Went for a ride today

[parb]

Took the wrong line and got stuck halfway up in some soft stuff. Had to reverse down and take the line to the right where I was successful and made it up.
About 26 degrees angle so a bit steep. Very soft with a few rocks to get your attention...

California's Sierras just south of lake tahoe, eagle rock.

New video · Sunday, Dec 7 🎬

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photos.app.goo.gl

And this was the successful line up to the top.

New video · Sunday, Dec 7 🎬

Tap to view!

photos.app.goo.gl

I have M/T tires that are really grippy but it was soft and steep, wasn't happening without a bit more momentum. I was tempted but decided it was best to do another line.

 

[ADVAW8S]

Were you running it triple lock, the first time? Had similar experience, tried double locked. Reverse down, went triple and walked up.

 

[parb]

This was triple locked both times. Soft ground with rocks so i had to.

 

[parb]

TL;DR

This is physics and physics I have an affinity for. So I researched the math a bit.
Thanks btw. these are the kind of fun math problems I enjoy.

The math is straightforward, larger brake disc creates a stronger lever for torque which directly translates into a mechanical advantage. The relative increase in stopping torque is 17.08%% for the power brake big brake kit vs stock Ineos brakes. You'll have to do the math for other vendors system (That is straight up math on facts I know using just physics and known simple math.

So at the same brake pressure you get 17% increase in counter torque. But this compounds as you brake, this math compounds as the car dissipates energy from motion into energy as heat while at the same time stopping faster at the same time faster through increased counter torque and controlled thermal management.

It's really lovely, the counter torque is really what the people report as pedal feel. Eg 17% more counter torque at the same press on the pedal.

net 17% increase in counter torque against the wheel just on the disc itself. A meaningful and significant increase. And probably better since the product of counter torque is energy, which has to be dissipated through heat and I didn't even include that in the math yet.

The more I'm looking at that math the more sold I am on the benefit for cars with weight on them. It looks meaningful, particularly as you're out in the edge of heat management and counter torque performance envelope.

Long version;
Here is a simplified text book definition of the underlying math. I didn't work in all aspects of it, piston size and pad calculations and thermal energy management, but that math works out and I'll probably do it at some point when I have the kit for my own enjoyment.

Braking is the process of converting the kinetic energy of a moving vehicle into thermal energy (heat) through friction.

Improved braking performance generally means two things:
Greater Stopping Power (Mechanical): The ability to generate more braking torque to decelerate the wheel faster.

Better Heat Management (Thermal): The ability to absorb and dissipate the immense heat generated without "fading" (losing effectiveness).
Larger discs and pads address both of these aspects, but in different ways.
Part 1: The Mechanical Advantage (Torque)

The primary mechanical benefit of a larger brake system comes specifically from a larger diameter brake disc (rotor).

When you press the brake pedal, hydraulic pressure forces the brake pads to clamp onto the spinning disc. This creates friction. However, friction alone doesn't stop the car; torque stops the car.

Torque is a twisting force. To stop a spinning wheel, you need to apply a counter-torque.

The Physics of Braking Torque
Imagine trying to stop a spinning bicycle wheel with your hand. If you grab the hub (the center), it is very difficult to stop. If you grab the tire (the outer edge), it is much easier. This is the principle of leverage.
The brake caliper applies a clamping force at a certain distance from the center of the wheel hub. This distance is the "effective radius."

The Formula for Braking Torque (T):
T = F_friction * r_eff
T = Braking Torque (The twisting force that stops the wheel rotation).
F = The total friction force generated between the pads and disc.
r_eff = The effective radius (the distance from the center of the wheel hub to the center point where the pads contact the disc).

We can break down F further:
Where:
mu * F_clamp
(mu) = Coefficient of friction between the pad material and the disc material.
F_clamp = The normal force (clamping force) exerted by the caliper pistons onto the pads.

