What are the typical range of maximum braking friction forces?

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.

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(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.

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.

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.

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.