 45 mph.

diamacleod

A useful linkL the square root relation between coasting speed and gradient can be verified using the calculator for terminal velocity in downhill coasting at Analytic Cycling. To use it, remember to enter zero power for the rider.

https://www.analyticcycling.com/DiffE...lope_Page.html




Another blog has a graph that clearly shows the square root dependence, though it wrongly gives the equation for terminal velocity as linear with slope (a typo I assume)
The Penultimates Cycling Team: The Physics of Cycling II, or Why I am Awesome on the Downhill

cyccommute 

The problem here is that you are attributing too much braking power to the wind. Yes, it has an effect. But the magnitude just isn't that great. I'm not talking about braking 10 times to a full stop on some 10 mile long descent but only controlling your speed to some speed the rider feels save with. Simply put, the wind will not do all the braking. It will do some of it, of course. And, if you have a long enough run-out at the bottom of the hill, you might have time to let the wind do all the braking but in the real world where you don't have long run-outs and 10 mile long descents include curves, your brakes will have to handle the job.

True, partly. If let the bike go faster and you have the nerve to go around corners faster, you will brake less. But if you have to slow or stop, the faster you go the more you will have to brake and the more heat you will generate. Again, aerodynamic drag will handle a portion of the speed control but it may not be enough of a portion to keep you from running off the road if you are coming into a corner too hot. Nor will it be enough if you have to stop completely.

It's a matter of magnitude and how much you consider to be "much higher". Sure, aerodynamic drag reduces the amount of time you have to use your brakes and the amount of heat generated. But I simply disagree with you as to how much that is. If you are in a tuck and squeezing every last mph out of a downhill...say doing 80 mph down a hill..., sitting up suddenly and increasing the drag isn't going to suddenly scrub half the speed off. I doubt that it will scrub a quarter of the speed off and it's going to take a long time to do even that.

Bike Gremlin

My point is you should wait until reaching the top speed that it is safe before braking, instead of braking before that. The more speed you let the bike gain, the more braking is done by the wind. If we're talking about minimal brake heating.

gsa103

The thing with aerodynamic drag is that its like a drag brake. It won't slow you instantly, but integrated over a long period of time, it has a huge effect. It won't scrub it instantly, but over the course of ~10s it will slow you dramatically.

Getting back to your point about kinetic energy. Starting at the top of the hill.
We'll assume 6% grade, 100m elevation, 100 kg (rider + bike). At the top of the hill, the rider has 98 kJ of potential energy.

If the rider applies, no brakes and simply coasts hitting a terminal velocity of 60.4 km/h.(Bicycle Speed (Velocity) And Power Calculator, [drops, 90kg rider, 10kg bike, 0W, 0 wind]
At the bottom of the hill, the rider has a kinetic energy of 14.1 kJ.

That means that the aerodynamic drag has dissipated the equivalent of 84 kJ of energy. Which is far more than the brakes would have to do to stop the rider at the bottom. If you don't like the calculated value for terminal velocity choose a different one, it doesn't really matter. Any realistic terminal velocity will result in the majority of the energy being dissipated by aero drag.

In the situation I just pointed, if the rider's potential energy were converted into kinetic energy, the rider would reach the bottom of the hill at 160 kph (~100 mph). Clearly, that's absurd, since no rider gets going that fast.

diamacleod

Wheel stress and braking force
The original post referred to spoke breakage, but mechanical stress has not been given much attention in this thread. It is tempting to suppose that braking on a steep descent stresses the wheels, but this may wrong. Theoretically, under admittedly drastic simplifying assumptions, *steady* braking to maintain a target downhill speed produces no wheel stress, and stresses on the wheel are actually reduced on a steep fast descent.
Suppose first for simplicity that the rim is absolutely rigid. The effect of gravity is then to decenter the circular rim slightly relative to the dropout, and the pattern of tension in the spokes follows from that. If you are parked by applying rim brakes on a downhill grade, both the downhill component of the gravitational force and the orthogonal ‘into the ground’ component are cancelled by an uphill braking force applied to the rim (that is divided equally between the brake pad contact patch and the tire contact patch). So the decentering force and the resulting spoke tensions are all the same as with no brakes on the flat. With slow downhill progress, the pattern of spoke tensions moves around the wheel without changing otherwise, and this happens in the same way whether the descent is gradual or steep.

On a fast descent, the uphill braking force is handed over in part to air resistance, and is not (fully) applied to the rim. The ‘into the ground’ force is reduced by the cosine of the angle of descent. So the variation in spoke tension is slightly reduced accordingly (until in the limit of vertical free fall, spoke tension becomes uniform).


If we relax the rigidity assumption, it’s clear that subtle flattening of the rim near the contact patch reduces the tension of just a few spokes there, with only a slight compensatory increase of tension at the top and sides (see for instance, Jobst Brandt). That effect survives on downhills, with the cosine reduction alleviating it slightly. The tangential braking forces applied to the rim will not cause deformation, so they add no stress to what is captured in the rigid rim model.

This analysis suggests that risks to the wheel from heavy braking on steep descents are not mechanical but thermal (like the blowout risk).

gregf83 

I agree the aero drag will not help in stopping you but that's not the point. The idea of letting your speed rise is you don't need to use your brakes until you need to slow down for a corner. The majority of time you won't be on your brakes at all and you'll either be at terminal velocity or accelerating to terminal velocity.