Combining these, we get the complete braking torque formula:
T = mu * F_clamp * r_eff

How a Larger Disc Improves Performance
Looking at the formula above, if we keep the brake pedal pressure constant (meaning F_clamp is unchanged) and use the same pad material (mu is unchanged), the only way to increase torque (T) is to increase the effective radius r_eff.

A larger diameter disc moves the caliper further away from the wheel hub, increasing r_eff.
Conclusion 1: A larger disc provides a longer lever arm, generating more stopping torque for the same amount of pedal effort. This is the most direct mechanical advantage.


Part 2: The Thermal Advantage (Heat Management)
This is where both larger discs and larger pads play a crucial role.
The amount of energy a braking system must handle is enormous. The kinetic energy (KE) of a moving vehicle is defined by:

KE = 1/2 *(mv(pov 2))

Where m is vehicle mass and v is velocity. Because velocity is squared, slowing down from 100 mph takes four times as much energy as slowing down from 50 mph.
All of this energy must be turned into heat instantly. If brake components get too hot, three bad things happen (known as "brake fade"):
The coefficient of friction (\mu) drops significantly (pad fade).
The brake fluid boils, introducing compressible air bubbles into the lines (fluid fade).
The metal rotors warp or crack.
Larger components combat this through Heat Capacity and Heat Dissipation.
A. Thermal Mass (Heat Capacity)
Think of the brake disc as a heat sink or a "thermal battery." It needs to absorb the initial intense burst of heat generated during a panic stop.
The relationship between heat absorbed (Q) and temperature rise (Delta T) is:

Q = m_disc * c * Delta T

Where:
Q = Heat energy absorbed (equal to the KE lost by the car).
m_disc = The mass (weight) of the brake disc.
c = Specific heat capacity of the disc material (usually cast iron or carbon ceramic).
Delta T = The change in temperature of the disc.

If we rearrange to solve for temperature rise:

Delta T = Q / (m_disc * c)

How Larger Components Help:
A larger diameter disc is physically heavier; it has more mass (m_disc). According to the formula, if you increase the mass in the denominator, the temperature rise (Delta T) becomes smaller for the same amount of braking energy (Q).

Conclusion 2: Larger, heavier discs act as larger heat reservoirs, keeping peak temperatures lower during a single hard stop.

B. Heat Dissipation (Surface Area)
Once the heat is absorbed into the disc, it must be shed into the surrounding air to prepare for the next stop. The rate at which a surface cools is roughly proportional to its surface area exposed to the air.
The formula for convective heat transfer (Newton's Law of Cooling) is complex, but simplified it is proportional to:

Heat Transfer Rate proportionalTo A_surface * (T_disc - T_air)

Where A_surface is the surface area of the disc and pads.

How Larger Components Help:
Larger Discs: Have more surface area exposed to the air, allowing them to radiate and convect heat away faster.

Larger Pads: Have a larger backing plate surface area to transfer heat into the caliper and surrounding air.

Conclusion 3: Larger surface areas allow the braking system to cool down faster between stops, preventing heat buildup over multiple braking events (e.g., driving down a mountain or racing on a track).

Part 3: The "Larger Pad" Misconception
It is a very common misconception that larger brake pads create more friction simply because they are bigger. This is false based on standard friction models.

As shown in Part 1, the friction formula is F_friction = mu * F_clamp.
Notice that surface area (A) is not in this formula.

If you apply 1000 lbs of clamping force to a small pad, you get the same friction force as applying 1000 lbs to a giant pad.

So, why use larger pads?
Larger pads are necessary to support the increased capabilities of a larger disc system for two reasons relating to pressure and wear:

Pressure (P) is Force (F) divided by Area (A):
P = F_clamp / A_pad

Lower Operating Pressure: By spreading the clamping force over a larger area, the pressure per square inch on the pad material is reduced. Lower pressure means less mechanical stress on the pad material.

Thermal Distribution: A larger pad spreads the intense heat generation over a wider area of the disc surface, preventing localized "hot spots" that can lead to disc warping and uneven friction material deposits.

Longevity: A larger pad simply has more physical material available to wear down, meaning it lasts longer under heavy use.

