F1 Braking at high speeds.

[747heavy]

timbo,

First of all, brakes are sized by regulations.

Brakes are not sized by regulations, wheel dimensions are. Wheel diameter limits obviously limit the max diameter of the brake rotor, but it does not limit the thickness of the rotor. And as I noted, carbon brakes are mostly sized by thermal mass. So an increase in thickness is just as effective as an increase in diameter.

Riff_Raff
Sorry, this statement of yours, as far as the regulations go, is incorrect.
(with appologies)

11.3 Brake discs and pads :
11.3.1 No more than one brake disc is permitted on each wheel.
11.3.2 All discs must have a maximum thickness of 28mm and a maximum outside diameter of 278mm.
11.3.3 No more than two brake pads are permitted on each wheel.

@ Mep
Brake "boosters" are not permitted by the regulations as well

11.5.2 No braking system may be designed to increase the pressure in the brake calipers above that achievable
by the driver applying pressure to the pedal under static conditions.

[xpensive]

If we begin with breaking power being proportional to speed, provided that braking torque is constant, said power will conclusively be three times higher at 300 km/h than the same at 100 km/h.

While the passing air's ability to convey, through forced convection if you wish, the same power to the atmosphere,
will have a squared relation to the speed, why said ability is nine times higher at 300 in the same speed-comparison.

[mep]

First of all welcome back 747heavy, the forum is really missing something without you.

If the brake system is working properly, it is no problem for a driver to create enough braking force to lock the tires at any speed. The brake pedal/master cylinder/caliper piston is a very efficient mechanism for multiplying force

Its multiplying force but not power. You can increase the pedal way but there are set limits. I would like to know what maximum power a human can generate with one leg.

as an example braking into Turn 5 in Abu Dhabi, the cars will reach ~5g decceleration with an pedal force of ~149 kgf (1462N)needed.

149kg sounds very much but those 5g’s definitely help the drivers to achieve the force.
This let me think about if its possible to just put a high mass on the pedal, give the brakes some initial bite so mass inertia will produce enough force to stop the car (more a theoretical thing of course).

[Formula None]

149kg sounds very much but those 5g’s definitely help the drivers to achieve the force.
This let me think about if its possible to just put a high mass on the pedal, give the brakes some initial bite so mass inertia will produce enough force to stop the car (more a theoretical thing of course).

Tungsten brake pedals, I like it. Can I pick those up at JC Whitney?



Seems like it would work, but the pedal return would also be affected negatively, right?

[autogyro]

Movable ballast in the drivers overall trouser leg would help

[xpensive]

...
149kg sounds very much but those 5g’s definitely help the drivers to achieve the force.
...

Not really mep, just imagine Norbert Haug standing on one leg?

[riff_raff]

"Riff_Raff
Sorry, this statement of yours, as far as the regulations go, is incorrect.
(with appologies)"

747heavy,

Thank you for the correction. I should have checked the rule book first.

As for the braking torque the driver can create, you can make a quick estimate. Assume that the brake hydraulic circuit is pressurized to say 1000(?)psi by the driver pushing on the brake pedal. And that pressure then acts on each caliper's four 40mm(?) pistons. That gives about 7800lbf clamping on the rotor. If that 7800lbf clamping is applied at a moment radius of say 4 inches, and assuming a Mu of .16(?) between pads and rotor, that gives a braking torque of around 416 ft-lb per wheel. If you assume a wheel speed, you can then calculate a power term based on that torque.

Just a rough guess.

riff_raff

[xpensive]

@riff_raff; Interesting line of thinking, how about taking it the other way?

- Beginning with going from 300 to 100 km/h within 2.5 seconds, that's a deccelleration of 22.2 m/s^2 (2.27 g).
- Force is mass times accelleration, why the above takes 15.6 kN of braking force for the 700 kg car without air resistance.
- A wheel diameter of 660 mm and 15.6 kN of force, means a total braking torque of 5.13 kNm, or 1280 Nm per wheel.
- Braking torque of 5.13 kNm gives a braking power of 1300 kW at 300 km/h and 433 at 100, averaging 870 (1200 Hp).
- If the pads act on an a mean radius of 130 mm and CoF is 0.16, clamping force would need be 61.7 kN or 12800 lbf.