Slowing down from a terminal velocity of 80kph is not going to cause aluminum rims to overheat since there isn't enough energy available. Dragging your brakes for 10 min to limit your speed to 50kph will inject a lot of heat into your rims.

Bike Gremlin

Quite right. While disc brakes are another matter (more stress to fork bearings, axle and spokes), rim brakes don't stress the spokes when braking, just heat the rim.

cyccommute 

That's only if you don't have to use the brakes for some reason. I'll agree that using the wind to help control speed is a wise thing to do. I practice that myself. I let the bike go as fast as it wants to...and often push it to go faster...then use the brakes quickly and only as much as necessary.

But I also know that the faster the bike goes, the more heat will be generated when I inevitably have to use the brakes. Dragging the brakes all the way down the hill is worse for overheating the rims but that's because riders who do this are never allowing the brake system to lose the heat that's being put into the system. It's not because they are putting more heat into the system, only that they are keeping what heat the generate at the rim.

Going faster doesn't put less heat into the rims when the brakes are used. The amount of heat put into the system is proportional to the square of the speed of the bike and rider. It's still possible to overheat the system by going faster if you have to use the brakes.

My point is not about the aerodynamic drag but about the times when you need to use the brakes. When you apply your brakes it becomes a simple energy conversion problem. The kinetic energy is converted to heat energy which, of course, heats up the brake system. The amount of energy converted is equal to the square of the speed differential. The more you have to slow, much more heat is generated from a higher speed.

You are underestimating the amount of energy that is available. It would, of course, depend on how much you have to slow down. If you have to slow only a small amount to, for example, go around a corner, the heat generated is going to be small. But if you have to slow down to a stop, the amount of heat generated is going to be much greater.

Yes, dragging your brakes down a hill will inject a lot of heat into your rims but braking hard from 80 kph to 50 kph will also inject a lot of heat into the rims. That's where you are going wrong in your assumption. The difference between the two is that one is a heat load over long period of time and the other is a short burst of intense heat.

Think of it this way: if you were traveling at 80kph and had to brake to 50 kph and keep the speed there for some reason would the brake system be cooler than if you kept the brakes on for 10 minutes to keep the bike at 50 kph down the same hill?

Pulse braking...i.e. getting short intense braking followed by coasting...works not because it generates less heat but because you are allowing the system to radiate the heat away during those coasting periods. The amount of heat generated is the same or even higher but the rider is managing the heat better.

gsa103

Yes, by a very large margin.

gregf83 

Good example. According to Kreuzotter.de to travel 80kph a 90kg (rider + bike) requires about 2400W of power to overcome wind and rolling resistance while riding in the drops. To generate that power at 80kph requires a slope of 12.7% assuming no wind. Riding down that same hill at 50kph, gravity will provide 1550W that needs to be dissipated by braking and aero resistance. Unfortunately, it only requires 620W to overcome the wind and rolling resistance at 50kph. The difference (1550 - 620) 930W will be converted by braking into heat. After 10 min that's over 560,000J of heat injected into the rims.

To summarize the two scenarios for a descent down a 12.7% hill:
A) Ride your brakes and limit your speed to 50kph => dissipate 560,000J of heat over 10 min
B) Coast at an 80kph terminal velocity and slow from 80 to 50kph at the end => 13,500J into the rims.

While you're pouring 900+W into your rims for 10min they will reach an equilibrium temperature but it's going to be fairly high. A bike rim is not an ideal shape for an effective heat sink.

This is also incorrect. Rims shed heat via convection and radiation to the air. Whether the pads are touching the rims or a mm or two away from the rims has an insignificant effect on the amount of heat they dissipate. Not braking keeps from adding more heat but it doesn't lower the heat being dissipated.

diamacleod

In #50 I suggested a different reason why intermittent braking can be helpful: for a given net propulsive energy, average speed is greatest--or total descent time is least—when the speed is constant. So you can increase descent time--which is also the time that the rim has to cool down--by braking strongly and infrequently rather than dissipating the same amount of energy in steady braking. But this benefit is small unless you drastically reduce your speed each time, and then coast back slowly to your chosen max speed. If the braking pulses are breif and come often enough that speed is almost constant, the benefit is negligible.

Bike Gremlin

The higher the top allowed speed, the more sense it makes to use the wind drag in order to keep the brakes cooler - since the energy absorbed by the wind drag increases exponentially with the speed increase.

Using wind to slow the rider down keeps the brakes doing total work - lower, as well as shorter total braking time. As well as more cooling time. So I think it's a better option for keeping them cooler.

There might be a flaw in my logic. The fact you're the only member disagreeing with the theory, doesn't automatically mean you're wrong. We could all be missing something. I really respect your posts and knowledge, but fail to see your point of view on this topic as a correct one. Still doesn't mean I'm right. The only way to be 100% sure is to borrow a speedometer and a laser thermomether and do a test on a local hill.

willbur

Not trying to do a Gotcha but the pressure increases 48 psi (Guy-Lussac Gas Law)