 

[parb]

@Clark Kent I recomputed the math because something didn't look right. It's 17% counter torque increase. Quite substantial increase

Buzz, alcon and power brake all have the same benefit since they all use roughly the same size disc.

I'm not surprised that those who get the kits say that the brake performance is increased, its a substantial improvement just based on how the math computes for counter torque

 

[Tazzieman]

TL;DR

This is physics and physics I have an affinity for. So I researched the math a bit.
Thanks btw. these are the kind of fun math problems I enjoy.

The math is straightforward, larger brake disc creates a stronger lever for torque which directly translates into a mechanical advantage. The relative increase in stopping torque is 17.08%% for the power brake big brake kit vs stock Ineos brakes. You'll have to do the math for other vendors system (That is straight up math on facts I know using just physics and known simple math.

So at the same brake pressure you get 17% increase in counter torque. But this compounds as you brake, this math compounds as the car dissipates energy from motion into energy as heat while at the same time stopping faster at the same time faster through increased counter torque and controlled thermal management.

It's really lovely, the counter torque is really what the people report as pedal feel. Eg 17% more counter torque at the same press on the pedal.

net 17% increase in counter torque against the wheel just on the disc itself. A meaningful and significant increase. And probably better since the product of counter torque is energy, which has to be dissipated through heat and I didn't even include that in the math yet.

The more I'm looking at that math the more sold I am on the benefit for cars with weight on them. It looks meaningful, particularly as you're out in the edge of heat management and counter torque performance envelope.

Long version;
Here is a simplified text book definition of the underlying math. I didn't work in all aspects of it, piston size and pad calculations and thermal energy management, but that math works out and I'll probably do it at some point when I have the kit for my own enjoyment.

Braking is the process of converting the kinetic energy of a moving vehicle into thermal energy (heat) through friction.

Improved braking performance generally means two things:
Greater Stopping Power (Mechanical): The ability to generate more braking torque to decelerate the wheel faster.

Better Heat Management (Thermal): The ability to absorb and dissipate the immense heat generated without "fading" (losing effectiveness).
Larger discs and pads address both of these aspects, but in different ways.
Part 1: The Mechanical Advantage (Torque)

The primary mechanical benefit of a larger brake system comes specifically from a larger diameter brake disc (rotor).

When you press the brake pedal, hydraulic pressure forces the brake pads to clamp onto the spinning disc. This creates friction. However, friction alone doesn't stop the car; torque stops the car.

Torque is a twisting force. To stop a spinning wheel, you need to apply a counter-torque.

The Physics of Braking Torque
Imagine trying to stop a spinning bicycle wheel with your hand. If you grab the hub (the center), it is very difficult to stop. If you grab the tire (the outer edge), it is much easier. This is the principle of leverage.
The brake caliper applies a clamping force at a certain distance from the center of the wheel hub. This distance is the "effective radius."

The Formula for Braking Torque (T):
T = F_friction * r_eff
T = Braking Torque (The twisting force that stops the wheel rotation).
F = The total friction force generated between the pads and disc.
r_eff = The effective radius (the distance from the center of the wheel hub to the center point where the pads contact the disc).

We can break down F further:
Where:
mu * F_clamp
(mu) = Coefficient of friction between the pad material and the disc material.
F_clamp = The normal force (clamping force) exerted by the caliper pistons onto the pads.

Combining these, we get the complete braking torque formula:
T = mu * F_clamp * r_eff

How a Larger Disc Improves Performance
Looking at the formula above, if we keep the brake pedal pressure constant (meaning F_clamp is unchanged) and use the same pad material (mu is unchanged), the only way to increase torque (T) is to increase the effective radius r_eff.

A larger diameter disc moves the caliper further away from the wheel hub, increasing r_eff.
Conclusion 1: A larger disc provides a longer lever arm, generating more stopping torque for the same amount of pedal effort. This is the most direct mechanical advantage.


