# Braking force required on the disc brake pads

#### GuiliuG

Hello,
I'm currently working on the design of an e-caliper and I need to this effect to determine the linear force at which the pads have to be pressed against the disc to slow down a car swiftly.

The result of my calculations is that the pad of a single front brake has to be pushed with a linear force of 35.51 kN (!) in order to bring to a halt (deceleration force of 1.2 G) a relatively weighty car (gross weight of 2353 kg) driving on the freeway at 130 km/h.

Is that a realistic value or is there something I've done completely wrong in my calculations ? It seems to be very high considering it's for a single braking unit.

I'd appreciate if you can take a gander at the attached picture and tell me if my calculations are right. (my apologies for the English typos in the text, it's just a draft and needs to be refined anyway).

Thanks in advance.

Guillaume.

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#### anorlunda

Mentor
Gold Member
I'm sorry you're not getting any answers. It may well be that we have nobody here experienced in similar brake calculations.

Also, when I viewed the picture you posted and zoomed in far enough for my bad eyes to read it, it was not legible.

#### jrmichler

Science Advisor
I can't read your picture either. I ran a similar calculation for my truck (a little less than 5000 lbs with normal load), some guesstimated dimensions, and 1 G deceleration. I got 4900 lbs clamping force on a pair of front brake pads. Remember that you are squeezing two pads together, with the pads on opposite sides of the brake disk, and each front brake does about 35% (or a little less) of the total braking. That sounds about right because that would be about 300 lbs on the brake pedal without power boost. That 300 lbs agrees with a test I did several years ago.

#### GuiliuG

Hi,
I'm sorry for the picture, I didn't notice it was so blurry. I've converted it to a pdf file, it's a lot clearer.
Ok, so yes, your number seems to be pretty close to what I get.
Not sure what you mean exactly with the two pads thing. If it's a floating caliper, possibly just one cylinder will be used to exert that force. But fundamentally, it doesn't change anything as you can have one very large cylinder whose cross-section is identical to two smaller ones and the force generated will therefore be identical for an equivalent pressure.

I didn't know in which section of the forum to post my question. Could you potentially move it to another section if you think it could target a more appropriate audience ?

Thanks.

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#### jrmichler

Science Advisor
Hi,
If it's a floating caliper, possibly just one cylinder will be used to exert that force. But fundamentally, it doesn't change anything as you can have one very large cylinder whose cross-section is identical to two smaller ones and the force generated will therefore be identical for an equivalent pressure.
Be careful. One piston in a floating caliper has the same braking force as two pistons of the same diameter on opposite sides of the brake disk. Any practical disk brake requires equal and opposite forces on both sides of the brake disk. You should consider adding a diagram to show this, and to show that opposed pistons act the same as one piston in a floating caliper.

#### jim hardy

Science Advisor
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Is that a realistic value or is there something I've done completely wrong in my calculations ?

I'd cross check for reasonableness with a back calculation from the practical side.

Here's a motorhead-oriented tech article that suggests typical pressure of 1200 psi from the master cylinder.
https://www.markwilliams.com/braketech.html

Size of the caliper piston of course depends on the vehicle size.
Here's an aftermarket caliper catalog showing pistons ranging from roughly 4 to 6 square inches

so i'd estimate you need to produce about 1200 psi X 4 to 6 in2 = 4800 to 7200 pounds of force, or 21,350 to 32,025 Newtons?
Your 35k wasn't a bad estimate.

old jim

#### GuiliuG

Hi,
Thanks for the confirmation. It's great to see my estimation was accurate.

@jrmichler : Not sure to get your point. What I was saying is that if there's just one piston of diameter 40 mm (floating) or two pistons of diameter 20 mm (fixed), the force produced for a given pressure will be identical. However, the advantage of a fixed setup is that you can use two pistons of diameter 40 mm and hence multiply by twofold the force produced for a given pressure.

#### jrmichler

Science Advisor
What I was saying is that if there's just one piston of diameter 40 mm (floating) or two pistons of diameter 20 mm (fixed), the force produced for a given pressure will be identical. However, the advantage of a fixed setup is that you can use two pistons of diameter 40 mm and hence multiply by twofold the force produced for a given pressure.
Wrong and wrong. Force is pressure times area, and one 40 mm diameter piston has the same area as four 20 mm diameter pistons. Also, two 40 mm pistons in a fixed caliper have exactly the same force as one 40 mm piston in a floating caliper.

