# What are the typical range of maximum braking friction forces?

There are many non-bike applications which benefit from using bicycle hardware. Case in point, there is a long and rich tradition of using bike parts in aviation. One particular area of interest is using bike brakes as a weight- and cost-optimized braking solution for many types of spinning shafts.

However, before prototyping with these parts it would be good to understand what are their operational limits.

While there are some good references for the maximum deceleration which can be reasonably attained by a bicycle (it seems that ~0.7g is a typical value), this does not address what are the brakes' ultimate force limits.

I am interested in understanding what are industry ball-park figures for the `friction force on disc` as in the below image. Please note that this is separate from the braking force a rider experiences, which is the `road pushes tyre`/`tyre pushes road`.

## Questions

1. What are typical ranges for maximum bike rim and disc braking forces?
2. How is this affected by off-the-shelf brake pad material?
3. What is the force difference (absolute or relative) between hydraulic and cable actuated disc brakes?
4. What is the force difference (absolute or relative) between hydraulic and cable actuated rim brakes?

P.S. I realize that torque is ultimately what is being transmitted through the wheel to the tire, but the torque arises because of drag forces generated by the brake. So if your answer gives torque specs please reference the diameter so I can convert to force.

• Ask folks who have done a head-over from braking with the front brakes. – Daniel R Hicks Oct 21 '20 at 1:47
• An endo is totally possible if your front brakes grab hard. TBH a heavier rider is more likely to endo than a light rider on the same bike at the same speed. Remember the rider is generally above the brake/axle so theres a moment of rotation as well. – Criggie Oct 21 '20 at 2:22
• Theres a lot of questions here - try reducing it to one. SE isn't a good format for large volume answers all about a topic. – Criggie Oct 21 '20 at 2:23
• Answering your questions in the context of cycling will not provide good data for the non-bicycle application you have in mind. You need to experiment with your use case to find the answers you seek. If your use case is aviation then people familiar with using bike parts in aviation would be a better group to query. – David D Oct 21 '20 at 13:52
• @DavidD The braking friction forces are independent of the application. They are only a function of the pad compound, the pad forces, and the braked material (steel or aluminum). I agree I will need experimentation to find if the final components can make an appropriate solution, but that is independent of the forces involved. For instance, imagine if someone wanted to repurpose a Ti bolt from a seat post. How strong the bolt is is independent of how it was marketed and sold. – Kenn Sebesta Oct 21 '20 at 16:02

What is the relative difference between hydraulic and cable actuated disc brakes?
What is the relative difference between hydraulic and cable actuated rim brakes?

Generally speaking - hydraulic brakes offer more modulation. There's more "hand-feel" between the initial bite and through to full-on braking.

Whether its disk or rim is less important - the rim brake can be treated as a really big rotor.

Cable actuated disk calipers tend to be single sided, so they have one moving pad and one stationary pad. These can work fine, but they have to flex the rotor subtly when braking. A hydraulic disk caliper, and any kind of rim brake will bring both pads to the braking surface.

A good rim brake can easily exceed a mediocre disk brake of any style. Which is better between a good rim brake and a good disk caliper is still undecided. Disks have a weight and aero penalty, but don't depend on a planar rim.

• @Craggie, thanks for posting but this doesn't answer the question. The part of "relative difference" is referencing the maximum braking forces, not the user's feeling. I have updated the question to clarify this. – Kenn Sebesta Oct 21 '20 at 3:24
• To clarify why this answer isn't headed in the right direction, the question is to document (approximate) values for the friction forces which arise from braking. These forces are a function of how hard the pad is pushed into the rim compared to the pad's friction coefficient. – Kenn Sebesta Oct 21 '20 at 3:26
• @KennSebesta there such a spectrum in quality of the styles of brakes in question, I don't think it's possible to give a definitive answer to your question. You might find this interesting nonetheless enduro-mtb.com/en/best-mtb-disc-brake-can-buy – Paul H Oct 21 '20 at 3:43
• @PaulH That link is great! I would definitely consider it definitive. In general, I would be surprised if there's more than a factor of 2 when dealing with reputable parts. Manufacturers are likely designing and testing against a spec somewhere so they'll come in relatively the same. The typical bike + rider is gonna be close to 100kg, but not 50 and not 200. One things we can learn from the link is that 75Nm on a 180mm rotor is a decent median value for good gear. – Kenn Sebesta Oct 21 '20 at 4:16
• FTR, “which is better between a good rim brake and a good disk caliper is still undecided” only applies to ideal dry conditions. I don't think anybody would dispute that rim brakes are much more of a hassle in the wet, let alone in mud, as well as in various maintenance senses – that's why MTB uses now almost exclusively disks. – leftaroundabout Oct 22 '20 at 22:17