Part 2: The Thermal Advantage (Heat Management)
This is where both larger discs and larger pads play a crucial role.
The amount of energy a braking system must handle is enormous. The kinetic energy (KE) of a moving vehicle is defined by:

KE = 1/2 *(mv(pov 2))

Where m is vehicle mass and v is velocity. Because velocity is squared, slowing down from 100 mph takes four times as much energy as slowing down from 50 mph.
All of this energy must be turned into heat instantly. If brake components get too hot, three bad things happen (known as "brake fade"):
The coefficient of friction (\mu) drops significantly (pad fade).
The brake fluid boils, introducing compressible air bubbles into the lines (fluid fade).
The metal rotors warp or crack.
Larger components combat this through Heat Capacity and Heat Dissipation.
A. Thermal Mass (Heat Capacity)
Think of the brake disc as a heat sink or a "thermal battery." It needs to absorb the initial intense burst of heat generated during a panic stop.
The relationship between heat absorbed (Q) and temperature rise (Delta T) is:

Q = m_disc * c * Delta T

Where:
Q = Heat energy absorbed (equal to the KE lost by the car).
m_disc = The mass (weight) of the brake disc.
c = Specific heat capacity of the disc material (usually cast iron or carbon ceramic).
Delta T = The change in temperature of the disc.

If we rearrange to solve for temperature rise:

Delta T = Q / (m_disc * c)

How Larger Components Help:
A larger diameter disc is physically heavier; it has more mass (m_disc). According to the formula, if you increase the mass in the denominator, the temperature rise (Delta T) becomes smaller for the same amount of braking energy (Q).

Conclusion 2: Larger, heavier discs act as larger heat reservoirs, keeping peak temperatures lower during a single hard stop.

B. Heat Dissipation (Surface Area)
Once the heat is absorbed into the disc, it must be shed into the surrounding air to prepare for the next stop. The rate at which a surface cools is roughly proportional to its surface area exposed to the air.
The formula for convective heat transfer (Newton's Law of Cooling) is complex, but simplified it is proportional to:

Heat Transfer Rate proportionalTo A_surface * (T_disc - T_air)

Where A_surface is the surface area of the disc and pads.

How Larger Components Help:
Larger Discs: Have more surface area exposed to the air, allowing them to radiate and convect heat away faster.

Larger Pads: Have a larger backing plate surface area to transfer heat into the caliper and surrounding air.

Conclusion 3: Larger surface areas allow the braking system to cool down faster between stops, preventing heat buildup over multiple braking events (e.g., driving down a mountain or racing on a track).

Part 3: The "Larger Pad" Misconception
It is a very common misconception that larger brake pads create more friction simply because they are bigger. This is false based on standard friction models.

As shown in Part 1, the friction formula is F_friction = mu * F_clamp.
Notice that surface area (A) is not in this formula.

If you apply 1000 lbs of clamping force to a small pad, you get the same friction force as applying 1000 lbs to a giant pad.

So, why use larger pads?
Larger pads are necessary to support the increased capabilities of a larger disc system for two reasons relating to pressure and wear:

Pressure (P) is Force (F) divided by Area (A):
P = F_clamp / A_pad

Lower Operating Pressure: By spreading the clamping force over a larger area, the pressure per square inch on the pad material is reduced. Lower pressure means less mechanical stress on the pad material.

Thermal Distribution: A larger pad spreads the intense heat generation over a wider area of the disc surface, preventing localized "hot spots" that can lead to disc warping and uneven friction material deposits.

Longevity: A larger pad simply has more physical material available to wear down, meaning it lasts longer under heavy use.

I'm 2 wines into a plane flight, that's overwhelming for me

 

[landmannnn]

TL;DR

This is physics and physics I have an affinity for. So I researched the math a bit.
Thanks btw. these are the kind of fun math problems I enjoy.

The math is straightforward, larger brake disc creates a stronger lever for torque which directly translates into a mechanical advantage. The relative increase in stopping torque is 17.08%% for the power brake big brake kit vs stock Ineos brakes. You'll have to do the math for other vendors system (That is straight up math on facts I know using just physics and known simple math.