Six tons of clamping force gives 123 Bar over 50 cm^2 (four times 40 mm dia pots), then the 149 kg of driver's force would call for brake-cylinder area of only 0.83 cm^2, or a diameter of 10.3 mm. Does this logic seem at all reasonable to you?

[747heavy]

just to put some numbers out there:

line pressure in the brake lines is normally around ~1000-1400 psi (65-97 bar)
brake pedal ratio is 1/3 to 1/4
CoF in the brakes ~0,25 when braking starts and for the most part when heat built up is finished 0,5
when speeds drop under ~120 km/h the CoF increases to ~ 0,8 which is in part the reasons for "lock ups" at lower speeds, as the drivers need to adjust for this change.
by the rules you can have max. 6 pistons per caliper.
the dia varies between ~25-45 mm
bore of the mastercylinder is ~12,7-15,88mm , can be changed depending on drivers preference

hope this helps to futher our little calculation exercise

[xpensive]

Guess I we weren't that far off then, were we?

Anyway, a CoF of 0.25 would reduce clamping force to 39.5 kN, why hydraulic pressure (with the same 50 cm^2 piston area) should be about 80 Bar, which in turn would cause a load of 1400 N on a on a 15 mm (1.8 cm^2) master cylinder.

Finally, a brake pedal ratio of 1/3 to 1/4 means a "pedal-force" of 350 to 470 N, far less than Norbert, more like Danica?

[747heavy]

I think in terms of deceleration vs. pedal force you are very close.

from the Brembo data from the last race:

[xpensive]

Than-x jumbo, a great way to begin this fine Saturday, what this forum should be all about actually.

[Tec5nical]

If the brakes needs more heat to build up friction the drivers can simply let the brakes grind little bit before braking.

Maybe this just won't be enough: grinding does not heat them up enough.
and of cource by doing so, you won't have your valuable peak acceleration possible, in full throttle periods, altough ginding them up would be benefical in braking and compensate the time.
I feel like it just would be like making braking part start sooner, with the same amount of braking power generated on both approaches

So, by attacking on the brakes at the maximum speed, first you don't have enough temperature to fully produce braking, second, donforce is at the highest point possible. By losing speed, you will have less DF and more braking force available. So these would be the reasons for the possibility of not locking up at very hish speeds. If so, there should be a speed threshold where depending on track's surface, tyre situation and available downforce, the least weighted tyre(s) starts to lock up.

According to Brembo, there's a braking zone at Spa where the car goes from 328 (91) to 167 km/h (46 m/s) in 1.40 sec, which gives an average of 3.25 g with a peak of 5.06 g. Braking power is 2350 kW, which, if equal during the entire sequence, means that braking force goes from 25.8 to 50.6 kN

expensive, I feel like braking power doesn't do so much with braking force generated and drivers' input braking force, as these deal with the static equilibrium, rather work done from calipers onto disks.

Maybe the drivers just can't produce enough power/force to lock the tyres at highest speeds.

I think you should also consider that, there is a maximum amount possible for the driver to finally put onto brake disks, as "All brake calipers must be made from aluminium materials with a modulus of elasticity no greater than 80Gpa."

[xpensive]

At 328 km/h, the air-resistance in itself is worth pretty much the same as your engine power deccelleration-wise,
lifting-off 550 kW gives you a braking force of 6 kN, 18 kNm if you wish to translate it to an equivalent "braking torque."

6000 N of force on a 700 kg object, means a deccelleration of 8.57 m/s^2 ( 0.87 g), this from air-resistance only!

Think about it.

[JohnsonsEvilTwin]

At 328 km/h, the air-resistance in itself is worth pretty much the same as your engine power deccelleration-wise,
lifting-off 550 kW gives you a braking force of 6 kN, 18 kNm if you wish to translate it to an equivalent "braking torque."

6000 N of force on a 700 kg object, means a deccelleration of 8.57 m/s^2 ( 0.87 g), this from air-resistance only!

Think about it.

So a brick of an aero design would be better under braking?