Calculate piston areas. Make the free body diagram.

• Ketch22

#### GuiliuG

Sorry, I was clearly in the clouds this morning. First, I messed up with area and diameter. And secondly I messed up with reactive force. It's obvious indeed that the forces are identical between the two setups mentioned.

Thanks anyway.

#### Tom Kunich

Wrong and wrong. Force is pressure times area, and one 40 mm diameter piston has the same area as four 20 mm diameter pistons. Also, two 40 mm pistons in a fixed caliper have exactly the same force as one 40 mm piston in a floating caliper.

Calculate piston areas. Make the free body diagram.
Don't forget that the stopping force is strongly effected by the coefficient of friction of the brake pads and disk material.

#### jack action

Science Advisor
Gold Member
Although I agree with the magnitude of the value for the linear pad force, I don't agree with the way it is found in the attached document.

You only need 3 relationships to determine the linear pad force (based on figure 3.1 of the document):
$$F_t = \mu_t N$$
$$F_d r = F_t R$$
$$F_d = 2 \mu_p F_p$$
Where:
• $N$ is the normal force acting on the wheel (i.e. the weight);
• $r$ is the disc radius;
• $R$ is the tire radius;
• $\mu_t$ is the tire friction coefficient;
• $\mu_p$ is the pad friction coefficient.
Note the factor of $2$ in the 3rd equation. That is because there are 2 pads acting on the disc, one on each side (that is important), even though they depends on the same linear force.

Note also that the maximum force that can be produced by a tire is determined by tire coefficient of friction and the weight of that tire. A braking system should be able to handle that maximum tire force (i.e. the tire should be the weakest point).

It doesn't matter how fast the car is going or how much inertia it has: The limiting factor is the tire friction force in all cases. The amount of energy the car has (i.e. speed and inertia) will only influence the heat absorption/dissipation of your brake design (which is mostly independent of the braking force).

Also, the deceleration is solely dependent on the tire friction force. So to get a high deceleration, you necessarily need a high friction coefficient. If we set $F_t = ma$ and $N = mg$, it can be shown that $\mu_t = \frac{a}{g}$, which basically means that to get a 1.2 G deceleration, you need a 1.2 tire friction coefficient. That is independent of the weight or speed of the vehicle.

Therefore, based on the 3 equations, the linear pad force is:
$$F_p = \frac{N}{2}\frac{\mu_t}{\mu_p}\frac{R}{r}$$
Assuming typical values:
• $\mu_p$: 0.35-0.42 $\approx$ 0.38;
• $\mu_t$: 0.9-1.0 $\approx$ 0.95 (typical passenger tires);
• $\frac{R}{r} \approx$ 2.75;
• $N$: about 35% of the total weight ($W$) for one front wheel during braking (but it could be different with some vehicle designs).
Thus:
$$F_p = \frac{0.35W}{2}\frac{0.95}{0.38}2.75$$
$$F_p = 1.2W$$

And for a 2353 kg car, you need a 27.7 kN linear pad force on the front caliper.

#### GuiliuG

Hi,
Yes indeed my previous calculations were erroneous. I wanted to edit my previous post to append the correct calculations but apparently it's the forum policy to prevent anyone to edit a message after 24 hours, so I left it that way.

Actually, I got a decent order of magnitude by pure luck. Indeed, the integral to determine the braking distance traveled was completely wrong (don't know what I did there :/ ). Here is the correct equation : I didn't think further as I thought the value made sense because I had read a bunch of articles referring to a stopping distance of around 110 meters at 130 km/h on a dry asphalt road but those distances always account for the driver's reaction time. Note to myself : always double check your calculations.

As the braking distance calculated was twice as much as the correct one (remember I was using a deceleration of 1.2 G instead of 0.95 G), it compensated for not dividing by a factor 2 when calculating the squeezing force required.

Anyway, thanks all for your help, I finally got it right.

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#### Tom.G

Science Advisor
Don't forget that the brakes must also hold the vehicle immobile at full throttle! That's a safety feature that needs to be addressed.

#### cjl

Science Advisor
That's very easy for most brake systems. Pretty much every car made today has brakes that will easily overpower the engine in basically every circumstance.

#### Tom.G

Science Advisor
That's very easy for most brake systems. Pretty much every car made today has brakes that will easily overpower the engine in basically every circumstance.
Agreed.
The OP started out with...
I'm currently working on the design of an e-caliper...
That's why I mentioned it.

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