So at the same brake pressure you get 17% increase in counter torque. But this compounds as you brake, this math compounds as the car dissipates energy from motion into energy as heat while at the same time stopping faster at the same time faster through increased counter torque and controlled thermal management.

It's really lovely, the counter torque is really what the people report as pedal feel. Eg 17% more counter torque at the same press on the pedal.

net 17% increase in counter torque against the wheel just on the disc itself. A meaningful and significant increase. And probably better since the product of counter torque is energy, which has to be dissipated through heat and I didn't even include that in the math yet.

The more I'm looking at that math the more sold I am on the benefit for cars with weight on them. It looks meaningful, particularly as you're out in the edge of heat management and counter torque performance envelope.

Long version;
Here is a simplified text book definition of the underlying math. I didn't work in all aspects of it, piston size and pad calculations and thermal energy management, but that math works out and I'll probably do it at some point when I have the kit for my own enjoyment.

Braking is the process of converting the kinetic energy of a moving vehicle into thermal energy (heat) through friction.

Improved braking performance generally means two things:
Greater Stopping Power (Mechanical): The ability to generate more braking torque to decelerate the wheel faster.

Better Heat Management (Thermal): The ability to absorb and dissipate the immense heat generated without "fading" (losing effectiveness).
Larger discs and pads address both of these aspects, but in different ways.
Part 1: The Mechanical Advantage (Torque)

The primary mechanical benefit of a larger brake system comes specifically from a larger diameter brake disc (rotor).

When you press the brake pedal, hydraulic pressure forces the brake pads to clamp onto the spinning disc. This creates friction. However, friction alone doesn't stop the car; torque stops the car.

Torque is a twisting force. To stop a spinning wheel, you need to apply a counter-torque.

The Physics of Braking Torque
Imagine trying to stop a spinning bicycle wheel with your hand. If you grab the hub (the center), it is very difficult to stop. If you grab the tire (the outer edge), it is much easier. This is the principle of leverage.
The brake caliper applies a clamping force at a certain distance from the center of the wheel hub. This distance is the "effective radius."

The Formula for Braking Torque (T):
T = F_friction * r_eff
T = Braking Torque (The twisting force that stops the wheel rotation).
F = The total friction force generated between the pads and disc.
r_eff = The effective radius (the distance from the center of the wheel hub to the center point where the pads contact the disc).

We can break down F further:
Where:
mu * F_clamp
(mu) = Coefficient of friction between the pad material and the disc material.
F_clamp = The normal force (clamping force) exerted by the caliper pistons onto the pads.

Combining these, we get the complete braking torque formula:
T = mu * F_clamp * r_eff

How a Larger Disc Improves Performance
Looking at the formula above, if we keep the brake pedal pressure constant (meaning F_clamp is unchanged) and use the same pad material (mu is unchanged), the only way to increase torque (T) is to increase the effective radius r_eff.

A larger diameter disc moves the caliper further away from the wheel hub, increasing r_eff.
Conclusion 1: A larger disc provides a longer lever arm, generating more stopping torque for the same amount of pedal effort. This is the most direct mechanical advantage.


Part 2: The Thermal Advantage (Heat Management)
This is where both larger discs and larger pads play a crucial role.
The amount of energy a braking system must handle is enormous. The kinetic energy (KE) of a moving vehicle is defined by:

KE = 1/2 *(mv(pov 2))

Where m is vehicle mass and v is velocity. Because velocity is squared, slowing down from 100 mph takes four times as much energy as slowing down from 50 mph.
All of this energy must be turned into heat instantly. If brake components get too hot, three bad things happen (known as "brake fade"):
The coefficient of friction (\mu) drops significantly (pad fade).
The brake fluid boils, introducing compressible air bubbles into the lines (fluid fade).
The metal rotors warp or crack.
Larger components combat this through Heat Capacity and Heat Dissipation.
A. Thermal Mass (Heat Capacity)
Think of the brake disc as a heat sink or a "thermal battery." It needs to absorb the initial intense burst of heat generated during a panic stop.
The relationship between heat absorbed (Q) and temperature rise (Delta T) is:

Q = m_disc * c * Delta T

Where:
Q = Heat energy absorbed (equal to the KE lost by the car).
m_disc = The mass (weight) of the brake disc.
c = Specific heat capacity of the disc material (usually cast iron or carbon ceramic).
Delta T = The change in temperature of the disc.