(How to brake properly on your MTB | Brake Power Meter)

Key stats from the video:

• On a 1km trail with 122m elevation loss, average braking power (while using the brakes, not averaged over the entire ride) was 441 watts for a novice rider, and 643 watts for an advanced rider.
• Peak braking power was close to 2000 watts for either rider.
• The total energy dissipated was around 31 kilojoules for both riders.
• Front/rear brake bias was around 2:1 for the novice and 3:1 for the advanced rider.

Also do consider that higher power isn’t always the end be all goal. For example, Magura disc brakes are very powerful, but also have ridiculously tight pad clearance. You will need to true and replace brake rotors far more often compared to other brands.

Edit: Here’s another two videos related to this subject. I won’t bother writing out the key facts for these as they’re far less scientific in nature.

• Just wanted to drop a comment here and say thanks for the video. I wrote the guy who made it, he clearly has done a fair amount of engineering in this domain and might have some representative values. – Kenn Sebesta Oct 21 '20 at 17:43
• I'd just like to add that power and force are different things. – ojs Oct 22 '20 at 10:13
• @ojs Agreed, but the power is certainly still a useful metric. – MaplePanda Oct 22 '20 at 14:47

There's more to braking than just the brakes.

Taken to an extreme, standard bike brakes likely can't throw a 300kg rider over the handlebars

No they can't, but that's because the tires will lose traction and skid, at which point any additional brake power is useless.

Tandem bikes have twice the mass to stop, and they use the same brakes as normal bicycles for rapid stops. Even with that extra mass, with properly setup brakes, you can skid the rear wheel if you grab that lever hard enough and don't care about possibly losing control.

In short, brakes are designed to deliver as much power as the tires are expected to be able to grip, and no more. Anything more requires additional rigidity/material which is effectively dead weight.

• A bike that can throw a 75kg rider over the bars can also throw a 300kg rider over the bars, provided sufficient brakes: The force that the tires transmit grows linearly with the weight of the rider. They do not limit braking force, they limit the angle at which you can apply force on the tire without slipping it. As such, the question whether your front wheel slips or blocks first is only down to frame geometry and tire material, not on rider weight. – cmaster - reinstate monica Oct 21 '20 at 8:02
• @cmaster-reinstatemonica but the tandem example has a much longer wheelbase, and the C of G is much further from the front axle, so it is a good demonstration – Chris H Oct 21 '20 at 9:52
• @ChrisH Yes, indeed. I would never expect a tandem to topple over, even when it's ridden by two kids. The angle between the front tire contact patch, the center of gravity, and the road is just too low. I was commenting on the first three paragraphs before the mention of tandems. (Btw: That's precisely why I prefer riding bikes with a really long wheel base. Skidding your front tire is just so much more healthy than going over the bars...) – cmaster - reinstate monica Oct 21 '20 at 10:13
• @cmaster-reinstatemonica fair enough. My tourer (XL frame size) has one of the longest wheelbases on the market. I have braked hard enough to lift the back wheel, but the only time I went OTB was hitting a rock – Chris H Oct 21 '20 at 10:27
• Thanks for putting the time into an answer, but unfortunately this answer is about bike operation, whereas the question was about bike component forces. I have updated the question to make it clear that while the mechanisms in question come from the bike industry the end application is not bicycles. – Kenn Sebesta Oct 21 '20 at 12:13

You can estimate a figure for your two forces -- "friction force on disc" and "tire pushes road" -- from physical considerations only.