If we rearrange to solve for temperature rise:

Delta T = Q / (m_disc * c)

How Larger Components Help:
A larger diameter disc is physically heavier; it has more mass (m_disc). According to the formula, if you increase the mass in the denominator, the temperature rise (Delta T) becomes smaller for the same amount of braking energy (Q).

Conclusion 2: Larger, heavier discs act as larger heat reservoirs, keeping peak temperatures lower during a single hard stop.

B. Heat Dissipation (Surface Area)
Once the heat is absorbed into the disc, it must be shed into the surrounding air to prepare for the next stop. The rate at which a surface cools is roughly proportional to its surface area exposed to the air.
The formula for convective heat transfer (Newton's Law of Cooling) is complex, but simplified it is proportional to:

Heat Transfer Rate proportionalTo A_surface * (T_disc - T_air)

Where A_surface is the surface area of the disc and pads.

How Larger Components Help:
Larger Discs: Have more surface area exposed to the air, allowing them to radiate and convect heat away faster.

Larger Pads: Have a larger backing plate surface area to transfer heat into the caliper and surrounding air.

Conclusion 3: Larger surface areas allow the braking system to cool down faster between stops, preventing heat buildup over multiple braking events (e.g., driving down a mountain or racing on a track).

Part 3: The "Larger Pad" Misconception
It is a very common misconception that larger brake pads create more friction simply because they are bigger. This is false based on standard friction models.

As shown in Part 1, the friction formula is F_friction = mu * F_clamp.
Notice that surface area (A) is not in this formula.

If you apply 1000 lbs of clamping force to a small pad, you get the same friction force as applying 1000 lbs to a giant pad.

So, why use larger pads?
Larger pads are necessary to support the increased capabilities of a larger disc system for two reasons relating to pressure and wear:

Pressure (P) is Force (F) divided by Area (A):
P = F_clamp / A_pad

Lower Operating Pressure: By spreading the clamping force over a larger area, the pressure per square inch on the pad material is reduced. Lower pressure means less mechanical stress on the pad material.

Thermal Distribution: A larger pad spreads the intense heat generation over a wider area of the disc surface, preventing localized "hot spots" that can lead to disc warping and uneven friction material deposits.

Longevity: A larger pad simply has more physical material available to wear down, meaning it lasts longer under heavy use.

Some detailed physics there, but I will challenge the torque improvement in the real world.

Normally when braking hard we push the pedal at about 30kg. On race cars it is 2 or 3 times that.

If we want 17% more torque we can just press the brake pedal 17% harder.

Bigger brakes is all about heat management as you detailed.

 

[Clark Kent]

@Clark Kent I recomputed the math because something didn't look right. It's 17% counter torque increase. Quite substantial increase

Buzz, alcon and power brake all have the same benefit since they all use roughly the same size disc.

I'm not surprised that those who get the kits say that the brake performance is increased, its a substantial improvement just based on how the math computes for counter torque

I commend you for your tenacity. When it comes to subjects like this I contribute my thoughts then move on. I've lost enthusiasm for the debate that typically follows. Took me a while. A slow learner apparently

 

[parb]

Some detailed physics there, but I will challenge the torque improvement in the real world.

Normally when braking hard we push the pedal at about 30kg. On race cars it is 2 or 3 times that.

If we want 17% more torque we can just press the brake pedal 17% harder.

Bigger brakes is all about heat management as you detailed.

Agreed, you've hit part 2 of the equation, and solidly land in the thermal management problem from Newton's second law.

But now you're also touching on the linearity of the brake booster and human perception of force (which I can't quantify precisely without measurements which I don't have but I'll outline it).

Remember that this energy is based off mass multiplied by velocity. A heavier car is really, really disadvantaged trying to dissipate themselves energy.

My car weighs roughly 800lbs more than a stock car or about 14% more weight.
Using Newton's second law of physics I have to apply 15.2% more force to stop at the same distance. That means that the brake system must absorb 15% more energy and dissipate this as heat.

But now we get into a second problem, the math above is straight up linear. Your applying force through a brake booster. These are force multiplier, letting you apply a 10lbs force and multiplying it into a 50lbs force as an example.

However, these are not fully linear, especially not when mass changes require more force than the stock weight design parameters. If you get close to the boost knee of the power assist you are now applying energy (force) into the braking system with less multiplication from the booster. This only happens at the edge of forces applied which now are 15% more just to be equal.

From literature (its amazing, I found a treasure trove of research on this) getting closer to the boost knee is what most drivers actually perceive as the problem with braking heavier cars. As you get into the boost knee your 15% extra power on the disc becomes 2-3X more power from the foot into the brake pedal since you now have to apply direct energy into the brake system. Can't quantify it without measurements so take that for what it is.

Remember these forces only apply when braking hard, lightly braking doesn't have a difference in feel, we are only talking about the outer bands of the forces in play -heavy braking.

Now we get to the biomechanical problem. Applying force through the leg and the knee is also not linear, a human barely notices extra force applied at light changes but at the outer edge of their power applied we think it's much heavier than it actually is.

A 10% increase in foot force at the lower part of our muscle band is barely noticed. the difference between 5lbs of force applied by the foot vs 5.5lbs is barely noticeable. 100 and 110lbs is noticed as 10% (depending on your natural muscle mass and positioning of the body in the vehicle).

But at the outer bands of forces applied, like during hard braking it feels like 20-25% more force because the muscle gets closer to it's limit. Apply a bit of loss in power boost loss and it feels like 2-3X harder.

This is what literature says is what actually happens.

Fascinating stuff, but I think my adventures into brake forces have come to an end. It's interesting but not that interesting. The reports of better brake feel and performance bears out in math.

The 15% increase in mechanical torque stopping power completely negates the 15% increase in weight. If the better thermals remove fade then I have a superior system at the same pedal feel as a stock vehicle.

I'm getting the bigger brake kit. I think it's going to work out. Math, physics and psychophysics (the human perception of force) is on my side

 

[landmannnn]

Agreed, you've hit part 2 of the equation, and solidly land in the thermal management problem from Newton's second law.

But now you're also touching on the linearity of the brake booster and human perception of force (which I can't quantify precisely without measurements which I don't have but I'll outline it).

Remember that this energy is based off mass multiplied by velocity. A heavier car is really, really disadvantaged trying to dissipate themselves energy.

My car weighs roughly 800lbs more than a stock car or about 14% more weight.
Using Newton's second law of physics I have to apply 15.2% more force to stop at the same distance. That means that the brake system must absorb 15% more energy and dissipate this as heat.

But now we get into a second problem, the math above is straight up linear. Your applying force through a brake booster. These are force multiplier, letting you apply a 10lbs force and multiplying it into a 50lbs force as an example.

However, these are not fully linear, especially not when mass changes require more force than the stock weight design parameters. If you get close to the boost knee of the power assist you are now applying energy (force) into the braking system with less multiplication from the booster. This only happens at the edge of forces applied which now are 15% more just to be equal.

From literature (its amazing, I found a treasure trove of research on this) getting closer to the boost knee is what most drivers actually perceive as the problem with braking heavier cars. As you get into the boost knee your 15% extra power on the disc becomes 2-3X more power from the foot into the brake pedal since you now have to apply direct energy into the brake system. Can't quantify it without measurements so take that for what it is.

Remember these forces only apply when braking hard, lightly braking doesn't have a difference in feel, we are only talking about the outer bands of the forces in play -heavy braking.