First, if we know the mass of the rider and the deceleration, we can calculate the net braking force on the rider/bicycle combination. This force comes from the "normal force" of the tires against the ground, equivalent to your "tire pushes road" force, but that's not important at this stage. Using your figure of 0.7G of deceleration, we can calculate acceleration in m/s^2 as (0.7*9.81)=6.9 m/s^2 of acceleration (or deceleration if you prefer; sign is not important here).

To calculate the total braking force, we simply use Newton's law Force = mass * acceleration. Assuming a 100kg rider, and rounding, the braking force = 100kg * 6.9 m/s^2 = 687 Newtons.

Assuming all of the braking force comes from the front wheel, which is a reasonable assumption for most short-wheelbase bicycles under maximum braking, we already have the value for "tire pushes road"...it's 687 Newtons.

To calculate the "friction force on disc", we simply multiply by the ratio of the effective disc diameter and tire diameter. If it's a bicycle with 26-inch (660mm) wheels and a 160mm front disc (effective diameter ~150mm) then the "friction force on disc" will be approximately 687 Newtons * (660mm / 150mm) = 3023 Newtons. This corresponds to about 680 pounds-force.

We can see that a larger brake disc reduces the "friction force on disc", as does a smaller wheel, but neither has any impact on "tire pushes road".

Now, as far as "industry standard figures", I have no knowledge of industry practice in this area, and it is a matter of engineering and testing to come up with design factors such as that. I agree that the best way to find that out, aside from running your own failure tests on components, would be to discuss with actual component engineers, but it will be difficult to find the right person who will work with you.

• I appreciate the effort, but this doesn't address the question. The question is about the absolute maximum force the brakes components can make, in contrast to this answer which (correctly) analyzes the holistic system force for a typical bike. – Kenn Sebesta Jan 26 at 1:21
• By following this exercise to calculate what forces that brakes achieve in real life, we can estimate or establish a bound for the forces. – BetterSense Jan 26 at 2:06
• I agree, insofar as through this exercise we establish only a lower bound. There are times when establishing a lower bound can answer the question "is it strong enough?", but the OQ is "where are the typical limits?" With that in mind, I hope you can appreciate where I'm coming from when I say the answer does not address the question. – Kenn Sebesta Jan 26 at 2:45

Kenn. I’m a bike enthusiast, a proud nerdy mech. engineer, and a (non-current) PPL who now runs a bike shop, so your question interests me. I am led to believe that Shimano employs more engineers per capita than any other bike industry company. If this is true, one way to get definitive numbers would be to seek out Shimano's centre of brake design expertise to get their numbers, e.g. typical hydraulic pressures reached by hand levers, (my guess is 1500 psi or 100 bar.), friction coefficients of various compounds with steel discs, and caliper forces and fluid volumes? Also maybe some heat dissipation and hose expansion data. As the saying goes, one measurement is worth a thousand opinions. Also, anecdotes are not data.

IMHO cable actuated brakes have almost no hope of repeatable reliable performance in a non-bicycle application, hence the complete dominance of hydraulics in automotive and motorbike braking use for many decades.

BTW I’d definitely stick with mineral oil rather than DOT fluid. Citroen pioneered that in the 1960s.

• This post has been flagged as "Not an Answer". Portions of it, like directing OP to contact Shimano, are better suited for comments rather than answers. If possible, please edit your answer to more directly address the question asked by the original post. – Gary.Ray Jan 26 at 0:07
• +1 for being a pilot, but I agree with @Gary.Ray that this would be better as a comment. I'm not at all suggesting you should call Shimano, but if you did and reported that here as absolute/relative numbers then it would be an answer. – Kenn Sebesta Jan 26 at 1:24
• Kenn. You’re getting a wide range of views from bike enthusiasts here, that’s nice. However most do not give you the real engineering data that you are after. Hence instead of going with a load of “made up” stuff and anecdotes, my answer is to suggest you seek some real expertise, e.g. from Shimano. Unless a brake technical expert from Shimano or a peer volunteers an answer on here (unlikely), then you’ll need to make contact yourself. You could then share the knowledge here fore future readers :) – Henry S Jan 26 at 10:53