Now we get to the biomechanical problem. Applying force through the leg and the knee is also not linear, a human barely notices extra force applied at light changes but at the outer edge of their power applied we think it's much heavier than it actually is.

A 10% increase in foot force at the lower part of our muscle band is barely noticed. the difference between 5lbs of force applied by the foot vs 5.5lbs is barely noticeable. 100 and 110lbs is noticed as 10% (depending on your natural muscle mass and positioning of the body in the vehicle).

But at the outer bands of forces applied, like during hard braking it feels like 20-25% more force because the muscle gets closer to it's limit. Apply a bit of loss in power boost loss and it feels like 2-3X harder.

This is what literature says is what actually happens.

Fascinating stuff, but I think my adventures into brake forces have come to an end. It's interesting but not that interesting. The reports of better brake feel and performance bears out in math.

The 15% increase in mechanical torque stopping power completely negates the 15% increase in weight. If the better thermals remove fade then I have a superior system at the same pedal feel as a stock vehicle.

I'm getting the bigger brake kit. I think it's going to work out. Math, physics and psychophysics (the human perception of force) is on my side

Nothing wrong with your decision.

I think you need to revisit the maths on the torque change though. It's closer to 5% (316mm rotors with 53mm tall pads vs 337mm rotors with 64mm pads)

 

[parb]

316mm is the front rotor diameter in a stock vehicle, that is a 158mm radius

on the powerbrake i have two sources of truth, one is 174mm radius and the other is 185mm radius.
I based my calculations off a 185mm radius. if its 174mm then the math is just a bit above 10% (i didn't redo it yet).

As i pondered your observation i went and looked at their website for some data. As now was on my laptop and bigger screen than on my tablet i could see the brake pistons a bit more clearly. Take a look at that lovely design, last night i read about the torsional forces (physics) being applied on the pad which also makes that a place where torque is applied -and how piston size counters this.

What i think is pretty cool is their cylinder arrangement. look how they they use differential bores to counter the effects of torsional forces applying torque to the pad itself in order to fix the problem of "mushy" brake pedal feel (or the effect of uneven pressure of the pad across the entire surface of the pad).

if you zoom in a bit you'll notice that the pistons are assymetrically sized and lined up differently along the rotational surface area of the pad.

So what is the physics?
The leading edge grab: The point where the spinning disc first enters the caliper and touches the pad is called the "leading edge." Due to the friction drag, the pad naturally wants to pivot, driving its leading edge very hard into the disc face. This is sometimes called a "self-energizing" or "servo" effect.

Gets us to uneven pressure: Even if the pistons behind the pad are applying equal force, this self-energizing effect means the actual pressure between the pad and disc is much higher at the leading edge than at the trailing edge.

Ending up in brake taper: Because the leading edge is working much harder, it gets much hotter and wears down significantly faster than the trailing edge. Over time, the brake pad turns into a wedge shape.

Force is simple: Pressure multiplied by area.

Eg, the hydraulic pressure coming from the master cylinder is the same everywhere inside the caliper. Therefore, to change the force applied to different parts of the pad, we must change the piston area (diameter).

By staggering the piston sizes, this design "tunes" the clamping force along the length of the pad.

In this differential bore 6-piston caliper (3 pistons on each side of the disc), the setup becomes a gradient along the path of rotation:

• Position 1 (Leading Edge - Rotor enters here): Smallest Piston.Because the natural friction drag is already forcing this part of the pad hard into the rotor, it needs the least amount of hydraulic help. The small piston applies less clamping force here to prevent over-braking at the leading edge.
• Position 2 (Middle): Medium Piston or the same size as position 1.
• Position 3 (Trailing Edge - Rotor exits here): Largest Piston.By the time the rotor reaches the back of the pad, the self-energizing effect is gone. To ensure the back of the pad is doing its fair share of the work, a large piston is used to apply maximum clamping force here.

Fun isn't it

 

[Tazzieman]

But what if I choose to facilitate my velocitous extramuralisation?